Benjamin Girard

Wolfgang A Schmid

INVERSE ZERO-SUM PROBLEMS FOR CERTAIN GROUPS OF RANK THREE

Keywords: 2010 Mathematics Subject Classification. 11B30, 11P70, 20K01 finite abelian group, zero-sum sequence, inverse problem, Erdős-Ginzburg-Ziv constant, inductive method

The inverse problem associated to the Erdős-Ginzburg-Ziv constant and the η-constant is solved for finite abelian groups of the form C 2 ⊕ C 2 ⊕ C 2n where n ≥ 2 is an integer.

Introduction

For (G, +, 0) a finite abelian group with exponent exp(G) the Erdős-Ginzburg-Ziv constant, denoted by s(G), is defined as the smallest nonnegative integer ℓ such that each sequence over G of length at least ℓ has a subsequence of length exp(G) whose sum is 0. The η-constant, denoted by η(G), is defined in the same way except that the length of the subsequence is at most exp(G) and at least 1.

These constants have been studied since the early 1960s; for an overview we refer to the survey article [START_REF] Gao | Zero-sum problems in finite abelian groups: a survey[END_REF], in particular to Sections 6 and 7. Their exact values are only known for groups of rank at most two (see [START_REF] Gao | Zero-sum problems in finite abelian groups: a survey[END_REF]Theorem 6.5] and [START_REF] Reiher | On Kemnitz conjecture concerning lattice-points in the plane[END_REF][START_REF] Savchev | Kemnitz conjecture revisited[END_REF] for key-contributions), and for a few special types of other groups. On the one hand, the approach used for groups of rank two can be adapted to apply to some other groups (see [START_REF] Luo | Short zero-sum sequences over abelian p-groups of large exponent[END_REF][START_REF] Roy | On zero-sum subsequences in a finite abelian p-group of length not exceeding a given number[END_REF]). On the other hand, there are various results for homocyclic groups, that is, groups of the form C r n where C n denotes a cyclic group of order n and groups that resemble (we refer to [START_REF] Fan | On the Erdős-Ginzburg-Ziv constant of finite abelian groups of high rank[END_REF]Theorem B] for an overview). There is also considerable work on bounds; we refer to [START_REF] Fox | Erdős-Ginzburg-Ziv constants by avoiding three-term arithmetic progressions[END_REF] and the references therein for recent results on upper bounds, and we refer to [START_REF] Edel | Sequences in abelian groups G of odd order without zero-sum subsequences of length exp(G)[END_REF] for lower bounds.

Only recently the exact values were determined for groups of the form C 2 ⊕ C 2 ⊕ C 2n (see [START_REF] Fan | Remarks on tiny zero-sum sequences[END_REF]Theorem 1.3] and [4, Theorem 1.2.1]); this was further extended to groups of the form C 2 ⊕ C 2m ⊕ C 2mn in [START_REF] Girard | Direct zero-sum problems for certain groups of rank three[END_REF]. Moreover, for C 2 ⊕ C 2 ⊕ C 2 ⊕ C 2n the value of the η-constant is known for all n ≥ 1, and the value of the Erdős-Ginzburg-Ziv constant is known for n ≥ 36 (see [START_REF] Fan | On the Erdős-Ginzburg-Ziv constant of groups of the form C r 2 ⊕ Cn[END_REF]Theorem 1.2.2]).

The associated inverse problems consist in determining all sequences of length s(G)-1 (respectively η(G)-1) that have no zero-sum subsequence of length exp(G) (respectively of length at most exp(G) and at least 1). For η(G) the inverse problem is solved for groups of rank at most two. For s(G) the inverse problem is solved for cyclic groups, and for groups of rank at most two there is a well-supported conjecture and partial results towards this conjecture are known (see [START_REF] Schmid | Restricted inverse zero-sum problems in groups of rank 2[END_REF]). For recent results for certain nonabelian groups see [START_REF] Oh | On Erdős-Ginzburg-Ziv inverse theorems for Dihedral and Dicyclic groups[END_REF].

In the current paper, we solve the inverse problems associated to η(G) and to s(G) for groups of the form C 2 ⊕C 2 ⊕C 2n . To solve the inverse problem associated to η(G), which is done in Theorem 4.1, we use that the structure of minimal zero-sum sequences of maximal length is known for this type of groups (see Theorem 4.2). We recall that a similar approach was used to solve the inverse problem associated to η(G) for groups of rank two (see [START_REF] Gao | On long minimal zero sequences in finite abelian groups[END_REF]Theorem 10.7] or [START_REF] Reiher | A proof of the theorem according to which every prime number possesses property B[END_REF]Section 11.3]). Then, the result for the η-constant is used to solve the problem for the Erdős-Ginzburg-Ziv constant (see Theorem 5.1). Other tools used in the proof are well-known results on the inverse problem for cyclic groups that we recall in Section 3 as well as ideas used in the proofs of the corresponding direct results mentioned above.

Conjecturally, there is a tight link between these two constants, namely Gao conjectured (see [START_REF] Gao | Zero-sum problems in finite abelian groups: a survey[END_REF]Conjecture 6.5]) that s(G) = η(G) + exp(G) -1 holds for all finite abelian groups. Furthermore, a close link can also be noted for the structure of extremal sequences in cases where the inverse problem is solved. A close link can also be observed in the current case, yet the exact link is more complex to describe than for groups of rank at most two and for homocyclic groups; we discuss this in more detail after Theorem 5.1.

Preliminaries

The notation used in this paper basically matches the one used in [START_REF] Gao | Zero-sum problems in finite abelian groups: a survey[END_REF][START_REF] Geroldinger | Additive group theory and non-unique factorizations[END_REF][START_REF] Geroldinger | Non-unique factorizations. Algebraic, combinatorial and analytic theory[END_REF][START_REF] Grynkiewicz | Structural Additive Theory[END_REF]. For definiteness, we give a brief summary.

In this paper, intervals are intervals of integers, specifically [a, b] = {z ∈ Z : a ≤ z ≤ b}.

Let (G, +, 0) be a finite abelian group. For g in G, let ord(g) denote its order in G. For a subset A ⊂ G, let A denote the subgroup it generates; A is called a generating set if A = G. Elements g 1 , . . . , g k ∈ G are called independent if k i=1 a i g i = 0, with integers a i , implies that a i g i = 0 for each i ∈ [1, k]; a set is called independent when its elements are independent. We refer to an independent generating set as a basis.

The exponent of G is the least common multiple of the orders of elements of G, it is denoted by exp(G); the rank of G, denoted by r(G), is the minimum cardinality of a generating subset of G. For n a positive integer, C n denotes a cyclic group of order n.

By a sequence over G we mean an element of the free abelian monoid over G, denoted by F (G). We use multiplicative notation for this monoid; its neutral element is denoted by 1. For a sequence S = g∈G g vg where v g is a nonnegative integer, for each g ∈ G, we denote by:

• v g (S) = v g the multiplicity of g in S. • h(S) = max{v g (S) : g ∈ G} the height of S. • σ(S) = g∈G v g (S)g the sum of S. • |S| = g∈G v g (S) the length of S. • supp(S) = {g ∈ G | v g (S) > 0} the support of S.
Moreover, for h ∈ G, one denotes by h + S the sequence where each element in S is translated by h, that is, h + S = g∈G (h + g) vg .

When one writes S = g 1 • • • g ℓ with g i ∈ G, then the g i are not necessarily distinct, yet they are determined uniquely up to ordering. A sequence for which the g i are pairwise distinct, equivalently v g (S) ≤ 1 for each g ∈ G, is called squarefree. A nonempty sequence over G of length at most exp(G) is called short.

Let S be a sequence over G. A divisor T of S in F (G) is called a subsequence of S; by ST -1 we denote the sequence such that T (ST -1 ) = S. Subsequences T 1 , . . . , T k of S are called disjoint if the product

T 1 • • • T k is also a subsequence of S.
An element s ∈ G is a subsum of S if s = σ(T ) for some 1 = T | S. The sequence S is called a zero-sum free sequence if 0 is not a subsum. If σ(S) = 0, then S is called a zero-sum sequence; if, in addition, one has σ(T ) = 0 for all proper and nonempty subsequences T of S, then S is called a minimal zero-sum sequence.

For L a set of integers, we set

Σ L (S) = {σ(T ) : 1 = T | S with |T | ∈ L}.
For Σ Z ≥1 (S) we just write Σ(S), and for Σ {k} (S) we write Σ k (S). Moreover, s L (G) is the smallest nonnegative integer ℓ (if one exists) such that for each sequence S ∈ F (G) with |S| ≥ ℓ, one has 0 ∈ Σ L (S); if no such integer exists, then one sets s L (G) = ∞. The constants we mentioned in the introduction are natural special cases; the choice of exp(G) in the definitions is natural, for example it is the smallest integer for which the respective constants are finite.

• The Davenport constant, denoted by

D(G), is s Z ≥1 (G). • The Erdős-Ginzburg-Ziv constant, denoted by s(G), is s exp(G) (G). • The η-constant, denoted by η(G), is s [1,exp(G)] (G).
The Davenport constant of G is also equal to the maximal length of a minimal zero-sum sequence over G.

Results for cyclic groups and auxiliary results

The η(G) and s(G) invariants for cyclic groups are well-known and the inverse problems are solved. Indeed, in the case of cyclic groups also the structure of sequences considerably shorter than η(G) -1 and s(G) -1 without the respective zero-sum subsequences is known (for example, see [START_REF] Gao | On the structure of long zero-sum free sequences and n-zero-sum free sequences over finite cyclic groups[END_REF][START_REF] Savchev | Long zero-free sequences in finite cyclic groups[END_REF][START_REF] Savchev | Long n-zero-free sequences in finite cyclic groups[END_REF][START_REF] Yuan | On the index of minimal zero-sum sequences over finite cyclic groups[END_REF]). We only recall a special case we need for our proofs. While the result below is formulated for short zero-sum subsequences, since this fits the current context, having no short zerosum subsequence and having no nonempty zero-sum subsequence are equivalent for cyclic groups, and the result is usually phrased in the latter form.

The following result is a direct consequence of [START_REF] Savchev | Long zero-free sequences in finite cyclic groups[END_REF]Theorem 8] (also see [START_REF] Grynkiewicz | Structural Additive Theory[END_REF]Theorem 11.1]). In the former case Σ(T ) = H \ {0, -g}, in the latter case Σ(T ) = H \ {0} (except for n = 3 where the latter case coincides with the former).

For the following results see [START_REF] Savchev | Long n-zero-free sequences in finite cyclic groups[END_REF], especially the discussion after Corollary 7 there.

Theorem 3.2. Let H ≃ C n where n ≥ 3 is an integer.

(1) Every sequence T of length |T | = s(H) -1 = 2n -2 not containing any short zero-sum sequence has the form T = g n-1 h n-1 where g, h ∈ H and ord(gh) = n. (2) Every sequence S of length |T | = s(H) -2 = 2n -3 not containing any short zero-sum sequence has the form:

(a) T = g n-1 h n-2 where ord(g -h) = n or (b) T = g n-1 h n-3 (2h -g) where ord(g -h) = n.
In the former case Σ n-2 (T ) = H \{-g-h}, in the latter case Σ n-2 (T ) = H (except for n = 3 where the latter case coincides with the former).

We give two results on groups of the form C r 2 that are needed for the proofs of our main results. For the following well-known result see, e.g., [START_REF] Freeze | Remarks on a generalization of the Davenport constant[END_REF]Theorem 7.2].

Lemma 3.3. For r ≥ 2, one has s [1,3] (C r 2 ) = 1 + 2 r-1 .
We also need a result on zero-sum sequences of length 4. Lemma 3.4. Let S be a squarefree sequence of length 5 over C 3 2 . Then S has a unique zero-sum subsequence of length 4.

Proof. Since the property is invariant under translation of the sequence, we can assume that 0 | S. The four nonzero elements in S cannot be contained in a proper subgroup, and thus S contains three independent elements f 1 , f 2 , f 3 .

Let g denote the fourth nonzero element. Either g = f 1 +f 2 +f 3 and the sequence gf 1 f 2 f 3 is the unique zero-sum subsequence of length four, or g is the sum of two of the independent elements, say g = f 1 + f 2 , and 0gf 1 f 2 is the unique zero-sum subsequence of length four.

Inverse problem associated to

η (C 2 ⊕ C 2 ⊕ C 2n ) for n ≥ 2
We determine the structure of all sequences over C 2 ⊕ C 2 ⊕ C 2n , for n ≥ 2, of length η (C 2 ⊕ C 2 ⊕ C 2n ) -1 = 2n + 3 that do not have a short zero-sum subsequence. The case n = 1, that is, C 3 2 , is different yet well-known and direct. For completeness we recall that η(C 3 2 ) -1 = 7, and the only example of a sequence of length 7 without zero-sum subsequence of length 2 is the squarefree sequence of all nonzero elements; in fact, this is true for any group of the form C r 2 , because the only zero-sum sequence of length at most exp(G) are 0 and

g 2 for g ∈ C r 2 . Theorem 4.1. Let G ≃ C 2 ⊕C 2 ⊕C 2n , where n ≥ 2. A sequence S over G of length |S| = η(G) -1 = 2n + 3 contains no short zero-sum subsequence if and only if there exists a basis {f 1 , f 2 , f 3 } of G, where ord(f 1 ) = ord(f 2 ) = 2 and ord(f 3 ) = 2n, such that S is equal to one of the following sequences: (η1) f 2n-1-2v 3 (f 3 + f 2 ) 2v+1 f 2 (af 3 + f 1 ) (1 -a)f 3 + f 2 + f 1 with v ∈ [0, n -1] and a ∈ [2, n -1]. (η2) f 2n-1 3 (af 3 + f 2 ) (1 -a)f 3 + f 2 (bf 3 + f 1 ) (1 -b)f 3 + f 1 with a, b ∈ [2, n -1] and a ≥ b. (η3) 2n+1 i=1 (f 3 +d i )f 2 f 1 with S ′ = 2n+1 i=1 d i ∈ F ( f 1 , f 2 ) and σ(S ′ ) / ∈ supp(S ′ ).
For n = 2, in fact only the last case can occur as [2, n-1] is empty. To prove this result we use that the structure of all minimal zero-sum subsequences of maximal length is known (see [START_REF] Schmid | The inverse problem associated to the Davenport constant for C 2 ⊕ C 2 ⊕ C 2n and applications to the arithmetical characterization of class groups[END_REF]Theorem 3.13]).

Theorem 4.2. Let G ≃ C 2 ⊕ C 2 ⊕ C 2n , where n ≥ 1. A sequence S over G is a minimal zero-sum sequence of length |S| = D(G) if and only if there exists a basis {f 1 , f 2 , f 3 } of G, where ord(f 1 ) = ord(f 2 ) = 2 and ord(f 3 ) = 2n
, such that S is equal to one of the following sequences:

(D1) f v3 3 (f 3 + f 2 ) v2 (f 3 + f 1 ) v1 (-f 3 + f 2 + f 1 ) with v i ∈ N odd, v 3 ≥ v 2 ≥ v 1 and v 3 + v 2 + v 1 = 2n + 1. (D2) f v3 3 (f 3 + f 2 ) v2 (af 3 + f 1 )(-af 3 + f 2 + f 1 ) with v 2 , v 3 ∈ N odd v 3 ≥ v 2 and v 2 + v 3 = 2n and a ∈ [2, n -1]. (D3) f 2n-1 3 (af 3 + f 2 )(bf 3 + f 1 )(cf 3 + f 2 + f 1 ) with a + b + c = 2n + 1 where a ≤ b ≤ c, and a, b ∈ [2, n -1], c ∈ [2, 2n -3] \ {n, n + 1}. (D4) f 2n-1-2v 3 (f 3 + f 2 ) 2v f 2 (af 3 + f 1 ) (1 -a)f 3 + f 2 + f 1 with v ∈ [0, n -1] and a ∈ [2, n -1]. (D5) f 2n-2 3 (af 3 + f 2 ) (1 -a)f 3 + f 2 (bf 3 + f 1 ) (1 -b)f 3 + f 1 with a ≥ b and a, b ∈ [2, n -1]. (D6) 2n i=1 (f 3 + d i )f 2 f 1 where S ′ = 2n i=1 d i ∈ F ( f 1 , f 2 ) with σ(S ′ ) = f 1 + f 2 .
As above, for n ≤ 2, only some of these cases occur, namely the first and the last. We now prove our result.

Proof of Theorem 4.1. Let G ≃ C 2 ⊕ C 2 ⊕ C 2n where n ≥ 2, and let H ≃ C n be the subgroup of G such that G/H ≃ C 3 2 .
Note that there is indeed a unique such subgroup H. We use the inductive method with

H ֒→ G π → G/H.
Let S be a sequence over G such that |S| = η(G) -1 = 2n + 3 that contains no short zero-sum subsequence.

First, let us prove that S contains no element g in H. Indeed, if it were the case, setting g = T 0 , we have ST -1

0 whose length is |ST -1 0 | = 2n+ 2 = η(G/H)+ 2(n-3) would contain n-2 disjoint nonempty subsequences T 1 , . . . , T n-2 such that |T i | ≤ 2 and σ (T i ) ∈ H for all i ∈ [1, n -2]. In particular, S = T 0 T 1 • • • T n-2 S 0 where S 0 has length |S 0 | ≥ 6.
Now, note that all elements of π(S 0 ) must be nonzero and distinct, otherwise we could extract from S 0 yet another subsequence

T n-1 | S 0 satisfying |T n-1 | ≤ 2 and σ (T n-1 ) ∈ H. Then, we have T = n-1 i=0 σ(T i ) is a sequence over H of length |T | = n, so it contains a nonempty zero-sum subsequence, say i∈I σ(T i ) = 0 with ∅ = I ⊂ [0, n -1]. Thus, i∈I T i is a subsequence of S with sum 0 and length at most 2|I| ≤ 2n.
However, a squarefree sequence of at least 6 nonzero elements over G/H ≃ C 3 2 has a zero-sum subsequence of length at most 3 (see Lemma 3.3). Applying this to π(S 0 ) gives that S 0 contains a subsequence

T n-1 of length |T n-1 | = 3 such that σ (T n-1 ) ∈ H. Then, again T = n-1 i=0 σ(T i ) is a sequence over H of length |T | = n.
It contains a nonempty zero-sum subsequence, and whence n-1 i=0 T i contains a nonempty zero-sum subsequence. Since | n-1 i=0 T i | ≤ 2n, it follows that this is a short zero-sum subsequence. Since n-1 i=0 T i is a subsequence of S, we see that S contains a short zero-sum subsequence, which is a contradiction. Thus, we know that S contains no element in H. Similarly as above, since

|S| = 2(n -3) + 9 ≥ η(G/H) + 2(n -3), there exist n -2 disjoint subsequences T 1 , . . . , T n-2 of S such that σ (T i ) ∈ H and |T i | ≤ 2 for all i ∈ [1, n -2]; yet, since S contains no element from H in fact |T i | = 2 for all i ∈ [1, n -2]. In particular, S = T 1 • • • T n-2 S 0 where S 0 has length |S 0 | = 7.
We now try to obtain further information on both S 0 and the sequences T i for i ∈ [1, n-2]. On the one hand, at most one of the elements in π(S 0 ) has multiplicity at least 2 and none has multiplicity at least 4. To see this it suffices to note that otherwise we could extract from S 0 two disjoint subsequences T n-1 and T n of length 2 whose sums are in H. Then, arguing as above, S would contain a short zero-sum subsequence, a contradiction. In particular, we have | supp(π(S 0 ))| ≥ 5.

On the other hand, For n = 2, this directly yields that σ(U ) = g where g is the nonzero element of H. For n = 3, we first infer that H \ -(Σ (T ) ∪ {0}) = ∅, which by Theorem 3.1 implies that T = g n-2 for some g ∈ H with ord(g) = n, and σ(U ) = g. It follows that there exists an element g ∈ H satisfying ord(g) = n such that S can be decomposed as

T = n-2 i=1 σ(T i ) is a sequence over H of length |T | = n -2,
T 1 • • • T n-2 S 0 where σ(T i ) = g for all i ∈ [1, n -2], and for each U | S 0 such that |U | ≤ 4 and σ (U ) ∈ H one has σ(U ) = g.
Case 1. π(S 0 ) is squarefree. Then, S 0 itself is squarefree and supp(π(S 0 )) = G/H \{0}. We now fix a basis {e 1 , e 2 , e 3 } of G/H; this quotient group is isomorphic to C 3 2 . For every set ∅ = I ⊂ {1, 2, 3}, we write e I = i∈I e i and we denote by a I the unique element of S 0 such that π(a I ) = e I . We have U 0 = a {1,2} a {2,3} a {1,3} , U k = a i a j a {i,j} and V i = a {i,j} a {i,k} a j a k for {i, j, k} = {1, 2, 3} are subsequences of S 0 satisfying |U k | = 3 and σ(U k ) ∈ H as well as |V i | = 4 and σ(V i ) ∈ H. In addition, we have

U 0 U 1 U 2 U 3 = V 1 V 2 V 3 .

By the argument above it follows that σ(U

k ) = σ(V i ) = g for all i, k ∈ [1, 3], which yields 4g = σ(U 0 U 1 U 2 U 3 ) = σ(V 1 V 2 V 3 ) = 3g. It follows that g = 0, which is a contradiction.
Case 2. π(S 0 ) is not squarefree. We recall that there is a unique element with multiplicity at least 2. In this case, S 0 contains a subsequence

T n-1 | S 0 such that |T n-1 | = 2 and σ (T n-1 ) ∈ H. We get again σ(T n-1 ) = g. Thus S = T 1 • • • T n-1 S ′ 0 where σ (T i ) = g for all i ∈ [1, n -1] and S ′ 0 has length |S ′ 0 | = 5.
We know that π(S ′ 0 ) is squarefree and does not contain 0. By Lemma 3.4, S ′ 0 contains a unique subsequence U = stuv | S ′ 0 such that |U | = 4 and σ (U ) ∈ H. First, we have σ(U ) = s+t+u+v = g. Now, setting S ′ 0 = U w, and since π(s) + π(t), π(s) + π(u), π(s) + π(v) are nonzero and pairwise distinct elements of the set G/H \ {0, π(s), π(t), π(u), π(v)} containing π(w), exactly one of them, say π(s) + π(t), is equal to π(w). Thus π(w) = π(s) + π(t) = π(u) + π(v). Now, U 1 = wst and U 2 = wuv are two subsequences of

S ′ 0 satisfying |U 1 | = |U 2 | = 3 and σ(U 1 ), σ(U 2 ) ∈ H. This yields σ(U 1 ) = w + s + t = g and σ(U 2 ) = w + u + v = g, so that s + t = u + v. Since also s + t + u + v = g, it follows that 2w = g and w = s + t = u + v. In addition, V = T 1 • • • T n-1 U is a zero-sum subsequence of S of length |V | = 2(n -1) + 4 = 2n + 2 = D(G).
Since S contains no short zero-sum subsequence, it cannot be decomposed into two zero-sum subsequences, as one of them would be short. That is, V is a minimal zero-sum sequence over G of length |V | = D(G).

Let us now give a summary of everything we proved so far. For every sequence S over G such that |S| = 2n + 3 and containing no short zero-sum subsequence, there exists an element g ∈ H with ord(g) = n such that S can be decomposed as

T 1 • • • T n-1 S ′ 0 where T 1 , . . . , T n-1 are subsequences of S that satisfy |T i | = 2 and σ (T i ) = g for all i ∈ [1, n -1]
, and where S ′ 0 is a sequence of length |S ′ 0 | = 5. Moreover, in every such decomposition of S, the sequence π(S ′ 0 ) is squarefree and does not contain 0, more precisely the sequence S ′ 0 is squarefree and can be uniquely decomposed as S ′ 0 = U w where U | S ′ 0 satisfies |U | = 4 and σ(U ) ∈ H. Also, we can write U = stuv so that the equalities s

+ t = u + v = w and σ(U ) = 2w = g hold. Finally, V = T 1 • • • T n-1 U is a minimal zero-sum sequence over G of length |V | = 2n + 2 = D(G).
In particular, using Theorem 4.2, we know there exists a basis {f 1 , f 2 , f 3 } of G, where ord(f 1 ) = ord(f 2 ) = 2 and ord(f 3 ) = 2n, such that V has one out of six possible forms. We treat each of these cases separately; we recall that for n = 2 only the first and the last can occur, so that for the others we can assume that n ≥ 3.

If V is of type (D1), then in order to have a support of size at least five, S ′ 0 must contain one copy of every member of supp(V ). Since the four elements of supp(V ) sum up to 2f 3 ∈ H, we obtain

U = f 3 (f 3 + f 2 )(f 3 + f 1 )(-f 3 + f 2 + f 1
), but it is easily seen that U cannot be decomposed as the product of two sequences of length 2 having the same sum.

If V is of type (D2), then in order to have a support of size at least five, S ′ 0 must contain one copy of every member of supp(V ). Since the four elements of supp(V ) sum up to 2f 3 ∈ H, we obtain

U = f 3 (f 3 + f 2 )(af 3 + f 1 )(-af 3 + f 2 + f 1 ) with a ∈ [2, n -1].
In addition, U can be decomposed as the product of two sequences of length 2 having the same sum only if 1a = 1 + a (mod 2n), that is to say only if a ∈ {0, n}, which is a contradiction.

If V is of type (D3), then in order to have a support of size at least five, S ′ 0 must contain one copy of every member of supp(V ). Since the four elements of supp(V ) sum up to 2f 3 ∈ H, we obtain If V is of type (D4), then in order to have a support of size at least five, S ′ 0 must contain at least four elements of supp(V ). First, note that S ′ 0 contains at most one copy of f 3 and at most one copy of f 3 + f 2 . In other words, T 1 • • • T n-1 contains at least 2n -2v -2 copies of f 3 and at least 2v -1 copies of

U = f 3 (af 3 + f 2 )(bf 3 + f 1 )(cf 3 + f 2 + f 1 ) with a+b+c = 2n+1
f 3 + f 2 . Assume that there exists no i ∈ [1, n -1] satisfying T i = f 2 3 or T i = (f 3 + f 2 ) 2 . Since (2n -2v -2) + (2v -1) = 2n -3 ≥ n, there exists i ∈ [1, n -1] satisfying S i = f 3 (f 3 + f 2 ) so that g = 2f 3 + f 2 . Yet, this is not an element of H, which is a contradiction. It follows that T i = f 2 3 or T i = (f 3 + f 2 ) 2 for at least one i ∈ [1, n -1] which yields g = 2f 3 . In particular, for every i ∈ [1, n -1] such that T i contains (f 3 + f 2 ), we have T i = (f 3 + f 2 ) 2 so that T 1 • • • T n-1 contains
an even number of copies of f 3 + f 2 . Therefore, the number of copies of f 3 + f 2 contained in S ′ 0 , which is at most one, must be zero. We thus obtain

U = f 3 f 2 (af 3 +f 1 ) (1-a)f 3 +f 2 +f 1 with a ∈ [2, n -1].
Since 1 + a = 1a (mod 2n) and 1 + (1a) = a (mod 2n), there is only one possible decomposition of U as the product of two sequences of length 2 having the same sum, which yields w = f 3 + f 2 . Therefore,

S = f 2n-1-2v 3 (f 3 + f 2 ) 2v+1 f 2 (af 3 + f 1 ) (1 -a)f 3 + f 2 + f 1 with v ∈ [0, n -1] and a ∈ [2, n -1]
. It remains to check that this sequence contains no short zero-sum subsequence. Indeed, since V is a minimal zero-sum sequence over G, any short zero-sum subsequence S ′ | S must satisfy (

f 3 + f 2 ) 2v+1 | S ′ . Now, if S ′ contains neither (af 3 + f 1 ) nor ((1 -a)f 3 + f 2 + f 1 ), we must have S ′ = f 2n-1-2v 3 (f 3 + f 2 ) 2v+1 f 2 so that |S ′ | = 2n + 1 > 2n which is a contradiction. If S ′ contains either (af 3 + f 1 ) or (1 -a)f 3 + f 2 + f 1 then it must contain both of them, which yields S ′ = f 2n-2-2v 3 (f 3 + f 2 ) 2v+1 (af 3 + f 1 ) (1 -a)f 3 + f 2 + f 1 so that |S ′ | = 2n + 1 > 2n, which is a contradiction.
If V is of type (D5), then in order to have a support of size at least five, S ′ 0 must contain at least four elements of supp(V ). Since S ′ 0 contains at most one copy of f 3 , the sequence T 1 • • • T n-1 contains at least 2n -3 ≥ n copies of f 3 . This implies that every T i contains at least one copy of f 3 and that at most one of them is different from f 2 3 . This yields g = 2f 3 so that T i = f 2 3 for every i ∈ [1, n -1]. Therefore, the elements (af 3 + f 2 ), (1a)f 3 + f 2 , (bf 3 + f 1 ) and

(1b)f 3 + f 1 belong to S ′ 0 and since their sum is equal to 2f 3 ∈ H, we have

U = (af 3 +f 2 ) (1-a)f 3 +f 2 (bf 3 +f 1 ) (1-b)f 3 +f 1 with a ≥ b and a, b ∈ [2, n-1].
Now, let w = αf 3 + d where α ∈ [0, 2n -1] and d ∈ f 1 , f 2 . On the one hand, 2w = g = 2f 3 yields α ∈ {1, n + 1}. On the other hand, since w must be equal to the sum of two elements of U , we have d ∈ {0, f 1 + f 2 }.

In case d = 0, we get that α = 1 which yields w = f 3 so that

S = f 2n-1 3 (af 3 + f 2 ) (1 -a)f 3 + f 2 (bf 3 + f 1 ) (1 -b)f 3 + f 1 with a ≥ b and a, b ∈ [2, n -1]
. It remains to check that this sequence contains no short zero-sum subsequence. Indeed, since V is a minimal zero-sum sequence over G, any short zero-sum subsequence

S ′ | S must satisfy f 2n-1 3 | S ′ , which implies |S ′ | = 2n so that S ′ = f 2n 3 ∤ S, a contradiction. In the case d = f 1 + f 2 and α = 1, we have w = f 3 + f 2 + f 1 . Since a + b = (1 -a) + (1 -b) (mod 2n) if and only if a + b = n + 1 = 1 (mod 2n), we have a + (1 -b) = (1 -a) + b (mod 2n) that is to say a = b. We thus obtain S = f 2n-2 3 (af 3 + f 2 ) (1 -a)f 3 + f 2 (af 3 + f 1 ) (1 -a)f 3 + f 1 (f 3 + f 2 + f 1 ) with a ∈ [2, n -1]. Yet, S ′ = f 2n-2a-1 3 (af 3 + f 2 )(af 3 + f 1 )(f 3 + f 2 + f 1 ) | S is a nonempty zero-sum subsequence of length |S ′ | = 2n -2a + 2 ≤ 2n.
In the case d = f 1 + f 2 and α = n + 1, we have w = (n + 1)

f 3 + f 2 + f 1 . Since a+(1-b) = (1-a)
S = f 2n-2 3 (af 3 +f 2 ) (1-a)f 3 +f 2 (n+1-a)f 3 +f 1 (a-n)f 3 +f 1 (n+1)f 3 +f 2 +f 1 with a ∈ [2, n-1]. Yet, S ′ = f 2n-2a-1 3 (af 3 +f 2 ) (a-n)f 3 +f 1 (n+1)f 3 +f 2 +f 1 | S is a nonempty zero-sum subsequence of length |S ′ | = 2n -2a + 2 ≤ 2n. If V is of type (D6), we have V = 2n i=1 (f 3 + d i )f 2 f 1 where 2n i=1 d i ∈ F ( f 1 , f 2 ) is such that 2n i=1 d i = f 1 + f 2 . Then w = f 3 + d for some d ∈ f 1 , f 2 so that we can write S = 2n+1 i=1 (f 3 + d i )f 2 f 1 where S ′ = 2n+1 i=1 d i ∈ F ( f 1 , f 2 ).
It is easily seen that such a sequence S contains no short zero-sum subsequence if and only if

S ′ = 2n+1 i=1 d i ∈ F ( f 1 , f 2 )
contains no zero-sum subsequence of size 2n, that is to say if and only if σ(S ′ ) / ∈ supp(S ′ ).

Inverse problem associated to

s (C 2 ⊕ C 2 ⊕ C 2n ) for n ≥ 2
We turn to the inverse problem associated to s

(C 2 ⊕ C 2 ⊕ C 2n ) for n ≥ 2. Again, the case n = 1, that is, C 3 2 ,
is different yet well-known and direct; s(C 3 2 ) -1 = 8 and the only example of a sequence of length 8 without zero-sum subsequence of length 2 is the squarefree sequence of all elements, because the only zero-sum sequences of length exp(G) are g 2 for g ∈ C r

2 . The proof of this result uses Theorem 4.1.

Theorem 5.1. Let G ≃ C 2 ⊕ C 2 ⊕ C 2n , where n ≥ 2. A sequence S over G of length |S| = s(G) -1 = 4n + 2 contains no zero-sum subsequence of length exp(G)
if and only if there exist a basis {f 1 , f 2 , f 3 } of G, where ord(f 1 ) = ord(f 2 ) = 2 and ord(f 3 ) = 2n, and an f ∈ G such that -f + S is equal to one of the following sequences:

(s1) 0 2α+1 f 2n-2α-1 2 f 2n-1-2β 3 (f 3 + f 2 ) 2β+1 (af 3 + f 1 ) (1 -a)f 3 + f 2 + f 1 with a ∈ [2, n -1] and α, β ∈ [0, n -1]. (s2) 0 2n-1 f 2n-1 3 (af 3 + f 2 ) (1 -a)f 3 + f 2 (bf 3 + f 1 ) (1 -b)f 3 + f 1 with a, b ∈ [2, n -1] and a ≥ b. (s3) 0 2α+1 f 2β+1 1 f 2γ+1 2 2n+1 i=1 (f 3 + d i ) where α, β, γ ∈ [0, n -1] are such that α + β + γ = n -1, S ′ = 2n+1 i=1 d i ∈ F ( f 1 , f 2 ) and σ(S ′ ) / ∈ supp(S ′ ).
As in the result for the η-constant, for n = 2 only the last case can occur. Before giving the proof of this result we put the result in context and derive two corollaries. The first is about the height of these extremal sequences, that is, the maximal multiplicity of an element in these sequences.

Corollary 5.2. Let G ≃ C 2 ⊕ C 2 ⊕ C 2n , where n ≥ 2. For every sequence S over G of length |S| = s(G) -1 without zero-sum subsequence of length exp(G) one has h(S) ≥      2n+3 3 if n ≡ 0 (mod 3) 2n+1 3 if n ≡ 1 (mod 3) 2n+5 3 if n ≡ 2 (mod 3)
and these bounds are attained.

Proof. For sequences of the second type in Theorem 5.1, it is clear that h(S) = 2n -1, and the claim follows. For sequences of the first type in Theorem 5.1, considering for example 0 and f 2 we see that h(S) > 2n/2; again the claim follows.

Proof of Theorem 5.1.

Let G ≃ C 2 ⊕ C 2 ⊕ C 2n
, where n ≥ 2. As in the previous proof, we use the inductive method with

H ֒→ G π → G/H
where H is the unique cyclic subgroup of order n such that G/H ≃ C 3 2 . Let S be a sequence over G such that |S| = 4n + 2 that contains no zero-sum subsequence of length 2n. Since |S| = 2(2n -4) + 10 ≥ s(G/H) + 2(2n -4), there exist 2n -3 disjoint subsequences S 1 , . . . , S 2n-3 of S such that

|S i | = 2 and σ (S i ) ∈ H for all i ∈ [1, 2n -3]. In particular, S = S 1 • • • S 2n-3 S 0 where S 0 has length |S 0 | = 8.
On the one hand, all elements of π(S 0 ) have multiplicity at most three and at most one of them has multiplicity at least two, otherwise we could extract from S 0 two disjoint subsequences S 2n-2 and S 2n-1 satisfying |S 2n-2 | = |S 2n-1 | = 2 and σ (S 2n-2 ) , σ (S 2n-1 ) ∈ H so that S would contain a zero-sum subsequence of length 2n, a contradiction. In particular, we have | supp(π(S 0 ))| ≥ 6.

On the other hand, T = 2n-3 i=1 σ(S i ) is a sequence over H of length |T | = 2n-3, containing no zero-sum subsequence of length n, otherwise S would contain a zerosum subsequence of length 2n, a contradiction.

By Theorem 3.2 we get that |H \ (-Σ n-2 (T )) | ≤ 1. Since | supp(π(S 0 ))| ≥ 6 it follows by Lemma 3.4 that one can extract from S 0 a subsequence U | S 0 such that |U | = 4 and σ (U ) ∈ H. Since S contains no zero-sum subsequence of length 2n, it follows that σ (U ) is an element of H \ -(Σ n-2 (T )). Since we saw that |H \ -(Σ n-2 (T )) | ≤ 1, it follows that this element is uniquely determined and |H \ (-Σ n-2 (T )) | = 1.
By Theorem 3.2 we get that there exist two elements g, h ∈ H satisfying ord(hg) = n such that, by relabeling the S i for i ∈ [1, 2n -3] if necessary, S can be decomposed as S 1 • • • S 2n-3 S 0 where σ(S i ) = g for all i ∈ [1, n -1] and σ(S i ) = h for all i ∈ [n, 2n -3]. In particular, H \ (-Σ n-2 (T )) = {g + h}. We now distinguish cases according to the cardinality of | supp(π(S 0 )|.

Case 1. | supp(π(S 0 ))| = 8. In this case, let x be any element of S 0 and S ′ 0 = S 0 x -1 . Note that S ′ 0 is a subsequence of S 0 of length |S ′ 0 | = 7 and that π(S ′ 0 ) consists of seven distinct elements of G/H. Now, let q, r be two distinct elements of S ′ 0 and decompose S ′ 0 r -1 in such a way that S ′ 0 r -1 = qstuvw, where

π(q) + π(s) = π(t) + π(u) = π(v) + π(w) = π(r) + π(x). It is easily seen that π(s) + π(t) + π(v) and π(s) + π(t) + π(w) are two distinct elements of the set (G/H) \ π(S ′ 0 r -1 ) = {π(r), π (x)
}, so that one of them, say π(s) + π(t) + π(w), is equal to π(r). It follows that U 1 = sqtu, U 2 = sqvw, U 3 = strw, U 4 = suvr and U 5 = tuvw are five subsequences of S ′ 0 of length 4 with sum in H. If n = 2, let y be the only element of H satisfying ord(y) = n. Since S contains no zero-sum subsequence of length 4, we have σ(U i ) = y for every i ∈ [START_REF] Edel | Sequences in abelian groups G of odd order without zero-sum subsequences of length exp(G)[END_REF][START_REF] Fox | Erdős-Ginzburg-Ziv constants by avoiding three-term arithmetic progressions[END_REF]. Therefore, we have σ(U 1 ) + σ(U 2 )σ(U 3 )σ(U 4 ) = 0 which is equivalent to 2q = 2r. Since the argument applies to any three distinct elements q, r, x of S 0 , we obtain that 2a ∈ H is constant over all elements a of S 0 . Now, σ(U 1 ) + σ(U 2 )σ(U 5 ) = y readily gives 0 = 2(2q) = 2s + 2q = y, which is a contradiction.

If n ≥ 3, then σ(U i ) ∈ H \ (-Σ n-2 (T )) = {g + h} for all i ∈ [START_REF] Edel | Sequences in abelian groups G of odd order without zero-sum subsequences of length exp(G)[END_REF][START_REF] Fox | Erdős-Ginzburg-Ziv constants by avoiding three-term arithmetic progressions[END_REF]. Therefore, σ(U i ) = g + h for all i ∈ [START_REF] Edel | Sequences in abelian groups G of odd order without zero-sum subsequences of length exp(G)[END_REF][START_REF] Fox | Erdős-Ginzburg-Ziv constants by avoiding three-term arithmetic progressions[END_REF] which yields σ(U 1 ) + σ(U 2 )σ(U 3 )σ(U 4 ) = 0 which is equivalent to 2q = 2r. Since the argument applies to any three distinct elements q, r, x of S 0 , we obtain that 2a ∈ H is constant over all elements a of S 0 . The equality σ(U 1 ) + σ(U 2 )σ(U 5 ) = g + h readily gives 4q = 2(s + q) = g + h so that 4a = g + h for any element a of S 0 .

Since supp(π(S 0 )) = G/H, for any i ∈ [1, 2n -3] and any a i | S i , there exists an element a of S 0 such that π(a) = π(a i ). Setting T 0 = S 0 a -1 a i , T i = S i a -1 i a and T j = S j for all j ∈ [1, 2n -3] such that i = j, we obtain a new decomposition of S having the form

S = T 1 • • • T 2n-3 T 0 ,
where there exist two elements g ′ , h ′ ∈ H satisfying ord(h ′g ′ ) = n such that, by relabeling the T i for i ∈ [1, 2n -3] if necessary, we have σ(T i ) = g ′ for all i ∈ [1, n -1] and σ(T i ) = h ′ for all i ∈ [n, 2n -3]. Using the same argument as above, we obtain that 2x ∈ H is constant over all elements x of T 0 , and that 4x = g ′ + h ′ for any x of T 0 . Since at least seven elements of T 0 and S 0 are equal, it follows that g ′ + h ′ = g + h, and 4a = 4a i .

Case 1.1. n is odd. Since 4a = 4a i , it follows that 2a = 2a i , and since moreover π(a) = π(a i ), we get that a = a i . Since a i was chosen arbitrarily, it follows that supp(S) = supp(S 0 ). In particular, this implies that S i = a 2 i for each i ∈ [1, 2n -3]; to see this just recall that σ(S i ) ∈ H implies that both elements in S i have the same image under π.

We recall that 2a is constant over all elements in S 0 , and thus in S. Yet, this would mean that σ(S i ) is the same for all i ∈ [1, 2n -3], that is, g = h, a contradiction.

Case 1.2. n is even. Since we assumed n ≥ 3, we have n ≥ 4 (and this is actually all that we use). Applying the replacement above with some i ∈ [n, 2n -3], it can be seen that h ′ = h (here we use n ≥ 4, so that the original sequence contains h not only once). Moreover, it follows that S i = a 2 where a is again the element of S 0 such that π(a i ) = π(a).

In particular, h = σ(S i ) = 2a. Since we already know that 4a = g + h, we obtain g + h = 4a = 2(2a) = 2h, that is, g = h, a contradiction. Case 2. | supp(π(S 0 ))| = 7. The argument is similar to the preceding case. There exists a subsequence S ′ 0 | S 0 of length |S ′ 0 | = 7 such that π(S ′ 0 ) consists of seven distinct elements of H. Now, let q, r be two distinct elements of S ′ 0 . We denote by γ the only element in G/H \ {π(S ′ 0 )} and decompose S ′ 0 r -1 in such a way that S ′ 0 r -1 = qstuvw, where π(q) + π(s) = π(t) + π(u) = π(v) + π(w) = π(r) + γ. It is easily seen that π(s)+π(t)+π(v) and π(s)+π(t)+π(w) are two distinct elements of the set G/H \ π(S ′ 0 r -1 ) = {π(r), γ}, so that one of them, say π(s) + π(t) + π(w) is equal to π(r). It follows that U 1 = sqtu, U 2 = sqvw, U 3 = strw, U 4 = suvr and U 5 = tuvw are five subsequences of S ′ 0 of length 4 with sum in H. If n = 2, let y be the only element of H satisfying ord(y) = n. Since S contains no zero-sum subsequence of length 4, we have σ(U i ) = y for every i ∈ [START_REF] Edel | Sequences in abelian groups G of odd order without zero-sum subsequences of length exp(G)[END_REF][START_REF] Fox | Erdős-Ginzburg-Ziv constants by avoiding three-term arithmetic progressions[END_REF]. Therefore, we have σ(U 1 ) + σ(U 2 )σ(U 3 )σ(U 4 ) = 0 which is equivalent to 2q = 2r. Since the argument applies to any two distinct elements of S ′ 0 , we obtain that 2a ∈ H is constant over all elements a of S ′ 0 . Now, σ(U 1 ) + σ(U 2 )σ(U 5 ) = y readily gives 0 = 2(2q) = 2(s + q) = y, which is a contradiction.

We can thus assume that n ≥ 3, so that σ(U i ) ∈ H \ (-Σ n-2 (T )) = {g + h} for all i ∈ [START_REF] Edel | Sequences in abelian groups G of odd order without zero-sum subsequences of length exp(G)[END_REF][START_REF] Fox | Erdős-Ginzburg-Ziv constants by avoiding three-term arithmetic progressions[END_REF]. Therefore, σ(U i ) = g + h for all i ∈ [1, 5] which yields σ(U 1 ) + σ(U 2 )σ(U 3 )σ(U 4 ) = 0 which is equivalent to 2q = 2r. Since the argument applies to any two distinct elements of S ′ 0 , we obtain that 2a ∈ H is constant over all elements a of S ′ 0 . The equality σ(U 1 ) + σ(U 2 )σ(U 5 ) = g + h readily gives 4q = 2(s + q) = g + h so that 4a = g + h for any element a of S ′ 0 . Now, let a be an element of S ′ 0 and let V 1 be a subsequence of S 0 such that |V 1 | = 6 and σ(V 1 ) ∈ H; such a sequence exists as π(S 0 ), not being squarefree, has a zero-sum subsequence of length 2, and the remaining squarefree sequence of length 6 has a zero-sum subsequence of length 4 by Lemma 3.4.

If n = 3, then σ(V 1 ) ∈ H \{0} = {2h+g, h+2g}. If n ≥ 4, then σ(V 1 ) ∈ H \(-Σ n-3 (T )) = {2h+g, h+2g} also. In all cases, we obtain either 2(2h + g) = 2σ(V 1 ) = 6(2a) = 3(4a) = 3(g + h) or 2(h + 2g) = 2σ(V 1 ) = 6(2a) = 3(4a) = 3(g + h) which both imply g = h, a contradiction.
Case 3. supp(π(S 0 )) = 6. We get that π(S 0 ) consists of one element repeated three times and five other elements; recall that at most one element of π(S 0 ) has multiplicity greater than one. Thus S 0 contains a subsequence S 2n-2 | S 0 such that |S 2n-2 | = 2 and σ (S 2n-2 ) ∈ H. It follows that, whatever n ≥ 2 we consider, there exist two elements g, h ∈ H satisfying ord(h-g) = n such that S can be decomposed as

S 1 • • • S 2n-2 S ′ 0 where σ(S i ) = g for all i ∈ [1, n-1], σ(S i ) = h for all i ∈ [n, 2n-2], and S ′ 0 has length |S ′ 0 | = 6; furthermore, supp(π(S 1 • • • S 2n-2 )) ∩ supp(π(S ′ 0
)) = ∅. Now, let us prove that, after a translation by an element of G, the sequence S can be decomposed as 0T

(f ) = π(a i ). If i ∈ [1, n-1]
, then replacing S i by f a i or f (g-a i ), we easily infer that f = a i = g-a i .

In particular, 2f = g, so that We now know that for every sequence S over G such that |S| = 4n + 2 and containing no zero-sum subsequence of length 2n, there is an f ∈ G such that -f + S can be decomposed as 0T

-f + S = 0(-f + S 1 ) • • • (-f + S n-1 )V where V has length |V | = 4n + 2 -2(n -1) -1 = 2n + 3 = η(G) -1. We set T i = -f + S i for each i ∈ [1, n -1]. Since for every t ∈ [1, 2n -1], the sequence 0T 1 • • • T n-1 contains a zero-
• • • T n-1 V where |T i | = 2 and σ(T i ) = 0 for all i ∈ [1, n -1],
and where V , which has length |V | = 2n + 3 = η(G) -1, contains no short zero-sum subsequence.

We recall that S has a zero-sum subsequence of length 2n if and only if -f +S has a zero-sum subsequence of length 2n. Thus, -f + S has no zero-sum subsequence of length 2n. To simplify notation, we assume without loss that f = 0.

On the one hand, note that Σ [1,2n-1] (V ) = G\{0}. On the other hand, we assert that: if x | T i for some i ∈ [1, n -1], then x / ∈ -Σ [2,2n-1] (V ). To see this note that otherwise V would contain a subsequence ). The former is a consequence of Theorem 4.1 and the latter follows by noting that we need at least ord(f 3 ) = 2n elements in a subsequence of 2n+1 i=1 (f 3 + d i ) whose sum is in f 1 , f 2 .

V ′ | V of length |V ′ | = k ∈ [2,

Theorem 3 . 1 . 1 )( 2 )

 3112 Let H ≃ C n where n ≥ 3 is an integer. (Every sequence T of length |T | = η(H) -1 = n -1 not containing any short zero-sum sequence has the form T = g n-1 where ord(g) = n. Every sequence S of length |T | = η(H) -2 = n -2 not containing any short zero-sum sequence has the form: (a) T = g n-2 where ord(g) = n, or (b) T = g n-3 (2g) where ord(g) = n.

  containing no nonempty zero-sum subsequence, otherwise S would contain a short zero-sum subsequence, a contradiction. By Theorem 3.1 we get |H \ -(Σ (T ) ∪ {0}) | ≤ 1. Since | supp(π(S 0 ))| ≥ 5, one can extract from S 0 a subsequence U | S 0 such that |U | ≤ 4 and σ (U ) ∈ H (see Lemma 3.3); in fact, we could even find a sequence of length at most 3. Since S contains no short zero-sum subsequence, it follows that σ (U ) ∈ H \-(Σ (T ) ∪ {0}).

  where a ≤ b ≤ c, and a, b ∈ [2, n-1], and c ∈ [2, 2n-3]\{n, n+1}. Now, U can be decomposed as the product of two sequences of length 2 having the same sum only if 1 + a = b + c (mod 2n) or 1 + b = a + c (mod 2n) or 1 + c = a + b (mod 2n); recall that a + b + c = 1 (mod 2n). This is possible only if a ∈ {0, n} or b ∈ {0, n} or c ∈ {0, n}, which yields a contradiction.

  +b (mod 2n) if and only if a = b which implies a+1-b = 1 = n+1 (mod 2n), we have a + b = (1a) + (1b) (mod 2n) if and only if a + b = n + 1 (mod 2n) . We thus obtain

  • • T n-1 V where |T i | = 2 and σ (T i ) = 0 for all i ∈ [1, n -1], and where V , which has length |V | = 2n + 3 = η(G) -1, contains no short zero-sum subsequence. Since supp(π(S 1 • • • S 2n-2 )) ∩ supp(π(S ′ 0 )) = ∅, there exist an element f of S ′ 0 and an element a i of some S i , where i ∈ [1, 2n -2], such that π

2 f 2n- 1 -2v 3 (f 3 + f 2 ) 2 ) 3 (f 3 + f 2 ) 3 (f 3 + f 2 ) 3 ((f 3 +

 21332233233233 2n -1] such that x = -σ(V ′ ). If k were odd, then the subsequence S ′ = xV ′ j∈J T j | S, where J is any subset of [1, n -1] \ {i} satisfying |J| = n -(k + 1)/2 ∈ [0, n -2] would have length |S ′ | = 2n and sum zero, a contradiction. If k were even, then the subsequenceS ′ = 0xV ′ j∈J T j | S, where J is any subset of [1, n -1] \ {i} satisfying |J| = n -(k + 2)/2 ∈ [0, n -2] would have length |S ′ | = 2n and sum zero, a contradiction. Therefore, any x ∈ supp(T 1 • • • T n-1 ) is either zero or an element of G \ {0} \ (-Σ [2,2n-1] (V )) = -Σ [1,2n-1] (V ) \ (-Σ [2,2n-1] (V )) =supp(V ). Since x | T i if and only if -x | T i , we obtain supp(T 1 • • • T n-1 ) ⊂ {0} ∪ supp(V ) ∩ (-supp(V )) .Finally, using Theorem 4.1, we know that there exists a basis {f 1 , f 2 , f 3 } of G, where ord(f 1 ) = ord(f 2 ) = 2 and ord(f 3 ) = 2n, such that has one out of three possible forms.If V is of type (η1), then supp(T 1 • • • T n-1 ) ⊂ {0} ∪ supp(V ) ∩ (-supp(V )) = {0, f 2 },so that we haveS = 0 2α+1 f 2n-2α-2 2v+1 (af 3 + f 1 ) (1a)f 3 + f 2 + f 1 with α, β ∈ [0, n -1] and a ∈ [2, n -1]. It remains to check that such a sequence contains no zero-sum subsequence of length 2n. Since Σ(02α+1 f 2n-2α-2 ⊂ {0, f 2 }, it suffices to check that 0 / ∈ Σ [1,2n] f 2n-1-2v 2v+1 (af 3 + f 1 ) (1a)f 3 + f 2 + f 1 and that f 2 / ∈ Σ [1,2n-1] (f 2n-1-2v 2v+1 (af 3 + f 1 ) (1a)f 3 + f 2 + f 1 .The former is a consequence of Theorem 4.1 and the latter follows by noting that if a subsequence with sum f 2 contains (af 3 + f 1 ) or (1a)f 3 + f 2 + f 1 , then it contains both, so that we need at least ord(f 3 ) = 2n elements in this subsequence.If V is of type (η2), then supp(T 1 • • • T n-1 ) ⊂ {0} ∪ supp(V ) ∩ (-supp(V )) = {0},so that, up to translation by an element of G, we haveS = 0 2n-1 f 2n-1 af 3 + f 2 ) (1a)f 3 + f 2 (bf 3 + f 1 ) (1b)f 3 + f 1 with a, b ∈ [2, n -1] and a ≥ b. Such a sequence contains no zero-sum subsequence of length 2n indeed. If V is of type (η3), then supp(T 1 • • • T n-1 ) ⊂ {0} ∪ supp(V ) ∩ (-supp(V )) = {0, f 1 , f 2 },so that, up to translation by an element of G, we have S = 0 2α+1 f d i ) where α, β, γ ∈ [0, n -1] are such that α + β + γ = n -1, and S ′ = 2n+1 i=1 d i ∈ F ( f 1 , f 2 ) and σ(S ′ ) / ∈ supp(S ′ ). It remains to check that such a sequence contains no zero-sum subsequence of length 2n. Since Σ(0 2α+1 f 2β+1 1 f 2γ+1 2 ) = {0, f 1 , f 2 }, it suffices to check that 0 / ∈ Σ [1,2n] ( 2n+1 i=1 (f 3 + d i )) and that f 1 , f 2 / ∈ Σ [1,2n-1] ( 2n+1 i=1 (f 3 + d i )

  sum subsequence of length t, it follows that V has no short zero-sum subsequence; otherwise -f + S and thus S would have a zero-sum subsequence of length 2n.If i ∈ [n, 2n-2], the argument is analogous, just replacing g by h and considering the sequence S n , . . . , S 2n-2 .

Acknowledgments

The authors thank the referee for a careful reading and useful remarks that helped to improve the paper.

For sequences of the third type in Theorem 5.1, considering 0, f 1 , and f 2 we see that h(S) ≥ 2⌊(n -1)/3⌋ + 1, which yields the claimed bounds.

To see that the bounds are attained we consider the sequence, where {f 1 , f 2 , f 3 } is a basis of G with ord(f 1 ) = ord(f 2 ) = 2 and ord(f 3 ) = 2n,

which is of the form given in Theorem 5.1 as the sum of f 2α+1 1

An interesting aspect is that for C 2 ⊕ C 2 ⊕ C 2n there are extremal sequences of height significantly below exp(G)/2, contrary to all results established so far (see the discussion in [START_REF] Schmid | Restricted inverse zero-sum problems in groups of rank 2[END_REF], in particular Corollary 3.3 there, where the still stronger conjecture that the height is always exp(G)-1 was refuted). This seems noteworthy as there is a well-known technical result (see [START_REF] Gao | On zero-sum subsequences of restricted size II[END_REF]Proposition 2.7] or also [START_REF] Schmid | Restricted inverse zero-sum problems in groups of rank 2[END_REF]Lemma 4.3]) that implies that if h(S) ≥ ⌊(exp(G) -1)/2⌋ for each sequence S over G of length s(G) -1 without zero-sum subsequence of length exp(G), then Gao's conjecture s(G) = η(G) + exp(G) -1 that we recalled in the introduction holds true for the group G. Thus, it seems interesting that despite the relatively low height of some extremal sequences over C 2 ⊕ C 2 ⊕ C 2n Gao's conjecture still holds. An explanation for this can be given by a more detailed analysis of the structure of extremal sequences and a recent refinement of the above mentioned technical result (see Lemma 5.4 below).

with nonnegative integers u, v, w, and

Proof. This follows rather directly by comparing the sequences given in Theorems 4.1 and 5.1. For a sequence S of the form given in point (s1) the sequence T is of the form given in (η1) and likewise for the other points; for the sequences in (s1) and (s2) only more special forms of sequences C arise.

We note that the sequences (f + C), with C as in the corollary above, have the property jf ∈ Σ j (f + C) for each j ≤ |f + C| thus the following lemma (see [START_REF] Girard | Direct zero-sum problems for certain groups of rank three[END_REF]Lemma 4.4]) is applicable. After this discussion we now turn to the proof of the theorem itself.


Margaux Renvoisé

Ludovic Bonhomme

Marlene Davanture

Michel Zivy

Claire Lemaire
email: claire.lemaire@i2bc.paris-saclay.fr

Marlène Davanture

Phosphoproteomic Analysis of Isolated Mitochondria in Yeast

Keywords: Saccharomyces cerevisiae, mitochondria, phosphoproteome

ou non, émanant des établissements d'enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

1.Introduction

Phosphorylation is the most frequent post-translational modification: 30% of the proteins are phosphorylated at some point during their cell cycle (1). This is a reversible process which varies especially during environmental changes (2, 3), the cell cycle (4), stress, or a nutritional change (5), suggesting a role of phosphorylation in environmental adjustment. A large number of phosphorylated mitochondrial proteins have been identified and phosphorylation appears involved in various functions such as transcription, apoptosis or energy metabolism (6). As mitochondria are a small cellular compartment, their isolation is necessary for the analysis of their phosphoproteome. This is illustrated by the work of Amoutzias et al. (7), who summarized several phosphoproteomic studies performed on yeast cells: they listed more than 10 000 sites of phosphorylation but only about one hundred were located on mitochondrial proteins. Phosphoproteomic analysis were developed using either labelling techniques with 32 P (8), detection with phosphoantibodies, fluorescent labelling with ProQ Diamond (9) or phosphorylated peptide enrichment [START_REF] Lee | Mitochondrial phosphoproteome revealed by an improved IMAC method and MS/MS/MS[END_REF]11). In all the cases, the identification of the phosphorylated residue required analysis by mass spectrometry. Analysis by mass spectrometry allowed a significant improvement of the knowledge of mitochondrial phosphoproteome in various organisms. However, phosphorylated proteins remain less abundant than their non-phosphorylated counterparts and the ionization of phosphopeptides is less effective than non-phosphorylated peptides, making their detection by mass spectrometry more difficult. Phosphoproteome analysis therefore requires specific analysis techniques, in particular enrichment in phosphopeptides upstream of the LC-MS / MS. The most widely used methods of enrichment are the Immobilized Metal ion Affinity Chromatography (IMAC) [START_REF] Lee | Mitochondrial phosphoproteome revealed by an improved IMAC method and MS/MS/MS[END_REF] which is based on the high affinity of phosphates to certain trivalent metal ions, and metal oxide chromatography (MOC) (11) using the titanium dioxide (TiO2) or zirconium dioxide (ZrO2). To optimize this enrichment step, the samples are prefractionated, either by SDS-PAGE, or by chromatography such as Strong Cation Exchange Chromatography (SCX) or hydrophilic interaction liquid chromatography (HILIC) [START_REF] Zarei | Comparison of ERLIC-TiO2, HILIC-TiO2, and SCX-TiO2 for Global Phosphoproteomics Approaches[END_REF]. These different strategies are optimal for the analysis of phosphorylated serines and threonines. Because of their presence in small quantities, phospho-tyrosines are often analysed after specific enrichment such as immunoprecipitation using an antibody against phospho-tyrosine (13). Histidine phosphorylation is not detected by standard LC-MS / MS analyzes, because they are generally conducted at acidic pH, at which phosphorylation is labile for this residue. The detection of this phosphorylated residue thus requires specific analysis strategy [START_REF] Besant | Detection and Analysis of Protein Histidine Phosphorylation[END_REF]. Finally, quantitative phosphoproteomic analysis can be performed using i) labelling methods either metabolic labelling (SILAC) [START_REF] Ong | Stable Isotope Labeling by Amino Acids in Cell Culture, SILAC, as a Simple and Accurate Approach to Expression Proteomics[END_REF] or chemical labelling (iTRAQ [START_REF] Boja | Quantitative mitochondrial phosphoproteomics using iTRAQ on an LTQ-Orbitrap with high energy collision dissociation[END_REF], multiplex stable isotope dimethyl labelling [START_REF] Boersema | Multiplex Peptide Stable Isotope Dimethyl Labeling for Quantitative Proteomics[END_REF]) ii) Single Reaction Monitoring/ Multiple Reaction Monitoring (SRM/MRM) analysis [START_REF] Lam | An MRM-based workflow for quantifying cardiac mitochondrial protein phosphorylation in murine and human tissue[END_REF].

In this chapter, we present two techniques (presented schematically in fig. 1) that we have carried out on isolated mitochondria to analyse protein phosphorylation state. The first one relies on Blue native and Sodium Dodecyl Sulfate -PolyAcrylamide Gel Electrophoresis (2D -BN /SDS-PAGE) associated to Pro-Q Diamond staining (sections 3.1 and 3.3 to 3.5). The second one is quantitative and uses multiplex stable isotope dimethyl labelling followed by fractionation by SCX and phosphopeptides enrichment by IMAC before LC-MS/MS analysis (sections 3.1, 3.2 and 3.6 to 3.9). This second approach allowed us to produce the first extensive study providing quantitative data on phosphosite responses to different carbon substrates in the yeast Saccharomyces cerevisiae [START_REF] Renvoisé | Quantitative variations of the mitochondrial proteome and phosphoproteome during fermentative and respiratory growth in Saccharomyces cerevisiae[END_REF].

Fig. 1: Scheme of two experimental approaches for the phosphoproteomic analysis of isolated mitochondria. Starting with pure yeast mitochondria, the qualitative approach permits to reveal the presence of phosphoproteins in the different complexes separated by BN-PAGE, while the quantitative approach allows the analysis of phosphopeptide abundance variations.

Materials

Prepare all solutions using ultrapure water (sensitivity 18 MΩ cm at 25°C).

Cell culture

Yeast WT strain was grown in galactose, glucose or lactate media (see Note 1).

2D-electrophoresis -Second dimension: SDS-PAGE

1. Acrylamide/Bis solution: 40% solution 37.

Methods

Preparation of mitochondria by differential centrifugation

Yeast wild type strain is grown either in 2% glucose medium or 2% galactose medium to an OD600nm = 6, or 2% lactate medium to an OD600nm = 4.5.

The following protocol is the one developed by Meisinger et al. [START_REF] Meisinger | Purification of Saccharomyces cerevisiae Mitochondria Devoid of Microsomal and Cytosolic Contaminations[END_REF] with slight modifications.

1. Harvest the cells by centrifugation at 3000g for 10 min at 4°C and wash with distilled water 2. Resuspend the cells in 2mL/ g (wet weight) DTT buffer ( see Note 5) and shake at 30°C, 20 min in an incubator (Infors -120rpm) 3. Centrifuge at 3000g , 5 min 4. Wash the pellets in zymolyase buffer 5. Centrifuge at 3000g , 5 min 6. Resuspend the pellets in 7mL/g (wet weight) of zymolyase buffer 7. Digest the cell wall enzymatically with Zymolyase-20T at 3mg/g of cells at 30°C for about 1H. (see Note 6) 8. Centrifuge at 3000g , 5 min at 4°C 9. Resuspend the pellets in 6.5mL/g (wet weight) of homogeneisation buffer (see Note 7) 10. Homogenize with 5 pestle strokes in a glass-teflon potter 11. Centrifuge at 1500g , 5 min at 4°C to eliminate cell debris and nuclei 12. Keep the supernatant and centrifuge at 4000g , 5 min at 4°C 13. Keep the supernatant and centrifuge at 12000g, 15 min at 4°C in order to harvest the crude mitochondria.

Preparation of mitochondria purified on sucrose gradient

Steps 1 to 4 follow the protocol developed by Meisinger et al. [START_REF] Meisinger | Purification of Saccharomyces cerevisiae Mitochondria Devoid of Microsomal and Cytosolic Contaminations[END_REF].

1. Resuspend the crude mitochondria pellet in 0.5 mL of SEM buffer containing 1mM PMSF, to have a final concentration of mitochondria of 5 mg/mL approximately.

2. Homogenize the mitochondria with 10 up-and-down strokes in a Potter homogenizer and load on a three-step sucrose gradient (1.5mL 60%, 4mL 32%, 1.5mL 23% and 1.5mL 15%).

3. Centrifuge the sucrose gradient at 134 000 g at 4°C during 1h in a SW41 Ti swinging bucket. 4. Collect the purified mitochondria at 60%-32% interface and resuspend in 1mL of SEM buffer containing 1mM PMSF to have a final concentration of mitochondria of 1 mg/mL approximately. 5. Measure the protein concentration of purified mitochondria using the 2-D Quant Kit with BSA as a standard. 6. Aliquot the solution of purified mitochondria in 2mL safe lock tubes.

7. Centrifuge the aliquots at 14000g at 4°C during 15 min; throw the supernatants. 8. Then, in order to avoid degradation of proteins and to limit dephosphorylation events, precipitate the proteins by adding 2 mL of TCA 10% in each tube. 9. Homogenize the protein solutions by inversion of the tubes and incubate at 4°C during 15 min. 10. Centrifuge the protein solutions at 14000g at 4°C during 15 min; throw the supernatants. 11. Add 2 mL of acetone (-20°C) in every tube to eliminate the excess of TCA and transfer the protein solutions in Corex tubes. 10. Homogenize the protein solutions and centrifuge at 14000 g at 4°C during 15 min; Throw the supernatants. 11. Dry the protein precipitates and store at -20°C.

BN-PAGE/ SDS -2D electrophoresis

Method for non-denaturing gels was initially developed in (21)

Solubilization of mitochondria

1. Prior to their solubilisation, treat the mitochondria with DNAse: resuspend in order to have a final concentration of mitochondria equal to 1 mg/mL in DNAse solution. Incubate the sample 30 min at 25°C. 2. Centrifuge the sample at 13 000g at 4°C during 15 min. 3. Resuspend the pellet of mitochondria in solubilisation buffer and incubate at 4°C during 30 min. 4. Centrifuge the sample at 100 000g and at 4°C during 30 min. Throw the pellet. 5. Add Serva Blue G in the supernatant in order to have a final concentration of 0.25%. (See Note 8).

Preparation of BN-PAGE

1. Prepare a 5%-10% acrylamide gradient resolving gel in a 8 cm x 6.5 cm x 0.10 cm gel cassette. 2. Load a 3% acrylamide stacking gel on the resolving gel. Then immediately insert a 10well gel comb. 3. Load the samples in the wells: the volume of samples is calculated to have 25μg of proteins/lane. After the loading of samples, completely fill the wells with cathode buffer A. 4. Perform electrophoresis at 4°C. 5. During the first electrophoresis period, cathode buffer is buffer A. Perform electrophoresis at 100V constant during 1h then at 200V constant during 1h more.

6. During the second electrophoresis period, cathode buffer is buffer B. Perform electrophoresis at 200V until the dye front has reached the bottom of the gel.

Preparation of the sample for the second dimension

Immediately following BN-PAGE, cut a lane of interest and incubate it in 60mM Na2CO3, 60 mM DTT and 2% SDS during 30 minutes.

Preparation of SDS-PAGE

1.

Prepare a 10%-20% acrylamide gradient resolving gel in a 8 cm x 5 cm x 0.15 cm gel cassette.

2.

Load a 5% acrylamide stacking gel on the polymerised resolving gel. Insert a preparative comb containing a wide single well and a narrow reference. 3. Carefully load one BN-PAGE lane, previously incubated as described in 2.3.3., on the gel to avoid any bubble and pour the resolving gel on the lane. Load the protein standards in the reference well. 4. Perform electrophoresis at 4°C and 120V constant until the dye front has reached the bottom of the gel.

ProQ Diamond staining (see Fig.2)

1. Following electrophoresis, incubate the gel in the fixation solution, during 1h at least. 2. Then, wash 3 times with water; each wash lasting 15 minutes. 3. Incubate with the Pro-Q Diamond during 2h in the darkness. 4. Incubate three times in the discoloration solution in the darkness; each incubation lasting 30 minutes. 5. Wash 2 times with water in the darkness; each wash lasting 5 minutes. 6. Phosphoproteins are then detected by fluorescence detection: the excitation maximum of Pro-Q Diamond is at 550 nm and its emission maximum at 580 nm.

Sypro Ruby staining

1. Following Pro-Q Diamond detection, incubate the gel in Sypro Ruby during 1 night in the darkness. 2. Incubate three times in the discoloration solution in the darkness; each incubation lasting 30 minutes. 3. Wash 2 times with ultrapure in the darkness, each washing step lasting 5 minutes.

4. Proteins are then detected by fluorescence detection: the two excitation maxima of Sypro Ruby are at 280 and 450 nm and its emission maximum at 610 nm.

Preparation and labeling of mitochondrial peptides for quantitative mass spectrometry experiment

Suspension

1. Dry the protein precipitates in corex tubes under vacuum during 15 min. 2. Resuspend the proteins in 400 µL of ZUT buffer in global, mix by vortex and apply short spin centrifuge (i.e. 200 rpm). Additional 200 μL of ZUT buffer can be added if necessary (see Note 9). 3. Collect the protein solutions and measure the protein concentration with the 2D Quant kit 3.6.2. Alkylation 1. Add 1M iodoacetamide in each sample to have a final concentration of iodoacetamide equal to 40 mM. 2. Incubate the samples 1h in darkness.

In-solution trypsin digestion

1. Dilute the samples with 7 volumes of 50mM of ammonium bicarbonate. 2. Add trypsin in samples in order to have a trypsin/protein ratio of 1/30. Before its addition in samples, resuspend the trypsin with 50mM ammonium bicarbonate (200µL ammonium bicarbonate for 40µg of trypsin). 3. Incubate the samples overnight at 37°C. 4. Stop the digestion by addition of acid in order to reach pH 2. 5. Dry the samples under vacuum using a speed vac.

Stable isotope on column labelling

The multiplex stable isotope dimethyl labelling procedure is those developed by Boeserma et al. [START_REF] Boersema | Multiplex Peptide Stable Isotope Dimethyl Labeling for Quantitative Proteomics[END_REF] ! CAUTION Formaldehyde solutions and formaldehyde vapors are toxic; place the vacuum manifold in a fume hood. The vacuum manifold solvent liners should be inert and disposable to prevent carry-over between experiments.

1. Samples are suspended with formic acid 5% (approximately pH 2). 2. Labelling of samples is made on Sep-Pak C18 columns (see 2.7.2). The pressure must not exceed 5 mmHg (vacuum manifold system). Columns are prepared with 2mL of acetonitrile and 2x2mL of buffer A. 3. Samples are loaded on columns 4. Columns are washed with 2 mL of buffer A. 5. Peptides fixed in the columns are labelled with 7 x 1 mL of the appropriate labelling solution. For complete labeling, make sure that step takes at least 10 min. 6. Columns are washed with 2mL of buffer A. 7. Labelled peptides are eluted with 2 x 500µL of buffer B. 8. Labelled peptides issued from different samples are mixed at a 1:1 ratio and dried by vacuum centrifugation.

1. Before the phosphopeptides enrichment, PHOS-Select Iron affinity gel is prepared as following:

1.1. Homogenize the resin in glycerol 1.2. Centrifuge at 500 g and at room temperature during 2 min: supernatant is thrown 1.3. Resuspend the resin in the Equilibration buffer and washed 4 times with this buffer 2. Resuspend each fraction with 300 µL of equilibration buffer and add 80 µL of resin. Incubate 1hour with mixing at room temperature (end over-end rotation recommended). 3. Load each fraction on one spin column and make a short spin centrifugation until 5000g. 4. Wash the spin column two times with 200µL of equilibration buffer and one time with 200µL of ultrapure water (short spin centrifugation until 5000g). 5. Elute the phosphopeptides with 2 x 30 µL of elution buffer: make a first short spin centrifugation until 2000g and a second one until 14000g (see Note 12).

Phosphopeptide analysis LC-MS/MS

1. Resuspend the fractions in 20 μl of buffer G. 2. Load 4 μL at 7.5 μL/min on the pre-column cartridge with buffer H. 3. Desalt the peptides with solvent I during 3 min. 4. Load the peptides on the analytical column and separate by a linear gradient from 5% to 30% buffer J during 37 min at 300nl/min. 5. Analyse the eluted peptides with mass spectrometer using Xcalibur 2.1 interface. 6. Data-dependent acquisition of peptide ions included a full MS scan covering the 300 to 1400 mass-to charge ratios (m/z) with a resolution of 70000 and a MS/MS step (normalized collision energy: 30%; resolution: 17500). The MS/MS step was reiterated for the 8 major ions detected during the full MS scan. The dynamic exclusion was set to 45 s.

Bioinformatics Data Analysis

1. Conversion of raw files to mzML format ( ProteoWizard software V3.0.3706). 2. Database searches process: use the XtandemPipeline (http://pappso.inra.fr/bioinfo/xtandempipeline/) to run database search with the X!Tandem search engine [START_REF] Craig | TANDEM: matching proteins with tandem mass spectra[END_REF] and to manage redundancy (grouping of phosphopeptides and proteins harboring the same phosphosite(s)). Export data in MassChroQ format 3. Quantification: use the MassChroQ software [START_REF] Valot | MassChroQ: A versatile tool for mass spectrometry quantification[END_REF] to quantify phosphopeptides according to their extracted ion current (XIC). Align LC-MS runs performed on the same SCX fraction number. 4. Computation of relative quantitative changes: for each phosphopeptide, compute the ratios between light, intermediary and heavy isotopes (if one of the isotopes is chosen as a reference, 2 ratios are computed, the reference being the common denominator). Normalize the ratios by dividing ratios by the median value of all phosphopeptide ratios within each triplex or duplex. Use log-transformed ratios for subsequent statistic tests. 5. When protein abundances were analysed in the same samples, phosphopeptide intensities can also be normalized by protein amounts, to take into account their possible variation in the computation of phosphopeptide quantitative changes.

Conclusion

The methods presented in this chapter are complementary means of studying the mitochondrial phosphoproteome. The qualitative method allows the analysis of the relation between the state of phosphorylation of proteins and their interaction with other proteins within a complex or within complexes interacting together (supercomplexes, for example). The quantitative method allows a comprehensive study on the state of phosphorylation of mitochondrial proteins independently from their possible interactions with each other, but allows the analysis of the effects of different conditions (carbon substrate, aerobic / anaerobic) or mutants. 12. The elution step in the IMAC enrichment protocol must be done quickly: in the presence of elution buffer, resin is not stable more than 5 minutes.

Fig. 2 :

 2 Fig.2: 2D BN /SDS-PAGE electrophoresis. Mitochondria were separated in the first dimension on BN-PAGE after digitonin solubilisation. Mitochondrial complexes (monomeric [V1] and dimeric [V2] complex V) and supercomplexes (IV-III2-IV and III2-IV1) subunits are then separated in a second dimension on SDS-PAGE and phosphoproteins were detected after ProQ Diamond staining.

2.2 Preparation of mitochondria by differential centrifugation

  Mix 1.3 mL of acrylamide:bis-acrylamide solution and 1.67 mL of gel buffer (x3). Complete to 5 mL with H2O. Add 2.5 µL of Temed and 25 µL of 10% ammonium persulfate. 5. Stacking gel (3%): Mix 375 µL of acrylamide: bis solution and 1.67 mL of gel buffer (x3). Complete to 5 mL with H2O. Add 5 µL of Temed and 50 µL of 10% ammonium persulfate. 6. Cathode buffer A: 15 mM Bis-Tris, 50 mM Tricine. The pH is adjusted to 7.0 with a concentrated HCl solution at 4°C. The buffer is filtered on Millipore 0.45 µm filter. Then 0.02% (w/v) Serva Blue G-250 is added and the buffer is sonicated in a bath sonicator to solubilize the Serva Blue G-250. 7. Cathode buffer B: same as cathode buffer A except that no Serva Blue G-250 is added. 8. Anode buffer: 50 mM Bis-Tris. The pH is adjusted to 7.0 with a concentrated HCl solution at 4°C. The buffer is filtered on 0.45 µm filter. 9. Molecular weight markers: High Molecular Weight Calibration Kit for Native Electrophoresis.

	4. 10% acrylamide solution:
	1. Medium 1: 1.25% (w/w) bacto yeast extract, 2.5% (w/w) bactopeptone. Weigh
	5g of bacto yeast extract and 10g of bactopeptone. Complete to a volume of
	400mL with water. Autoclave at 120°C, 20 min.
	2. 10% galactose: weigh 10g of galactose. Complete to a volume of 100mL with
	water. Autoclave at 120°C, 20 min.
	3. 10% glucose: weigh 10g of glucose. Complete to a volume of 100mL with
	water. Autoclave at 120°C, 20 min.
	4. Galactose medium: 1% (w/w) bacto yeast extract, 2% (w/w) bactopeptone, 2%
	(w/w) galactose. Just before use, mix 400mL of medium 1 and 100mL of 10%
	galactose.
	5. Glucose medium: 1% (w/w) bacto yeast extract, 2% (w/w) bactopeptone, 2%
	(w/w) glucose. Just before use, mix 400mL of medium 1 and 100mL of 10%
	glucose.
	6. Lactate medium: 0.5% (v/v) lactic acid, 0.75% (w/w) bacto yeast extract, 0.75%
	(w/w) bactopeptone. Sample 5.9 mL of lactic acid. Add 800 mL of water.
	Adjust the pH to 4 with a solution of 10N KOH. Then add 7.5g of bactopeptone
	and 7.5g of bacto yeast extract. Check the pH to be 4.5. Add water to a volume
	of 1L. Autoclave at 120°C, 20 min.
	1. DTT buffer: 100 mM TRIS-H2SO4 pH 9.4 and 10 mM DTT (see Note 2).
	2. 0.1M Potassium phosphate buffer pH 7.4: mix 80.2 mL of 1M K2HPO4 and
	19.8 mL of 1M KH2PO4. Complete to a volume of 1L with water.
	3. Zymolyase buffer: 1. 2M sorbitol, 20mM potassium phosphate, pH7.4.
	4. Zymolyase-20T.
	5. PMSF 100 mM: weigh 17.4mg of PMSF in 1 mL of ethanol. (see Note 3 )
	6. Homogenization buffer: 0.6M sorbitol, 10 mM TRIS-HCl pH7.4, 1 mM EDTA,
	PMSF 1mM.

  Stacking gel (5%): Measure 1.5 mL of acrylamide: bis-acrylamide solution and 3 mL Tris-HCl 0.5M pH 6.8. Add 20 µL of Temed and 200 µL of 10% ammonium persulfate. Complete with H2O to a volume of 12 mL.

	8. Intermediate labelling solution: Add 5% (v/v) formaldehyde 2 D 4%, 5%
	(v/v) cyanoborohydride 0.6M in phosphate buffer pH 7.5.
	9. Heavy labelling solution: Add 5% (v/v) formaldehyde 2 D, 13 C 4%, 5% (v/v)
	cyanoborodeutéride 0.6M in phosphate buffer pH 7.5.
	10. Buffer A: 0.6% acetic acid in ultrapure water.
	11. Buffer B: 0.6% acetic acid in 80% (v/v) acetonitrile.
	12. Sep-Pak C18 3 cc Vac RC Cartridge 500mg.
	13. Vacuum manifold system.
	14. Vacuum centrifuge (Speed vac concentrator).
	! CRITICAL Labelling reagent mixtures should be kept at 4°C and not stored
	longer than 24 h to ensure labelling efficiency.
	2.9 Two-steps phosphopeptides enrichment
	2.9.1 Strong cation exchange chromatography fractionation
	1. Buffer C: 30% (v/v) acetonitrile, 0.5% (v/v) formic acid in ultrapure water (pH 2.5)
	2. Buffer D: 30% (v/v) acetonitrile, 0.5% (v/v) formic acid and 540 mM ammonium
	formate in ultrapure water (pH 4.7)
	3. 2. Buffer F (elution buffer): 30% (v/v) acetonitrile, 0.4M ammonium hydroxide in
	water.
	3. PHOS-Select iron affinity gel.
	4. SigmaPrep Spin-Columns.
	5. Centrifuge.
	6. Vacuum centrifuge (Speed vac concentrator).
	2.10 LC-MS/MS analysis
	1. Buffer G: 2% acetonitrile, 0.05% trifluoroacetic acid and 0.05% formic acid. 7. Tris-glycine (x5) buffer: Tris 250 mM and glycine 1.92M pH 8.8. Weigh 2. Buffer H: 2% acetonitrile and 0.1% formic acid. 60.5g Tris and 288g of glycine. Complete with H20 to a volume of 2L. pH 3. Buffer I: 0.1% AF H2O 100%. will be around 8.8 without any adjustement. 4. Buffer J : 0.1% AF CH3CN 100%. 8. Cathode Buffer: 50 mM Tris, 384 mM Glycine, 0.1% SDS, 1 mM EDTA: 5. NanoHPLC : NanoLC-Ultra system. dilute x5 Tris-glycine (x5) buffer and add 0.1% SDS (w/v) and 1 mM 6. Mass spectrometer: Q-Exactive mass spectrometer with a nano-electrospray EDTA. interface. 9. Anode Buffer: 50 mM Tris, 384 mM Glycine; dilute x5 Tris-glycine (x5) 7. Pre-column cartridge: PepMap 100 C18 (particles of 5μm, 100 μm i.d., 10 mm buffer length). 10. Molecular weight markers: Precision Plus Protein Standards, dual color. 8. Analytical column: PepMap 100 C18 (particles of 3 μm; 75 μm i.d., 150 mm
	length).

5:1. 2. Resolving gel buffer: 3M Tris-HCl pH8.8. Weigh 363 g Tris. Add water to a volume of 700 mL. Mix and adjust pH to 8.8 with HCl. Complete to 1L with water. Filter through a 0.45 µm filter. Store at 4°C. 3. Stacking gel buffer: 0.5M Tris-HCl, pH 6.8. Weigh 60.5g Tris and prepare a 1L solution as in previous step. Filter through a 0.45 µm filter. Store at 4°C. 4. 10% acrylamide solution: Measure 2.5 mL of acrylamide: bis-acrylamide solution and 1.25 mL of Tris-HCl 3M pH8.8. Add 3 µL of Temed and 30 µL of 10% ammonium persulfate.

Complete with H20 to a volume of 10 mL. 5. 20% acrylamide solution: Measure 5 mL of acrylamide: bis-acrylamide solution and 1.25 mL of Tris-HCl 3M pH8.8. Add 2.5 µL of Temed and 22.5 µL of 10% ammonium persulfate. Complete with H20 to a volume of 10 mL. 6. HPLC: Ultimate LC system combined with a Famos autosampler and a Switchos II microcolumn switch system. 4. Collector: Probot system. 5. SCX column: Zorbax 300 SCX, 2.1mm ID x 150 mm.

2.9.2 Enrichment of phosphopeptides by Immobilized Metal ion Affinity Chromatography (IMAC)

1. Buffer E (equilibration buffer): 30% (v/v) acetonitrile, 250 mM acetic acid in water.

4 .

 4 Notes1. Cells were grown in liquid medium with shaking (130 rpm) at 28°C in a standard incubator. 2. DTT buffer has to be prepared ex-temporarily. 3. PMSF is added in buffer just before its use; its stability in aqueous buffer solutions doesn't exceed 1h. 4. Be careful: methanol can't be substituted by ethanol. 5. Pre-equilibrate the DTT buffer at 30°C before use. 6. The enzymatically cell wall digestion is followed by measurement of OD at 600nm, until it reaches 10% of the initial value. 7. Precool the homogenisation buffer at 4°C. 8. Solubilized mitochondria can be stored at -20°C before loaded on BN-PAGE gel. 9. To simultaneously analyse different samples, concentrations of each sample are compared in order to have the same quantity of proteins in the different samples. If samples' volumes are not the same, they will be adjusted with ZUT buffer. 10. Start degassing buffer C and buffer D 30 min before the start of SCX chromatography. 11. Purge channels A and B, respectively connected to buffer C and buffer D, of the SCX chromatograph before its use.


Murad Abuaisha

David W Eaton

Jeffrey Priest

Ron Wong

Review of hybrid continuum/discontinuum methods for geomechanical modelling of hydraulic fracture growth

This paper presents a comparison between two newly developed methods to describe Hydraulic Fracturing (HF) in petroleum industry. The first method is called Finite-Discrete Element Method (FDEM) where a combination of the FEM and DEM is considered. The second approach is based only on the FEM where a fully coupled system consisting of the standard equations of motion to model the rock matrix, and a parallel plate flow simplification (i.e. lubrication theory) to model the flow in the fracture is discussed. The evolution of fractures is tracked by using remeshing. This paper discusses, as well, the issues involved in the development of the two methods, both from a fundamental theoretical point of view along with some related algorithmic considerations essential for the efficient numerical solution of largescale industrial problems.

Introduction

Considering the discontinuous nature of rock, there is a set of rock engineering problems that aims to maximize the formation of new discontinuities. Examples of these problems include block-caving mining and blasting. In contrast, another set of rock mechanics problems aims to limit failure of rock and the formation of discontinuities. Examples of this set include stability assessment of open pit and natural slopes, dams, and the stability of underground openings. In these cases, the existing rock fabric and discontinuities should be considered as part of the stability assessment [START_REF] Mahabadi | Y-Geo: a new combined finite-discrete element numerical code for geomechanical applications[END_REF]. Numerical modeling techniques have been used to investigate these problems. The most commonly used techniques are continuum approaches, including finite-difference methods (FDM), finite-element methods (FEM), and boundary-element methods (BEM). These methods have been successfully applied to the assessment of global behavior of rock masses and the analysis of stress and deformation. However, explicit representation of fractures and fracture growth is not straightforward in these methods, mainly because of their continuum assumptions. For instance, in FDM, these assumptions require the functions to be continuous across neighboring cells or grid points. For FEM, the continuum assumptions permit the elements to undergo only small strains. Therefore, even when fracturing is allowed, largescale opening, sliding, or complete detachment of elements is not possible. Also, a large number of fractures may cause the FEM stiffness matrix to be ill-conditioned. Fracture analysis using BEM has also been limited to isolated, non-interacting cracks. Furthermore, BEM is restricted in handling material heterogeneity and nonlinearity (i.e., plasticity). The limitations of continuum approaches motivated the development of discrete element methods (DEMs). Although continuum models are based on constitutive laws, DEMs are based on interaction laws. Also, unlike continuum techniques, the contact patterns of the DEM system can continuously change as the system deforms [START_REF] Mahabadi | Y-Geo: a new combined finite-discrete element numerical code for geomechanical applications[END_REF].

Theory The FDEM/ Y-Geo code

The numerical methodology adopted for this study, namely FDEM, uses continuum mechanics principles and DEM techniques to describe the elastic deformation and the material failure process, respectively. Starting from a continuum representation by finite elements of the solid region in question, progressive fracturing is allowed to take place according to some fracturing criterion, thereby forming discrete elements, which may be composed of one or more deformable finite elements. Subsequent motion of these discrete elements and further fracturing of both remaining continuum and previously created discrete elements is then modeled. This evolution process is continued until either the system comes to rest or up to the time of interest. A combined FDEM simulation comprises a large number of interacting bodies, each with a separate finite-element mesh. To ensure that no bodies overlap at any time, an efficient treatment of contact mechanics is required. From an algorithmic point of view, contact is treated by two processes: contact detection and contact interaction [START_REF] Munjiza | The Combined Finite-Discrete Element Method[END_REF]).

By discretizing discrete elements into finite elements, FDEM is able to model both continuum and discontinuum behavior, thus capturing the whole loading and failure path and the progressive damage process of fractured rocks. A unique feature of FDEM is the ability to model the transition from continuum to discontinuum by means of fracture and fragmentation processes [START_REF] Munjiza | Combined single and smeared crack model in combined finite-discrete element analysis[END_REF]). The key processes in FDEM include contact detection, interaction, and friction between discrete elements, elastic deformation of finite elements, and fracture of finite elements. Y-Geo is a new numerical code for geomechanical applications based on the combined finite-discrete element method (FDEM). Several algorithmic developments have been implemented in Y-Geo to specifically address a broad range of rock mechanics problems. These features include (1) a quasi-static friction law, (2) the Mohr-Coulomb failure criterion, (3) a rock joint shear strength criterion, (4) a dissipative impact model, [START_REF] Settgast | Simulation of hydraulic fracture networks in three dimensions utilizing massively parallel computing resources[END_REF] an in situ stress initialization routine, (6) a material mapping function (for an exact representation of heterogeneous models), and (7) a tool to incorporate material heterogeneity and transverse isotropy [START_REF] Mahabadi | Y-Geo: a new combined finite-discrete element numerical code for geomechanical applications[END_REF]).

The FEM with remeshing/ GEOS code

The FEM with remeshing is considering realistic simulations of hydraulic fracturing where several phenomena are taken into consideration. Such phenomena include: 1. fracture interaction in the presence of three-dimensional heterogeneous properties, 2. competing paths for existing flowpaths, and, 3. effects of changes to the stress field resulting from the development of fluid pressure field on fracture propagation [START_REF] Settgast | Simulation of hydraulic fracture networks in three dimensions utilizing massively parallel computing resources[END_REF]).

To account for such phenomena, this method provides a fully coupled system consisting of the standard equations of motion to model the rock matrix, and a parallel plate flow simplification (i.e. lubrication theory) to model the flow in the fracture. The evolution of a fracture proceeds along element interfaces, and fluid elements are inserted at the newly formed crack faces. Thus, as the fracture grows, the mesh topology for both the fluid and solid meshes evolves.

Though not totally appropriate, an explicit method is chosen to run the time integration scheme. A standard Newmark method is applied to the finite element equations, while a standard forward Euler method is applied to the finite volume equation to describe fluid diffusion. A rupture criterion is first specified and evaluated on every internal face. Upon completion of a closed path of surfaces where either the rupture criterion is satisfied, or the surface is external to the body, a new set of nodes, edges, and faces are generated, and the mesh topology is updated. To account for geologic materials high nonlinearities, the stress intensity factor is calculated at the crack tip for Mode I and Mode II of fracture propagation. The energy release rate is estimated by multiplying the field of displacement by the nodal forces at the crack tip. The energy release rate is then converted to the stress intensity factor. Fracture radius and wellbore aperture are plotted versus time, and the aperture and pressure profiles at the end of the simulation are plotted against the radial coordinate.

Examples

Conclusions

A preliminary evaluation of a hybrid continuum-discontinuum numerical code to simulate hydraulic fracture propagation has been carried out. The FDEM code was enhanced with a computational module dedicated to the simulation of fluid-pressure-driven fracturing. The approach was validated by reproducing injection pressure responses in homogeneous and isotropic conditions.

A fully coupled Finite Element/Finite Volume approach to modeling the evolution of hydraulic fractures using the GEOS code is presented as well. This approach has been verified against accepted solutions for the viscosity and toughness dominated regimes for radial fracture propagation and for the viscosity dominated regime for lateral fracture propagation.

Future studies are ongoing to verify and apply both Y-Geo and GEOS for modeling networks of intersecting fractures, variations in the in-situ stress field, variations in material properties, and non-linear as well as anisotropic material response. The effect of possible thermal strains, due to temperature change during the injection of cold fluid, as well as fluid leak off from the fracture surfaces are to be considered.

shown in Figure 1 (viscosity dominated) and Figure 2 (toughness dominated). In these figures, the wellbore aperture (Aperture0) is defined by measuring the aperture at the fluid element closest to the injection point. The fracture radius (Radius) is calculated by taking the square root of the total area of the all fluid elements divided by p to yield an average radius. The radial coordinate is taken as the distance between the center of a fluid element and the injection point normalized by the fracture radius. Here the solution for both the viscosity dominated and toughness dominated solutions are included for both cases for reference. In Figure 1, the plots of wellbore aperture and fracture radius versus time are in excellent agreement with the asymptotic solutions, showing a slight over-estimation of the wellbore aperture, and a slight underestimation of the fracture radius. The plots of aperture versus the radial coordinate show a small overestimation (~10%) of the aperture along the crack for both the Cartesian and radial meshes. The plot of fluid pressure versus radial coordinate shows excellent agreement along the majority of the radius, with greater divergence near the crack tip. This is due to the unbounded negative pressures at the crack tip in the asymptotic solutions, while the GEOS simulations do not allow negative fluid pressure to be applied. The GEOS assumption is more physically correct, while allowing negative pressure is more mathematically convenient. The result of this difference in representation is that the pressures directly behind the tip in the GEOS simulation will be higher than the analytical solution.

Figure 2 shows a comparison of the GEOS simulation for a propagating fracture in the toughness-dominated regime. The toughness-dominated simulations are not a constant fluid pressure case where the flow of fluid does not have a bearing on the solution. The plots of wellbore aperture versus time indicate that GEOS agrees with the asymptotic solution, with the GEOS simulation exhibiting a slightly lower value of aperture on average. In the radial mesh, there is a slight oscillation at 180 seconds that correlates with the fracture front passing through a mesh transition zone, while the Cartesian mesh tracks the asymptotic solution in a more continuous manner. The plot of fracture radius indicates that the GEOS solution is close to the asymptotic solution, giving a slightly higher value on average.

Figure 1 .

 1 Figure 1. Material failure modelling in FDEM. (a) Conceptual model of a tensile crack in a heterogeneous rock material. (b) Theoretical FPZ model. (c) FDEM implementation of the FPZ using triangular elastic elements and four-noded crack elements to represent the bulk material and the fracture, respectively. (d) FDEM representation of a fracturable body with continuum triangular elements and embedded crack elements indicated in grey and pink, respectively.

Figure 2 .Figure 3 .

 23 Figure 2. Developed injection pressure as function of time for the Schla and BDS models. Schla and BDS models refer to different values of the far field stresses; see (Lisjak et al., [2014]) for more details.

Figure 1 :

 1 Figure 1: Comparison of GEOS against asymtotic solutions for viscosity dominated case. Fracture radius and wellbore aperture are plotted versus time, and the apeture and pressure profiles at the end of the simulation are plotted against the radial coordinate.

Acknowledgements

This work was funded by the Microseismic Industry Consortium and a Collaborative R&D grant from the Natural Sciences and Engineering Research Council of Canada.


Murad S Abuaisha

David W Eaton

Jeffrey Priest

Ron Wong

Simulating hydraulic fracturing using finite-discrete

published or not. The documents may come

Introduction

Solutions of stress distribution around a borehole provide a starting point for analysis of hydraulicfracturing (HF) operations. In the simplest cases, these solutions are based upon assumptions of elastic, isotropic and homogeneous media; however, the presence of pre-existing fractures and weak bedding planes render such assumptions questionable, motivating a more general numerical simulation approach. Here, we make use of a technique (the Y-Geo code) that combines both the Discrete Element Method (DEM) and the Finite Element Method (FEM). The Y-Geo approach has the capabilities of simulating fluid-driven fractures in jointed/pre-fractured rock masses. This contribution is organized as follows. First, two HF tests are simulated in an elastic continuous medium, considering both isotropic (Shmax = Shmin) and anisotropic (Shmax >> Shmin) far-field stress states. Next, tests are performed with different configurations of rock joints that intersect the wellbore. Thirdly, the presence of lateral stress gradients is considered by introducing joints in the medium that do not intersect the wellbore.

Method

The Y-Geo uses continuum mechanics principles and DEM techniques to describe the elastic deformation and the material failure process, respectively. Starting from a continuum representation by finite elements of the solid region in question, progressive fracturing is allowed to take place according to some fracturing criterion, thereby forming discrete elements, which may be composed of one or more deformable finite elements. Subsequent motion of these discrete elements and further fracturing of both remaining continuum and previously created discrete elements is then modeled [START_REF] Munjiza | Combined single and smeared crack model in combined finite-discrete element analysis[END_REF]), [START_REF] Munjiza | The Combined Finite-Discrete Element Method[END_REF]) and [START_REF] Mahabadi | Y-Geo: a new combined finite-discrete element numerical code for geomechanical applications[END_REF]). Fluid injection and pressure-driven fracture propagation are captured by a simplified approach based on the principle of mass conservation for a compressible fluid injected into a deformable solid. The model is hydro-mechanically coupled exclusively in the sense that variations in cavity volume, due to either rock elastic deformation or fracturing, affect the pressure of the compressible fluid, which, in turn, affects rock deformation and failure, Fig. 1.

Examples

Fluid-driven fracture topology is studied in isotropic and anisotropic far-field stress states. The effect of pre-existing joints on fracture evolution is also addressed. Figure 2 shows the fracture trajectories for the two cases of isotropic and anisotropic stress states around a vertical wellbore. These trajectories are plotted in the field of velocity magnitude as an indication of the elastic wave spreading in the medium due to fracturing. To study the effect of man-created joints on fracking pressure as well as fracture topology, the previous simulations, typically for the anisotropic stress state being the most common in field, are repeated by considering several defect configurations around the wellbore, Fig. 3.

Figure 4 shows fracture trajectories for three cases of defect orientations at distance 0.3L (L is the defect length) from the wellbore. Apparently, the presence of pre-existing defects close to the wellbore creates a perturbation to stress field (lateral stresses) which affects fracture evolution/topology.

Conclusions

The Y-Geo code is capable of tracking fluid-driven fractures topology in homogeneous and prefractured media. Consistent with HF observations, the study has shown that an isotropic stress state favours complex fracture growth compared with anisotropic stress state. Defects that are misaligned with the maximum stress produce fractures that initiate along the defect but subsequently rotate into the maximum stress orientation. The only scenarios considered that generated a simple bi-wing fracture; involve either wellbore defects combined with an isotropic far-field stress state or wellbore defects in the direction of maximum stress in anisotropic stress state. Stress gradients, introduced here by joints offset from the wellbore, lead to asymmetric fracture growth with a tendency for fractures to initiate in a direction away from the defect.

Figure 2 .

 2 Figure 2. HF modelling using the FDEM. A tensile mode I fracturing is shown due to excessive fluid injection in the borehole. Only elements in concern are shown for simplicity.

Figure 2 .

 2 Figure 2. Fracture topology/trajectory for the cases of isotropic and anisotropic stress states and for two different time steps. The trajectories are plotted in the field of velocity magnitude.

Figure 3 .

 3 Figure 3. Fracture trajectories for different defect configurations at the wellbore. The trajectories are plotted in the field of velocity magnitude.

Figure 4 .

 4 Figure 4. Fracture trajectories for three cases of defect configurations at time steps of 1.26, 1.38 and 1.68 ms and at some distance from wellbore. The trajectories are plotted in the field of velocity magnitude.

Acknowledgements

This work was funded by the Microseismic Industry Consortium and a Collaborative R&D grant from the Natural Sciences and Engineering Research Council of Canada.