Vanishing Immanants

We classify all the irreducible characters of a symmetric group such that the induced immanant function $d_{\chi}$ vanishes identically on alternate matrices with the entries in the complex field.


Introduction
Let C be the field of the complex numbers, S n be the permutation group acting on the set {1, . . ., n}, let M n (C) denote the space of all n-by-n matrices with entries in C, and Notice that immanant generalizes the permanent, per(A) = σ∈Sn n i=1 a iσ(i) , where χ = 1 is the principal character, and the determinant, det(A) = σ∈Sn ε(σ) n i=1 a iσ(i) , where χ = ε is the alternating character of S n .
Since computing an immanant d χ may not be always easy, one may hope to find a suitable linear map T that converts d χ into an immanant d χ ′ .Given a subspace V ⊆ M n (C), we say that a linear transformation T : V → V converts an immanant d χ into an immanant d χ ′ on V if d χ ′ (T (X)) = d χ (X) for all X ∈ V .In the case where χ = χ ′ , one says that these maps preserve the immanant d χ .
The maps which convert/preserve immanants have been investigated in many papers.First, in 1994, Duffner [8] characterized linear maps which preserve the immanant d χ on M n (C), n ≥ 3, where χ / ∈ {1, ε} and proved that linear preservers of immanants are always non-singular.The main result shows that, for n ≥ 4 and χ / ∈ {1, ε}, the group of such linear preservers is generated by permutation of rows, permutation of columns, transposition, and a Hadamard multiplication with a matrix that has some additional properties.In 1997, Coelho and Duffner [1], showed that the same result is true for linear immanant preservers on S n (C), the space of symmetric n-by-n matrices where again χ / ∈ {1, ε}.The only difference is that permutation of rows and columns should be done in unison.In 2012, they [6] extended this result to the alternate matrices and classified the linear maps which preserve the immanant d χ on A n (C) (either n ≥ 6, or n = 4 and χ = (2, 2)), where χ / ∈ {1, ε, (n − 1, n), (2, 1 n−2 )}.The result is exactly the same as for linear immanant preservers on S n (C), except for immanants which vanish identically on A n (C) (clearly any linear map will preserve such immanants).
Next, in 1998, Coelho and Duffner [3] proved that there is no linear map T : M n (C) → M n (C), n ≥ 3 that converts an immanant d χ into an immanant d χ ′ , where χ ̸ = χ ′ .In 2003, they [4] extended this result to the symmetric matrices.They proved that there is no linear map T : S n (C) → S n (C), n ≥ 3 that converts an immanant d χ into an immanant d χ ′ , where χ ̸ = χ ′ .In 2021, Duffner et al. [10] extended this result to the alternate matrices.They proved that there exists no map T : A n (C) → A n (C), n ≥ 6, satisfying d χ (A + αB) = d χ ′ (T (A) + αT (B)), α ∈ C, where χ and χ ′ are not proportional on the subset P n of all permutations of S n with no cycles of odd length in the decomposition into the product of disjoint cycles (i.e., there is no 0 ̸ = β ∈ C such that χ ′ | Pn = β.χ|Pn ).Also, they characterized such maps T if χ and χ ′ are proportional on P n , but the induced immanants do not vanish identically on A n (C), and proved that T is bijective and linear.Then, they extended [11] this result to the alternate matrices of order 4 and proved that there are no linear maps T : A 4 (C) → A 4 (C) that converts an immanant d χ into an immanant d χ ′ for all matrices A ∈ A 4 (C), where χ, χ ′ ∈ {1, ε, (2, 2)} are two distinct irreducible characters of S 4 .
In 2006, Coelho and Duffner [5] proved that if χ and χ ′ are arbitrary irreducible complex characters of S n and T : M n (C) → M n (C), n ≥ 3, is a surjective map satisfying the condition d χ (A + αB) = d χ ′ (T (A) + αT (B)), for all A, B ∈ M n (C) and all α ∈ C, then T is linear.Their main theorem combined with known results on linear preservers/converters of immanants, allows us to characterize the pairs (χ, χ ′ ) for which such maps exist and, in the cases they exist, to obtain the respective description.
In 2008, Kuzma [13], following the idea of [16], showed that if F is any field with at least n + 1 elements (n ≥ 3) and χ, χ ′ are irreducible characters of S n , then the immanant converter on matrix pencils T : is automatically linear and bijective.His arguments rely on the property that for every pair of integers (i, j) ∈ {1, . . ., n} there exists a permutation σ ∈ S n with σ(i) = j and χ(σ) ̸ = 0 ∈ F. Note that this is automatically satisfied in fields of characteristic 0. However if characteristic of the field is nonzero, it is not obvious that such a permutation σ exists.We postpone this question and also give the correction of the main result of [13] in Addendum (section 6).This was extended, in 2014, by Coelho et al. [7] to matrix pencils consisting of symmetric matrices.Namely, following [5,13], they proved that the maps T : S n (C) → S n (C) satisfying the condition d χ (A + αB) = d χ ′ (T (A) + αT (B)), for any fixed irreducible characters χ, χ ′ : S n → C, any A, B ∈ S n (C), and any α ∈ C, are automatically linear and bijective.In 2017, Duffner and Guterman [9] extended this result and studied converters between d χ and d ε = det, where unlike [7] they considered matrix pencils spanned by singular symmetric matrices.
As already remarked, some immanants vanish identically on A n (C).In [6, Corollary 2.2], it was shown that for every even integer n, the immanant induced by a triangular character vanishes identically on A n (C).In fact, the triangular character is the only character which vanishes on all the conjugacy classes whose cycle decomposition contains at least one transposition (see [2]).
The main goal in this paper is to find all the irreducible characters such that the induced immanant function d χ vanishes identically on A n (C).Obviously, for A ∈ M n (C) we have Hence, if n is odd, then d χ (A) = 0 for any A ∈ A n (C).So, for finding the immanants which vanish identically, it is enough to consider the irreducible characters of S n in the case when n is even.In this case, Duffner et al. [10,Proposition 2.11] proved the following result.
. So, for finding the vanishing immanants, it is enough to find the irreducible characters which are identically zero for any permutation σ ∈ P n .
Recall that irreducible characters of S n are in bijective correspondence with partitions of integer n (see [14,Theorem 2.4.6]).Also, partitions are usually visualized by their Young diagrams.With this in mind, our first main result is contained in Theorems 5.3 and 5.7, where by using the recursive Murnaghan-Nakayama rule, we find all the Young diagrams with induced characters vanishing identically on P n .The second main result is Corollary 5.9 which classifies the immanants vanishing identically on alternate matrices.

Preliminaries
A decomposition λ of an integer n ≥ 1 is a sequence (λ 1 , . . ., λ t ) where λ i ∈ N and t i=1 λ i = n.A partition λ of an integer n is a decomposition (λ 1 , . . ., λ t ) of n where for some r, s ∈ {1, . . ., t}, then (λ 1 , . . ., λ r , . . ., λ s , . . ., λ t ) will be written as (λ 1 , . . ., λ s−r+1 r , . . ., λ t ) for simplicity.A shape λ is a geometric object which consists finitely many one by one square cells arranged in left-justified rows and top-justified columns where adjacent cells are edge-connected.Shapes are in bijective correspondence with decompositions of integer n, i.e., the number of cells in a shape.This correspondence is given by where λ i is the number of cells in a row i and t is the number of rows of λ.We also use the notation If the number of cells in consecutive rows of a shape is decreasing, we call it a valid shape or a (Young) diagram.These are in bijective correspondence with partitions of integer n.Notice that a Young diagram λ = (λ 1 , . . ., λ t ) is a shape such that A rim-hook ζ of a Young diagram λ consists of a chain of consecutive edge-connected cells by taking all the cells always leftwards or downwards such that the obtained shape λ \ ζ is also a Young diagram.If all the cells of ζ are only in one row (respectively, one column), we call it a horizontal rim-hook (respectively, vertical rim-hook ).
From now on, we simply say diagram instead of Young diagram.

Destructibility of diagrams
Let λ be a diagram.A domino D in λ is a pair of vertically or horizontally connected cells of λ that intersect in a common edge, i.e., form a 2 × 1 or 1 × 2 rectangle.We call these two cells a vertical domino or a horizontal domino, respectively.A domino rim-hook is a rim-hook of length 2. Notice that it coincides with a single vertical or horizontal domino, respectively.
The diagram λ is called destructible if there exists a recursive procedure R λ which removes all the cells from λ such that at each step we remove a single domino rim-hook (where a rim-hook refers to a diagram from this particular step, and so it is different at each step).If there is no such recursive procedure, we call the diagram indestructible.For example, the diagram λ = (2n) is a destructible diagram, since it consists of n horizontal dominoes.Examples of indestructible diagrams will be provided in the sequel.
Second, if one of D, E is vertical, in rows i − 1 and i, and another is horizontal, in row j > i, then, by a similar argument, is again a valid shape, and the same conclusion holds.Likewise, we argue if j < i − 1 or if D, E are both vertical (and The following lemma is crucial for our subsequent investigation.Lemma 3.2.Let λ be a diagram with |λ| = 2n, and let D be one of its domino rim-hooks.Then λ is destructible if and only if λ \ D is destructible.
Proof.Assume λ \ D is destructible and let is a recursive procedure for λ.Therefore λ is destructible.Conversely, assume λ is destructible.We use induction on the number n of dominoes in λ.The case n = 1 is trivial.If n = 2, then λ can take one among the five different forms (4), (3,1), (2, 2), (2, 1, 1) and (1, 1, 1, 1).By removing a single domino rim-hook from any one of these, we get either a diagram (2) or a diagram (1, 1), both of which are obviously destructible.This proves the base of induction.
2. Assume that at the first step none of the cells of domino D is removed.By Lemma 3.1, D is also a domino rim-hook of λ (1) .Since λ (1) is destructible with |λ (1) | = 2n, by the induction, λ (1) 3. Assume that D is a horizontal domino rim-hook and at the first step, domino E removed a single cell from D. Then E was vertical.Let D belong to a row i.Then λ i ≥ λ i+1 + 2, and hence E cannot belong to rows i and i + 1.Therefore, since λ \ E is a valid shape, (3.2) Let E ′ be the vertically placed domino in the last cells of rows i − 1 and i of λ (1) = λ \ E.
is a valid shape, and hence E ′ is a domino rim-hook of a destructible diagram λ (1) with n dominoes.Notice that (3.3) can also be written as (λ \ D) \ D ′ , where D ′ is horizontal domino occupying the last two cells of row i − 1, and since also destructible, and hence there exists a recursive procedure 4. Assume that D is a vertical domino rim-hook and at the first step, domino E removed a single cell from D. Then E was horizontal.Let D belong to rows i − 1 and i.Then λ i−1 = λ i ≥ λ i+1 + 1, and hence E cannot belong to row i − 1.Therefore, since λ \ E is a valid shape, (3.4) Let E ′ be the horizontally placed domino in the last two cells of row i − 1 of λ (1) = λ \ E. By (3.4), is a valid shape, and hence E ′ is a domino rim-hook of a destructible diagram λ (1) with n dominoes.Notice that (3.5) can also be written as (λ \ D) \ D ′ , where D ′ is vertical domino occupying the last cells of rows i − 1 and i, and since (3.5) is a valid shape, also destructible, and hence there exists a recursive procedure λ is triangular if and only if there is no domino rim-hook in λ.
Proof.A vertical domino rim-hook requires two consecutive rows of equal length, while a horizontal domino rim-hook requires that there exist two consecutive rows whose lengths differ by at least two, but neither is possible if λ = ▽ m 1 .Now, assume that λ is not triangular.Then there exist two consecutive rows i and i + 1 Then there exists a vertical domino rim-hook in rows m − 1 and m. □ Clearly if there is no recursive procedure, the diagram must be indestructible.Hence, by Lemma 3.3 we can state the following corollary: 1 , where m > 1, is indestructible.We generalize the above corollary in the following lemma that classifies the indestructible diagrams.(i) λ is indestructible.
(ii) λ is triangular or else there exists a recursive procedure of domino rim-hook removal which transforms λ into a triangular diagram.(iii) Every recursive procedure of domino rim-hook removal applied on λ will eventually end up in a triangular diagram.
(iv) There exists a triangular diagram ▽ m 1 such that every recursive procedure of domino rim-hook removal will transform λ into ▽ m 1 .Proof.(i) ⇒ (iii).Suppose that there exists a recursive procedure of domino rim-hook removal applied on λ which does not end up in a triangular diagram.Then, by Lemma 3.3, at each step there exists a domino rim-hook until removing all the cells from λ. Therefore λ is destructible, contradicting (i).
(iii) ⇒ (ii) is obvious.(ii) ⇒ (i).If λ is triangular, then by Corollary 3.4, λ is indestructible.Now, assume that there exists a recursive procedure of domino rim-hook removal which transforms λ into a triangular diagram.Since, by Lemma 3.3, the transformed triangular diagram is indestructible, by recursively using Lemma 3.2, λ is also indestructible.
(iii) ⇒ (iv).We recall that λ consists of cells indexed by left-justified rows and topjustified columns where adjacent cells are edge-connected.We denote the cells of row i in the diagram λ by C i1 , C i2 , . . ., C iλ i here C ij belongs to row i and column j and the top-most row and the left-most column have index 1.With this notation each vertical/horizontal domino can be written as a set for some i and j.Let m, m ′ > 1, where m < m ′ , be two distinct integers such that two recursive procedures of domino rim-hook removal R λ and R ′ λ applied on λ will eventually end up in triangular diagrams ▽ m 1 and ▽ m ′ 1 , respectively.By reversing the procedure R λ , the diagram λ can be obtained from ▽ m 1 by adding domino rim-hooks.We then apply the procedure R ′ λ to see that the diagram ▽ m ′ 1 can also be obtained from ▽ m 1 by adding domino rim-hooks.Then ▽ m ′ 1 \ ▽ m 1 must be covered with non-overlapping, i.e., disjoint dominoes.We will show that this is impossible.Notice that C 1m ′ , the last cell of first row in ▽ m ′ 1 \ ▽ m 1 , is also the only cell in its last column, and therefore it can only be covered with a horizontal domino , the last cell of second row, cannot be covered with a vertical domino }.This allows us to recursively proceed forwards and show that the last cell of each succeeding row can only be covered with a horizontal domino.This is a contradiction, because the last row in ▽ m ′ 1 \ ▽ m 1 contains only one cell and it cannot be covered with a horizontal domino.□ Now, we give the prototypical example of indestructible diagrams, which generalizes triangular ones from Corollary 3.4. .Then, by Lemma 3.5, λ is again indestructible.
Remark 3.7.By reversing the procedure described in Lemma 3.5, we obtain one possibility to construct all indestructible diagrams.We just start with a triangular diagram and recursively add dominoes in such a way that at each step, the newly added domino becomes a domino rim-hook for the obtained diagram.Notice that by applying this procedure to a single vertical or horizontal domino as a starting point, one can also build all possible destructible diagrams.
Another possibility to obtain all (in)destructible diagrams will be given in the next section.

Two rules which preserve the (in)destructibility of diagrams
In this section, we state and prove two main rules about destruction diagrams.If D is a horizontal (respectively, vertical) domino, then we define its transpose to be a vertical (respectively, horizontal) domino.For any diagram λ, its transpose, λ T = (γ 1 , . . ., γ s ) is a diagram where γ i is the length of the i'th column of λ, that is, λ T is a diagram obtained from λ by interchanging the rows and columns of λ.It is easy to see that γ i = |{j : λ j ≥ i}|.Therefore the partition determined by λ T is the conjugate partition of the one determined by λ.Now, we have the second rule: Now, as we mentioned in the previous section, using the above two rules, other possibilities are given to obtain all (in)destructible diagrams.Proof.Using Corollary 4.1 we can transfer recursively horizontal/vertical domino rimhooks that are in rows 2, 3, . . .and place them horizontally to the first row.If all the dominoes from other rows can be transferred to the first row, then the obtained diagram is (2n).But, if after some steps there is no way to transfer domino rim-hooks from other rows to the first row, then there is no way to remove domino rim-hooks from those rows.In this case, by Lemma 3.3 there exists an integer m such that the obtained diagram is of the form (2n − m(m+1) 2 , ▽ m 1 ).If m = 1, the obtained diagram is (2n−1, 1), so that by using the second rule, we transpose into (2n − 1, 1) T = (2, 1 2n−2 ).Then using Corollary 4.1, after n − 1 times removing the vertical domino rim-hooks from the first column and placing them horizontally to the first row, we again obtain the diagram (2n).Since this is clearly destructible and the two rules preserves destructibility, our initial diagram λ is also destructible.We remark that the above corollary does not always hold in indestructible case.For example, it can be easily checked that two indestructible diagrams λ = (6, 3, 2, 1) = (6, ▽ 3 1 ) and γ = (9, 2, 1) = (9, ▽ 2 1 ) with |λ| = |γ| = 12 cannot be changed from one to another.However, it may still hold in some cases.For example, the two indestructible diagrams λ = (6, 4, 3, 2, 1) = (6, ▽ 4  1 ) and γ = (13, 2, 1) = (13, ▽ 2 1 ) with |λ| = |γ| = 16 can be changed from one to another as follows (first we transpose, and then we transfer domino rim-hooks to the first row): 1 ) ↔ (▽ 5 1 , 1) ↔ (7, 4, 3, 2) ↔ (9, 4, 3) ↔ (11, 4, 1) ↔ (13, 2, 1).

Characters induced by (in)destructible diagrams
In this section we prove that the characters induced by indestructible diagram vanish identically on P 2n ⊆ S 2n , and the characters induced by destructible diagram are always nonzero on P 2n (recall that P 2n denotes the set of permutations from S 2n whose cycle type contains only cycles of even length).
Our main tool will be the recursive Murnaghan-Nakayama rule (see [ We will also rely on the following result.Let us now focus on irreducible characters of S 2n that correspond to destructible diagrams (see Theorem 5.7 below).Recall from Remark 3.7 that every destructible diagram can be recursively built up by adding vertical/horizontal dominoes to its current border.It follows that every destructible diagram can be covered (perhaps in more than one way) with horizontal and vertical dominoes.For example, the diagram (2, 2) has two different tilings D v D v and D h D h , where D v and D h are vertical and horizontal dominoes, respectively.
To be more precise, a (domino) tiling T of a diagram λ is a placement of dominoes that covers all the cells of the diagram perfectly (i.e., no overlaps, no diagonal placements, no protrusions off the diagram).We mention in passing that, alternatively, one can define a tiling to be a partition of the cells of λ into a disjoint union of pairs of cells which share common edge (c.f.(3.6)).Proof.Let λ = (λ 1 , . . ., λ t ) be a destructible diagram and T be one of its tilings.As in the proof of Lemma 3.5, we denote the cells of row i in λ by C i1 , C i2 , . . ., C iλ i , so each vertical/horizontal domino can be written as (3.6).
Since λ is destructible, by Lemma 3.5, it cannot be triangular.So, the difference between two consecutive rows is not always 1, here for simplicity, we consider λ appended with a virtual (empty) last row containing no cells.
Assume first there exists a row k such that the difference between two consecutive rows, (5.1) Without loss of generality, we may assume that k is the smallest one.In this case, the two cells D = {C k(λ k −1) , C kλ k } form a horizontal domino rim-hook.If D is a part of the tiling T we are done.Otherwise, we must have k ≥ 2 and the cell C kλ k must be covered with a vertical domino Now, if λ k−1 = λ k , since by (5.1) we have λ k ≥ λ k+1 + 2, then D v is a domino rim-hook, and again we are done.The only possibility left to consider is that In this case C (k−1)λ k−1 , the last cell of row k−1, cannot be covered with a horizontal domino D h , because D h intersects the vertical domino D v in the cell Recall that k was the first row satisfying (5.1), and consequently d i = λ i − λ i+1 ∈ {0, 1} for every i < k − 1. Therefore we can recursively proceed backwards to show that the last cell of each preceding row can only be covered with a vertical domino until either we hit a row with d i = 0, or else d i = 1 for 1 ≤ i ≤ k − 2 and we reach a first row.In the former case we stopped at row i + 1 and cover a cell C (i+1)(λ i+1 −1) with a vertical domino, and hence its edge-adjacent cell C (i+1)λ i+1 can only be covered with a vertical domino is actually a domino rim-hook, and we are done once again.In the latter case, the cell C 1λ 1 (the last cell of the first row) would also have to be covered with a vertical domino, a contradiction.Therefore, the latter case is impossible.
Assume the last d i ≤ 1 for every i, and there exists k ′ < t with We may suppose that k ′ is the largest one with this property.Notice that now λ t = 1 and that is, each of the last two rows contains a single cell which form a vertical domino rim-hook D v .Clearly, the only cell of the last row cannot be covered with a horizontal domino, and therefore D v ∈ T .So, again we are done.Finally, let k , then the last two cells of rows k ′ and k ′ + 1 form a vertical domino rim-hook.If it is a part of the tiling T , we are done.Otherwise, the last cell of row k ′ + 1 must be covered with a horizontal domino Therefore, C, the last cell of row k ′ +2, cannot be covered with a vertical domino, because it intersects D h in the cell C (k ′ +1)(λ k ′ +1 −1) .So, also C is covered with a horizontal domino , C} ∈ T. This allows us to recursively proceed forwards and show that the last cell of each succeeding row can only be covered with a horizontal domino.This is a contradiction, because the last row contains only one cell and it cannot be covered with a horizontal domino.□ The parity is a mapping π : Z → {−1, 1} defined by π(k) = (−1) k .We also introduce the parity of a diagram λ = (λ 1 , . . ., λ t ) by Lemma 5.5.Let λ = (λ 1 , . . ., λ t ) be a destructible diagram covered with dominoes.Then the parity of v, the number of vertically placed dominoes, satisfies Proof.For i ∈ {1, . . ., t − 1}, let z i be the number of vertically placed dominoes which intersect both rows i and i + 1.Let also m i be the number of vertically placed dominoes that intersect row i with one among its two cells, i ∈ {1, . . ., t}.Clearly, (5.2) and moreover, ( Next notice that λ i − m i equals the number of horizontally placed dominoes in row i, so it must always be an even integer.Therefore, 1 = π(λ i − m i ) = π(λ i )/π(m i ), and so From (5.4) and (5.2) we have π(λ 1 ) = π(m 1 ) = π(z 1 ).Moreover, from (5.4) and ( 5.

□
Remark 5.6.Let a destructible diagram λ = (λ 1 , . . ., λ t ) be covered with dominoes and let m i be as above.Then v and h, the numbers of vertically/horizontally placed dominoes in any tiling of λ, satisfy Namely, we have The second equation follows from 2(v + h) = |λ|, because each domino occupies two cells.
Given a tiling T of a destructible diagram λ, the number of different ways to recursively remove all the dominoes from λ, where at each step we remove a single domino rim-hook by following the tiling T , will be denoted by |T | (see Fig. 3).
Theorem 5.7.Let λ be a destructible diagram of |λ| = 2n, and let T be the set of all possible tilings of λ.If the cycle type of a permutation ρ ∈ S 2n is (2, . . ., 2), then Proof.Let ρ = ρ 1 . . .ρ n be a decomposition of ρ into disjoint cycles, each of order two.By Murnaghan-Nakayama rule, χ λ (ρ) = D 1 (−1) ll(D 1 ) χ λ\D 1 (ρ\ρ 1 ), where the sum runs over all domino rim-hooks D 1 of λ.By Lemma 3.2, the diagram λ \ D 1 is still destructible, so it again contains at least one domino rim-hook.This allows us to recursively continue the Murnaghan-Nakayama rule on each summand χ λ\D 1 (ρ \ ρ 1 ).After r steps we get where the summation is over all dominoes D 1 , . . ., D r such that for each i ∈ {1, . . ., r}, domino D i is a domino rim-hook of λ \ i−1 j=1 D j .Since λ is destructible, we will remove all the cells from λ after n steps and get Notice that more than one summand may correspond to the same tiling T .This happens when during the recursive Murnaghan-Nakayama procedure we encounter a step where we have more than one possibility to remove domino rim-hook by following the given tiling.It follows that in (5.5) there are exactly |T | summands corresponding to the tiling T .
Conversely, given any tiling T of λ, it contains, by Lemma 5.4, at least one domino rim-hook.This allows us to recursively remove all the dominoes from T in such a way that at each step one of the current domino rim-hooks is removed.If we ignore the sign, this corresponds exactly to one of the summands of (5.5) obtained after finishing the recursive Murnaghan-Nakayama procedure.This shows that the equation (5.5) equals to π(λ) |T |, where we sum up over all the tilings T of λ, as claimed.□ Recall that P 2n is the subset of all permutations in the symmetric group S 2n with no cycles of odd length in the decomposition into the product of disjoint cycles; that is, the set of all permutations of even length.Now, we have the following corollary:  Proof.(iii) ⇒ (i).Let A be an alternate matrix in M n (C), and A T be its transpose.If n is odd, then d χ (A) = 0. Now, let n be even and χ be induced by an indestructible diagram.Then, by Theorem 5.3, χ vanishes identically on every permutation ρ ∈ P n .Therefore, by Proposition 1.1, d χ vanishes identically on every alternate matrix.(i) ⇒ (ii) is trivial since J ⊕ • • • ⊕ J is an alternate matrix.(ii) ⇒ (iii).Let n be even and let Notice that if (i, j) / ∈ {(2k − 1, 2k), (2k, 2k − 1); 1 ≤ k ≤ n 2 }, then a ij = 0. So, in computing the immanant of A, every summand corresponding to a permutation σ with σ(2k − 1) ̸ = 2k or σ(2k) ̸ = 2k − 1 for some k ∈ {1, . . ., n 2 } is multiplied by n i=1 a iσ(i) = 0. Therefore Let us finish with a remark that Corollary 5.8 is not true in other fields.For example in fields of characteristic 2, character induced by a destructible diagram λ = (4, 4) vanishes identically on P 8 .We do not know what happens in fields of other characteristic.

Addendum
In [13], the following result was stated.Theorem 6.1.[13, Theorem 1] Let n ≥ 3 be an integer, let F be a field with |F| ≥ n + 1, and let χ, χ ′ be two irreducible complex characters of the symmetric group S n .Suppose Φ : M n (F) → M n (F) is any map with the property (6.1) Then Φ is automatically linear and bijective.
However, the arguments which prove this result work only under the additional assumption that for every pair of integers (i, j) ∈ {1, . . ., n} there exists a permutation σ ∈ S n with σ(i) = j and with χ(σ) ̸ = 0 ∈ F (this fact is needed in the proof of [13,Lemma 4], see also Example 6.5 below).
With this additional assumption on the character χ, all the arguments in [13] are then valid, and hence prove the following (which corrects Theorem 6.1).Theorem 6.2.Let n ≥ 3 be an integer, let F be a field with |F| ≥ n + 1, and let χ, χ ′ be two irreducible complex characters of the symmetric group S n such that for every pair of integers (i, j) ∈ {1, . . ., n} there exists a permutation σ ∈ S n with σ(i) = j and χ(σ) ̸ = 0 ∈ F. Suppose Φ : M n (F) → M n (F) is any map with the property Then Φ is automatically linear and bijective.
In fact, as we show next, this additional assumption is automatically satisfied if char(F) ̸ = 2.To make the proof easier, we introduce an additional terminology.
A rim-hook is complete if it is of maximal possible length.Such a rim-hook consists of a chain of consecutive edge-connected cells which start at the last cell of the first row in a diagram and continue all the time leftwards or downwards ending up at the last cell of the first column.Since the chain starts at the last cell of the first row and ends up at the last cell of the first column, we have moved s − 1 times leftwards and r − 1 times downwards, where r is the number of rows and s is the number of columns in our diagram.By adding also the starting cell of the chain, we see that the length of a complete rim-hook is (s − 1) + (r − 1) + 1 = s + r − 1.Notice that the length of a complete rim-hook is 1 if and only if |λ| = 1.Lemma 6.3.Let n ≥ 3.If χ is an irreducible character of S n , then for every (i, j) ∈ {1, . . ., n} there exists a permutation σ ∈ S n such that σ(i) = j and χ(σ) ∈ {−2, −1, 1, 2}.
Proof.Let λ be a diagram with |λ| = n ≥ 3 such that χ = χ λ , and let be the recursive procedure starting with a diagram λ whereby at step i, a complete rimhook is removed from the obtained diagram λ (i−1) , i ∈ {1, . . ., j}, until we reach a diagram λ (j) which is its own rim-hook; that is At each step i, the length of the complete rim-hook of λ (i) equals ℓ i = s i + r i − 1, where s i and r i denote the number of columns and rows of λ (i) , respectively.Since for any i, s i+1 < s i and r i+1 < r i , then ℓ i+1 < ℓ i .
In the remaining cases we will find a single permutation σ to do the job.In fact, it suffices to give its cycle type so that cycle lengths are not increasing and the first one has length at least two, while the last one has length one.If j > 0, we will start by cycle lengths ℓ 1 , . . ., ℓ j .Since at each step i ∈ {1, . . ., j}, there exists only one complete rim-hook (i.e., of length ℓ i ), then by Murnaghan-Nakayama rule (Theorem 5.1) any such permutation σ will satisfy χ λ (σ) = ±χ λ (σ \ {σ 1 , . . ., σ j }), where |σ i | = ℓ i , i ∈ {1, . . ., j}.
Email address: bojan.kuzma@famnit.upr.si 0} be the subspace of the alternate matrices over C. The immanant associated with the irreducible character χ of S n is the function d χ : M n (C) → C given by d χ (A) = σ∈Sn χ(σ) n i=1 a iσ(i) .

Lemma 3 . 1 .
Let λ be a diagram with two disjoint domino rim-hooks D and E. Then E is a domino rim-hook of λ \ D.

Lemma 4 . 3 (
Second rule).A diagram λ is destructible if and only if its transpose, λ T , is destructible.Sketch of the proof.A horizontal (respectively, vertical) domino of λ is a vertical (respectively, horizontal) domino of λ T .Moreover, for any rim-hook D of λ, we have (λ \ D) T = (λ \ D) T and D T , the transpose of D, is a domino rim-hook of λ T .□

Lemma 4 . 4 . 2 , ▽ m 1 )
Let λ be a diagram with |λ| = 2n.Then λ is destructible if and only if the diagrams λ and (2n) can be transformed to each other by applying the above two rules.Moreover, λ is indestructible if and only if there exists an integer m > 1 such that the diagrams λ and (2n − m(m+1) can be transformed to each other by applying the above two rules.

2 , ▽ m 1 ) 2 , ▽ m 1 )Corollary 4 . 5 .
If m > 1, then, by Example 3.6, the obtained diagram (2n − m(m+1) is indestructible.Since the first rule preserves indestructibility, our initial diagram λ is also indestructible.By reversing the above procedure, we can transform the diagram (2n) to any destructible one, and can transform the diagrams (2n − m(m+1) , for suitable m > 1, to any indestructible one.□Let λ and γ be two destructible diagram with |λ| = |γ|.Then diagrams λ and γ can be obtained from each other by applying the above two rules.

Lemma 5 . 4 .
Every tiling of a destructible diagram contains a domino rim-hook.

Corollary 5 . 9 .
Let d χ : M n (C) → C be an immanant induced by an irreducible character χ of S n .Then the following are equivalent.(i)d χ vanishes identically on a subspace A n (C) of alternate matrices.(ii) Either n is odd, or n is even and d χ (J ⊕ • • • ⊕ J) = 0 where J = ( 0 1 −1 0 ).(iii) Either n is odd, or n is even and χ is induced by an indestructible diagram.

Figure 3 .
Figure 3.There are |T | = 7 possibilities to recursively remove rim-hook dominoes by following a fixed tiling T of diagram (4, 4, 3, 1) (c.f.Theorem 5.7 and the text immediately before it).The domino rim-hooks of each obtained diagram are shaded.