The complexity of the fermionant, and immanants of constant width

where S n denotes the symmetric group, i.e., the group of permutations of n objects, and c(π) denotes the number of cycles in π. Chandrasekharan and Wiese [10] showed that the partition functions of certain fermionic models in statistical physics can be written as fermionants with k = 2. This raises the interesting question of whether the fermionant can be computed in polynomial time.

Since (-1) n+c (n) is also the parity of π, the fermionant for k = 1 is simply the determinant which can of course be computed in polynomial time. But this appears to be the only value of k for which this is possible: Theorem 1. For any constant k > 2, computing Ferm k for the adjacency matrix of a planar graph is #P-hard under Turing reductions. Indeed, recent results of Goldberg and Jerrum [14] imply that for k > 2 the fermionant is hard to compute even approximately. Unless NP ⊂ RP, for k > 2 there can be no “fully polynomial randomized approximation scheme” (FPRAS) that computes Ferm k to within a multiplicative error 1 + in time polynomial in n and 1/ with probability at least, say, 3/4.

For the physically interesting case k = 2, we have a slightly weaker result. Recall that ⊕P is the class of problems of the form “is |S| odd,” where membership in S can be tested in polynomial time.

Theorem 2. Computing Ferm 2 for the adjacency matrix of a planar graph is ⊕P-hard. [23] and more generally Toda’s theorem [20], the polynomial hierarchy [18] reduces to ⊕P under randomized polynomial-time reductions. Therefore, if Ferm 2 is in P then the polynomial hierarchy collapses, and in particular NP ⊆ RP.

Theorems 1 and 2 imply new hardness results for the immanant. Given an irreducible character χ λ of S n , the immanant Imm λ of a matrix A is

In particular, taking λ to be the parity or the trivial character yields the determinant and the permanent respectively. Since the determinant is in P but the permanent is #P-hard [21], it makes sense to ask how the complexity of the immanant varies as λ’s Young diagram ranges from a single row of length n (the trivial representation) to a single column (the parity). Strengthening earlier results of Hartmann [15], Bürgisser [9] showed that the immanant is #P-hard if λ is a hook or rectangle of polynomial width,

where w = Ω(n δ ) for some δ > 0. Recently Brylinski and Brylinski [7] improved these results by showing that the immanant is #P-hard whenever two successive rows have a polynomial “overhang,” i.e., if there is an i such that λ i -λ i+1 = Ω(n δ ) for some δ > 0. The case where λ has large width but small overhang, such as a “ziggurat” where λ i = wi + 1 and n = w(w + 1)/2, remains open.

At the other extreme, Barvinok [3] and Bürgisser [8] showed that the immanant is in P if λ is extremely close to the parity, in the sense that the leftmost column contains all but O(1) boxes. Specifically, [8] gives an algorithm that computes Imm = O(n 2c ) respectively, giving a polynomial-time algorithm.

Any function of the cycle structure of a permutation is a class function, i.e., it is invariant under conjugation. Since any class function is a linear combination of irreducible characters, the fermionant is a linear combination of immanants. If k is a positive integer, it has nonzero contributions from Young diagrams whose width is k or less:

where λ ranges over all Young diagrams with depth at most k, and d λ denotes the number of semistandard tableaux of shape λ and content in {1, . . . , k}. To see this, first note that the class function k c(π) is the trace of π’s action on ( k ) ⊗n by permutation, i.e.,

By Schur duality (see e.g. [13]) the multiplicity of λ in ( k ) ⊗n is d

We then transform k c(π) to (-1) n (-k) c(π) by tensoring each λ with the sign representation, flipping it to its transpose λ T , which has width at most k.

Since there are O(n k-1 ) = poly(n) Young diagrams of width k or less, for any constant k there is a Turing reduction from the fermionant Ferm k to the problem of computing the immanant Imm λ where λ is given as part of the input, and where λ has width at most k. Then Theorems 1 and 2 give us the following corollary. Corollary 3. For any constant integer k, the problem of computing the immanant Imm λ A as a function of A and λ is #P-hard under Turing reductions if k ≥ 3, and ⊕P-hard if k = 2, even if λ is restricted to Young diagrams with width k or less.

In particular, unless the polynomial hierarchy collapses, it cannot be the case that Imm λ is in P for all λ of width 2. This is somewhat surprising, since these diagrams are “close to the parity” in some sense.

We conjecture that the fermionant is #P-hard (as opposed to just ⊕P-hard) when k = 2. Moreover, we conjecture that the immanant is #P-hard for any family of Young diagrams of depth nn δ , or equivalently, any family with a polynomial number of boxes to the right of the first column: Conjecture 4. Let λ(n) be any family of Y

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