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There are a number of problems in combinatorial representation theory and algebraic geometry for which no positive formula is known. There are several examples I am thinking of, but let me take computing Kronecker coefficients as my example. Usually, the notion of "positive formula" is not precisely defined in combinatorics, but it roughly means "a description as the cardinality of seem reasonably explicit set". Recently, I've been talking to Jonah Blasiak, and he's been convincing me that the right definition of "positive formula" is #P. I'm going to assume that, on this site, I don't need to define #P.

Buergisser and Ikenmeyer show that Kronecker coefficients are #P hard. (They are also always positive, because they are tensor product multiplicities.) But I am reasonably sure that no one knows a way of computing them which even gets them into #P.

So, suppose that I were to actually make an attempt at proving Kronecker coefficients aren't in #P. I assume that what I would do is assume some complexity theoretic conjecture and then reduce Kronecker product to some other problem which is known to be complete for a class larger than #P.

What conjecture might I assume, and what problem might I try to reduce to?

ADDED: As has been pointed out in the comments, Buergisser and Ikenmeyer show that Kronecker coefficients are in Gap-P, which is pretty close to #P. So it sounds like the questions I should be asking are (1) What are some Gap-P-complete problems I could plausibly reduce to and (2) what are the prospects of showing that Gap-P is not #P? I guess (2) should break up into two parts (2a) do experts believe these classes are different? and (2b) are there any likely strategies to prove it?

I hope that this much editing of the question is not frowned on.

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    $\begingroup$ Welcome to cstheory! (I added counting-complexity and lower-bounds to the question). $\endgroup$
    – Kaveh
    Sep 11, 2011 at 13:53
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    $\begingroup$ @Kaveh Bürgisser and Ikenmeyer show that computing Kronecker coefficients is in GapP. David, are Kronecker coefficients always non-negative integers? $\endgroup$
    – Tyson Williams
    Sep 11, 2011 at 14:39
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    $\begingroup$ Yes. They are multiplicites of tensor products, so they are always nonnegative. $\endgroup$
    – David E Speyer
    Sep 11, 2011 at 15:11
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    $\begingroup$ You have a problem in GapP and you want to prove that it is outside #P. An obvious approach is to show that the problem is GapP-complete under functional (Levin) reducibility, which will imply that the problem is outside #P assuming #P≠GapP. $\endgroup$
    – Tsuyoshi Ito
    Sep 11, 2011 at 17:34
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    $\begingroup$ What I wrote in my previous comment is incorrect, because any problem in GapP is functional reducible to #P (if I am not mistaken this time). In other words, the difference between #P and GapP is too delicate to handle by using functional reducibility. $\endgroup$
    – Tsuyoshi Ito
    Sep 11, 2011 at 20:26

3 Answers 3

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I'd suggest looking at properties of #P functions that are different than Gap-P functions. For example, determining if a #P function is zero is in co-NP. If you could show determining whether the Kronecker coefficients is zero is UP-hard then you would have "Kronecker coefficients in #P implies UP in co-NP", an unlikely conclusion.

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GapP is exactly the closure of #P under subtraction. On the other hand, #P is not closed under subtraction unless UP=PP. I believe that answers your questions.

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    $\begingroup$ If you voted it down, at least explain why it is wrong.. Thanks $\endgroup$
    – Tayfun Pay
    Sep 13, 2011 at 0:09
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    $\begingroup$ I agree. As far as I can tell the answer makes two correct statements and answers the original question (although my searching revealed that UP=PH is the desired conditional?) $\endgroup$
    – Suresh Venkat
    Sep 13, 2011 at 13:23
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    $\begingroup$ @Suresh: How does this post answer the original question? The question is not about a GapP-complete problem. $\endgroup$
    – Tsuyoshi Ito
    Sep 13, 2011 at 14:06
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    $\begingroup$ part (2) in the update asks: "what are the prospects of GapP not being equal to #P". this answer points out that unless a collapse happens, #P is not closed under subtraction and so there's no point even talking about equality. $\endgroup$
    – Suresh Venkat
    Sep 13, 2011 at 15:09
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    $\begingroup$ @Suresh: This is the paper. M.Ogiwara & L. Hemachandra. “A complexity theory for feasible closure properties.” Journal of Computer and System Sciences Volume 46 Pages 295-325. 1993. $\endgroup$
    – Tayfun Pay
    Sep 14, 2011 at 23:44
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The question of computing Characters of irreducible representations of the symmetric group might be a natural candidate.

I think Charles Hepler shows it is Gap-P complete, but I'm not sure: for a link to his Master's thesis, see https://dspace.ucalgary.ca/handle/1880/45530?mode=full

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