I came across this preprint that claims in Lemma 1.1 that Gelfand's trick (also known as Gelfand's lemma) only works in characteristic zero:

Let $H < G$ be finite groups. Suppose we have an anti-involution $\sigma : G \rightarrow G$ that preserves all H double-cosets. Then over algebraically closed fields of characteristic zero (G, H) is a Gelfand pair.

It is not obvious to me where the characteristic of the ground field being zero is used in the proofs from Lang's $SL_2(\mathbb{R})$ book (Theorem 1 and Theorem 3 in Chapter IV), the introduction in this preprint, the last slides in these slides, or anything else that I've seen.

What causes Gelfand's trick to fail in positive characteristic? In this setting, the groups are finite but I would also like to know the answer for the more general versions of Gelfand's trick (i.e. also for locally compact groups with compact subgroups or even reductive groups over local fields with closed subgroups).

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    $\begingroup$ The proof of Lemma 1 in math.toronto.edu/murnaghan/courses/mat445/GelfandPairs.pdf uses that the group algebra $\mathbb{C}G$ is semisimple. I guess, thatGelfand's lemma works over alg. closed fields those characteristic do not divide the order of $G$ $\endgroup$ – tj_ Jun 26 '19 at 3:53
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    $\begingroup$ Ah, so it is actually in the equivalence between the Hecke algebra being commutative and the multiplicity one statement? Indeed, the use of semisimplicity (Maschke's theorem) is also present in Theorem 2 in Chapter IV of Lang's $SL_2(\mathbb{R})$ and in Proposition 6 here: mathematics.stanford.edu/wp-content/uploads/2013/08/…. $\endgroup$ – ferrari Jun 26 '19 at 9:45
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    $\begingroup$ In the case of finite groups in characteristic zero or coprime to |G|, it is Frobenius reciprocity which makes the Gelfand pair condition equivalenet to the Hecke algebra being commutative. In other characteristics, Frobenius reciprocity is replaced with Nakayama reciprocity, which is more subtle. $\endgroup$ – Geoff Robinson Jun 26 '19 at 9:54
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    $\begingroup$ I assume that by Nakayama reciprocity, you mean the statement of Frobenius reciprocity for the induction and restriction functors? Then, is the Gelfand pair condition is equivalent to the Hecke algebra being commutative over any characteristic when G is a compact (or locally compact) group? $\endgroup$ – ferrari Jun 26 '19 at 10:59
  • $\begingroup$ It (the last quesstion) is not immediate ( if even true) as far as I can see. Yes, that is what I mean by Nakayama reciprocity. $\endgroup$ – Geoff Robinson Jun 26 '19 at 11:23

Let me try to muddle my way through. In positive characteristic the Gelfand trick will still tell you that the Hecke algebra of $H$-biinvariant functions $A=Func(H\backslash G/H,k)$ is commutative.

Then, by general nonsense, the Hecke algebra is isomorphic to $End_G(P)$, where $P=k(G/H)=Ind_H^G (k_{triv})$ is the permutation representation.

Since $P$ is not semisimple, its structure can still be complicated. The sort of information you need to conclude that it is a Gelfand pair is that $Soc(P)$ or $P/Rad(P)$ is multiplicity-free. There is no way you can conclude that about $Soc(P)$.

But, here, Doc, is a happy ending for your opera: it follows from Nakayama Lemma that $$End(P/Rad(P))\cong A/Rad(A),$$ forcing the semisimple module $P/Rad(P)$ to be multiplicity free.

Let $V$ be an irreducible $G$-module. By Frobenius-Nakayama-Fudd Reciprocity, $$ Hom_H(k_{triv},V)=Hom_G(P,V)=Hom_G(P/Rad(P),V) $$ must be at most one-dimensional! And so is $$ Hom_H(V,k_{triv})=Hom_H(k_{triv},V^\ast). $$

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  • $\begingroup$ Sorry, what is $\text{Soc}(P)$? $\endgroup$ – ferrari Jul 9 '19 at 15:27
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    $\begingroup$ sets? I will change to Func $\endgroup$ – Bugs Bunny May 27 at 6:11
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    $\begingroup$ The isomorphism claim involving the Nakayama Lemma is not true for all $P$. It's still not clear how the isomorphism for $P = \mathrm{Ind}_H^G (k_{\mathrm{triv}})$ (or all finitely generated $P$?) in particular should follow from the Nakayama Lemma. $\endgroup$ – preamble Sep 5 at 13:04
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    $\begingroup$ See this thread for a finite-dimensional counterexample where $End(P/Rad(P))\not\cong A/Rad(A)$. There must be some other special circumstance or condition on $P$ for the isomorphism to hold? $\endgroup$ – preamble Sep 5 at 17:29
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    $\begingroup$ Googling around, I found this paper that looks like it does the problem, at least for finite groups. $\endgroup$ – preamble Sep 14 at 13:49

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