3 added 1187 characters in body; edited body; added 4 characters in body

I don't remember the

The proof of this fact is given as follows (see http://www-math.mit.edu/~etingof/tenscat.pdf, but or Akhil's comments below):

$Hom_H(V\otimes H,W)=Hom_H(H,^\ast V\otimes W) = \widetilde{^\ast V\otimes W}$

On the other hand, $Hom_H(\tilde{V}\otimes H,W) = \tilde{V}\otimes Hom_H(H,W) = \widetilde{V\otimes W}$,

where $\tilde{M}$ means we forget the module $M$ down to a vector space, which we use as a multiplicity space (just because the direct sum decomposition I asserted originally isn't canonically given, you just know that there's this multiplicity space)

(above we took right duals since I didn't assume $H$ is commutative or co-commutative; for $C[G]$ there is no need to distinguish.) One could (and should) be uncomfortable that we got duals on the one hand and not on the other. However, the standard representation for $S_n$ is special in that it is isomorphic to its own dual, by sending $e_i$ to $e^i$ (the point is that the standard rep for $S_n$ has a basis build into its definition).

The general fact above about Hopf algebras is used to relate Frobenius-Perron dimension for representations of Hopf algebras to ordinary dimension of the underlying vector space. If needed, I can try to find a reference for this; indeed the regular representation is the unique eigenvector which realizes the Frobenius Perron dimension as an eigenvalue.

Okay so now we are considering $\mathfrak{h}\otimes \mathbb{C}[S_n/S_{\pi}]\to \mathbb{C}[S_n/S_{\pi}]$. This should be is then isomorphic to $(\mathfrak{h}\otimes \mathbb{C}[S_n])\otimes_{S_\pi}\mathbf{1}$, where we tensor the trivial $S_\pi$-module on the right. I'm asserting This is because $C[S_n]$ is a $S_n-S_\pi$ bi-module, so that taking this commutes the map$\mathfrak{h}\otimes \mathbb{C}[S_n] \to \mathfrak{h}\otimes \mathbb{C}[S_n/S_\pi]$ given by right multiplying with taking tensor productthe symmetrizer $a_\pi=\sum_{g\in S_\pi} g$ is an $S_n$-morphism, and allows us to identify $\mathfrak{h}\otimes \mathbb{C}[S_n]\otimes_{S_\pi}\mathbf{1}$ with $\mathfrak{h}\otimes \mathbb{C}[S_\pi]$.Morally, this is just because $S_\pi$ acts on the right, while the tensor product above other action is relative to on the leftaction.

This

Together with the Fact, this implies that $\mathfrak{h}\otimes \mathbb{C}[S_n/S_\pi]$ is in fact just isomorphic to $\mathbb{C}[S_n/S_\pi]^{\oplus dim(\mathfrak{h})},$ which I should really write as $\mathbb{C}[S_n/S_\pi]\otimes \tilde{\mathfrak{h}}$, where $\tilde{\mathfrak{h}}$ means it's just a vector space, which we use as a multiplicity space (just because the direct sum decomposition I asserted isn't canonically given, you just know that there's this multiplicity space).tilde{\mathfrak{h}^\ast}$. I haven't really proved that this last paragraph is what happens, but once one has applied "Fact" above, this seems like the only natural guess. I imagine verifying it would be pretty straightforward. What do you think? 2 edited an error where indicated I suspect I know the answer, but I don't yet have a proof (not because I think it would be hard to prove, but because I didn't try really; when you see my guess, you'll likely want to believe it). The answer is stated not in the basis of simples, because I didn't compute the decomposition of$\mathbb{C}[S_n/S_{pi}]$. However, it is stated in the tensor category S_n-mod, so that given that decomposition, you can easily adjust what I write here. Fact: Let H be a finite dimensional semi-simple Hopf algebra (e.g. H=\mathbb{C}[S_n]), and let$V\in H$-mod be an irrep. Let us regard H as an H-module via the left action. Then$V\otimes H\cong H^{\oplus dim(V)}$. I don't remember the proof of this fact, but it is used to relate Frobenius-Perron dimension for representations of Hopf algebras to ordinary dimension of the underlying vector space. If needed, I can try to find a reference for this. Okay so now we are considering$\mathfrak{h}\otimes \mathbb{C}[S_n/S_{\pi}]\to \mathbb{C}[S_n/S_{\pi}]$. This should be isomorphic to$(\mathfrak{h}\otimes \mathbb{C}[S_n])^{inv}$, mathbb{C}[S_n])\otimes_{S_\pi}\mathbf{1}$, where we take invariants with respect to tensor the trivial $S_\pi$ acting S_\pi$-module on the right. I'm asserting that taking$S_\pi$-invariants this commutes with taking tensor product, because$S_\pi$acts on the right, while the tensor product above is relative to the left action. [edited an error from preceding paragraph] This implies that$\mathfrak{h}\otimes \mathbb{C}[S_n/S_\pi]$is in fact just isomorphic to$\mathbb{C}[S_n/S_\pi]^{\oplus dim(\mathfrak{h})},$which I should really write as$\mathbb{C}[S_n/S_\pi]\otimes \tilde{\mathfrak{h}}$, where$\tilde{\mathfrak{h}}$means it's just a vector space, which we use as a multiplicity space (just because the direct sum decomposition I asserted isn't canonically given, you just know that there's this multiplicity space). Well, now we have this function$c: \mathfrak{h}\to \mathbb{C}$. We will project$\mathbb{C}[S_n/S_\pi]^{\oplus dim(\mathfrak{h})}$(or rather$\mathbb{C}[S_n/S_\pi]\otimes \tilde{\mathfrak{h}}$) to$\mathbb{C}[S_n/S_\pi]$by just applying$c$to the multiplicity space. I haven't really proved that this is what happens, but once one has applied "Fact" above, this seems like the only natural guess. I imagine verifying it would be pretty straightforward. Note that it doesn't seem to matter how$\mathbb{C}[S_n/S_{\pi}]$decomposes into simples, since they all get lumped together. What do you think? 1 I suspect I know the answer, but I don't yet have a proof (not because I think it would be hard to prove, but because I didn't try really; when you see my guess, you'll likely want to believe it). The answer is stated not in the basis of simples, because I didn't compute the decomposition of$\mathbb{C}[S_n/S_{pi}]$. However, it is stated in the tensor category S_n-mod, so that given that decomposition, you can easily adjust what I write here. Fact: Let H be a finite dimensional semi-simple Hopf algebra (e.g. H=\mathbb{C}[S_n]), and let$V\in H$-mod be an irrep. Let us regard H as an H-module via the left action. Then$V\otimes H\cong H^{\oplus dim(V)}$. I don't remember the proof of this fact, but it is used to relate Frobenius-Perron dimension for representations of Hopf algebras to ordinary dimension of the underlying vector space. If needed, I can try to find a reference for this. Okay so now we are considering$\mathfrak{h}\otimes \mathbb{C}[S_n/S_{\pi}]\to \mathbb{C}[S_n/S_{\pi}]$. This should be isomorphic to$(\mathfrak{h}\otimes \mathbb{C}[S_n])^{inv}$, where we take invariants with respect to$S_\pi$acting on the right. I'm asserting that taking$S_\pi$-invariants commutes with taking tensor product, because$S_\pi$acts on the right, while the tensor product above is relative to the left action. This implies that$\mathfrak{h}\otimes \mathbb{C}[S_n/S_\pi]$is in fact just isomorphic to$\mathbb{C}[S_n/S_\pi]^{\oplus dim(\mathfrak{h})},$which I should really write as$\mathbb{C}[S_n/S_\pi]\otimes \tilde{\mathfrak{h}}$, where$\tilde{\mathfrak{h}}$means it's just a vector space, which we use as a multiplicity space (just because the direct sum decomposition I asserted isn't canonically given, you just know that there's this multiplicity space). Well, now we have this function$c: \mathfrak{h}\to \mathbb{C}$. We will project$\mathbb{C}[S_n/S_\pi]^{\oplus dim(\mathfrak{h})}$(or rather$\mathbb{C}[S_n/S_\pi]\otimes \tilde{\mathfrak{h}}$) to$\mathbb{C}[S_n/S_\pi]$by just applying$c$to the multiplicity space. I haven't really proved that this is what happens, but once one has applied "Fact" above, this seems like the only natural guess. I imagine verifying it would be pretty straightforward. Note that it doesn't seem to matter how$\mathbb{C}[S_n/S_{\pi}]\$ decomposes into simples, since they all get lumped together.

What do you think?