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Let $n \geq 7$ and $V = \mathbb{C}^n$ be the standard representation for $S_{n+1}$, the symmetric group of cardinal $(n+1)!$

Let $k$ be an integer such that $2 \leq k \leq n$. Is it true or false that $\bigwedge^k V$ does not appear as an irreducible sub-representation of $\mathrm{Sym}^k V$? I am looking for a reference which would either prove or give counter-examples to this.

Thanks in advance!

EDIT : Is there an online software (analogous to LIE) that would compute the decompositions into irreducible representations of the tensor product of two ireps for the symmetric group?

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Let $W = \mathbb C^n$, with the obvious representation of $S_n$.

Lemma Let $\lambda$ be a partition of $n$. The multiplicity of the irreducible representation $V_\lambda$ inside $W^{\otimes k}$ is given by the number of ways of obtaining the partition $\lambda$ by starting with the partition $[n]$, and then alternating between removing a box from the Ferrers diagram and adding a box to the Ferrers diagram $k$ times.

Proof Induction on $k$, the base case $k=0$ being trivial. We have $$ W^{\otimes (k+1)} = W^{\otimes k} \otimes \mathrm{Ind}_{S_{n-1}}^{S_n} \mathbb C = \mathrm{Ind}_{S_{n-1}}^{S_n} \mathrm{Res}^{S_n}_{S_{n-1}} W^{\otimes k},$$ see e.g. Exercise 3.16 of Fulton-Harris. Thus the result follows from Pieri's rule. QED

Let $V = W/\mathbb C$ be the irreducible representation corresponding to the partition $[n-1,1]$. Then $\wedge^k V$ is also irreducible, with corresponding partition $[n-k,1,1,\ldots,1]$, for $k\leq n-1$. By the preceding lemma, it is easy to see that the multiplicity of $\wedge^k V$ inside $W^{\otimes k}$ is exactly $1$ for $k \leq n-1$.

It follows in particular that $\wedge^k V$ occurs with multiplicity $0$ inside $\mathrm{Sym}^k(W)$ for $k=2,\ldots,n-1$. Indeed $\wedge^k(V) \subset \wedge^k(W)$ and $\wedge^k(W) \cap \mathrm{Sym}^k(W) = \{0\}$ for $k \geq 2$, where the intersection is taken inside of $W^{\otimes k}$. In particular this representation doesn't occur inside $\mathrm{Sym}^k(V)$, either.


Also, here is some SAGE code that will compute things for you.

s = SymmetricFunctions(QQ).schur() V = s([n-1,1]) ext = s([1]*k) sym = s([k]) print ext.inner_plethysm(V) ## decomposition of \wedge^k(V) print sym.inner_plethysm(V) ## decomposition of Sym^k(V)

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  • $\begingroup$ Thanks for your answer and for the Sage code! I also discover the program SCHUR (schur.sourceforge.net) and using it I also noticed the same thing as you did : $\bigwedge^k V$ seems to appear with multiplicty one in $V^{\otimes k}$ for $k \leq \dim V$. But I had absolutely no idea how to prove it. So you say the key is Pieri's rule. $\endgroup$ – Libli Oct 24 '17 at 19:37
  • $\begingroup$ Since I am a bit of a beginner in rep-the, I want to be sure I understood your answer. You say that $V_{\lambda} \otimes_{S_n} \mathbb{C}^n = V_{\lambda} \otimes_{S_n} \mathrm{Ind}_{S_{n-1}}^{S_n} \mathbb{C}$. And the RHS is equal to $\mathrm{Res}_{S_{n-1}}^{S_n} V_{\lambda} \otimes_{S_{n-1}} \mathbb{C}$ by adjunction. $\endgroup$ – Libli Oct 24 '17 at 20:02
  • $\begingroup$ But to get $\mathrm{Res}_{S_{n-1}}^{S_n} V_{\lambda}$ you take the sum over all ways of removing one box of the Ferrer diagram of $\lambda$. And then to get $V_{\mu} \otimes_{S_{n-1}} \mathbb{C}$, you apply Pieri's rule to find that it is the sum over all ways of adding one box to the Ferrer diagram of $\mu$. Is it correct? $\endgroup$ – Libli Oct 24 '17 at 20:05
  • $\begingroup$ @Libli I wrote out a more detailed argument. I'm not really using adjunction - I use the formula $V \otimes \mathrm{Ind}\, W = \mathrm{Ind}(\mathrm{Res}(V) \otimes W)$ for representations $V$, $W$ of a group $G$ and a subgroup $H$ respectively. I don't really know what to call the formula (which is why I wrote "induction-restriction"); in algebraic geometry you would call this the "projection formula". $\endgroup$ – Dan Petersen Oct 25 '17 at 12:26
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Here is another attempt. It follows from Enumerative Combinatorics, vol. 2, Exercise 7.73, that the Frobenius characteristic of the representation $\mathrm{Sym}^k V$ is the coefficient of $q^k$ in the power series $$ (1-q)\sum_{\lambda\vdash n+1}s_\lambda(1,q,q^2,\dots)s_\lambda. $$ (The factor $1-q$ appears because Exercise 7.73 deals with the symmetric algebra of the defining representation of degree $n$. The defining representation is the sum of $V$ and the trivial representation.) By Corollary 7.21.3, $$ s_\lambda(1,q,q^2,\dots) = q^{b(\lambda)} +\ \mbox{higher order terms}, $$ where $b(\lambda) = \sum (i-1)\lambda_i$. Hence the least $k$ for which an irrep indexed by $\lambda$ appears in $\mathrm{Sym}^k V$ is $b(\lambda)$. For $\bigwedge^k V$ we have $b(\lambda) =b(n-k+1,1^k) = {k+1\choose 2}>k$ for $k>1$.

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