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Let $V$, $\dim V=n-1$ be the standard representation of the symmetric group $S_n$ and let $V'= \langle x_1,x_2,\ldots,x_n \rangle$ be its natural representation. Then ( see. Fulton, Harris, 4.19) we have $$ {\rm Sym}^2 V=U \oplus V \oplus V_{(n-2,2)}, $$ where $U$ is the trivial representation and $V_{(n-2,2)}$ is a representation that correspond to a partition $(n-2,2).$

Question 1. What is the decomposition $ {\rm Sym}^2 V'? $

Question 2. Can we indicate in explicit way a basis for each irreducible component of this decomposition?

My try is as follows. Since $V'=U \oplus V$ then $$ {\rm Sym}^2 V'={\rm Sym}^2U \oplus {\rm Sym}^2 V \oplus U \cdot V. $$ I think that ${\rm Sym}^2U \cong U $ and $U \cdot V \cong V.$ Then $$ {\rm Sym}^2 V'={\rm Sym}^2U \oplus {\rm Sym}^2 V \oplus U \cdot V=2 U \oplus 2 V \oplus V_{(n-2,2)}. $$ Am I right?

If yes, then what be answer for the second question? I know that $U=\langle x_1+x_2+\cdots+x_n \rangle$ and $V= \langle x_1-x_2,x_1-x_3,\ldots,x_1-x_n \rangle$, but I don't know what is the basis for its realisation in ${\rm Sym}^2V'$ as polynomials of degree $2.$

Edit.

$2U=\langle x_1^2+x_2^2+\cdots+x_n^2 \rangle \oplus \langle x_1 x_2+x_1 x_2 +\cdots+ x_{n-1} x_n\rangle $

Butwhat is the basis of $V_{(n-2,2)}$ and $V$ realised in ${\rm Sym}^2 V'?$

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2 Answers 2

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This problems have been solved in a more general setting in the two following papers

1- P.P. Nikitin et al., On the decomposition of tensor representations of symmetric groups

2- P.P. Nikitin, A Realization of the Irreducible Representations of Sn Corresponding to 2-Row Diagrams in the Space of Square-Free Symmetric Forms

In what follows I will answer both questions only in part, as a complete treatment requires tools more advanced then the ones employed in the question. For detailed proofs I raccomend the previous papers and B.E. Sagan, The Symmetric Group.

We suppose to work on a field $\mathbb{K}$ with characteristic $0$.

Question 1:

Notice that $Sym^2 V'= \langle x_i x_j \rangle_{1 \leq i, j\leq n}$ and $Sym^2 V' = Sym^2 V'_{(2)} \oplus Sym^2 V'_{(1, 1)}$ where $$ Sym^2 V'_{(2)} = \langle x_i x_i \rangle_{1 \leq i \leq n} $$ and $$ Sym^2 V'_{(1, 1)} = \langle x_i x_j \rangle_{1 \leq i < j \leq n}. $$ Note that $Sym^2 V'_{(2)} \cong V'$ as $S_n$-representatations by the bijection $x_ix_i \mapsto x_i$. Hence, $Sym^2 V'_{(2)} \cong V' = U \oplus V$.

While $Sym^2 V'_{(1, 1)} \cong M^{(n-2, 2)}$, where $M^{(n-2, 2)}$ is the vector space generated by the formal linear combinations of $(n-2, 2)$-tabloids (see Definition 2.1.5 in Sagan). Thanks to Theorem 2.11.2 in Sagan, we get the decomposition $$ M^{(n-2, 2)} = U \oplus V \oplus V_{(n-2, 2)} $$ where we explicitly compute Kostka numbers (Definition 2.11.1 in Sagan).

Question 2:

Note that $U$ and $V$ can be representated as elements in $Sym^2 V'_{(2)}$ as follows. $$ U \cong \langle \sum_{i = 1}^n x_ix_i \rangle $$ and $$ V \cong \langle x_ix_i - x_1x_1 \rangle_{2 \leq i \leq n}. $$ While a base for the decomposition of $Sym^2 V'_{(1, 1)}$ is $$ U \cong \langle \sum_{1 \leq i < j \leq n} x_ix_j \rangle, $$ $$ V \cong \langle \sum_{1 \leq i < j \leq n} (a^k_i - a^k_j)x_ix_j \rangle_{2 \leq k \leq n} $$ where $a^k = e_k - e_1$ and $e_i$'s form the canonical base for $\mathbb{K}^n$ and $$ V_{(n-2, 2)} = \langle x_i x_j - x_1 x_j - x_i x_n + x_1 x_n \rangle_{1 < i < j < n} \oplus \langle x_{k+1} x_k - x_{1} x_k - x_{k+1} x_{n} + x_1 x_n \rangle_{1 < k < n-1}. $$ To see that such bases generates the proper spaces, look at the cited papers.

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  • $\begingroup$ Thank you!........................................... $\endgroup$
    – Leox
    Jul 21, 2023 at 18:01
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  1. You're right, that's the decomposition.
  2. As you mention, $Sym^2(U\oplus V)=Sym^2U\oplus (U\cdot V) \oplus Sym^2V$. From this you can immediately see that $Sym^2(U)$ corresponds to $\langle (x_1+...+x_n)^2\rangle$, and $U\cdot V$ to $(x_1+...+x_n)\langle x_1-x_2,...,x_{n-1}-x_n \rangle$. The copies of the trivial and standard representations inside $Sym^2(V)$ correspond to $U\cong\langle \sum_{i,j}(x_i-x_j)^2\rangle$ and $V\cong \langle (nx_i-\sum_{j=1}^nx_j)^2-(nx_{i+1}-\sum_{j=1}^nx_j)^2 \rangle_{i=1}^{n-1}$. For the copy of $V_{(n-2,2)}$ in theory you can apply the proof of Maschke's theorem to find the complement to $U\oplus V$ in $Sym^2(V)$.
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