Decomposition of tensor power of symmetric square Studying some representation theory I came up with the following problem.
We work over a field of characteristic $0$. Let $V$ be the standard representation of $\mathrm{GL}_n$ and let $W$ be the representation $(\mathrm{Sym}^2(V))^{\otimes n}$.
Is it possible to describe the weights of the irreducible components of $W$ (after choosing the usual Borel and so on)? In particular I would like to know if the representation $\det(V)^{\otimes 2}$ appears in $W$.
 A: The answer is yes.
Let $e_1,\ldots, e_n$ be the standard basis of $V$. Consider the morphism
$$
f \colon \det(V) \to V^{\otimes n}
$$
given by
$$
f(e_1 \wedge \cdots \wedge e_n) = \sum_{\sigma \in S_n}(-1)^{\varepsilon(\sigma)}e_{\sigma(1)} \otimes \cdots \otimes e_{\sigma(n)},
$$
where $S_n$ is the symmetric group on $n$ letters and $\varepsilon(\sigma)$ is the parity of $\sigma$. This gives a morphism
$$
f^{\otimes 2} \colon \det(V)^{\otimes 2} \to V^{\otimes 2n}=(V^{\otimes 2})^{\otimes n}.
$$
Using the natural projection $V^{\otimes 2} \to \mathrm{Sym}^2(V)$, we get a morphism
$$
g \colon \det(V)^{\otimes 2} \to (\mathrm{Sym}^2(V))^{\otimes n},
$$
that it is easily checked to be injective.
I think that moreover $\det(v)^{\otimes 2}$ appears with multiplicity $1$ (I checked this using a computer up to $n=10$), but I didn't tried to prove it.
A: Just an addendum to Ricky's answer: the multiplicity is indeed 1 which can be proved as follows.
An occurrence of ${\rm det}(V)^{\otimes 2}$ inside $({\rm Sym}^2(V))^{\otimes n}$ is the same thing as a nonzero joint multilinear ${SL}_n$-invariant of $n$ quadratic forms $Q^{(1)},\ldots,Q^{(n)}$ in $n$ variables. By the first fundamental theorem of classical invariant theory, this must be a linear combination of expressions (after choice of coordinates) of the form
$$
\sum_{i_1,\ldots, i_{2n}=1}^{n}
\epsilon_{i_1,\ldots,i_n}\ \epsilon_{i_{n+1},\ldots,i_{2n}}
\ Q_{i_{\sigma(1)},i_{\sigma(2)}}^{(1)}
\ Q_{i_{\sigma(3)},i_{\sigma(4)}}^{(2)}
\cdots
\ Q_{i_{\sigma(2n-1)},i_{\sigma(2n)}}^{(n)}
$$
where $\sigma$ is a permutation of $\{1,\ldots,2n\}$.
Here the $Q_{i,j}^{(a)}$ denote the matrix elements of the quadratic forms and $\epsilon_{i_1,\ldots,i_n}$ is completely antisymmetric with the normalization $\epsilon_{1,\ldots,n}=1$.
If you take a symmetric matrix $A$ and a skew-symmetric $B$ then $\sum_{i,j}A_{ij}B_{ij}=0$ because you are contracting two symmetic indices with two antisymmetric ones. The same is true if $A$ and $B$ are tensors with more indices that are frozen.
Thus the above expression is zero for all permutations $\sigma$ which send two elements of the same block of the partition $\{\{1,\ldots,n\},\{n+1,\ldots,2n\}\}$ to the same block of the partition $\{\{1,2\},\{3,4\},\ldots\{2n-1,2n\}\}$.
It is then easy to see that all you get are multiples of the expression corresponding to say the permutation $\sigma$
defined by
$$
\sigma(i)=2i-1\ \ ,\ \ \sigma(n+i)=2i
$$
for $1\le i\le n$.
Moreover, this invariant is not zero because when specializing to all
quadratics being equal to say $Q$ this gives the non identically vanishing polynomial $n!\ {\rm det}(Q)$.
Note that the above permutation $\sigma$ is not the only that works. There are $2^n\ n!^2$ permutations which satisfy the combinatorial requirement I mentioned but their corresponding invariants differ by a $\pm 1$ factor.
Finally, as remarked by Darij, this easily generalizes to occurrences of ${\rm det}(V)^{\otimes k}$ inside $({\rm Sym}^k(V))^{\otimes n}$. 
A: By Pieri's formula, a partition with $2n$ elements in $n$ rows, corresponding to a representation of $GL_n$, occurs in this representation with multiplicity equal to the number of ways of obtaining that partition by starting with the empty partition and $n$ times adding two elements, no two in the same column.
For the determinant squared, which corresponds to a partition with $2$ columns of length $n$, this occurs with multiplicity exactly one. 
