Let $E$ and $F$ be $n$-dimensional representations of $S\_n$ of the form $V\oplus\mathbb{C}$, where $V$ is the standard representation. I just wonder if there is a formula to decompose $S^k(E\otimes F)\otimes\Lambda^m(E\otimes F)$ into irreducible modules of $S\_n\times S\_n$. Many thanks.



Symmetric group module decomposition
This is very far from a complete answer, but it should give enough results to enable the irreducible constituents to be computed for some small values of $k$, $m$ and $n$. I would be amazed if there was an easy formula for the decomposition. To clarify the notation, I'll write $\boxtimes$ for the outer tensor product, so if $U$ and $V$ are representation of $S\_n$ then $U \boxtimes V$ is a representation of $S\_n \times S\_n$.


The question asks for the inner tensor product of $S^k( E\boxtimes F)$ and $\Lambda^m(E \boxtimes F)$ where $E$ and $F$ are the natural $n$-dimensional permutation representations of $S\_n$. The symmetric power and exterior powers can be decomposed using the plethystic identities


$$ S^k(E \boxtimes F) = \sum\_\mu \Delta^\mu(E) \boxtimes \Delta^\mu (F)$$


$$ \Lambda^m (E \boxtimes F) = \sum\_\mu \Delta^\mu(E) \boxtimes \Delta^{\mu'}(F)$$


where the sums all over all partitions $\mu$ of $k$ or $m$ respectively, $\Delta^\mu(E)$ is the image of $E$ under the Schur functor for the partition $\mu$, and $\mu'$ is the conjugate partition to $\mu$. One source for these identities is this [article](https://link.springer.com/article/10.1007/s10801-010-0238-4) by Loehr and Remmel (see bottom of page 191). They are generalizations of the isomorphism of Schur functors used by S. Carnahan in this [Mathoverflow answer](https://mathoverflow.net/questions/77948/an-isomorphism-of-2-schur-modules).


We then need to understand $\Delta^\mu(E)$ as a representation of $S\_n$. I believe this is very tricky, and only a few special cases are known. If $\mu = (k)$ then $\Delta^k(E) = S^k E$: in this case the character can be written as a sum of Young permutation characters, and the irreducible constituents determined by Young's rule. (There are also relevant results from invariant theory.) If $\mu = (1^k)$ then $\Delta^k(E) = \Lambda^k(E)$ and the character is well known to be


$$ \chi\_{\Lambda^k(E)} = \chi^{(n-k+1,1^{k-1})} + \chi^{(n-k,1^k)} $$


using the standard notation.


Assuming that the characters of $S^k(E\boxtimes F)$ and $\Lambda^m(E \boxtimes F)$ have been expressed as sums of irreducible characters of $S\_n \times S\_n$, the identity
$$ (U \boxtimes V) \otimes (U' \boxtimes V') \cong (U \otimes U') \boxtimes (V \otimes V') $$
then reduces the problem to decomposing various inner tensor products of representations of $S\_n$ into irreducible representations. This is yet another very tricky problem and few general results are known. [This paper](http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=ICN&pg5=TI&pg6=ALLF&pg7=REFF&pg8=ET&review_format=html&s4=bessenrodt&s5=kronecker&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=2&mx-pid=1835510) by Bessenrodt and Kleshchev is a good introduction. One very special case is tensor products with the $(n-1)$-dimensional standard representation: for this the identity


$$ \chi^\lambda \chi^{(n-1,1)} = \rm{Ind}^{S\_n} \bigl(\rm{Res}\_{S\_n-1} \chi^\lambda \bigr) - \chi^{(n-1,1)} $$


combined with the ordinary branching rule is often useful.


To give a small example, suppose we want to compute the character of $S^2(E \boxtimes F) \otimes (E \boxtimes F)$. The character of $S^2 E$ is the sum of the Young permutation characters
$ \pi^{(n-2,2)} + \pi^{(n-1,1)} $
and so
$$\chi\_{S^2 E} = 2\chi^{(n)} + 2\chi^{(n-1,1)} + \chi^{(n-2,2)}.$$
By the result on exterior powers mentioned above, $\chi\_{\Lambda^2 E} = \chi^{(n-1,1)} + \chi^{(n-2,1,1)}$. This gives enough to write $S^2(E \boxtimes F)$ as a sum of irreducible characters of $S\_n \times S\_n$, and then the restriction / induction trick can be used to finish the calculation.


