End of standard representation of $SL_n$

Question

Let $V \cong \mathbb{C}^n$ be the standard representation of $SL_n(\mathbb{C})$. One can consider $End(V) \cong V \otimes V^*$ as a representation of $SL_n(\mathbb{C})$.

Is it known how $End(V)$ decomposes as $SL_n(\mathbb{C})$-representation? More precisely, is it isomorphic to $\mathbb{C} \oplus W$ with $W$ some irreducible $SL_n(\mathbb{C})$-representation?

Any reference would be very welcome.

In case of a positive answer, I am also interested in the setting where $\mathbb{C}$ is replaced by a more general field of characteristic $0$ or even a division algebra.

In case of a negative answer, I would like to know if there is some other irreducible $SL_n(\mathbb{C})$-module $M$ for which $End(M) \cong \mathbb{C} \oplus W$.

In the case of $SL_2(\mathbb{C})$ it is known that all irreducible representation are isomorphic to a symmetric power $S^mV$ of the standard representation. Moreover, $S^mV \otimes S^mV^* \cong S^mV \otimes S^mV \cong S^{2m}V \oplus \ldots \oplus S^2V \oplus \mathbb{C}$. Thus in this case only the standard representation, i.e. $m=1$, has the desired property.

Answer 1

To elaborate a bit on what has been said in the comments (working over $\mathbb{C}$, or any algebraically closed field of characteristic $0$), but using the language of highest weights, note that the standard representation $V$ in terms of irreducible highest weight modules is $V(\varpi_1)$, corresponding to the first fundamental weight, while $V^\ast$ is the module $V(\varpi_{n-1})$. Then the tensor product is, as pointed out in the comments and in the linked related question, $$ V(\varpi_1) \otimes V(\varpi_{n-1}) \cong V(\varpi_1+\varpi_{n-1}) \oplus V(0), $$ and $V(\varpi_1+\varpi_{n-1})$ is isomorphic to the adjoint representation, while $V(0)=\mathbb{C}$ is the trivial representation. To see this, note that any other irreducible component of the tensor product must have highest weight less than $\varpi_1+\varpi_{n-1}$ in the dominance order on the weight lattice; but this is equal to the highest root, and there are no other dominant weights less than this for $SL_n$ except for $0$. The trivial module always appears in $\text{End}(M)$ for any module $M$, so we know it appears in this case and gives the only other possible component in the tensor product.

Perhaps (to me) the more interesting part of your question is: could there be another module $M$ with $$\text{End}(M) \cong V(\varpi_1+\varpi_{n-1}) \oplus V(0)$$ (again working over $\mathbb{C}$)? There are likely easier ways to see it, but this follows from a very nice result of C.S. Rajan (here) which says (briefly) that tensor products of irreducible finite-dimensional modules can be "uniquely factored": in this case, we get $$ \text{End}(M)=\text{End}(N) \implies M \cong N \text{ or} M \cong N^\ast$$ for any irreducible, finite-dimensional modules $M$ and $N$.

Answer 2

Question: "Is it known how $End(V)$ decomposes as $SL_n(\mathbb{C})$-representation? More precisely, is it isomorphic to $\mathbb{C} \oplus W$ with $W$ some irreducible $SL_n(\mathbb{C})$-representation? Any reference would be very welcome."

Answer: If $V:=\mathbb{C}^n$ there is a trace map $tr: End_{\mathbb{C}}(V) \cong V^* \otimes V \rightarrow \mathbb{C}$ defined by $tr(f \otimes v):=f(v)$. The map $tr$ is a map of representations and $ker(tr) \cong \mathfrak{sl}(V)$ is the Lie algebra of matrices with trace zero. This Lie algebra is simple hence it has no submodules, and you get a decomposition into irreducible modules

$$End(V) \cong \mathfrak{sl}(V) \oplus T$$

where $T \cong \mathbb{C}$ is the trivial module. A good introductory book is the book of Fulton/Harris on representation theory. This book has many instructive examples.

"In the case of $SL_2(\mathbb{C})$ it is known that all irreducible representation are isomorphic to a symmetric power $S^mV$ of the standard representation. Moreover, $S^mV \otimes S^mV^* \cong S^mV \otimes S^mV \cong S^{2m}V \oplus \ldots \oplus S^2V \oplus \mathbb{C}$. Thus in this case only the standard representation, i.e. $m=1$, has the desired property."

Answer: In the case of $SL_2$ it follows $\mathfrak{sl}(V) \cong Sym^2(V) \cong Sym^2(V^*)$.