You can find the decomposition of $S^kV$ and $\Lambda^k V$ into a direct sum of irreducible representations in Table 5 in the book: Onishchik and Vinberg, Lie Groups and Algebraic Groups, Springer-Verlag 1990.

Instead of $U(n)$ you can consider its complexification ${\rm GL}(n,{\Bbb C})$. Then both $S^k V$ and $\Lambda^k V$ are irreducible.

For ${\rm SO}(n)$, the representations $\Lambda^k V$ are irreducible except for $k=n/2$, while $S^k V$ are reducible for $k>1$ (indeed, for $k=2$ we know that $S^2 V$ contains the trivial representation).

For ${\rm Sp}(2n)$, the representations $S^k V$ are irreducible, while $\Lambda^k V$ are reducible for $1<k<2n-1$ (indeed, for $k=2$ we know that $\Lambda^2 V$ contains the trivial representation).

You can find the decomposition of $S^kV$ and $\Lambda^k V$ into a direct sum of irreducible representations in Table 5 in the book: Onishchik and Vinberg, Lie Groups and Algebraic Groups, Springer-Verlag 1990.

Instead of $U(n)$ you can consider its complexification ${\rm GL}(n,{\Bbb C})$. Then both $S^k V$ and $\Lambda^k V$ are irreducible.

For ${\rm SO}(n)$, the representations $\Lambda^k V$ are irreducible for $k\neq n/2$, while $S^k V$ are reducible for all $k>1$ (indeed, for $k=2$ we know that $S^2 V$ contains the trivial representation).

For ${\rm Sp}(2n)$, the representations $S^k V$ are irreducible, while $\Lambda^k V$ are reducible for $1<k<2n-1$ (indeed, for $k=2$ we know that $\Lambda^2 V$ contains the trivial representation).

You can find the decomposition of $S^kV$ and $\Lambda^k V$ into a direct sum of irreducible representations in Table 5 in the book: Onishchik and Vinberg, Lie Groups and Algebraic Groups, Springer-Verlag 1990.

Instead of $U(n)$ you can consider its complexification ${\rm GL}(n)$. Then both $S^k V$ and $\Lambda^k V$ are irreducible.

For ${\rm SO}(n)$, $\Lambda^k V$ are irreducible, while $S^k V$ are not (indeed, $S^2 V$ contains the trivial representation).

For ${\rm Sp}(2n)$, $S^k V$ are irreducible, while $\Lambda^k V$ are not (indeed, $\Lambda^2 V$ contains the trivial representation).

You can find the decomposition of $S^kV$ and $\Lambda^k V$ into a direct sum of irreducible representations in Table 5 in the book: Onishchik and Vinberg, Lie Groups and Algebraic Groups, Springer-Verlag 1990.

Instead of $U(n)$ you can consider its complexification ${\rm GL}(n,{\Bbb C})$. Then both $S^k V$ and $\Lambda^k V$ are irreducible.

For ${\rm SO}(n)$, the representations $\Lambda^k V$ are irreducible except for $k=n/2$, while $S^k V$ are reducible for $k>1$ (indeed, for $k=2$ we know that $S^2 V$ contains the trivial representation).

For ${\rm Sp}(2n)$, the representations $S^k V$ are irreducible, while $\Lambda^k V$ are reducible for $1<k<2n-1$ (indeed, for $k=2$ we know that $\Lambda^2 V$ contains the trivial representation).