Why is $\operatorname{Sym}^k(F^n)$ irreducible?

Question

Consider a field $F$ and the standard representation of $SL_n(F)$ on $F^n$. Let $k$ be an integer. Then $SL_n(F)$ acts on $\operatorname{Sym}^k(F^n)$. Why is this latter representation irreducible? (I think it is true but I may be wrong).

Answer 1

First, the result is false in positive characteristic: if $\text{char}(F) = p$ and $p | k$ then $\text{Sym}^k(F^n)$ contains a subrepresentation given by the $p^{th}$ power of $\text{Sym}^{ \frac{k}{p} }(F^n)$ in $\text{Sym}(F^n)$.

The result is true in characteristic zero. If $e_1, ... e_n$ is the standard basis of $F^n$, then we can identify $\text{Sym}^k(F^n)$ with homogeneous polynomials of degree $k$ in the variables $e_i$. We want to show that if $f(e_1, ... e_n) \in \text{Sym}^k(F^n)$ is nonzero, then the $\text{SL}_n(F)$-subrepresentation it spans is the whole thing; call it $V$. This is straightforward when $k = 1$. When $k > 1$, write $$f(e_1, ... e_n) = \sum e_1^{k_1} e_2^{k_2} g_{k_1 k_2}(e_3, ... e_n) \in V$$

and note that for any $t \in F$ we have $$f(te_1, t^{-1} e_2, ... e_n) = \sum t^{k_1 - k_2} e_1^{k_1} e_2^{k_2} g_{k_1 k_2}(e_3, ... e_n) \in V.$$

By taking suitable linear combinations of these for varying $t$ (this is essentially Lagrange interpolation and fails if $F$ is finite) it follows that $$\sum_{k_1 - k_2 = r} e_1^{k_1} e_2^{k_2} g_{k_1 k_2}(e_3, ... e_n) \in V$$

for all $r$. In other words we can isolate the terms in $F$ where $k_1$ differs from $k_2$ by a chosen constant. Applying the above to all pairs of variables $e_i, e_j$ we may isolate a unique term $$e_1^{k_1} ... e_n^{k_n} \in V$$

by specifying the differences between its exponents (this determines the term uniquely because the total degree is fixed to be $k$). Now note that for any $t \in F$ and any $i, j$ we have $$e_1^{k_1} ... (e_i + t e_j)^{k_i} ... e_j^{k_j} ... e_n^{k_n} \in V$$

and by taking appropriate linear combinations of these vectors (this step fails if $F$ has positive characteristic and $\text{char}(F) | k_i$) it follows that $$e_1^{k_1} ... e_i^{k_i - s} ... e_j^{k_j + s} ... e_n^{k_n} \in V$$

for any $0 \le s \le k_i$. Applying the above to all pairs $i, j$, we conclude that we can in fact obtain any monomial of degree $k$, and the conclusion follows.