Space of symmetric $n$-tensors spanned by elements of form $\underbrace{v \otimes \ldots \otimes v}_{n}$?

Question

Let $k$ be a field of characteristic $0$, and let $V$ be a finite dimensional vector space over $k$. Consider the space of symmetric $n$-tensors,$$S^nV = (V \otimes V \otimes \ldots \otimes V)^{S_n},$$where $\otimes = \otimes_k$ and the symmetric group $S_n$ acts on the $n$-fold tensor product$$V^{\otimes n} = V \otimes V \otimes \ldots \otimes V$$by permuting the factors. How do I see that $S^nV$ is spanned by elements of the form$$\underbrace{v \otimes \ldots \otimes v}_{n}, \quad v \in V?$$I've seen some proofs online, but they're way too terse or long and not easily understandable to a beginner like me. Can someone give me a straight to the point sketch of a proof that's clear?

Answer 1

Since you have proofs available, perhaps the most useful thing is for you to see the general pattern. Of course, the statement is trivial for $n = 1$.

Let $e_1,\dots,e_d$ form a basis of $V$, so that the vectors of the form $e_{i_1} \vee \cdots \vee e_{i_n}$ space $V$. For $n = 2$, we see that for $i \neq j$, we have $$ e_i \vee e_j = (e_i + e_j) \otimes (e_i + e_j) - e_i \otimes e_i - e_j \otimes e_j $$ so that we hit every element of that spanning set. Similarly, for $n = 3$, we have $$ e_i \vee e_j \vee e_j + e_i \vee e_i \vee e_j = \\ (e_i + e_j) \otimes (e_i + e_j) \otimes (e_i + e_j) - e_i \otimes e_i \otimes e_i - e_j \otimes e_j \otimes e_j\\ e_i \vee e_j \vee e_j - e_i \vee e_i \vee e_j = \\ (e_i - e_j) \otimes (e_i - e_j) \otimes (e_i - e_j) - e_i \otimes e_i \otimes e_i - e_j \otimes e_j \otimes e_j $$ (note: your coefficients may vary) At this point, we see that we have all elements of the form $u_1 \vee u_2 \vee u_2$ in the span.

From there, we can use the idea from $n = 2$: $$ e_i \vee e_j \vee e_k = e_i \vee (e_j + e_k) \vee (e_j + e_k) - e_i \vee e_j \vee e_j - e_i \vee e_k \vee e_k $$ Perhaps now you can see that pattern which will persist.