# Symmetric power of vector space

Let $V$ be a vector space over a field $k$ of char. zero and denote by $Sym^n_k V$ its $n$-th symmetric power over $k$.

Now I simply want to know what

$Hom_k(V,Sym^n_k V)$

is for $n \geq 2$. To be more precise: might it be zero? Can one say anything more about this $Hom$?

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These are just vector spaces, right? You have a formula for the dimension of $\operatorname{Sym}^n V$. – Dylan Moreland May 10 '12 at 6:12
How can a Hom between two vector spaces of positive dimension possibly be zero? Do you perhaps want the space of intertwining operators as $\text{GL}(V)$ representations? – Qiaochu Yuan May 10 '12 at 6:29
of course, sorry for this stupidity. actually i was considering modules over $k$ a ring of char. zero, and interested in some natural (unequal zero) morphisms. but then already an anwer was made and so i didn't want to edit the question. in fact, i didn't really know what kind of naturality i precisely wanted, so again, sorry for this. – Cyril May 10 '12 at 7:06

As Dylan Moreland points out, this is a strange question. Unless you are asking for some kind of natural isomorphisms we are in vector space and so you should do as the vector space theorists do--count dimensions. Namely, if $\dim_k V=m$ then $\dim_k \text{Sym}^n(V)={m+n-1 \choose n}$. Thus, $\dim_k \text{Hom}_k(V,\text{Sym}^n(V))=m{m+n-1 \choose n}$.
You should note that one of the things that makes vector spaces so incredibly nice is that the statement $\text{Hom}_k(V,W)$ is zero is almost absurd. Vector space theory is just studying finite sets and maps between those finite sets. Thus, the statement "Do there exist vector space maps $V\to W$?" is the same thing as asking "Do there exist SET maps $X\to Y$?" where $X$ and $Y$ are the bases of $V$ and $W$.