Definition of symmetric power of a linear represenation I'm reading Kowalski's Representation theory, and there's a part about the symmetric and antisymmetric powers of a representation, and I'd like to ask a question about those.
So there's a proposition in the book that says if we have a representation $\rho: G \to GL(E)$ and a covariant functor $T:E \mapsto T(E)$, $\Phi \mapsto \Phi_{*}$, from $Vect_{k}$ to itself, then we can construct a representation $T(\rho): G \to GL(T(E))$ naturally. This is obvious and I completely understand this statement.
Then, when defining the symmetric and antisymmetric powers of a representation, the author simply states that there is a covariant functor that maps $E$ to $\mathrm{Sym}^{m}(E)$ (which is, as far as I understand, the set of all symmetric $m$-linear maps $f: E \times \dots \times E \to k$), and thus we can define the $m$-th symmetric power of a representation $\rho: G \to GL(E)$, calling it for example $\mathrm{Sym}^m \rho: G \to GL(\mathrm{Sym}^{m}(E))$. However, I don't know of any covariant functors from $E$ to $\mathrm{Sym}^{m}(E)$; all I know is a contravariant functor: $$ E \mapsto \mathrm{Sym}^{m}(E), (\Phi: E_{1} \to E_{2}) \mapsto (\Phi^{*}: \mathrm{Sym}^{m}(E_{2}) \to \mathrm{Sym}^{m}(E_{1}))$$
given by $\Phi^{*}(f)(v_{1},...,v_{m}) = f(\Phi(v_{1}),...,\Phi(v_{m}))$.
My question is: What is the symmetric power of a representation?
Here are some of my observations:


*

*If there were a natural covariant functor from $E$ to $\mathrm{Sym}^{m}(E)$, i.e. if someone knows of one, I'd appreciate a clear formulation of one, which would answer my question.

*Is there a way to define a representation with regard to a contravariant tensor? Here's my attempt in this particular case:


$$\mathrm{Sym}^{m}\rho: G \to \mathrm{Sym}^{m}(E), \mathrm{Sym}^{m}\rho(g)f(v_{1},...,v_{m}) = f(\rho(g^{-1})v_{1},...,\rho(g^{-1})v_{m}).$$ This is indeed a homomorphism, and the reason I had to put $g^{-1}$ instead of $g$ was because if I had put $g$, I'd have $\mathrm{Sym}^{m}\rho(gh) = \mathrm{Sym}^{m}\rho(h) \circ \mathrm{Sym}^{m}\rho(g)$. Is this the standard definition?
I would appreciate any details with regards to what is standard in this case.
 A: What you've defined is not the symmetric power, since as you say it's contravariant. The $n^{th}$ symmetric power $S^n(V)$ of a vector space $V$ is the quotient of the $n^{th}$ tensor power $V^{\otimes n}$ (which is also covariant) by the action of the symmetric group $S_n$ permuting the factors. It has the following universal property: every symmetric multilinear map $V^n \to W$ factors uniquely through a linear map $S^n(V) \to W$. So there is a universal symmetric multilinear map
$$V^n \to S^n(V).$$
What you call the symmetric power is the dual of $S^n(V)$. 
The symmetric powers are the graded components of the symmetric algebra
$$S(V) = \bigoplus_{n \ge 0} S^n(V)$$
and in particular there's a bilinear map $S^n(V) \times S^m(V) \to S^{n+m}(V)$ giving the symmetric algebra the structure of a commutative algebra. If $e_1, \dots e_n$ is a basis for $V$, $S^n(V)$ has a basis consisting of all monomials in the $e_i$, thought of as variables, of degree $n$, and $S(V)$ is the polynomial algebra on the $e_i$. 
As for your second question, yes, and that's how you do it. See dual representation. 
