Why is the identity representation of $SL(2,k)$ isomorphic to its dual representation?

By identity representation I mean the representation sending each element of $SL(2,k)$ to itself. Is there a simple way to see this isomorphism? I feel like I am missing something incredibly basic here.

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Call $V$ that representation. The wedge product in the exterior algebra gives you a map $\mu:V\otimes V\to\Lambda^2V$, which is a morphism of $\mathrm{SL}(2,k)$-modules. Now $\Lambda^2V$ is one-dimensional, and the induced action of $\mathrm{SL}(2,k)$ on it is trivial (precisely because the elements of the group have determinant one!), so we can identify it with the trivial module $k$. Partially transposing $\mu$ (that is, considering its image under the natural isomorphism $\hom_k(V\otimes V,k)\to\hom_k(V,V^*)$), we obtain a map $\mu^t:V\to V^*$ which is also of $\mathrm{SL}(2,k)$-modules. Since $\mu$ is non-degenerate, $\mu^t$ is an isomorphism.
Alternatively, the module $V$ is simple and, for reasonable characteristics (I guess this means in this case different from $2$...) and reasonable notions of what a representation is (see Matt's comment below), the group $\mathrm{SL}(2,k)$ has exactly one simple module of dimension $2$. Since $V^*$ is also simple and of dimension $2$, so we must necessarily have $V\cong V^*$.
Dear Mariano, Regarding your last paragraph, you have to be a little careful in what category you view $SL(2,k)$. E.g. if $k$ is a finite field, then for each automorphism $\sigma$ of $k$, the twist $V^{\sigma}$ is a two-dimensional rep. of $SL(2,k)$ over $k$, and these are non-isomorphic for different $\sigma$. (Here I am regarding $SL(2,k)$ simply as an abstract group.) Similarly if we regard $SL(2,\mathbb C)$ as a real Lie group, it has two continuous $2$-dimensional reps., the standard one and its complex conjugate. Best wishes, –  Matt E Feb 10 '11 at 6:33
I'm working on a problem for homework that's similar to this, however, I don't know of a natural isomorphism $hom_k(V \otimes V, k) \rightarrow hom_k (V,V^*)$. It appears to be standard, would I be able to find this in the literature on wedge products? –  Heidi Feb 24 '11 at 5:57
@Heidi: you surely know of a natural isomorphism $\hom_k(V\otimes W,U)\to\hom_k(V,\hom_k(W,U))$. Now take $W=V$ and $U=k$. –  Mariano Suárez-Alvarez Feb 24 '11 at 6:52