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.
 A: 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 $V\cong V^*$.
A: Mariano's answer is the more general and enlightening one, but it is interesting to see this concretely.
An isomorphism between between the natural representation of $SL_2$ and its dual simply requires the existence of a matrix $S \in GL_2$ with the property that $SAS^{-1} = (A^{-1})^T$, for any $A \in SL_2$. (Then $S$ will be the intertwiner between the action of $SL_2$ on $\mathbb{C}^2$, and the action of $SL_2$ on $(\mathbb{C}^2)^*$, where we have a dual basis.)
One can check that this matrix works, $S = $
\begin{bmatrix}
    0 & 1 \\
    -1 & 0
  \end{bmatrix}
