Casimir Element of $\mathfrak{sl}_{2}(\Bbb{C})$ I am trying to show that the Casimir element of $\mathfrak{g} = \mathfrak{sl}_{2}(\Bbb{C})$, which is has standard generating set $h,e,f$, with respect to the bilinear form $(x,y) = tr(xy)$ is $C = \frac{1}{2} h^2 + fe + ef$. From my understanding, I need to use the bilinear form to find the corresponding dual basis $\{h^*,e^*,f^*\}$ of $\frak{g}^*$, and then the Casimir element will be $C = h h^* + e e^* + f f^*$. 
However, I'm having trouble computing this dual basis; actually, I don't really know how to compute. Also, how do I interpret, e.g., $h h^*$. Isn't this technically an abuse of notation? How do you multiply a matrix and a linear functional? 
 A: Think of $h^*$, $e^*$, and $f^*$ as being elements of $\mathfrak{g}$, and not the dual. The bilinear form allows you to make this identification. What this means is that instead of $h^*$ being a functional, it is the element of $\mathfrak{g}$ with the property that $\mathrm{tr}(hh^*) = 1$, and $\mathrm{tr}(eh^*) = \mathrm{tr}(fh^*) = 0$.
If you compute the trace form on the standard basis
$$ h = \begin{pmatrix} 1&0\\0&-1 \end{pmatrix}, \quad
   e = \begin{pmatrix} 0&1\\0&0 \end{pmatrix}, \quad
   f = \begin{pmatrix} 0&0\\1&0 \end{pmatrix}, $$
you get that
$$
\begin{align*}
\mathrm{tr}(he) &= 0 & \mathrm{tr}(hf) &= 0 & \mathrm{tr}(h^2) &= 2 \\
\mathrm{tr}(e^2) &= 0 & \mathrm{tr}(ef) &= 1 & \\
\mathrm{tr}(f^2) &= 0
\end {align*}
$$
Now you just do a little linear algebra. You write $h^* = ah + be + cf$ for some coefficients $a,b,c$. Then you write the equations $\mathrm{tr}(hh^*) = 1$, and $\mathrm{tr}(eh^*) = \mathrm{tr}(fh^*) = 0$. This gives you a $3 \times 3$ linear system:
$$
\begin {align*}
1 &= \mathrm{tr}(h(ah+be+cf)) = 2a + 0b + 0c \\
0 &= \mathrm{tr}(e(ah+be+cf)) = 0a + 0b + c \\
0 &= \mathrm{tr}(f(ah+be+cf)) = 0a + b + 0c \\
\end {align*}
$$
This immediately shows you that $a = \frac{1}{2}$, and $b=c=0$, and so $h^* = \frac{1}{2}h$. Do this similarly for $e^*$ and $f^*$ to get the other coefficients.
