Find all invariant symmetric bilinear forms of $\mathfrak {sl}(2)$ (as defined in the previous exercise; assume that the field $k$ is of characteristic zero.)
Here is the exercise before it:
Let $L$ be a Lie algebra and $\rho: L \to \mathfrak{gl}(V)$ a finite-dimensional representation of L. Define a symmetric bilinear form on $L$ by $$\langle x, y \rangle_{\rho} = tr (\rho(x) \rho(y))$$
Where $\text{ tr }$ denotes the trace of endomorphisms.
$(a)$ Prove that this form is invariant, i.e., we have $$\langle [x,y], z \rangle_{\rho} = \langle x,[y,z] \rangle_{\rho}$$
for all $x,y,z \in L.$
Some prior knowledge:
I know that the Lie algebra $\mathfrak{gl}(2) = L(M_2(k))$ of $2 \times 2$-matrices with complex entries is four-dimensional and that the four matrices $ X = \begin{matrix}
0 & 1 \\
0 & 0
\end{matrix}, Y= \begin{matrix}
0 & 0\\
1 & 0
\end{matrix}, H = \begin{matrix}
1 & 0\\
0 & -1
\end{matrix}, I = \begin{matrix}
1 & 0\\
0 & 1
\end{matrix}$ form a basis of $\mathfrak {gl}(2).$ Their commutators are $$[X,Y] = H, [H, X] = 2X, [H, Y] = -2Y,$$ and $$[I,X] = [I, Y] = [I, H] = 0. \quad \quad \quad (3.1)$$
The matrices of trace zero in $\mathfrak {gl}(2)$ form the subspace $\mathfrak {sl}(2)$ spanned by the basis $\{X, Y, H\}$ Relations $(3.1)$ show that $\mathfrak {sl}(2)$ is an ideal of $\mathfrak {gl}(2)$ and that there is an isomorphism of Lie algebras $$\mathfrak {gl}(2) \cong \mathfrak {sl}(2) \oplus kI.$$
Then I know also some information about the envelopping algebra $U$ of $\mathfrak {sl}(2).$
Still, I do not know how can I find all invariant symmetric bilinear forms of $\mathfrak{sl}(2)$ specifically I know that it is the subspace of all matrices of trace zero in $\mathfrak{gl}(2).$ Also, what is $\rho$ in my case here?
Best Answer
It's enough that the field characteristic is $\ne 2$.
Let $\beta$ be an invariant symmetric bilinear form on $\mathfrak{sl}(2)$. \begin{align} 0 &= \beta(h, 0) = \beta(h, [x,x]) = \beta([h, x], x) = \beta(2x, x)\\ 0 &= \beta(h, 0) = \beta(h, [y,y]) = \beta([h, y], y) = \beta(-2y, y)\\ 0 &= \beta(y, 0) = \beta(y, [x,x]) = \beta([y, x], x) = \beta(-h, x)\\ 0 &= \beta(x, 0) = \beta(x, [y,y]) = \beta([x, y], y) = \beta(h, y)\\ \beta(2x, y) &= \beta([h, x], y) = \beta(h, [x,y]) = \beta(h, h) \end{align}
So $\beta(x,x) = \beta(y,y) = \beta(h, x) = \beta(h, y) = 0$ and $\beta(h,h) = 2\beta(x,y)$. Ordering the basis as $(x, h, y)$ and writing in a matrix, the matrix of $\beta$ is a scalar multiple of \begin{pmatrix} 0 & 0 & 1\\ 0 & 2 & 0\\ 1 & 0 & 0 \end{pmatrix}
and in fact, this is a valid invariant symmetric binlear form. It's the one associated with $\text{tr}(\phi(x)\phi(y))$ where $\phi$ is the representation of $\mathfrak{sl}(2)$ on a two dimensional space.