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I have a bilinear form $$\sigma_V:L\times L\rightarrow \mathbb{k}$$ $$\sigma_V(x,y)=tr(\rho_V(x)\rho_V(y)) \forall x,y \in L$$ and am looking for the matrix of this form. I am in the algebra $$L= \mathfrak{sl} _2(\mathbb{k})$$ in the standard basis $(x,y,h)$ and the representation $V=\mathbb{k}^2$ which is the standard representation of $\mathfrak{sl}_2(\mathbb{k})$. I know $\sigma_V$ is symmetric and L-invariant but am not sure how to go about finding its matrix? I found lots of information on the Killing form but as $V$ is not the adjoint representation it seems this is not the Killing form. Am I wrong? How do I go about finding the matrix?

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  • $\begingroup$ The standard representation can be regarded as a map $L \to \mathfrak{gl}(V)$, and so we can view its images as endomorphisms of $V$. $\endgroup$ – Travis Oct 26 '15 at 9:54
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For a simple Lie algebra, any two non degenerate invariant bilinear forms are multiples of each other. Indeed, the Killing form is just $4$ times the trace form for $\mathfrak{sl}_2(K)$. In general we have $tr(ad(x)ad(y))=2n \cdot tr(x)tr(y)$ for the Lie algebra $\mathfrak{sl}_n(K)$. So the matrix of $\sigma_V$ above is just the matrix of the Killingform multiplied by $\frac{1}{4}$. This can be verified directly, of course, with $\rho(x)=\begin{pmatrix} 0 & 1 \cr 0 & 0 \end{pmatrix}$, $\rho(y)=\begin{pmatrix} 0 & 0 \cr 1 & 0 \end{pmatrix}$ and $\rho(h)=\begin{pmatrix} 1 & 0 \cr 0 & -1 \end{pmatrix}$. For example, $\sigma_V(x,x)=tr(\rho(x)\rho(x))=tr(0)=0$, $\sigma_V(x,y)=tr(\rho(x)\rho(y))=tr(diag(1,0))=1$, etc.

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