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On page 245 of Weibel's An introduction to homological algebra, he assigns the following as an exercise:

The image of $c_M$ in the matrix ring $\operatorname{End}_k(M)$ is $r/m$ times the identity matrix.

Here's how the stuff in the exercise is defined:

  • $\mathfrak{g}$ is a semisimple Lie algebra over a field $k$ of characteristic 0.
  • $M$ is an $m$-dimensional $\mathfrak{g}$-module (not assumed irreducible).
  • $r=\dim (\operatorname{im}(\rho))$, where $\rho\colon \mathfrak{g}\to \operatorname{End}_k(M)$ is the structure map.
  • $c_M$ is the Casimir operator for $M$:

... $\mathfrak{g}\cong \mathfrak{h}\times \ker(\rho)$, $\mathfrak{h}\subseteq\mathfrak{gl}_m$, and the bilinear form $\beta$ on $\mathfrak{h}$ is nondegenerate by Cartan's criterion 7.8.4. Choose a basis $\{e_1,\ldots,e_r\}$ of $\mathfrak{h}$; by linear algebra there is a dual basis $\{e^1,\ldots,e^r\}$ of $\mathfrak{h}$ such that $\beta(e_i,e^j)=\delta_{ij}$. The element $c_M=\sum e_ie^i\in U\mathfrak{g}$ is called the Casimir operator for $M$;...

My Question

I seem to get a different answer when I compute this for $\mathfrak{g}=\mathfrak{sl}_2$.

Let $M$ be the irreducible representation of $\mathfrak{sl}_2$ with highest weight $2$. Then $\rho$ is injective, and therefore (using the usual basis $h,x,y$ of $\mathfrak{sl}_2$) we get $$c_M=\frac{h^2}{2}+h+2yx.$$ Computing the action of $c_M$ on a highest weight vector $v$ of $M$, we get $$c_Mv = \left(\frac{h^2}{2}+h+2yx\right)v = 4.$$ However, $r=\dim(\mathfrak{sl}_2)=3$ and $m=\dim(M)=3$, so the exercise from Weibel would predict that $$c_Mv = v.$$

What am I missing?

Avi Steiner
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1 Answers1

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I think you miscalculated the dual basis of (x,y,h). I find $(\frac{y}{4},\frac{x}{4},\frac{h}{8})$ for the dual basis, and so $c_M=\frac{yx}{4}+\frac{h}{4}+\frac{h^2}{8}$

  • The Casimir i gave is given explicitly in an example in Weibel. – Avi Steiner Oct 15 '20 at 03:46
  • But even if I miscalculated the Casimir, a similar argument as with the heights weight two guy shows that for highest weight n, one gets $cv = (n^2/2 + n)v$. This $n^2/2+n$ grows with the dimension $m$ of the module, whereas $r/m$ shrinks. – Avi Steiner Oct 15 '20 at 03:50
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    Actually the Casimir in your question is in Exercice 7.8.3, right ? Weibel takes here the representation $\mathfrak{sl}_2 \to \mathrm{End}(k^2)=\mathfrak{gl}_2$, not the weight 2 irreducible representation. –  Oct 15 '20 at 04:04
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    Also I realize the form $\beta$ in the book is not the Killing form of $\mathfrak{sl}_2$... The dual basis I gave in my answer was for the adjoint representation –  Oct 15 '20 at 04:14