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Let L be a Lie algebra, H a CSA of L, and $(V, \phi)$ be an irreducible representation of $L$. For any non-negative integer $k$, define the trace polynomial $f_k:L \rightarrow F$ by $f_k(z) = Tr(\phi(z)^k)$. Now, take $L = \mathfrak{sl}(2, F)$, $(V, \phi) = (L, ad)$ (the adjoint representation). Let {x, h, y} be the usual basis for $L$ and {x*, h*, y*} be its dual basis. I am trying to recover the result on p.132 in Humphreys' book, which gives us that $$f_2 = (h^\ast)^2 + x^\ast y^\ast$$, whose restriction to $H$ is $f_2|_H = (h^\ast)^2$. However, the matrix representation of ad(h) with respect to the given basis is

$$ \begin{pmatrix} 2 & 0 & 0 \newline 0 & 0 & 0 \newline 0 & 0 & -2 \end{pmatrix} $$

Hence, for every $t \in F, f_2(th) = Tr(ad(th)^2) = 8t^2$, so $f_2|_H = 8(h^*)^2$, but this doesn't coincide with the previous result. Do I make any mistake? Shouldn't I use the adjoint representation?

Trace polynomials of sl(2,F)

Edited: I have uploaded the page where Humphrey talks about this.

Jacob Lee
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  • @DietrichBurde, Hi. I don't understand the scaling part. Indeed, he identified $h^\ast$ with $\frac{h}{8}$ via Killing form, so are you saying this scaling has something to do with the calculation here? If possible, would you please explain more? Thank you – Jacob Lee Sep 18 '23 at 17:40
  • I don't have Humphreys book here, but "dual basis" in which sense? See here. – Dietrich Burde Sep 19 '23 at 08:12
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    I also don't have the book to hand but this is certainly a matter of scaling, whether in the definition of the dual basis, or in some other definition/convention he is using. If you run your calculations to completion, you get simply $f_2 = 8((h^)^2 + x^ y^*)$ – Callum Sep 19 '23 at 13:59
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    Although it's not directly related to the point here, it can't be said often enough that there is not "the" CSA of a Lie algebra, rather there are infinitely many CSAs in general. – Torsten Schoeneberg Sep 19 '23 at 14:45
  • @TorstenSchoeneberg Thank you for pointing out. – Jacob Lee Sep 19 '23 at 17:57
  • @DietrichBurde I think the dual basis is in the sense that h(h) = 1 and h(x)=h(y)=0. Same for x and y* – Jacob Lee Sep 19 '23 at 17:59
  • Reading the page you linked it feels likely to me that Humphreys is being (deliberately) vague here with the term "trace polynomial" as he doesn't specifically call it $f_2$ there. It is reasonable to call any rescaling of a trace polynomial also a trace polynomial as the important thing about these is that they are invariant and that is preserved by scaling. – Callum Sep 20 '23 at 11:40

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