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