0

For $\mathfrak{g}$ a semisimple Lie algebra, $\mathfrak{h}$ a choice of Cartan subalgebra, $\Phi$ the set of roots, there is the integrality property that if $\alpha$ is a root then $c \alpha \in \Phi$ if and only if $c = \pm 1$. I am trying to understand the proof of the 'only if' direction as it appears in my textbook Goodman-Wallach (although Humphreys has a similar proof) (I have also seen the proof here, but I want to do it the way it is presented in my book).

To begin with, $s(\alpha) = \text{span}\{e_\alpha, f_\alpha, h_\alpha\} \cong \mathfrak{sl}_2$ is an isomorphic copy inside $\mathfrak{g}$. Construct $$ M_\alpha = \mathbb{C}h_\alpha + \sum_{c \neq 0} \mathfrak{g}_{c\alpha}.$$ Then the eigenvalues of $h_\alpha$ on $M_\alpha$ (under the adjoint representation) are 0 and $2c$. Since $M_\alpha$ is completely reducible and the weights of $\mathfrak{sl}_2$-irreps are known, we conclude that every $2c$ must be an integer, so $c$ is an integral multiple of 1/2. By the classification of $\mathfrak{sl}_2$-irreps (eigenvalues are integers which differ by 2), this then tells us that the set $\{ 2c : \mathfrak{g}_{c\alpha} \in M_\alpha\} $ either consists of all even or all odd integers. In particular, there are two cases: either every $c$ is an integer, or else every $c$ is of the form $p+1/2$ for some positive integer $p$ (they are very similar, so I am posting them as one question).

In the first case, my textbook says that since $s(\alpha)$ contains the zero eigenspace in $M_\alpha$ (I agree with this), then it follows that $c\alpha$ is not a root for any integer $c$ for which $\vert c \vert > 1$. I don't see how this conclusion follows at all. I understand that $s(\alpha)$ is one such irreducible component of $M_\alpha$, and that its eigenvalues are $0, \pm 2$. I also recognise that this exhausts the occurrences of the zero weight, since 0 has multiplicity one in $M_\alpha$. However I don't understand how either of these statements imply the conclusion.

In the second case, if $(p+1/2)\alpha \in \Phi$, then the eigenvalues of $h_\alpha$ are $2p+1, 2p-1, \dots, 3, 1$ on $M_\alpha$, and then somehow this implies $(1/2)\alpha$ is also a root. I also don't see how this follows, although maybe if my first question is answered then this will also follow.

mi.f.zh
  • 1,309

1 Answers1

1

Your agrument goes wrong where you say "... by the classification of $\mathfrak{sl}_2$-irreps ...". What you can conclude there is that $M_\alpha$ is a direct sum of finitely many non-trivial irreducible representations. Now consider irreducible components of odd dimension. On these, $h_\alpha$ has only even eigenvalues and always has a one-dimensional zero-eigenspace. Since the zero-eigenspace in $M_\alpha$ is one-dimensional and $s(\alpha)\subset M_\alpha$ is one irreducible component, you conclude that there may not be any other odd dimensional irreducible components. Hence in all other irreducible components $h_\alpha$ has only odd eigenvalues. In particular, the eigenspace for the eigenvalue $4$ in $M_\alpha$ has to be trivial, so $\mathfrak g_{2\alpha}=\{0\}$ and hence $2\alpha$ cannot be a root.

But now if there were any even dimensional irreducbible components, each of them would have a non-trivial eigenspace for the eigenvalue $1$. But this would imply that $\mathfrak g_{\alpha/2}$ would be non-zero and hence $\alpha/2$ would be a root. But this is a contradiction, since from above, this would imply that $2(\alpha/2)=\alpha$ is not a root.

Andreas Cap
  • 22,380