# Cartan Weyl Basis : Only elements in the CSA have eigenvalue $0$

Suppose we have a cartan subalgebra (CSA) $\mathfrak{h} \subset \mathfrak{g}$. Assign a basis to this $\{H_{i}\}_{i=1...r}$ such that $ad_{H_{i}}$ are all simultaneously diagonalized [ $\implies ad_{H}$ diagonal $\forall H \in \mathfrak{h}$] (which we know we can do from linear algebra). Want to find set of eigenvectors

• Why is it true that the only eigenvectors with eigenvalue $0$ are the basis elements $H_{i}$ of $\mathfrak{h}$?

• Why is it true that the only eigenvectors with eigenvalue $0$ are the elements of $\mathfrak{h}$ only?

Attempt at Solution:

• If $Y \in \mathfrak{g}$ is an eigenvector with eigenvalue $\lambda =0$ then: $$ad_{H_{i}}(Y) = [H_{i},Y] = 0 \; \forall i \implies ad_{Y}(H) =0$$

• If we can also show that $ad_{Y}$ is diagonalizable, then it follows that $Y \in \mathfrak{h}$, but why is this true? Note that we can't assume the Cartan-Weyl basis, since we need this property for the basis to be valid?

EDIT: Definition of CSA:

For $\mathfrak{h} \subset \mathfrak{g}$ to be a CSA, the following must be true: $$H \in \mathfrak{h} \implies ad_{H} \; \text{is diagonalizable}$$ $$H,H' \in \mathfrak{h} \implies [H,H'] = 0$$ $$\text{if} \; X \in \mathfrak{g} \; \text{is ad diagonalizable and } [X,H] =0 \; \forall H \in \mathfrak{h} \implies X \in \mathfrak{h}$$

• I think your definition of CSA will only work (in the sense that the statement is correct) if you assume the Lie algebra to be semisimple (or at least reductive). Otherwise, consider the Lie algebra of strictly upper triangular matrices and note that $\{0\}$ forms a CSA, since no non-zero element is diagonalizable (since it acts nilpotently but is not $0$). But everything is an eigenvector with eigenvalue $0$ for this CSA. – Tobias Kildetoft Apr 16 '18 at 9:38
• Sorry, the above was a bit silly. The CSA will be the center, which has dimension $1$. But the same argument still works. – Tobias Kildetoft Apr 16 '18 at 9:42

Your statement is not true, because every element of $\mathfrak{h}$, i.e. not only the basis elements $H_i$ but every linear combination of those, is an eigenvector of eigenvalue $0$ (to all the $ad_{H_i}$ simultaneously).
Now, to show that these are indeed the only elements of $\mathfrak{g}$ which are eigenvectors to the eigenvalue $0$ to all $ad_{H_i}$, just notice that from your first attempt it follows that for any such eigenvector $Y$, we have $[\mathfrak{h}, Y] = 0$ or in other words, $Y$ is contained in the centraliser and hence in the normaliser of $\mathfrak{h}$. But by definition of CSA, a CSA $\mathfrak{h}$ equals its own normaliser. Hence $Y\in \mathfrak{h}$.
• Ooops, sorry, just corrected. What I'm confused about is why only we can't have an element of the LA which has eigenvalue 0 but isn't $ad$ diagonlizable - i.e has eigenvalue 0 but isn't part of the CSA? – thesundayscientist Apr 15 '18 at 22:19
• ? Doesn't my argument show that any such eigenvector $Y$ is in $\mathfrak{h}$? (And then from this it might follow that it will be ad-diagonalisable, but that is secondary and has nothing to do with the argument.) – Torsten Schoeneberg Apr 15 '18 at 22:22
• My understanding was that by definition for an element of the LA to be in the CSA you required it to both be $ad$ diagonalizable and in the centralizer. Else it's not possible to form the Cartan-Weyl basis - you can't simultaneously diagonalize them if they aren't individually diagonalizable? – thesundayscientist Apr 15 '18 at 22:28