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Let $\mathfrak{g}$ be a Lie algebra and $\mathfrak{h}\subseteq \mathfrak{g}$ be a nilpotent subalgebra such that for every $H \in \mathfrak{h}$, the adjoint map $ad(H): \mathfrak{g} \rightarrow \mathfrak{g}$ is diagonalizable. Does it follow that the set of endomorphisms $ad(\mathfrak{h})$ is simultaneously diagonalizable?

Marc van Leeuwen
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Tim kinsella
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  • @MarcvanLeeuwen It's interesting: the book I'm reading is the only place I've found the word "diagonable" and I sort of like it because its less cumbersome. But it may be better to use the more standard "diagonalizable" here. – Tim kinsella Jun 12 '14 at 08:18
  • This would obviously be false if there were $H_1,H_2\in\mathfrak h$ such that $\def\ad{\operatorname{ad}}\ad([H_1,H_2])\neq 0$, since $\ad(H_1)$ and $\ad(H_2)$ would not commute. However, I think that elements of a nilpotent subalgebra are necessarily ad-nilpotent, in which case your hypothesis implies that $\ad(H)=0$ for all $H$; then the result is trivially true. – Marc van Leeuwen Jun 12 '14 at 08:21
  • @MarcvanLeeuwen I'm not sure that they are ad-nilpotent. Consider a 1 dimensional subalgebra generated by something which isn't ad-nilpotent. Or am I confused? – Tim kinsella Jun 12 '14 at 08:24
  • Yes, now I see that Googling "diagonable" gets a few hits (and the question "did you mean diagonal?"). But I had never seen it before, so it must be a minority dialect. – Marc van Leeuwen Jun 12 '14 at 08:24
  • Your example is convincing; I guess it is I who am confused. – Marc van Leeuwen Jun 12 '14 at 08:26
  • @MarcvanLeeuwen You are probably thinking of the result that nilpotent and "all elements ad-nilpotent" are equivalent, but this fails for subalgebras when we consider the ad as being on the larger algebra. – Tobias Kildetoft Jun 12 '14 at 08:37
  • @MarcvanLeeuwen It seems like your first comment must be the right track for a counter example. Thank you. – Tim kinsella Jun 12 '14 at 08:52

1 Answers1

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Yes, and you don't even need to assume that $\mathfrak h$ is nilpotent.

Lemma: Let $\mathfrak h$ be a subalgebra of a Lie algebra $\mathfrak g$ (finite dimensional over a field $K$) such that for each $h \in \mathfrak h$, $ad(h) : \mathfrak g \rightarrow \mathfrak g$ is diagonalisable. Then $\mathfrak h$ is abelian. (Consequently, all its elements are simultaneously diagonalisable.)

Proof (Here I gave section 8.1 of Humphreys' Introduction to Lie Algebras and Representation Theory as reference for this. I do not have that reference handy right now, but I think the following works):

Let $x \in \mathfrak h$. Since $\mathfrak h$ is invariant under $ad(x)$, the restriction $ad_{\mathfrak h}(x)$ is also diagonalisable, i.e. $\mathfrak h$ has a basis $y_1, ..., y_n$ such that for each $1 \le i \le n$, there exists $a_i \in K$ with $[x,y_i] = a_i \,y_i$. Then for each $i$, the subspace $\mathfrak h_i :=\mathrm{span}(x,y_i)$ is also invariant under $ad(y_i)$, which by assumption is diagonalisable, hence again its restriction to $\mathfrak h_i$ is diagonalisable as well. But if any $a_i \neq 0$, then $x,y_i$ would be a basis of $\mathfrak h_i$, and the matrix of $ad_{\mathfrak h_i}(y_i)$ in that basis would be $\pmatrix {0 & 0\\-a_i&0}$, contradiction. So all $a_i =0$.

Note: If we already know that $\mathfrak h$ is nilpotent, we can shorten the above proof by saying that -- not $ad(x)$, but -- $ad_{\color{red}{\mathfrak h}}(x)$ is diagonalisable (as above, because it's the restriction of a diagonalisable map to an invariant subspace) and at the same time nilpotent (by Engel's theorem on the Lie algebra $\mathfrak h$, as alluded to in comments), which forces it to be zero.

Note also that the lemma becomes false if we try to generalize from $ad$-diagonalisable to $ad$-semisimple elements. I blundered about this in my very first MSE question, an easy counterexample is $\mathfrak g = \mathfrak h = \mathfrak{su}_2$ (or any compact simple real Lie algebra).

Torsten Schoeneberg
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  • Thanks for the proof! I have a question, did you intend to write ${\rm{ad}}{{\frak{h}}_i}$ on the last line of the proof? I guess it's not ${\rm{ad}}{{\frak{h}}_i}(y_i)$. – ChoMedit Nov 04 '21 at 03:21
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    I really mean the map $ad_{\mathfrak h_i}(y_i): \mathfrak h_i \mapsto \mathfrak h_i$ which sends $h\in \mathfrak h_i$ to $[y_i,h]$. (I.e. I would call $(ad_{\mathfrak h_i}(y_i))(h):=[y_i,h]$.) I know the notation is cumbersome, and sometimes people would put the "first argument" $y_i$ in the subscript, but I needed to put the subalgebra to which we restrict somewhere ... – Torsten Schoeneberg Nov 04 '21 at 14:45