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I'm currently trying to learn about regular elements of a Lie algebra but i'm finding the definition quite abstract and can't seem to find many examples anywhere.

One thing i'm really unsure about is that I have read that in $\mathfrak{sl}(3)$ a regular element is $X= E_{1,2} + E_{2,3}$ where $E_{x,y}$ corresponds to the matrix with a 1 in position (x,y) and 0's elsewhere. The centraliser of this group is given by $\text{span} \{E_{1,3} \}$. A theorem then claims that this set, which is equal to $\mathfrak{g}^0(X)$ is a Cartan subalgebra. But I on't see how this is true?

I was wondering if someone could give me an example of how to determine the regular elements of a given Lie algebra, say $\mathfrak{sl}(3)$.

glS
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Aran
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1 Answers1

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There is apparently more than one definition of "regular element".

The definition that relates to Cartan subalgebras is: $x \in \mathfrak{g}$ is regular if $$\mathfrak{g}_0(x) = \{y \in \mathfrak{g}: \; \mathrm{ad}(x)^n(y) = 0 \; \mathrm{for} \; \mathrm{some} \; n\}$$ has minimal dimension among all elements of $\mathfrak{g}.$ In this case, $\mathfrak{g}_0(x)$ is a Cartan subalgebra. This is not the centralizer.

Your matrix $$X = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$$ does not satisfy this criterion. Even though its centralizer is $2$-dimensional (which looks like the right number), the kernel of $\mathrm{ad}(X)^2$ is larger ($5$-dimensional).

In fact the regular elements of $\mathfrak{sl}(n,\mathbb{C})$ are exactly the matrices with no repeated eigenvalues.

user347160
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  • Okay I think this makes sense. My interest in the regular elements is stemming from me wanting to understand the principal $\mathfrak{sl}(2)$ subalgebra a lie algebra, which is defined such that the element of this subalgebra are regular. I'd be correct in assuming this uses a different definition of regular element? – Aran Jun 12 '16 at 06:36
  • @Aran Yes, the principal subalgebra's $e$ and $f$ are "regular nilpotent" elements, which is not regular in this sense of their generalized $\mathrm{ad}$-eigenspace being a CSA. – user347160 Jun 12 '16 at 07:31
  • Wait, what is this definition? Basing on the text 'Lie Algebras and Lie groups III' they seem to define them the same? That is, the same as your definition above. – Aran Jun 12 '16 at 10:40
  • Okay, so from what I understand, the regular they should be talking about just means the above definition but taking n to only equal 1. Is that correct? That would mean the textbook is wrong, but would make more sense. Also, I found online that for the X you have defined above, its centralizer is a cartan subalgebra given X is regular. Would I be correct in assuming they took the wrong definition for regular and that that's not true? – Aran Jun 12 '16 at 12:42
  • @Aran Some sources like the Wikipedia article define regular elements only looking at the centralizer. This is not the right definition if you want a Cartan subalgebra. – user347160 Jun 12 '16 at 16:16
  • But in both cases, the matrix $X$ defined above is not regular? You have shown that its not regular in the above sense, and if we consider over the complex numbers, $X$ is not diagonalizable and going by the wiki page, isn't regular? – Aran Jun 13 '16 at 01:55
  • @Aran Well it's regular in the sense that its centralizer is as small as possible. The wiki page only asks for a single Jordan block per eigenvalue. Just realize that this isn't the same definition as your theorem about CSAs asks for – user347160 Jun 13 '16 at 02:02
  • Where can I read this result you mention about regular elements of $\mathfrak {sl} (n, \mathbb{C})$ ? – JKEG Oct 02 '20 at 03:16