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Let $G$ be a connected reductive group over an algebraically closed field and consider a semisimple element $s \in G$ and let $L$ be a Levi subgroup containing $s$.

My question is about the two ways we can look at $C_L(s)^\circ$. On the one hand, it is the connected centralizer of $s$ in the Levi subgroup $L$, on the other hand, it is a Levi subgroup in the connected centralizer $C_G(s)^\circ$.

The Dynkin diagram of $C_{G}(s)^\circ$ can be obtained from the extended Dynkin diagram of $G$ by removing nodes. By removing appropriate nodes from the Dynkin diagram of $C_{G}(s)^\circ$, we obtain the Dynkin diagram of the Levi subgroup $C_{L}(s)^\circ$ of $C_{G}(s)^\circ$.

On the other hand, the Dynkin diagram of $C_L(s)^\circ$ can be obtained from the extended Dynkin diagram of $L$ and the Dynkin diagram of $L$ can be obtained by removing nodes from the Dynkin diagram of $G$.

For example let $G$ have type $B_{10}$ and choose $s$ such that $C_{G}(s)^\circ$ has type $D_4 \times B_6$, then I can find a Levi subgroup of $C_{G}(s)^\circ$ of type $D_4 \times A_2 \times B_3$. This Levi is of the form $C_L(s)^\circ$ for some Levi $L$ of $G$.

However, even though there exists a Levi subgroup of $G$ of type $A_2 \times B_7$ and one of its subgroups of maximal rank has the desired type $A_2 \times D_4 \times B_3$, the bases of the corresponding root systems seem to be completely different to me. In fact, I am not even sure if that subgroup of type $A_2 \times D_4 \times B_3$ lies in $C_G(s)$ or not.

Also, it seems to me that the other Levi subgroups of $G$ do not even have a subgroup of the desired type.

Did I do anything wrong here? If not, why is it not possible for me to obtain the same basis of the Levi of $C_{G}(s)^\circ$ when considering it as a maximal rank subgroup of a Levi subgroup of $G$?

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    $\begingroup$ I'm having a little trouble with the question. Are you asking why it's not the case that a specific $D_4 \times A_2 \times B_3$ subgroup of $B_{10}$ embeds in both a $D_4 \times B_6$ and an $A_2 \times B_7$ subgroup? Is there any reason to expect that it would? $\endgroup$ – LSpice Mar 17 '17 at 17:14
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    $\begingroup$ @LSpice : To clarify my question, I have a Levi of $C_G(s)^\circ$ of type $D_4 \times A_2 \times B_3$, I would like to know how I have to choose $L$ so that this Levi is $C_L(s)^\circ$. You seem to have a suggestion for $L$, however, I do not see how removing nodes from the extended Dynkin diagram of type $A_6 \times B_3$ could give me type $A_2 \times D_4 \times B_3$. $\endgroup$ – Matthias Klupsch Mar 17 '17 at 19:11
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    $\begingroup$ My $L$ was total nonsense (wrong type, and wrong realisation of that type!). I have posted an answer that I think shows that your expectations are correct, but that you have to conjugate your choice of $L$ (or of $s$) a little to realise them. $\endgroup$ – LSpice Mar 17 '17 at 20:01
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    $\begingroup$ @JimHumphreys, $C_G(s)^\circ \cap L = C_G(s)^\circ \cap C_G(Z(L)^\circ)$ is the centraliser in $C_G(s)^\circ$ of the torus $Z(L)^\circ$, and the torus lies in $C_G(s)^\circ$ since $s$ lies in $L$; so the intersection is connected, hence equals $C_L(s)^\circ$, which is therefore a Levi subgroup of $C_G(s)^\circ$. $\endgroup$ – LSpice Mar 20 '17 at 1:59
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    $\begingroup$ @JimHumphreys: There is no need to associate a $\lambda$ to the semisimple element $s$. The torus to be considered is $Z(L)^\circ$. Its centralizer in $G$ is a Levi subgroup of $G$, namely $L$, and its centralizer in $C_G(s)^\circ$ is also a Levi subgroup of $C_G(s)^\circ$. By the above argument, it turns out to be $C_L(s)^\circ$. $\endgroup$ – Matthias Klupsch Mar 21 '17 at 6:25
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In the Bourbaki numbering ($\alpha_i = \epsilon_i - \epsilon_{i + 1}$ for $i < 10$ and $\alpha_{10} = \epsilon_{10}$), I believe that you can take $s = \alpha_1^\vee(-1)\alpha_3^\vee(-1)$ (with connected centraliser of type $D_4 \times B_6$, where the base for the $D_4$ piece is $\{\alpha_3, \alpha_2, \alpha_1, -\mu\}$ and that for the $B_6$ piece is $\{\alpha_5, \dotsc, \alpha_{10}\}$) and $L$ to be the centraliser of the image of $\alpha_5^\vee + 2\alpha_6^\vee + 3(\alpha_7^\vee + \dotsb + \alpha_{10}^\vee)$, which, as you predicted, is of type $A_2 \times B_7$. The base for the $A_2$ piece is $\{\alpha_5, \alpha_6\}$, and that for the $B_7$ piece is $\{\alpha_1, \alpha_2, \alpha_3, \alpha_4 + \dotsb + \alpha_7, \alpha_8, \alpha_9, \alpha_{10}\}$.

I found Carter's paper Conjugacy classes in the Weyl group (MR) very useful for understanding this kind of calculation with what he calls admissible diagrams (Section 4 in his paper). Section 4 in my paper On counting orbits in root systems gives some amateurish examples.

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  • $\begingroup$ Thank you for your answer and the references. It is a bit surprising for me that one has to consider conjugates. I mean, I only used the Dynkin diagram and not the root system to construct maximal rank subgroups and Levi subgroups, but when trying to make these constructions compatible, I have to use more knowledge about the root system. $\endgroup$ – Matthias Klupsch Mar 20 '17 at 7:46
  • $\begingroup$ @MatthiasKlupsch, part of the power of the Dynkin diagram is that it literally knows everything about the original root system; there is no difference between using knowledge of the root system, and 'just' using the Dynkin diagram (although there may seem to be to those of us who are not, say, Dynkin :-) ). Incidentally, it has always been one of my favourite trivia (though I don't know it firsthand) that Dynkin created these diagrams as part of an REU-type experience when he was asked to help organise the mess of then-ad hoc information about disparate root systems. $\endgroup$ – LSpice Mar 20 '17 at 16:07
  • $\begingroup$ A Dynkin diagram contains the same amount of information as the associated root system as long as we keep in mind the interpretation of nodes and edges. My hope was that it would be possible to construct all the objects of interest in the given situatuion by manipulating the Dynkin diagram as a graph without using the associated interpretation during the process. I see now that this does not work. $\endgroup$ – Matthias Klupsch Mar 20 '17 at 20:47
  • $\begingroup$ @MatthiasKlupsch, I don't mean to be argumentative, but I would argue that "manipulating the Dynkin diagram as a graph" is just what your post did: using just the graph, you were able to determine the types of $C_G(s)^\circ$, $L$, and $C_L(s)^\circ$. It was only when it came to writing down a particular choice of basis that you were forced to deal with information beyond the root diagram; but, as you say, as soon as you attempt to do that you are attaching extra information to the nodes and edges, and it is perhaps no surprise that you have to do something with that information. $\endgroup$ – LSpice Mar 20 '17 at 21:57
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    $\begingroup$ Yes, you are right, this is what I was trying to do and I was able to determine the type of $L$ given the one of $C_{L}(s)^circ$ in this example. But this was only because I could look at all possible types of Levi subgroups and see that there was only one of them where the type of $C_{L}(s)^circ$ could occur. In general, these manipulations only seem to find candidates for the types of Levi subgroups. I believe I just had my hopes too high. $\endgroup$ – Matthias Klupsch Mar 24 '17 at 8:34

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