Let $\varphi \colon G \to \tilde{G}$ be a group homomorphism between two real semisimple Lie groups. For example, $\varphi$ could be an inclusion of a subgroup $G \subseteq \tilde{G}$. Let $\mathfrak{g}, \tilde{\mathfrak{g}}$ denote the Lie algebras with Cartan decompositions $\mathfrak{g} = \mathfrak{k} \oplus \mathfrak{p} , \tilde{\mathfrak{g}} = \tilde{\mathfrak{k}} \oplus \tilde{\mathfrak{p}}$. Let $\mathfrak{a}, \tilde{\mathfrak{a}}$ be maximal abelian subspaces of $\mathfrak{p}, \tilde{\mathfrak{p}}$. Then we have root space decompositions \begin{align} \mathfrak{g} &= \mathfrak{g}_0 \bigoplus_{\alpha \in \Sigma} \mathfrak{g}_\alpha \\ \tilde{\mathfrak{g}} &= \tilde{\mathfrak{g}}_0 \bigoplus_{\alpha \in \tilde{\Sigma}} \tilde{\mathfrak{g}}_\alpha \\ \end{align} Let us assume that $G$ and $\tilde{G}$ are compatible, in the sense that $D_0\varphi(\mathfrak{p}) \subseteq \tilde{\mathfrak{p}}$ and $D_0\varphi(\mathfrak{a}) \subseteq \tilde{\mathfrak{a}}$.

What can we say about the root systems $\Sigma, \tilde{\Sigma}$?

Concretely, this amounts to:

  1. Do we have $D_0\varphi \mathfrak{g}_0 = \tilde{\mathfrak{g}}_0 \cap D_0\varphi (\mathfrak{g})$?
  2. Is there a relation between $\Sigma$ and $\tilde{\Sigma}$?
  3. Do we get a group homomorphism between the Weyl groups $W \to \tilde{W}$?
  4. Looking at the hyperplanes $M_\alpha = \{H \in \mathfrak{a} \colon \alpha(H) = 0\}$, $\tilde{M}_\alpha = \{H \in \tilde{\mathfrak{a}} \colon \alpha(H) = 0\}$ is it true that $D_0\varphi^{-1}\tilde{M}_\alpha \cap \mathfrak{a}$ is always a hyperplane $M_\beta$ for some $\beta \in \Sigma$ or all of $\mathfrak{a}$?
  5. Do any of the above questions hold when $\varphi$ is just an inclusion? Or a surjection?

Experiments on representations $\operatorname{SL}(2,\mathbb{R}) \subseteq \operatorname{SL}(n,\mathbb{R})$ and $\operatorname{Sp}(2\cdot 2, \mathbb{R}) \subseteq \operatorname{SL}(4,\mathbb{R})$ seem to suggest that questions 1, 3 and 4 have a positive answer.

  • $\begingroup$ Please ask one question at a time. $\endgroup$
    – Shaun
    Aug 23, 2023 at 10:52
  • $\begingroup$ Please provide context with your questions. $\endgroup$
    – Shaun
    Aug 23, 2023 at 10:52
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    $\begingroup$ I see. I'm sorry. I have now upvoted, and retract my close vote. I have also edited the question in the hope that others will not make the same mistake. $\endgroup$
    – Shaun
    Aug 23, 2023 at 12:28
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    $\begingroup$ Consider the inclusion of split forms $SO(p,p+1) \subset SL(2p+1)$: For general $p$, it is not so easy to relate the root systems $B_p$ and $A_{2p}$. But cf. math.stackexchange.com/a/3972084/96384 and math.stackexchange.com/a/3300945/96384, and links from there. $\endgroup$ Aug 23, 2023 at 15:07
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    $\begingroup$ @TorstenSchoeneberg I have posted an answer which I think shows this (and thus answers this question in the inclusion case) if both groups are split. $\endgroup$
    – Callum
    Aug 24, 2023 at 16:23

1 Answer 1


Here's the inclusion case when both groups are split (following up on my comment):

Firstly, let's observe a few things. Let $\mathfrak{g} \leq \tilde{\mathfrak{g}}$, $\mathfrak{p} \leq \tilde{\mathfrak{p}}$, $\mathfrak{a} \leq \tilde{\mathfrak{a}}$. Since we are focusing on the split case, we can simply assume $\mathfrak{g}_0 = \mathfrak{a}$, $\tilde{\mathfrak{g}}_0 = \tilde{\mathfrak{a}}$ are Cartan subalgebras. We can quickly see that in this case $\mathfrak{g}_0 = \mathfrak{g} \cap \tilde{\mathfrak{g}}_0$. I'm not sure that holds in more generality though. For a start, consider the degenerate case that $\mathfrak{g}$ is compact when $\mathfrak{p},\mathfrak{a} = 0$ and $\mathfrak{g}_0 = \mathfrak{g}$.

Then we can show:

$$ \bigoplus_{\alpha \in \Sigma} \mathfrak{g}^\alpha = \mathfrak{g} \cap \bigoplus_{\beta\in \tilde{\Sigma}} \tilde{\mathfrak{g}}^\beta. $$

To do this, write $X \in \mathfrak{g}^\alpha$ as $X= \tilde{H} + \sum_{\beta \in \tilde{\Sigma}} X_\beta$ for $\tilde{H} \in \tilde{\mathfrak{g}}_0$. Then take any $H \in \mathfrak{g}_0$ so $[H,X] = \sum_{\beta \in \tilde{\Sigma}} \beta(H)X_\beta$ and $X - \frac{1}{\beta'(H)}[H,X] = \tilde{H} + \sum_{\beta \in \tilde{\Sigma}} \left(1 - \frac{\beta(H)}{\beta'(H)}\right)X_\beta$ for some $\beta' \in \tilde{\Sigma}$ with $\beta(H') \neq 0$. This new thing is still in $\mathfrak{g}$ but we have reduced the terms in the sum and we can keep going recursively until we reach $\tilde{H}$ so that $\tilde{H} \in \mathfrak{g}_0$ but $X \in \mathfrak{g}^\alpha$ then implies $\tilde{H} = 0$. Thus $\bigoplus_{\alpha \in \Sigma} \mathfrak{g}^\alpha \subset \mathfrak{g} \cap \bigoplus_{\beta\in \tilde{\Sigma}} \tilde{\mathfrak{g}}^\beta$ and we have equality by considering dimensions.

In light of that, $$\sum_{\beta \in \tilde{\Sigma}} \alpha(H) X_\beta = \alpha(H)X = [H,X] = \sum_{\beta \in \tilde{\Sigma}} \beta(H) X_\beta. $$

Therefore for each $X_\beta \neq 0$, $\beta(H) = \alpha(H)$. In particular, any such $\beta$ differ by an element of $\operatorname{ann}(\mathfrak{g}_0) = \{f \in \tilde{\mathfrak{g}}_0^*|f|_{\mathfrak{g}_0} = 0\}$ and at least 1 such $\beta$ must exist.

Now to what this means for the root systems. We have a natural identification of $\mathfrak{g}_0^* \cong \tilde{\mathfrak{g}}_0^*/\operatorname{ann}(\mathfrak{g}_0)$ and thus a quotient map $\pi:\tilde{\mathfrak{g}}_0^* \to \mathfrak{g}_0^*$. Then what we have shown is that $\Sigma \subset \pi(\tilde{\Sigma})$ i.e. that $\Sigma$ is a subset of a quotient of $\tilde{\Sigma}$. Note that this includes all sorts of possibilities. If $\mathfrak{g}_0 = \tilde{\mathfrak{g}}_0$ we have the maximal rank subalgebras which are given by root subsystems (see here) and more general subsystems. We can also find foldings of Dynkin diagrams and even the $B_3 \to G_2$ pseudo-folding (as discussed here).

The only thing which I don't quite know is how to classify what kinds of quotients are acceptable. Note that the quotient doesn't have to be reduced here (even though both our groups are split) as in the case Torsten mentions of $\operatorname{SO}(p,p+1) \subset \operatorname{SL}(2p+1)$ where the quotient is of type $BC_p$ and we take the $B_p$ inside it.

Note, moreover, that we can conclude that for the root hyperplanes we have $M_\alpha = \tilde{M}_\beta \cap \mathfrak{g}_0$, for each $\beta$ with $\pi(\beta) = \alpha$.

If there is a Weyl group relationship I would expect it to be the other way round. Something like $\psi:\tilde{W} \to W$ with $\psi(s_\beta) := s_{\pi(\beta)}$ (with $\psi(s_\beta)= \mathrm{id}$ if $\pi(\beta) = 0$, etc.). But I cannot see if this defines a valid homomorphism.

Edit: I'm not sure how well this argument would extend to the non-split case as it does depend on the property $\mathfrak{g}_0 = \mathfrak{g} \cap \tilde{\mathfrak{g}}_0$ which I am not convinced of in general.

As an example suggested by Torsten, I calculated what some of the quotient relations for $E_6$ from $A_{26}$ would be. Note, I obtained these by first, finding an explicit representation of a Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{e}_6$ as a subalgebra of the usual diagonal Cartan subalgebra $\tilde{\mathfrak{h}}$of $\mathfrak{sl}_{27}$. Then I compared roots of the larger algebra restricted to $\mathfrak{h}$ (actually I computed their values on several randomly chosen elements) and noted which ones were the same. Note that roots in $A_{26}$ are always of the form $\alpha_i + \alpha_{i+1} + \cdots + \alpha_{i+j}$ where $\alpha_i$ are simple roots.

Then denoting by $\alpha \equiv \beta$ the roots agreeing on $\mathfrak{h}$, I obtained the following relations

  • $ \alpha_{1} \equiv \alpha_{4} \equiv \alpha_{7}$
  • $ \alpha_{2} \equiv \alpha_{5} \equiv \alpha_{8}$
  • $ \alpha_{10} \equiv \alpha_{13} \equiv \alpha_{16}$
  • $ \alpha_{11} \equiv \alpha_{14} \equiv \alpha_{17}$
  • $ \alpha_{19} \equiv \alpha_{22} \equiv \alpha_{25}$
  • $ \alpha_{20} \equiv \alpha_{23} \equiv \alpha_{26}$
  • $ \alpha_{8} + \alpha_{9}+ \alpha_{10}+ \alpha_{11} \equiv \alpha_{17} + \alpha_{18}+ \alpha_{19}+ \alpha_{20}$
  • $ \alpha_{6} + \cdots + \alpha_{10} \equiv \alpha_{15} + \cdots + \alpha_{19}$
  • $ \alpha_{3} + \cdots + \alpha_{9} \equiv \alpha_{15} + \cdots + \alpha_{21}$
  • $ \alpha_{2} + \cdots + \alpha_{9} \equiv \alpha_{17} + \cdots + \alpha_{24}$
  • $ \alpha_{7} + \cdots + \alpha_{14} \equiv \alpha_{16} + \cdots + \alpha_{23}$
  • $ \alpha_{4} + \cdots + \alpha_{12} \equiv \alpha_{17} + \cdots + \alpha_{25}$
  • $ \alpha_{1} + \cdots + \alpha_{15} \equiv \alpha_{3} + \cdots + \alpha_{9}$
  • $ \alpha_{1} + \cdots + \alpha_{13} \equiv \alpha_{2} + \cdots + \alpha_{10}$

This seems like enough relations but I don't know if there is some redundancy. I certainly didn't do an exhaustive search. Note there is a very distinct pattern in the first ones but the later ones are less clear to me. Also I've written these all as positive roots and mostly as relations between sums of the same length (because of the way I found them) but possibly there are better ways to write these.

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    $\begingroup$ Thank you so much. I am digesting the answer. One thing I noted is that for $\tilde{\mathfrak{g}}_0 \cap \mathfrak{g} = \mathfrak{g}_0$, it suffices that $\mathfrak{g}$ is split. The inclusion $\tilde{\mathfrak{g}}_0 \cap \mathfrak{g} \subseteq \mathfrak{g}_0$ always works (without assumptions on split..). $\endgroup$ Aug 25, 2023 at 7:55
  • $\begingroup$ Great. Just to check we're on the same page: considering that the split form of $E_6$ has a $27$-dimensional representation, this means $E_6$ can be realized as a subset of some quotient of $A_{26}$, right? I would like to see that ... $\endgroup$ Aug 25, 2023 at 17:16
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    $\begingroup$ @TorstenSchoeneberg I've added an edit with the quotient relations that I can see in that case. $\endgroup$
    – Callum
    Aug 25, 2023 at 23:45
  • $\begingroup$ You seem to argue about the case when $\mathfrak{g}$ is compact and split. But that means $\mathfrak{g}=0$, no? $\endgroup$ Aug 29, 2023 at 14:43
  • $\begingroup$ @Strichcoder Sorry if my wording is confusing. I mean that extending beyond the split case introduces some concerns. In particular, you can see it breaks down completely in the compact case. It would be natural to exclude the compact case when talking about restricted root systems anyway but I think examples on that side of the spectrum are likely to break as well. $\endgroup$
    – Callum
    Aug 29, 2023 at 15:16

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