Let $\mathfrak{g}_2$ / $\mathfrak{b}_3$ be the simple complex Lie algebra of type $\mathsf{G}_2$ / $\mathsf{B}_3$ (the latter is also known as $\mathfrak{so}_7$).

How is the embedding $\mathfrak{g}_2\subseteq\mathfrak{b}_3$ explicitly defined on Chevalley generators? Is there more than one such embedding? If yes, I am looking for the description of the one which is probably uniquely determined by the requirement that $\mathfrak{g}_2\cap\tilde{\mathfrak{p}}_1=\mathfrak{p}_1$, i.e. that the maximal parabolic subalgebras associated to the first simple root, respectively, are preserved.

More details and background. Let $\mathfrak{h}$ / $\tilde{\mathfrak{h}}$ be a Cartan subalgebra of $\mathfrak{g}_2$ / $\mathfrak{b}_3$ with associated root system $R$ / $\tilde{R}$. Let $h_\alpha,x_\alpha$ / $\tilde{h}_{\tilde{\alpha}},\tilde{x}_{\tilde{\alpha}}$ be the Chevalley generators of $\mathfrak{g}_2$ / $\mathfrak{b}_3$ where $\alpha\in R$ / $\tilde{\alpha}\in\tilde{R}$.

Can you described the emebedding $\mathfrak{g}_2\subseteq\mathfrak{b}_3$ in terms of a map $$ h_\alpha\mapsto\sum_{\tilde{\alpha}}\lambda_{\tilde{\alpha}}\tilde{h}_{\tilde{\alpha}}+\sum_{\tilde{\alpha}}\lambda_{\tilde{\alpha}}'\tilde{x}_{\tilde{\alpha}}\quad,\quad x_\alpha\mapsto\sum_{\tilde{\alpha}}\mu_{\tilde{\alpha}}\tilde{h}_{\tilde{\alpha}}+\sum_{\tilde{\alpha}}\mu_{\tilde{\alpha}}'\tilde{x}_{\tilde{\alpha}}\,? $$ If there's more than one possibility, I am again looking for the one which preserves the parabolic subalgebras as described above.

Remark. According to my considerations, it seems impossible to arrange the embedding such that $\mathfrak{g}_2\cap\tilde{\mathfrak{h}}=\mathfrak{h}$ holds. But I might be mistaken.

References. The answer by Kanakoglou to this question lists relevant references. Among them, I find only the paper by A. N. Minchenko online. But it assumes as "well-known" how the embedding is defined. I want to add to Kanakoglou's list the following reference

MR0476234 (57 #15807) Reviewed Group theory in non-linear problems. Lectures presented at the 2nd NATO Advanced Study Institute on Mathematical Physics, held in Istanbul, August 7–18, 1972. Edited by A. O. Barut. NATO Advanced Study Institutes Series C, Vol. 7. D. Reidel Publishing Co., Dordrecht-Boston, Mass., 1974. vii+281 pp. ISBN: 90-277-0412-0 00A10 (22-06 81.00)

And in this boook the paper: Bruno Gruber, The semisimple subalgebras of the algebra B3 (SO(7)) and their inclusion relations (pp. 231–247)

which seems to contain the answer but is unavailable to me. Indeed, consider the following image from the google preview of this volume. enter image description here If someone can explain the notation in this picture to me, or tell me where I find a free version of Gruber's paper, I will be very grateful. Thank you in advance!

EDIT. I want to add some interpretation of the picture. $(1,-1,0)$ is the short simple root, and $(-1,2,1)$ is the long simple root of $\mathsf{G}_2$. $(1,-1,0)$ and $\left(-\tfrac{1}{3},\tfrac{2}{3},\tfrac{1}{3}\right)$ appear to be the corresponding simple coroots. If we compute with the notation of the picture, we find \begin{align} \left[(0,1,-1),E_{(0,1,-1)}\right]&=2E_{(0,1,-1)}\,,\\ \left[f\left(-\tfrac{1}{3},\tfrac{2}{3},\tfrac{1}{3}\right),E_{(0,1,-1)}\right]&=-E_{(0,1,-1)}\,,\\ \left[(0,1,-1),f\left(\tilde{E}_{(-1,2,-1)}\right)\right]&=-f\left(\tilde{E}_{(-1,2,-1)}\right)\,,\\ \left[f\left(-\tfrac{1}{3},\tfrac{2}{3},\tfrac{1}{3}\right),f\left(\tilde{E}_{(-1,2,-1)}\right)\right]&=\tfrac{2}{3}f\left(\tilde{E}_{(-1,2,-1)}\right)\,. \end{align} The first two brackets are correct, the last two are correct up to a factor of $3$. What does this mean? How can we multiply with $3$ in the last two equations?


1 Answer 1


Definitely there is more than one embedding, for you can get any conjugate of ${\mathfrak g}_2$. The variety of all such embeddings is actually a seven-dimensional sphere ${\mathbb S}^7$.

One nice way to see ${\mathfrak g}_2$ inside ${\mathfrak b}_3$ is to use triality for ${\mathfrak d}_4$. Namely, you have three non-equivalent 8-dimensional irreducible representations of ${\mathfrak d}_4$. If you compute the stabilizer of a vector in general position, you get ${\mathfrak b}_3$. Now if you intersect two such ${\mathfrak b}_3$ corresponding to non-equivalent representation, you get ${\mathfrak g}_2$.

  • 1
    $\begingroup$ Explicit formulae you can see at M.M. Postnikov, "Lectures in Geometry: Lie Groups and Lie Algebras" (I don't know if there a free online copy in English). $\endgroup$ Sep 29, 2018 at 15:08
  • $\begingroup$ Thank you very much @Victor Petrov, I need explicit formulas as described in the question for the embedding preserving the parabolic subalgebras. I will check the book you recommend. It is much more likely to be available in the library as the literature discussed in my post. $\endgroup$ Sep 29, 2018 at 15:13
  • $\begingroup$ I think I can extract the needed formulas from Postnikov, Lecture 14, pp. 260, and Knapp, Lie groups beyond an introduction, Example 2, pp. 127. I don't see however how my formulas are related to those given by Gruber (cf. picture and edit below). $\endgroup$ Oct 3, 2018 at 15:13
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    $\begingroup$ I don't understand Gruber's notation, but the embedding is very easy, as per Victor's answer: if $\alpha_1,\alpha_2$ are the simple roots in ${\mathfrak g}_2$ (with $\alpha_1$ short) and $\beta_1,\beta_2,\beta_3$ are the roots in ${\mathfrak b}_3$ then $e_{\pm\alpha_2}\mapsto e_{\pm\beta_2}$, $e_{\alpha_1}\mapsto e_{\beta_1}+e_{\beta_3}$ and $e_{-\alpha_1}\mapsto e_{-\beta_1}+2e_{-\beta_3}$. (The multiple $2$ is to ensure that $[e_{\alpha_1}, e_{-\alpha_1}]=\alpha_1^\vee$.) The image of any other element of ${\mathfrak g}_2$ can now be computed by taking commutators. $\endgroup$
    – Paul Levy
    Oct 5, 2018 at 8:57
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    $\begingroup$ I haven't checked this explicitly, but there must be an element of the normalizer of $\widetilde{\mathfrak h}$ which sends my embedding onto yours. The benefit of mine is that it embeds the standard Borel subalgebra of ${\mathfrak g}_2$ into the standard Borel subalgebra of ${\mathfrak b}_3$. $\endgroup$
    – Paul Levy
    Oct 8, 2018 at 6:45

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