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Let $H$ be a simple algebraic group of type $\mathbf{G}_2$ over $\mathbb{C}$. Let $\rho$ be the standard 7-dimensional complex representation $$ \rho\colon H=\mathbf{G}_2\to \mathrm{SO}_7.$$ We consider the corresponding homomorphism $$H\to \mathrm{Spin}_7=G.$$ We regard $H$ as a subgroup of $G$. Let $T_H\subset H$ and $T_G\subset G$ be compatible maximal tori. We obtain a homomorphism of cocharacter groups $$\rho_*\colon X_*(T_H)\to X_*(T_G).$$ We choose compatible Borel subgroups $B_H$ and $B_G$ such that $T_H\subset B_H\subset H$ and $T_G\subset B_G\subset G$, then we obtain bases consisting of simple coroots in $X_*(T_H)$ and $X_*(T_G)$. Since we have bases, we can associate with $\rho_*$ an integral $3\times 2$ matrix.

Question. How can one compute this matrix?

Note that I need this matrix only modulo 2.

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    $\begingroup$ Probably you want the matrix, and not a "here's how I would try to do it" comment. Nonetheless: here is how I would try to do it. For each of the minimal parabolic groups $P_H$ properly containing $B_H$, and for each of the maximal parabolic subgroups $P_G$ properly contained in $G$, $P_H/B_H$ is a $\mathbb{P}^1$ that maps to $G/P_G$. It is possible to compute the degree on this $\mathbb{P}^1$ of the ample generator of the Picard group of $G/P_G$ by localization with respect to the $2$ fixed points for the action of the maximal torus. Of course I am not saying that is easy . . . $\endgroup$ Aug 3, 2015 at 17:31
  • $\begingroup$ Thank you, Jason. I indeed want "here is how I would try to it", also for some orthogonal representation of another group $H$. However, I have in mind another kind of approach: to use somehow tables from Bourbaki or from Onishchik and Vinberg, and I don't know how to use those tables for my question... $\endgroup$ Aug 3, 2015 at 17:49

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Like Jason I'd be more inclined to look at this kind of embedding in a geometric or conceptual way. But for your purposes a concrete description seems needed. For this it might be simpler to work in the Lie algebra setting. I don't see immediately what the group viewpoint does for you, since the centerless group $G_2$ embeds in either the special orthogonal group or its slightly elusive simply connected covering group Spin. The basic roots-and-weights technology here depends just on the Lie algebra embedding. For this to be made concrete, however, you'd have to line up the two Cartan subalgebras (Lie algebra of maximal tori) in a compatible way and relate the two root systems via simple roots.

A description of the embedding here is given (in an adapted version going back to one of the Paris seminars) in the first part of section 19.3 of my 1972 Lie algebra textbook. (Needless to say, I haven't spent a lot of time with this material since then.) The main deficiency in the details written down is that a choice of simple roots for $G_2$ isn't made explicit (though it is in Bourbaki). So you'd have to work that out further. Since the weight lattice and root lattice of $G_2$ coincide, passage to the duals can be done over $\mathbb{Z}$.

ADDED: With apologies for the delay in filling in some details, I still have some doubts about the basic setting here. On the level of root lattices it seems fairly clear geometrically. For the embedding $\mathfrak{g}_2 \hookrightarrow \mathfrak{so}_7$, one can work over $\mathbb{R}$ (since the Lie algebras are split). Here the bigger root lattice lives in a 3-dimensional euclidean space with standard orthonormal basis $\varepsilon _1, \varepsilon_2, \varepsilon_3$.

For $B_3$ there are 9 positive roots. Two simple roots are long: $\alpha_1 = \varepsilon_1 - \varepsilon_2$ and $\alpha_2 = \varepsilon_2 - \varepsilon_3$. A third simple root is short: $\alpha_3=\varepsilon_3$. Then the root lattice for $G_2$ lies in the plane through 0 defined by the condition that coordinates sum to 0. Here you can take a short simple root $\alpha = \alpha_1$ and a long simple root $\beta = \varepsilon_2 + \varepsilon_3 -2\varepsilon_1$. Then you can compute your $2 \times 3$ or $3 \times 2$ matrix in terms of these bases, keeping in mind that for $G_2$ the root lattice equals the weight lattice and maps into the root lattice for $B_3$ (which however has index 2 in its weight lattice). My quick arithmetic wasn't reliable, but while the matrix you get depends on the choice of bases, the relationship between the two root systems depends only on the Cartan matrices.

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  • $\begingroup$ Thank you, Jim! However, I would appreciate if you add details of your calculation. Do you use the Cartan's matrices for $H$ and for $G$? I would like to repeat the calculation starting from an orthogonal representation of another group. $\endgroup$ Aug 4, 2015 at 5:03
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    $\begingroup$ If you are happy to work on the level of algebras, then you can find more specific examples of these matrices in the source code of the branching_rule function of sagemath system. E.g. here is the linear mapping that corresponds to $\mathfrak{so}_7 \to \mathfrak{g}_2$ branching: github.com/sagemath/sage/blob/master/src/sage/combinat/… If the developers tell you how they computed them, please share their wisdom here with us. $\endgroup$ Aug 5, 2015 at 8:03
  • $\begingroup$ Thank you, @VítTuček! Lie algebras are OK. Hopefully I will learn to compute these things in a week, after my coauthor returns from vacations. $\endgroup$ Aug 5, 2015 at 13:44

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