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All groups here are compact semisimple Lie groups. Out of laziness I will use $B_7$ to mean $Spin(15)$.

Suppose that one has a group $H$ and a subgroup $G$. The embedding determines the decomposition of $Res^H_G(V)$ into $G$-irreducibles for $V$ a representation of $H$; certainly different embeddings can lead to different decompositions of $Res^H_G(V)$.

What I am wondering is whether knowing the decomposition determines the embedding.

Here is a specific instance of why I am not sure whether this is true or not. Let $H = B_7$ and $G = G_2$ and consider the embedding of $G_2$ such that the 15-dimensional irreducible representation of $B_7$ restricts to the direct sum of the 14-dimensional adjoint representation with the trivial representation. To describe the restriction of an arbitrary character, it is enough to describe the image of the fundamental weights under the restriction since the restriction of an arbitrary weight is then a linear combination of these images.

Let $\omega_1,\ldots,\omega_7$ be fundamental weights of $B_7$ with $\omega_1$ the highest weight of the 15-dimensional representation and $\omega_7$ the highest weight of the 128-dimensional spinor representation. Define $V_i$ to be a fundamental $B_7$-module with highest weight $\omega_i$. Finally, let $\eta_1$ be the fundamental weight of $G_2$ which is the highest weight of the 7-dimensional representation of $G_2$ and $\eta_2$ the fundamental weight which is the highest weight of the 14-dimensional adjoint representation which I will call $W$.

EDIT: To help clarify the situation, here are the weight diagrams of these representations, truncated at the zero weight:


Since $V_k \cong \Lambda^kV_1$ for $k=1\ldots 6$, if $Res^{B_7}_{G_2}(V_1) = W\oplus 1$ then one can map the weight $\omega_k$ to the sum of the $k$ top-most weights in the weight diagram of $W$ for $k=1\ldots 6$ (actually it may map to the sum of any $k$ weights of $W$; we choose the topmost weights so that the restriction takes highest weights to highest weights). The four top-most weights of $W$ are totally ordered by their height, so there is no ambiguity in choosing the sums of the top-most weights.

$\omega_1\mapsto \eta_2$

$\omega_2\mapsto 3\eta_1$

$\omega_3\mapsto 4\eta_1$

$\omega_4\mapsto 3\eta_1+\eta_2$

However, when one reaches the 5th level of the weight diagram of $W$, the diagram branches (both $-3\eta_1+2\eta_2$ and $2\eta_1-\eta_2$ sit at this level) and so there are two possible choices for where to map $\omega_5$:

$\omega_5\mapsto 3\eta_2$


$\omega_5\mapsto 5\eta_1$

After this, there are no further ambiguities; $\omega_6$ maps to $2\eta_1+2\eta_2$ which is the sum of the 6 highest weights, and subsequently one works out that $\omega_7$ maps to $\eta_1+\eta_2$. This leads to the actual question.

Question: Does the choice of how one maps $\omega_5$ affect the conjugacy of the embedding of $G_2$ in $Spin(15)$?

It is not hard to check that either choice of how to map $\omega_5$ leads to isomorphic decompositions of the restriction to $G_2$ of an arbitrary $B_7$ representation. However, what I cannot decide is whether the two different ways of mapping $\omega_5$ to weights of $G_2$ correspond to two embeddings of $G_2$ (call them $G_2(A)$ and $G_2(B)$) which are not conjugate to one another in $B_7$, and yet $Res^{B_7}_{G_2(A)}(V)\cong Res^{B_7}_{G_2(B)}(V)$ for every representation $V$ of $B_7$.

Since the restriction is really just a projection of the weight space of $B_7$ onto a subspace which is isomorphic to the weight space of $G_2$, I feel that there should be no real difference between the two choices of where to send $\omega_5$ (indeed there are lots of choices of where to map the $\omega_i$'s, and certainly not all choices lead to non-conjugate $G_2$'s) but I cannot adequately convince myself that the two choices really do correspond to conjugate embeddings.

Motivation: Recently I have been working on behaviors of certain representations upon restriction to a subgroup. Most of the cases I have dealt with have been simple to deal with, so I decided to try a more complicated example. In order to embed $G_2$ in $B_7$ one can of course go through the 'obvious' inclusion chain: $G_2\hookrightarrow B_3\hookrightarrow B_4\hookrightarrow B_5\hookrightarrow B_6\hookrightarrow B_7$. It is clear (to me at least) that all such embeddings of $G_2$ into $B_7$ are conjugate to one another. On the other hand, since the adjoint representation of $G_2$ is orthogonal, one can embed $G_2$ directly in $B_7$ leading to the above question. This is the simplest example of the more general embeddings I am thinking about, each involving restrictions to representations with branching in their weight diagrams.

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This is a substantial-looking (and lengthy) question but still hard for me to sort out, partly due to some loose or unfamiliar language used such as "four top-most weights". – Jim Humphreys Mar 2 '11 at 13:40
So if one starts at the highest weight $\eta_2$ and then draws out the weight diagram by subtracting off simple weights, the second highest weight is $3\eta_1-\eta_2$, then $\eta_1$ then $-\eta_1+\eta_2$; then the branching occurs at the next level. So these first four weights are the 'top-most' weights I refer to. The point is that the weights are totally ordered when subtracting off the first three simple weights; and then the branching occurs in the diagram when one goes to subtract a fourth simple weight. – ARupinski Mar 2 '11 at 15:00

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