5 votes
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Endomorphisms of a Lie algebra representation

Things are not quite as easy. If you have non-zero eigenvectors for different weights, then the sum of these spaces is not an eigenspace. Moreover, any homomorphism of $\mathfrak g$-modules preserves ...
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4 votes
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Why should I expect the generators of Lie Groups to be closed under the commutator?

So the proper formal answer should be that the vector fields on a Lie group have a natural Lie bracket (as do all vector fields on a general manifold). Then the Lie algebra can be identified with the ...
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4 votes

Commutator subgroup of connected group.

The argument is a bit more complicated than what you suggest. For each $k$, there is a continuous map $f:G^{2k}\to G$ sending $(g_1,g_1',\dots,g_k,g_k')$ to $[g_1,g_1']\dots[g_k,g_k']$. So for each $...
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3 votes

Decomposing a Lie Group representation into irreds using Lie Algebra representations.

Here is the answer in the form of a two-part exercise. Exercise 1a: Let $G$ be a connected Lie group, $\rho : G \to GL(V)$ be a finite-dimensional representation of $G$ over $\mathbb{R}$ or $\mathbb{...
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2 votes
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Image of the adjoint representation of a Lie algebra

No, this is not true. It would say that a discrete subgroup $K\subset G$ of a connected Lie group would automatically lie in the kernel of $Ad_G$ and hence in the center of $G$. But this is not true ...
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1 vote

Decomposing a Lie Group representation into irreds using Lie Algebra representations.

If someone is interested here an answer based on Qiaochu Yuan‘s suggestions and definitions: „Exercise 1a“: $\forall g \in G \ \ \forall w \in W \ \ \forall X \in \mathfrak{g} $ $ \rho(g)w \in W \...
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1 vote
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Prove that this action admits three orbits

You have already identified the 3 orbits of $T$ on $C$ by the way you have written $C$. Namely, the two points of $N(T)/T$ are both $T$-fixed, and so constitute 2 orbits. For the 3rd, note that $$ \...
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