# Tag Info

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### The natural representation of the real group $G=SO(2)$ on $V=\Bbb R^2$ is irreducible.

A unitary representation requires that the representation space is a vector space over the complex numbers, which $\Bbb{R}^2$ is not. Thus, you cannot use $(a)$. This is not a minor annoyance, it is ...
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### Prove that $\exists g \in SO(n)$ such that $gv_i = w_i$ for $i = 1, 2$ where all these vectors are O.N.

$v_1, v_2$ can be completed to an orthonormal basis (of $\Bbb{R}^n$) $v_1, v_2, \ldots, v_n$. Now put the $v_i$ as column vectors in a matrix $V$. Then $V\in O(n)$. If $\det(V)=-1$ you can invert ...
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### dual representation map as a linear map

Every linear map $f:V\to W$ induces a linear map $W^*\to V^*$. This linear map is denoted by any of the following symbols: $f^*,f^t,f’$, and is called the dual map or adjoint map or transpose map or ...
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### Irreps of crystallographic groups

I'll stick to the notation from the ncatlab article linked in the post. $S$ is the crystallographic group, $N \cong \mathbb{Z}^n$ is the normal subgroup of translations, and $G = S/N$ is the point ...
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### Prove that every finite subgroup of $SL_2(\mathbb{C})$ is conjugate to a finite subgroup of $SU_2$

There is such a Moore's theorem from 1898 (Eliakim Hastings Moore: 1862-1932). Every finite group $G\leq GL_n(\mathbb{C})$ is conjugate to a subgroup of the unitary group $U_n$. In your case $n=2$. ...
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### $G$-equivariant map between Lie group representations

It depends on the context, but the answer to your first question should be yes if you are thinking of complex semi-simple Lie groups. Indeed, this should be true in any setting where your ...
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### Commutator of a connected semi simple Lie group

The commutator subgroup $[G,G]$ is a normal subgroup of $G$. And $G/[G,G]=G^{ab}$ is abelian. So suppose that $G$ is semisimple and connected. Then $G^{ab}$ is a semisimple Lie group. So \$ G^...

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