4
votes
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Product of spherical tensors
I denote the $(2j+1)$-dimensional representation by $S^{2j}\mathbb{C}^2$ (physicists, I believe, just write the dimension $\mathbf{2j+1}$ in bold).
Let $M_j=S^{2j}\mathbb{C}^2 \otimes S^{2j}\mathbb{C}^...
- 643
4
votes
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How do projective representations map elements? Are they mulivalued?
I have rewritten my answer.
Let $V$ be a vector space on a field $k$. Let us recall that if $V$ is a vector space, $\mathbb{P}(V)$ is the set of equivalence classes of vectors up to the multiplication ...
- 2,347
4
votes
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Dimension of $\mathfrak{sp}(2n, \mathbb{R})$
OP should probably recheck their calculation (3). According to Wikipedia it should be
$$D~=~-A^T,\qquad B~=~B^T,\qquad C~=~C^T,$$
leading to
$$\dim \mathfrak{sp}(2n, \mathbb{R})~=~n^2+2\frac{n(n+1)}{2}...
- 12k
3
votes
G abelian Lie group, $(\pi, V)$ a finite-dim unitary rep. $\exists $ mutually orthog. 1-dim invariant linear subspaces s.t $V= \bigoplus_i V_i$
Let $\{\pi(g):g\in G\}\subset GL(V)$. These are unitary operatots that also commute ($\pi(g_1)\pi(g_2)=\pi(g_1g_2)=\pi(g_2g_1)=\pi(g_2)\pi(g_1))$.
Moreover, as they are unitary, they are also ...
- 4,431
3
votes
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Isomorphism between $\mathbb{G}_m\times SL(n)$ and $GL(n)$
Yes, this holds quite generally.
Let us assume that $G$ is an algebraic group i.e., a finite type group scheme over some field $k$. There is a natural map
$$p\colon G^\mathrm{der}\times Z(G)\to G,\...
- 53.3k
3
votes
Accepted
Is there a way to convert an element of $\mathfrak{so}(3)/\mathfrak{so}(2)$ into a "geometric" tangent vector on $S^2$?
Consider $\Bbb S^n\subseteq \Bbb R^{n+1}$, write $\Bbb R^{n+1} = \Bbb R^n \times \Bbb R$, and fix the north pole $p = (0,1)$. Note that $T_p(\Bbb S^n) = \Bbb R^n\times \{0\}$. Then ${\rm SO}(n+1)$ ...
- 76.5k
3
votes
Reference for universal cover of $\mathrm{SL}_2(\mathbb{R})$
I doubt this will answer all the parts of your question, but the following paper on Geometries of 3-manifolds by Peter Scott has a nice description of $\tilde SL_2(\mathbb{R})$.
Algebraically, it can ...
- 733
2
votes
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Explicit expression for simple roots of the root systems $A_2$, $B_2$ and $G_2$ in 2D
To get this off the unanswered list:
An explicit realization of a root system of type $A_n$ in $n$-dimensional Euclidean space (that is, with coroots / reflections defined to match the standard ...
- 25.4k
2
votes
Lie algebras are to Lie groups what Lie modules are to …?
This is not the common definition of a Lie module. Usually, a Lie module (aka Lie algebra representation) is defined as a module on which a Lie algebra acts (in a suitable way). A Lie algebra can be ...
2
votes
Accepted
Is the fundamental group of a closed orientable hyperbolic $3$-manifold isomorphic to a non-discrete subgroup of $\mathrm{PSL}(2, \mathbb{C})$?
Here is a construction, but at this moment I do not have a proof of nondiscreteness (in general).
Let $\Gamma< SL(2, {\mathbb C})$ be a cocompact discrete subgroup (a uniform lattice): The ...
- 94.1k
2
votes
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Understanding the Geometry Spawned from Quotient Spaces $GL^+(4,R)/SO(3,1)$, $GL^+(4,R)/Spin(3,1)$, and $GL^+(4,R)/Spin^c(3,1)$
The group $Spin(3,1)$ is nothing but $SL(2, {\mathbb C})$. The latter has a natural embedding in $GL(4, {\mathbb R})$. The subgroup
$SL(2, {\mathbb C})< GL(4, {\mathbb R})$ preserves two things:
...
- 94.1k
2
votes
Accepted
The order in which diffeomorphisms should be applied
Introducing extra notation for the intermediate steps of the computation and worrying about the point evaluations at $z\in S^1$ separately might make things clearer.
Let $g:=\pi(\xi_2)f$. Applying the ...
- 2,536
2
votes
Accepted
Extensions of simple Lie groups with a topological splitting
Yes, indeed, if $H$ is abelian then the Lie group central extensions of $G$ by $H$ are classified by the 2ned continuous cohomology $H^2_c(G; H)$. A good survey can be found in
Stasheff, James D., ...
- 94.1k
2
votes
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Set of positive definite Hermitian matrices as quotient
In general, when a group acts on a space $\Omega$, for any $x\in\Omega$, the fibers of the map $G\to\mathrm{Orb}(x)$ given by $g\mapsto gx$ are the cosets of $\mathrm{Stab}(x)$. Thus, $g\mathrm{Stab}(...
- 2,497
2
votes
Accepted
Representation theoretic condition to show an inner product vanishes
Consider the representation $ (W^* \otimes R \otimes W, \lambda^* \otimes \omega \otimes \lambda) $. The map
$$
W^* \otimes R \otimes W \to \mathbb{C}
$$
given by
$$
w_1 \otimes T \otimes w_2 \mapsto ...
- 7,218
2
votes
Accepted
details in the proof: G a commutative Lie group and $(\pi, V)$ a finite-dim unitary representation of G. Then $\pi$ is irreducible iff dim $V = 1$
A representation assumes that the map $\pi:G\times V\to V$ is continuous. So the composition $V\to \{g\}\times V\subseteq G\times V \xrightarrow{\pi} V$ is continuous. The topologies taken are the ...
- 2,794
2
votes
Accepted
Why is $ V \longrightarrow V: v \mapsto \pi(g)(v)$ an endomorphism of $G$-modules?
Lets write down explicitly:
$$f:V\to V$$
$$f(v)=\pi(g)(v)$$
Or simply $f=\pi(g)$. It is linear because $\pi(g)$ is. It is a bijection because $\pi(g^{-1})$ is inverse of $\pi(g)$.
So the only question ...
- 40.1k
2
votes
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A question related to the action of a group on the real projective space
The action $G \times \mathbb{P} \to \mathbb{P}$ is what we might expect: $(g, \mathbb{R}v) \mapsto \mathbb{R}gv$, but we have to verify that it's well-defined. For this consider any nonzero multiple $...
- 23.4k
1
vote
Accepted
How can I show this property about lie algebras and lie subgroups?
Continuing off of your approach, it suffices to show that $\overline{\mathfrak h}$ is commutative. Consider any $X,Y \in \overline{\mathfrak h}$. $e^{sX}$ and $e^{tY}$ commute for all $s,t \in \Bbb R$ ...
- 222k
1
vote
Accepted
Homomorphisms $U(1) \to SU(2)$
The matrices in the image commute and thus are simultaneously diagonalizable, so indeed the same $h$ works for all.
- 28.5k
1
vote
Accepted
Show that $\lambda:G \to \mathbb{R}^\ast$ where $R_h^\ast \mu = \lambda (h) \mu$ is smooth
The multiplication map $m\colon G\times G\to G$, $m(g,h) = gh$, is smooth. This means that it induces a smooth bundle map $Dm\colon T(G\times G)\to TG$. In particular, $R_h(g)=m(g,h)$ and so for $X\in\...
- 112k
1
vote
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 ...
- 2,604
1
vote
Accepted
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 ...
- 2,794
1
vote
Accepted
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 ...
- 51.2k
1
vote
Understanding Representation theory
We know the group $G=\mathrm{SL}(2,\mathbb{R})$ acts on $\mathbb{R}^2$. Thus, it acts on the real projective line $\mathbb{RP}^1$. With the congruence $\big[\begin{smallmatrix} x \\ y \end{smallmatrix}...
- 2,497
1
vote
Accepted
Isomorphic connected Lie subgroup
It seems to me that your approach is too complicated. By construction, the homomorphism $\Phi:K\to H$ has bijective derivative. But surjectivity of the derivative and connectedness of $H$ imply that $\...
- 20.1k
1
vote
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 ...
- 11.1k
1
vote
Accepted
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 $n=2$.
...
- 9,363
1
vote
Accepted
$G$-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 ...
- 1,539
1
vote
Accepted
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^...
- 7,218
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