New answers tagged lie-groups
12
The comment of @abx is part of a general picture. For $n >7$ the alternating group $A_{n}$ has no non-trivial complex irreducible character of degree less than $n-1$, so that for $n > 7$, $A_{n}$ is not isomorphic to any subgroup of ${\rm GL}(n-3, \mathbb{C}).$
Furthermore, if $n > 7$ is also even, then the double cover of $A_{n}$ (which, by a ...
2
Maybe these arguments are of interest to you. It is known that for any compact symmetric space $M$ the tangent bundle $TM$ possesses a canonical structure of a complex
manifold. Multiplication by $-1$ on $TM$ is an antiholomorphic involution; its set of fixed points is $M$, when identified with the zero section
of $TM$ (see e.g. Thm 2.5a of Szőke, R.: ...
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Your tangent space is too big. It has rank $\dim G + \dim V$ but it should have rank $\dim M = \dim G - \dim G_\beta + \dim V$.
I think that the tangent space at $[g, v]$ is actually $T_{g} (G/G_\beta) \times T_vV$. On the first factor you have a symplectic form $\omega_\beta$ and on the second factor the symplectic form $\omega_v$. On the whole tangent ...
1
What about the formula $(D_X\varphi)(g)=\frac{\rm d}{{\rm d}t}\Bigl\vert_{t=0}\varphi(g\exp_G(tX))$ for $\varphi\in C^\infty(G)$, $g\in G$, and $X\in \mathfrak g$, where $G$ is a Lie group with its Lie algebra $\mathfrak g$? See for instance Eq. (5) in Ch. II of the book by S.Helgason, "Differential geometry, Lie groups, and symmetric spaces".
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The connected Lie groups whose points are separated by the finite-dimensional complex representations are exactly the linear Lie groups, for instance by Th. 5.3 in Beltiţă and Neeb - Finite-dimensional Lie subalgebras of algebras with continuous inversion.
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A finite-dimensional representation of a Lie algebra is in particular a homomorphism of finite-dimensional Lie algebras. Hence your question is answered in the affirmative e.g. by Th. 3.27 in F. Warner's book "Foundations of differential geometry and Lie groups".
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$\DeclareMathOperator\SO{SO}\DeclareMathOperator\U{U}\DeclareMathOperator\O{O}$Denote by $\U'(n)$ the normalizer of $\U(n)$ in $\mathrm{GL}_{2n}(\mathbf{R})$. It is not hard to see that $\U(n)$ has index 2 in $\U'(n)$, which is generated by $\U(n)$ and by the coordinate-wise complex conjugation. Moreover, $\U'(n)$ is maximal in $\O(2n)$ (if $n\ge 2$).
In $G=\...
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The weak order of a dihedral group looks like a polygon (see e.g. Figure 3.1 in the book "Combinatorics of Coxeter groups" by Björner and Brenti). Hence there are 2 reduced decompositions of $w_0$ (= maximal chains from bottom to top) in these cases.
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I think the answer to the first question is yes and the answer to the second one is no:
Yes, the quotient is an orbifold. The action of the finite group $G_x$ in a neighbourhood of $x$ can be linearized (at least if the action is by diffeomorphisms, I don't know about $C^0$ regularity), and the quotient $M/G$ is locally modelled on $G_x \backslash T_xM / T_x ...
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This is maybe easier to see if you think about weights in terms of the usual Cartan of $\mathfrak{so}_{n}$. Choose a basis $e_i$ for $\mathbb{C}^{n}$ and the form so that the dual basis is $e_i^*=e_{n+1-i}$. The Cartan is then diagonal matrices so that $a_{n+1-i}=-a_i$, i.e $\mathbb{C}^m$ where $m=\lfloor \frac{n}{2}\rfloor$ and the dual Cartan can be ...
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Your answer is correct for $n$ odd. For $n$ even you instead need the sum of the coefficients of the last two weights $a_{l-1} + a_{l}$ to be even. Either way you are effectively asking for the non-trivial element of the centre of $\mathrm{Spin}(n)$ to act trivially. Since it acts as $-1$ on the spin representation (or on each half spin representation for $n$...
2
For $n$ even, $M$ admits such an action. Indeed, the antipodal map of the even-dimensional sphere is orientation reversing, so you can realize $M$ as the quotient
$$\langle \gamma \rangle \backslash \left(S^n\times \mathbb R\right)$$
where $\gamma = (-\mathrm{Id}, 1)\in O_{n+1} \times \mathbb R$.
Since $\langle \gamma\rangle$ is central in $O_{n+1} \times \...
0
The crucial point is that the eigenvectors are not unique, as David Handelman commented. There is a gauge freedom since you can change their phase at will. If you want the map $U\to(D,V)$ to be bijective, you must fix the gauge. For example, you may require that the first element of every eigenvector is real and positive. In terms of real dimension, you then ...
2
@MartinSkilleter has posted an answer in the comments. I'll summarise here an elementary proof; it is almost the same as in @MartinSkilleter's link Extensions of tori by tori are tori, just written slightly differently.
Let $k$ be the field of definition. Since $k$ has characteristic $0$, the group scheme $T$ is smooth. (In general, we could observe that $...
3
It is possible to have a non-trivial, closed, connected Lie subgroup $K$ of a compact, connected Lie group $G$ such that no element of $\mathfrak k$ is regular in $\mathfrak g$. For example, consider $K = \operatorname{SU}_2$ embedded in $G = \operatorname{SU}_4$ as $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \mapsto \begin{pmatrix} a &&&...
5
The (orientation-preserving) isometry group $G=PSL(2,{\bf C})$ acts on the bundle of (positively oriented) orthonormal frames on ${\bf H}^3$.
An ad hoc argument using the Iwasawa decomposition $G=KAN$ shows that this action is transitive. Indeed, considering the upper half space model and fixing the standard frame in $(0,0,1)$ you can first use the action of ...
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I am just posting my final comment as one answer. As the OP notes, there are answers that go through the classification, and there may be a best answer that uses very little. The answer here relates the post to the theory of the automorphism group schemes of semisimple algebraic groups.
For a pinned semisimple group scheme $G$ over a base scheme $S$, there ...
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Another approach.
To show it's impossible (the rank can't be 1), it is enough to show this when the field (assumed of char 0) is algebraically closed, and in turn it's enough to show the result in case $\mathfrak{g}$ is simple. If $x$ has $\mathrm{ad}(x)$ of rank 1, $x$ has centralizer of codimension 1. It is known (see e.g. this MathSE answer) that $\...
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$\DeclareMathOperator\ad{ad}\DeclareMathOperator\rk{rk}$I claim that it is impossible to have $\rk(\ad_x) = 1$.
I'll assume that $\mathfrak g$ is $k$-split, which seems to be OK since you are interested in the case $k = \mathbb C$. (Or you could just tensor up to $\overline k$, which does not change the rank of the adjoint operator.) Let $\mathfrak h$ be a ...
1
The isometry group of a compact Riemannian manifold is always compact, so the answer to this question doesn’t really depend on the fundamental group at all: if $G$ is noncompact, then it doesn’t act isometrically on a compact $M$ for any choice of metric, but if $G$ is compact, then we can get an invariant Riemannian metric using the usual averaging trick.
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First, some remarks that may help with literature-searching.$\newcommand{\fsnorm}[1]{{\Vert#1\Vert}_{\rm B}}$
$\newcommand{\supnorm}[1]{{\Vert#1\Vert}_\infty}$
The algebra you have denoted by $A_0(G)$ is known as the Fourier--Stieltjes algebra of $G$, and is usually denoted by $B(G)$, so I will do that from now on. $B(G)$ has been much studied: it turns out ...
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