18
votes

Accepted

### A question about Lie groups

A Lie algebra $\frak g$ determines a simply connected Lie group. Let's call it $G$. It is the only simply connected Lie group with that Lie algebra. More precisely, if $H$ is a simply connected Lie ...

13
votes

### Representation theory of $\mathrm{GL}_n(\mathbb{Z})$

For $n=2$, these groups are very close to free groups, so their representation theory is not very tractable. For $n \geq 3$, however, there is a beautiful story arising from Margulis superrigidity. ...

6
votes

### Is $SO(N)$ a sphere?

Spheres are simply connected but $\operatorname{SO}(3)$ is not simply connected. Hence they cannot be homeomorphic, and a fortiori they cannot be isometric.
If $a$ is a unit quaternion, i.e. a ...

4
votes

### What's an example of an $n$-step nilpotent Lie algebra whose centre is not $g^{(n)}$?

Any direct sum of a $n$-step nilpotent Lie algebra and a nonzero abelian Lie algebra satisfies the required property.
Acknowledgement: After posting this answer, I realized that this class of examples ...

4
votes

Accepted

### Holomorphic discrete series vs. discrete series

The holomorphic discrete series are precisely parametrized by the $\lambda$ as indicated, which satisfy the following additional condition. Let
$\Delta^+=\{\alpha\in\Delta\mid \langle\lambda,\alpha^\...

3
votes

### Conjugacy of Cartan subgroups in $\mathrm{GL}(n)$

A. Borel, Linear Algebraic Groups, 2nd edition, Grad. Texts in Math., 126, Springer-Verlag, New York, 1991. xii+288 pp.
ISBN:0-387-97370-2, p. 48 section IV, 11.3 Corollary
The maximal tori in $G$ ...

2
votes

### Lie group framing and framed bordism

Lie group framing is a reference to the group action. A Lie group $G$ acts on the left of $G$ by the map
$$(g,h) \longmapsto gh.$$
Similarly there are actions on the right, and conjugation actions, ...

2
votes

### Holomorphic discrete series vs. discrete series

If I remember correctly, then holomorphic discrete series are a subset of the unitarizable highest (or lowest) weight modules which are Verma modules. There are several parametrizations of these ...

2
votes

Accepted

### Are there natural isomorphisms $S^{(2,1)}(k^{m+1})\cong k^2\otimes W$?

This is not a complete answer, but it strongly suggests where to look. Since $\binom{m+2}{3}_q$ is the $q$-character of $\mathrm{Sym}^3\mathrm{Sym}^{m-1}(E)$, working over the complex numbers we have
$...

1
vote

### On Haar measure and Spherical measure

Yes, (of course, up to scalar multiples), the $U(n)$-invariant measure on $S^{2n-1}$ must be the same as the $O(2n)$-invariant measure on $S^{2n-1}$... for example, because the $U(n)$-invariant ...

1
vote

### Does the 3875-dimensional rep of $E_8$ have a solution to $x\star x=0$?

If you want to construct the multiplication table i suggest you lok at the explicitly definned product in the article
https://www.cambridge.org/core/services/aop-cambridge-core/content/view/...

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