14

In response to Qiaochu's question in the comments, Fong and Sourour prove in their paper The group generated by unipotent operators that every element of $\mathrm{SL}_n(\mathbb C)$ is a product of three unipotent matrices.
Edit: Sourour proves in the paper A factorization theorem for matrices that a nonscalar matrix in $\mathrm{SL}_n(\mathbb C)$ is indeed a ...

11

If you only need four factors then you can have much more than that. Namely, you can take as the unipotent matrices upper and lower uni-triangular matrices. It was noted by G. Strang in "Every unit matrix is a LULU" that for any field $K$ anfy $A\in M(n,K)$ can be presented as $A=U_1L_1U_2L_2$, where $U_i$ are upper uni-triangular and $L_i$ are ...

10

I have not seen this result before, but here is a proof. Let $e_{ij}$ be the matrix with a $1$ in position $(i,j)$ and all other entries $0$. Let $x_i(t)$ and $y_i(t)$ be the Chevalley generators $\mathrm{Id} + t e_{i (i+1)}$ and $\mathrm{Id} + t e_{(i+1) i}$.
Since being unipotent is unchanged by conjugacy, and every matrix in $\mathrm{SL}_n$ is conjugate ...

6

If you just want an explicit realization the of outer automorphisms of $\mathrm{Spin}(8)$, here is one, assuming that you know about the algebra of octonions $\mathbb{O}$, the unique $8$-dimensional (and hence nonassociative) inner product algebra over $\mathbb{R}$ with positive definite inner product.
The subgroup $\mathrm{Spin}(8)\subset{\mathrm{SO}(8)}^3$ ...

4

See below a detailed version of the comment of @LSpice. (Edited taking into account a comment of @YCor.) This is an answer to the question on homomorpisms of real algebraic groups.
Proposition.
Let $\varphi\colon G\to K$ be a homomorphism of real algebraic groups, where $G$ is a connected semisimple real algebraic group without compact factors and $K$ is a ...

4

This is only a partial answer, and it's based on YCor's comment about the groups that act transitively on spheres. What is missing, as YCor commented, is knowing that if $G\subset\mathrm{GL}(n,\mathbb{R})$ acts transitively on $\mathrm{Gr}(k,\mathbb{R}^n)$ for some $k$ with $1<k<n$, then it acts transitively on $\mathrm{Gr}(1,\mathbb{R}^n)$.
Here is ...

3

Proposition. Equivalences ($G$ connected Lie group):
(i) $G$ has a faithful finite-dimensional continuous unitary representation;
(ii) $G$ is locally isomorphic to some compact Lie group;
(iii) $G$ is direct product of some Euclidean group (=$\mathbf{R}^d$ for some $d$) with a compact Lie group.
$\bullet$ Indeed (iii) clearly implies (i).
$\bullet$ ...

3

$\DeclareMathOperator\Image{Image}$Here's the most general result I think I can muster. I've split it out into a second answer in order to keep the answer to the original question more self-contained.
Theorem 1: Let $G$ be a compact Lie group, let $X$ be a $G$-space, and let $E$ be a spectrum. Then the following hold, where $F$ ranges over finite subgroups ...

2

Since a theorem of Borel (which you quote in the comment) tells you that this regular map is dominant for every $w$, its differential is surjective on a Zariski-dense open subset $U_w$.
Hence there intersection (over all $w\neq 1$) of these subsets $U_w$ is a $\mathrm{G}_\delta$-dense subset of full measure.

2

Probably someone will step in with a more detailed answer soon... but here are some comments.
I think the line you are looking for between easy and hard could be the following:
The algebraic representations of a split reductive algebraic group $G$, and
The representations of some associated Lie groups $G(\mathbb R)$ or $G(\mathbb C)$ (or indeed p-adic ...

2

I think an elementary argument is that if such a homomorphism exists, one can pull back the Cartan-Killing form (or an extension of that, in case the compact group has positive dimensional center) to an Ad-invariant definite inner product on the Lie algebra of a quotient of the original group. This quotient is then compact.

2

Found the proof! It's done using the integral definition of $T$:
$$
T(v) = \int_0^1 R(su) ds = \lim_{n\rightarrow \infty} \frac{1}{n}\sum_{i=1}^n R\left(\tfrac{i}{n}v\right)
$$
So for any vectors $X$ and $Y$:
\begin{align*}
&\biggl|\left[\mathrm{D}_v \left(R(v)X\right)\right]Y - \left[\mathrm{D}_u \left(R(u)X\right)\right]Y\biggr| = \biggl|\left[R(u)X\...

2

I would like to know an example of an abelian Lie group A where the statement is wrong (and why).
There is no such example because the statement is true for all $A$.
(This also implies trivial answers for the other two questions: always and no.)
Any abelian Lie group $A$ fits into an exact sequence
$$1→π_1(A)→U→A→π_0(A)→1,$$
where $π_0(A)$ denotes the ...

1

Looks like I found the reference. Thanks YCor!
https://www.ams.org/journals/proc/1999-127-11/S0002-9939-99-04959-X/S0002-9939-99-04959-X.pdf

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