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Lie Groups are Groups that are additionally smooth manifolds such that the multiplication and the inverse maps are smooth.

5 votes

Center of a compact group

This is not true. Take $\mathrm{U}(2)$, for example. It is not the direct product of its center (the matrices of the form $\mathrm{e}^{i\theta} \mathrm{I}_2$) with the simple part $\mathrm{SU}(2)$. …
Robert Bryant's user avatar
2 votes
Accepted

How to compute SE(2) group exponential and logarithm?

Just computing the exponential gives $$ \exp\left( \begin{matrix} 0 & 0 & 0 \cr x & 0 & -t\cr y & t & 0 \end{matrix} \right) = \left( \begin{matrix} 1 & 0 & 0 \cr x\frac{\sin t}{t}-y\frac{1{-}\cos t}{ …
Robert Bryant's user avatar
2 votes
Accepted

A lie Subgroup of SO(4n)

One way to make the subgroup $\mathrm{Sp}(n)\cdot\mathrm{Sp}(1)\subset\mathrm{SO}(4n)$ explicit is to think of $\mathbb{R}^{4n}$ as $\mathbb{H}^n$, i.e. as columns of quaternions of height $n$, where …
Robert Bryant's user avatar
10 votes

Do all closed connected subgroups of $SO(2n+1)$ embed into $SO(2n)$?

As an answer to your weaker version: No. For example, $\mathrm{G}_2$ is a closed proper subgroup of $\mathrm{SO}(7)$, but it is not isomorphic to any subgroup of $\mathrm{SO}(6)$. Of course, this f …
Robert Bryant's user avatar
8 votes
Accepted

Rank of a Lie subgroup generated by two Lie subgroups

No. A simple example is to let $G = \mathrm{SO}(6)$, which has rank $3$. Let $v_1$ and $v_2$ be unit orthogonal vectors in $\mathbb{R}^6$ and let $H_1\simeq \mathrm{SO}(5)\subset G$ be the stabilize …
Robert Bryant's user avatar
10 votes

Exponential map

Sure. Let $G$ be the group of diffeomorphisms of the real line. Then $\frak{g}$ is the Lie algebra of vector fields on the real line. However, many vector fields on the real line cannot be exponent …
Robert Bryant's user avatar
4 votes
Accepted

geodesics on $G/K$ which are not the orbits of a 1-parameter subgroup of $G$

Perhaps I'm misunderstanding your question, but what about the following example? Let $G = \mathrm{SO}(3)$ and let $K=\{e\}$ be the identity subgroup. Then $G/K = \mathrm{SO}(3)$ and $G_v = K$ for a …
Robert Bryant's user avatar
2 votes
Accepted

Stationary curves on homogeneous spaces

This is not really an answer as much as it is a caution that the problem, even with the extra assumptions I was able to solicit from the OP, is not going to have a very nice answer unless one adds som …
Robert Bryant's user avatar
12 votes
Accepted

Unusual decomposition of 3x3 real symmetric matrices - is this possible?

Unfortunately, the answer is 'no'. You are basically asking whether you can simultaneously diagonalize two quadratic forms in three variables, and the answer is that, 'generically' you can (and you a …
Robert Bryant's user avatar
6 votes
Accepted

Normal form for trace-free real cubic forms in 3 variables under SO(3)-action?

One way to do this is to reduce this to understanding sets of points on the $2$-sphere up to a certain equivalence. If you let $\mathcal{H}_k$ denote the homogeneous polynomials $p$ of degree $k$ on …
Robert Bryant's user avatar
6 votes

Reference request: Calculation in exceptional Lie groups

But, can't you read this off the table of simple roots for each of the exceptional Lie algebras? For example, $\mathrm{G}_2$ contains $\mathrm{SU}(3)$ as a subgroup and the maximal torus of $\mathrm{ …
Robert Bryant's user avatar
10 votes
Accepted

Finite-dimensional subgroups of diffeomorphism groups

I guess that the answer to your first question is no, based on the following: If the union of the $G_i$ were dense in $G=\mathrm{Diff}(M)$, then, presumably, for $i$ sufficiently large, the action of …
Robert Bryant's user avatar
11 votes
Accepted

A primitive to $\mathrm{Vol}(x)\pm \mathrm{Vol}(y)$ on $(\mathbb{S}^n \times \mathbb{S}^n)\b...

NB: I have edited my answer to remove the comment at the beginning about correcting the original sign of the OP in the formula for $H$, since the OP has now corrected the erroneous sign in the questio …
Robert Bryant's user avatar
25 votes

Left invariant metric on ${\rm SL}_n(\mathbb{R})$

The OP specifically asked for a(n ordinary) metric, not a Riemannian metric. While Misha and Paul have given good answers, I think that it's worth pointing out that, if one just takes an arbitrary le …
Robert Bryant's user avatar
7 votes

Appearances of 'exotic' compact Lie Groups

There won't be any `right' answer here because there are many different ways that one can come across groups in studying various algebra problems, but here are a few examples: Maybe the most famous e …

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