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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
Accepted
Why a nilpotent Lie group must be a matrix group?
Another reference which was not yet mentioned, I think, is the article of M. Moskowitz, "Faithful Representations and a local property of Lie groups", Math. Z. $143$, 1975.
There the question is disc …
9
votes
Accepted
Is the unitary matrix group path-connected?
Every matrix Lie group is a smooth manifold, hence it is path-connected if and only if it is connected.
And $U(n)$ is compact and connected as a topological space (any unitary matrix can be diagonali …
6
votes
semidirect product
Since a Lie group is semisimple if and only if its Lie algebra is semisimple, we may show
that the semidirect sum of semisimple Lie algebras is semisimple. Indeed, if $L$ is a semidirect sum of two se …
3
votes
Bound on the index of an abelian subgroup in discrete subgroup of the euclidean group?
Every discrete group of Euclidean isometries acts properly discontinuously and cocompactly on some subspace of $\mathbb{R}^n$ (a result of Bieberbach). If we are in the crystallographic case, then the …
7
votes
real representation of real semi simple Lie algebra
A reference for this is also "Lectures on Real Semisimple Lie Algebras and Their Representations" by A. L. Onishchik. A first result here is:
any irreducible real representation $\rho\colon \mathbb{g …
18
votes
How many three dimensional real Lie algebras are there?
There are already uncountably many isomorphism classes of $3$-dimensional real Lie algebras. In fact, there are $1$-parameter families of $3$-dimensional solvable Lie algebras. The classification has …
2
votes
A question on Lie algebras
They are the compact Lie algebras. Note that are two different definitions in the literature. One is that a compact Lie algebra is the Lie algebra of a compact Lie group. This includes tori, and the K …
2
votes
Can this Lie group written as a direct product?
A. L. Onishchik has classified decompositions $G=G_1G_2$ of redcutive Lie groups (e.g., see his article
"Decompositions of reductive Lie groups" in Math. USSR-Sbornik, Vol. 9 (1969), No. 4). We have
$ …
4
votes
Maximal subgroups of semisimple Lie groups
Maximal (closed) subgroups of semisimple Lie groups have been
first classified by E.B. Dynkin (see the reference above). There is indeed a way to "reduce"
the problem to the case of a simple Lie group …
3
votes
Reductive space & Reductive Lie algebra
It depends on the definition of a reductive space $G/H$. Some authors require that
in addition to $\mathfrak{g}=\mathfrak{h}+\mathfrak{m}$, $[\mathfrak{h},\mathfrak{m}]\subseteq \mathfrak{m}$, we hav …
2
votes
Determining the Lie algebra elements exponentiating to the center of a Lie group
I have only some simple remarks here, valid also for non-compact Lie groups.
Let $G$ be a connected Lie group with Lie algebra $\mathfrak{g}$. A first remark is that for $X\in Z(\mathfrak{g})$ and $Y\ …
5
votes
How can I tell if a group is linear?
Here are some more purely group theoretical conditions. This is also not a complete answer, since it gives just some necessary conditions for certain groups to be linear.
Schur: Suppose that $G$ is …
6
votes
Complete classification of six dimensional non-semi simple Lie algebra
There are exactly four non-solvable complex Lie algebras of dimension $6$ with nontrivial Levi decomposition, namely
\begin{align*}
\mathfrak{sl}_2(\Bbb C) & \ltimes_{\phi} \Bbb C^3\\
\mathfrak{sl}_2( …
10
votes
Square root in complex reductive groups
As the comment shows the answer is negative in general. Perhaps it is worth to mention that for connected compact Lie groups the answer is yes, because its
exponential map is surjective. In general, i …
6
votes
Is there a Levi decomposition for Lie group and algebraic group?
Question $1$: The theorem of Mostow says that every connected algebraic group $G$ over a field $K$
of characteristic zero has a Levi decomposition. This means, $G$ has a reductive algebraic subgroup $ …