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Questions tagged [lie-groups]

A Lie group is a group (in the sense of abstract algebra) that is also a differentiable manifold, such that the group operations (addition and inversion) are smooth, and so we can study them with differential calculus. They are a special type of topological group.

Consider using with the (group-theory) tag.

Lie groups are groups that are also differentiable manifolds that represent the best developed theory of continuous symmetry of mathematical objects.

Examples of lie groups are:

1) The Euclidean space $\mathbb{R}^n$ under addition is a lie group.

2) The special orthogonal group of real orthogonal matrices with determinant $1$ (note that $n=3$ is the rotation group in $\mathbb{R}^3$).

3) The spin group, which is the double cover of the special orthogonal group such that $\exists$ a sequence of lie groups:

\begin{equation*} 1\to Z_2\to~\text{Spin}(n)\to SO(n)\to 1. \end{equation*}

Note that it has dimension $\frac{n(n-1)}{2}.$

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If a group is isomorphic to a Lie group will that group also be a Lie group?

So I was working through some exercises on Lie groups and I was wondering if the group isomorphisms carry any of the differentiable structure with them. Explicitly, if two groups, $G$ and $H$ are isomorphic and $G$ is a Lie group of dimension n can…
EHH
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A question on the codimension of $O(n)$ and $SO(n)$ reative to $GL(n,R)$

This seems like a very silly way to control the website, because as a new user it will not let me make comments. So apologies in advance to reference a previous question, but given the warning not to post answers that are new questions, this is my…
Flint72
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The product of two Lie groups

The Wikipedia entry on Lie Groups states (under the "Construction" section): "The product of two Lie groups is a Lie group." There is no further references or explanations about this statement. What is the "product" that this refers to? For…
Adi Shavit
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In Lie groups, why is left translation a diffeomorphism?

The map $G \times G \rightarrow G$ defined as $(x,y) \mapsto xy$ is differentiable. The left translation is $L_x(y) = xy$. To show that it is a diffeomorphism, we need it to be a bijection, that is both differentiable and has a differentiable…
user124398
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A question on Lie groups and subgroups

I have the following exercises: (a) Verify directly that the group $SO(n)$ has a natural Lie group structure. (b) Given this structure, prove that $SO(n) \subset GL(n;\mathbb{R})$ is a Lie subgroup. (c) $GL(n;\mathbb{R})$ lies as an open subset…
user 3462
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Stuck on an example of why a Matrix lie algebra needs to be closed with respect to GL(n,C)

Can someone please help me on this example. $$ G = \left\{ \begin{pmatrix} { e }^{ it } & 0 \\ 0 & { e }^{ 2\pi it } \end{pmatrix},\quad t\in\mathbb{R} \right\} $$ G is a matrix group, but not a matrix Lie group, why not? I cannot think of a…
Harch
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How to show that $\mathbb{Z}[1/p]$ is a discrete subgroup of $\mathbb{R} \times \mathbb{Q}_p$?

By definition, a discrete subgroup $\Gamma$ of a Lie group $G$ is a subgroup such that there is some open neighborhood $U$ of the identity $e$ such that $\Gamma \cap U = \{e\}$. How to show that $\mathbb{Z}[1/p]$ is a discrete subgroup of…
LJR
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All compact, connected, complex Lie groups are Abelian.

Here I found one interesting result about connected, complex Lie groups. In the 1 answer, the author claims that this fact is a consequence of the maximus modulus principle, but I can not understand why this is so.
AndrewGap
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Does every Lie group have a Cartan subgroup?

I recently came across the concept of Cartan subgroup while studying physics. I took the definition from Wikipedia that a Cartan subgroup is a Lie sub-group whose Lie algebra is a Cartan subalgebra. We know Cartan subalgebras exist for…
Rescy_
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are all $G/H$ embeddable in a linear representation?

Say $G$ is a connected Lie group. The orbits in any linear representation are of the form $G/H$, where $H$ is the stabilizer of any element of the orbit. Which $G/H$ appear this way? edit: It seems the compact case works whenever $G/H$ is nice…
wzzx
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Can maximal compact subgroups of subgroups be extended to maximal compact subgroups of the ambient Lie group?

Given a Lie group $G$ and a closed subgroup $U \subset G$: Is it always possible to choose a maximal compact subgroup $K \subset G$ of $G$ such that $K \cap U$ is again a maximal compact subgroup of $U$? Or equivalently: Given a maximal compact…
Akim Eismann
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Show that the kernel of the adjoint representation of a connected Lie group is its center.

Let $G$ be a connected Lie group with Lie algebra $\mathfrak g$. I'm trying to proof the following assertion: the kernel of the adjoint representation of $G$ coincides with its center. What I have done so far: Let's recall that $Ad: G \to Aut(G)$…
user 242964
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Using the Cayley-Hamilton Theorem to rewrite formally a power matrix series.

They tell that if $f(X)=\sum_{k}\alpha X^{k}$ is a power matrix series then use the Cayley-Hamilton theorem to rewrite formally $f(X)$ as : $$ f(X) = c_{0}(X)+ c_{1}(X)X+\cdots+c_{n-1}(X)X^{n-1}$$ Where $c_{j}(X)$ are (multiple) power series in…
Lucía Osorio
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Conjugacy class in a group Lie

I have one question: Let $G$ is a group Lie and $H$ is closed subgroup. Let $M=\cup g^{-1}Hg < G$. Is it true that $M$ is manifold? What is the dimension of $M$? upd. Let $G$ is a compact. I have a hypothesis that, in this case $M$ is manifold, and…
Alex-omsk
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$\mathrm{Sp}(4, \mathbb{C})$ is simply connected

I'm trying to prove that $\mathrm{Sp}(4, \mathbb{C})$ is simply connected. Note that it is a group of complex $4\times 4$ matrices $A$ satisfying $A^{T}JA = J$, where $$ J = \begin{pmatrix} O & I_{2} \\ -I_{2} & O \end{pmatrix} $$ What I've got so…
Seewoo Lee
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