92
votes
Accepted
What is Lie Theory/ a Lie Group, simply?
This was going to be a comment, but it got too long.
I don't know how "simple" this is, but the 5 word summary is "a group with manifold structure". Or perhaps if you're a ...
- 36.7k
40
votes
Accepted
Under what conditions is the exponential map on a Lie algebra injective?
There is a complete characterization, in a large part due to Dixmier and Saito (both independently in 1957):
If $G$ is a real (finite-dimensional) Lie group with Lie algebra $\mathfrak{g}$, then ...
- 17.7k
36
votes
Accepted
Can every manifold be turned into a Lie group?
There is an easy counterexample: $S^2$ cannot be given a Lie group structure (this is a consequence of the hairy ball theorem). The problem with your construction is that it doesn't offer how to ...
- 2,067
32
votes
Accepted
When will two isomorphic Lie algebras have the same representation?
Short answer: If two Lie algebras are isomorphic, they have "the same" complex representations. A real semisimple Lie algebra and its complexification also have "the same" complex representations, but ...
- 25.3k
32
votes
Accepted
Do Lie algebras "know" about their Lie groups?
In principle it is possible to determine every (edit: connected) Lie group having Lie algebra some finite-dimensional real Lie algebra $\mathfrak{g}$. As radekzak says in the comments, first we find ...
- 415k
29
votes
Accepted
Is the Lie Algebra of a connected abelian group abelian?
Yes, and connectedness is not necessary. I know three proofs:
Proof 1
When $G$ is abelian, the inverse map
$$i:G\to G,\quad g\mapsto g^{-1}$$
is a group homomorphism. Hence, its differential at $1\...
- 19.2k
28
votes
Can every manifold be turned into a Lie group?
Lie groups as manifolds, are very special, owing to the group operations. Basically, "what happens at the identity" determines what happens everywhere. And this means that the tangent bundle $T G$ is ...
- 28.8k
27
votes
Prove that $SL(2,\mathbb{R})$ acts transitively on the upper half plane
Let $z=x+iy$ be a given point in the upper half plane. Then the $SL(2,\mathbb{R})$ matrix
$$
\begin{pmatrix}
\sqrt{y}&x/\sqrt{y} \\
0&1/\sqrt{y}
\end{pmatrix}
$$
maps $i$ to $z$. Since $z$ ...
- 371
26
votes
Why do we care about two subgroups being conjugate?
This answer isn't of the form you asked for, but I'm posting it because it's been very helpful to me. I apologize if this is the sort of answer you're trying to avoid.
Let's start with a very ...
- 4,780
25
votes
Accepted
Homotopy groups U(N) and SU(N): $\pi_m(U(N))=\pi_m(SU(N))$
You are correct. This follows from the fact that $U(N)$ is diffeomorphic to $S^1\times SU(N)$. To see this, consider the map $f : U(N) \to S^1\times SU(N)$ given by $A \mapsto (\det A, \operatorname{...
- 97.9k
24
votes
Are all Lie groups Matrix Lie groups?
As other answers mention, it is not true that any Lie group is a matrix group; counterexamples include the universal cover of $SL_2(\mathbb{R})$ and the metaplectic group.
However it is true that all ...
- 3,917
24
votes
Accepted
Can you give me an example of topological group which is not a Lie group.
Another example is the rationals under addition with the subspace topology induced from inclusion in $\mathbb R$. Since the rationals are countable, they can't be a manifold of dimension exceeding $0$,...
- 27k
24
votes
Can every manifold be turned into a Lie group?
To add to the previous answers, topological groups have abelian fundamental groups.
$G$ is Topological $\implies$ $\pi_1(G,e)$ is Abelian
Orientable surfaces of genus at least two are not ...
- 1,437
24
votes
Accepted
Why is enveloping algebra called enveloping algebra?
1. The very first thing that is crucial to have in mind is that every associative algebra (where we have a vector space addition $+$, and a multiplication $\cdot$) can be turned into (or let's better ...
- 25.3k
22
votes
Quaternion–Spinor relationship?
Some background in Lie theory first. To any Lie group $G$ one may associate its lie algebra $\mathfrak{g}$. Among all isomorphism classes of Lie groups $G$ with a given lie algebra $\mathfrak{g}$, ...
- 151k
22
votes
Accepted
Are all Lie groups Matrix Lie groups?
Not all Lie groups are matrix groups. Consider the metaplectic group. From wikipedia:
The metaplectic group $M_{p_2}(\mathbb{R})$ is not a matrix group: it has no faithful finite-dimensional ...
- 1,017
22
votes
Why do we care about two subgroups being conjugate?
$H_1$ and $H_2$ are conjugate as subgroups iff $G/H_1$ and $G/H_2$ are isomorphic as $G$-sets.
Edit: Two related settings where this condition shows up are Galois theory and covering space theory. ...
- 415k
22
votes
Differential of the multiplication and inverse maps on a Lie group
One of Tu's previous exercises, ex. $8.7^*$ shows that for $M,N$ manifolds, by defining $\pi_1:M\times N\to M$ and $\pi_2: M\times N \to N$, one can show for $(p,q)\in M\times N$ that $$\pi_{1*}\times\...
- 4,305
21
votes
Accepted
Does a Lie group's group structure (not Lie group structure) determine its topology?
$\mathbb{R}^n$ and $\mathbb{R}^m$ are abstractly isomorphic (assuming the axiom of choice) for $n \neq m$ but not homeomorphic and so not isomorphic as topological groups.
I think this might be the ...
- 415k
20
votes
prove $RP^3\cong SO(3)$
Each rotation in $\Bbb R^3$ is characterized by an "oriented axis" $v\in S^2$ and an angle $\varphi\in [0,\pi]$ and the only relations are $(v,\pi)=(-v,\pi)$ and $(v,0)=(w,0)$ for each $v,w\in S^2$. ...
- 11.4k
18
votes
Prove that the manifold $SO(n)$ is connected
We have that $S^n \cong SO(n+1)/SO(n)$.
Using the fact that:
If $H <G $ is a closed subgroup and both $H$ and $G/H$ are connected,
then $G$ is connected,
the claim follows by induction.
- 33.9k
18
votes
Accepted
Relations between center (fundamental group) and (co)root and weight lattices for Lie groups
To see such relations more clearly, one should first of all avoid identifying the weight space $\Lambda\otimes\Bbb R$ with its dual space. Even in the semisimple case, an invariant inner product is ...
- 114k
18
votes
Accepted
Is there a way to associate a Lie algebra to the group of diffeomorphisms?
This question certainly is too broad to be answered completely. Before talking about infinite dimensional Lie groups you need a concept of infinite dimensional manifolds. As you note correctly, you ...
- 20k
18
votes
Accepted
Lie algebra: intuition of "Lie Algebra is tangent space of corresponding Lie Group"?
Intuitively, you can think of the tangent space of a surface at a point as the space of all "directions" you can move from that point (with all velocities) while staying on the surface. That ...
- 2,215
18
votes
Accepted
Name and layperson's explanation for an E8 group diagram.
The diagram you posted is not a Dynkin diagram, but is the projection of the convex hull of the root system $E_8$ in the Coxeter plane, i.e. the invariant plane for the action of the Coxeter element ...
- 5,041
17
votes
Accepted
Prove that the manifold $SO(n)$ is connected
Given two vectors $v,w$ of same length (say of length $1$) in $\mathbb{R}^n$, let's first verify that there is indeed a path in $\gamma(t) \in SO(n)$ so that $\gamma(t)v$ "starts off" at $v$ and "...
- 186
17
votes
Why do we care about two subgroups being conjugate?
I will not attempt a complete answer, but just as an example, if you are studying finite simple groups $G$ (which lots of people do), then one of the first questions you might ask about $G$ is: what ...
- 87.8k
17
votes
Accepted
What is symplectic geometry?
Quick "fake" answer: In classical mechanics one usually describes a particle measuring its position $q_1, \dots, q_n$ and momentum $p_1, \dots, p_n$. To describe how these change one needs to ...
- 3,098
17
votes
Accepted
Example of a representation of $U(1)$ with $n>2$
Some remarks on language
A representation of a group $G$ is indeed what you describe: a linear map $R$ from $G$ to a group $GL(n, \mathbb{C})$ of matrices. Most of the time, the vector space on which ...
- 10.3k
16
votes
Why do we care about two subgroups being conjugate?
I don't understand why "conjugacy" is an equivalence relation we care about, beyond the fact that it is stronger than "abstractly isomorphic."
An observational fact about conjugacy: things are ...
- 151k
Only top scored, non community-wiki answers of a minimum length are eligible