Isomorphism between groups and their Lie Algebra

raopeng
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I must apologize if this question sounds dump but if an isomorphism is established between two groups, is it true that their lie algebra is an isomorphism too? My idea is that since the tangent space is sent to tangent space also in the matrix space, both groups' lie algebra will be isomorphism also. For example we have: SU(2) × SU(2) ≈ SO(4) and su(2) × su(2) ≈ so(4).
 
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Hi raopeng,

Could you explain a little bit more?

Best regards

GoodSpirit
 
For example we have an isomorphism between two groups, and we know that Lie Algebra of a group is the commutator of matrix of its tangent space at the identity. So if there exists an isomorphism, wouldn't that entail that, since the tangent space is sent to tangent space under the mapping, their lie algebra is isomorphic too? I have this idea when trying to establish an isomorphism between su(2) x su(2) ≈ so(4) and there is an isomorphism between SU(2) x SU(2) ≈ SO(4)
 
A lie group isomorphism f between lie groups G and H will have full rank, so that the corresponding map between lie algebras df is an isomorphism. However, the converse is false, two lie groups can have the same lie algebra but be non isomorphic lie groups.
Groups that are locally diffeomorphic at the identity will have isomorphic lie algebras
 
raopeng said:
I must apologize if this question sounds dump but if an isomorphism is established between two groups, is it true that their lie algebra is an isomorphism too? My idea is that since the tangent space is sent to tangent space also in the matrix space, both groups' lie algebra will be isomorphism also. For example we have: SU(2) × SU(2) ≈ SO(4) and su(2) × su(2) ≈ so(4).

you need to show that df[X,Y] = [df(X),dfY)] which implies that the linear isomorphism of left invariant vector fields is a Lie algebra homomorphism.
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...

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