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Lie algebras are algebraic structures which were introduced to study the concept of infinitesimal transformations. The term "Lie algebra" (after Sophus Lie) was introduced by Hermann Weyl in the 1930s. In older texts, the name "infinitesimal group" is used. Related mathematical concepts include Lie groups and differentiable manifolds.
25
votes
Accepted
What is the homomorphism between the third exterior and third symmetric power of the adjoint...
There is a natural homomorphism that I am pretty sure is not zero in general that can be described most easily using the dual language of the Lie algebra $\frak g$:
Recall that there is a differen …
3
votes
Accepted
Smallest subalgebra of $\mathfrak{su}(4)$ arrising from a control problem on $SU(4)$
The answer depends on the values of the constants $J^x$, $J^y$, and $J^z$. Here is what direct computation yields:
If $J^x=J^y=J^z=0$, so that $A=0$, then $B_1$ and $B_2$ span a $2$-dimensional abel …
12
votes
Accepted
A Manifold for which $\chi^{\infty}(M)$ is rich
The answers to your questions are 'no' and 'yes'.
In the first place, there is no finite dimensional manifold whose Lie algebra of vector fields contains all of the Lie algebras ${\frak{sl}}(n,\mathb …
10
votes
Three-dimensional simple Lie algebras over the rationals
Here is my own favorite way of understanding this problem: Let $\bigl(L,[,]\bigr)$ be a $3$-dimensional Lie algebra over a field $K$ (assumed of characteristic $0$ for my comfort, though this probabl …
15
votes
Accepted
Maurer-Cartan structure equation derivation
What you are asking for is an introduction to the theory of $G$-structures, for which the (Maurer-Cartan) structure equations are a basic tool.
There are many sources for this material, starting, o …
8
votes
When is the Lie algebra of automorphisms of a geometrical structure finite-dimensional?
The general procedure for deciding the 'local' version of this question, at least in the real-analytic connected case, was certainly known to Élie Cartan and, probably known to Lie in some form. The …
5
votes
Accepted
Manifold_Lie algebra compatibility
A simple example is to let $M=S^2$ and let $L$ be the nonabelian Lie algebra of dimension $2$. If such an $\alpha$ existed, its range would be a rank-1 subbundle $L\subset TS^2$, but this cannot exis …
5
votes
Curves on $SU(4)$ whose adjoint action on $\mathfrak{su}(4)$ integrates to $0$
It is not clear what you mean by 'find' (give an explicit parametrization?) and 'curve' (continuous? differentiable? smooth? closed?). Certainly, it is possible to 'write down' many curves satisfying …
3
votes
Accepted
Special Riemannian connections?
The answer is 'no, not always'.
Here's an example: Let $E\to M$ be an oriented Riemannian $3$-plane bundle over $M$, with inner product $g$. Then there is a well-defined bilinear cross-product op …
2
votes
Accepted
Largest subgroup of $SU(n)$ for which the adjoint action preserves specific inner product on...
This is really an extended comment, but, because it's too long to put into a comment box and because it may help answer some of the OP's questions, I'm putting it here.
If one endows $\mathrm{SU}(n)$ …
4
votes
Dimension of Span of Adjoint orbit in $\mathfrak{su}(n)$
The span is the same for all $t>0$. The reason is that the curve $\gamma(s) = \mathrm{Ad}_{e^{sA}}(B)$ is a real-analytic curve in a vector space $V$ (in this case, $V={\frak{su}}(n)$). If $\lambda: …
7
votes
Accepted
Trivialization of the vector bundle of centralizers of regular elements
The answer is 'yes', and it follows from Theorem 0.1 in B. Kostant, "Lie group representations on polynomial rings" (American Journal of Mathematics, 85 (1963), 327–404). Here is the argument/constru …
3
votes
Accepted
Simple identity on Lie algebras in a note of Koszul
Are you sure about the factor of $3$ in Koszul's formula? The computation below does not give such a factor, and it gives a stronger result.
Let $X_i$ be any basis of the left-invariant vector field …
2
votes
Accepted
Right Invariant Randers metrics
You are asking about a particular case of the general right invariant Lagrangian for curves on a Lie group. This is a well-known story, but I can summarize it here:
Let $G$ be a Lie group with Lie a …
10
votes
Accepted
SU(6) -> SU(3) branching rule
You appear to have made a mistake in your calculation of the branching rules. The answer given in the wiki is correct, but it seems that you are using the 'wrong' subgroup of $\mathrm{SU}(6)$. Perha …