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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.
4
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1
answer
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A Manifold for which $\chi^{\infty}(M)$ is rich
Is there a manifold $M$ for which $\chi^{\infty} (M)$, the lie algebra of smooth vector fields on $M$ contains all finite dimensional Lie algebras(Up to isomorphism)?
A weaker question:
Is there a …
1
vote
1
answer
138
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Manifold_Lie algebra compatibility
In this question we try to improve some parts of this post as follows:
What is an example of a manifold $M$ and a Lie algebra $L$ (with the same dimension) such that $M$ does not admit …
3
votes
1
answer
667
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A particular embedding of a Lie group in Euclidean space
I apologize in advance if my question is elementary.
Before I present my question I mention my motivation:
Motivation:
A Lie group is a manifold. At the same time it is a Riemannian manifold equippe …
1
vote
1
answer
304
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A Lie algebra associated to a symplectic manifold
Let $(M,\omega)$ be a symplectic manifold of dimension $2n$ with the volume form $\omega^n.$
In this question we associate a Lie algebra $L(M,\omega)$ to $(M,\omega)$. Then we are interested …
1
vote
2
answers
313
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A question on involutions on the Lie algebra of vector fields
Edite According to the essential comment of Ian Agol I revise the question as follows
For a smooth manifold $M$, is there a non identity involution $\theta$ on the lie algebra $\chi^{\infty}(M)$ s …
8
votes
2
answers
373
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The minimum codimension of Lie subalgebra of $\chi^{\infty}(M)$
Assume that $M$ is an arbitrary manifold.
Is there a Lie subalgebra of $\chi^{\infty}(M)$, the space of smooth vector fields on $M$, whose codimension is equal to one?
If not, what is a counter …
0
votes
0
answers
143
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A question on lie groups( Lie algebras)
What is an example of a compact Lie group $G$ which is not isomorphic to a product of two nontrivial lie groups and satisfies the following property:
There are two non zero vector fields $X, Y \in …
0
votes
2
answers
284
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A question on Lie algebras
To what extent, the following types of Lie algebras are classified :
Those Lie algebras $L$ such that every Lie Group $G$ with $Li(G)\sim L$, is necessarily compact.
0
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1
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166
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Comparison of two infinite dimensional Lie Algebras
Is there an example of a real analytic (compact) manifold $M$ such that the following two lie algebras are isomorphic Lie algebras:
$\chi^{\infty}(M)$, the Lie algebra of all smooth vector fields o …
1
vote
0
answers
136
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Alternative Lie bracket on $\chi^{\infty}(M)$ where $M$ is a Riemannian manifold with a symp...
Let $(M,\omega, g)$ be a Riemannian symplectic manifold. The Poisson bracket of two functions $f,g$ is denoted by $\{f,g\}$. The Hamiltonian vector field and gradient vector field …
4
votes
1
answer
284
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A different Lie algebra structure on $\chi^{\infty}(\mathbb{R}^{2})$
In this question $\chi^{\infty}(\mathbb{R}^{2})$ or $\chi^{\infty}(S^{2})$ is the space of all smooth vector fields on the plane or sphere. A limit cycle for a vector field $X$ is an isolated closed …
7
votes
1
answer
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The Hessian of invariant functions on a Lie group
Assume that $G$ is a Lie group with Lie algebra $\mathfrak{g}$. We fix an invariant Riemannian metric on $G$ and fix its corresponding $LC$ connection.
Consider the natural right action of …
1
vote
0
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164
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Rotation number for homeomorphisms of a Lie group other than $S^1$
Let $G$ be a Lie group whose Lie algebra is $\mathfrak{g}$ with exponential map $\exp:\mathfrak{g}\to G$.
For what kind of Lie group $G$ the standard process of definition of rotation number …
1
vote
2
answers
144
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Does an analytic tensorial Lie structure on $S^2$ gives a fiberwise Abelian Lie algebra stru...
Motivated by the answer to this question we ask:
Is it true to say that for every real analytic tensorial Lie algebra structure $\alpha$ on $\chi^{\infty}(S^2)$, all fibers are necessarily Abe …
0
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Does an analytic tensorial Lie structure on $S^2$ gives a fiberwise Abelian Lie algebra stru...
No. there is an analytic $(1,2)$ tensor on $S^2$ which satisfies the Jacobi identity. Its restriction to the equator is abelian. It is non abelian at points out of equator.
The construction is simi …