In these two questions we try to find a relation between vector bundle theory and Lie algebras:(Note that we identify a finite dimensional Lie algebra L with R^n or C^n)
1)Assume that $L$ is a n dimensional real (complex) Lie algebra such that O(n) (U(n)) is a subgroup of Aut(L). Is L necessarily an Abelian Lie algebra? Of course every Lie algebra with this property satisfies the following:
The structure group of every vector bundle E over compact space X can be reduced to Aut(L).
2) Assume that E is a n-dimensional vector bundle over a compact space X such that for every n- dimensional Lie algebra L, the structure group of E can be reduced to Aut(L). Is E a trivial vector bundle? I confess that perhaps this second question is very broad and general, so we can consider this second part in low dimensional case.
This is a good question with more than one possible answer.
Lie groups on vector bundles are considered in
Y. Kosmann-Schwarzbach and K. C. H. Mackenzie, DIFFERENTIAL OPERATORS AND ACTIONS OF LIE ALGEBROIDS, Contem. Math. 2000:
http://www.math.polytechnique.fr/cmat/kosmann/diff.pdf
In particular Lie algebroids and Lie groups on vector bundles are considered. Actions of Lie groups on a vector bundle give rise to representations in its space of sections by semi-linear isomorphisms. See Section 1, starting on page 4, on Lie algebroids on vector bundles (see Def. 1.3 on page 4 and Theorem 1.5 on page 5). See Section 5, starting on page 16, on Lie groups on vector bundles.
Lie algebroids on a vector bundle are defined by this generalized Nijenhuis-Richardson algebra are considered in
Qunfeng Yang, Some Graded Lie Algebra Structures Associated with Lie Algebras and Lie Algebroids, Ph.D. thesis, University of Toronto, 1999:
https://tspace.library.utoronto.ca/bitstream/1807/13323/1/NQ41350.pdf
A good introduction to Lie groups on vector bundles is given in
Eckhard Meinrenken, Group actions on manifolds, 2003:
http://www.math.toronto.edu/mein/teaching/action.pdf
Dear Ali Taghavi,
I sent you interesting article.
https://math.berkeley.edu/~jawolf/publications.pdf/paper_038.pdf
Regards, Shafagat
Dear Ali:
The answer to your first question is Yes: only abelian Lie algebras can have O_n (similar: U_n) as group of automorphisms. Let us consider the real case. The Lie bracket on L = R^n is a linear map F : L \wedge L \to L (where L\wedge L denotes the antisymmetric tensors or endomorphims on L), but on the other hand, L\wedge L is the Lie algebra of O_n (antisymmetric matrices). If O_n acts by automorphisms, then F is equivariant with respect to O_n, i.e. F transforms the O_n-representation on L\wedge L into the one on L. But the first one is the adjoint representation on the Lie algebra L\wege L = o_n which is irreducible. Thus F is either zero or it has zero kernel (the kernel would be subrepresentation, but there is none in an irreducible representation). Hence L\wedge L must be a subrepresentation of L. Counting dimensions, this is possible only if dim L = 3. Then there is in fact one such map F, the cross product (vector product) on L = R^3 (= the Lie algebra of SO_3). But O_3 is not in the automorphism group of the cross product since reflections do not preserve orientation, note that (a,b,a x b) is an oriented orthonormal basis if a,b is an orthonormal 2-frame. Thus F = 0.
Best regards
Jost
ِDear Prof. Peters
Thank you very much for your very interesting answer and very helpful links.
Dear Prof. Mahmudova Thank you very much for your very interesting link.
Dear Prof. Eschenburg thank you very much for your very elegant argument.
I thank you all, for your attention.
This idea is associated with tangent bundles and abelian Lie algebra structures..
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