How Is U(n) Embedded in O(2n) as a Subgroup?

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The discussion focuses on the embedding of the Lie algebra of U(n) into the Lie algebra of O(2n). It explains that this embedding arises from identifying complex matrices with real matrices through an isomorphism between C^n and R^2n. The conversation highlights that U(n) can be represented as a subgroup of O(2n) and provides references to symplectic geometry literature for further reading. Additionally, a question is raised about the existence of an element in O(2n) that maintains the structure of the Lie algebra under the adjoint action, with a tentative conclusion that such an element may not exist. The thread emphasizes the mathematical intricacies of this embedding and its implications in the context of Lie algebras.
timb00
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Hi everybody,

I hope that I chose the right Forum for my question. As the title might suggest, I am interested in the embedding of the Lie algebra of U(n) into the Lie Algebra of O(2n). In connection with this it would be interesting to understand the resulting embedding of U(n) in O(2n). I tried to find something in the literature, but I didn't succeed.

Since the application is far away I can't really explain the reason for my question. Eventually I start with a matrix which is an Element of Lie(U(n)) and want to understand how this matrix is embedded in Lie(O(2n)).

Here O(2n) means the the symmetry group of a scalar product on a real vector space of dimension 2n.

Maybe someone of you knows a good book or a publication that is related to this embedding and explains it.

Thanks for reading and perhaps for answering,

Timb00
 
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Hello timb00. This folllows from the natural identification of C with R² that sends a+ib to (a,b). First extend this to an isomorphism (of real vector spaces) btw C^n with R^2n. Then through conjugation by this iso, endomorphisms of C^n can be identified with endomorphisms of R^2n.* This defines an embedding of Mat(n,C) in Mat (2n,R). It is not hars to show that U(n) lands precisely in GL(n,C) n O(2n,R) = GL(n,C) n SP(2n,R) = O(2n) n Sp(2n,R). Therefor as a lie subalgebra of o(2n,R), u(n) is o(2n,R) n sp(2n,R).

*You should find that a complex nxn matrix with ij-th entry is a+ib corresponds to the real 2nx2n matrix obtained by remplacing a+ib by the 2x2 sub matrix

a -b
b a

P.S. A source for this stuff is the symplectic geometry books by McDuff-Salamon and by Anna Cana da Silva. The later is free online on her website.
 
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Hi,

thank you quasar987 and dextercioby for your replay. I will have a look on the sources you recommended to me. Till now I only though about it using Dynkin diagrams.

Maybe a last question. If u is an element of the lie algebra of U(n), say an n x n skew hermitian matrix. Exist there an element of O of O(2n) such that the adjoint action Ad(O)u = OuO^T is an element of the lie algebra of O(2n)?

During I am writting it, I think the answer is: No! But maybe someone has an explanation why ?
 

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