U(n) as subgroup of O(2n)

[timb00]

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

 

[quasar987]

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.

 

[dextercioby]

Since the name is not properly spelled, here's the direct link http://www.math.ist.utl.pt/~acannas/

 

[timb00]

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 ?

 

[blue_raver22]

n

Hello Timb00n,

I can provide some insight on your question. The embedding of the Lie algebra of U(n) into the Lie algebra of O(2n) is a well-studied topic in mathematics. It is known as the Cartan embedding, named after the French mathematician Élie Cartan who first studied it in the early 1900s.

The resulting embedding of U(n) in O(2n) can be understood in terms of the root system of O(2n). The root system is a set of vectors that determine the structure of the Lie algebra. In the case of O(2n), the root system consists of n short roots and n long roots, which correspond to the positive and negative roots of the Lie algebra.

The embedding of U(n) in O(2n) can be described in terms of these roots. The short roots of O(2n) correspond to the roots of U(n), while the long roots correspond to the negative roots of U(n). This means that the Lie algebra of U(n) is embedded in the Lie algebra of O(2n) as a subalgebra.

As for literature, there are many books and publications that discuss this embedding in detail. Some good resources include "Lie Algebras in Particle Physics" by Howard Georgi and "Lie Groups, Lie Algebras, and Representations" by Brian C. Hall.

I hope this helps answer your question and provides a starting point for further exploration. Good luck with your research!