Thread: What is the dimension of U(n), the group of all n x n unitary matrices?

I am not sure how to find the dimension of U(n).

Suggestions i have been given are: ((n^2) - n)) / 2

or: 2(n^2)


But i don't know which (if either) of these is correct, and how they are found.

Thanks for any help

Originally Posted by Siknature

I am not sure how to find the dimension of U(n).

Suggestions i have been given are: ((n^2) - n)) / 2

or: 2(n^2)


But i don't know which (if either) of these is correct, and how they are found.

Thanks for any help

The tangent space at 0 is the space of antisymmetric matrices, which has dimension (half of the off-diagonal entries can be arbitrarily chosen). So this must be the dimension of .

Originally Posted by Laurent

The tangent space at 0 is the space of antisymmetric matrices, which has dimension (half of the off-diagonal entries can be arbitrarily chosen). So this must be the dimension of .

Thanks for your reply.

How do you find the tangent space at a point?

I'm interested in finding the tangent space at the identity.

Originally Posted by Siknature

Thanks for your reply.

How do you find the tangent space at a point?

I'm interested in finding the tangent space at the identity.

Yes, in fact I mean "tangent space at the identity" in my post...

To get the tangent space at , differentiate at the equation defining implicitly the manifold (here, ) : it is the space of matrices such that . If , this reduces to . (More formally, the tangent space at A is the kernel of the differential at A of a (submersive?) map such that the manifold is a level set of (at least locally at ))