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

1. ## 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

2. 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 $\frac{n(n-1)}{2}$ (half of the off-diagonal entries can be arbitrarily chosen). So this must be the dimension of $U(n)$.

3. Originally Posted by Laurent
The tangent space at 0 is the space of antisymmetric matrices, which has dimension $\frac{n(n-1)}{2}$ (half of the off-diagonal entries can be arbitrarily chosen). So this must be the dimension of $U(n)$.

How do you find the tangent space at a point?

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

4. Originally Posted by Siknature
To get the tangent space at $A$, differentiate at $A$ the equation defining implicitly the manifold (here, $A^TA=I$) : it is the space of matrices $H$ such that $A^T H+H^T A=0$. If $A=I$, this reduces to $H=-H^T$. (More formally, the tangent space at A is the kernel of the differential at A of a (submersive?) map $\phi$ such that the manifold is a level set of $\phi$ (at least locally at $A$))