U(n) is the group of all n x n unitary matrices.
Say TEU(n) is the tangent space at the identity. (E just refers to "at the identity").
For n = 2, how do i show that the following 2 x 2 matrices make a basis of TEU(2)?:
s0 = (first row: i, 0 /second row:0 , i)
s1 = (first row: i , 0 /second row:0, -i)
s2 = (first row: 0,1 / second row: -1,0 )
s3 = (first row:0 , i / second row: i , 0)
Thanks for any help
I am sure that s0 and s1 do belong to the tangent space.
For example s1 = (first row: i, 0 /second row:0 , -i)
So s1 transposed is the same.
Then taking the conjugate i get: (first row:-i, 0 /second row:0, i)
This is clearly equal to -s1 which is required for it to be skew hermitian.
Check s0 in the same way and it can be seen that s0 also skew hermitian.
Maybe you thought the rows were columns?
Nevertheless, i believe these matrices must make a basis of TEU(2) but i dont know how to do this.
From Wikipedia, the free encyclopedia
In linear algebra, a square matrix with complex entries is said to be skew-Hermitian or antihermitian if its conjugate transpose is equal to its negative. That is, the matrix A is skew-Hermitian if it satisfies the relation
where denotes the conjugate transpose of a matrix.
Oh, I see where the confusion is, and I'm the one which first forced it since I didn't pay attention to the fact that skew-hermitian is the complex equivalent to the real skew-symmetry, , and all the time along I was thinking thus of orthogonal matrices instead of complex ones.
Anyway, and as for your question: are a basis for this tangent space since these are four -lin. indenpendent skew-hermitian matrices, and 4 is the dimension of and of its tangent space at the identity.