# Thread: Basis of the tangent space of unitary matrices

1. ## Basis of the tangent space of unitary matrices

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

2. Originally Posted by Siknature
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

This can't be right, imo, since the tangent space at the identity of the unitary matrices are the skew-symmetric, or anti-symmetric(anti-hermitian, in the complex case) matrices, and both $s_0\,,\,s_1$ aren't skew-symmetric...

Tonioi

3. Originally Posted by tonio
This can't be right, imo, since the tangent space at the identity of the unitary matrices are the skew-symmetric, or anti-symmetric(anti-hermitian, in the complex case) matrices, and both $s_0\,,\,s_1$ aren't skew-symmetric...

Tonioi
I think you must have misunderstood because i dont think that what you have said is true.

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.

Thanks

4. Originally Posted by Siknature
I think you must have misunderstood because i dont think that what you have said is true.

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.

Thanks

Why would you take the conjugate? Are you doing inner spaces linear algebra? This belongs to Lie groups and stuff, and a matrix $A$ is skew-symmetric iff $A^T=-A$ , without conjugates and stuff, and this forces the diagonal to be all zeroes...

Tonio

5. Originally Posted by tonio
Why would you take the conjugate? Are you doing inner spaces linear algebra? This belongs to Lie groups and stuff, and a matrix $A$ is skew-symmetric iff $A^T=-A$ , without conjugates and stuff, and this forces the diagonal to be all zeroes...

Tonio
Skew-Hermitian matrix

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.[1] That is, the matrix A is skew-Hermitian if it satisfies the relation
where denotes the conjugate transpose of a matrix.

6. Originally Posted by Siknature
Skew-Hermitian matrix

Anyway, and as for your question: $s_0\,,\,s_1\,,\,s_2\,,\,s_3$ are a basis for this tangent space since these are four $\mathbb{R}$-lin. indenpendent skew-hermitian matrices, and 4 is the dimension of $U(2)$ and of its tangent space at the identity.