# linear algebra – dimension of a unitary matrix

Suppose I have a hilbert space $$V$$ of dimension $$N = 7$$ which carries a unitary representation $$rho$$ of $$C_ {2v}$$ group of points.

In particular, suppose $$rho$$ divides the space $$V$$ so that $$V cong 2A + B_1 + 2B_2 + 2B_3$$.

The variety of unit operators $$U$$ sure $$V$$ commute with $$rho$$ should have a dimension $$2 ^ 2 + 2 ^ 2 + 2 ^ 2 + 1 ^ 2 = 13$$ since such $$U$$ can be broken in $$U_A + U_ {B_1} + U_ {B_2} + U_ {B_3}$$ where each component acts only on the corresponding subspace.

What is the size of the variety of "superoperators" $$tilde {U}$$ formed of the $$U$$ by the rule
$$X to tilde {U} X: = U X U ^ Dagger + U & # 39; {} ^ Dagger$$

Or $$U # stag { pi} {2}} U ^ dagger _ { frac { pi} {2}}$$ and $$O _ { frac { pi} {2}}$$ is the element of a representation $$rho & # 39;$$ of $$C_ {4v}$$ group of points corresponding to $$90 ^ circ$$ rotation.

$$rho$$ is obtained from $$rho & # 39;$$ by restriction of $$C_ {4v}$$ to the subgroup $$C_ {2v}$$.

I understand that the dimension can not be bigger than $$13$$ but it's difficult for me to know if the operation of "quadrature" $$U to U ( cdot) U ^ Dagger$$ or the $$90 ^ circ$$ The "averaging" serves to reduce the dimensionality of the space of possibilities $$tilde {U}$$.

Also what is the dimension of the space of $$tilde {U}$$ if we only allow it to operate on the operator space totally symmetrical with respect to $$rho$$ (that is to say, symmetry $$A$$)?
I believe this is equivalent to defining the $$tilde {U}$$ to be instead
$$X to tilde {U} X: = sum_ {n = A, B_1, B_2, B_3} U_n X U_n ^ dagger + U & nbsp; _n X U _n {} ^ dagger$$