# What is the set of fixed point under the action of T in this examole

Let $$M= \lbrace g\lambda g^{-1} ,$$ $$g \in SU(3)\rbrace$$ such that $$\lambda = diag(i\lambda_1, i\lambda_2, i\lambda_3)$$, $$\lambda_i \in \mathbb{R}$$ and $$\lambda_1 > \lambda_2 > \lambda_3 .$$

Let $$T= \lbrace diag (t,t,t') , t,t' \in U(1)\rbrace$$ be the subgroup of SU(3) of diagonal matrices such that the first and the second entries are equal. Consider the action of T on M by conjugation.

What is the set $$M^T=\lbrace m \in M , t.m=m \forall t \in T\rbrace$$ of fixed point in $$M$$ under the action of T on M ?

I greatly appreciate any help in this.

• Please note that askers are expected to provide context for their posts, as is explained here. For example, it would be helpful if you could answer any of the following by editing your post accordingly. Where did you encounter this problem? What are your thoughts on the problem? What have you tried so far? Jan 19 at 16:44
• Actually, the fact that the fixed point set $M^T$ is finite was used in a problem in equivariant cohomology to calculate the integral of an equivariant differential form. But i'm not able to figure out what is the elements of this set . My thoughts so far is that perhaps this has to do with the weyl group of SU(3) and with permutations !
– asma
Jan 19 at 17:15
• It is not true that $M^T$ is finite, maybe you're missing some details of the question. In any case, are you able to characterize the elements of the set $M$? Jan 19 at 17:26
• The only information I have about M is that it is a coadjoint orbit of SU(3) of dimension 6
– asma
Jan 19 at 17:31
• The elements of $M$ are the skew-Hermitian matrices. This is indeed a space of dimension $6$. Jan 19 at 17:33

Hint: We have $$A \cdot B = B$$ if and only if $$AB = BA$$, where $$AB$$ denotes the matrix product of the matrices $$A$$ and $$B$$. Now, consider the specific matrix $$A_0 = \pmatrix{-1 \\ & -1\\ && 1} \in T.$$ If $$B$$ is any matrix such that $$A_0B = BA_0$$, what can we say about the matrix $$B$$? Show that if $$B$$ is of this required form, then we have $$AB = BA$$ for every matrix $$A \in T$$.
Then, note that the set $$M$$ can be nicely characterized with the help of the spectral theorem.