TITLE: Number of fixed points of torus action over partial flag variety
QUESTION [7 upvotes]: Consider $g\in U(n)$ and $t\in T$, where $T$ is the diagonal maximal torus in $U(n)$. 
Some common manifolds may be obtained as quotients of the $U(n)$ like the complex grassmannian, $Gr(k,n)=U(n)/U(k)\times U(n-k)$ and the manifold of complete flags in $\mathbb{C}^n$, $F_n=U(n)/T$, and there is an induced action of $T$ coming from the action of $U(n)$ in them. Counting the fixed points of this action yields ${n\choose k}$ and $n!$, respectively. These numbers are precisely $|w_n|/(\,|w_{n-k}|\cdot|w_{k}|\,)$ and $|w_n|$, respectively, where $|w_k|$ denotes the order of the Weyl group of $U(k)$, i.e., $w_k=N_{U(k)}(T)/T$.
From this there is a natural candidate to generalize the result for partial flag varieties $U(n)/U(k_1)\times\cdots\times U(k_d)$, where $k_1+\cdots+k_d=n$: the number of fixed points of this action will be
$$
\frac{|w_n|}{|w_{k_1}|\cdots|w_{k_d}|}=\frac{n!}{k_1!\cdots k_d!}=\frac{|N_{U(n)}(T)/T|}{|N_{U(k_1)}(T_1)/T_1|\cdots|N_{U(k_d)}(T_d)/T_d|},
$$
where $T_i$ is the diagonal maximal torus in $U(k_i)$.
I would like some help on how to show this, in other words, how to show that there is a bijection between 
$$
\text{$g\in U(n)$ such that $g^*tg\in U(k_1)\times\cdots U(k_d)$ for all $t\in T$ (fixed points)} 
$$
and 
$$
\frac{N_{U(n)}(T)/T}{N_{U(k_1)}(T_1)/T_1\times\cdots\times N_{U(k_d)}/T_{k_d}}=\frac{N_{U(n)}(T)}{N_{U(k_1)}(T_1)\times\cdots\times N_{U(k_d)}}.
$$
I'v tried brute forcing some conditions on the matrices $g$, yet I don't see anything that is of help, so I think a conceptual proof may be the way to go.

REPLY [3 votes]: Let me propose an absolutely different attempt to this problem.
The idea is to use Lefschetz fixed point theorem.
Recall that it states if $f \colon X \to X$ is an automorphism of a smooth manifold with all fixed points isolated, then
$$ 
\sum_{f(x) =x} i_x = \sum_{k=0}^{n} (-1)^k\operatorname{Tr} f^*|_{H^k(X, \mathbb{R})}
$$
Here $i_x$ are the indices of fixed points.
In our case the automorphism is homotopic to identity (since $T$ is connected), so the RHS of the Lefschetz formula is just the Euler characteristic of your partial flag variety.
One can check up that for any $t \in T$ the map induced by left multiplication by $t$ in fact has only isolated fixed points with all the indices equal to $1$.
Therefore, LHS of the Lefschetz formula is precisely the number of the fixed points.
Thus it is rest to compute the (topological) Euler characteristic of a partial flag variety.  Recall, that any partial flag variety can be seen as $G/P$, where $G$ is a complex algebraic group and $P \subset G$ is a parabolic subgroup. It has cellular Schubert decomposition, that is $G/P = \bigsqcup B_w$ , where all $B_w$ are contractible submanifolds of even dimension numerated by $w \in W/W_P$.  Here $W$ is the Weyl group of $G$ and $W_P$ is the Weyl group of $P$. These are the so-called Schubert cells.
This implies that 
$$b_k(G/P) = \begin{cases} 
0, \text{ if $k$ is odd,}\\ 
\text{number of Schubert cells of real dimension $k$ otherwise} 
\end{cases} $$
and $\chi(G/P) = \sum_{w \in W/W_P} 1 = |W/W_P|$, just as predicted by your hypothesis.