Consider a complex Grassmannian $Gr_{k}(C^{n})$, which is a symmetric space with symmetry group $U(n)$ (i.e. unitary group). Consider a subspace $S_{0}$ of the Grassmannian determined by the canonical inclusion $C^{k}\rightarrow C^{n}$. The isotropy group of $S^{0}$ is the group $H=U(k)\times U(n-k)$.

My question is: why the isotropy group of $S^{0}$ is the group $H=U(k)\times U(n-k)$? I have a hunch on why there is a sign "$\times$" in the expression of $H$. But that's it. I have no clue on why the isotropy group is the way it is.

The isotropy group of group $G$ for element $x\in G$ is the group: $$I_{x}=\left \{ g\in G:gx=x \right \}$$

For example if we have any invertable a matrix as an element $x$, then the equality $Ax=x$ is satisfied by the unit matrix. But this logic does not lead to anything like the group $H$ in the $S_{0}$ case.


You're writing the Grassmannian as a homogeneous space of the unitary group, so the isotropy group of the point $S_0$ represented by the $k$-dimensional subspace $\mathbf{C}^k \times\{\mathbf{0}^{n-k}\} \subset \mathbf{C}^n$ is the group of unitary transformations leaving $\mathbf{C}^k \times\{\mathbf{0}^{n-k}\}$ invariant, hence leaving its unitary complement $\{\mathbf{0}^k\} \times \mathbf{C}^{n-k} \simeq \mathbf{C}^{n-k}$ invariant.


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