Tangent space of $U(n+1)/U(1)\times U(n)$ Let $U(n)=\{X\in gl(n,\mathbb{C})|XX^*=I_n\},$ where $gl(n,\mathbb{C})$ is the set of all $n\times n$ matrices whose entries are complex numbers, $X^*=\bar{X}^T$ and $I_n$ is the identity matrix of order $n$.
I want to compute the tangent space $T_0\left(U(n+1)/U(1)\times U(n)\right),$ where $0=[I_n]$ is the coset of $I_n$ in $U(n+1)/U(1)\times U(n).$ I don't know how to compute it. But I know the result is $$T_0\left(U(n+1)/U(1)\times U(n)\right)=\left\{\left.\begin{pmatrix}0&a\\-a^*&0\end{pmatrix}\right|a\in\mathbb{C}^n\right\}$$
 A: The natural interpretation of $T_0(U(n+1)/(U(1)\times U(n))$ is as the quotient $\mathfrak u(n+1)/(\mathfrak u(1)\times\mathfrak u(n))$ of the Lie algebras. This comes from the fact that the projection $p:U(n+1)\to U(n)/(U(1)\times U(n))$ is a submersion, so its derivative at $I_n$ defines a surjection $\mathfrak u(n+1)\to T_0(U(n+1)/(U(1)\times U(n))$ and the kernel is clearly the indicated Lie subalgebra. Now you could take any linear subspace in $\mathfrak u(n+1)$ and restrict the derivative to this subspace to obtain an identification with $T_0(U(n+1)/(U(1)\times U(n))$.
The space you indicate in the question is such a complement and it is a particularly natural choice, since it is invariant under the action of the subgroup $U(1)\times U(n)$ (coming from the restriction of the adjoint action of $U(n+1)$. This also has the advantage that the canonical action of $U(1)\times U(n)$ on the quotient $\mathfrak u(n+1)/(\mathfrak u(1)\times\mathfrak u(n))$ is isomrphic to the action on that specific.
