The Grassmannian $\mathrm{Gr}(k, V)$ is the space of all $k$-dimensional subspaces of a (real or complex) vector space $V$. I am interested in the following strategy of proving $\mathrm{Gr}(k, V)$ is a manifold.
Let us suppose that $V$ has the structure of an inner product space. Then each subspace $W \subseteq V$ can be uniquely identified with the orthogonal projection $P : V \to V$ whose image is $W$. Hence, we can view $\mathrm{Gr}(k, V)$ as the space of orthogonal projections $$ \mathrm{Gr}(k, V) = \{P : V \to V \mid P^2 = P, P^* = P\}. $$ The first condition $P^2 = P$ ensures $P$ is a projection, and the second condition $P^* = P$ ensures $P$ is orthonormally diagonalizable.
With this formulation, we see that $\mathrm{Gr}(k, V)$ can be viewed as the preimage of $0$ under the map $F : \mathrm{Sym}_n \to \mathrm{Sym}_n$ given by $$ F(A) = A^2 - A, $$ where $\mathrm{Sym}_n$ is the set of self-adjoint matrices. In particular, if the differential $dF$ had constant rank near $\mathrm{Gr}(k, V)$, then it would follow automatically that $\mathrm{Gr}(k, V)$ has the structure of a smooth manifold.
It is easy to show that $dF : \mathrm{Sym}_n \to \mathrm{Sym}_n$ is given by $$ dF(B) = AB + BA - B. $$ However, I'm not sure how to find the rank of $dF$. Does anyone know how I might proceed? Or is this not an effective strategy of showing that the Grassmannian is a manifold?