# Cover of a Grassmannian by an open set

I am reading this document here and in exercise 1, the author asks to show the Grassmannian $G(r,d)$ in a $d$ dimensional vector space $V$ has dimension $r(d-r)$ as follows. For each $W \in G(r,d)$ choose $V_W$ of dimension $d-r$ that intersects $W$ trivially, and show one has a bijection

$$\{ \text{Subspaces of dimension r that intersect V_W trivially}\} \leftrightarrow \operatorname{Hom}(W,V_W).$$

Now I can set up a bijection as follows. For each $U$ on the left, we have $U \oplus V_W = V$. Thus any $w \in W$ can be written uniquely as $u + v_w$ for $u\in U$ and $v_w \in V_W$ and so we obtain a linear map $T : W \to V_W$ by sending $w$ to this $v_w$. Conversely for any linear map $T \in \text{Hom}(W,V_W)$ we have a subspace on the left consisting of vectors $w + T(w)$ for $w \in W$.

My question is: The author has not defined why the set on the left hand side above is an open set. Presumably he meant that if we look in $\Bbb{P}(V)$, the set of all linear varieties of dimension $r -1$ that do intersect a given linear variety of dimension $d-r - 1$ is closed (in the Zariski topology), but why is this so?

-

One way to think about this is to pass to the frames (so we're looking at a Stiefel variety). Fix a basis $u_{r+1},\dots,u_d$ for $V_W$. The set of frames $v_1,\dots,v_r\in V$ so that $v_1,\dots,v_r,u_{r+1},\dots,u_d$ form a linearly independent set is an open set ($\det\ne 0$) in the space of all $r$-frames. This descends, taking quotients, to an open set in $G(r,d)$.