Let $k$ be an algebraically closed field, and let $r\le n$ be positive integers. Let $\mathrm{Grass}(r,n)$ be the projective variety of all $r$-dimensional planes in $k^n$. Notice that $\mathrm{Grass}(1,n)\cong\mathbb{P}^{n-1}(k)$.
For $r>1$, we have $\dim(\mathrm{Grass}(r,n))=r(n-r)<r(n-1)=\dim(\mathbb{P}^{n-1}(k))^{\times r}$ from which it follows that $\mathrm{Grass}(r,n)\not\cong(\mathbb{P}^{n-1}(k))^{\times r}$. However, can we see $\mathrm{Grass}(r,n)$ as a subvariety of $(\mathbb{P}^{n-1}(k))^{\times r}$?
Invoking the axiom of choice, if for each $r$-dimensional plane $W$ we choose a basis $\{w_1,\ldots,w_r\}$, then the map $W\mapsto ([w_1],\ldots,[w_r])$ is an injective mapping of sets, but unless there is some coherence to how the bases are chosen as $W$ ranges over all $r$-dimensional planes, this map will not identify $\mathrm{Grass}(r,n)$ with a subvariety of $(\mathbb{P}^{n-1}(k))^{\times r}$. Is there some natural way to choose a basis for an $r$-dimensional plane $W$, or is there another way to view $\mathrm{Grass}(r,n)$ as a subvariety of $(\mathbb{P}^{n-1}(k))^{\times r}$?