# Corresponding a vector subspace to a point of the space.

Grassmannians by definition are the parameter spaces for linear subspaces, of a given dimension, in a given vector space $W$. If $G$ is a Grassmannian, and $V_g$ is the subspace of $W$ corresponding to $g$ in $G$, this is already almost the data required for a vector bundle: namely a vector space for each point $g$, varying continuously.

I don't understand the sentence " $V_g$ is the subspace of $W$ corresponding to $g$ in $G$". Indeed I don't understand the notion of corresponding a particular subspace $V_g$ to a point $g$ using the word "The subspace $V_g$ of $W$... This means there is a unique particular subspace $V_g$ for each $g$ !! Thank you for explaining this correspondence!

A Grassmannian $G$ parametrizes the $n$-dimensional subspaces of $W$, and $g\in G$ is the parameter. So you can consider $V_g=V(g)$ to be a function from $G$ to the set of $n$-dimensional subspaces of $W$.
An alternative definition would have $G$ actually be the space of $n$-dimensional subspaces of $W$, and then the subspaces $V_g$ would be the elements of $G$. This can sometimes be a more intuitive definition, but is possibly less helpful if you are trying to build a bundle using $G$.
• Could you please explain the meaning of the notion of a Grassmannian parametrizes the $n-$dimensional subspces of $W$. What do you mean by parametrises ? Commented Mar 4, 2015 at 12:03
• @palio: In the same way that $[0,2\pi)$ parametrizes the circle via the map $t\mapsto (\cos t,\sin t)$ by identifying $t\in [0,2\pi)$ with a point in the circle, every element $g\in G$ is identified with a particular subspace of $W$. Commented Mar 4, 2015 at 12:10
• Yes but here you have two well defined sets: $[0,2\pi)$ and the circle $S^1$ and a well defined map. I see that the analogous of $S^1$ is the set $B$ of all $n-$dimensional subspaces of $W$ but what is exactly $G$ that you take as the analogous of $[0,2\pi)$ and the map that takes an element of $G$ to a unique well defined $n-$dimensional subspace of $W$ ? Commented Mar 4, 2015 at 12:32
• @palio: The simplest example would probably be the real projective line, where we could parametrize the $1$-dimensional subspaces of $\mathbb{R}^2$ by $t\in[0,\pi)\mapsto\mathrm{span}(\cos t,\sin t)$. But if you think of the Grassmannian as simply being the set of $n$-dimensional subspaces of $W$, then the parametrization is trivial--just $V_g=g$. Commented Mar 4, 2015 at 12:51
• @palio: According to the definition in your question, the parameter space $[0,\pi)$ would be the Grassmannian. Other definitions would say that the set of subspaces $\mathbb{R}P^1$ is the Grassmannian. But since they are in bijection, there is no conflict between the two sets' definitions. Commented Mar 4, 2015 at 13:00
It's slightly awkward wording: One way to think of this is that the points $g$ in a Grassmannian are the subspaces of $W$ of a given dimension; in this interpretation, $g$ and $V_g$ are just different names for the same object, and we might choose to use the former when we want to emphasize that the object is an element in some set, and the latter when we want to emphasize that the object is itself a vector (sub)space in its own right.