# Construction of Grassmannian

I am trying to understand what is a Grassmannian.

Starting with the projective space $\mathbb{R}P^n$ = {lines in $\mathbb{R}^{n+1}$} and the grassmannian $G_r(k,n)$ = {$k-dimentional\space subspace\space E \subseteq\mathbb{R}^n$}. Moreover I am told that $\mathbb{R}P^n = G_r(1,n+1)$ which implies that {lines} compose a 1-dimentional subspace. So here is my first question although elementary:
1) How does one prove that the set of lines is a 1-dimensional subspace? (just a hint/idea) (I know how to prove that lines in $\mathbb{R}^2$ are a subspace but do not understand why they are 1-dimentional say in $\mathbb{R}^n$)

2) What are examples of k-dimentional subspaces in $\mathbb{R}^n$ with $k\le n$, how many are there (in $\mathbb{R}^2$ there are 3)

3) Can we define the grassmannian as the quotient space of some manifold? Such as $S^n/\sim \space with\space x\sim\pm x$ for $\mathbb{R}P^n$

4) How do we explain $dim(G_r(k,n))=k(n-k)$? would it be linked with the set of linear maps $L(\mathbb{R}^k,\mathbb{R}^{n-k})$

Thank you for any insight, somehow i did not find projective spaces too hard to understand but this is not breaking through...


1. "Subspace" is meant in the sense of elementary linear algebra, a non-empty subset of a vector space that is closed under addition and under scalar multiplication. A line through the origin is a one-dimensional subspace.

(Incidentally, the Grassmannian you ask about is the unoriented Grassmannian. There is also an oriented Grassmannian, whose elements are oriented subspaces of fixed dimension. The oriented Grassmannian of lines in $\Reals^{n+1}$ is the $n$-sphere: Each oriented line through the origin contains a unique "positive" unit vector, and conversely each unit vector determines a unique oriented line through the origin.)

2. If $n \leq 3$, the only proper, non-trivial subspaces of $\Reals^{n}$ have dimension $1$ (i.e., are lines) or codimension $1$ (i.e., are planes in $\Reals^{3}$).

The "smallest" Grassmannian that is not (effectively) a projective space is the Grassmannian $\Gr(2, 4)$ of planes in $\Reals^{4}$.

3. As a homogeneous space, $\Gr(k, n) \simeq O(n)/O(k) \times O(n-k)$. Each orthonormal frame in $\Reals^{n}$ (i.e., each element of the orthogonal group $O(n)$) determines the $k$-plane spanned by (say) its first $k$ elements. The isotropy group of $\Reals^{k} \simeq \Reals^{k} \times \{0\} \subset \Reals^{n}$ is $O(k) \times O(n-k)$.

Alternatively, there is a bundle over $\Gr(k, n)$, called a Stiefel manifold, whose fibre over a $k$-plane is the set of orthonormal bases of that subspace. The Grassmannian $\Gr(k, n)$ may be viewed as the quotient of the bundle of orthonormal $k$-frames (a Stiefel manifold) obtained by mapping an orthonormal frame to its linear span.

4. As you say, the dimension of $\Gr(k, n)$ may be linked with the set of linear maps $L(\Reals^{k}, \Reals^{n-k})$. Geometrically, a curve through $\Reals^{k} \subset \Reals^{n}$ is a "smooth perturbation" of the standard $k$-plane in $\Reals^{n}$. Infinitesimally, a smooth perturbation amounts to a linear map $T$ from the standard $k$-plane to its orthogonal complement (via the graph of $T$).

All these constructions can be carried out using general linear frames rather than orthonormal frames. Symbolically, replace each $O$ with $GL$. (One advantage of using orthonormal frames is that compactness of the Grassmannians follows from compactness of the orthogonal group. One advantage of working in the general linear setting is that the complex Grassmannians are easily seen to be holomorphic manifolds, not merely smooth manifolds.)

If you need more detailed information, e.g., the cohomology ring, Milnor and Stasheff's Characteristic Classes is (for good reason) a standard reference.

• The third point is slightly beyond my range for now, however for the 4th point: This smooth perturbation you mention, is it well defined? Would it be the projection from $R^k$ to the subspace (would this be isomorphic? and why?) and then from the a projection subspace to $R^k perp$. And would the subspace here be some graph? Could you possible elaborate a bit further on this, i am still confused – rannoudanames Jan 24 '17 at 0:31
• I was vague; the smooth perturbation is not well-defined, but the velocity (time derivative) at $0$ can be identified with the tangent space to the Grassmannian. Explicitly, you can identify a $(n - k) \times k$ matrix $A$ with the one-parameter subgroup$$\exp t\left[\begin{array}{@{}cc@{}}0 & -A^{T} \\ A & 0\end{array}\right] \subset O(n).$$ – Andrew D. Hwang Jan 24 '17 at 1:07
• thank you for an example of application! I am going to look into the references you mentioned. – rannoudanames Jan 24 '17 at 1:17