Does $Graff(n,\mathbb{R}^{\infty})$ generate all $n$-dimensional closed Riemannian manifolds $M$? How does one generate all possible $n$-dimensional simply connected closed Riemannian manifolds $M$ from the affine Grassmannian $Graff(n,V)$? Would $Graff(n,\mathbb{R}^{\infty})$ suffice? (It seems like we need more for it to be closed and simply connected.) Thanks in advance!

There are lots of simply connected, closed manifolds which are not Grassmannians of $k$ planes in some $\mathbb{R}^n$.

One way to see this is as follows: the dimension of $G(k,n)$ is $k(n - k)$.

Solving for $k(n -k) = 8$, we get $(k,n) \in \\{ (1,9), (2,6), (4,6), (8,9) \\}$

However, here are five non-homeomorphic $8$-manifolds that are closed and simply connected:

1) $S^2 \times S^2 \times S^2 \times S^2$

2) $S^8$.

3) $S^3 \times S^3 \times S^2$

4) $S^4 \times S^2 \times S^2$

5) $S^6 \times S^2$.

Thus, at least one of these is not homeomorphic to a Grassmannian.

To check that these five products of spheres are not homeomorphic to each other, you can use the Poincare polynomial: <