# Fundamental groups of Grassmann and Stiefel manifolds

Could someone provide details on how to compute fundamental groups of real and complex Grassmann and Stiefel manifolds?

Grassmanians are homogeneous spaces.

In the real case, you have the oriented Grassmanian $G^0(k,\mathbb{R}^n)$ of oriented $k$-planes in $\mathbb{R}^n$ is diffeomorphic to $SO(n)/\left(SO(k)\times SO(n-k)\right)$ where $SO(n)$ is the collection of $n\times n$ special orthogonal matrices.

Likewise, the nonoriented Grassmanian $G(k,\mathbb{R}^n)$ of nonoriented $k$-planes in $\mathbb{R}^n$ is diffeomorphic to $SO(n)/S(O(k)\times O(n-k))$.

Finally, the complex Grassmanian, $G(k,\mathbb{C}^n)$, is diffeomorphic to $SU(n)/(SU(k)\times SU(n-k))$ where $SU(n)$ denotes the $n\times n$ special unitary matrices.

(Some slight modifications may be necessary when $k=0$ or $k=n$).

Once you have written them like this, you have a general theorem that given compact Lie groups $G$ and $H$, then $H\rightarrow G\rightarrow G/H$ is a fiber bundle. In particular, we can use the long exact homotopy sequence.

It follows immediately that the complex Grassmanian is simply connected because $SU(n)$ is both connected and simply connected.

In the real case, a bit more work needs to be done. For the oriented Grassmanian, it's enough to note that the canonical map $SO(k)\rightarrow SO(n)$ is a surjection on $\pi_1$ as soon as both $n$ and $k$ are bigger than 1, (isomorphism when $n,k>2$) and is always an isomorphism on $\pi_0$. Thus, the real oriented Grassmanian is simply connected.

This also gives the answer for the unoriented real Grassmanian because there is a natural double covering $G^0(k,\mathbb{R}^n)\rightarrow G(k,\mathbb{R}^n)$ given by forgetting the orientation. Hence, the real unoriented Grassmanian has $\pi_1=\mathbb{Z}/2\mathbb{Z}$.

Alternatively, note that the induced map from $S(O(k)\rightarrow O(n-k))$ to $SO(n)$ is an isomorphism on $\pi_1$, but that $S(O(k)\times O(n-k))$ has more than one component.

• Thanks for the detailed answer. I hope one can compute fundamental group of Stiefel manifolds in a similar way. – google Oct 3 '11 at 15:05
• @google, Stiefel manifolds are also homogeneous spaces—look in the corresponding Wikipedia page. The same argument will deal with them. – Mariano Suárez-Álvarez Oct 3 '11 at 15:51
• @google: I forgot about the Stiefel Manifolds bit - sorry! But Mariano is exactly correct. – Jason DeVito Oct 3 '11 at 16:56
• Re: Isomorphism at $\pi_1$ level of a map $SO(k)\rightarrow SO(n)$: My question: $SO(2)=S^1$ and $SO(3)=\mathbb RP^3$. So, I don't see how any map $SO(2)\hookrightarrow SO(3)$ induces an isomorphism at $\pi_1$ level! Is there more to be said about $k,n$ for such a map be a $\pi_1$-isomorphism? – Karthik C Jul 29 '13 at 5:21
• @Aneesh: That should read "bigger than 2" instead of "bigger than 1.". When $k=2$, it's a surjection, which is still good enough for the rest of the argument. – Jason DeVito Jul 29 '13 at 19:19