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

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up vote 14 down vote accepted

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.

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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-Alvarez 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
Many thanks to all. – google Oct 4 '11 at 4:32
@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

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