Let $G_k$ be the infinite rank $k$ Grassmannian. For $n>k$, is $\pi_n(G_k)$ trivial? Phrased differently, is every rank $k$ vector bundle on an $n$-sphere, for $n>k$, trivial? (This is motivated by the $k=1$ case, where $G_1=\Bbb{RP}^\infty$ is aspherical.)

up vote 8 down vote accepted

There is a fibre bundle $$O(n)\to V_k(\mathbb{R}^\infty)\to G_k(\mathbb{R}^\infty)$$ where $V_k(\mathbb{R}^\infty)=\cup_nV_k(\mathbb{R}^n)$ with $V_k(\mathbb{R}^n))=\{k \text{ frames in } \mathbb{R}^n\}$.

In particular, the total space is contractible, so the long exact sequence in homotopy groups gives us isomorphisms $$\pi_i(O(n))\cong \pi_{i+1}G_k(\mathbb{R}^\infty)$$

Now the homotopy groups of $O(n)$ is at least as hard as the homotopy groups of $S^{n-1}$ via the fiber bundle

$$O(n-1)\to O(n)\to S^{n-1}$$

but if we work with stable homotopy groups (so that $\pi_i(O(n))$ becomes independent of $n$), there is a very nice description using Bott's Periodicity Theorem: the homotopy groups are 8-periodic ($\mathbb{Z}/2,\mathbb{Z}/2,0,\mathbb{Z},0,0,0,\mathbb{Z}$). So to answer your question: $\pi_i(G_k)$ is in general not trivial for $i>k$.

  • Thanks! It seemed unlikely things would be so simple. – Mike Miller Oct 21 '14 at 22:37

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.