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

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