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Let $M$ be a closed connected oriented smooth manifold and $\mathrm{Diff}_{+}(M)$ the group of orientation preserving diffeomorphisms of $M$ endowed with the compact-open topology. Pick a base point $x_{0} \in M$ and consider the evaluation map $$\mathrm{ev}\colon \mathrm{Diff}_{+}(M) \to M, \quad \mathrm{ev}(g) = g(x_{0}).$$

I would like to know when the evaluation map admits a global section, i.e. a continuous map $s\colon M \to \mathrm{Diff}_{+}(M)$ such that $\mathrm{ev} \circ s = \mathrm{id}_{M}$.

Examples:

  1. If $M = G$ is a Lie group, then a section $s$ exists: for $x\in G$ put $s(x) := \text{left multiplication by }x\cdot x_{0}^{-1}$.
  2. However, for $M = S^{2}$, the 2-sphere, such a section $s$ cannot exist: it would induce an injective map on homotopy groups $s_{*}\colon \pi_{k}(S^{2}) \to \pi_{k}(\mathrm{Diff}_{+}(S^{2}))$, but $\pi_{2}(S^2) \cong \mathbb{Z}$ while $$\pi_{2}(\mathrm{Diff}_{+}(S^{2})) \cong \pi_{2}(\mathrm{SO}(3)) = 0.$$

It is probably hard to decide on the existence of a section for general $M$, so let us restrict to spheres $S^{n}$:

Question: For which $n\in \mathbb{N}$ does the evaluation map $\mathrm{ev}\colon \mathrm{Diff}_{+}(S^{n}) \to S^{n}$ admit a continuous global section?

Since $S^1$ and $S^3$ are naturally Lie groups, the section exists in thsese two cases by the first example above. We can analogously use octonionic multiplication to construct a global section for $S^{7}$. Are there any other cases than $n=1,3,7$?

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3 Answers 3

up vote 12 down vote accepted

If a section $\sigma : M \rightarrow \text{Diff}_{+}(M)$ to $\text{Diff}_{+}(M) \rightarrow M$ exists, then $M$ must be parallelizable (i.e. the tangent bundle of $M$ must be trivial). Indeed, if $\vec{e}_1,\ldots,\vec{e}_n$ is a basis for the tangent space of $M$ at $x_0$, then $\sigma(x)_{\ast}(e_1),\ldots,\sigma(x)_{\ast}(e_n)$ is a basis for the tangent bundle to $M$ at $x \in M$. These vary continuously with $x$, and thus define a trivialization of the tangent bundle of $M$.

For spheres, a theorem that was proved independently in

Kervaire, Michel A. Non-parallelizability of the n-sphere for n>7. Proc. Natl. Acad. Sci. USA 44 (1958), no. 3, 280–283.

and

Bott, R.; Milnor, J. On the parallelizability of the spheres. Bull. Amer. Math. Soc. 64 1958 87–89.

says that $S^k$ is only parallelizable when $k=1,3,7$, so your examples are the only ones.

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Evaluation maps are studied in rational homotopy theory, see e.g. Evaluation maps in rational homotopy by ‎Félix and Lupton. Here is a sample result (see Corollary 1.9 of the above paper):

Let $\mathrm{Aut}_0(M)$ denote the identity component of $\mathrm{maps}(M,M)$. A section of $\mathrm{Diff}_0\to M$ followed by the inclusion into $\mathrm{Aut}_0(M)$ gives a section of the evaluation map $e: \mathrm{Aut}_0(M)\to M$, so the latter is surjective in rational homology. Suppose $M$ is simply-connected. If $e$ is surjective on rational homology, then $M$ is rationally homotopy equivalent to a product of odd-dimensional spheres.

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To add to Andy Putman's answer: An interesting modification of the question is to replace diffeomorphisms by homotopy equivalences. Thus, for a given based manifold or CW complex $(X,p)$, is there a continuous map $F:X\times X\to X$ such that $F(p,x)=x$ for all $x$ and $x\mapsto F(x,y)$ is a homotopy equivalence for all $y$? If $X$ is connected then it is enough if $x\mapsto F(x,p)$ is a homotopy equivalence, and without loss of generality that homotopy equivalence is the identity. So the question is, is there is a continuous multiplication on $X$ for which $p$ is a two-sided identity element.

The answer is no if $\pi_1(X)$ is not abelian, or if the canonical action of $\pi_1(X)$ on some $\pi_n(X)$ is nontrivial, or if some Whitehead product in $X$ is nontrivial. This rules out some parallelizable manifolds for the original question.

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Sorry, I could not stop myself from correcting the spelling of my last name :). –  Andy Putman Apr 30 at 23:34
    
Oh, I'm so sorry. I usually know how to spell your name. –  Tom Goodwillie May 1 at 0:07
    
This is very nice! –  Oldřich Spáčil May 1 at 12:55

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