# Tag Info

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Here is a proof that no sphere (of dimension $>0$) admits an averaging function. Suppose there is an averaging function $f:S^n\times S^n\to S^n$. Let $T(x_0,x_1,x_2,\dots,x_n)=(-x_0,-x_1,x_2,\dots,x_n)$; then $T:S^n\to S^n$ is homotopic to the identity and satisfies $T(T(x))=x$. Now consider the map $g:S^n\to S^n$ given by $g(x)=f(x,T(x))$. Since $T$ ...

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Such functions are called means. The earliest paper on the subject that I’ve seen is G. Aumann, Über Räume mit Mittelbildungen, Mathematische Annalen (1943), Vol. 19, 210-215. He shows inter alia that no $S^k$ has a mean; that the only $2$-dimensional manifold with a mean is the open disk; and that if $X$ has a mean, then so does every retract and every ...

19

Edit: This post has been edited to take Eric Wofsey's comments into account. The end result is the following theorem: Suppose $X$ is a closed manifold of positive dimension. Then $X$ does NOT admit an averaging function. If an averaging $f:X\times X\rightarrow X$ exists, it induces a map $f_\ast: \pi_k(X)\oplus \pi_k(X)\rightarrow \pi_k(X)$. ...

7

The generator of $H^2(S^2\times S^4)$ is $\pi^*(\alpha)$ where $\pi:S^2\times S^4\rightarrow S^2$ is the projection on the first factor, and $\alpha$ is a generator of $H^2(S^2)$. Then $\gamma:=\pi^*(\alpha)\smile\pi^*(\alpha)=\pi^*(\alpha\smile\alpha)$ by functoriality. But $\alpha\smile\alpha=0$ as $H^4(S^2)=0$, so $\gamma=0$ in $H^4(S^2\times S^4)$. ...

6

Okay, it actually is pretty simple. We want to argue that the map $H_k(A)\to H_k(S^n)$ is trivial. This follows because it factors through the contractible space $S^n\setminus\{p\}$ for any point $p\notin A$.

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