Let $M$ be a topological space. Define an averaging function as a continuous map $f:M \times M \to M$ which satisfies $f(a,b) = f(b,a)$ for all $a,b \in M$ and $f(a,a) = a$ for all $a \in M$.

These seem like reasonable properties for a function which deserves the name "average" to have, since they are enjoyed by the usual averaging functions on $\mathbb{R}^n$.

One can rule out an averaging function on both $S^1$ and $S^2$ by some cohomological arguments. I have not worked out whether any sphere admits an averaging function, but I suspect not.

Euclidean spaces all do (given by the usual average).

Does anyone have a characterization of spaces which admit an averaging function?

Steven Gubkin
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    It comes up every so often that someone wants to average two angles, or average points on the globe, for one purpose or another. I usually try to reason with them first that there is no canonical average between $0$ and $\pi$, and when that fails I intimidate them with cohomology. I would like a complete listing of spaces with averages, since averaging seems like something people often want to do, but the structure of their space might not allow it. – Steven Gubkin Aug 24 '15 at 00:43
  • Now, what made you think of this problem? :) – John Hughes Aug 24 '15 at 00:52
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    At least for spaces with a Reimannian metric (surely you only want to consider those!), it might also be reasonable to say that if there's a unique shortest geodesic from $a$ to $b$, then $f(a, b)$ should lie on that geodesic. And it's probably also reasonable to insist that $f$ be continuous in its arguments, at least on the set of pairs for which there's a unique geodesic; I think your cohomology proof relies on this hidden assumption. – John Hughes Aug 24 '15 at 01:02
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    I may be missing some subtle issue, but what about taking any point $c \in M$ and defining $$f(a,b) = \left\{ \begin{array}{ll} a &\mbox{ if $a=b$} \\ c & \mbox{ if $a\neq b$} \end{array} \right.$$ Also, in what sense is this an "averaging"? – Michael Aug 24 '15 at 01:04
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    I think that @StevenGubkin meant to implicitly include the idea that $f$ should be continuous. It's "proto-averaging" in the sense that the usual notion of "midpoint" in $R^n$ satisfies these two properties, and you'd like any averaging map on any space to do the same. So having such a function is a bare minimum for having a true "averaging" function, which probably should have lots of other properties as well. If you can show that a space admits no function of Steven's kind, you can stop right there. – John Hughes Aug 24 '15 at 01:13
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    @StevenGubkin, that's a reasonable strategy, most people are indeed intimidated by cohomology. – Spooky Aug 24 '15 at 01:28
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    Yes, I should have required continuity! Otherwise, no need to have it be a topological space at all. @JohnHughes I agree that property is nice for metric spaces, but I am mostly interested in just the topological question (at least for starters). – Steven Gubkin Aug 24 '15 at 02:25
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    A trivial but perhaps helpful reformulation is that $X$ has an averaging function iff the diagonal is a retract of $X\times X/\Sigma_2$ – Eric Wofsey Aug 24 '15 at 02:25
  • For a related post see http://mathoverflow.net/questions/203390/polygons-with-centroid-at-origin-and-vertices-on-compact-codimension-one-submani – Ali Taghavi Feb 11 '16 at 09:38

4 Answers4


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 $T$ is homotopic to the identity, $g$ is homotopic to $x\mapsto f(x,x)=x$, and so $g$ has degree $1$. But note that $$g(T(x))=f(T(x),T(T(x)))=f(T(x),x)=f(x,T(x))=g(x),$$ so $g$ factors through the quotient $S^n\to S^n/{\sim}$, where $\sim$ identifies $x$ with $T(x)$. It is easy to see that $S^n/{\sim}\cong S^n$ and the quotient map has degree $2$. Thus $g$ must have even degree, which is a contradiction.

On the other hand, here are some spaces that have means. First, suppose $M=|X|$ is the geometric realization of a countable contractible simplicial set. The functor $X\mapsto X\times X/\Sigma_2$ commutes with geometric realization for countable simplicial sets, so we get a CW-complex structure on $M\times M/\Sigma_2$ such that the diagonal $M\to M\times M/\Sigma_2$ is a subcomplex. Since $M$ is contractible, it follows by obstruction theory that $M$ is a retract of $M\times M/\Sigma_2$. But such a retraction is exactly a mean on $M$. (The restriction to countable simplicial sets is just so that the product has the right topology; if you work in the category of compactly generated spaces rather than all spaces, you can remove the countability hypothesis. I also suspect that this argument works for arbitrary contractible CW-complexes (not just realizations of simplicial sets), but in that case it is not obvious how to get a CW-structure on $M\times M/\Sigma_2$ with the diagonal as a subcomplex).

There are also some spaces with means with more interesting homotopical properties. For instance, let $(A_n)_{n\geq 0}$ be a sequence of countable $\mathbb{Z}[1/2]$-modules (as above, the countability hypothesis can be dropped if you work with compactly generated spaces). Consider the bounded chain complex of $\mathbb{Z}[1/2]$-modules whose objects are the $A_n$ and whose maps are all zero. Via the Dold-Kan correspondence, we obtain from this chain complex a countable simplicial $\mathbb{Z}[1/2]$-module $X$ such that $\pi_n(X)=A_n$. The geometric realization $M=|X|$ is then a topological $\mathbb{Z}[1/2]$-module with $\pi_n(M)=A_n$. We can then define a mean on $M$ by $f(a,b)=(a+b)/2$.

Eric Wofsey
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  • Thanks! I had (apparently different) ad hoc cohomological proofs for $S^1$ and $S^2$. This nicely takes care of the spheres. – Steven Gubkin Aug 24 '15 at 02:26
  • This is really great. You guys made it hard to figure out which answer to accept, but I am going with this one. I will try to see if I can construct a mean on the whitehead manifold along these lines. – Steven Gubkin Aug 24 '15 at 12:40

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 component of $X$.

Three relatively recent papers that I was able to find fairly quickly:

Mirosław Sobolewski, Means on chainable continua, Proc. Amer. Math. Soc., Vol. 136, Nr. 10, October 2008, 3701-7, shows that a chainable continuum that admits a mean is homeomorphic to the interval.

T. Banakh, R. Bonnet, W. Kubiś, Means on scattered compacta, Topological Algebra and its Applications (2014), Vol. 2, Nr. 1, 5-10, construct a scattered compact space admitting no mean.

Section 6 of Janusz J. Charatonik, Selected problems in continuum theory, Topology Proceedings, Vol. 27, Nr. 1, 2003, 51-78, lists some known results and open questions concerning means on continua.

All have some references that may be of interest.

I’ve a faint recollection that in the 1970s or 1980s Jan van Mill or some of the Dutch topologists around him did some work on means, but I can’t at the moment locate any of it.

Brian M. Scott
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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 $f_\ast: \pi_k(X)\oplus \pi_k(X)\rightarrow \pi_k(X)$.

Observation 1: $f_\ast(g,g) = g$. Proof: Consider the composition $X\rightarrow X\times X\rightarrow X$, where the first map is the diagonal embedding. By assumption, this composition is the identity. Now, apply functoriality.

Observation 2: $f_\ast(g,h) = f_\ast(h,g)$. Proof: Consider the composition $X\times X\rightarrow X\times X\rightarrow X$ where the first map swaps the two factors and apply functoriality.

These two observations together have very strong consequences. For example:

Corollary: Suppose $G$ is a group with a homomorphism $f: G\times G\rightarrow G$ satisfying both observations. Then $G$ is abelian and every element of $G$ is twice another element of $G$.

Proof: First we show that every element of $G$ is twice another, or, in multiplicative notation, that for any $g\in G$, $g = h^2$ for some $h\in G$. In fact, setting $h = f(g,e)$, we see $h^2 = f(g,e)f(g,e) = f(g,e)f(e,g) = f(g,g) = g$.

Now we show that $G$ is abelian. We have \begin{align*} [f(g,1)f(h,1)]^2 &= f(gh, 1)^2 \\ &= f(gh,1)f(1,gh) \\ &=f(gh,gh)\\ &= gh \\ &= f(g,1)^2 f(h,1)^2\end{align*} so $[f(g,1)f(h,1)]^2 = f(g,1)^2 f(h,1)^2$. Writing these out an canceling, this reduces to $f(h,1)f(g,1) = f(g,1)f(h,1)$. Replacing $h$ and $g$ with $h^2$ and $g^2$, this becomes $hg = gh$, so $G$ is abelian. $\square$

Now, it is easy to see that in a cyclic group, every element is twice another iff the group has odd order. It follows that a finitely generated abelian groups for which every element is twice another is a direct sum of cyclic groups of odd order. In particular, it is finite.

Theorem: If $X^n$ is a closed manifold of positive dimension $n > 0$, then $X$ does not admit an averaging function.

Proof: Since $X$ is closed, $\pi_1(X)$ is finitely generated, and hence finite for all $k$. Since $\pi_k(X)$ is finitely generated as a $\pi_1(X)$ module, it follows that $\pi_k(X)$ is is finitely generated for all $k$. Pass to the universal cover $\tilde{X}$, which is closed because $\pi_1(X)$ is finite. By the Hurewicz Theorem mod $\mathcal{C}$), it follows that all the homology groups of $\tilde{X}$ are finite. But, by Poincare duality, $H_n(\tilde{X})\cong \mathbb{Z}$. This is a contradiction, unless $n = \dim X = 0$. $\square$

These ideas can also be used to show that $\mathbb{R}^2$ is the only surface (no boundary) admitting an averaging function.

Sketch: We have already ruled out closed manifolds, so we assume $X$ is a non-compact manifold. Then, it is relatively well known that $X$ deformation retracts onto its $1$-skeleton. If follows that $\pi_1(X;\mathbb{Z})$ is a free abelian group. This contradicts the corollary, unless $\pi_1(X;\mathbb{Z})$ is trivial. This, in turn implies that $X$ deformation retracts onto its $0$-skeleton, that is $X$ is contractible. This implies $X$ is homeomorphic to $\mathbb{R}^2$.

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Jason DeVito
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    So, having ruled out many manifolds from admitting averaging functions, the natural question is: Does anything other than Euclidean space support an averaging function? Note that if a compact manifold admits an averaging function, then every $\pi_k$ with $k\geq 2$ is a finite group of odd order. Can this actually occur? – Jason DeVito Aug 24 '15 at 04:20
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    This is very nice. I agree that it seems more an more unlikely that any manifold other than a Euclidean space supports an averaging function. In some of Brian Scott's resources there are constructions of spaces with means which are not manifolds. It would be a pretty nice way to characterize Euclidean space among manifolds, if true. – Steven Gubkin Aug 24 '15 at 10:08
  • I wonder if some Eilenburg MacLane spaces for odd groups admit means. – Steven Gubkin Aug 24 '15 at 10:22
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    @StevenGubkin: Yes, for any finite abelian group $G$ of odd order and any $n$ there is a $K(G,n)$ which admits a mean; see the update to my answer for a stronger result. There are also other (non-compact) manifolds with means, since my answer implies any contractible triangulable manifold has a mean. – Eric Wofsey Aug 24 '15 at 11:25
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    @JasonDeVito: Your argument shows not just that every element of $G$ has a square root, but that it is abelian and a $\mathbb{Z}[1/2]$-module. Indeed, $gh=f(gh,e)^2=(f(g,e)f(h,e))^2$ implies $f(g,e)$ and $f(h,e)$ commute, and hence so do $g$ and $h$, and then $g\mapsto f(g,e)$ makes $G$ a $\mathbb{Z}[1/2]$-module. – Eric Wofsey Aug 24 '15 at 11:39
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    It follows that no closed manifold admits a mean, since its universal cover would be a simply connected closed manifold all of whose homotopy groups are all finite of odd order. But this implies the homology groups are also all finite (by the Serre spectral sequence, for instance), and this is incompatible with Poincare duality. – Eric Wofsey Aug 24 '15 at 11:58
  • @EricWofsey : $\:$ $\mathbb{R}^0$ is a closed manifold that admits a mean. $\;\;\;\;$ –  Aug 24 '15 at 13:46
  • @RickyDemer: Yes, I meant positive-dimensional closed manifold. – Eric Wofsey Aug 24 '15 at 13:55
  • @Eric: Very nice. – Jason DeVito Aug 24 '15 at 13:58
  • Please, please update the answers (the parts that have not already been incorporated). – Danu Aug 25 '15 at 00:49
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    I don't think it's true that $X$ closed is enough to imply that its homotopy groups are finitely generated, but by the Hurewicz theorem mod C we're okay using the additional assumption that $\pi_1(X)$ is finite and looking at the universal cover. – Qiaochu Yuan Aug 25 '15 at 04:15
  • @Qiaochu: You're right, of course. Thanks for catching that! (Fixed now) – Jason DeVito Aug 25 '15 at 04:46

The paper

B.Eckmann, Räume mit Mittelbildungen, Comment.Math.Helv. 28(1954), 329-340

contains most of the answers given so far. The results of this paper are summarized in the following very interesting note of Benno Eckmann:

B. Eckmann, Social choice and topology.

where he defines $n$-means as symmetric functions $\prod_{i=1}^nX\to X$ such that precomposing with the diagonal gives the identity. The means in this mathoverflow question are, in this new language, $2$-means. He proves:

Theorem 6: If a finite polyhedron admits $n$-means for some $n$, then it is contractible.

Theorem 7 + Theorem 4: A polyhedron $X$ admits $n$-means for all $n$ if and only if either it is contractible or it has the homotoy type of a product of rational Eilenberg Mac Lane spaces.

Theorem 2. If a space $X$ admits an n-mean for some n ≥ 2 then $\pi_1(X)$ is Abelian, multiplication by n is an automorphism of $\pi_k(X)$ for all $k$, and $\pi_k(X)$ is uniquely divisible by $n$.

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  • Thank you. This is an incredibly well written paper, and certainly answers most my questions. I have switched my accepted answer to yours, because I think you are unlikely to get as many upvotes as the older answers. – Steven Gubkin Dec 30 '15 at 17:14