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Motivated by this paper and its economics motivations, we recall that a social choice among $n$ objects is a continuous function $$f:\overbrace{M\times M\times\cdots\times M}^{\text{$n$ times}}\to M$$

which satisfy the following conditions:

1) $f(x,x,\ldots,x)=x$

2) $f$ is unchanged under all permutations $\sigma \in S_{n},$ the symmetric group on $n$ elements.

Perhaps the following could be considered as a social model for the above mathematical problem.

A client has to choose one item among $n$ items $(x_{1},x_{2},\ldots,x_{n})$. His preference function is denoted by the above $f$.

Now it is natural that we assume that the client faces with n different items, so $x_{i} \neq x_{j}$, $\forall i \neq j$. So we consider the ordered configuration space $$F_{n}(M)=\{(x_{1},x_{2},\ldots,x_{n})\in M^{n}\mid x_{i} \neq x_{j},\;\forall i \neq j\} $$.

There is an obvious action of the symmetric group $S_{n}$ on $F_{n}(M)$.

Now the following question can be counted as an equivariant analogy of the social choice problem:

Assume that $M$ is a manifold which is acted by the symmetric group $S_{n}$. Is there always an equivariant continuous map $f:F_{n}(M)\to M$? If not, for what type of manifolds the answer is affirmative? What type of algebraic topological obstructions would appear?

Can we find an economics interpretation for this equivariant version?

Example: For $M=S^{1},\; n=2$, the symmetric group $S_{2}$ can act on $S^{1}$ by complex conjugation $z\mapsto \bar{z}$ Then the function $f(z,w)=z\bar{w}$ satisfies $f(z,w)=\overline{f(w,z)}$.

In this example let's replace the conjugate action by antipodal action. Then $f(z,w)= (z-w)/|z-w|$ is a map $f:F_{2}(S^{1})\to S^{1}$ which satisfy $f(z,w)=-f(w,z)$?

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    $\begingroup$ I do not understand the bit about a client "choosing one item among $n$ items". Are you saying that for every $(x_1,\ldots,x_n)$ there exists an $i$ such that $f(x_1,\ldots,x_n)=x_i$ ? $\endgroup$
    – Steven Landsburg
    Commented Oct 13, 2016 at 22:05
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    $\begingroup$ I think Ali must have gotten the following standard model a bit twisted? $M$ is a space of possible preferences, and there are $n$ agents each of whom, $i$, has a preference $x_i \in M$. A social choice function $f: M^n \to M$ maps $(x_1,\dots,x_n)$ to a single group preference $f(x_1,\dots,x_n) \in M$. Condition (1) is unanimity (if everyone prefers $x$, the group chooses $x$) and condition (2) is anonymity (rearranging the agents' names doesn't matter). But I don't see how to reconcile this with the client-item story or the ordered configuration space. $\endgroup$
    – usul
    Commented Oct 13, 2016 at 23:25
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    $\begingroup$ @MichaelGreinecker: Suppose $n=2$, $M$ is the positive real numbers, $x$ is the amount I'd like our government to spend on mathematical research and $y$ is the amount you'd like our government to spend on mathematical research. Then it seems entirely reasonable to hope that the amount $f(x,y)$ that the government actually spends is a continuous function of $x$ and $y$. Once you admit that continuity is a reasonable requirement, you are forced to consider the topology on $M$. That seems like a pretty sensible motivation. $\endgroup$
    – Steven Landsburg
    Commented Oct 14, 2016 at 17:30
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    $\begingroup$ Dear @AliTaghavi, what is unclear to me is why the manifold of preferences $M$ should come with an action of the symmetric group? $\endgroup$
    – Mark Grant
    Commented Oct 16, 2016 at 9:51
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    $\begingroup$ @AliTaghavi: Regarding the assumption that $M$ is a manifold: That depends on the context. For voters in an election, for example, there is usually a finite number of preferences. But it is easy to imagine situations where the set of preferences naturally form a metric space, or even a manifold (as in Steven Landsburg's explanation regarding govt spending on Maths above). $\endgroup$
    – Mark Grant
    Commented Oct 22, 2016 at 9:41

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I'll attempt an answer to the mathematical question, without discussing the motivation. As I understand it, $M$ is a manifold with $S_n$-action, and we are asking whether there exists an $S_n$-equivariant map $f:F_n(M)\to M$, where the action on $F_n(M)$ is by permutation of coordinates (and in particular has nothing to do with the action on $M$).

As a first observation, note that if $x\in M$ is a fixed point $x$, then such an $f$ exists; just map everything to $x$.

A general necessary condition for the existence of a $G$-map $f:X\to Y$ between $G$-spaces is given by the Faddell-Husseini index, as decribed for example in this paper:

Pavle V. M. Blagojević, Wolfgang Lück, Günter M. Ziegler, Equivariant Topology of Configuration Spaces, J. Topology 8 (2015) pp 377–413, doi:10.1112/jtopol/jtu029, arXiv:1207.2852.

Given a $G$-space $X$ and a commutative ring $R$, the Faddell-Husseini index of $X$ is the ideal in $H^*(BG;R)$ defined by $$ \operatorname{Index}_G(X;R):=\ker(p^*:H^*(BG;R)\to H^*(EG\times_G X;R)), $$ where $p:EG\times_G X\to BG$ is the projection of the Borel fibration. Then it is easy to see that if there exists a $G$ map $f:X\to Y$ then $$ \operatorname{Index}_G(Y;R)\subseteq \operatorname{Index}_G(X;R) $$ must hold. You can sometimes rule out existence of equivariant maps using this property. The index of the configuration space $F_n(M)$ is probably difficult to compute in general, but is done in the linked paper for $M=\mathbb{R}^d$.

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  • $\begingroup$ Thank you very much for your very interesting answer and very helpful link. $\endgroup$
    – Ali Taghavi
    Commented Oct 15, 2016 at 12:31
  • $\begingroup$ It seems that the link no longer works. Here is arXiv link and pdf on the author's website. $\endgroup$
    – Martin Sleziak
    Commented Aug 15, 2019 at 6:52

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