Motivated by [this paper and its economics motivations](http://www.sciencedirect.com/science/article/pii/S0723086904800161), 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](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/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?
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> 


**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 $f(z,w)=-f(w,z)$?



An equivariant social choice in Mathematical economics
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](https://doi.org/10.1112/jtopol/jtu029), arXiv:[1207.2852](https://arxiv.org/abs/1207.2852).
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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 $M=\mathbb{R}^d$. 


