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.

This is a  mathematical modeling of the following economic situation:

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?


**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 the later example let's replace the conjugate action by antipodal action. Is there  a  continuous  map $f:F_{2}(S^{1})\to S^{1}$  which satisfy $f(z,w)=-f(w,z)$?