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Edit: According to the comment of Steven Landsburg, I edit the lines with economics interpretation of the "Social choice problem". I thank him for his comment.

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

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

Edit: According to the comment of Steven Landsburg, I edit the lines with economics interpretation of the "Social choice problem". I thank him for his comment.

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

Edit: According to the comment of Steven Landsburg, I edit the lines with economics interpretation of the "Social choice problem". I thank him for his comment.

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

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.

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 $$F_{n}(M)=\{(x_{1},x_{2},\ldots,x_{n})\in M^{n}\mid x_{i} \neq x_{j},\;\forall i \neq j\} $$.

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

Edit: According to the comment of Steven Landsburg, I edit the lines with economics interpretation of the "Social choice problem". I thank him for his comment.

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