Let $X$ be a space. The symmetric group $\Sigma\_{n+1}$ acts on the function space
$$
X^{\Delta^n}
$$
of continuous maps from the standard $n$-simplex to $X$. The action is induced
by permuting the vertices. 


Let $\Sigma\_{n+1}$ act on $\Bbb Z$ by means of the sign representation.


Then the singular $n$-cochains 
$$
S^n(X) := \text{map}(X^{\Delta^n},\Bbb Z)
$$
inherits a $\Sigma\_{n+1}$-action given by conjugating functions. 


**Definition.** 
The *skew $n$-cochains* on $X$ is given by the invariants 
$$
S^n(X)^{\Sigma\_{n+1}}
$$
that is, by equivariant functions $X^{\Delta^n} \to \Bbb Z$.


Then an elementary calculation shows that the usual singular coboundary operator $\delta$
maps $S^n(X)^{\Sigma\_{n+1}}$ into $S^{n+1}(X)^{\Sigma\_{n+2}}$.


So we get a cochain complex:
$$
S^0(X) \overset \delta \to \cdots \overset \delta \to S^n(X)^{\Sigma\_{n+1}} \overset \delta \to S^{n+1}(X)^{\Sigma\_{n+2}} \overset \delta \to \cdots
$$
Define the *skew cohomology* of $X$ to be the cohomology of this cochain complex.


**Questions:** *What is it? What properties does it have? Has it ever before been studied?*



"Skew Cohomology" of a Space
I believe it should be exactly the same as ordinary singular cohomology. It should define a cohomology theory for the exact same reason that usual singular cohomology does (the usual proof of excision by subdivision seems to work since the cosubdivision of a $\Sigma\_n$-invariant cochain can be checked to still be $\Sigma\_n$-invariant; the usual proof of homotopy invariance doesn't work because the usual triangulation of a product $\Delta^n\times I$ is incompatible with the $\Sigma\_n$ action, but all you need to do is geometrically construct a different, finer triangulation that is more symmetric), and on a point it can be computed explicitly.


As a heuristic argument for why this should be true at least rationally, note that de Rham cohomology already has this $\Sigma\_n$-invariance built in, in the requirement that differential forms be alternating.


