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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?

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  • $\begingroup$ I think I remember reading somewhere that the dual construction on homology was actually the first version of singular homology that was defined, but it was awkward to work with because the chain groups had 2-torsion. $\endgroup$
    – Eric Wofsey
    Mar 4, 2011 at 2:36
  • $\begingroup$ My recollection is that Lefschetz failed to order the vertices of his simplices. This is where the 2-torsion problems arise. But I don't see how this is dual to my construction... $\endgroup$
    – John Klein
    Mar 4, 2011 at 3:32
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    $\begingroup$ If you consider chains on simplices which are oriented but whose vertices are unordered and declare that the same simplex with opposite orientation is the negative of the original simplex, this should be exactly the same as taking $\Sigma_n$-coinvariants on the usual chain group. $\endgroup$
    – Eric Wofsey
    Mar 4, 2011 at 3:43
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    $\begingroup$ That is, you can make $\Sigma_{n+1}$ act on $S_n(X)$ by permuting the basis (the singular simplices) as you describe, and then twist this by tensoring with the sign representation on $\mathbb Z$. The coinvariants $S_n(X)_{\Sigma_{n+1}}$ form a chain complex, and its $\mathbb Z$-dual is your complex of invariant cochains. But the $2$-torsion (which arose from the fact that some basis elements fixed by some odd permutations) had no effect on the dual. $\endgroup$
    – Tom Goodwillie
    Mar 4, 2011 at 4:54
  • $\begingroup$ Ah yes, I see that. $\endgroup$
    – John Klein
    Mar 4, 2011 at 5:06

2 Answers 2

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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.

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  • $\begingroup$ Consider the barycentric subdivision of the $n$-simplex, and choose one of the smaller $n$-simplices $\Delta^n_F$ as your favorite. Then a skew singular simplex $\Delta^n\to X$ is the same as an ordinary singular simplex $\Delta^n_F \to X$. $\endgroup$
    – Jeff Strom
    Mar 4, 2011 at 2:25
  • $\begingroup$ @Jeff: what you wrote does not make sense to me: I am defining a cohomology theory not a homology theory. $\endgroup$
    – John Klein
    Mar 4, 2011 at 2:54
  • $\begingroup$ @Eric: your last sentence is actually what motivated my construction. That is, the skew cochain complex is closer in spirit to the de Rham complex than the total singular cochain complex is. $\endgroup$
    – John Klein
    Mar 4, 2011 at 2:58
  • $\begingroup$ @John: I was really just musing in the general direction of Eric's response, thinking that it was kind of a nice idea. $\endgroup$
    – Jeff Strom
    Mar 4, 2011 at 14:12
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I do not have the book at hand to check, but maybe you do...

Loday defines in his book on Cyclic homology what a crossed simplicial group is and for each such gadget a corresponding homology theory. There is, for example, the "cyclic" crossed simplicial group, and the corresponding homology theory is cyclic homology. Well, there is a crossed simplicial groups---call it $\Delta S$---built from symmetric groups, and it turns out that the corresponding homology theory coincides with Hochschild homology. I am pretty sure your complex is the analogue of Connes' $\lambda$-complex corresponding to the homology theory for $\Delta S$, which is quasi-isomorphic to the complex which defines the actual $\Delta S$-homology rationally and, therefore, should just give you simplicial homology over $\mathbb Q$.

Integrally, you should not take invariants in each degree, but consider a double complex, each of whose columns are the complex which computes $H^\bullet(S_{n+1},\mathord-)$, much as the double complex which computes cyclic homology has columns which compute $H^\bullet(C_{n+1},\mathord-)$.

(I would love to have an example where taking invariants integrally does not work, by the way!)

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