# Has this chain complex associated with a simplicial complex been studied before?

I have stumbled upon a construction that has probably been noticed before, and I wonder if anyone can point me to a reference.

Suppose that $K$ is a simplicial complex. Let $P(K)$ be the free abelian group on the set of simplices of $K$, all dimensions together. View it as a Grothendieck group of subcomplexes, so that for example the boundary $\partial \sigma$ of a simplex $\sigma$ appears in it as an alternating sum of the proper faces of $\sigma$. The open simplices $int (\sigma) =\sigma-\partial \sigma$ form another basis. Do not think about orientations.

Let $I$ be the involution of $P(K)$ given by $\sigma\mapsto (-1)^m\ int\ (\sigma)$ where $m$ is the dimension of $\sigma$.

It appears that the homology of the $2$-periodic chain complex given by $1-I:P(K)\to P(K)$ and $1+I:P(K)\to P(K)$ is isomorphic to the mod $2$ homology of $K$ made $2$-periodic.

EDIT: In case the slightly vague reference to a "Grothendieck group" is confusing, let me be more explicit: $P(K)$ has a basis consisting of the simplices of the complex. If $\sigma$ is a $p$-simplex of $K$, then the map $I$ takes the basis element $\sigma$ to the sum of $2^{p+1}-1$ terms $(-1)^{dim\ \tau}\tau$, one term for each simplex $\tau$ that is contained in $\sigma$, including $\sigma$ itself. Note that I never mentioned an orientation or an ordering of vertices.

-
Could you clarify the definition of $P(K)$ some more? I'm not sure what you mean when you say that the boundary of $\sigma$ appears "in it" as an alternating sum. In short, how is $P(K)$ different from the direct sum of all the simplicial chain groups $C_m(K)$? – Vidit Nanda Mar 21 '14 at 18:57
Your $I$ is $K$ of Verdier duality. Thus, the homology of $1-I$ is about self-dual sheaves. Cross-referencing $L$-theory and 2-periodic 2-torsion suggests the Rothenberg exact sequence for change of decoration. – Ben Wieland Mar 31 '14 at 18:39
Oh, I didn't know what Verdier duality is. I suppose I'm looking at cosheaves rather than sheaves but it's otherwise the same. This formalism was serving me nicely even though I wasn't using any derived functors as far as I can tell. – Tom Goodwillie Apr 1 '14 at 22:49
There is an obvious contravariant equivalence between sheaves and cosheaves on a fixed stratification with values in perfect complexes. Verdier duality is best thought of as a weird covariant equivalence, $F\mapsto\{U\mapsto H_c^*(U;F)\}$. How did you get $I$ (or $1-I$) without derived functors? Maybe you should have had them. – Ben Wieland Apr 2 '14 at 22:11
Ben, it's just what I said. And then you can take a direct limit over subdivision to get something that depends on a PL space instead of a simplicial complex. – Tom Goodwillie Apr 3 '14 at 23:26