# Poincaré quasi-isomorphism

Suppose we have a simplicial combinatorial manifold (just a triangulated manifold) and its Poincaré dual cell complex. Corresponding homology simplicial and homology cell complexes are quasi-isomorphic of cause. But this quasi-isomormohism as it usually quoted from Solomon Lefschetz "Alg Topology" book is transcendental. Can we have a reasonable formula for such a quasi-isomorphism (for homology over a good field at least)?

Update 1) sorry for Russian math-slang use of "transcendental" here "transcendental"="non-constructive" 2) among the others one motivation is to see really the Poincare duality for simplicial chains and cochains of the given triangulation of a manifold.

UUpdate The problem has a nice very canonical solution, with Laplas operator Green function heat kerenel etc. It allows to solve some problems. The preprint(s) are in preparation.

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Are you asking about the map of chain complexes whose domain is the simplicial homology chain complex and whose range is the chain complex for the cohomology of the dual polyhedral complex? I'm not sure what you don't like about the formula. –  Ryan Budney Dec 21 '09 at 0:20
I'm not getting the question either. What is "transcendental" about this Poincare isomorphism? –  Greg Kuperberg Dec 21 '09 at 1:11
I am asking about explicit formulas for two maps between simplicial homology chain complex and cell homology chain of the dual polyhedral complex forming quiasi-isomorphism –  Nikolai Mnev Dec 21 '09 at 10:02
To Greg -- may be misleading my use of "Poincare" in my question. Nothing transcnental in standard isomormphism of homology and cohomology complexes of dual. The isomorphism is transcendental betwin homology and homology of dual cell complex (or cohomology and cohomology of dual cell complex) –  Nikolai Mnev Dec 21 '09 at 10:09
It sounds like he is using "transcendental" to mean "non-canonical"? I was thrown because there is no evident connection to transcendental numbers like $\pi$. –  Greg Kuperberg Dec 21 '09 at 17:48

Both the cell complex, $C$, and the dual cell complex $C'$ are refined by the first barycentric subdivision $BC$. There are maps $C \to BC$ and $C' \to BC$, sending a cell $\sigma$ to the sum of all cells of the same dimension contained in $\sigma$; these maps are both quasi-isomorphisms.

So, if you allow me to formally invert quasi-isomorphisms, I'm done.

Is the question whether there is an honest map of chain complexes between $C$ and $C'$, without subdividing?

UPDATE Here is something you can do, and something you can't do.

With $C$ and $BC$ as above, and $r : C \to BC$ the refinement map, there is a homotopy inverse $s: BC \to C$. (More precisely, $C \to BC \to C$ is the identity, and $BC \to C \to BC$ is homotopic to the identity.) Working the same trick with $r' : C' \to BC$, we get quasi-isomorphisms between $C$ and $C'$ which are homotopy inverse to each other. As you will see, however, this construction is very nongeometric and inelegant.

Construction: Let $q:BC \to Q$ be the cokernel of $C \to BC$. An easy computation checks that each $Q_i$ is free. Since $C \to BC$ is a quasi-isomorphism, $Q$ is exact. An exact complex of free $\mathbb{Z}$ modules must be isomorphic to a direct sum of complexes of the form $\cdots \to 0 \to \mathbb{Z} \to \mathbb{Z} \to 0 \to \cdots$. Choose such a decomposition of $Q$, so $Q_i = A_{i+1} \oplus A_{i}$ and the map $Q_i \to Q_{i-1}$ is the projection onto $A_{i}$.

Now, consider the map $q_i^{-1}(A_i) \to A_i$ in degree $i$. This is surjective, and $A_i$ is free, so choose a section $p^1_i$. We also define a map $p^2_i$ from the $A_{i+1}$ summand of $Q_{i}$ to $BC_i$ by $p^2_i = d p^1_{i+1} d^{-1}$. In this way, we get maps $p_i = p^1_i \oplus p^2_i: Q_{i} \to BC_i$ which give a map of chain complexes.

We note that $qp: Q \to Q$ is the identity. Therefore, $1-pq$, a map from $BC \to BC$, lands in the subcomplex $C$ and gives a section $s:BC \to C$. Proof of the claim about homotopies will be provided on request.

On the other hand, here is something you can't do: Get the quasi-isomorphism to respect the symmetries of your original space. For example, let $C$ be the chain complex of the cube, and $C'$ the chain complex of the octahedron. I claim that there is no quasi-isomorphism $C \to C'$ which commutes with the group $S_4$ of orientation preserving symmetries.

Consider what would happen in degree $0$. A vertex of the cube must be sent to some linear combination of the vertices of the octahedron. By symmetry, it must be set to $$a (\mbox{sum of the "near" vertices}) + b (\mbox{sum of the "far" vertices})$$ for some integers $a$ and $b$. But then the map on $H_0$ is multiplication by $3(a+b)$, and cannot be $1$.

I imagine you want something stronger then my first answer, but weaker than my second. I am not sure what it it, though.

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Dear David, you got the point -- i need the explicit formula for the invertion of your chain maps to form a quasyi-isomorpism . And this in general is transcnental for BC->C' I hope very much that there should be a nice and scientific one. –  Nikolai Mnev Dec 21 '09 at 14:15
Dear David, yes your construction is of cause what one have in mind. While, I need a finite formula, this means that choices have to be fixed somehow respecting combinatorics. I hope that this can be done in a nice way. –  Nikolai Mnev Dec 21 '09 at 19:31
If $c$ is the cochain that is equal to 1 on some (oriented) simplex $s$ and is zero elsewhere, then the dual of $c$ is the barycentric star of $s$ i.e. the union of the simplices of the barycentric subdivision of dimension equal to the codimension of $s$ and passing through the barycenter of $s$; we consider this as an element of the simplicial complex of the barycentric subdivision. The orientation of the barycentric star is determined by the orientations of $s$ and of the manifold.