When studying a new cohomology, one has to see what goes over. One of the nice identities in simplicial cohomology is the Alexander duality. It has little to do with Poincare duality but it would be nice if something similar would work for connection cohomology. So far this has not worked yet for connection cohomology and this is not a surprise as Alexander duality is more homotopical than topological and connection cohomology is topological in nature, and not invariant under homotopy. We see that especially in the discrete, the dual of a low dimensional space is a high dimensional simplicial complex. Here is something about combinatorial Alexander duality. It is a really cute result and illustrates the work of a fascinating and very unusual topologist.

James Waddell Alexander proved a beautiful duality result. He was not only a great algebraic topologist, but a mountaineer, making over 200 climbs in the Alps. [Climbing seems a bit to be in the nature of topologists, George de Rham, Hassler Whitney or John Milnor come in mind. See a photo diary from 2014, where we follow some of the footsteps of George de Rham and Jean Piaget in the alps. There are some nice quotations here about de Rham climbing. Having climbed the Baltschieder Stockhorn myself, I like the following quotations of Milnors on that page.]. |

The Alexander duality gives a duality $b_k(M)=b_{d-1-k}(S-M)$. It relates the the Betti numbers of a manifold $M$ inside a $d$-sphere $S$ with the Betti numbers of of the complement $S-M$. Alexander proved it in 1922. There is a biography of Alexander at Bulletin 2001 [PDF].

A short proof of a combinatorial version was proven by Bjoerner and Tancer PDF. They state Kalai (1983) and Stanley (1982) as the first sources of a combinatorial result. The combinatorial result holds for abstract finite simplicial complexes. There, if the complex $G$ has $n$ zero dimensional elements $V$, it tells that for $k=1,…,n-1$, the Betti numbers are related by the formula

For any finite abstract simplicial complex G, one has $b_k(G) = b_{n-3-k}(G^*)$, where $G^*$ is the dual complex, the complex generated by the complements of $G$. |

Of course, as stated, the result assumes that $n \geq 5$. Indeed, for the example with $n=4$ given by $G=C_4 = \{1,2,3,4,12,23,34,41\}$ with Betti numbers $(1,1,0,0,…)$, we have $ G^*=\{1,2,3,4,13,24 \}$ which has Betti numbers $b=(2,0,…)$. The case of $C_5$ is already interesting as we have then a cohomological duality between a circle and the complement of a circle in a $3$-sphere.

We have a 6 line code to compute a basis for the cohomology groups of a complex and so the Betti numbers. One can easily check experimentally the Alexander duality to hold.

If you want to do that, here is the procedure to generate the complement:

GenerateComplex[A_]:=Module[{G=A}, Do[G=Union[G,Delete[Subsets[A[[k]]],1]],{k,Length[A]}]; Union[Map[Sort,G]]]; DualComplex[G_]:=Module[{s={}}, V=Union[Flatten[G]]; F=GenerateComplex[{V}]; F=Complement[F,{V}]; Do[x=F[[k]]; y=Complement[V,x]; If[Not[MemberQ[G,y]],s=Append[s,x]],{k,Length[F]}]; Sort[Map[Sort,s]]]; G=GenerateComplex[{{1,3,2},{1,4}}]; GG=DualComplex[G];

Now, you can add the 6 line code and compute the Betti numbers of both complexes.

The Bjoerner and Tancer paper is elegant but one might wonder whether there is not a pure linear algebra proof as the Betti numbers are just kernels of explicit matrices. The question is whether there exists an explicit correspondence between the harmonic k-forms of G and the harmonic n-3-k forms in $G^*$. So, given a harmonic k form, is there a direct way to get a harmonic n-3-k form? We have not yet experimented with it, but one thing to try is just to extend a function to the complement of the faces in the sphere and check that this function is harmonic. We have to try it out. If it worked, then the Bjoerner-Tancer proof could be simplified even more.

[Update: April 15: it might actually be very easy: take a harmonic p-form f, build the usual Hodge dual $f^*(x)$ on every face of the $n-2$ dimensional sphere defined by the complete complex with $n$ elements (it defines a (n-1) dimensional simplex which becomes a (n-2) dimensional sphere if the set $\{1,2,3,,…,n\}$ is removed (the interior of the tetrahedron). Given a harmonic p-form, we get so a harmonic (n-3-p)-form $f^*$. The map $f \to f^*$ is an isomorphism. The Alexander duality result follows. It is the magic of Hodge again.]

You might wonder where the “3” in the n-3-k comes from. It is very natural and explained in the Bjoerner-Tancer paper. If we take away the (n-1) simplex in the simplicial complex, we get a (n-2) dimensional sphere. Every simplex x in G has a set of dual simplices obtained by intersecting the complement $x^*$ with the faces of that sphere. Now, p-forms are described by simplices with p+1 elements. If it pairs with a q-form with q+1 elements, then (p+1)+(q+1) = n-1, explaining q=n-p-3.

Alexander also discovered the Alexander horned sphere in 1924. It is an important construction as it shows that the complement of a 3 ball within a 3 sphere does need to have a non-simply connected outside. The example highlights the difference between topological and smooth manifolds. When doing it in the discrete, we are essentially doing the smooth case, where the Jordan-Brouwer-Schoenfliess separation theorem holds: See Jordan Brouwer Schoenfliess in the discrete.