# Planar duality generalized to embedded simplicial complexes

Let $$K$$ be a finite $$d$$-dimensional simplicial complex embedded in $$\mathbb{R}^{d+1}$$. The setting of this question is simplicial homology with coefficients over $$\mathbb{Z}_2$$. By Alexander duality $$K$$ partitions $$\mathbb{R}^{d+1}$$ into $$\beta_d + 1$$ connected components, where $$\beta_d$$ is the $$d$$th Betti number of $$K$$. We define $$G$$ to be the dual graph of $$K$$. $$G$$ is the graph whose vertices are the connected components of $$\mathbb{R}^{d+1}$$ and whose edges are the $$d$$-dimensional simplices of $$K$$. Two vertices are adjacent if and only if their corresponding connected components share a common $$d$$-simplex in the intersection of their boundaries.

It is easy to see that there is a one-to-one correspondence between $$d$$-dimensional cycles in $$K$$ and minimal edge cuts in $$G$$. This is analogous to cycle/cut duality in planar graphs.

What does a cycle in $$G$$ correspond to in $$K$$? My intuition is that it should be something analogous to a cut in a graph. In terms of homology a graph cut is a set of edges whose removal increases the rank of the zeroth homology group of the graph (because this homology group counts the number of connected components.)

Is the dual of a cycle in $$G$$ a set of $$d$$-dimensional simplices in $$K$$ whose removal increases the rank of the $$(d-1)$$th homology group? This seems likely to me, but I do not know how to prove it.

• I fixed an error in my answer below. Is anything missing / are you looking for something different? – Geva Yashfe Apr 19 '20 at 16:39
• This answer is exactly what I wanted. Thanks a lot, the answer will be very useful to me. – Will Apr 19 '20 at 18:38
• You're welcome, it was a very nice question. In retrospect I'm surprised that I haven't seen it before. – Geva Yashfe Apr 19 '20 at 18:51

Like you guessed, the dual of a cycle in $$G$$ is a set of $$d$$-simplices in $$K$$ whose removal increases the rank of the $$(d-1)$$-th homology group.

In the following I use that $$H_i(X;\mathbb{Z}_2) \cong H^i(X;\mathbb{Z}_2)$$. It might have been better to keep track of the distinction...

Consider the closures of the connected components of $$K$$'s complement in $$\mathbb{R}^{d+1}$$, $$\{X_1,\ldots,X_n\}$$, where $$n=\beta_d(K)+1$$. These are $$(d+1)$$-dimensional simplicial complexes,and their boundaries are $$d$$-dimensional.

Edit: It is of course not true that the connected components of the complement are simplicial complexes, although if the embedding is a PL embedding in $$S^{d+1}$$ instead then they can be triangulated. I mistakenly answered this more combinatorial variant of the question.

Nevertheless the method of proof works if we think of the components of $$\mathbb{R}^{d+1}\setminus K$$ as being topological subspaces of $$\mathbb{R}^{d+1}$$, with the following addition: note that no more than two connected components can contain a single $$d$$-dimensional simplex of $$K$$ in their common boundary. A reference is Daverman and Venema, "Embeddings in Manifolds," corollary 7.1.2 and the preceding proposition (the section is "Codimension-one separation properties," accessible from Google Books).

Let us consider the boundaries of each pair of distinct $$X_i,X_j$$ as though they were disjoint: we want to think of them as different simplicial complexes.

Let $$C$$ be a cycle in $$G$$. We want to know what happens when the $$d$$-simplices $$C\cap K"$$ is removed from $$K$$. Since $$K$$ is the Alexander dual of $$\bigsqcup_{i=1}^n X_i,$$ we can determine this by looking at what happens to the $$\{X_i\}$$ instead. Removing a $$d$$-simplex from $$K$$ is equivalent to gluing some pair along it, say $$X_1,X_2$$, so we can use the Mayer-Vietoris sequence for reduced homology: $$\ldots\rightarrow H_k(X_1 \cap X_2) \rightarrow H_k(X_1)\oplus H_k(X_2) \rightarrow H_k(X_1 \cup X_2) \rightarrow H_{k-1}(X_1\cap X_2) \rightarrow \ldots$$

Here, the intersection $$X_1 \cap X_2$$ is the intersection after the gluing. Hence it is a $$(d-1)$$-dimensional simplex on the boundary, and its reduced homology is zero in all dimensions. Thus the first homology of $$X_1 \cup X_2$$ is just the direct sum $$H_1(X_1)\oplus H_1(X_2)$$.

Now we can remove, one by one, the edges (or $$d$$-simplices) in a simple cycle of $$G$$. Let us remove the last pair of edges at once; this means we are gluing the connected component $$X_t$$ dual to the last vertex in the cycle by both of its edges simultaneously. The other piece glued to $$X_t$$ is the union of components corresponding to the other vertices of the cycle, let's call it $$Y$$. The reduced Mayer-Vietoris sequence in low degrees gives us $$H_1(X_t \cap Y) \rightarrow H_1(X_t)\oplus H_1(Y) \rightarrow H_1(X_t \cup Y) \rightarrow \tilde{H}_0(X_t \cap Y) \overset{j}{\rightarrow} \tilde{H}_0(X_t) \oplus \tilde{H}_0(Y).$$ Here the intersection $$X_t \cap Y$$ is a disjoint union of two simplices, which has $$H_1 = 0$$ but $$\tilde{H}_0 = \mathbb{Z}_2$$. However, the map $$j$$ in the sequence above is $$0$$. So the rank of $$H_1(X_t \cup Y)$$ is exactly $$\mathrm{rk}(H_1(X_t)\oplus H_1(Y)) + \mathrm{rk}(\tilde{H}_0(X_t \cap Y)) = \mathrm{rk}(H_1(X_t))+\mathrm{rk}(H_1(Y))+1.$$

By induction on the length of the cycle, and using that the first homology of a space with several connected components is the direct sum of their first homologies, we see that the result of cutting out the dual of a cycle of $$G$$ from $$K$$ increases the first homology of $$\mathbb{R}^{d+1}\setminus K$$ by $$1$$. Alexander duality (this time applied to the complement) gives the result on $$H_{d-1}(K)$$.

• The procedure used in the above can also be used to show that any inclusion-minimal set of edges in $G$ such that the removal of the corresponding simplices in $K$ increases the rank of $H_{d-1}(K)$ is a cycle: Observe that removing the simplices corresponding to an acyclic set of edges does not change $H_{d-1}(K)$, and that a cycle does. Thus if a set of edges increases the rank of $K$'s $(d-1)$-th homology it contains a cycle; this cycle is inclusion-minimal among such sets. – Geva Yashfe Apr 18 '20 at 15:12