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.

  • $\begingroup$ I fixed an error in my answer below. Is anything missing / are you looking for something different? $\endgroup$ Apr 19, 2020 at 16:39
  • $\begingroup$ This answer is exactly what I wanted. Thanks a lot, the answer will be very useful to me. $\endgroup$
    – Will
    Apr 19, 2020 at 18:38
  • $\begingroup$ You're welcome, it was a very nice question. In retrospect I'm surprised that I haven't seen it before. $\endgroup$ Apr 19, 2020 at 18:51

1 Answer 1


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 $H_{d-1}(K)$.

  • $\begingroup$ 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. $\endgroup$ Apr 18, 2020 at 15:12

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