Intersection of the realizations of simplicial complexes

Question

I have great difficulties with an exercise regarding the shape of the intersection of the realizations of two simplicial complexes.

Given a finite simplicial complex K in $R^n$, let |K| $\subset R^n$ denote its realization, i.e. the union of the simplices contained in K.

Let K and L be finite simplicial complexes in $R^n$. Show that $|K| \cap |L|$ is the realization of some simplicial complex in $R^n$.

I have tried to solve this as follows:

Since |K| and |L| are the unions of the simplices in K and L respectively, we have:

$|K|=\bigcup\limits_{\Delta \in K} \Delta$

$|L|=\bigcup\limits_{\Delta^\prime \in L} \Delta^\prime$

Then by De Morgans Laws we get:

$|K| \cap |L|=|K| \cap \bigcup\limits_{\Delta^\prime \in L} \Delta^\prime =\bigcup\limits_{\Delta^\prime \in L} (|K| \cap \Delta^\prime)= \bigcup\limits_{\Delta^\prime \in L} (\bigcup\limits_{\Delta \in K} \Delta \cap \Delta^\prime) $

This suggests that the simplicial complex I'm looking for consists of those simplices that are generated as the intersection of simplices in K and simplices in L. However in order to proceed I have to prove that for every $\Delta \in K$ and every $\Delta^\prime \in L$ the intersection $\Delta \cap \Delta^\prime$ is also a simplex. I know that if $v_0,...,v_k$ are the vertices of $\Delta$ and $w_0,...,w_l$ are the vertices of $\Delta^\prime$ I have

$\Delta=conv(v_0,...,v_k)$

$\Delta^\prime=conv(w_0,...,w_l)$

where $conv(...)$ denotes the convex hull of the vertices. I think it is possible to show that there are index sets $I \subset \{0,1,...,k\}, J \subset \{0,1,...,l\}$ such that

$\Delta \cap \Delta'=conv((v_i)_{i \in I}=conv((w_j)_{j \in J}$.

But I do not see how to give a formal proof of that. I have found a similar question, but since the answers assume a definition I'm not working with, it unfortunately doesn't help me.

Answer 1

You don't want to prove that $\Delta \cap \Delta'$ is a simplex for each $\Delta \in K$ and $\Delta' \in L$, because it's false.

For a counterexample, take a regular hexagon in $\mathbb R^2$ with vertices $A,B,C,D,E,F$, let $K=\Delta$ be the 2-simplex with vertices $A,C,E$, and let $L=\Delta'$ be the 2-simplex with vertices $B,D,F$, and so $\Delta \cap \Delta'$ is a smaller regular hexagon. Of course that smaller hexagon is the realization of a simplicial complex, in fact any regular hexagon is the realization of a simplicial complex with one extra vertex in the center, and six 2-simplices altogether.

You should be able to find a simplicial complex structure on $|K| \cap |L|$ by carefully keeping this subdivision concept in mind. It'll be messy, if you want to do it with full rigor. But the key idea should be that the intersection of any simplex $\Delta$ of $K$ and any simplex $\Delta'$ of $L$ can be subdivided into a simplicial complex. Of course you have to do these subdivisions over all choices of $\Delta$ and $\Delta'$, to make sure that they all fit together into a simplicial complex structure on $|K| \cap |L|$.

Using the assumption that $K$ and $L$ are finite simplicial complexes, I think the best approach to writing down a rigorous proof is probably as a double induction. For the primary induction, you start with $L = \emptyset$ and then add one simplex of $L$ at a time, writing $L$ as the union of subcomplexes $$L_1 \subset L_2 \subset \cdots \subset L_I $$ where each $L_i$ is obtained from $L_{i-1}$ by adding a single simplex $\tau_i$. Assuming that $K$ and $L_{i-1}$ have been subdivided so that their intersection is a subcomplex of each, then you add $\tau_i$. Now you do a secondary induction on the dimension of the skeleta of $K$, writing $K$ as the union of its skeleta $K^{(0)} \subset K^{(1)} \subset \cdots \subset K^{(d)}=K$. You first subdivide $\tau_i$ at each point where its interior intersects $K^{(0)}$, after which $K^{(0)} \cap L_i$ will be a subcomplex of each of $K^{(0)}$ and $L_i$. In the induction step, assuming that $K^{(j)} \cap L_i$ is a subcomplex of each of $K^{(j)}$ and $L_i$, you then further subdivide $\tau_i$ so that $K^{(j+1)} \cap L_i$ is a subcomplex of each of $K^{(j+1)}$ and $L_i$.