Alright 3rd time's the charm - editing again to put all my cards on the table.
Consider a PL $n$-manifold $M$. Define the complexity $c(M)$ of $M$ to be the minimum number of $n$-simplices needed to triangulate $M$.
For $n>2$ getting lower bounds on $c(M)$ is a difficult problem. I am interested in the particular case where $M$ is a closed, orientable, simply-connected 4-manifold. It is conjectured that $c(M)$ is bounded below by $2\chi(M)-2$. To prove this, I wonder if we can take a "local-to-global" type argument in the sense that I will explain as follows.
Fix a triangulation $\mathcal{T}$ of $M$. For a vertex $V$ of $\mathcal{T}$, consider the link of $V$: $\mathrm{lk}(V)$ -- this is the boundary of a regular neighbourhood of $V$. Suppose the $f$-vector (the vector whose $i$th entry is the number of $i$-simplices in the given complex) of $\mathrm{lk}(V)$ is $(v,e,f,t)$. Now consider the star of $V$, which is a regular neighbourhood of $V$. This can be realised as $\mathrm{st}(V)\cong C(\mathrm{lk}(V))$, where $C(K)$ is a cone over over $K$. If we write the $f$-vector of $\mathrm{st}(V)$ as $(v',e',f',t',p')$, then since the star is the cone of the link, we have $$(v',e',f',t',p')=(v+1,e+v,f+e,t+f,t),$$ (this can be readily seen by considering how the cone over something is constructed) where the $v,e,f,t$ come from the $f$-vector of $\mathrm{lk}(V)$.
Now note the following:
- Since $\mathrm{lk}(V)$ is a closed 3-manifold, we have that $f=2t$.
- The number of tetrahedra $t$ in $\mathrm{lk}(V)$ is equal to the degree of $V$ in $\mathcal{T}$.
- Hence, the number of pentachora (4-simplices) $p'$ in $\mathrm{st}(V)$ is also equal to $\deg(V)$.
Let me now define the quantity $X(K):=v-e+f$ for any $n$-dimensional complex $K$ with $n\geq 2$. From the $f$-vector of $\mathrm{st}(V)$, we have that $$X(\mathrm{st}(V))=(v+1)-(e+v)+(f+e)=f+1=2t+1=2p'+1=2\deg(V)+1.$$
The lower bound on $c(M)$ I desire could be shown (amongst many other equivalent formulations) if one can show that $$X(\mathcal{T})\leq 2P(\mathcal{T})+1,$$ where $P(\mathcal{T})$ is the number of pentachora in $\mathcal{T}$ (the lower bound comes from bounding the Euler characteristic of $M$, $\chi(M)=X(\mathcal{T})-T(\mathcal{T})+P(\mathcal{T})$)
Now we arrive at the heart of the title of this post: We can see that the "$X(K)\leq 2P(K)+1$" bound holds locally on the neighbourhoods of the vertices of $\mathcal{T}$ --- which brings us to my question can we extend this local condition to a global one? i.e. can we use the fact that for all vertices $V$ of $\mathcal{T}$ we have $X(\mathrm{st}(V)\leq 2P(\mathrm{st}(V))+1$ to establish the same $X$ bound on the entirety of $\mathcal{T}$?
Summing $X$ over all $V$ seems promising, and since each pentachoron has 5 vertices, we can divide by 5 to account for (at least some of?) the overcounting in the sum (though I'm not sure whether this accounts for all of the overcounting when summing up the $X$ quantities..?) - but I'm not entirely sure how to then tie this sum back to $\chi(M)$, etc.
Or whether we could argue something like "X holds on the vertices, so extending linearly over the simplices of $\mathcal{T}$... something something... the bound holds on all of $\mathcal{T}$"??
Anyway, I think this pretty much covers all the details for my particular case... but again, happy to elaborate on anything if need be.
I will also add finally that all of this conjecture is backed up by extensive "experimental" evidence (i.e. that the $X(\mathcal{T})\leq 2P(\mathcal{T})+1$ bound holds, $2\chi\leq c(M)$, etc.).
