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I need help with a result about barycentric subdivision someone already asked about here. However, the part I've been struggling with is the one the original poster saw as trivial, statement (c).

The idea is the following: given a simplicial complex $K$ of dimension $n$ and $K^1$ its first barycentric subdivision, the diameter of any simplex of $K^1$ is, at most, $\frac{n}{n+1}$ times as large as the diameter of the largest simplex in $K$.

I've been following the proof in the following book (it's Lemma 6.4):

Armstrong, M.A., Basic topology., Undergraduate Texts in Mathematics. New York etc.: Springer-Verlag. xii, 251 p. DM 69.00 (1990). ZBL0711.55001.

The author states that the diameter of a simplex is the length of its longest edge. Then, as a consequence of one of the previous statements, proves that any edge $\sigma$ of $K^1$ has two barycenters as its endpoints, say the barycenter of $A$ and the barycenter of $B$, where $B$ is a face of $A$. As a result, the edge is fully contained in $A$ and its barycenter is one of its end-points. The other must be a point in the boundary of $A$.

And this is the step I have trouble with. He states that: $$\text{length}(\sigma)\leqslant\frac{k}{k+1}\text{diameter}(A)$$ where $k$ is the dimension of $A$.

This step must be really easy but I can't seem to prove it. If someone could lend a hand I'd be very grateful!

Short Version:

I need to prove that the distance from the barycentre of any $k$-simplex $A$ to a point of its boundary is $\leqslant \frac{k}{k+1}\text{diameter}(A)$.

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  • $\begingroup$ Poirot would have known. The little grey cells. $\endgroup$ – Will Jagy Apr 16 '18 at 20:46
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It suffices to prove it when $\sigma$ goes from the barycenter of $A$ to one of the vertices of $A$. Let the vertices be $x_0,\ldots,x_k$ with barycenter $b=\frac{1}{k+1}\sum_{i=0}^k x_i$. Then

$$\|b-x_0\| = \|\frac{1}{k+1}\sum_{i=1}^k (x_i-x_0)\|\leq \frac{k}{k+1}\max_i\|x_i-x_0\|.$$

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  • $\begingroup$ Wow, thanks! However, how could I prove that the vertices are indeed the points farthest away from the barycenter? In other words, why does it suffice to prove it for the vertices? $\endgroup$ – Hercule Poirot Apr 17 '18 at 20:07
  • $\begingroup$ If the distance from $b$ to $x_1,\ldots,x_k$ is at most $d$, then the distance from $b$ to any convex combination $\sum_it_ix_i$, where $\sum t_i=1$ satisfies $$\|b-\sum t_ix_i\|=\|\sum t_i(b-x_i)\|\leq \sum t_id = d.$$ $\endgroup$ – Michal Adamaszek Apr 17 '18 at 20:28
  • $\begingroup$ Perfect! Thank you very much! Everything's clear now. $\endgroup$ – Hercule Poirot Apr 17 '18 at 21:03

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