Barycentric subdivisions of simplices yield a simplicial complex

The following interesting result (in particular parts (b) and (d)) is stated either as a obvious fact or as an exercise in several books on algebraic topology:

The barycenter $b_\sigma$ of an $n$-simplex $\sigma$ with vertices $v_0, \ldots, v_n$ is the point $$\frac{1}{n+1}(v_0 + \cdots + v_n).$$ Show that:

(a) For any strictly increasing sequence $\sigma_0 \subset \sigma_1 \subset \ldots \subset \sigma_k$ of faces of $\sigma$, the barycenters $b_{\sigma_0}, b_{\sigma_1}, \ldots, b_{\sigma_k}$ form the vertices of a $k$-simplex.

(b) These simplices form a simplicial complex, whose topological space is $\sigma$. This is called the barycentric subdivision of $\sigma$.

(c) The diameter of any simplex in the barycentric subdivision of $\sigma$ is at most $\frac{n}{n+1}$ times as large as the diameter of $\sigma$.

(d) If $K$ is any simplicial complex, the barycentric subdivisions of its simplices form a simplicial complex $K'$ with $|K'| = |K|$ (here $|K|$ is the geometric realization of $K$).

I recently proved part (c), but I am not seeing how the proofs of (a), (b), and (d) would proceed. I was wondering if anyone visiting today knows how to deduce any of the parts besides (c), and if so, would be willing to prove any items or give hints towards proving it. Any constructive responses would be greatly appreciated!

• For (a) you need only prove they sequence of barycentres are in general position, right? i.e. you can't find one of the points in the convex hull of the others. (b) by (a) you've got a bunch of simplices. Check their union is the entire big simplex, and they intersect in simplices. – Ryan Budney Feb 26 '12 at 10:22
• Have you looked at Hatcher's Algebraic Topology? I seem to recall it discussing barycentric subdivision. You might find it useful. – Neal Mar 1 '12 at 12:23

First, let us fix some definitions.

Definition: an $n$-simplex is a $n$-dimensional polytope which is the convex hull of it's $n+1$ vertices. Importantly, no vertex is contained in the convex hull of any other vertices.

Definition: a simplicial complex $K$ is a set of simplicies such that any face of a simplex of $K$ is also in $K$ and the intersection of any two simplicies $\sigma_1, \sigma_2$ is a face of $\sigma_1$ and $\sigma_2$.

Part (a): In order to show that these vertices actually form the vertices of simplex, we must check that no vertex is contained in the convex hull of any of the others. Suppose that $b_k=(x_{0,k},x_{1,k},\cdots,x_{n,k})$ (in $v_i$ coordinates) is contained in the convex hull of the other $b_j$. This would mean that there's a nontrivial linear dependence relation on the the set $\{b_j\}$.

But if we have some linear relation $\sum_{i=0}^n a_ib_i=0$, we can replace the $b_i$ by their defining linear combinations $\frac{1}{n+1}(v_0+\cdots+v_n)$ and obtain a linear relation on the $v_i$. But this is clearly ridiculous as the vectors $v_i$ are linearly independent since we assumed they were the vertices of a simplex. Some quick calculation shows that if the coefficients of $v_i$ are all zero, then so too are the coefficients $a_i$.

Part (b): For this, we need to check that the union of these simplicies is the entire space and that any two simplicies intersect in simplices.

To do the first, I'll give a method for determining which simplex a given point lies in. Represent an arbitrary point in our simplex as the linear combination of $v_i$, ie $x=a_0v_0+\cdots+a_nv_n$. I claim that it lies in the simplex determined by $b_{\sigma_0},\cdots,b_{\sigma_n}$ where $\sigma_i$ is the simplex defined by taking the greatest $i+1$ elements of the set $\{v_j\}$ given an order by $v_i\geq v_j$ if $a_i\geq a_j$ and ties broken by the lexographic ordering. Note that this always returns an answer- so every point in our original simplex is in at least one simplex in the barycentric subdivision.

Next, we need to show that if any collection of simplicies intersect, they intersect in a simplex. Since a simplex is determined entirely by its vertices, this means that the intersection set is determined by intersection of the vertices of the simplicies in common. But this means that the intersection of simplicies is exactly a simplex with vertices which are the intersection of the vertex sets of the simplicies we're intersecting. (Since these form a subset of the vertices of a simplex, they again determine a simplex.)

Part (d): Apply (b) to see that the barycentric subdivision forms a simplicial complex. The geometric realizations are the same because the behavior on the barycentric subdivision is still totally determined by the behavior of the points that were inherited from the original simplex. This is a moral-of-the-story answer, because I'm not sure what definition of geometric realization you're working with (the one I know is defined as a functor from simplicial sets to compactly generated hausdorff topological spaces- if you post your definition and you're still interested, I can add more to this.)

Finally, I've noticed from your recent questions that you seem to be in the midst of self-studying a lot of the foundations for algebraic topology. I highly recommend actually taking a course in the subject and discussing issues with your professor and coursemates- live interaction with peers and teachers is a much more reliable and useful tool than consulting stackexchange every time you have a potential issue.