Deeper meanings of barycentric subdivision

Question

I just want to ask if there is any deeper motivation or clear geometric "sense" behind the barycentric subdivision. Some friend asked me about this a few months ago, looking back the section at Hatcher, I still feel quite confused. I remember one friend told me combinatorically one can do this from posets back to posets, but this does not give me any way to "understand" it properly. In some books (Bredon, for example), the author use excision property as one of the axioms, I'm wondering "where they came from, why they make any sense?".

Answer 1

There can be many reasons for subdividing simplices, barycentrically or otherwise.

For a simplicial complex (triangulated space) there are the simplicial homology groups. These are known to be isomorphic to the singular homology groups, therefore (1) invariant under homeomorphism, and in particular (2) invariant under (not necessarily barycentric) subdivision. Before the invention of singular homology, I believe that (1) was unknown. Fact (2) was a key part of the theory. Subdivision is important simply because even if your space is made out of simplices you will sometimes care about subsets which are only unions of simplices after you cut the space up finer. In simplicial homology, excision is an easy algebraic fact, stemming from the fact that when a complex is a union of two subcomplexes then every simplex is in one or the other (or both).

In singular theory, as you know, invariance under homeomorphism is a triviality but excision requires some work. The point is that when a space is a union of two open sets then (bad news) not every singular simplex is in one or the other but (good news) simplices can be systematically replaced by combinations of smaller simplices to show that this does not matter. This is where subdivision is used, and there is no reason it has to be barycentric. It's like with the fundamental group: you might explore a space by using maps of a standard unit interval into it, but in proving the Seifert-Van-Kampen Theorem you might want to subdivide that interval into little pieces.

Barycentric subdivision also rises in PL (piecewise linear) topology in one other specific technical way that has nothing much to do with homology: regular neighborhoods. In a finite simplicial complex $K$, the smallest neighborhood of a given subcomplex $L$ that is itself a subcomplex does not in general have $L$ as a deformation retract, but this becomes true if you first barycentrically subdivide twice.

And in the interplay between categories and simplicial constructions barycentric subdivision turns up in various ways.

ADDED in response to Hatcher's answer and its comment thread:

Yes, there is a way of extending to all $n$ the pattern that begins: cut a segment in half, cut a triangle into four equal pieces using midpoints of edges ... It is sometimes called "edgewise subdivision", I believe. It may be realized for simplicial sets as follows: A simplicial set is a functor $\Delta^{op}\to Set$ where $\Delta$ is the category of standard nonempty ordered finite sets; its subdivision is obtained by composing with (the opposite of) the functor $\Delta\to\Delta$ which takes an ordered set to two copies of that set laid end to end. This leaves the realization unchanged. Applied to a standard $n$-simplex, it gives a certain subdivision with $2^n$ pieces. If $n>2$ then the pieces are not all the same shape. If $n=3$ you get a tetrahedron cut into four scaled-down models of itself sitting in the corners and four more whose union is an octahedron; these four all share an edge, the only internal edge that there is. It's not immediately clear to me what diameter estimate is available for the pieces.

This can be generalized so that you now cut an edge into $k$ equal pieces and a triangle into $k^2$ congruent pieces (almost half of which are upside down) and in general cut an $n$-simplex into $k^n$ pieces. This $k$-fold edgewise subdivision plays a role in the area of cyclic homology and related things: when a simplicial set $X$ has the kind of extra structure that makes it a cyclic set (a suitable action of a cyclic group of order $m$ on the set $X_{m-1}$ for all $m>0$) then its realization has an action of the circle group, and to make the action of the subgroup of order $k$ appear as a simplicial action you can do the $k$-fold edgewise subdivision described above.

There is also another edgewise subdivision. In this one the $1$-simplex is cut in half as before and the $2$-simplex is cut into four pieces in the following way: join the middle vertex to the midpoint of the opposite side, and join that midpoint to the midpoints of both of the other sides. This construction corresponds to the functor $\Delta\to\Delta$ that takes an ordered set to two copies of the same laid end to end but with the order reversed in one copy.

See also my answer to the recent MO question "Endofunctors of the Simplex Category".

The second edgewise subdivision that I described can be used to analyze the relationship between two definitions of algebraic $K$-theory: Quillen's $Q$-construction is essentially a subdivision of Waldhausen's $S$-construction.

Answer 2

In the other replies there has been some mention of alternative methods for subdivision besides barycentric subdivision, but these are rarely encountered in algebraic topology. What are some of these other methods, in fact? Preferably they should be natural and canonical, not based on random choices. I dimly recall seeing somewhere (in a paper of Quillen or Segal?) a subdivision method generalizing the simple idea of subdividing a triangle into four triangles by adding new vertices at the midpoints of the three edges, but the generalization to higher dimensions isn't obvious. Does anyone know a reference for this? Another approach might be to use the canonical subdivision of an n-simplex into n+1 cubes, one at each vertex of the simplex, then subdivide each cube into small cubes in the obvious way, then subdivide the small cubes into simplices in some natural way. This seems a bit cumbersome, however.

A drawback of barycentric subdivision is that it takes some work to show that sufficiently many iterations of barycentric subdivision produce arbitrarily small simplices. It would be nice to have a subdivision method for which this was obvious.

Answer 3

The following paper includes a very detailed and elegant description of how to construct edgewise subdivisions that subdivide a $d$-simplex into $k^d\cdot d$-simplices all of the same volume and shape characteristics, for every integer $k\geq 1$

Edgewise Subdivision of a Simplex H. Edelsbrunner and D. R. Grayson DISCRETE AND COMPUTATIONAL GEOMETRY Volume 24, Number 4, 707-719

Answer 4

You can think of a simplex as a finite ordered list (i.e., the vertices). The simplices of its barycentric subdivision are the lists of subsets of the first list, ordered by inclusion.

Answer 5

This is really just a comment/question for Tom (or for any other knowledgable topologist), but it has got far too long. It's also an attempt made from a position of almost complete ignorance to (re)construct an alternative foundation for singular homology :-)

Is the following a correct description of his first "edgewise decomposition" of the $3$-simplex?

Let our standard tetrahedron have vertices (0,0,0), (1,0,0), (1,1,0) and (1,1,1) in $\mathbb{R}^3$. So it consists of the $(x,y,z)$ with $1\ge x\ge y\ge z\ge0$. We split this into eight small tetrahedra. To avoid fractions, let's first double the size so we split the tetrahedron with vertices (0,0,0), (2,0,0), (2,2,0) and (2,2,2) into tetrahedra with vertices:

• (0,0,0), (1,0,0), (1,1,0) and (1,1,1), (*)
• (1,0,0), (2,0,0), (2,1,0) and (2,1,1), (*)
• (1,0,0), (1,1,0), (2,1,0) and (2,1,1),
• (1,0,0), (1,1,0), (1,1,1) and (2,1,1),
• (1,1,0), (2,1,0), (2,2,0) and (2,2,1), (*)
• (1,1,0), (2,1,0), (2,1,1) and (2,2,1),
• (1,1,0), (1,1,1), (2,1,1) and (2,2,1),
• (1,1,1), (2,1,1), (2,2,1) and (2,2,2). (*)

Then the starred tetrahedra are those in the "corners" of the large tetrahedron while the other four share the "internal" edge from (1,1,0) to (2,1,1).

How can this be generalized? Take as the standard $n$-simplex $\Delta_n$ that defined in $\mathbb{R}^n$ by the inequalities $1\ge x_1\ge x_2\ge\cdots\ge x_n\ge0$. Then the unit cube can be decomposed as $n!$ copies of $\Delta_n$ obtained by coordinate permutations. Then $\mathbb{R}^n$ itself can be decomposed into copies of $\Delta_n$ by translating the decomposition of the unit cube by vectors in the integer lattice. Call this our standard decomposition of $\mathbb{R}^n$.

We can now decompose our standard simplex $\Delta_n$ into $k^n$ congruent simplices each similar to $\Delta_n$. Again it's more convenient to scale $\Delta_n$ by a factor of $k$ and then decompose into simplices congruent to $\Delta_n$. But $k\Delta_n$ is a union of $k^n$ simplices in the standard decomposition of $\mathbb{R}^n$, and this does it.

What I haven't worked out yet, is if there an analogue of the barycentric chain maps $S$ and $T$ (in Hatcher's notation) and what those should be. If so then this decomposition would provide an alternative approach to the excision theorem in singular homology. There would be a couple of advantages to this over the classical approach.

• The standard simplex $\Delta_n$ lives in $\mathbb{R}^n$ rather than $\mathbb{R}^{n+1}$.

• Repeated subdivision of the standard simplex results in simplices similar to the original, as opposed to repeated barycentric subdivision which produces a plethora of different-shaped simplices. The technical advantage here is that we immediately see what the diameter of the simplices in our subdivision are, rather than have the bound of the factor $n/(n+1)$ occurring in the barycentric subdivision which takes some effort to prove.

On the other hand, there are disadvantages: for instance, most of the faces of $\Delta_n$ are not congruent to $\Delta_{n-1}$.

Is there some reference where these ideas are fully worked out?

Answer 6

I won't address the second part of your question, since I'm not sure it's well-defined. But the first (as I understand it, "where does barycentric subdivision come from?") is a good question.

If you've read Hatcher's proof of the excision theorem, you'll remember he defines, for an open cover $\mathcal{U}$ of $X$, the chain complex $C^\mathcal{U}(X)$ to be the subcomplex of $C(X)$ given by singular simplices whose images are contained in an element of $\mathcal{U}$. He shows that the inclusion $C^\mathcal{U}(X)\to C(X)$ is a homotopy equivalence using barycentric subdivision---and excision, in the second form he states it, is obvious enough for the homology of $C^\mathcal{U}(X)$.

So there are two questions:

(1) What is the motivation for introducing the complex $C^\mathcal{U}(X)$?

(2) Why do we want barycentric subdivision to prove the homotopy equivalence?

Question (2) is easy enough -- we don't actually need barycentric subdivision, we just need something like it. We want to be able to send an arbitrary simplices $\sigma$ to sums of simplices contained within elements of $\mathcal{U}$, such that the sum in question is homologous to $\sigma$ (i.e. the boundaries cancel out). The obvious thing to do is to apply the Lebesgue number lemma, so we need some way of making simplices smaller by some definite factor. Furthermore, we need the map in question to be a chain map (to commute with the boundary map), which means it has to be built up inductively -- $\partial S=S\partial$ means that restricting the subdivision of an $n+1$-simlex to its faces must give the same subdivision as simply subdividing the faces. Barycentric subdivision is an obvious way to do this. You can motivate it yourself by trying to come up with a subdivision satisfying these two criteria for $1$-simplices and $2$-simplices; I bet you'll come up with barycentric subdivision. But it's by no means necessary -- there are any number of similar subdivisions. (You might for example define singular homology via cubes; then there's an extremely obvious subdivision!)

Question (1) is much deeper. Of course, excision in the second form Hatcher gives it suggests an approach like this -- but the intuition is a bit more exciting than that. What this approach means is that homology can be computed "locally"! Introducing singular cohomology, this idea leads directly to sheaf cohomology -- motivationally, if not historically.

Answer 7

Not sure whether this is an answer to the question but thought of sharing what I feel makes the idea of barycentric subdivision very natural.

On metric spaces one has the crutch of the notion of distance to make sense of what is "small" and then things like Lebesgue Number help one get covers with as small open sets as possible. This notion of smallness gets hard to emulate on general topological spaces and here barycentric subdivision helps get across that by exploiting together the benefit of having a notion of smallness in Euclidean space and the notion of continuity. If something is small in the euclidean norm (as available on the simplex) it will map to something small in an arbitrary topological space under a continuous map.

Further barycentric subdivision sort of very optimally exploits the notion of convexity unlike say thinking in terms of cubes. I am not sure how to make it precise but using cubes instead of tetrahedrons in 3D will bring in many more maps than necessary.

Homology in some sense depends on lower amount of information than a priori it seems to need. Like instead of all continuous maps from the simplex to the space if one takes only those maps which are non-degenerate on some face of the simplex, even then one gets the same homology theory. (This is what Massey does). This kind of think might get harder to see with out barycentric subdivision.

I am not sure what sense it would make if one needed to do infinite barycentric subdivision since in that "limit" one will land up looking at maps from 0 volume subsets of euclidean space to the topological space and won't these simply not allow non-trivial continuous maps from them? I am not sure how to make it precise.

In "reasonable" topological spaces for any open cover chosen on it, the corresponding cover gotten on the simplex has a Lebesgue number which is a finite number. Hence one can do a finite number of barycentric subdivisions till each piece has diameter less than this and hence a finite number of barycentric subdivisions should be enough.

It would be illuminating if someone can explain why this might fail for things like fractals. And if it does then how does one go around it?

Also the notion of barycenter coincides with the idea of "center of mass" if one can attach equal masses to all the vertices of the simplex. This gives quite an intuitive help. I guess should be cases where this identification is more tangibly fruitful.