Video (Introduction to subdivision)
Consider the following pictures:
On the left we have the -simplex , divided into two pieces. In Lemma 10.27 we showed that if and are joinable paths in , then . Equivalently, if we start with a path and split it in the middle to get two paths and , then . Thus, there is a sense in which subdivision of paths acts as the identity in homology.
In the middle picture we have divided the simplex into pieces. In the right hand picture, and the interactive demonstration, we have divided into pieces. It will again turn out that this kind of subdivision acts as the identity in homology. To prove this, we need to study the combinatorics of the subdivision process.
We subdivide by introducing a new vertex in the middle (which we call the barycentre), giving two copies of . To subdivide , we first subdivide each of the edges in the same way as , giving edges altogether. We then take each of these subdivided edges and connect it to the barycentre of ; this divides into copies of . Next, we divide all of the faces of in the same way as , giving triangles on the surface of . We connect all of these to the barycentre of ; this divides into copies of . We can continue in the same way to divide into copies of . We now start to make this more formal.
The barycentre of is the point (so , for example). We will write instead of if there is no danger of confusion.
Given any linear -simplex , we define
More generally, if with and each being a linear -simplex, we define .
It is possible to define for nonlinear -simplices, but a little work is required to check that the resulting maps are always continuous. We do not need the general case so we omit it.
Let be a linear combination of linear -simplices in with . Then .
We can easily reduce to the case where is a single linear -simplex, say . Let be the same as except that is omitted, so , so . On the other hand, we have . For the initial term in we omit the and we have a sign ; this just gives us . For each of the remaining terms in we omit the appearing in position of , and multiply by ; this gives , which cancels with a term in . Putting everything together gives as claimed. ∎
We now want to define certain elements for all . The idea is that we subdivide into smaller copies of as sketched previously, and take to be the sum of these smaller copies with suitable -signs to make the orientations match up correctly. We can mark some points in and as follows:
It will work out that
The general picture is as follows.
We start with . Now suppose that we have and we have already defined an element which is a -linear combination of linear simplices. For we have an affine map and thus a chain , which is again a -linear combination of linear simplices. We put
This defines for all by recursion. Next, suppose we have a space and a map , so . The map then gives a map , and we define .
The map sends , and to , and respectively. It follows that
After expressing and in the same way, we obtain the advertised formula for :
Suppose we have a path . We identify with as usual, so the points , and become , and respectively. The map is thus , and the map is . This means that , where and . In other words, is the second half of and is the reverse of the first half of , so .
An alternative approach is as follows. Let be a permutation of . For we put . This gives a linear -simplex . It can be shown that . We could instead have taken this formula as the definition of ; that would make some things easier and some other things harder.
For any and any we have in .
We can easily reduce to the case where , or in other words . We then have
∎
If we let denote the identity map considered as an element of , then we have , and therefore .
We have . By definition is linear and has . It follows that as claimed. ∎
The map is a chain map.
We must show that for all and all we have in . If then and so the claim is clear. For we will argue by induction. We can easily reduce to the case where , or in other words . We then have , and is a chain map, so . On the other hand, we have
It will therefore be enough to prove that .
We are assuming inductively that for all . We can take , so ; it follows that , or in other words . We also know from Lemma 18.3 that . As and this can be rewritten as , as required. ∎
The chain map is chain-homotopic to the identity.
We define chains recursively as follows. We start with . Now suppose that and we have already defined . For we have a face inclusion , using which we can form . We put
and ; this completes the recursion step.
Next, given we note that and so . We define , and extend this linearly to get . We will prove that this gives the required chain homotopy.
As a first step, we will reformulate the definition of . We have so
It follows that
We will now prove by induction that for all spaces and all . When the claim is just that , which is true. Suppose we have proved the claim for . We can then apply it to the element ; we find that
Using and we can rewrite this as
or in other words . We can therefore take in Lemma 18.3 to get . After recalling our formula above for and the definition we get
Now suppose we have a map . We apply to the above equation, noting that and and and . We get
or equivalently . We have proved this for , but it follows by linearity for all , as required. ∎
Let be a linear simplex. We define
and we call this the diameter of . More generally, given a chain we put .
If is a -linear combination of linear simplices then we have .
In the case all diameters are zero so the claim is clear. We can therefore assume that and argue by induction. The claim involves the number ; from the second form it is clear that .
We can easily reduce to the case where is a single linear simplex, say . Put , so for all . Put , which is the barycentre of . For any we can write as . Using this, we get . In the sum on right hand side, the term for is zero and the other terms have norm at most ; it follows that . More generally, consider a point , say with and . We can write as , so
Now let be a simplex occuring in . Then for some occuring in for some , so the vertices of are the vertices of together with . It is clear that so by induction we have . Also, from the discussion above, any vertex in has distance at most from . It follows that has diameter at most as required. ∎