## Short Exact Sequences

We have slowly introduced the tools needed to understand modern physics and we are almost there before seeing them in action. One particular mathematical pattern proves to be very useful and is one of those cases where you need to recognize at first sight: a short exact sequence

We have seen exact sequences before in the homology and cohomology posts where a set of groups $$G_1,...,G_n$$ are linked by maps and the image of one map is the kernel of the next map. Then you can only "hop twice" an element through the sequence before you end up in the identity element of the group. The geometric interpretation is that "the boundary of a boundary is zero".

So what happens when the exact sequence is short? The best way to get this is to work the problem from the other end. Suppose we have two sets A and C (I am skipping B on purpose). We can then construct the Cartesian product AxC and for each element from AxC we can understand A as an equivalence class. Then C ~ AxC/A and we can introduce a short exact  sequence:

$$0 \rightarrow A \rightarrow (A \times C) \rightarrow C \rightarrow 0$$

Let's first clarify what is with the zeros at each end. The first zero means that the map between A and AxC is injective (one-to-one). This is because the first zero can only be mapped to the zero element of A and that in turn must be the only element which maps to the zero element of AxC. Similarly the last zero element implies that the map between AxC and C is surjective (onto).

Let us now introduce our element B:

$$0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$$

Here is the problem: given A and C, what can we say about B?

If A, B, C are groups, here is an example:

• A = Z (integers)
• B = Z (integers)
• C = {0, 1} = Z/2Z (equivalence classes of odd and even numbers, or Z modulo even numbers)
where

• The first map is the multiplication by 2
• The second map is the reduction modulo 2: for any $$m \in Z, m=2k+p$$ we take k=0.
So is C ~ B/A:  Z/2Z = {0,1} ~ Z/Z?

Z/Z = {m in Z | m=0} = {0} so the answer is no.

In general knowledge of A and C does not determine B. This is because A and C can interlock in a nontrivial way and B is not necessarily AxC!

So how about some fancy short exact sequences:

$$0 \rightarrow S_1 \rightarrow S_3 \rightarrow S_2 \rightarrow 0$$

$$0 \rightarrow S_3 \rightarrow S_7 \rightarrow S_4 \rightarrow 0$$

Those are two of the celebrated Hopf fibrations. I the first case, S3 is a sphere in 4 dimensions and it can be decomposed into an S2 (the ordinary sphere in 3 dimensions) where in each point you have a circle (S1). The picture at the top of this post is a stereographic projection of the rigamarole S3 Hopf fibration because S3 cannot be plotted directly.

The real interesting use of short exact sequences however occurs in physics in gauge theory in fiber bundles.

In this case the short exact sequence pattern is:

$$0 \rightarrow F (fiber) \rightarrow E (total space) \rightarrow B (base) \rightarrow 0$$

In particular Connes' non-commutative geometry model of the Standard Model is best understood if you work out first a simpler unphysical proposal which gives rise to a short exact sequence. We'll get there...