Exact sequence (Quotient space)

Exact sequence (Quotient space)
If X is a space and A is a nonempty closed subspace that is a deformation retract of some neighborhood in X, then there is an exact sequence
\begin{aligned}  \dots\to\widetilde{H}_n(A)\xrightarrow{i_*}\widetilde{H}_n(X)\xrightarrow{j_*}\widetilde{H}_n(X/A)\xrightarrow{\partial}\widetilde{H}_{n-1}(A)&\xrightarrow{i_*}\widetilde{H}_{n-1}(X)\to\dots\\  &\dots\to\widetilde{H}_0(X/A)\to 0  \end{aligned}
where i is the inclusion A\to X and j is the quotient map X\to X/A.

Reduced homology of spheres (Proof)
\widetilde{H}_n(S^n)\cong\mathbb{Z} and \widetilde{H}_i(S^n)=0 for i\neq n.

For n>0 take (X,A)=(D^n,S^{n-1}) so that X/A=S^n. The terms \widetilde{H}_i(D^n) in the long exact sequence are zero since D^n is contractible.

Exactness of the sequence then implies that the maps \widetilde{H}_i(S^n)\xrightarrow{\partial}\widetilde{H}_{i-1}(S^{n-1}) are isomorphisms for i>0 and that \widetilde{H}_0(S^n)=0. Starting with \widetilde{H}_0(S^0)=\mathbb{Z}, \widetilde{H}_i(S^0)=0 for i\neq 0, the result follows by induction on n.

Persistent Homology Algorithm

Algorithm for Fields
In this section we describe an algorithm for computing persistent homology over a field.

We use the small filtration as an example and compute over \mathbb{Z}_2, although the algorithm works for any field.
A filtered simplicial complex with new simplices added at each stage. The integers on the bottom row corresponds to the degrees of the simplices of the filtration as homogenous elements of the persistence module.

The persistence module corresponds to a \mathbb{Z}_2[t]-module by the correspondence in previous Theorem. In this section we use \{e_j\} and \{\hat{e}_i\} to denote homogeneous bases for C_k and C_{k-1} respectively.

We have \partial_1(ab)=-t\cdot a+t\cdot b=t\cdot a+t\cdot b since we are computing over \mathbb{Z}_2. Then the representation matrix for \partial_1 is
\displaystyle M_1=\begin{bmatrix}[c|ccccc]  &ab &bc &cd &ad &ac\\ \hline  d & 0 & 0 & t & t & 0\\  c & 0 & 1 & t & 0 & t^2\\  b & t & t & 0 & 0 & 0\\  a &t &0 &0 &t^2 &t^3  \end{bmatrix}.

In general, any representation M_k of \partial_k has the following basic property: \displaystyle \deg\hat{e}_i+\deg M_k(i,j)=\deg e_j provided M_k(i,j)\neq 0.

We need to represent \partial_k: C_k\to C_{k-1} relative to the standard basis for C_k and a homogenous basis for Z_{k-1}=\ker\partial_{k-1}. We then reduce the matrix according to the reduction algorithm described previously.

We compute the representations inductively in dimension. Since \partial_0\equiv 0, Z_0=C_0 hence the standard basis may be used to represent \partial_1. Now, suppose we have a matrix representation M_k of \partial_k relative to the standard basis \{e_j\} for C_k and a homogeneous basis \{\hat{e}_i\} for Z_{k-1}.

For the inductive step, we need to compute a homogeneous basis for Z_k and represent \partial_{k+1} relative to C_{k+1} and the homogeneous basis for Z_k. We first sort the basis \hat{e}_i in reverse degree order. Next, we make M_k into the column-echelon form \tilde{M}_k by Gaussian elimination on the columns, using elementary column operations. From linear algebra, we know that rank M_k=rank B_{k-1} is the number of pivots in the echelon form. The basis elements corresponding to non-pivot columns form the desired basis for Z_k.

Source: “Computing Persistent Homology” by Zomorodian & Carlsson

Universal Property of Quotient Groups (Hungerford)

If f:G\to H is a homomorphism and N is a normal subgroup of G contained in the kernel of f, then f “factors through” the quotient G/N uniquely.Universal Property of Quotient

This can be used to prove the following proposition:
A chain map f_\bullet between chain complexes (A_\bullet, \partial_{A, \bullet}) and (B_\bullet, \partial_{B,\bullet}) induces homomorphisms between the homology groups of the two complexes.

The relation \partial f=f\partial implies that f takes cycles to cycles since \partial\alpha=0 implies \partial(f\alpha)=f(\partial\alpha)=0. Also f takes boundaries to boundaries since f(\partial\beta)=\partial(f\beta). Hence f_\bullet induces a homomorphism (f_\bullet)_*: H_\bullet (A_\bullet)\to H_\bullet (B_\bullet), by universal property of quotient groups.

For \beta\in\text{Im} \partial_{A,n+1}, we have \pi_{B,n}f_n(\beta)=\text{Im}\partial_{B,n+1}. Therefore \text{Im}\partial_{A,n+1}\subseteq\ker(\pi_{B,n}\circ f_n).

How to calculate Homology Groups (Klein Bottle)

This post will be a guide on how to calculate Homology Groups, focusing on the example of the Klein Bottle. Homology groups can be quite difficult to grasp (it took me quite a while to understand it). Hope this post will help readers to get the idea of Homology. Our reference book will be Hatcher’s Algebraic Topology (Chapter 2: Homology). I will elaborate further on the Hatcher’s excellent exposition on Homology.

This is also Exercise 5 in Chapter 2, Section 2.1 of Hatcher.

The first step to compute Homology Groups is to construct a \Delta-complex of the Klein Bottle.

klein bottle

One thing to note for \Delta-complexes, is that the vertices cannot be ordered cyclically, as that would violate one of the requirements which is to preserve the order of the vertices.

The key formula for Homology is: \boxed{H_n=\ker\partial_n/\text{Im}\ \partial_{n+1}}.

We have \ker\partial_0=\langle v\rangle, the free group generated by the vertex v, because there is only one vertex!

Next, we have \partial_1(a)=\partial_1(b)=\partial_1(c)=v-v=0. Thus \text{Im}\ \partial_1=0.

Therefore H_0=\ker\partial_0/\text{Im}\ \partial_1=\langle v\rangle /0\cong\mathbb{Z}.

Next, we have \ker\partial_1=\langle a,b,c\rangle. \partial_2U=a+b-c, \partial_2L=c+a-b. To learn more about calculating \partial_2, check out the diagram on page 105 of Hatcher.

We then have \text{Im}\ \partial_2=\langle a+b-c, c+a-b\rangle=\langle a+b-c, 2a\rangle, where we got 2a from adding the two previous generators (a+b-c)+(c+a-b).

Thus H_1=\ker\partial_1/\text{Im}\ \partial_2=\langle a,b,c\rangle/\langle a+b-c, 2a\rangle=\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}.

To intuitively understand the above working, we need to use the idea that elements in the quotient are “zero”. Hence a+b-c=0, implies that c=a+b, thus c can be expressed as a linear combination of a, b, thus is not a generator of H_1. 2a=0 implies that a+a=0, which gives us the \mathbb{Z}/2\mathbb{Z} part.

Finally we note that \ker\partial_2=0, and also for n\geq 3, \ker\partial_n=0 since there are no simplices of dimension greater than or equal to 3. Thus, the second homology group onwards are all zero.

In conclusion, we have H_k(K)=\begin{cases}\mathbb{Z}&k=0\\    \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}&k=1\\    0&\text{otherwise}    \end{cases}