### Homological Algebra and Morse Theory

The principles of Homological Algebra as used in Topology will be developed here.

Let $R$ be a ring and $L,M$ and $N$ be right (or left) $R$-modules. Then the sequence $L\xrightarrow{f} M\xrightarrow{g} N$ is exact iff $\text{im}(f)=\text{ker}(g)$.

More generally, $\dots \xrightarrow{f_{n+1}}M_n\xrightarrow{f_n}M_{n-1}\xrightarrow{f_{n-1}}M_{n-2}\dots$ is an exact complex if im$(f_i)=$ker$(f_{i-1})$.

The $n^{\text{th}}$ homology $H_n$ is $\frac{\text{ker}(f_n)}{\text{im}(f_{n-1})}$.

Let $x_0,x_1,x_2,\dots,x_n$ be points in $\Bbb{R}^n$. The set of points satisfying $\sum\limits_{i=0}^n{\lambda_ix_i}$ where $\sum\limits_0^n{i}=1$ forms an $n$-simplex.

The demarcation of a topological space $X$ as a simplicial complex is called the triangulation of $X$. A simplicial complex satisfies the following two properties:

1. Any face of a simplex in $X$ is also a simplex in $X$.
2. The intersection of two simplices $P$ and $Q$ is a face of both the simplices.

For example, dividing the unit square into two triangles (cutting along the diagonal) would form a simplicial complex. However, Cutting the square into small rectangles would not form a simplicial complex, as the small rectangles are not simplices.

For a triangulated space $X\subseteq\Bbb{R^n}$, we can form a free $\Bbb{Z}$-module generated by the $n$-simplices. For example, for the unit square, we can form a free abelian group generated by the two triangles that the unit square has been divided into.

Now we move on to the signs. It is easy to see that $[12]=-[21]$. In general, we have

$[x_0,x_1,x_2,\dots,x_n]=(-1)^i [x_{\sigma^{-1}{(0)}},x_{\sigma^{-1}{(1)}},x_{\sigma^{-1}{(2)}},\dots,x_{\sigma^{-1}{(n)}}]$,where $i$ is the number of transpositions.

Now we shall talk about the mapping $\partial_n:C_n\to C_{n-1}$. The simplex $r[x_0,x_1,x_2,\dots,x_n]$ is mapped to $r(\sum\limits_{i=0}^n{(-1)^i [x_0,x_1,\dots,\hat x_i,\dots,x_n]})$, where $r\in\Bbb{Z}$.

It is easy to see, on working out some examples on your own, that $\partial_{n-1}\circ\partial_{n}=0$. I recommend working on $I^2$, as it gives the simplest triangulation.

The $n^{\text{th}}$ homology $H_n$ here is equal to $\frac{\text{ker}(\partial_n)}{\text{im}(\partial_{n-1})}$.

We know move on to the following important proposition: $\sum\limits_{i=-N}^N{(-1)^i\dim C_i}=\sum\limits_{i=-N}^N{\dim H_i}$. It is important to note that dimension implies the number of generators of the free module, and not any other geometric interpretation that you may have in mind.

Proof: $\dim H_i=\dim \ker\partial_i-\dim\text{im }\partial_{i+1}$. So for example if we had $\sum\limits_{i=-2}^2{\dim H_i}$, this would be equal to

$\dim \ker\partial_{-2}-0-(\dim\ker\partial_{-1}-\dim\text{im }\partial_{-2})+\dim\ker\partial_0-\dim\text{im }\partial_{-1}-(\dim\ker\partial_1-\dim\text{im}\partial_0)+\dim\ker\partial_2-\dim\text{im}\partial_1$.

If we assume that $\partial_2$ maps $C_2$ to $0$, we can conclude that $\dim\ker\partial_2=\dim C_2$. Also, we know that $\dim\ker\partial_i+\dim\text{im}\partial_i=\dim C_i$. Hence, we get $\dim C_{-2}-\dim C_{-1}+\dim C_0-\dim C_1+\dim C_2$. This method can now be generalised for any $N$.

Now we come to the Euler characteristic. Let $H_i(M)$ be the $i^{\text{th}}$ homology group of $M$. The Euler characteristic is defined as $\sum\limits_{i=0}^m (-1)^i \dim H_i(M)$.

How would you interpret this statement? Take the unit square, and divide it into triangles (or triangulate it in any other fashion you like). The $\Bbb{Z}$-modules $C_i$ for $i>2$ and $i<0$ are all $0$, as these are no $i$-simplices in the unit square for those values of $i$. Hence, the Euler characteristic of the unit square would be #(Triangles)-#(Edges)+#(Vertices), where # denotes cardinality. The Euler characteristic of the closed interval $[0,1]$ would be $1$.

Morse Theory

Now we come to Morse Theory.

Let $M$ be a $k$-manifold, and let $f:M\to\Bbb{R}$ be a twice differentiable function. The Hessian $H_f$ would be

$\begin{pmatrix} \frac{\partial^2 f}{\partial x_1^2}&\frac{\partial f^2}{\partial x_2\partial x_2}&\dots &\frac{\partial f^2}{\partial x_1x_k}\\ \vdots&\vdots&\ddots&\vdots\\\frac{\partial^2 f}{\partial x_n \partial x_1}&\frac{\partial f^2}{\partial x_n\partial x_2}&\dots &\frac{\partial f^2}{\partial x_n^2}\end{pmatrix}$

Since this matrix is symmetric, as a result of the Spectral Theorem it has $n$ mutually orthonormal eigenvectors.

A critical point $p$ of a Morse function $f$ is that where the derivative is $0$. A non-degenerate critical point is that where the derivative is $0$, but the Hessian is non-singular.

The index of a Morse function $f$ at a non-degenerate critical point would be the number of negative eigenvalues of the Hessian at that point.

Now we come to a very interesting lemma of Morse Theory. Say $f:M\to \Bbb{R}$ is a function, of which $p$ is a non-degenerate critical point. Then there is a neighbourhood $U$ of $p$ in $M$ such that $f$ can be written as $f(p)-x_1^2-x_2^2-\dots-x_i^2+x_{i+1}^2+\dots+x_k^2$, where $i$ is the index of the critical point. For instance, if $f(x,y)=x^2-2y^2-2x$, then $(1,0)$ is a non-degenrate critical point with one negative eigenvalue. Hence, there exists a neighbourhood of $(1,0)$ where $f=-1-x^2+y^2$.

Now we move on to homotopy types. Let $X$ and $Y$ be two topological spaces. We say that $X$ and $Y$ are of the same homotopy type if there exist functions $f:X\to Y$ and $g:Y\to X$ such that $gf\sim 1_X$ and $fg\sim 1_Y$. For example, $[0,1]$ is homotopy equivalent to $[0,2]$.

Theorem: If $f$ is a smooth real-valued function on a manifold $M$, and if $a$ and $b$ are points such that $a and $f^{-1}[a,b]$ has no critical points, then $f^{-1}(-\infty,a]$ is diffeomorphic to $f^{-1}(\infty,b]$.

Experimenting with functions from $\Bbb{R}$ to $\Bbb{R}$ should convince you of this fact. It is important to note that there not being a critical point between $a$ and $b$ is not a necessary condition for $f^{-1}(-\infty,a]$ to be diffeomorphic to $f^{-1}(-\infty,b]$. Experimenting further should convince you of this too.

Now we move on to a rather interesting theorem. If $f$ is a smooth real valued function on a manifold $M$, and $p$ a non-degenerate critical point with index $i$, if $f^{-1}([f(p)-\epsilon,f(p)+\epsilon])$ is compact with exactly one critical point, then for all $\epsilon$ sufficiently small, $f^{-1}(-\infty,f(p)-\epsilon)$, with an $i$-cell attached appropriately, is of the same homotopy type as $(-\infty,f(p)+\epsilon)$.

Take $y=x^2$ for instance. Clearly, $0$ is a critical point, and the hessian here is $2$; hence no negative eigenvalues. The theorem above says that a point (attaching a $0$-simplex to an empty set as $f^{-1}(-\infty,-\epsilon)$ for $\epsilon>0$ is empty) is homotopic to the bounded $1$-manifold $f^{-1}(-\infty,\epsilon)$.

Here, it is instructional to note that two manifolds of the same homotopy type will have the same Euler characteristic. The number of $i$-simplices may not remain the same, however. For instance, the unit disc is homotopic to the unit square. One way to think about his is that whenever you’re adding vertices, you’re also adding faces and edges; and somehow the alternating sum always remains constant.

It is also important to note that if $f:M\to\Bbb{R}$ is a Morse function, then the Euler characteristic of $M$ is the same as the Euler characteristic of $f^{-1}(\Bbb{R})$. They’re obviously the same, and hence have the same Euler characteristic.

If you’ve convinced yourself of the fact above, we will move on to some interesting formulations.

Let $i_{f,j}$ be the number of critical points of $f$ with index $i$. Then $\sum\limits_{j=0}^n{(-1)^j i_{f,j}}=\chi(M)$. In other words, I am saying that the number of $j$-simplices in $M$ is equal to the number of critical points of Morse index $j$. How come that is true? Assume that $a\in\Bbb{R}$ is the lowest critical value (we’re assuming that it exists). Now keep hopping one at a time until you reach the greatest critical value. After doing this, we now have a manifold $M$, which has the same Euler characteristic as $M$ attached with certain simplices at critical points; a quotient space of the disjoint union of $M$ and some simplices. Let us call the latter $M'$. Using the standard formula for Euler characteristic, we see that $\sum\limits_{j=0}^n{(-1)^j i_{f,j}}=\chi(M)$. This is the Poincare-Hopf conjecture!