I am studying Morse Homology using the book "Lectures on Morse Homology" by Augustin Banyaga and David Hurtubise and I am having a hard time to understand how to count flow lines with sign. Now I am trying to understand the example 7.7 of the circle. I did all the calculation to find the critical points and to verify that they are non-degenerated. Also, I had no difficulty in visualizing who are the stable and unstable manifolds.

My problem is to visualize the gradient flow, the choice of orientation and how to count the flow lines.

If anyone could help me. Thank you very much.


1 Answer 1


Let's solve the issues one by one. Let's also use the case of a deformed sphere with the height function as the Morse function, which is similar to the circle but has more structure so that we understand things better. (The case of the circle is given right after this example of the sphere.)

enter image description here

Gradient flow.

If you know how to visualize the gradient, the gradient flow is just given by "following the gradient". It is usual in Morse theory to talk about the flow of $-$gradient instead of the gradient, so when we say "gradient flow" from now on, we actually mean the flow of $-$gradient. Some flow lines for the (minus, just to emphasize one last time) gradient flow are shown in the picture below. (One thing to notice is that the flow lines never actually hit the critical points, they are just asymptotic to them.)

enter image description here

These curves $c$ satisfy the differential equation $\dot{c} = -\nabla f \circ c$, which is to say precisely that at each point of them, the velocity is given by $-\nabla f$.


This is somewhat messy. We proceed by steps:

  1. First, orient each unstable manifold. Arbitrarily, however you want.
  2. This determines an orientation on each intersection of unstable and stable manifolds, as we shall see.
  3. The next step lies on how to count the flow lines. When $\operatorname{ind}_x=\operatorname{ind}_y+1$, if the orientation above coincides with the direction of the gradient flow, then we let the sign of counting be positive. Else, it is negative.

So, consider the following chosen orientations. (On the two-dimensional unstable manifolds, the first vectors with respect to the ordering $(e_1,e_2)$ that gives the orientation are the "horizontal" ones.)

enter image description here

Now, proceeding to step $2$, we need to know how those orientations induce an orientation on each $\mathcal{M}(x,y)=W^u(x)\cap W^s(y)$. Denote the tangent space to $W^u(x), W^s(x)$ and $\mathcal{M}(x,y)$ by $T^u$, $T^s$ and $T^{u,s}$ respectively.

Take a point $p \in \mathcal{M}(x,y)$. Then, $T^{u}(p)=T^{u,s}(p)\oplus N^{u,s}(p)$, where $N^{u,s}(p)$ is the orthogonal complement of $T^{u,s}(p)$ with respect to the induced metric on $\mathcal{M}(x,y)$ obtained from the fixed metric of our manifold. By taking $p$ close to $y$ in the same connected component of $\mathcal{M}(x,y)$ as the original $p$ and noting that $N^{u,s}(p)$ will be identified with the tangent space to the unstable manifold of $y$, we have that $N^{u,s}(p)$ has a specified orientation. Therefore, since $N^{u,s}(p)$ and $T^u(p)$ are oriented, so is $T^{u,s}(p)$. (In the case where $N^{u,s}(p)=0$, we just orient $T^{u,s}(p)$ with the same orientation as $T^u(p)$.)

We should arrive at the situation described in the figure below.

enter image description here

Counting the flow lines.

Now, for each connected component of $\mathcal{M}(x,y)$ (i.e., each flow line) when $\operatorname{ind}_x=\operatorname{ind}_y+1$, we will see if the orientation induced above for the path concerning that connected component coincides or not with the orientation given by the direction of the flow. If it coincides, we count that flow line with a positive sign. If it doesn't, it is counted with a negative sign. Explicitly, our boundary map on the chain complex where each group is the free abelian group generated by the critical points is defined on the critical points (the generators) as $$\partial(x)=\sum_y n(x,y) y, $$ where $y$ runs through the critical points of one lower index and $n(x,y)= \sum_C \operatorname{sign}(C)$, where $C$ runs over all connected components of $\mathcal{M}(x,y)$ and $\operatorname{sign}(C)$ is as described above.

So, denoting the free abelian group generated by a set $S$ as $F(S)$, we get that the morse complex is $$F(\{\color{red}\bullet, \color{green}\bullet \}) \stackrel{\partial_2}{\to} F(\{\color{blue}\bullet\}) \stackrel{\partial_1}{\to} F(\{\bullet\}),$$ where $$\partial_2 =\begin{cases} \color{red}\bullet \mapsto -\color{blue}\bullet, \\ \color{green}\bullet \mapsto -\color{blue}\bullet \end{cases}$$ and $$\partial_1(\color{blue}\bullet)=\bullet-\bullet = 0.$$ As expected, we get $H^{Morse}_2(M)=\mathbb{Z}=H^{Morse}_0(M)$ and $0$ for all other degrees.

The case of the circle.

In the case of the circle, it then just suffices to verify if the direction of the flow coincides with the orientation chosen for $S^1$ minus the south pole on each side of the circle minus the south and north pole. But it is clear that it will coincide on one side, and not coincide on the other. Therefore, denoting the top critical point as $p$ and the bottom one as $q$, the boundary map of the chain complex $$F(p) \to F(q)$$ will be $p \mapsto q-q=0$, thus yielding the homology of the circle.

Another good example for grasping the concepts is $\mathbb{R}P^2$. (Note that we do not need an orientation for the ambient manifold whatsoever, only of the unstable manifolds, which are disks and thus orientable.)

Take $X=\{(x,y,z) \mid \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\}$. The function $f:x \mapsto \Vert x\Vert^2$ restricts to a Morse function. Since it is invariant under the antipode map, it induces a Morse function on the quotient $X/\mathbb{Z}_2 \simeq \mathbb{R}P^2$. We now draw the critical points, together with chosen (arbitrary) orientations for the unstable manifolds. (Take careful notice that we are identifying antipodal points! We will also be more explicit with the ordering of the vectors that determine the orientation now, since it may be confusing otherwise.)

enter image description here

Now we induce the orientations on each $\mathcal{M}(x,y)$ when the indices of $x,y$ differ by one.

enter image description here

The Morse complex will then be $$F(\{\color{blue}\bullet\}) \stackrel{\partial_2}{\to} F(\{\color{red}\bullet\}) \stackrel{\partial_1}{\to} F(\{\color{green}\bullet\}),$$ where $$\partial_2:\color{blue}\bullet \mapsto \color{red}\bullet+\color{red}\bullet = 2\cdot\color{red}\bullet$$ and $$\partial_1:\color{red}\bullet \mapsto \color{green}\bullet-\color{green}\bullet = 0,$$ thus yielding $H_1^{Morse}(\mathbb{R}P^2)=\mathbb{Z}_2$, $H_0^{Morse}(\mathbb{R}P^2)=\mathbb{Z}$ and $0$ for all other degrees.

  • $\begingroup$ Hi @AloizioMacedo, thank you very much for the beautiful answer. I have a question. Why $N^{u,s}(p)$ will be identified with the tangent space to the unstable manifold of $y$? And in the last picture of the sphere, you put colors on the arrows :), the blue ones are the tangent space of the unstable manifold of the saddle point? And the pink one is the $T^{u,s}(p)$? $\endgroup$
    – Jude
    Oct 23, 2019 at 12:42
  • $\begingroup$ @Mariana Question 1: Underlying this is the fact that at $y$, the tangent space of the entire manifold is split onto the negative part of the Hessian (which gives rise to the unstable manifold) and the positive part (which gives rise to the stable manifold). Both are orthogonal to each other, since the Hessian is self-adjoint. So, you take the negative part and, for instance, make a parallel transport of an orthogonal basis for it near the critical point. This is the identification of $N^{u,s}(p)$ with the tangent space to the unstable manifold at $y$. $\endgroup$
    – Aloizio Macedo
    Oct 23, 2019 at 14:27
  • $\begingroup$ So, actually, there is a small detail here: by this construction, $N^{u,s}(p)$ is not necessarily actually the orthogonal complement of $T^{u,s}(p)$. It is just a complement, but this is enough for orienting the tangent spaces of $\mathcal{M}(x,y)$. (It will be actually the orthogonal complement if the gradient is in standard form near $p$, i.e., if the Morse chart can be taken as to be an isometry with the standard metric of $\mathbb{R}^n$. That, or if we change the gradient to a pseudo-gradient vector field such that it is in standard form near the critical points.) $\endgroup$
    – Aloizio Macedo
    Oct 23, 2019 at 14:28
  • $\begingroup$ Regarding Question 2, the colors are given as follows: On the first image (where we orient the unstable manifolds), I chose colors for the critical points. Then, for each critical point, the orientation chosen for the unstable manifold of that critical point receives that color. Later, I chose colors for the connected components of each $\mathcal{M}(x,y)$. Then, the orientation$^*$ that is assigned to each such component via the process I mentioned receives the color of that component. $\endgroup$
    – Aloizio Macedo
    Oct 23, 2019 at 14:34
  • $\begingroup$ $^*$ That orientation is given with help of the vectors with dots and dashes, which are the ones that give rise to the complement $N^{u,s}(p)$ via transportation of the tangent space of the unstable manifold of the critical point of lower index. $\endgroup$
    – Aloizio Macedo
    Oct 23, 2019 at 14:34

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