Vector field on 3-sphere

Let $V$ be a vector field on $S^3$ such that its singularity points, namely the points at which the vector field vanishes, are only sinks or sources (i.e. the field is converging or diverging). Is there an upper bound on the number $N$ of sinks and sources? Is this number 2? Is it possible that N=1?

There are examples of non-singular fields (Hopf fibering) for which N=0 and simple examples with N=2 (gradient of height function). I am asking if there are other possibilities.

The analogous question on $S^2$ has negative answer by the Poincaré-Hopf theorem which states that N=2.

Here is my attempt: If there is an integral curve of the vector field which joins a source A with a sink B then every integral curve from A must end in B (intuitively, this follows from an argument on connectedness of the boundary of a small ball centered at A) thus in this case N=2, and we are left with the possibility that all the sinks and sources are 'separated' by invariant sets, but is this really possible?

• I think your intuition is incorrect. For instance, the gradient vector field by the canonical Morse-Smale function on a torus. There are one source, one sink and two saddle sigularities. One side, there exists an integral curve connecting the source and the sink. On the other hand, there also exists an integral curve connecting the source and a saddle singularity. Sep 27, 2013 at 7:46
• Thank you, I now see my mistake. While the set of directions from A whose flow lines end at B must be open, similarly defined sets of directions for which the endpoints of the flow lines is a saddle C can be closed. Sep 27, 2013 at 17:37

3 Answers

This is not possible if the sources and sinks form a Morse decomposition. That means that there are no other invariant sets of the flow of the vector field. This happens if one can find a function $f:S^3\rightarrow\mathbb{R}$ that strictly decreases along the flow, except at the sources and sinks. For example if the flow is the gradient flow of a function (not necessarily Morse). The Morse-Conley relations state that, for a Morse decomposition of the three sphere

$$\sum P_t(S_i)=P_t(S^3)+(1+t)Q_t$$

Where $P_t(S^3)=1+t^3$ is the Poincaré polynomial of the three sphere, $P_t(S_i)$ are the Poincaré polynomials of the Conley indices of the invariant sets, and $Q_t$ is a polynomial with non-negative coefficients. The Poincaré polynomial of a source is $t^3$ and of a sink it is $1$. Convince yourself that it is not possible to solve the equation, if there is more than one source or sink.

However, it is possible if the sources and sinks do not form a Morse decomposition. I think I found an example of a vector field on $S^3$ with two sources, two sinks, and one periodic orbit. The idea is to start with the standard heightfunction, and change the sink into two sinks, a source and a periodic orbit, using a Hopf bifurcation.

Replace the a neighbourhood of the source of the standard height function by the vector field whose flow lines are like this:

The vector field has two sources, two sinks and a periodic orbit. Of course one can iterate this procedure around the sinks to obtain a vector field with $n$ sinks, $n$ sources and $n-1$ periodic orbits

• Thank you! Your example is fairly intuitive. Can you please suggest a possibly introductory reference in which the Conley indices are given for the most common invariant sets? Sep 27, 2013 at 17:43
• Conley's monograph, and the paper by Salamon: "Connected simple systems and the Conley index of isolated invariant sets" are very nice. Sep 29, 2013 at 20:19
• The Conley index is defined as the homotopy type of a space. Roughly this is an isolating neighborhood of the invariant set modulo its exit set, the points where the flow points outward. The homological Conley index is the homology of this space (modulo the base point). For an index $k$ critical point of a Morse function, the critical point has a Conley index of a $k$-sphere, which more or less is the computation I did above. Sep 29, 2013 at 20:23
• can we reconstruct the example you provided via the following process. put $M=S^2$ Assume that $X$ is a nice vector field on $M$ we define the vector field $Y$ on $S^2\times [-1,1]$ with $(t^2-1)X \partial_M+(t^3-t)\partial_t$ This vector field vanishes on the boundary of the cylinder so it defines a vector field on the suspension of $S^2$? Feb 18, 2021 at 2:53
• BTW as another question: Is there a hamiltonian field on $\mathbb{R}^4$ such that $S^3$ is invariant under the flow of Hamiltonian and this restricted flow has the desired properties(asked by OP)? Feb 18, 2021 at 2:57

The answer is that $\{N\} = 2\mathbb{N}$ such that both of the number of sinks and sources are $0.5 N$. Poincare-Hopf Theorem promises that $\{N\} \subset 2\mathbb{N}$ and the fact both of the number of sinks and sources are $0.5 N$.

Lemma 3.4 of this paper based on the main Theorem of this paper ensures that for any $K\in \mathbb{N}$, there exists a smooth flow whose singularities are composed of $K$ sinks and $K$ sources.

Actually, we can construct such a flow directly.

Claim: For any $K\in \mathbb{N}$, there exist a smooth flow $\phi_t$ on $S^3$ whose limit set is composed of $K$ sinks, $K$ sources and $2K$ saddle closed orbits.

Here, I only provide the main ideas.

'Proof:'

• Suppose $W$ is a genus $K$ handlebody with $K$ small holes. $\partial W =\Sigma \cup S_1 \cup \dots \cup S_K$ where $S_i$ is a 2-shpere and $\Sigma$ is a genus $K$ closed orientable surface.
• To prove the claim, one only need to construct a smooth flow $\psi_t$ on $W$ such that, (1) $\psi_t$ is transverse inward to $\Sigma$ and outward to $S_i$. (2) The limit set of $\psi_t$ is composed of $K$ saddle closed orbits.

(If this is true, one can glue $W$ and $W'$ along $\Sigma$ (in some sense Heegaard splitting), then glue some neighborhoods of sinks and sources to form $\phi_t$ on $S^3$ in the Claim.Here $W'$ is $W$ with converse orientation).

• For any $k\in \mathbb{N}$, there exists $V_k$ with smooth flow $\psi_t^k$ such that,

(1) $\partial V_k = \Sigma_k^+ \cup T^+ \cup \Sigma_{k+1}^- \cup S^-$. $\Sigma_k^+$ and $\Sigma_{k+1}^-$ are genus $k$ and genus $k+1$ surfaces correspondingly. $T^+$ and $S^-$ are a torus and a shpere correspondingly.

(2) The topology of $V_k$ is very well, see

.

(3)$\psi_t^k$ is transverse inward to $\Sigma_k^+$ and $T^+$ and outward to $\Sigma_{k+1}^-$ and $S^-$.

(4)The maximal invariant set of $\psi_t^k$ is a saddle closed orbit $\gamma_i$.

• Then we can glue $(V_k, \psi_t^k)$ ($k\in \{1,\dots, (N-1)\}$) together by attaching $\Sigma_{k}^-$ to $\Sigma_{k}^+$ suitablely. We can obtain $(W,\psi_t)$ which we expected.

• Thank you very much for the reply, there are some advanced notions that I would need to absorbe before I could understand it. Both your example and Rot's involve some closed curve. Do you expect that every example must have one? Sep 28, 2013 at 18:00
• @Ettore, I think there exists some new limit set except for N=2. In some sense, it is the simplest case that the new limit set is composed of some circles (limit circles, or closed orbits). I don't know whether we can construct such examples without closed orbits. Such a construction seems intersting and I guess it is possible (See Krystyna Kuperberg's famous conterexample to the Seifert conjecture: jstor.org/stable/2118623). Sep 29, 2013 at 2:31

Suppose a vector field V on S3 has K > 0 source singularities and L sink singularities, and no others. Then by the Poincaré-Hopf index theorem, K - L = 0, so K = L.

Now for each singularity, remove a small neighborhood chosen so that its boundary 2-sphere is transverse to V. Let this new manifold-with-boundary be called W. Its boundary ∂W equals 2K copies of the 2-sphere.

Thus V restricted to W is a nowhere-vanishing vector field that points inward on K boundary components (∂-W) and outward on the remaining K of them (∂+W).

By the relative version of Poincaré-Hopf we must have 𝜒(∂-W) = 𝜒(W). In other words, K × 2 = 0 - 2K, and so K = 0, contradiction.

• I don’t understand your final deduction. Shouldn’t relative Poincare—Hopf give $\chi(\partial_-W)=\chi(W)$, which leads to no contradiction because both sides are equal to $2K$?
– HJRW
Feb 18, 2021 at 8:31
• The statement of the relative PH is incorrect. $S^3$ clearly admits a vector field with one sink and one source: The gradient flow of the height function is an example. Then $\partial_-W$ is a two sphere, which has Euler characteristic $2$. But $S^3$ has Euler characteristic 0. Feb 18, 2021 at 8:57
• In the last Euler characteristic equation I erroneously wrote S^3 where I meant to write W. Now fixed. Feb 18, 2021 at 20:09
• HJRW: One side is equal to 2K and the other side is equal to -2K, implying K = 0. Feb 18, 2021 at 20:16
• The formula for the euler characteristic of W is not correct. The euler characteristic of the three sphere minus two discs is two, not minus two. Feb 18, 2021 at 20:25