Existence of a vector field with one singularity on a surface

Question

In a paper I'm reading, it is stated that "It is known that on a compact, connected, oriented two dimensional manifold, there exists a vector field with only one singularity".

Where can I find a proof of this fact?

Answer 1

The statement actually holds in any dimension! There are references to this appearing in Hopf and Alexandroff's Topologie (1935), but I'm sure it was known earlier.

Claim. Every compact, connected, smooth manifold admits a vector field with at most one singularity.

Proof. By compactness and genericity, we can find a vector field with finitely many singularities, all of which lie in the interior of the manifold. Connectedness lets us find an embedded graph (tree, in fact) whose vertices are the singularities. A normal neighborhood of this tree is a disk. As in the following lemma, we can use Alexander's trick to redefine the vector field inside this disk so that it contains a unique singularity. $\square$

Lemma (Adaptation of Alexander's Trick). If $v$ is a smooth vector field on $D^n$ with no singularities on $\partial D^n$, then there exists another smooth vector field on $D^n$ that agrees with $v$ near $\partial D^n$ and has a unique singularity at $0 \in D^n$.

Proof of Lemma. By compactness, we can find $\epsilon \in (0,1)$ such that all singular points of $v$ lie in the interior of $D_\epsilon^n$. Fix a bump function $\rho$ whose value and derivatives vanish at $0$ (and only at $0$) and is identically 1 outside the interior of $D^n_\epsilon$. Define $w$ on $D^n$ by $$w(r,\theta) =\begin{cases} v(r,\theta) & r \geq \epsilon \\ \rho(r) \cdot v(\epsilon,\theta) & r \leq \epsilon.\end{cases}$$ This is continuous everywhere and is smooth near $0 \in D^n$ and $\partial D^n$. Since $D^n$ has a trivial tangent bundle, we can view $w$ as a continuous map $D^n \to \mathbb{R}^n$. As stated, this map is smooth near $\partial D^n$ and $0 \in D^n$, so there is a smooth approximation $w'$ that agrees with $w$ near $0$ and $\partial D^n$. In particular, we can assume no new zeroes have been introduced because $|w|$ was bounded below outside a neighborhood of $0 \in D^n$ and the smooth approximation can be taken to be arbitrarily small. $\square$