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I'd just like to know if proofs of the Hairy Ball Theorem along the following lines are well-known or even somewhere in the literature.

From a given vector field $V_1$ on $S^2$, form another, $V_2$, by rotating each vector 90 degrees. Wherever either field has a zero, both must. From any $x\in S^2$, form the integral curve $\tau_t$ for $V_1$ starting at $\tau_0=x$. Perhaps for large $t$, $\tau_t$ converges to $y$; then $y$ constitutes a zero of $V_1$. Regardless, by compactness of $S^2$, sequence $\tau_0, \tau_1, \tau_2,\ldots$ has a convergent subsequence, limit point $z$ say. Assume now that $\tau_t$ itself does not converge to $z$. With radius $r$ small enough, a ball $B(z,r)$ will have $V_1$ sufficiently close to constant inside that $\tau_t$ must leave $B(z,r)$ after each sufficiently close approach to $z$. Accordingly, we can form a simple closed curve $S$ by following $\tau_t$ from some exit of $B(x,r)$, at $b$ say, until its next entrance, say at $a$, finally connecting $a$ and $b$ within $B(x,r)$ with a great circle segment. $V_2$ flows transversally across $S$, whether inside or outside of $B(x,r)$. Since $S$ bounds two (closed) disks, the vector-field version of the Brouwer Fixed Point Theorem now forces a zero of $V_2$ (indeed at least one in each disk if a shared zero doesn't fall on $S$).

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    $\begingroup$ Nice. Your technique to obtain the simple closed curve $S$ is often called a "shadowing lemma" or a "closing lemma" in dynamics. $\endgroup$
    – Sam Nead
    Commented Dec 11, 2022 at 12:14
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    $\begingroup$ It seems a bit of a shame to invoke "the vector field version of the Brouwer FPT", which other proofs of the hairy ball theorem will give you for free. (I have Thurston's in mind.) Can you iterate the argument to obtain a nested sequence of closed discs, and maybe get a zero that way? $\endgroup$
    – HJRW
    Commented Dec 11, 2022 at 14:24
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    $\begingroup$ Can you iterate the argument to obtain a nested sequence of closed discs, and maybe get a zero that way? Yes, I had exactly that thought after I posted (but didn't edit just to keep things simple). The iteration will have to go on transfinitely, so there will be things to check at limit ordinals, but not bad. $\endgroup$
    – David Feldman
    Commented Dec 11, 2022 at 18:34
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    $\begingroup$ Can you iterate the argument to obtain a nested sequence of closed discs, and maybe get a zero that way? Thinking about it a little more, I so see a Zorn's Lemma phrasing that's quite simple. Will write it up. As Brouwer FPT is not intuitionistically valid (indeed Brouwer invented intuitionistic logic to capture this insight), there's got to be something wrong with any proof that looks too constructive. So an appeal to something like ZL seems natural. $\endgroup$
    – David Feldman
    Commented Dec 11, 2022 at 21:15
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    $\begingroup$ BFPT itself admits a proof along similar lines, as follows. Write $D$ for either disk that $S$ bounds. Order, by containment of the closed disks they bound within $D$, the family ${\cal F}$ of all simple closed curves that sit in $D$ and sit transverse to either $V_1$ or $V_2$ (or both). Should every descending chain of such curves have a lower bound, Zorn's Lemma would yield a minimal such curve, $S'$ say. From a point on $S'$ and the corresponding transverse vector field, the argument above yields a fixed point or exposes a violation of the minimality of $S'$. $\endgroup$
    – David Feldman
    Commented Dec 12, 2022 at 5:37

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