On grad school, science, academia, and also a problem on Riemann surfaces

I like mathematics a ton and I am not bad at it. In fact, I am probably better than many math graduate students at math, though surely, they will have more knowledge than I do in some respects, or maybe even not that, because frankly, the American undergrad math major curriculum is often rather pathetic, well maybe largely because the students kind of suck. In some sense, you have to be pretty clueless to be majoring in just pure math if you’re not a real outlier at it, enough to have a chance at a serious academic career. Of course, math professors won’t say this. So we have now an excess of people who really shouldn’t be in science (because they much lack the technical power or an at least reasonable scientific taste/discernment, or more often both) adding noise to the job market. On this, Katz in his infamous Don’t Become a Scientist piece writes:

If you are in a position of leadership in science then you should try to persuade the funding agencies to train fewer Ph.D.s. The glut of scientists is entirely the consequence of funding policies (almost all graduate education is paid for by federal grants). The funding agencies are bemoaning the scarcity of young people interested in science when they themselves caused this scarcity by destroying science as a career. They could reverse this situation by matching the number trained to the demand, but they refuse to do so, or even to discuss the problem seriously (for many years the NSF propagated a dishonest prediction of a coming shortage of scientists, and most funding agencies still act as if this were true). The result is that the best young people, who should go into science, sensibly refuse to do so, and the graduate schools are filled with weak American students and with foreigners lured by the American student visa.

I’ll transition now to a problem that I’ve been asked to solve. Its statement is the following:

Let $f$ be holomorphic on a simply-connected Riemann surface $M$, and assume that $f$ never vanishes. Then there exists $F$ holomorphic on $M$ such that $f = e^F$. Show that harmonic functions on $M$ have conjugate harmonic functions.

Every $p_0 \in M$ corresponds to an open connected neighborhood $U = \{p : \lVert F(p) - F(p_0) \rVert < F(p_0)\}$. Let $\{U_{\alpha}\}$ be the system consisting of these neighborhoods, $(\log F)_{\alpha}$ a continuous branch of the logarithm of $F$ in $U_{\alpha}$. From this arises a family $F_{\alpha} = \{(\log F)_{\alpha} + 2n\pi i, n \in \mathbb{Z}\}$.

In Schlag, there is the following lemma.

Lemma 5.5. Suppose $M$ is a simply-connected Riemann surface and

$\{D_{\alpha} \subset M : \alpha \in A\}$

is a collection of domains (connected, open). Assume further that these sets form an open cover $M = \bigcup_{\alpha \in A} D_{\alpha}$ such that for each $\alpha \in A$ there is a family $F_{\alpha}$ of analytic functions $f : D_{\alpha} \to N$, where $N$ is some other Riemann surface, with the following properties: if $f \in F_{\alpha}$ and $p \in D_{\alpha} \cap D_{\beta}$, then there is some $g \in F_{\beta}$ so that $f = g$ near $p$. Then given $\gamma \in A$ and some $f \in F_{\gamma}$ there exists an analytic function $\psi_{\gamma} : M \to N$ so that $\psi_{\gamma} = f$ on $D_{\gamma}$.

Using the families of analytic function $F_{\alpha}$ as given above, it is clear that near $p \in D_{\alpha} \cap D_{\beta}$, $(\log F)_{\alpha} + 2n_{\alpha}\pi i = (\log F)_{\beta} + 2n_{\beta}\pi i$ when $n_{\alpha} = n_{\beta}$, which means the hypothesis of Lemma 5.5 is satisfied by the above families.

I’ll present the proof of the above lemma here, to consolidate my own understanding, and also out of its essentiality in the construction of a global holomorphic function matching some function in each family. It does so in generality of course, whereas in the problem we are trying to solve it is on a specific case.

Proof. Let

$\mathcal{U} = \{(p, f) | p \in D_{\alpha}, f \in F_{\alpha}, \alpha \in A\} / \sim$

where $(p, f) \sim (q, g)$ iff $p = q$ and $f = g$ in a neighborhood of $p$. Let $[p, f]$ denote the equivalence class of $(p, f)$. As usual, $\pi([p, f]) = p$. For each $f \in F_{\alpha}$, let

$D'_{\alpha, f} = \{[p, f] | p \in D_{\alpha}\}$.

Clearly, $\pi : D_{\alpha, f}' \to D_{\alpha}$ is bijective. We define a topology on $\mathcal{U}$ as follows: $\Omega \subset D_{\alpha, f}'$ is open iff $\pi(\Omega) \subset D_{\alpha}$ is open for each $\alpha, f \in F_{\alpha}$. This does indeed define open sets in $\mathcal{U}$: since $\pi(D'_{\alpha, f} \cap D'_{\beta, g})$ is the union of connected components of $D_{\alpha} \cap D_{\beta}$ by the uniqueness theorem (if it is not empty), it is open in $M$ as needed. With this topology, $\mathcal{U}$ is a Hausdorff space since $M$ is Hausdorff (we use this if the base points differ) and because of the uniqueness theorem (which we use if the base points coincide). Note that by construction, we have made the fibers indexed by the functions in $F_{\alpha}$ discrete in the topology of $\mathcal{U}$.

The main point is now to realize that if $\widetilde{M}$ is a connected component of $\mathcal{U}$, then $\pi : \widetilde{M} \to M$ is onto and in fact is a covering map. Let us check that it is onto. First, we claim that $\pi(\widetilde{M}) \subset M$ is open. Thus, let $[p, f] \in \widetilde{M}$ and pick $D_{\alpha}$ with $p \in D_{\alpha}$ and pick $D_{\alpha}$ with $p \in D_{\alpha}$ and $f \in F_{\alpha}$. Clearly, $D'_{\alpha, f} \cap \widetilde{M} \neq \emptyset$ and since $D_{\alpha}$, and thus also $D'_{\alpha, f}$, is open and connected, the connected component $\widetilde{M}$ has to contiain $D'_{\alpha, f}$ entirely. Therefore, $D_{\alpha} \subset \pi(\widetilde{M})$ as claimed.

Next, we need to check that $M \setminus \pi(\widetilde{M})$ is open. Let $p \in M \setminus \pi(\widetilde{M})$ and pick $D_{\beta}$ so that $p \in D_{\beta}$. If $D_{\beta} \cap \pi(\widetilde{M}) = \emptyset$, then we are done. Otherwise, let $q \in D_{\beta} \cap \pi(\widetilde{M})$ and pick $D_{\alpha}$ containing $q$ and some $f \in F_{\alpha}$ with $D'_{\alpha, f} \subset \widetilde{M}$ (using the same “nonempty intersection implies containment” argument as above). But now we can find $g \in F_{\beta}$ with the property that $f = g$ on a component of $D_{\alpha} \cap D_{\beta}$. As before, this implies that $\widetilde{M}$ would have to contain $D'_{\beta, g}$ which is a contradiction.

To see that $\pi : \widetilde{M} \to M$ is a covering map, one verifies that

$\pi^{-1}(D_{\alpha}) = \bigcup_{f \in F_{\alpha}} D'_{\alpha, f}$.

The sets on the right-hand side are disjoint and in fact they are connected components of $\pi^{-1}(D_{\alpha})$.

Since $M$ is simply-connected, $\widetilde{M}$ is homeomorphic to $M$ (proof given in the appendix). We thus infer the existence of a globally defined analytic function which agrees with some $f \in F_{\alpha}$ on each $D_{\alpha}$. By picking the connected component that contains any given $D_{\alpha, f}'$ one can fix the “sheet” locally on a given $D_{\alpha}$.     ▢

By this, we can construct an analytic $F$ such that for all $\alpha$,

$f_{|U_{\alpha}} = (\log F)_{\alpha} + n_{\alpha} \cdot 2\pi i, \qquad n_{\alpha} \in \mathbb{Z}$.

from which follows $e^F = f$.

For the existence of harmonic conjugates, we do similarly. Take a connected open cover of $M$, $\{U_{\alpha}\}$ where each $U_{\alpha}$ is conformally equivalent to the unit disc, and $v_{\alpha}$ is a harmonic conjugate of $u$ in $U_{\alpha}$ (which exists uniquely up to constant on the unit disc. Let $F_{\alpha} = \{v_{\alpha} + c, \quad c \in \mathbb{R}\}$. Then by the same lemma, there exists $v$ such that for all $\alpha$,

$v_{|U_{\alpha}} = v_{\alpha} + c_{\alpha}, \quad \text{some } c_{\alpha} \in \mathbb{R}$

that is harmonic and conjugate to $u$ since it is the harmonic conjugate to $u$ on every element of the cover, again with choise of $c_{\alpha}$s to ensure that on intersection of cover elements there is a match.