What Remains to Be Proved

I once wrote that studying for twelve to fourteen hours a day had become routine. At the time, I treated endurance as proof. The more work I could carry, the more serious and capable I felt. The work taught me what I could do. It could not tell me what I was worth, though I often asked it to. I no longer need to organize my life around proving that I belong.

When I was a junior engineer, raising my hand for nearly everything helped me learn quickly. I wanted to be useful, but I also wanted to be seen as useful. Because I usually delivered, the habit looked like a strength. Yes became a reflex, and I considered the cost only after I had committed. There is now room between a request and my answer. I can pause and ask whether the work should be done, whether I am the right person to do it, and what I would have to set aside. Sometimes I accept the work. Sometimes I push back or suggest another path. Either way, the answer no longer has to prove that I belong.

My ambition is no longer limited to what I can accomplish myself. I still want to solve difficult problems and build systems that remain useful beyond my involvement. I also want to work in ways that make room for the judgment of other engineers, especially when it differs from mine. Much of this happens in ordinary conversation: talking through a design or technical problem and making time for us to teach one another.

The same need to prove myself shaped how I handled disagreement. When being right was tied to belonging, criticism could feel larger than the idea under discussion. I can be wrong without becoming a fraud, and I can revise a belief without diminishing myself. As people have come to trust my judgment, my words can carry farther than my own work. That gives me more reason to listen when evidence or another person challenges me.

That openness changes how I meet ideas outside work. A novel can change me before I can say what it has taught me. I can take a philosophy seriously without accepting it, and I can try to understand another person’s convictions without sharing them. Influence is not assent. I think of identity as a working model: stable enough to live by, but open to revision when it no longer fits reality. I still care about working hard and being capable. What has changed is what I ask the work to mean. Before I say yes, I can ask whether this is where my time and effort belong.

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A First Step into Algebraic Geometry

Mathematics builds on itself. If your foundation is brittle, you cannot grow tall. The following exercise, suitable for a first-year graduate student in algebraic geometry, illustrates this idea. It sits atop a tall hierarchy of prerequisites. Intuitively, it says that the graph of any polynomial is algebraically indistinguishable from the graph of a line.

Claim. The affine coordinate ring of the zeros of the polynomial $f(x,y)=y-g(x)$ is isomorphic to the polynomial ring in one variable over an algebraically closed field $k$.

Proof. Since $f$ is monic of degree one in $k[x][y]$, $f$ is irreducible. Since $k[x][y]$ is a unique factorization domain, the ideal $(f)$ generated by $f$ is prime, hence radical. By Hilbert's Nullstellensatz, the ideal $I(Z(f))$ of the zero set $Z(f)=Z((f))$ of $(f)$ is $\sqrt{(f)}=(f)$. By the universal property of polynomial rings, there exists a ring homomorphism $\phi:k[x][y]\to k[t]$ defined by $x\mapsto t$ and $y\mapsto g(t)$. Since $\phi(h(x))=h$ for any $h\in k[t]$, $\phi$ is surjective. If $h\in (f)$, then $$\begin{align}\phi(h)&=\phi(qf)\\&=\phi(q(y-g(x)))\\&=\phi(q)(\phi(y)-\phi(g(x)))\\&=\phi(q)(g(t)-g(t))\\&=0,\end{align}$$ so $h$ is an element of the kernel $K$ of $\phi$. Conversely, by the division algorithm, if $h\in K$, then $$\begin{align}0&=\phi(h)\\&=\phi(qf+r(x))\\&=\phi(q)\phi(f)+\phi(r(x))\\&=r(t),\end{align}$$ so $h=qf\in (f)$; hence $(f)=K$. By the first isomorphism theorem, $$\begin{align}k[x,y]/I(Z(f))&=k[x,y]/(f)\\&=k[x,y]/K\\&=k[x][y]/K\end{align}$$ is isomorphic to $k[t]$. $$\tag*{$\blacksquare$}$$

On the other hand, not every algebraic set has this property, as the following exercise illustrates.

Claim. The affine coordinate ring of the zeros of the polynomial $f(x,y)=xy-1$ is not isomorphic to the polynomial ring in one variable over an algebraically closed field $k$.

Proof. Suppose that $f$ is reducible. Then there exist non-units $g$ and $h$ such that $f=gh$. Since $$2=\deg(f)=\deg(gh)=\deg(g)+\deg(h),$$ we have $0<\deg(g)=\deg(h)=1$. So $g=ax+by+c$, $h=a'x+b'y+c'$, and $$gh=aa'x^2+a'bxy+a'xc+ab'xy+bb'y^2+b'cy+ac'x+bc'y+cc'.$$ Suppose, without loss of generality, that $a=0$. This forces $b\neq0$, $c=0$, $a'\neq0$, $b'=0$, and $c'=0$, but $cc'=0\neq-1$, which is a contradiction. Therefore, $f$ is irreducible.

By the same argument as in the previous claim, $k[x,y]/I(Z(f))=k[x,y]/(xy-1)$. Since $2=\deg(xy-1)>\deg(x-\lambda)=1$ for any $\lambda\in k$, $xy-1$ does not divide $x-\lambda$. So $x-\lambda\notin(xy-1)$; hence $\overline x\in k[x,y]/(xy-1)$ is non-constant, and $\overline x\cdot\overline y=\overline{x\cdot y}=\overline1$. So $\overline x$ and $\overline y$ are units. Therefore, since $k[t]$ has no non-constant units, $k[t]$ is not isomorphic to $k[x,y]/(xy-1)$. $$\tag*{$\blacksquare$}$$

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The Vitali Paradox of Measure

STEM fields are often interested in measuring regions. For example, an astrophysicist might be interested in measuring the volume of a star. The concept of measure is intuitive, and some of its features are self-evident. For example,

(i) if a region can be divided into countably many nonoverlapping subregions, then the measure of the region should be equal to the sum of the measures of the subregions;

(ii) if two regions are identical up to rotations, translations, and reflections, then their measures should be equal; and

(iii) the measure of a square (cube, hypercube, etc.) with side $1$ should be $1$.

We could begin to formalize this concept by attempting to measure all subsets of the set of real numbers $\mathbb{R}$, but it turns out that this is impossible, as I will now demonstrate.

Let us say that two real numbers are equivalent if their difference is rational. Construct a subset $N$ of the half-open interval $[0,1)$ in $\mathbb{R}$ such that no two distinct elements are equivalent (this is possible by the axiom of choice), and define

\begin{equation}
    N_r = \{x+r : x \in N \cap [0,1-r)\} \cup \{x+r-1 : x \in N \cap [1-r,1)\},
\end{equation}

where $r\in\mathbb Q\cap[0,1)$.

It follows that all $x \in [0,1)$ belong to exactly one $N_r$, namely $r = x-y$ or $r = x-y+1$, where $y \in N$ such that $x$ is equivalent to $y$. Otherwise, if $x \in N_r \cap N_s$, there exist $y, z \in N$ such that $x$ is equivalent to both $y$ and $z$, implying by transitivity that $y$ is equivalent to $z$, which is a contradiction. Thus, the $N_r$ are disjoint.

Then, by (i) and (ii), the measures of $N$ and $N_r$ are equal for all $r$. Moreover, by (iii) and (i),

\begin{equation}
    1=\mu\left(\left[0,1\right)\right)=\mu\left(\bigcup_rN_r\right)=\sum_r\mu\left(N_r\right)=\sum_r\mu\left(N\right),
\end{equation}

where $\mu(N)$ is the measure of $N$. However, the rightmost sum is either $0$ or $\infty$, which is a contradiction.

This problem is a consequence of our ambition to assign a measure to all of $2^{\mathbb R}$. Nevertheless, we can circumvent it by focusing on a particular subset of $2^{\mathbb R}$ rather than the whole, and this is much of the subject of measure theory.

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The Ehrenfest Theorem

While exploring the undergraduate Física Cuántica I course offered by the Pontificia Universidad Católica de Chile, I found a homework question asking to prove the momentum part of the theorem of Ehrenfest.

Although I know little to no quantum physics, I was curious whether, as a mathematician, I could understand and complete the proof.

I learned that the momentum part of the theorem of Ehrenfest relates the time derivative of the expectation value of the momentum $p$ to the expectation value of the force $F=-V'(x)$ on a massive particle moving in a scalar potential $V(x)$.

Mathematically,$$\frac{\mathrm{d}}{\mathrm{d}t}\langle p\rangle=-\langle V'(x)\rangle.$$In simpler terms, as I understand it, this means that the average momentum of a quantum particle changes over time much like that of a classical particle under the average force from the potential (cf. second Newton law in momentum form); it connects quantum behavior with classical physics, showing that quantum averages obey familiar classical laws.

I also learned that this observation is often misused, e.g., by overlooking the assumptions on $V$, which is why I emphasize the importance of truly understanding the underlying principles rather than merely applying formulas.

I did not know how $\langle\cdot\rangle$ is defined in this context, so I looked it up: if $A$ is an operator such that $A\psi\in L^2(\mathbb R)$, where $\psi$ is the wavefunction of the particle, then$$\langle A\rangle=\underbrace{\langle \psi, A\psi \rangle}_{\text{inner product}}=\int_{\mathbb R}\psi^*(x,t)A\psi(x,t)\,\mathrm{d}x.$$In simpler terms, $\langle A \rangle$ is the average value one would get by measuring $A$ on a system described by $\psi$, weighted by the probabilities in $\psi$.

We actually study $L^p$ spaces in mathematics quite thoroughly since they are prototypical Banach spaces of Lebesgue-measurable function classes with $p$-integrable absolute value. In particular, $L^2$ is also a Hilbert space, so this quantum-mechanical setting is really nice.

By the way, a wavefunction $\psi(x,t)$ describes the state of a quantum particle: for each fixed time $t$, it is a function of position in $L^2(\mathbb R)$, meaning it is square-integrable. Moreover, physical wavefunctions are normalized so that total probability equals $1$, and we assume sufficient decay at infinity so we end up with nice equations. 😄

In order to proceed with the proof, one must know both the Schrödinger equation$$i\hbar\partial_t\psi(x,t)=\left(-\frac{\hbar^2}{2m}\partial_x^2+V(x)\right)\psi(x,t)$$and the definition of the momentum operator $p=-i\hbar\partial_x$.

I am not a physicist, so I am not concerned with how these were derived, but they seem like facts every physicist should know by heart.

Anyway, we proceed with the proof; from here I will abuse notation modestly to avoid clutter.$$\begin{align}\frac{\mathrm{d}}{\mathrm{d}t}\langle p\rangle&= \frac{\mathrm{d}}{\mathrm{d}t}\int \psi^*p\psi\\&= \int \partial_t\psi^*\,p\psi + \int \psi^*p\,\partial_t\psi\\&= \frac{i}{\hbar}\int (T+V)\psi^*p\psi + \frac{1}{i\hbar}\int \psi^*p\,(T+V)\psi\\&= \underbrace{\frac{i}{\hbar}\int V\psi^*p\psi + \frac{1}{i\hbar}\int \psi^*pV\psi}_{\text{potential energy}}+ \underbrace{\frac{i}{\hbar}\int T\psi^*p\psi + \frac{1}{i\hbar}\int \psi^*pT\psi}_{\text{kinetic energy}}.\end{align}$$Here, $T=-(\hbar^2/2m)\partial_x^2$. This is often called the kinetic-energy operator.

We now tackle the potential-energy term:$$\begin{align}\frac{i}{\hbar}\int V\psi^*p\psi + \frac{1}{i\hbar}\int \psi^*pV\psi&= \frac{i}{\hbar}\int V\psi^*(-i\hbar\partial_x\psi) + \frac{1}{i\hbar}\int \psi^*(-i\hbar\partial_xV\psi)\\&= \int V\psi^*\partial_x\psi - \int \psi^*\partial_xV\psi\\&= \int V\psi^*\partial_x\psi - \int \psi^*(V'\psi+V\partial_x\psi)\\&= -\int V'|\psi|^2.\end{align}$$We now tackle the kinetic-energy term:$$\begin{align}\frac{i}{\hbar}\int T\psi^*p\psi + \frac{1}{i\hbar}\int \psi^*pT\psi&= \frac{i}{\hbar}\int \Big(-\frac{\hbar^2}{2m}\partial_x^2\psi^*\Big)(-i\hbar\partial_x\psi) + \frac{1}{i\hbar}\int \psi^*(-i\hbar\partial_x)\Big(-\frac{\hbar^2}{2m}\partial_x^2\psi\Big)\\&= -\frac{\hbar^2}{2m}\int \partial_x^2\psi^*\partial_x\psi + \frac{\hbar^2}{2m}\int \psi^*\partial_x^3\psi\\&= -\frac{\hbar^2}{2m}\int \partial_x^2\psi^*\partial_x\psi + \frac{\hbar^2}{2m}\int \partial_x^2\psi^*\partial_x\psi\\&= 0.\end{align}$$Note that we integrated the rightmost term above by parts twice, still assuming sufficient decay so that the boundary terms vanish.

Finally,$$\frac{\mathrm{d}}{\mathrm{d}t}\langle p\rangle=-\int V'|\psi|^2+0=-\langle V'(x)\rangle,$$which is what we wanted. This result extends analogously to $\psi(\cdot,t)\in L^2(\mathbb R^n)$.

 

This exercise humbled me. The mathematics of physics is not easy: it draws on a great deal of experimentally-derived knowledge, despite being modeled with elementary, yet nontrivial, tools like finite-dimensional Hilbert spaces.

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Munkres §54: The Fundamental Group of the Circle

Munkres §53: Covering Spaces ►

Even though it has been four years since I left academia, I still miss solving math problems. They always leave my brain completely fried. 🫠 Math is fairly non-intuitive.

Topological spaces are fascinating mathematical objects. What is most fascinating is that their definition is simple: If a set $X$ and a subset $\tau$ of its power set $2^X$ are such that $\varnothing, X \in \tau$, arbitrary unions of elements of $\tau$ are elements of $\tau$, and finite intersections of elements of $\tau$ are elements of $\tau$, then $(X, \tau)$ is a topological space. A variety of other objects emerge from this definition, and one of them is the homeomorphism, which is a continuous bijection from one space to another with a continuous inverse. If two spaces are homeomorphic, then we consider them to be the same space, but finding homeomorphisms is hard. We created a new object—called the fundamental group—to represent a property of topological spaces in terms of algebraic groups, which enriches our study of them, and in this post, we take a look at the fundamental group of the circle $S^1$. However, in order to do so, we must first take a look at a handful of new definitions and lemmas.

Definition. If $p: E \to B$ is continuous and surjective, $U \subset B$ is open, and $p^{-1}(U)$ is the union of disjoint, open sets $V_\alpha \overset{p}{\simeq} U$, then $U$ is said to be evenly covered by $p$.

Definition. If $p: E \to B$ is continuous and surjective, and for all $b \in B$, there exists an open neighborhood $U$ of $b$ that is evenly covered by $p$, then $p$ is said to be a covering map.

Definition. If $p: E \to B$, and $f: X \to B$, then $\tilde{f}: X \to E$ such that $p \circ \tilde{f} = f$ is said to be a lifting of $f$.

Lemma 1. If $p: E \to B$ is a covering map, $p(e_0) = b_0$, and $f$ is a path in $B$ beginning at $b_0$, then there exists a unique continuous lifting of $f$ beginning at $e_0$.

Proof. We spare the reader the pain of this intricate proof. $$\tag*{$\blacksquare$}$$

Lemma 2. If $p: E \to B$ is a covering map, $p(e_0) = b_0$, and $F: I \times I \to B$ is continuous such that $F(0,0) = b_0$, then there exists a unique continuous lifting of $F$ such that $\tilde{F}(0,0) = e_0$. If $F$ is a path homotopy, then $\tilde{F}$ is a path homotopy.

Proof. We once again spare the reader the pain of this intricate proof. $$\tag*{$\blacksquare$}$$

Lemma 3. If $p: E \to B$ is a covering map, $p(e_0) = b_0$, and $f$ and $g$ are path homotopic paths in $B$ beginning at $b_0$, then the unique continuous liftings of $f$ and $g$ beginning at $e_0$ are path homotopic.

Proof. If $F$ is the path homotopy between $f$ and $g$, then $F(0,0) = b_0$. Therefore—by Lemma 2—the unique continuous lifting of $F$ such that $\tilde{F}(0,0) = e_0$ is a path homotopy. Note that $\tilde{F}(x,0)$ is a lifting of $F(x,0) = f(x)$. Hence, $\tilde{F}(x,0) = \tilde{f}(x)$ since $\tilde{f}$ is unique. Similarly, $\tilde{F}(x,1) = \tilde{g}(x)$. $$\tag*{$\blacksquare$}$$

Definition. If $p: E \to B$ is a covering map, and $p(e_0) = b_0$, then $\phi: \pi_1(B, b_0) \to p^{-1}(b_0)$ defined by $[f] \mapsto \tilde{f}(1)$, where $\tilde{f}$ is the unique continuous lifting of $f$ beginning at $e_0$, is said to be the lifting correspondence derived from $p$. Note that $\phi$ depends on $e_0$ and is well-defined—by Lemmas 1 and 3.

Lemma 4. If $p: E \to B$ is a covering map, $p(e_0) = b_0$, and $E$ is path connected, then the lifting correspondence $\phi$ derived from $p$ is surjective. If $E$ is simply connected, then $\phi$ is bijective.

Proof. If $E$ is path connected and $e_1 \in p^{-1}(b_0)$, then there exists a path $\tilde{f}$ in $E$ beginning at $e_0$ and ending at $e_1$. Therefore, $p \circ \tilde{f} = f$ is a loop in $B$ at $b_0$, and $\phi([f]) = \tilde{f}(1) = e_1$.

If $E$ is simply connected, then suppose that $\phi([f]) = \phi([g])$ and $[f] \neq [g]$. If $\tilde{f}$ and $\tilde{g}$ are the unique continuous liftings of $f$ and $g$ beginning at $e_0$, then $\tilde{f}(1) = \tilde{g}(1)$. Therefore, there exists a path homotopy $\tilde{F}$ between $\tilde{f}$ and $\tilde{g}$, and $p \circ \tilde{F}$ is a path homotopy between $f$ and $g$, which is a contradiction. $$\tag*{$\blacksquare$}$$

Lemma 5. $p: \mathbb{R} \to S^1$ defined by $x \mapsto (\cos 2\pi x, \sin 2\pi x)$ is a covering map.

Proof. $p$ is clearly continuous and surjective. $U = \{(x, y) \in S^1 : x > 0\}$ is open. $p^{-1}(U)$ is the union of disjoint intervals $V_n = (n - 1/4, n + 1/4)$. $p|_{\bar{V}_n}$ is bijective onto $\bar{U}$ since it is clearly surjective and $\sin 2\pi x$ is strictly monotonic. $\bar{V}_n \overset{p}{\simeq} \bar{U}$ since $\bar{V}_n$ is compact and $\bar{U}$ is Hausdorff. Therefore, $V_n \overset{p}{\simeq} U$. Repeat this argument with the remaining $90^\circ$ rotations of $U$. $$\tag*{$\blacksquare$}$$

Theorem. The fundamental group of $S^1$ is isomorphic to the additive group of integers.

Proof. If $p$ is the covering map of Lemma 5, $e_0 = 0$, and $b_0 = p(e_0)$, then $p^{-1}(b_0) = \mathbb{Z}$, and—by Lemma 4—the lifting correspondence $\phi: \pi_1(S^1, b_0) \to \mathbb{Z}$ derived from $p$ is bijective since $\mathbb{R}$ is simply connected.

If $\tilde{f}$ and $\tilde{g}$ are the unique continuous liftings of $f$ and $g$ beginning at $e_0$, $\tilde{f}(1) = n$, and $\tilde{g}(1) = m$, then $\tilde{\tilde{g}} = n + \tilde{g}$ is a lifting of $g$ beginning at $n$ since $p(n + x) = p(x)$. Moreover, $\tilde{f} * \tilde{\tilde{g}}$ is the lifting of $f * g$ beginning at $e_0$, and $\tilde{\tilde{g}}(1) = n + m$. Therefore, $\phi([f] * [g]) = n + m = \phi([f]) + \phi([g])$.$$\tag*{$\blacksquare$}$$

Claim. If $p:E\to B$ is a covering map, $\alpha$ and $\beta$ are paths in $B$ such that $\alpha(1)=\beta(0)$, $\tilde\alpha$ and $\tilde\beta$ are liftings of $\alpha$ and $\beta$ such that $\tilde\alpha(1)=\tilde\beta(0)$, then $\tilde\alpha*\tilde\beta$ is a lifting of $\alpha*\beta$.

Proof.

$$\begin{align}p\circ\tilde\alpha*\tilde\beta(x)&=\begin{cases}p\circ\tilde\alpha(2x)=\alpha(2x)&0\leq x\leq1/2\\p\circ\tilde\beta(2x-1)=\beta(2x-1)&1/2<x\leq1\end{cases}\\&=\alpha*\beta(x).\tag*{$\blacksquare$}\end{align}$$

Claim. If $p:E\to B$ is a covering map, $E$ is path connected, and $B$ is simply connected, then $p$ is a homeomorphism.

Proof. The lifting correspondence $\phi: \pi_1(B, b_0) \to p^{-1}(b_0)$ derived from $p$ is surjective—by Lemma 4—since $E$ is path connected, and its domain is trivial since $B$ is simply connected. Therefore, $p$ is injective.

If $V\subset E$ is open and $b\in p(V)$, then there exists an open neighborhood $U$ of $b$ that is evenly covered by $p$. If, $e\in V$, $p(e)=b$, and $V_\alpha$ is the slice containing $e$, then $p(V\cap V_\alpha)\subset p(V)$ is open since $p|_{V_\alpha}$ is a homeomorphism. Therefore, $p(V)$ is a union of open sets. Hence, $p$ is open.$$\tag*{$\blacksquare$}$$

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