Geometricity and Galois actions on fundamental groups

A couple months ago I went to a fantastic conference on Arithmetic Topology at PIMS. The video from my talk is now up: enjoy!

I spend the first few minutes of the talk joking around about this cartoon:

Pictured: Wei Ho, Ravi Vakil, Benson Farb, Jordan Ellenberg, Kirsten Wickelgren, Soren Galatius, Will Sawin, and the Dragon Langlands

I'm a Numberphile!

The popular YouTube channel Numberphile has just released a video featuring yours truly — filming this with Brady Haran was a real pleasure!

A couple of quick mathematical comments: in the video, I compute the Dehn invariant of the cube with side length $1$ as $12\otimes \pi/2$ — astute YouTube commenters have already noticed that this is the same as zero! I never really discussed the group structure on $\mathbb\otimes \mathbb/2\pi$, so I resisted saying this in the video.

For a similar reason, the proof that the Dehn invariant is actually an invariant (under cutting and pasting) sketched in the video is not quite complete. The gap is that I never explain what happens with new edges added (in the middle of what used to be a face) when one makes a cut! The point is that new edges come in pairs with equal length and with dihedral angles which sum to $\pi$. So the contribution to the Dehn invariant is $\ell\otimes \theta_1+\ell\otimes \theta_2 =\ell\otimes \pi=0,$ where here I use that $\ell\otimes \theta=0$ if $\theta$ is a rational multiple of $2\pi$.

Anyway, hope you all enjoyed the video!

Arithmetic and Representations of Fundamental Groups

I gave a talk at the IPAM conference “Braids, Resolvent Degree, and Hilbert’s 13th Problem” this past Thursday. The conference was really inspiring — I enjoyed all the talks, but I’d in particular like to recommend Jesse Wolfson’s three talks for a really excellent overview of Hilbert’s 13th problem, Mark Kisin’s talk on recent progress he, Jesse, and Benson Farb have made on the problem, Laure Flapan’s talk for a fascinating overview of her work on Kodaira fibered surfaces (along with some intriguing open questions about them), and Alexander Duncan’s talk on the numerology of essential dimensions of small finite simple groups. Basically, I’m pretty into resolvent degree right now.

My talk, on “Arithmetic and Representations of Fundamental Groups” is below — it starts about 5 minutes in to the video.

Holding the p-adics in the palm of your hand (with thanks to Matt Kukla)

A few years ago I made a picture of the 3-adic numbers in answer to a MathOverflow question (using the free 3d modeling and rendering software Blender, which I highly recommend despite its steep learning curve). Recently I discovered that I lost the original file, so I recreated it, with the following result:

Really this is $\mathbb{Z}/243\mathbb{Z}$

After I posted the picture above on Twitter, Matt Kukla, a budding mathematician and student at the University of Maryland, asked for the source file, and today, he’s posted a beautiful 3D print of the 3-adic numbers!

EDIT 5/19/2019: If you’d like to try this yourself, here is the original .blend file and here is an exported .obj file.

More Tate curves, more problems

I’m giving an expository lecture on the Tate curve and non-archimedean geometry next Tuesday, November 6, as one of the preparatory lectures for Peter Scholze’s Chow Lectures in Leipzig. I thought the problem set for the lectures might be interesting to people, so here it is! Please let me know if you have any comments.

12 minutes of monodromy

Here’s a video of a (12 minute!) talk I gave on Friday about some of my recent work, mostly from this paper of mine. The talk is aimed at a general mathematical audience — please let me know if you have any questions!

The video has received exactly one comment so far:

A quick comment on recent RH news

I recently commented for this article in Gizmodo on Atiyah’s supposed proof of the Riemann Hypothesis. In the article I assert there is an internal inconsistency (in fact there are several) in his paper. This post is meant to direct people to the issue I had in mind.

Atiyah claims to construct a “weakly analytic” auxiliary function T. I do not claim to understand his construction. That said, he lists many properties of T; enough to see that no such function exists. Many commenters have focused on the claim that T is locally polynomial — this implies that it is globally polynomial, which is contradicted by, for example, property 2.5 of the paper in question. However, because Atiyah never defines the notion of “weak analyticity,” it is hard to be sure if this objection is valid.

That said, here are my tweets on a simpler issue, which I think cannot be dealt with by any reasonable relaxation of what it means to be a function:

It seems to me this contradiction is not really resolvable.

The Boston-Markin Conjecture for Three-Manifolds

A couple of months ago I was in Oberwolfach, where Tomer Schlank told me about the following conjecture:

Conjecture. (Boston-Markin) Let $G$ be a finite group.  Then there exists a Galois extension of $\mathbb{Q}$ with Galois group $G$ ramified over exactly $d(G)$ primes, where $d(G)$ is the minimal number of generators of $G^{\text{ab}}$.  (Here we violate convention and say that $d(\{1\})=1$.)

It's not hard to see that $d(G)$ is a lower bound for the minimal number of primes over which a $G$-extension ramifies, but I don't see much evidence for the conjecture (except for some results in the case of e.g. solvable groups).  Here is a topological analogue (translating the conjecture through the usual primes-vs.-knots dictionary).

Question. Let $G$ be a finite group and $M$ a $3$-manifold.  Does there exist a link $L$ in $M$ with $d(G)$ components, and a $G$-cover of $S^3$  branched only over $L$?

The case most analogous to the original conjecture is $M=S^3$ (or in general, the case where $M$ is simply connected), where one can see that $d(G)$ is a lower bound for the number of components of a link with the desired properties.  We weren't able to make much progress with $S^3$, but Tomer, Brian Lawrence, and I were able to show that the answer is yes if $G$ is a finite simple group and $M$ is $S^2\times S^1$, or indeed any surface bundle over a circle!  (Of course such manifolds have infinite fundamental group, so the problem is strictly easier in this case.)  If I have time, I'll sketch the argument in another post.  Please let me know if you have any thoughts about this question!

Guest Post: Ordinals and Hydras, by Brian Lawrence

A few days ago, the hydra game came up in discussion with Brian Lawrence; we'd both run into a somewhat mystifying analysis of the game via ordinals.  Brian and I came up with a much more down-to-earth analysis (which experts will recognize as the same as the usual one); Brian decided to write it up, and I asked him if I could post his write-up here.

Guest post by Brian Lawrence:

I've read in a number of places about the Hydra game, which one plays by applying a certain type of move to a rooted tree.  The theorem (surprising at first) is that the Hydra game always terminates: there is no infinite sequence of legal moves.  The theorem is used in mathematical logic as an example of a result that can only be proven in a sufficiently strong proof system.  I'm not an expert in logic, so I won't say any more about that.  Instead, I'd like to give a straightforward proof of the Hydra theorem.

The game is based on the mythical Hydra, a many-headed monster.  In myth, every time you cut off one of the Hydra's heads, it would grow two more.

Our Hydra is a (finite) rooted tree.  One performs a sequence of moves; each move is as follows.  Choose a leaf $x_0$ of the rooted tree, and remove it.  Then, let $x_1$ be the parent node of $x_0,$ and $x_2$ the parent node of $x_1$ (assuming it exists).  Let $S$ be the subtree consisting of $x_1$ and all its descendants, after $x_0$ has been removed.  When you remove $x_0,$ the Hydra chooses a $N > 0,$ and attaches $N$ new copies of $S$ as children of $x_2.$  If $x_0$ doesn't have a grandparent, the Hydra doesn't get to add anything to the tree.  (Note that the Hydra can choose a different $N$ at every step.)  The game ends when only a single vertex (the root) remains.

One move of the hydra game.

There's a link to a Java version at the bottom of this page.  (The in-website applet doesn't work for me, but the downloadable version does.)

Here's the remarkable fact: no matter how we choose which leaves to cut off, and no matter how large the Hydra chooses its positive integers $N,$ the game is guaranteed to terminate after finitely many steps.

Let's define the "depth" of a rooted tree to be the length of the longest path from root to leaf.  If a tree has only one vertex, it has depth $0;$ if all leaves are adjacent to the root, the tree has depth $1;$ and so forth.  I'll prove the theorem by induction on the depth, starting with the case of depth $2.$  (The theorem is trivial for trees of depth $0$ and $1.$)

For any rooted tree, call an "immediate subtree" the subtree consisting of $x$ and all its descendants, where $x$ is a neighbor of the root.

Now suppose we have a tree $T$ of depth at most $2.$  Its immediate subtrees have depth at most $1.$  For every integer $r \geq 0,$ call $T_r$ the tree of depth at most $1$ where the root has $r$ children.

The tree $T_4$

(The depth of $T_0$ is $0.$)  To describe our tree $T,$ we just have to describe its immediate subtrees; in fact, if $a_r$ is the number of immediate subtrees isomorphic to $T_r,$ then our tree $T$ is determined completely by the tuple $(a_0, a_1, a_2, \ldots).$

The tree corresponding to the tuple $(0, 2, 1, 1, 0, 0, \ldots)$.

Let $\mathcal{I}_2$ be the set of tuples $(a_0, a_1, a_2, \ldots)$ of nonnegative integers, such that all but finitely many $a_r$ are nonzero.  We impose the lexicographic total ordering on $\mathcal{I}_2:$ given $a = (a_0, a_1, a_2, \ldots)$ and $b =(b_0, b_1, b_2, \ldots)$ in $\mathcal{I}_2,$ take the largest $r$ such that $a_r \neq b_r,$ and define $a > b$ if and only if $a_r > b_r.$

We have described a bijection between trees of depth at most $2$ and tuples in $\mathcal{I}_2.$  The theorem (for depth $2$) now follows from the following two lemmas, which are left as an exercise to the reader.

Lemma. Any legal move changes a tuple $a \in \mathcal{I}_2$ to another tuple $b$ such that $b < a$ for our total ordering.

Lemma. The set $\mathcal{I}_2$ is well-ordered: it contains no infinite descending chain.

We will construct recursively a well-ordered set $\mathcal{I}_d$ which classifies the isomorphism types of trees of depth at most $d.$  Specifically, we will construct $\mathcal{I}_{d+1}$ as a set of tuples of nonnegative integers, where instead of indexing by $\mathbb{N}$ we index by $\mathcal{I}_d.$  Precisely, $\mathcal{I}_{d+1}$ is the set of functions $a: \mathcal{I}_d \rightarrow \mathbb{N}$ (we notate such a function $i \mapsto a_i$) such that $a_i = 0$ for all but finitely many $i \in \mathcal{I}_d.$

Again, we impose on $\mathcal{I}_{d+1}$ the lexicographic total ordering: given any $a, b \in \mathcal{I}_{d+1},$ choose $i \in \mathcal{I}_d$ maximal such that $a(i) \neq b(i);$ and take $a > b$ if and only if $a(i) > b(i).$  (Only finitely many $i$ satisfy $a(i) \neq b(i),$ so there is no difficulty in choosing the maximal one.)

Just like before, there is a bijection between elements of $\mathcal{I}_{d+1}$ and isomorphism types of trees of depth at most $d+1:$ given a tree $T,$ produce the tuple which records for each $i \in \mathcal{I}_d$ the number of copies of $T_i$ that are immediate subtrees of $T.$

The result now follows from the following two lemmas, the first of which we again leave as an exercise.

Lemma. Any legal move changes a tuple $a \in \mathcal{I}_{d+1}$ to another tuple $b$ such that $b < a$ for our total ordering.

Lemma. The set $\mathcal{I}_{d+1}$ is well-ordered: it contains no infinite descending chain.

Proof. Suppose $a^{(1)} \geq a^{(2)} \geq a^{(3)} \geq \cdots$ is an infinite descending chain in $\mathcal{I}_{d+1}.$  We want to show that eventually the chain stabilizes: there exists $N$ such that $a^{(n)} = a^{(N)}$ for $n \geq N.$

$($A remark about notation: $a^{(N)} \in \mathcal{I}_{d+1}$ is a tuple in our infinite descending chain of tuples.  For $i \in \mathcal{I}_d,$ the $i$-th entry of this tuple is $a^{(N)}_i.)$

Let $j \in \mathcal{I}_d$ be an index; we say that the sequence $(a^{(n)})$ stabilizes to the right of index $j$ if there exists a positive integer $N$ such that
$$a^{(n)}_i = a^{(N)}_i$$
whenever $i \geq j$ and $n > N.$

Since $\mathcal{I}_d$ is well-ordered, we can choose $j$ minimal such that our sequence stabilizes to the right of $j;$ also choose $N$ as above, so that $(a^{(n)})_i = (a^{(N)})_i$ whenever $i \geq j$ and $n > N.$

The tuple $a^{(N)}$ has only finitely many nonzero entries.  If $a^{(N)}_k = 0$ for all $k < j$, then the sequence stabilizes at $a^{(N)}$ and we are done.  Otherwise, choose $k$ maximal such that $k < j$ and $a^{(N)}_k > 0.$   (There are only finitely many $k$ with $a^{(N)}_k > 0,$ so again there is no harm in choosing the maximum.)

I claim that in fact the sequence stabilizes to the right of index $k$, contradicting our choice of $j.$  Consider an index $i \geq k.$  If $i \geq j$ then we know $(a^{(n)})_i = (a^{(N)})_i$ for $n > N.$  If $k < i < j$ then $a^{(N)}_i = 0;$ and by the lexicographic ordering on $\mathcal{I}_{d+1}$ we know that $a^{(n)}_i = 0$ for $n > N$ as well.  So we only need to worry about $i = k.$  Again by lexicographic ordering, the sequence
$$a^{(N)}_k, a^{(N+1)}_k, a^{(N+2)}_k, \ldots$$
is a nonincreasing sequence of natural numbers.  Any such sequence in $\mathbb{N}$ eventually stabilizes.  So our sequence $(a^{(n)})$ stabilizes to the right of $k$ as well, contradicting our choice of $j.$ $\blacksquare$

In closing I'll just mention that the sets $\mathcal{I}_d$ are examples of ordinals; interpreting them as such gives rise to the proofs linked to at the very beginning of this post.

Donate to MathOverflow

I just noticed that MathOverflow is soliciting donations at this link, or click here to donate directly.  If you haven't donated yet, please consider it -- the site is an important place for discussion, and I know that I benefitted from it quite a bit in the first couple years of grad school.  From the post on Meta.MathOverflow:

If you prefer not to donate via PayPal, the post discusses other options.

A non-prorepresentable deformation functor

I've been teaching a topics course on deformation theory this semester; so far we've covered quite a few interesting things -- the Tian-Todorov theorem and Green-Lazarsfeld's generic vanishing theorem, for example.  I just wanted to use this post to record an easy and fun example that I wasn't able to find in the literature.

Question: What is a good example of a smooth projective variety $X$ over a field $k$ such that the deformation function $\text{Def}_X$ is not pro-representable?  Here $$\text{Def}_X: \text{Art}/k\to \text{Sets}$$ is the functor sending a local Artin $k$-algebra $A$, with residue field $k$, to the set of isomorphism classes of flat $A$-schemes $Y$ equipped with an isomorphism $\phi: Y_k\overset{\sim}{\to} X$.

Answer: Let $X=\text{Bl}_Z(\mathbb{P}^2)$, where $Z$ is a set of $10$ points lying on a line.

It's not totally obvious, of course, that this works.  The idea of the proof follows.  Given a variety $X$, we let $$\text{DefAut}_X: \text{Art}/k\to \text{Sets}$$ be the deformation functor sending a local Artin $k$-algebra $A$, with residue field $k$, to the set of isomorphism classes of triples $Y, \psi, \phi$, where $Y$ is a flat $A$-scheme, $\phi: Y_k\overset{\sim}{\to}X$ is an isomorphism of the special fiber of $Y$ with $X$, and $\psi: Y\overset{\sim}{\to} Y$ is an automorphism of $Y$ which is the identity on $Y_k$.

Proposition.  Let $X$ be a proper variety.  The functor $\text{Def}_X$ is pro-representable if and only if the natural map $\text{DefAut}_X\to \text{Def}_X$ is formally smooth.  (Here the natural map simply forgets the automorphism $\psi$.)

This is somewhat standard; it's an easy application of Schlessinger's criteria, for example.  One way of saying this is that given a small extension $$0\to I\to B\to A\to 0,$$ and a deformation $Y$ of $X$ to $B$, there's a long exact sequence of sets $$0\to H^0(X, T_X)\otimes I\to \text{Aut}^0(Y)\to \text{Aut}^0(Y_A)\to H^1(X, T_X)\otimes I\to \text{Def}_X(B)\to \text{Def}_X(A)\to H^2(X, T_X)\otimes I.$$

Here $\text{Aut}^0(Y)$ is the set of automorphisms of $Y$ fixing $Y_k$.  The fact that this is a long exact sequence means that the first four non-zero terms are an exact sequence of groups; that $H^1(X, T_X)\otimes I$ acts transitively on the fibers of the map $\text{Def}_X(B)\to \text{Def}_X(A)$, with kernel exactly the image of $\text{Aut}^0(Y_A)$; and that an element of $\text{Def}_X(A)$ lifts to an element of $\text{Def}_X(B)$ if and only if it maps to zero in $H^2(X, T_X)\otimes I$.  Now formal smoothness of the map  $\text{DefAut}_X\to \text{Def}_X$ is the same as surjectivity of the map $\text{Aut}^0(Y)\to \text{Aut}^0(Y_A)$ for all small extensions and all choices of $Y$.  This is the same as faithfulness of the action of $H^1(X, T_X)\otimes I$ on the fibers of the map $\text{Def}_X(B)\to \text{Def}_X(A)$, which is a restatement of Schlessinger's pro-representability criterion in our setting.

So why does our example $X$ -- namely, $\mathbb{P}^2$ blown up at a bunch of points all lying on a line $\ell$ -- have non-prorepresentable deformation functor?  The point is that infinitesimal automorphisms of $X$ are precisely infinitesimal automorphisms of $\mathbb{P}^2$ fixing the line $\ell$.  But if one moves the points to general position, the infinitesimal automorphism group becomes trivial.  So the map $\text{DefAut}_X\to \text{Def}_X$ is far from smooth.  Pretty easy and cool.

Is there a reference with examples like these?  I'm sure they're well-known, but I wasn't able to find any smooth proper examples in the literature.

Hire me!

I'm currently on the market for tenure-track jobs in mathematics. The application process is somewhat formal -- I think it's hard, in a research statement, to get across my passion and excitement for mathematics (though hopefully this comes through in recommendation letters). This post is an informal pitch, and an explanation of what I'm thinking about and why I think I'll be a good colleague.  Of course, my CV is probably the best place to see my formal qualifications.

What I'm thinking about

I'm interested in pretty much all of algebraic and arithmetic geometry, and in many aspects of related fields (for example, parts of algebraic topology and representation theory). I think my colleagues will testify that I love talking about more or less anything they're thinking about, and that I'll happily think about any mathematical problem in a field related to algebraic geometry, very broadly construed. Thus far, I've written about:

My current projects are mostly related to Galois actions on fundamental groups, which is incredibly fun -- I expect that the techniques I've been developing will have really cool applications to Iwasawa theory and to questions surrounding the geometric torsion conjecture, for example. I'm also working on a direct computation (along the lines of work of Deligne, Anderson, and Ihara) of Galois actions on certain fundamental groups, which reveals some really beautiful structure.  If you're interested in more technical details of my work, you can read a draft of my research statement.

I'm thinking about a few more classical questions as well; for example, I'm working on a project with Alex Perry about the irrationality of low-degree hypersurfaces in projective space. I'm also thinking about some somewhat more speculative projects on Brauer groups, isomonodromic deformations, and p-adic Hodge theory.

My favorite theorem

I want to tell you a bit about the theorem I'm proudest of proving in the two years since finishing my PhD, as I think it gives a somewhat better idea of the things I'm thinking about than the vague remarks above.

Theorem (L-, Theorem 1.1.2) Let $X$ be a normal algebraic variety over a finitely-generated field $k$ of characteristic zero, and let $\ell$ be a prime.  Then there exists an integer $N=N(X, \ell)$ such that: if $$\rho: \pi_1^{\text{ét}}(X_{\bar k})\to GL_n(\mathbb{Z}_\ell)$$ is a non-trivial, semi-simple, continuous representation which extends to a representation of $\pi_1^{\text{ét}}(X_{k'})$ for some $k'/k$ finite, then $\rho$ is non-trivial mod $\ell^N$.

The $N$ in the theorem is effective; in particular, if $X=\mathbb{P}^1\setminus\{x_1, \cdots, x_m\}$, then $N(X, \ell)=1$ for almost all $\ell$.

So why should you care?  The first reason is that the main examples of representations satisfying the hypotheses (i.e. those which extend from the geometric fundamental group to the arithmetic fundamental group) are monodromy representations, e.g. on the cohomology of a family of varieties over $X$.  So this tells us a new structural result about such monodromy representations, which I view as a global analogue of Grothendieck's quasi-unipotent local monodromy theorem, related to e.g. the geometric torsion conjecture.  The second reason is that this is an anabelian result -- the statement is about the structure of the arithmetic fundamental group $\pi_1^{\text{ét}}(X)$.  So in fact it tells us something completely new about the Galois action on fundamental groups of arbitrary normal varieties.

I also think the proof is pretty awesome, but I won't test your patience by describing it here.  And the use of arithmetic/anabelian techniques like these to deduce fairly classical consequences (i.e. the application to monodromy representations) is pretty cool as well, I think.

Why I'm a good colleague

I probably shouldn't write this part, since I'm not my own colleague.  But let me give it a shot.  I take service to the department seriously -- in the two years I've been at Columbia, I've run two REUs (one with Daniel Halpern-Leistner and one with David Hansen).  As a graduate student, I helped organize GAeL XXII and XXIII. I've sat on the graduate admissions committee, one thesis defense, three oral exam committees, helped organize student seminars, and the algebraic geometry research seminar.  I am currently mentoring two undergraduate research projects, both of which are going swimmingly.  I think I've been good at these mentorship activities, and I really enjoy them -- watching the students grow mathematically has been incredibly rewarding.

I also regularly give expository talks, both to undergraduate and graduate students.  I take teaching seriously (if you'd like to read a more serious-minded document outlining my thoughts on teaching, check out a draft of my teaching statement), and I think the students like me and learn from my classes -- last year I taught Calculus III.  Some entertaining excerpts from my evaluations:

Of course I've largely focused on mathematical matters and service to the department, but I also think there are less tangible, non-mathematical aspects of being a good colleague.  It's hard to give concrete evidence that I possess these qualities, but here is a sample: I have many non-mathematical interests (for example, literature -- if you search this website sufficiently hard, you might find some non-mathematical writing of mine); I care a lot about the treatment and quality of life of graduate students; and I think that I've contributed positively to the social life of the department, both at Columbia and during my graduate student years at Stanford.

Please let me know if you want to hear more, and if you have a job opening!  Of course I'll be applying via mathjobs in the next couple of months, so if you're on an admissions committee, you may see some of this in a much more formal way soon.  Wish me luck!

The parity of zero, the primality of two, and other mysteries

From time to time, I try to speak or write about mathematics for general (non-mathematical) audiences.  If you've done this, you know it's pretty hard -- in large part because it's hard to know what people know, despite my best attempts to find out.

Enter Google Surveys.  For a pretty reasonable fee, it turns out anyone can run a survey through Google; the respondents are randomly selected and reweighted by demographics (age, gender, location).  So I decided to find out:  What percentage of Americans over the age of 18 know what a prime number is?  What about an even number?  I also tried to design the questions so they tested a bit more than basic knowledge; for example, I wanted to know whether the respondents knew that zero is even (a surprisingly controversial topic).

Here are the two surveys I ran, as they would appear to respondents.  Each survey received about 250 responses from randomly selected Americans over the age of 18.  (And cost me a well-spent $25.) Even numbers: You'd think this one would be pretty easy... I included 0 because I suspected it would be the most "difficult" number to identify as even; I included 774 to check that people know how to deal with large-ish numbers. 17 and 99 were supposed to be easy, whereas 257 was aimed at checking if people were simply looking for an even digit. Prime numbers: Maybe a little tougher. Here 57 is included in honor of Grothendieck. The order of the answers was reversed for a random half of the respondents. As I understand it, Google shows these questions on sites with some premium content -- users can take the survey in lieu of paying. Data You can download the raw data for the survey on even numbers here, and the survey on prime numbers here. The data includes the type of website on which the survey was taken (news, arts and entertainment, reference, etc.), the gender of the respondent, their approximate age, region within the US, whether they are browsing from a rural, suburban, or urban area, their approximate income, and the amount of time it took them to respond to the question. Google infers much of this data from the browsing habits of the user, though, so I don't know how reliable it is. Analysis So, what percentage of American know that 2 is a prime number? That zero is even? Well 8 is pretty even, I guess. The percentages indicate how many survey-takers thought the number in question was even. So about 75.7% of people think 8 is even (not bad!) but 774 is much harder. I don't know what was going on with the 0.8% of people who thought that 17 was even, but maybe this is an example of the Lizardman constant. Yeesh. (Note that the histogram above says that there were 199 respondents. In fact, there were 250, but because of the reweighing, the survey only had the power of the survey with 199 truly randomly-chosen respondents.) The good news is that more than 40% of survey-takers knew that 13 is prime; on the other hand, 17% thought that 9 is prime. That said, I founded it heartening that the top three answers were indeed the three primes. That's the wisdom of crowds for you. How did the respondents do overall? Below are graphs indicating what percentage of respondents got 0,1, ..., 6 answers correct on each survey. Even numbers: Not bad! In particular, more than half of the survey-takers were able to get 5 or 6 answers correct. Not too shabby! To get a perfect score, one had to identify zero as even, which only 24% of the respondents were able to do, so I think this is a pretty good result. Interestingly, about 2/3 of the people who correctly identified zero as even got perfect scores. The median number of correct answers was 5 out of 6; the mean was about 4.5. Prime numbers: Pretty tough. Identifying primes was evidently much harder. The median number of correct answers was 3 out of 6 (no better than chance), and the mean was about 3.6. I did do some more detailed analysis (e.g. breaking the results down by demographics, looking at the response time, etc.) but didn't find anything particularly interesting. But for your edification, here is a plot of median response time (in milliseconds) against the number of correct answers to the survey on even numbers. Hmm. There seems to be a weak relationship between time spent and the number of correct answers (though the people who answered almost everything wrong did so pretty slowly), but maybe this isn't surprising. Conclusions I actually found these results pretty heartening! My biggest worry is that I'm now addicted to polls, at$25 a pop.

I was a bit surprised how few respondents knew that 0 is even.  Parity is a concept which actually comes up in daily life -- for example, when one wants to know which side of the street a given address is on, or in certain regulatory questions.  I was also a bit surprised that it was so difficult to identify 2 as a prime.

Of course there are some problems with these polls.  The biggest, in my opinion, is that they don't let people indicate how sure they are -- one worry I have is that if people weren't sure if, say, 2 was prime, they'd just leave it blank.  So, for the sake of symmetry, I should really run another survey, asking people to identify non-prime numbers.  I suspect far fewer than 70% of respondents would say 2 is composite.  If I decide to run another survey, I'll post about it here, of course.

Please let me know if you download and do anything with the data!  (Data on even numbers, data on primes.)

Villani for Parliament!

The (relatively new) French party En Marche! (from which the winning presidential candidate, Macron, hails) has nominated spider-brooch wearing mathematician Cedric Villani as a parliamentary candidate!  For the full list of candidates, see this list.

Uniformization over finite fields

I just asked this question on MathOverflow.  It's basically idle curiosity, but I've now been idly curious about this for several years, so I figured I might as well ask it publicly.  Please let me know if you have any thoughts!

In short, the question is:  Do every two smooth projective curves over $\overline{\mathbb{F}_q}$ of genus at least $2$ have a finite etale cover in common?

See the question for more details and remarks.

Constructive Criticism

Here is a classical example of a non-constructive proof.

Thm 1. There exist irrational numbers $x,y$ such that $x^y$ is rational.

Proof.  If $\sqrt{2}^{\sqrt{2}}$ is rational, then we may take $x=y=\sqrt{2}$.  Otherwise, we may take $x=\sqrt{2}^{\sqrt{2}}, y=\sqrt{2}$; then $x^y=2$, which is certainly rational. $\blacksquare$

This proof is non-constructive in that it doesn't actually give a (proven) example of a pair $x,y$ with the desired property; it gives two possibilities (namely $(\sqrt{2}, \sqrt{2})$ or $(\sqrt{2}^{\sqrt{2}}, \sqrt{2})$).  Of course one can give more constructive proofs; for example, one can take $$x=\sqrt{2}, y=2\log_2(3),$$ and it's relatively easy to check that $x,y$ are irrational.

What if we ask that $x,y$ are transcendental, instead of irrational?

Thm 2. There exist transcendental numbers $x,y$ such that $x^y$ is rational.

Proof 1. Take $x=e, y=\ln 2. \blacksquare$

Of course it is well-known that $e, \ln 2$ are transcendental, but the proofs (especially in the latter case), are pretty non-trivial.  I recently ran across the following non-constructive proof of Thm 2, which is much easier.  Before I give the proof, I need a definition:

Definition. A real number is incomputable if there is no computer program (Turing machine) which prints out its decimal expansion.

There are of course lots of incomputable numbers, since there are only countably many computer programs.  Of course, any incomputable number is transcendental, since computer programs can approximate roots of polynomials with rational coefficients arbitrarily well.

Proof 2.  Let $x$ be any (positive) incomputable number.  Let $y=\log_x(2)$.  Then $y$ is incomputable, as otherwise $x$ would be computable; hence $x,y$ are transcendental, and by definition $x^y=2$. $\blacksquare$.

Proof 2 is even worse than our original proof of Theorem 1 -- not only is it non-constructive, the examples it produces cannot in principle be constructed!

The purpose of this post is to observe that with some mild modification, we can make a constructive version of Proof 2.  Instead of using the fact that algebraic numbers are computable, we can use the fact that they are efficiently computable.  In particular, if a number is algebraic, then the $n$-th digit of its decimal expansion may be computed in polynomial time in the number of digits of $n$.

Proof 3.  By the (proof of) the Time Hierarchy Theorem, there exists an(explicit) number $x$ so that the $n$-th digit of $x$ is not computable in polynomial time, but so that $x$ is computable.  (In particular, $x$ is computable but transcendental.)  Set $y=\log_x(2)$.  Then $x^y=2$, but $y$ is not computable in polynomial time, as otherwise $x$ would be as well, (as one may solve the equation $x^y=2$ efficiently, since logarithms and exponents may be computed efficiently).$\blacksquare$

This is the cheapest constructive proof of Theorem 2 that I know, though of course the $x, y$ which are produced can only be written down slowly.

The Lost World

I.  Opening the Airlock

While I was in Arizona last week (running a study group for this year's Arizona Winter School on perfectoid spaces), I took the opportunity to visit another world, 50 minutes outside of Tucson.  Dreamt up by a group of experimental theater performers and environmental scientists -- a group based at Synergia Ranch, whom some have referred to as a cult -- and funded by idiosyncratic billionaire Ed Bass, Biosphere 2 is an unbelievable achievement and a heart-breaking failure.  It housed two missions as an almost entirely closed ecosystem between 1991 and 1994, but due to mismanagement from its founders and Steve Bannon (now White House chief strategist), as well as sabotage from without, it was eventually opened to the outside world.  After five years of management by Columbia University (and some litigation between Columbia and its funder, Ed Bass), management was transferred to the University of Arizona, which now runs it as both an environmental research site and a tourist attraction.

The view as one approaches the complex

I wrote an earlier post on closed ecosystems, with Biosphere 2 as the prime example.  This post is a tour of Biosphere 2's history and facilities; I took all the pictures here except those explicitly marked to the contrary.  I've gotten most of my information on the incredible history of this place from Rebecca Reider's excellent book Dreaming the Biosphere: The Theater of All Possibilities, as the tour was almost devoid of historical information.  I suspect this is part of the University of Arizona's attempt to rebrand after the high-profile disasters of the original missions, but to me at least, the interest in the facility is largely in its history.  My impression after the visit was that little serious environmental research has taken place in Biosphere 2 over its 30-year history -- but the anthropological value of its story is immense.

This post will also be in part a review of Reider's book.  There are a few other books on the Biosphere, including The Human Experiment: Two Years and Twenty Minutes Inside Biosphere 2, by Jane Poynter, one of the original Biospherians, as well as multiple books by John Allen, the charismatic mastermind behind Biosphere 2 and the leader of Mission Control for the original team of biospherians.

II. Another Earth

Biosphere 2 was originally divided into five wilderness biomes: Rainforest, Savanna, Desert, Ocean, and Marsh, as well as the Intensive Agriculture Biome, which the University of Arizona has turned into the Landscape Evolution Observatory -- essentially a large research lab where scientists will eventually study how landscapes are influenced by biological processes.

The rainforest (pictured in the four photos above) seems to be the biome whose form most matches the designers' original vision.   It is, however, overrun by morning glories, which the biospherians introduced because of their beautiful purple flowers.  These vines are, however, quite invasive -- apparently the University of Arizona scientists now maintaining the biosphere must engage in a labor-intensive purge of the plants every few months.  This sort of unintended consequence is endemic in the biosphere.  Currently, the rainforest plays host to drought experiments, which study the water sources preferred by plants under varying conditions -- different water sources are labeled by deuterium and then tracked.

The savanna is pictured in the six photos above (two of which contain yours truly).  You may notice that it contains more trees than a typical savanna.  During the original two-year mission, an excess of microbes in the soil caused oxygen levels in the Biosphere to decline precipitously, in tandem with rising carbon dioxide levels.  In an attempt to ward off suffocation, the biospheres began planting trees in the savanna -- their attempt was not successful.  More on this later.

The ocean and marsh are adjacent to the savanna -- unfortunately the marsh was too far from the boardwalk (built by Columbia during their tenure managing the biosphere in order to open it up to tourists) to take any reasonable pictures.  I've included two pictures of the ocean above, which U of A scientists are currently modifying by increasing the salinity and decreasing the temperature (reading between the lines, it seems this is an attempt to reduce heating costs).  The ocean is home to several species of fish -- the tour guide blithely informed us that the managing scientists were "curious to see if they would survive the transition," which is a far cry from the biosphere founders' original ethos.  One interesting feature of the ocean is a "wave machine," which at regular intervals produces an ominous groan and a 3-to-4 inch high wave which traverses the ocean; it's hidden behind the rock wall on the right hand side of the pictures above.

The desert (pictured in the three photos above) is apparently more humid than it once was -- during the biospherians' original tenure, it was watered by fog machines.  Later management has replaced these with sprinklers, and our tour guide described being soaked to the bone during an early morning walk through the biome.  The biosphere is surrounded by desert, but the interior desert is of a visibly different character; it is cool and pleasant -- and utterly useless for supporting the biospherians.  It's a testament to the original designer's commitment to representing the entire earth, independent of its utility.

The desert also houses the entrance to the technosphere, 2 underground acres of machinery which supports the biosphere and which once supported its human inhabitants, including carbon dioxide scrubbers (which eventually proved insufficient to the task), heat exchangers, and, most impressively, the lungs of the biosphere.  As the building was originally almost entirely sealed, it would have been vulnerable to changes in pressure; in order to avoid blown out windows, broken seals, or other structural damage, the designers decided that some part of the building would have to expand and contract with changes in internal pressure.  Thus the lungs (one of which is pictured in the second photo above): two giant rooms with immense rubber roofs, holding up a 16-ton aluminum cap.  While the building is no longer sealed, the lungs are currently inflated by two fans -- and when one opens or closes a door anywhere in the facility, they visibly shrink or grow.  The name "technosphere," by the way, is a hint at the philosophy of the residents of Synergia Ranch, many of whom took part in the biosphere's design -- they felt that technology would eventually work in harmony with nature, and Biosphere 2 was meant to be a proof of concept.

Perhaps the most impressive aspect of the facility is only visible from Biosphere 1 -- the outside world.  The building is clearly the work of visionaries: massive, beautiful, and weird.  The six pictures above are views of different parts of the biosphere.  Note that the lung (in the fourth picture above) is contained in a dome. This is because the rubber ceiling was not made to withstand the elements.  The tour guide claimed that, covered, it is expected to last 99 years (from the original date of construction, circa 1990).  The fifth picture is of the exterior of the human habitat -- in the next section, on the fraught history of Biosphere 2, I'll include some pictures of its interior.

It seems that the original biospherians expected their two-year stay in the biosphere to be quite comfortable.  And indeed, no expense was spared in making sure this would be the case -- while of course the vast majority of the $150 million spent by Ed Bass on the construction of the project went to other things, he did not skimp on creature comforts. Below is a picture of the original biospherians' kitchen (complete with kitchen island), and a view of one of the individual two-story apartments each resident was housed in. Not lavish, but not bad. So when the original crew of eight -- Abigail Alling, Linda Leigh, Taber MacCallum, Mark Nelson, Jane Poynter, Sally Silverstone, Mark Van Thillo, and Roy Walford -- entered the biosphere, they had no idea what they were getting into. Some of the smaller problems included: the coral in the ocean bleached; the marsh water, filled with organics, mixed with the salty ocean water; the salinity of the water throughout the biosphere increased; and carbonic acid in the biosphere's artificial rain accumulated to such an extent that the biosphere experienced the equivalent of 200 years of acid rain weathering during its original two year mission. But when the biospherians began to starve, these concerns faded. The original plan was to survive on a diet of mostly vegetables, with some meat and animal products produced by the small amount of livestock kept in the intensive agriculture biome (pictured below). But an "unusually cloudy El Nino" limited the amount of sunlight available to the eighteen crops grown by the farmers; broad mites killed their potato crop; the lack of wind and insect life meant the biospherians had to pollinate their crops one flower at a time. They were working 70-hour weeks, essentially subsistence farming. And still, they lost weight. Some of the group begged mission control to send them food. Eventually, a year into the mission, they began to eat their emergency seed stock. Famine had entered their new world. Roy Walford, the group's doctor, referred to their food situation as a "healthy starvation diet," in what may have been a fit of wishful thinking. On average, the men in the group lost 18 percent of their body weight; the largest, Taber MacCallum, entered the biosphere at 208 pounds and left at 150. As most of their crops failed, they adopted a diet consisting mostly of beets, "lab-lab beans," and sweet potatoes, causing their skin to turn orange. Poynter (from whose book the title of this section is taken) claims in a video, currently playing on a loop in the biosphere's exhibit center, that the diet was low in calories but high in nutrition -- but I can't help but wonder if the interpersonal problems that eventually plagued the group were in part a result of their empty stomachs. The original location of the intensive agriculture biome, now the Landscape Evolution Observatory We have precise statistics on the weight loss the crew underwent because the biospherians tracked everything about their environment -- and they soon found that food wasn't the only thing they were running out of. Over the first year of their mission, the oxygen level in the biosphere's atmosphere dropped from 21% to 14% -- equivalent to an elevation of 17,000 feet. The crew members started to pant as they climbed the stairs; four of them developed sleep apnea; Roy Walford, their doctor, was affected so severely that he couldn't add a column of single digit numbers in his head. Even worse, the biospherians had no idea where their oxygen was going. They quickly figured out that microbes in the soil were metabolizing oxygen and excreting carbon dioxide at a faster rate than expected -- the microbes had discovered the rich manure in the intensive agriculture biome, and had started to multiply out of control. But for some reason the concentration of carbon dioxide in the atmosphere was increasing much faster than the concentration of oxygen was decreasing. Without knowing where the oxygen was going, the biospherians could not recover it. They began to plant trees in the savanna -- irreparably altering their precious wilderness biomes -- and stopped irrigating the farm, hoping that they could sequester some carbon dioxide. They grew as much plant matter as possible, reaped it, and stored it in the basement technosphere, trying to keep the $CO_2$ concentration down. Eventually, they discovered that some of the concrete in the biosphere had not fully cured before they sealed the biosphere. As it continued curing, it bonded with 7 tons of oxygen in the biosphere's atmosphere. But this discovery came too late; their funder, Ed Bass, was forced to send trucks of oxygen to the biosphere, and enough was injected into the West Lung to raise the oxygen levels in the facility to 19%. The crisis was over, but suffocation had done what starvation could not -- it had forced the biospherians to open their world to the atmosphere outside. The biospherians -- photo credit Biosphere 2, by way of the Huffington post Even after the oxygen concentration was raised to acceptable levels, the biosphere remained sick. During the long months of crisis, Mission Control -- led by John Allen, the charismatic leader of Synergia Farms and the man whose obsessions birthed the biosphere -- attempted to micromanage every aspect of the crew's response to the disasters they were facing. Four of the biospherians were loyal to management; the other four felt that they should be able to act more independently: after all, they were in another world. By the end of their first year, this split was impossible to ignore. "This is the only situation I have ever been in that would drive me to drink, except there is no drink in here," Linda Leigh wrote in her journal. When a tour guide leading a group of students around the outside of the facility asked Leigh (by telephone) what they would need to be trained in, in order to work inside the Biosphere, she responded: "They must learn to work with people they despise in order to get a job done." Poynter later commented, "People were fired and unfired so often that it was almost as if John [Allen] and Margret [Augustine, Biosphere 2 CEO and co-architect] considered firing merely an extreme form of ordering someone to go stand in a corner." Once, while she was walking up the stairs, Alling and Van Thillo spat in her face. The factionalism was not limited to the inside of the biosphere. During the famine, Tony Burgess, the designer of the desert, testified to the Scientific Advisory Committee about the biospherians' mental and physical states; contra Allen's wishes, he did not sugarcoat the situation. Allen later told him, apparently in all seriousness, that "betrayal is punished by the lowest depths of hell." The music video below, created by the biospherians during their two-year mission -- dreamed up by Roy Walford, the crews' doctor and the only biospherian who maintained contact with the other seven, until his death in 2004 -- may give some insight into the best of times in the biosphere. Even so, it is quite strange, and several of the participants are visibly underweight. IV. Two World Collide Spooked by the disasters befalling the biospherians, their funder, Ed Bass, called in a man whom he believed would manage the project with a steadier hand: Steve Bannon, then an investment banker and now chief White House strategist and adviser to the President. Bannon immediately began generating proposals to monetize the biosphere, including plans to open "Biosphere 3," a casino operated jointly with the Luxor in Las Vegas. Perhaps unsurprisingly, Biosphere 2 remained in the red. Under Bannon's direction, the second and final mission began. But tensions between the old management (Allen, Augustine, and the original crew of biospherians) on one hand, and the money -- Bass and Bannon -- on the other, soon reached a boiling point. On April 1, 1994, Bass and Bannon sent US Marshals to Biosphere 2 with a restraining order, removing the original management from the project. The seven biospherians of mission two, on hearing the news, thought it was an April Fools' joke. They were given the option to leave the project, but decided not to open the airlock. Two of the original crew -- Alling and Van Thillo -- snuck back onto the property under cover of night and, according to Reider, "smashed small glass safety panels to neutralize the Biosphere's air pressure, then threw open the airlock doors. They quickly left, then telephoned the biospherians, telling the crew that they now had the freedom to leave...however, the biospherians poked their heads out, closed the doors, and chose to go on with their mission...to Gaie [Alling] and Laser [Van Thillo], seeing bankers take over Biosphere 2 was like watching their world come under foreign occupation by an enemy." Alling and Van Thillo were arrested three days later, but were never charged. Allen, the poet, playwright, and dreamer who conceived of the biosphere, was included in Bass's restraining order and exiled from his promised land. He had mismanaged the project, but there is little to suggest that Bannon did any better. Allen had brought Biosphere 2 into existence out of sheer will and$150 million of his former friend Ed Bass's money; the two-year mission he oversaw reached its conclusion, albeit without meeting the parameters Allen had originally set for it.  Under new management, Biosphere 2 was not able to complete even the single year mission it attempted.

Where the old mission control had micromanaged the first crew of biospherians, the new residents of the biosphere found that Bannon and company could not be bothered to manage them at all.  When Reider interviewed him, Bannon asked, "What was being gained by locking these people up for a year?"  Six months into the second mission, the atmospheric concentration of $N_2O$ -- laughing gas -- in the biosphere exceeded safe limits, and the experiment was declared over.

As Biospheres 1 and 2 collided, the crew found themselves without a mission.  But management was concerned with consolidation, not vision.  Reider writes that Bannon and company "purged the staff of suspected loyalists" (an act with eerie echoes today), eventually asking a staff scientist "descended from the Cherokee medicine tradition, to ceremonially cleanse the place."  Eventually they were able to persuade Columbia University to manage the facility (on which \$200 million had already been spent), and Bannon washed his hands of it.

A record of Bannon's tenure at the biosphere, immortalized via litigation, can be found in this Motherboard article.  Unsurprising highlights include accusations of sexual harassment and threats of physical violence.

V. Expelled from the Garden

When I toured Biosphere 2 last week, I was struck by the extent to which the original purpose of the facility has been erased.  I was able to find two short videos, playing on a loop, which discussed the original mission; one in which Jane Poynter discusses the biospherians' "high-nutrition, low-calorie diet," and the other a video of the original crew members exiting the airlock after their two-year mission.  Of the two pamphlets I picked up at Biosphere 2, one devotes a single line to the 2.5 storied years the facility spent as a sealed ecosystem.  The other (which contains more information, as it is aimed at hearing-impaired visitors) does spend a paragraph on the original mission, referring to its end (due to nitrous oxide poisoning) as an "administrative decision ... made to change the direction of the program."

The tour guide made almost no reference to the original mission, except for an oblique reference to the surfeit of trees in the savanna, which, he quickly commented, "were there to control the carbon dioxide levels."  Indeed, he spent more time discussing the biosphere's high school summer science program than he did the facility's original purpose.  When I asked about the history of the institution, he more or less refused to comment.  Nonetheless, I highly recommend a visit if you are in the Tucson area.

Most of the information I gathered for this post came from Rebecca Reider's Dreaming the Biosphere, which starts slow but is overall excellent.

Many of the original group of biospherians have continued to have interesting lives.  For example, Poynter has given a TED Talk in which she summarizes her time in the biosphere and discusses her new company, Paragon Space Development, which hopes to use her expertise in building closed ecosystems on the moon and mars.  While the mission of the biosphere under the management of the University of Arizona seems unclear, they seem to be pivoting in a similar direction; for example, the lunar garden below is being exhibited in what used to be the biosphere's human habitat.

The lunar garden at Biosphere 2

Synergia Ranch, where the idea of the biosphere was hatched in the mind of John Allen, Margret Augustine, and their followers, now markets itself as a location for conferences and retreats; four of the original crew members still lived there, as of the publication of Reider's book.  The Ranch's website seems to be managed by Marie Harding, once Biosphere 2's Chief Financial Officer, and includes a link to her CV and several of her paintings.  Some, like the one below, clearly depict the interior of the biosphere.

Painting by Marie Harding - Synergia Ranch

Were the dreamers behind the biosphere foolish?  Was their effort wasted?  With their history almost entirely scrubbed from the facility, it's easy to think that the biospherians' legacy will fade into nothing -- that in the middle of the Arizona desert, a pyramid of glass will remain for a few decades, but will signify nothing but loss and inevitable mediocrity.

I don't think that's what Biosphere 2 represents, though.  Our world too is plagued by famine, strife, poor management, climate change, and selective memory -- some of it perpetrated by the same person who caused many of Biosphere 2's eventual problems. The difference is that our biosphere doesn't have an airlock.

Honestly, I prefer the bumbling madness of the Biosphere's first mission to the corporate shuffle that came afterwards; the biospherians' determination to stick it out even though they couldn't stand one more second of each others' company; the willpower that let them continue while they could barely breathe.  I think I understand the urge that made them try to retreat to a better world.

I opened this post with Shelley's "Ozymandias."  But I've always thought that the poem's fame undercuts itself.  We do remember Ozymandias, if only through Shelley.  And I think that even if we laugh at the biospherians, we can't help but recognize how our world echoes theirs; and our hearts can't help but ache for what they lost.

Sawin on Severi's Conjecture

One of my favorite questions is: for which $g, n, p$ is the moduli space of $n$-pointed genus $g$ curves $\mathscr{M}_{g,n, \mathbb{F}_p}$ unirational/uniruled?  Will Sawin has just posted a beautiful paper on the ArXiv answering this question in most cases, for $g=1$.  Indeed, he shows that for $n\geq p\geq 11, \mathscr{M}_{1, n, \mathbb{F}_p}$ is not uniruled... (more below the fold)