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Joints of varieties

September 12, 2020

The summer has come to an end. The temperature in Boston seems to have dropped all the sudden. Our fall semester has just started, though without all the new and returning students roaming around campus this time (most classes are online this term). I quite miss the view from my MIT office, and all the wonderful people that I would run into the department corridors.

The view from my (currently unoccupied) MIT office

I’ve been working from home for the past several months and connecting with my students via Zoom. In this and the upcoming several posts, I would like to tell you about some of the things that we have been working on recently.


With PhD student Jonathan Tidor and undergraduate student Hung-Hsun Hans Yu, we recently solved the joints problem for varieties by a developing a new extension of the polynomial method. (Also see my talk video and slides on this work.)

Jonathan Tidor and Hung-Hsun Hans Yu

I discussed the joints problem in an earlier blog post. Recall that the joints problem asks:

Joints problem: At most how many joints can $N$ lines in $\mathbb{R}^3$ can make?

Here a joint is a point contained in three non-coplanar lines.

A joint

Here is an example of a configuration of lines making many joints. Take k generic planes in space, pairwise intersect them to form \(N = \binom{k}{2}\) lines, and triplewise intersect to form \(J = \binom{k}{3} \sim \frac{\sqrt{2}}{3} N^{3/2}\) joints.

Planes, lines, and joints

The joints problem was raised in the early 1990’s. It is a problem in incidence geometry, which concerns possible incidence patterns of objects like points and lines. The joints problem has also received attention from harmonic analysts due to its connections to the Kakeya problem.

The big breakthrough on the joints problem was due to Guth and Katz, who showed that the above example is optimal up to a constant factor:

Joints theorem (Guth–Katz). $N$ lines in \(\mathbb{R}^3\) form $O(N^{3/2})$ joints.

Guth and Katz solved the problem using a clever technique known as the polynomial method, which was further developed in their subsequent breakthrough on the Erdős distinct distances problem. The polynomial method has since then grown into a powerful and indispensible tool in incidence geometry, and it also led to important recent developments in harmonic analysis.

I learned much of what I know about the polynomial method from taking a course by Larry Guth while I was a graduate student. Guth later developed the course notes into a beautifully written textbook, which I highly recommend. A feature of the expository style of this book, which I wish more authors would adopt, is that it frequently motivates ideas by first explaining false leads and wrong paths. As I was reading this book, I felt like that I was being taken on a journey as a travel companion and given a lens towards the thinking process, rather than somehow being magically teleported to the destination.

In my previous paper on the joints problem with Hans Yu, which I discussed in earlier blog post, we sharpened the Guth–Katz theorem to the optimal constant.

Theorem (Yu–Zhao). $N$ lines in \(\mathbb{R}^3\) have \(\le \frac{\sqrt{2}}{3} N^{3/2}\) joints.

Everything I have said so far are also true in arbitrary dimensions and over arbitrary fields, but I am only stating them in \(\mathbb{R}^3\) here to keep things simple and concrete.

Joints of flats

Our new work solves the following extension of the joints problem.

Joints problem for planes. At most how many joints can $N$ planes in \(\mathbb{R}^6\) make?

Here a joint is a point contained in a triple of 2-dimensional planes not all contained in some hyperplane.

The example given earlier for joints of lines extends easily to joints of planes.

Example. Take $k$ generic 4-flats in \(\mathbb{R}^6\), pairwise intersect them to form $N = \binom{k}{2}$ planes, and triplewise intersect them to form \(\binom{k}{3} = \Theta(N^{3/2})\) joints.

Joints of planes concerns incidences between high dimensional objects, which in general tend to be more intricate compared to incidence problems about points and lines.

I first learned about this problem from a paper of Ben Yang, who did this work as a PhD student of Larry Guth. Yang proved that $N$ planes in \(\mathbb{R}^6\) form at most $N^{3/2+o(1)}$ joints. His proof uses several important recent developments in incidence geometry, and in particular uses the polynomial partitioning technique of Guth and Katz (as well as subsequent extensions due to Solymosi–Tao and Guth). This is a fantastic result, though with two important limitations due to the nature of the method:

  1. There is an error term in the exponent $N^{3/2+o(1)}$ that is conjectured not to be there. One would like a bound of the form $O(N^{3/2})$
  2. The result is restricted to planes in \(\mathbb{R}^6\) and does not work if we replace \(\mathbb{R}\) by an arbitrary field. This is because the polynomial partitioning method uses topological tools, in particular the ham sandwich theorem, which crucially requires the topology of the reals.

Our work overcomes both difficulties. As a special case, we prove (here \(\mathbb{F}\) stands for an arbitrary field, and the hidden constants do not depend on \(\mathbb{F}\)):

Joints of planes (Tidor–Yu–Zhao). $N$ planes in \(\mathbb{F}^6\) have $O(N^{3/2})$ joints.

And more generally, our results hold for varieties instead of flats. Here a joint is a point contained in a triple of varieties, such that it is a regular point on each of the three varieties and the three tangent plane do not lie inside a common hyperplane.

Joints of varieties (Tidor–Yu–Zhao). A set of 2-dimensional varieties in \(\mathbb{F}^6\) of total degree $N$ has $O(N^{3/2})$ joints.

The above statements are special cases of our results. The joints problem for lines is known to hold under various extensions (arbitrary dimension and fields, several sets/multijoints, counting joints with multiplicities), and our theorem holds in these generalities as well.

Proof sketch of the joints theorem for lines

Here is a sketch of the proof that $N$ lines in \(\mathbb{R}^3\) make $O(N^{3/2})$ joints (following the simplifications due to Kaplan–Sharir–Shustin and Quilodrán; also see Guth’s notes for details of the proof).

The first two steps below are both hallmarks of the polynomial method, and they are present in nearly every application of the method. The third step is a more joints-specific argument.

  1. Parameter counting. The dimension of the vector space of polynomials in \(\mathbb{R}[x,y,z]\) up to degree $n$ is \(\binom{n+3}{3}\). We deduce that if there are $J$ joints and \(J < \binom{n+3}{3}\), then one can find a nonzero polynomial $g$ of degree $\le C J^{1/3}$ (for some constant $C$) that vanishes on all the joints (as there are more coefficients to choose from than there are linear constraints). Take such a polynomial $g$ with the minimum possible degree.

  2. Vanishing lemma. A nonzero single variable polynomial cannot vanish more times than its degree.

  3. If all lines have $> CJ^{1/3}$ joints, then the vanishing lemma implies that $g$ vanishes on all the lines. At each joint, $g$ vanishes identically along three independent lines, and hence its directional derivatives vanish in three independent directions, and thus the gradient \(\nabla g\) vanishes at the joint. Therefore, the three partial derivatives \(\partial_x g\), \(\partial_y g\), \(\partial_z g\) all vanish at every joint, and at least one of them is a nonzero polynomial and has smaller degree than $g$, thereby contradicting the minimality of the degree of $g$. Therefore, the initial assumption must be false, namely that some line has \(\le C J^{1/3}\) joints. We can then remove this line, deleting \(\le CJ^{1/3}\) joints, and proceed by induction to prove that $J = O(N^{3/2})$.

Some wishful thinking

It is natural to try to extend the above proof for the joints problem for planes. One quickly runs into the following conundrum:

How does one generalize the vanishing lemma to 2-variable polynomials?

The vanishing lemma is such an elementary fact about how polynomial behaves on a line. Yet it plays a crucial role in the polynomial method. We would be able to adapt the above proof to joints of planes if only something like the following would be true (they are not):

Wishful thinking 1. Show that every nonzero polynomial $g(x,y)$ of degree $n$ has $O(n^2)$ zeros.

Well, this is obviously nonsense. The zero set of $g(x,y)$ is typically a curve with infinitely many points.

How else might we try to extend the vanishing lemma to two dimensions? Bézout’s theorem is one such extension. It tells us that two polynomials $g(x,y)$ and $h(x,y)$ without common factors cannot have more than $(\deg g)(\deg h)$. So, perhaps, instead of trying to find one polynomial vanishing at all joints, we try to find a pair of polynomials.

Wishing thinking 2. Show that one can find two polynomials \(g, h \in \mathbb{R}[x_1, x_2, x_3, x_4, x_5, x_6]\) vanishing at all $J$ joints and $(\deg g)(\deg h) = O(J^{1/3})$, and such that $g$ and $h$ have no common factors when restricted to any of the given planes.

This approach is a natural one to try for the joints of planes problem, and is an approach that many people have thought about. It also seems extremely hard to get it to work (likely impossible). It seems closely related to the inverse Bézout problem, which already illustrates some of the difficulties encountered.

Instead of asking a polynomial to vanish at every joint, we can demand more. Namely that we can ask the polynomial to vanish at each joint to some high order (equivalently, asking some derivatives of the polynomial to vanish at the joints). This idea (sometimes called the “method of multiplicities”) has shown to be useful in a variety of contexts. Perhaps we can modify Wishing thinking 1 to account for these multiplicities?

Wishful thinking 3. Show that every nonzero polynomial $g(x,y)$ of degree $n$ can vanish to order $s$ at no more than $O(n^2/s^2)$ points.

The reason why one might hope for such a statement is that there are \(\binom{n+2}{2} \approx n^2/2\) degrees of freedom for choosing $g$, and this number is roughly the same as the number of constraints associated to forcing the polynomial to vanish to order $s$ at $n^2/s^2$ points.

However, Wishful thinking 3 is still too good to be true. For example, the polynomial $g(x,y) = y^s$ is a low-degree polynomial that vanishes to order $s$ at every point on the $x$-axis.

So what was wrong with the parameter counting intuition? It must be the case that some of the linear constraints associated to these high order vanishing constraints are linearly dependent.

Extending the polynomial method

So far all I have told you are what doesn’t work, and I haven’t said anything about what does work. Where we left off above is a good starting point of what we actually do, which I will not describe here in this blog post but instead refer you to either my talk or our paper for details. We have to account for the linear dependencies between these higher order vanishing conditions, and we come up with a process of assigning non-redundant vanishing conditions (each in the form of some higher order derivative operator evaluated at a joint). We prove a new vanishing lemma tailored for this joints problem. Rather vaguely, it has the following form:

A new vanishing lemma. Given a finite set of planes in \(\mathbb{R}^6\) forming some joints, and a positive integer $n$, and an integer (“handicap”) attached to each joint, we run a certain process that assigns, on each plane, a total of \(\binom{n+2}{2}\) derivative operators across the joints on this plane. Then for every nonzero polynomial \(g(x_1, \dots, x_6)\) of degree at most $n$, there is some joint $p$, three planes $F_1, F_2, F_3$ containing $p$, and a derivative operator $D_i$ assigned to $p$ on $F_i$ through our process for each $i =1,2,3$, so that $D_1 D_2 D_3 g(p) \ne 0$.

This new vanishing lemma overcomes the earlier difficulty and allows us to extend the polynomial method to solve the joints problem for planes.


Let me finish with an open problem: while we can prove the optimal constant for the joints theorem for lines, we do not yet know how to prove the optimal constant for the joints theorem for planes.

Conjecture. $N$ planes in \(\mathbb{F}^6\) have \(\le (\frac{\sqrt{2}}{3} + o(1)) N^{3/2}\) joints.



Ashwin Sah's new diagonal Ramsey number upper bound

May 20, 2020

Ashwin Sah

Ashwin Sah just proved a new upper bound to diagonal Ramsey numbers. See his preprint on the arXiv. This is the first improvement since Conlon’s upper bound published in Annals of Math in 2009, which in turn built on earlier work of Thomason (1988).

Obtaining asymptotics of Ramsey numbers is perhaps the central open problem of Ramsey theory in combinatorics. Research on this problem had led us to a lot of new mathematics that have proven to be useful elsewhere. Most notably, Erdős’ proof of a lower bound to Ramsey numbers (1947) initiated the probabilistic method, by now an indispensable method in combinatorics.

Ashwin’s improvement relies on new insights into graph quasirandomness properties.

Congratulations Ashwin! This is an impressive accomplishment.

Ashwin had just finished his undergraduate studies at MIT. I’m happy that he will be staying at MIT for his PhD. Ashwin already has an impressive list of papers. I had previously mentioned Ashwin on this blog reporting on our joint work (together with Mehtaab Sawhney and David Stoner) on our work on independent sets and the reverse Sidorenko inequality.



Joints tightened

November 21, 2019

I’m happy to announce a new paper titled Joints tightened coauthored with Hung-Hsun Hans Yu, an undergraduate student at MIT. In this paper, we determine the tight constant in the joints problem.

Hung-Hsun Hans Yu

The joints problem is a classic problem in incidence geometry. A typical question in incidence geometry concerns what kinds of configurations can be built using just points and lines.

Suppose you are allowed to draw L lines in space. What is the maximum possible number of points that lie on three lines (let’s call these points “triple intersections” for now)?

It turns out this is not a great question, since you can “cheat” by drawing a 2-dimensional picture from a grid as below. This configuration of L lines has \(\Theta(L^2)\) triple intersections, and you cannot do much better since L lines have at most $L^2$ intersections.

Not a joints configurations

The issue with the above example is that it is really a 2-dimensional configuration, whereas we really want to ask a 3-dimensional question. One way to make the problem much more interesting is only count “truly 3-dimensional” intersection points.

Given a collection of lines in space, we define a joint to be a point that is contained in three of our lines and not all lying on the same plane.

A joint

Joints problem. What is the maximum number of joints that can be formed by L lines in 3-dimensional space?

The joints problem was raised in early 1990’s, and it turned out to be a very interesting problem! Besides its intrinsic interest as a combinatorial geometry problem, it also caught the attention of analysts as the joints problem is connected to the Kakeya problem, a central problem in analysis (this connection was popularized by Wolff).

Here are some examples of configurations of lines that have a lot of joints.

Example 1. The easiest example to visualize is by considering a \(k \times k \times k\) grid of lines, which has \(L = 3k^2\) lines and \(J = k^3 = (L/3)^{3/2}\) joints:

A grid of joints

Example 2. Actually, there is another example that has even more joints (by a constant factor). Think about a tetrahedron, formed by 4 faces (planes), and whose edges form 6 lines, and the vertices form 4 joints. We can generalize the tetrahedron example by taking k generic planes in space, and have the planes pairwise intersect to make \(L = \binom{k}{2}\) lines, and triplewise intersect to make \(J = \binom{k}{3}\) joints. (This example appeared in the original paper on the joints problem.)

Planes, lines, and joints

In both examples, the number of joints is \(\Theta(L^{3/2})\). A matching upper bound on the number of joints turned out to be much less obvious. A stunning breakthrough took place in 2008 when Guth and Katz proved an \(O(L^{3/2})\) upper bound on the number of joints in using the polynomial method. Their solution was partly inspired by Dvir’s stunning solution to the finite field Kakeya problem, and it was also a precursor to Guth and Katz’s subsequent celebrated solution of the Erdős distinct distances problem.

Larry Guth conjectured in his book The Polynomial Methods in Combinatorics that the second example above is best possible. I first learned about this very nice conjecture as a graduate student in Guth’s class on the polynomial method.

The main result of our paper is a new improved upper bound on the joints problem, matching the constant in the second example above. More precisely, we prove that L lines in space make at most \(\frac{\sqrt{2}}{3}L^{3/2}\) joints, thereby confirming Guth’s conjecture up to a \(1+o(1)\) factor. Our results also hold in arbitrary dimensions and over arbitrary fields.

It was quite unexpected to us that such a tight bound can even be proved, especially given how long it took just to obtain an upper bound of the right order of magnitude on the joints problem. In pretty much all other classical problems in incidence geometry (i.e., the Szemerédi–Trotter theorem), the optimal constant factor is not known. In fact, our result might be the first one in incidence geometry that obtains a tight constant factor.



Equiangular lines with a fixed angle

July 30, 2019

I’m happy to announce our new paper Equiangular lines with a fixed angle joint with four MIT coauthors: Zilin Jiang, Jonathan Tidor, Yuan Yao, and Shengtong Zhang.

(L->R) Shengtong Zhang, Yuan Yao, Jonathan Tidor, Zilin Jiang, me

Zilin is an Instructor (postdoc), Jonathan is my PhD student who just finished his second year, and Yuan and Shengtong are undergraduates who just finished their second and first respective years.

Our paper solves a longstanding problem in discrete geometry concerning equiangular lines, which are configurations of lines in space that are pairwise separated by the same angle. How many such lines can exist simultaneously in a given dimension?

We determine, for every fixed angle, the maximum number of equiangular lines with the given angle in high dimensions.

Our paper builds on a long sequence of earlier ideas. There were some important progress on this problem by Igor Balla, Felix Dräxler, Peter Keevash, and Benny Sudakov, as featured in this earlier Quanta Magazine article written for a lay audience.

In the end we finished off the problem in a clean and crisp manner, in a 10-page paper with a self-contained proof.

It has been known that the study of equiangular lines is closed linked to spectral graph theory, which seeks to understand graphs via their spectrum (i.e., eigenvalues). We prove our main result via a new insight in spectral graph theory, namely that a bounded degree graph must have sublinear second eigenvalue multiplicity.



Impartial digraphs

June 27, 2019

I just finished a new paper, Impartial digraphs, coauthored with Yunkun Zhou, who just completed his undergraduate studies at MIT and will be moving to Stanford to start his PhD this fall.

with Yunkun Zhou (R) after his MIT commencement

The problem (along with the conjecture that we proved) was proposed by Jacob Fox, Hao Huang, and Choongbum Lee. Ron Graham had also told me about this problem with much enthusiasm during tea time one afternoon at the Berkeley Simons Institute a couple of years ago, after learning about it from Jacob.

I really enjoyed working on this paper with Yunkun. The problem is natural and simple to state, the answer appears somewhat unexpected (at least initially), and the solution ended up being quite tricky yet clean, employing a variety of tools in combinatorics, ranging from analytic methods (graph limits), algebraic considerations (polynomial identities and factorization), as well basic combinatorial techniques such as bijections.

We are going to talk about directed graphs (which graph theorists like to abbreviate as digraphs). These are graphs where every edge is given an orientation.

A tournament is a directed graph where there is a directed edge between every pair of vertices (think of it as recording the outcome of a round-robin tournament where every player plays against every other player).

Given a directed graph H (think of it as a pattern), we can ask: how many different copies of the pattern H appear in a given tournament?

Look at the following 4-vertex tournament H. It has the following curious property: all tournaments of a fixed size contains the same number of copies of H, no matter how the edges in the tournament are oriented.

Hmm, why is this true? And are there other directed graphs H with the same property, namely that all tournaments of a given size contain the same number of copies of H?

In the good old mathematical tradition, we came up with a name for directed graphs H with this curious property: we called them impartial (as in: impartial judges of a tournament).

Okay. So why is the above directed graph impartial? Are there any other impartial directed graphs? What are all the impartial directed graphs?

Here is a simple example of an impartial digraph:

Here is another one:

Indeed, given any 10-vertex tournament, we can count the number of copies of the first edge (it doesn’t depend on the tournament), and then count the number of copies of the second edge among the remaining 8 vertices (it still doesn’t depend on the tournament).

Once these two edges are placed in a tournament, there are two ways to join their tips, but they result in identical patterns (one flipped from the other):

It must then follow that this directed graph is impartial as well:

We can follow the same logic and continue this procedure, iterate, and build up even larger impartial directed graphs. Each step, take two copies of the previous tree (keeping the same edge orientations), and then add a new edge between a twin pair of vertices.

We call any directed graph that can building up as above a recursively bridge-mirrored tree.

The same argument as earlier shows that every recursively bridge-mirrored tree is impartial.

Are there other impartial digraphs? Did we miss any?

It turns out, there are actually no others. We have found them all, and that is the main result of our paper:

Theorem. A directed graph is impartial if and only if every component is recursively bridge-mirrored trees.

Just a word about how about this problem is related to a bigger picture. It is related to a number of other graph homomorphism inequalities that are central in extremal graph theory. It can be considered as the “equality case” for a directed analog of an open problem known as Sidorenko’s conjecture. See our paper for some further discussions and open problems.



A reverse Sidorenko inequality

September 26, 2018

Together with undergraduates Ashwin Sah, Mehtaab Sawhney, and David Stoner (the same team that proved Kahn’s conjecture on independent sets that I blogged about earlier, we are excited to announce our new paper A reverse Sidorenko inequality. This paper solves several open problems concerning graph colorings and homomorphisms, including one of my favorite problems regarding maximizing the number of q-colorings in a d-regular graph. I had previously blogged about some of these problems.

It has been truly a pleasure working with Ashwin, Mehtaab, and David on this project. I have learned so much from them.

Here are some highlights of our new results.

1. Graph colorings.

 Among d-regular graphs, which graph has the most number of proper q-colorings, exponentially normalized by the number of vertices of G?

We show that the extremal graph is the complete bipartite graph \(K_{d,d}\) (or disjoint unions thereof, as taking disjoint copies does not change the quantity due to the normalization).

2. Graph homomorphisms.

Among d-regular graphs, which triangle-free graph G has the most number of homomorphisms into a fixed H, again exponentially normalized by the number of vertices of G?

We show that the extremal graph G is the complete bipartite graph \(K_{d,d}\).

The triangle-free hypothesis on G is best possible in general. For certain specific H, such as \(K_q\), corresponding to proper q-colorings as earlier, the triangle-free hypothesis can be dropped. It remains a wide open problem to determine for which H can one drop the triangle-free hypothesis on G.

3. Partition functions of Ferromagnetic spin models.

Among d-regular graphs, which graph maximizes the log-partition function per vertex of a given Ferromagnetic spin model (e.g., Ising model)?

We show that the extremal graph is the clique \(K_{d+1}\).

For each setting, we establish our results more generally for irregular graphs as well, similar to our earlier work on independent sets.

Our results can be interpreted as a reverse form of the inequality in Sidorenko’s conjecture, an important open problem in graph theory stating a certain positive correlation on two-variable functions.

One can also view our results as a graphical analog of the Brascamp–Lieb inequalities, a central result in analysis.

This paper resolves one of my favorite open problems on this topic (the number of graph colorings). It also points us to many other open problems. Let me conclude by highlighting one of them (mentioned in #2 earlier). I’ll state it in a simpler form for d-regular graphs, but it can be stated more generally as well.

Open problem. Classify all H with the following property: among d-regular graphs G, the number of homomorphisms from G to H, exponentially normalized by the number of vertices of G, is maximized by \(G = K_{d,d}\).

Our results show that \(H = K_q\) works, even when some of its vertices are looped. Generalizing this case, we conjecture that all antiferromagnetic models H have the same property (though we know that this would not be the complete class, even if true).



The number of independent sets in an irregular graph

May 12, 2018

I am excited to announce our new paper, The number of independent sets in an irregular graph, coauthored with these three amazing collaborators:

Ashwin Sah, Mehtaab Sawhney, David Stoner

Ashwin is a freshman at MIT, Mehtaab is a sophomore at MIT, and David is a junior at Harvard.

The paper solves a conjecture made by Jeff Kahn in 2001 concerning the number of independent sets in a graph.

An independent set of a graph is a subset of vertices with no two adjacent.

Example: the complete list of independent sets of a cycle of length 4

Since many problems in combinatorics can be naturally phrased in terms of independent sets in a graph or hypergraph, getting good bounds on the number of independent sets is a problem of central importance.

Instead of giving the exact statement of the conjecture (now theorem), which you can find in the abstract of our paper, let me highlight a specific instance of a problem addressed by the theorem:

Question. Consider the family of graphs with no isolated vertices and

  • exactly a third of the edges have degree 3 on both endpoints, and
  • a third of the edges have degree 4 on both endpoints,
  • the remaining third of the edges have degree 3 on one endpoint and degree 4 on the other endpoint.

What is the smallest constant c such that every n-vertex graph satisfying the above properties has at most \(c^n\) independent sets?

In other words, letting \(i(G)\) denote the number of independent sets and \(v(G)\) the number of vertices of \(G\), maximize the quantity \(i(G)^{1/v(G)}\) over all graphs G in the above family.

(See the end of this post for the answer)

In summer 2009, as an undergraduate participating in the wonderful Research Experience for Undergraduates (REU) run by Joe Gallian in Duluth, MN, I learned about Kahn’s conjecture (somewhat accidentally actually, as I was working on a different problem that eventually needed some bounds on the number of independent sets, and so I looked up the literature and learned, slightly frustratingly at the time, that it was an open problem). It was on this problem that I had written one of my very first math research papers.

I had been thinking about this problem on and off since then. A couple years ago, Joe Gallian invited me to write an article for a special issue of the American Mathematical Monthly dedicated to undergraduate research (see my previous blog post on this article), where I described old and new developments on the problem and collected a long list of open problems and conjectures on the topic (one of them being Kahn’s conjecture).

This spring, I had the privilege with working with Ashwin, Mehtaab, and David, three energetic and fearless undergraduate students, finally turning Kahn’s conjecture into a theorem. Fearless indeed, as our proof ended up involving quite a formidable amount of computation (especially for such an innocent looking problem), and my three coauthors deserve credit for doing the heavy-lifting. I’ve been told that there had been many late night marathon sessions in an MIT Building 2 classroom where they tore apart one inequality after another.

This is certainly not the end of the story, as many more interesting problems on the subject remain unsolved—my favorite one being the analogous problem for colorings, e.g., instead of counting the number of independent sets, how about we count the number of ways to color the vertices of the graph with 3 colors so that no two adjacent vertices receive the same color? (See my previous blog post for some more discussion.)

Anyway, here is the the answer to the question above. The number of independent sets in an n-vertex graph satisfying the above properties is at most \(c^n\), where the optimal constant for c is \(15^{4/63} \times  23^{1/21} \times 31^{1/28} = 1.558\dots\), or, more helpfully, the maximum is attained by the following graph (in general, the theorem says that the maximizer is always a disjoint union of complete bipartite graphs):