Earlier this month, Hao Huang (who, incidentally, was a graduate student here at UCLA) gave a remarkably short proof of a long-standing problem in theoretical computer science known as the sensitivity conjecture. See for instance this blog post of Gil Kalai for further discussion and links to many other online discussions of this result. One formulation of the theorem proved is as follows. Define the -dimensional hypercube graph to be the graph with vertex set , and with every vertex joined to the vertices , where is the standard basis of .

Theorem 1 (Lower bound on maximum degree of induced subgraphs of hypercube)Let be a set of at least vertices in . Then there is a vertex in that is adjacent (in ) to at least other vertices in .

The bound (or more precisely, ) is completely sharp, as shown by Chung, Furedi, Graham, and Seymour; we describe this example below the fold. When combined with earlier reductions of Gotsman-Linial and Nisan-Szegedy; we give these below the fold also.

Let be the adjacency matrix of (where we index the rows and columns directly by the vertices in , rather than selecting some enumeration ), thus when for some , and otherwise. The above theorem then asserts that if is a set of at least vertices, then the minor of has a row (or column) that contains at least non-zero entries.

The key step to prove this theorem is the construction of rather curious variant of the adjacency matrix :

Proposition 2There exists a matrix which is entrywise dominated by in the sense that

Assuming this proposition, the proof of Theorem 1 can now be quickly concluded. If we view as a linear operator on the -dimensional space of functions of , then by hypothesis this space has a -dimensional subspace on which acts by multiplication by . If is a set of at least vertices in , then the space of functions on has codimension at most in , and hence intersects non-trivially. Thus the minor of also has as an eigenvalue (this can also be derived from the Cauchy interlacing inequalities), and in particular this minor has operator norm at least . By Schur’s test, this implies that one of the rows or columns of this matrix has absolute values summing to at least , giving the claim.

Remark 3The argument actually gives a strengthening of Theorem 1: there exists a vertex of with the property that for every natural number , there are at least paths of length in the restriction of to that start from . Indeed, if we let be an eigenfunction of on , and let be a vertex in that maximises the value of , then for any we have that the component of is equal to ; on the other hand, by the triangle inequality, this component is at most times the number of length paths in starting from , giving the claim.

This argument can be viewed as an instance of a more general “interlacing method” to try to control the behaviour of a graph on all large subsets by first generating a matrix on with very good spectral properties, which are then partially inherited by the minor of by interlacing inequalities. In previous literature using this method (see e.g., this survey of Haemers, or this paper of Wilson), either the original adjacency matrix , or some non-negatively weighted version of that matrix, was used as the controlling matrix ; the novelty here is the use of signed controlling matrices. It will be interesting to see what further variants and applications of this method emerge in the near future. (Thanks to Anurag Bishoi in the comments for these references.)

The “magic” step in the above argument is constructing . In Huang’s paper, is constructed recursively in the dimension in a rather simple but mysterious fashion. Very recently, Roman Karasev gave an interpretation of this matrix in terms of the exterior algebra on . In this post I would like to give an alternate interpretation in terms of the operation of *twisted convolution*, which originated in the theory of the Heisenberg group in quantum mechanics.

Firstly note that the original adjacency matrix , when viewed as a linear operator on , is a convolution operator

where

is the counting measure on the standard basis , and denotes the ordinary convolution operation

As is well known, this operation is commutative and associative. Thus for instance the square of the adjacency operator is also a convolution operator

where the convolution kernel is moderately complicated:

The factor in this expansion comes from combining the two terms and , which both evaluate to .

More generally, given any bilinear form , one can define the *twisted convolution*

of two functions . This operation is no longer commutative (unless is symmetric). However, it remains associative; indeed, one can easily compute that

In particular, if we define the twisted convolution operator

then the square is also a twisted convolution operator

and the twisted convolution kernel can be computed as

For general bilinear forms , this twisted convolution is just as messy as is. But if we take the specific bilinear form

then for and for , and the above twisted convolution simplifies to

and now is very simple:

Thus the only eigenvalues of are and . The matrix is entrywise dominated by in the sense of (1), and in particular has trace zero; thus the and eigenvalues must occur with equal multiplicity, so in particular the eigenvalue occurs with multiplicity since the matrix has dimensions . This establishes Proposition 2.

Remark 4Twisted convolution is actually just a component of ordinary convolution, but not on the original group ; instead it relates to convolution on a Heisenberg group extension of this group. More specifically, define the Heisenberg group to be the set of pairs with group lawand inverse operation

(one can dispense with the negative signs here if desired, since we are in characteristic two). Convolution on is defined in the usual manner: one has

for any . Now if is a function on the original group , we can define the lift by the formula

and then by chasing all the definitions one soon verifies that

for any , thus relating twisted convolution to Heisenberg group convolution .

Remark 5With the twisting by the specific bilinear form given by (2), convolution by and now anticommute rather than commute. This makes the twisted convolution algebra isomorphic to a Clifford algebra (the real or complex algebra generated by formal generators subject to the relations for ) rather than the commutative algebra more familiar to abelian Fourier analysis. This connection to Clifford algebra (also observed independently by Tom Mrowka and by Daniel Matthews) may be linked to the exterior algebra interpretation of the argument in the recent preprint of Karasev mentioned above.

Remark 6One could replace the form (2) in this argument by any other bilinear form that obeyed the relations and for . However, this additional level of generality does not add much; any such will differ from by an antisymmetric form (so that for all , which in characteristic two implied that for all ), and such forms can always be decomposed as , where . As such, the matrices and are conjugate, with the conjugation operator being the diagonal matrix with entries at each vertex .

Remark 7(Added later) This remark combines the two previous remarks. One can view any of the matrices in Remark 6 as components of a single canonical matrix that is still of dimensions , but takes values in the Clifford algebra from Remark 5; with this “universal algebra” perspective, one no longer needs to make any arbitrary choices of form . More precisely, let denote the vector space of functions from the hypercube to the Clifford algebra; as a real vector space, this is a dimensional space, isomorphic to the direct sum of copies of , as the Clifford algebra is itself dimensional. One can then define a canonical Clifford adjacency operator on this space bywhere are the generators of . This operator can either be identified with a Clifford-valued matrix or as a real-valued matrix. In either case one still has the key algebraic relations and , ensuring that when viewed as a real matrix, half of the eigenvalues are equal to and half equal to . One can then use this matrix in place of any of the to establish Theorem 1 (noting that Schur’s test continues to work for Clifford-valued matrices because of the norm structure on ).

To relate to the real matrices , first observe that each point in the hypercube can be associated with a one-dimensional real subspace (i.e., a line) in the Clifford algebra by the formula

for any (note that this definition is well-defined even if the are out of order or contain repetitions). This can be viewed as a discrete line bundle over the hypercube. Since for any , we see that the -dimensional real linear subspace of of sections of this bundle, that is to say the space of functions such that for all , is an invariant subspace of . (Indeed, using the left-action of the Clifford algebra on , which commutes with , one can naturally identify with , with the left action of acting purely on the first factor and acting purely on the second factor.) Any trivialisation of this line bundle lets us interpret the restriction of to as a real matrix. In particular, given one of the bilinear forms from Remark 6, we can identify with by identifying any real function with the lift defined by

whenever . A somewhat tedious computation using the properties of then eventually gives the intertwining identity

and so is conjugate to .

** — 1. The Chung, Furedi, Graham, and Seymour example — **

The paper of by Chung, Furedi, Graham, and Seymour gives, for any , an example of a subset of of cardinality for which the maximum degree of restricted to is at most , thus showing that Theorem 1 cannot be improved (beyond the trivial improvement of upgrading to , because the maximum degree is obviously a natural number).

Define the “Möbius function” to be the function

for . This function is extremely balanced on coordinate spaces. Indeed, from the binomial theorem (which uses the convention ) we have

More generally, given any index set of cardinality , we have

Now let be a partition of into disjoint non-empty sets. For each , let be the subspace of consisting of those such that for all . Then for any , we have

and the right-hand side vanishes if and equals when . Applying the inclusion-exclusion principle, we conclude that

and thus also (assuming )

so that

Thus, if denotes the set of those with , together with those with , then has to have two more elements than its complement , and hence has cardinality .

Now observe that, if with and , then , and if then unless . Thus in this case the total number of for which is at most . Conversely, if with and , then , and for each there is at most one that will make lie in . Hence in this case the total number of for which is at most . Thus the maximum degree of the subgraph of induced by is at most . By choosing the to be a partition of into pieces, each of cardinality at most , we obtain the claim.

Remark 8Suppose that is a perfect square, then the lower bound here exactly matches the upper bound in Theorem 1. In particular, the minor of the matrix must have an eigenvector of eigenvalue . Such an eigenvector can be explicitly constructed as follows. Let be the vector defined by settingfor some , , , and for all other (one easily verifies that the previous types of lie in ). We claim that

for all . Expanding out the left-hand side, we wish to show that

First suppose that is of the form (3). One checks that lies in precisely when for one of the , in which case

Since , this simplifies using (3) as

giving (5) in this case. Similarly, if is of the form (4), then lies in precisely when , in which case one can argue as before to show that

and (5) again follows. Finally, if is not of either of the two forms (3), (4), one can check that is never of these forms either, and so both sides of (5) vanish.

The same analysis works for any of the other bilinear forms in Remark 6. Using the Clifford-valued operator from Remark 7, the eigenfunction is cleaner; it is defined by

when is of the form (3), and

when is of the form (4), with otherwise.

** — 2. From induced subgraph bounds to the sensitivity conjecture — **

On the hypercube , let denote the functions

The monomials in are then the characters of , so by Fourier expansion every function can be viewed as a polynomial in the (with each monomial containing at most one copy of ; higher powers of each are unnecessary since . In particular, one can meaningfully talk about the degree of a function . Observe also that the Möbius function from the preceding section is just the monomial .

Define the *sensitivity* of a Boolean function to be the largest number for which there is an such that there are at least values of with . Using an argument of Gotsman and Linial, we can now relate the sensitivity of a function to its degree:

Corollary 9 (Lower bound on sensitivity)For any boolean function , one has .

*Proof:* Write . By permuting the indices, we may assume that contains a non-trivial multiple of the monomial . By restricting to the subspace (which cannot increase the sensitivity), we may then assume without loss of generality that . The Fourier coefficient of is just the mean value

of times the Möbius function , so this mean value is non-zero. This means that one of the sets or has cardinality at least . Let denote the larger of these two sets. By Theorem 1, there is an such that for at least values of ; since , this implies that for at least values of , giving the claim.

The construction of Chung, Furedi, Graham, and Seymour from the previous section can be easily adapted to show that this lower bound is tight (other than the trivial improvement of replacing by ).

Now we need to digress on some bounds involving polynomials of one variable. We begin with an inequality of Bernstein concerning trigonometric polynomials:

Lemma 10 (Bernstein inequality)Let be a trigonometric polynomial of degree at most , that is to say a complex linear combination of for . Then

Observe that equality holds when or . Specialising to linear combinations of , we obtain the classical Bernstein inequality

for complex polynomials of degree at most .

*Proof:* If one is willing to lose a constant factor in this estimate, this bound can be easily established from modern Littlewood-Paley theory (see e.g., Exercise 52 of these lecture notes). Here we use an interlacing argument due to Boas. We first restrict to the case when has real coefficients. We may normalise . Let be a real parameter in . The trigonometric polynomial alternately takes the values and at the values . Thus the trigonometric polynomial alternates in sign at these values, and thus by the intermediate value theorem has a zero on each of the intervals . On the other hand, a trigonometric polynomial of degree at most can be expressed by de Moivre’s theorem as times a complex polynomial in of degree at most , and thus has at most zeroes. Thus we see that has exactly one zero in each . Furthermore, at this zero, the derivative of this function must be positive if is increasing on this interval, and negative if is decreasing on this interval. In summary, we have shown that if and are such that , then has the same sign as . By translating the function , we also conclude that if and are such that for some , then has the same sign as .

If , then we can find such that and is positive, and we conclude that ; thus we have the upper bound

A similar argument (with now chosen to be negative) similarly bounds . This gives the claim for real-valued trigonometric polynomials . (Indeed, this argument even gives the slightly sharper bound .)

To amplify this to complex valued polynomials, we take advantage of phase rotation invariance. If is a complex trigonometric polynomial, then by applying Bernstein’s inequality to the real part we have

But then we can multiply by any complex phase and conclude that

Taking suprema in , one obtains the claim for complex polynomials .

The analogue of Bernstein’s inequality for the unit interval is known as Markov’s inequality for polynomials:

Lemma 11 (Markov’s inequality for polynomials)Let be a polynomial of degree . Then

This bound is sharp, as is seen by inspecting the Chebyshev polynomial , defined as the unique polynomial giving the trigonometric identity

Differentiating (6) using the chain rule, we see that

the right-hand side approaches as , demonstrating that the factor here is sharp.

*Proof:* We again use an argument of Boas. We may normalise so that

The function is a trigonometric polynomial of degree at most , so by Bernstein’s inequality and the chain rule we have

for all . This already gives Markov’s inequality except in the edge regions (since ). By reflection symmetry, it then suffices to verify Markov’s inequality in the region .

From (6), the Chebyshev polynomial attains the values alternately at the different points . Thus, if , the polynomial changes sign at least times on , and thus must have all zeroes inside this interval by the intermediate value theorem; furthermore, of these zeroes will lie to the left of . By Rolle’s theorem, the derivative then has all zeroes in the interval , and at least of these will lie to the left of . In particular, the derivative can have at most one zero to once to the right of .

Since , is positive at , and hence positive as since there are no zeroes outside of . Thus the leading coefficient of is positive, which implies the same for its derivative . Thus is positive when .

From (9) one has , hence by (7) we see that is also positive at . Thus cannot become negative for , as this would create at least two zeroes to the right of . We conclude that in this region we have

From (7) we have , and the claim follows.

Remark 12The following slightly shorter argument gives the slightly weaker bound . We again use the normalisation (8). By two applications of Bernstein’s inequality, the function has first derivative bounded in magnitude by , and second derivative bounded in magnitude by . As this function also has vanishing first derivative at , we conclude the boundsand thus by the chain rule

For , one easily checks that the right-hand side is at most , giving the claim.

This implies a result of Ehlich-Zeller and of Rivlin-Cheney:

Corollary 13 (Discretised Markov inequality)Let be a polynomial of degree . If

*Proof:* We use an argument of Nisan and Szegedy. Assume for sake of contradiction that , so in particular . From the fundamental theorem of calculus and the triangle inequality one has

By a rescaled and translated version of Markov’s inequality we have

which when inserted into the preceding inequality gives after some rearranging

and then after a second application of (11) gives

Comparing with (10), we conclude that

and the claim follows after some rearranging.

Nisan and Szegedy observed that this one-dimensional degree bound can be lifted to the hypercube by a symmetrisation argument:

Corollary 14 (Multidimensional Markov inequality bound)Let be such that and for . Then

*Proof:* By averaging over all permutations of the indices (which can decrease the degree of , but not increase it), we may assume that is a symmetric function of the inputs . Using the Newton identities, we can then write

for some real polynomial of degree at most , where

is the Hamming length of . By hypothesis, , , and , hence by the mean-value theorem . Applying Corollary 13 with , we obtain the claim.

Define the *block sensitivity* of a Boolean function to be the largest number for which there is an such that there are at least disjoint subsets of with for . We have

More precisely, the sensitivity conjecture of Nisan and Szegedy asserted the bound ; Huang’s result thus gives explicit values for the exponents. It is still open whether the exponent in this theorem can be improved; it is known that one cannot improve it to below , by analysing a variant of the Chung-Furedi-Graham-Seymour example (see these notes of Regan for details). *Proof:* The lower bound for is immediate from the definitions, since the sensitivity arises by restricting the in the definition of block sensitivity to singleton sets. To prove the upper bound, it suffices from Proposition 9 to establish the bound

Let . By hypothesis, there are and disjoint subsets of such that for . We may normalise , and . If we then define the pullback boolean function by the formula

then it is easy to see that , , and for . The claim now follows from Corollary 14.

Remark 16The following slightly shorter variant of the argument lets one remove the factor . Let be as above. We again may normalise and . For any , let be iid Bernoulli random variable that equal with probability and with probability . The quantityis a trigonometric polynomial of degree at most that is bounded in magnitude by , so by two applications of Bernstein’s inequality

On the other hand, for small , the random variable is equal to zero with probability and equal to each with probability , hence

and hence . Combining these estimates we obtain and hence .

Remark 17The sensitivity of a Boolean function can be split as , where is largest number for which there is an such that there are at least values of with . It is not difficult to use the case of Remark 3 to improve Corollary 9 slightly to . Combining this with the previous remark, we can thus improve the upper bound in Theorem 15 slightly to

## 56 comments

Comments feed for this article

26 July, 2019 at 7:18 pm

YI HAOImpressive. Solving harder problems with lesser mathematics.

27 July, 2019 at 10:17 pm

AnonymousBravo! More pride for UCLA

29 July, 2019 at 2:57 pm

AnonymousYour “lesser mathematics” comment implies the use of simpler mathematical methods to prove conjectures or to solve problems. And I strongly agree with you…

Relevant Reference Link:

https://en.wikipedia.org/wiki/Hilbert%27s_twenty-fourth_problem

26 July, 2019 at 7:32 pm

Anonymous“…gave a of a long-standing problem…”

could be revised to (you forgot to add, “proof” above)

“…gave a proof of a long-standing problem” in your first statement above.

[Corrected, thanks – T.]26 July, 2019 at 8:31 pm

LThis is more illuminating than the original proof. Thank you.

29 July, 2019 at 5:51 pm

AnonymousPlease don’t forget to add the illuminating questions/comments/explanations… of Prof. T. Tao, Prof. Huang and others too. We are very fortunate! Thank you very much! :-)

27 July, 2019 at 1:07 am

Spectral proofs of theorems on the boolean hypercube | Anurag's Math Blog[…] discussions on Huang’s proof of the sensitivity conjecture, see the following blog posts: Terry Tao, Gil Kalai, Scott Aaronson, Ken […]

27 July, 2019 at 1:20 am

Anurag BishnoiInterlacing methods in general have been used before, see for example this 1995 survey of Haemers: https://www.sciencedirect.com/science/article/pii/0024379595001992, where it goes beyond the Cauchy interlacing inequalities.

Also, using a pseudo-adjacency matrices instead of just the adjacency matrix, was already used by Rick Wilson in his 1984 paper where he proved the general Erdős-Ko-Rado theorem:

https://link.springer.com/article/10.1007/BF02579226

27 July, 2019 at 2:01 am

Hao HuangTerry: thanks for this great expository article! I have been following your amazing blog for 10+ years (well before taking your 254 sequence at UCLA), and it is certainly a great honor to see my name mentioned here:)

Let me try to give an alternative explanation of where the pseudo-adjacency matrix comes from, from a more combinatorial perspective. Recall that for the proof to work, the pseudo-adjacency matrix needs to have entries between -1 and 1, and its 2^{n-1}-th largest eigenvalue must be \sqrt{n}. Hadamard’s inequality tells us that the determinant of such matrix is at most the product of the lengths of its row vectors, which is upper bounded by \sqrt{n}^{2^{n-1}}. On the other hand, the constraint on the 2^{n-1}-th largest eigenvalue requires that all eigenvalues have absolute value at least \sqrt{n}, since the hypercube is bipartite and its spectrum is symmetric about 0. Therefore the determinant, which is the product of eigenvalues, is at least \sqrt{n}^{2^{n-1}}.

Now for the Hadamard inequality to be tight, the rows must be pairwise orthogonal. By decomposing the matrix into 2 by 2 blocks (each corresponding to a (n-1)-dimensional subcube), it is not hard to see that each (n-1)-cube also needs to have a good signing. And for example if we fix the signing of the top (n-1)-cube, and an arbitrary signing of the vertical perfect matching, then the signing of the bottom cube is fixed. When choosing all vertical edges to be positive, this is exactly the recursive construction in the paper.

I am quite interested to see that to what extent the same signing exists for other very symmetric graphs. Moreover, it seems that certain signing “compresses” the positive (negative) spectrum of the adjacency matrix, maybe this recursively constructed matrix is no more than composing a number of “compressing signing”?

29 July, 2019 at 11:41 am

Gil KalaiDear Hao, what do you mean by the last sentence?

29 July, 2019 at 4:59 pm

Terence TaoI can take a stab at this. The unsigned adjacency matrix has a rather “spread out” spectrum; the eigenvalues are all the signed sums of signs. These eigenvalues mostly cluster within a distance from the origin but the spectral radius is huge (equal to ). With the right signing of the matrix, however, the spectrum is “compressed” down to just and , which is what is needed for this application – very roughly speaking, the signs have somehow become “orthogonal” to each other (perhaps in some Clifford algebra sense) and hence their sums now always have “magnitude ” (by some sort of “Pythagorean theorem”). (Any symmetric signing of the matrix will preserve the first two moments, so this is the best compression of the spectral radius that one can hope for.) The compression is achieved by an iterative method in which the spectrally compressed matrix is constructed as a sort of skew tensor product of the matrix and an explicit matrix. So the proof can be ultimately boiled down to the properties of explicit matrices (another comment on this blog post reached a similar conclusion, where the matrices were Pauli matrices). This reminds me a bit of how hypercontractivity type inequalities can often be proven by iterating an elementary inequality involving a small number real variables (as few as two, in some cases), though in those cases the iteration is fairly straightforward and does not involve any “skewness”.

29 July, 2019 at 5:29 pm

Terence TaoOne can also interpret everything in terms of eigenvalues of sums of Hermitian matrices, a topic I am personally rather fond of :). If are Hermitian matrices that each have eigenvalues in , then the sum can have eigenvalues anywhere in the interval between and (though if one requires the to commute, then the eigenvalues have to be integers of the same parity as ). However, if one instead requires that the all

anticommute, then the spectrum of the sum has now been compressed into the set , because this matrix now squares to times the identity matrix (one can interpret the anticommutator identity as a sort of “orthogonality” between and ). This is the matrix analogue of the vector fact that the sum of unit vectors can have magnitude anywhere between and in general, but is forced to have magnitude if all the unit vectors are orthogonal to each other. So the game is all about weighting the graph to make the different edges anticommute with each other, which is what one sees in all the different interpretations of the argument appearing in this blog.29 July, 2019 at 9:48 pm

AnonymousIf this is the matrix analogue for orthogonality, is it follows from a more general underlying matrix analogue for inner product?

30 July, 2019 at 9:23 am

Terence TaoI think the role of the inner product here is played by the anticommutator of two Hermitian matrices : it is symmetric, bilinear, and is “positive definite” in the sense that is positive definite when is non-zero. The one catch is that the anticommutator is matrix-valued rather than scalar valued. But it is related to the more usual scalar (Frobenius) inner product by the identity . This analogy helps explain why anticommutation seems to be a matrix analogue of orthogonality, which is a nice insight that I was not consciously aware of until looking at this recent work.

Another relationship between the anticommutator and the Frobenius inner product is the cyclic symmetry (and similarly for other permutations, since both the anticommutator and Frobenius inner product are symmetric). In particular, if anticommute, then not only are they orthogonal to each other in the Frobenius sense, but also is Frobenius orthogonal to (and is similarly Frobenius orthogonal to ) for any matrix . So this is another way to connect anticommutation with orthogonality.

The analogy between anticommutators and inner products seems to break down a bit once one goes beyond the Hermitian setting though; one either has to abandon positive definiteness, or abandon (complex) bilinearity. But for real non-symmetric matrices one might possibly be able to work with as a substitute for the anticommutator.

1 August, 2019 at 12:35 am

TWhat operators preserve this inner product? Is there an analog of orthogonality here?

1 August, 2019 at 9:45 am

Terence TaoThe anticommutator on Hermitian matrices is equivariant with respect to unitary conjugation: . (Given that the anticommutator is matrix-valued rather than scalar-valued, equivariance is arguably a more natural notion of symmetry here than invariance.) I suspect these are basically the only (invertible) linear operations on Hermitian matrices that make the anticommutator equivariant.

1 August, 2019 at 5:46 pm

TCoordinatization of some space?

29 July, 2019 at 3:00 pm

Ferdinand IhringerToday this paper on the arXiv appeared [1] (by my old colleagues of the Discrete Math Group in Regina) which reminded me that there is a research area devoted towards the study of such weighted adjacency matrices. The are called multiplicity partitions and in this particular case it is an example for a multiplicity partition of the form [ 2^{n-1}, 2^{n-1} ] of the hypercube.

Now for the fun of it I was looking at some older results on this. It turns out that your matrix was already constructed in [2], Theorem 6.7 and Corollary 6.9, as an example for a graph with only two different eigenvalues. (The scaling is different to your matrix to keep the eigenvalues restricted to -1 and 1, but I guess that one can ignore this fact.)

[1] https://arxiv.org/abs/1907.11328

[2] https://arxiv.org/abs/1304.1205

29 July, 2019 at 3:22 pm

Terence TaoScott Aaronson noted in his own blog post on this result that this argument was a great example of the empirical phenomenon in mathematical proofs, and now I think the case is even stronger; the result can now be deduced in a paragraph from existing results in the literature, using a method that was also already in the literature, but the task of

findingthat short proofab nihilowas highly nontrivial!Another example of this is Dvir’s proof of the finite field Kakeya conjecture. I’m still kicking myself for coming up with half of that proof six years earlier with a coauthor (see Proposition 8.1 of this paper) and completely failing to see that the conjecture could then be resolved with an additional paragraph of effort [amusingly, in both cases the “additional effort” consists of “a tiny bit of undergraduate linear algebra”]. Of course, this doesn’t detract at all from Dvir’s accomplishment, instead it is another example of the empirical phenomenon. Perhaps when attributing credit to a result, one has to make a distinction between the “NP-credit” (the credit one would assign based on the optimised argument from the literature that one can construct

afterseeing a proof) for a result and the “P-credit” (the credit one would assign to each mathematician for their contribution in advancing the explicit state of knowledge on the problem [as opposed to what can be implicitly extrapolated from the total corpus of literature with the benefit of hindsight] closer to a solution (or pushing beyond a mere solution of the problem and gaining further insight and understanding around the problem and its connection with the rest of mathematics)).29 July, 2019 at 3:56 pm

Bizarre.This is a bizarre conclusion. Is the analogy ‘the problem is encodable somehow with a boolean formula with n variables with say n=100 and we are are finding some version of satisfiability’ and if so did it take 2^(c.100) time?

29 July, 2019 at 4:00 pm

Bizarre.Then the evidence is towards P=NP since apparently it did not take that long. He is probably the only a handful who persisted on this graph theoretic approach and that too for 5 years (and/but that too only for 5 years).

29 July, 2019 at 4:17 pm

AnonymousI think the analogy is that it often appears more difficult to generate a correct solution than to check or justify or reformulate the solution with the benefit of hindsight.

29 July, 2019 at 4:40 pm

Terence TaoPretty much, yes (though I would replace the exponential bound here with something more empirically realistic, such as a superpolynomial or even superquadratic bound). The proof already has been encoded into a single tweet, which is very roughly of the same order of difficulty as satisfying a boolean formula with variables, and can now be verified using results in the literature in less than an hour; yet, as you note, it took at least 5 years (=43800 hours) to locate this proof (actually a more accurate metric here would be the total amount of expert person-hours expended on the problem, which is more difficult to compute but similarly substantial, as Scott pointed out in his blog post).

29 July, 2019 at 6:07 pm

Bizarre.Don’t know about expert man hours. None of the experts suggested graph theory approach even though the strategy was pointed by Nisan in Scott`s MO problem years back. More realistic might be an year of intense effort (about 5000 hours) and also checkable by whom is 1/2 half (again by experts). If at all the moral seems to be be stay convinced in your belief on the conjecture (whether conjecture was correct) and be lucky in your toolkit.

29 July, 2019 at 9:22 pm

Alex JonesI think this P vs NP distinction strengthens Timothy Gowers’s claim that computers can have a meaningful impact on generating new mathematical proofs. It seems very reasonable (to me) that a smart computer can get that ‘NP-credit’ that Terry commented about above.

30 July, 2019 at 12:38 pm

Bizarre.FLT took 350 years. Assume 10 mathematicians each worked intensely 4000 hours everyday for 400 years then we have 16000000 person hours. 16000000/50 is roughly 300000. Would you agree 300000 bits is a reasonable measure to encode all of algebraic number theory, scheme theory, arithmetic geometry necessary to encode the proof of FLT notwithstanding the amount of essential information that was written down on these topics in 400 years? Just asking because it is tempting to understand the complexity of mathematics because it is one evidence shown in favor of P not NP inspite of the fact SAT has no evidence of being outside linear time as far as we know.

29 July, 2019 at 4:24 pm

Ferdinand IhringerRecently, Hajime Tanaka and I slightly modified a proof by Frankl from the 1980s on the independence number of the orthogonality graph [1] (the vertices are the hypercube and adjacency is Hamming distance n/2). It is all about writing down the right matrix. Frankl took an argument by Delsarte and showed it for , p an odd prime. Hajime and I added a divided by 2 into Frankl’s argument to cover p even.

I am sure that the complete solution will also be easy. Just that someone needs to find the right matrix and apply the right variation of one of the known rank arguments.

[1] https://arxiv.org/abs/1901.04860

30 July, 2019 at 12:29 am

AnonymousA reasonable “standard” measure for the complexity of a mathematical problem may be defined as a the smallest complexity (with respect to some fixed “standard” complexity measure) over all its possible proofs.

27 July, 2019 at 4:46 am

Huangs’ Breakthrough, Cvetković’s Bound, Godsil’s Question, and Sinkovic’s Answer | Ratio Bound – A Combinatorics Blog[…] ” and everyone is blogging about it (only a small selection): Anurag Bishnoi, Gil Kalai, Terry Tao, Fedya Petrov. Here I am jumping on this […]

27 July, 2019 at 11:00 am

Amazing: Hao Huang Proved the Sensitivity Conjecture! | Combinatorics and more[…] and many other share: Huang’s new method is very very promising. July 27: A great expository post on WN (What’s new) where Terry Tao presents the proof and describe a connection with Clifford […]

27 July, 2019 at 7:59 pm

kodluVery interesting… There are three formulae which give the “formula does not parse” error, and I have not been able to tell exactly why, since they look the same as other formulae which parse OK. I’ve tried both safari and chrome on iOS which give the same errors:

Section 1:

Then for any {J \subset \{1,\dots,k\}}, we have [the formula below doesn’t parse, maybe a missing }?]

\displaystyle \sum_{x \in \bigcap_{j \in J} V_j} \mu(x) = (-1)^{1_{n= \sum_{j \in J}|I_i|},

Remark 6:

Let {f \in \ell^2(E)} be the vector defined by setting [the formula below doesn’t parse, maybe a missing }?]

\displaystyle f(x) := (-1)^{\sum_{1 \leq j < j' \leq k} B( e_{i_j}, e_{i_{j'}} )

and further on

Since {B(e_{i_{j_0}}, e_{i_j}) = B(e_{i_j}, e_{i_{j_0}) + 1}, this simplifies using (3) as

The above doesn't parse either.

[Corrected, thanks – T.]27 July, 2019 at 8:41 pm

SIt is a pity that Oleksiy Klurman and Cosmin Pohoata are not receiving more credit since the method in Huang’s paper appeared first in a joint paper with them, https://arxiv.org/abs/1812.05989

27 July, 2019 at 10:42 pm

TAs Anurag mentioned in his comment, the use of pseudo-adjacency matrix is not new at all. The difficult step is to realize that it can be applied to this specific cube problem. You could also say that it is a pity that Wilson is not receiving more credit.

28 July, 2019 at 2:12 am

LIn this case the concerned are coauthors however the related paper does not appear as a reference in the celebrated proof paper. May be the author of the proof had a different path to realization.

29 July, 2019 at 2:34 pm

Ferdinand IhringerOr Cvetkovic (if we want to find earliest written precedent). His PhD from 1971 already had it. There are other theses from the 1970s with similar observations (maybe Lovasz’s, Delsarte’s and Haemer’s are the ones which I know). These methods are also often rediscovered as they are not too complicated linear algebra.

27 July, 2019 at 10:54 pm

SSure, but Wilson is retired whereas Oleksiy Klurman and Cosmin Pohoata are young mathematicians who are Huang’s coauthors.

28 July, 2019 at 8:44 am

Anonymousgrind up old mathematicians to feed the young

27 July, 2019 at 11:09 pm

La conjetura de la sensibilidad ha sido demostrada - La Ciencia de la Mula Francis[…] Los interesados en los detalles de la demostración, que es fácil de entender si ya se conocía su enunciado, así como los conceptos de subgrafo inducido y autovalor, pueden disfrutarla en Hao Huang, «Induced subgraphs of hypercubes and a proof of the Sensitivity Conjecture,» arXiv:1907.00847 [math.CO] (01 Jul 2019). Una presentación divulgativa para matemáticos en Scott Aaronson, «Sensitivity Conjecture resolved,» Shtetl-Optimized, 02 Jul 2019; sobre la relación entre el grado y la sensibilidad recomiendo Lance Fortnow, «Degree and Sensitivity,» Computational Complexity, 11 Jul 2019; las consecuencias del nuevo teorema no han tardado en llegar, como R. J. Lipton, K. W. Regan, «Discrepancy Games and Sensitivity,» Gödel’s Lost Letter and P=NP, 25 Jul 2019; y Terence TaoTwisted convolution and the sensitivity conjecture,» What’s new, 26 Jul 2019. […]

28 July, 2019 at 6:26 am

ghI wonder if there is some “homology” involved. Namely, for the standard adjacency matrix of the hypercube graph, counts the paths of length 2 from vertex to vertex , and hence because each vertex has n neighbors and because for there are two paths (a square just as in the case (0,0) to (1,1).

Now the pseudo-adjacency matrix considered in the proof similarly has (paths of length 2 returning to each contributing ) and ) because the contribution of the two paths from to now cancel each other out. So “the square spanned by and is filled in”.

28 July, 2019 at 9:42 am

Terence TaoI’ve added a new remark (Remark 6) in which the proof is reformulated without reference to a bilinear form or any arbitrary choice of signs, and the controlling matrix is now canonically defined using the Clifford algebra generated by with the relations , which abstract the two properties that make the matrix square to . (Basically, each edge is now weighted by a Clifford generator , rather than by a sign .) This is in some sense the “universal” version of Huang’s argument. Perhaps these Clifford generators also have some homological interpretation.

28 July, 2019 at 7:49 am

AnonymousThis short proof seem to be “from the Book” !

28 July, 2019 at 9:12 am

MichaelShortly before Remark 6, there is ‘Conversely, if x not in bigcap…’ should be bigcup

Near the end of the proof of Corollary 10, there is eqreef{disco} should be eqref{disco} ; then in the displayed equation, ‘frac’ is spelled out; then a paragraph with just the letter b.

[Corrected, thanks – T.]28 July, 2019 at 8:00 pm

LMost studies on spectral graph theory seem to be related to geometry (such as embedding type) and looking at Clifford algebra and related avatars there seems to be more essential reasoning in terms of purely geometric terms. Is there a way to look at things purely in geometric terms and interpret the result geometrically?

28 July, 2019 at 10:10 pm

Piotr SniadyThe explanation of Roman Karasev of the structure of the matrix can be rephrased in a slightly more concrete language as follows.

It will be convenient to identify the square matrices of size with the tensor power of square matrices of size .

Let and be Pauli matrices and let be the identity matrix.

Now let

,

,

,

…

;

( contains exactly one factor on -th position, all factors before it are equal to and all factors after it are equal to the identity matrix ).

Note that the matrices described above appear naturally in the context of quantum mechanics for the following reason: since Pauli matrices anticommute: , it follows that the matrices anticommute as well: whenever . Also, is the identity matrix.

Luckily, we have that is a (signed) adjacency matrix for the graph we are interested in.

From the anticommutation of the matrices it follows immediately that

.

29 July, 2019 at 7:51 am

FiodrI think and should be switched in the definition of .

29 July, 2019 at 8:15 am

Piotr SniadyOops, you are right, thanks!

[I have now swapped and in the original comment – T.]30 July, 2019 at 6:51 am

MikeRThis exposition by Wenjie Zheng is crystal clear. Not written at the same level, but definitely comprehensible.

http://www.zhengwenjie.net/huanghao/

30 July, 2019 at 7:20 am

AnonymousSimplicity is clear, wonderful, and essential! Thank you!

30 July, 2019 at 10:08 am

Karl MahlburgTerry,

Thanks for an extremely informative post! Huang’s proof is extraordinarily clean and concise, but I also appreciate all of the background material that you have provided.

I believe that I noticed a few minor typos, which are listed below.

1. The fourth displayed equation in Section 1 should be

This follows from applying the third displayed equation.

2. Near the end of the proof of Lemma 11 is the statement: “From (9) one has .” However, I believe this should instead be

The point is then that by (7)

so in all

3. In the proof of Theorem 15, the sentence before the first displayed equation says “it suffices from (9)”. However, the link actually points to Corollary 9, which I believe is the correct reference.

4. Towards the end of Remark 16, there is the statement “and equal to each with probability “. I think this should instead be “and equal to each “.

[Corrected, thanks – T.]1 August, 2019 at 4:26 am

chorasimilarityThe relation with the Heisenberg group suggests there might be a reformulation in terms of symplectic capacities?

21 August, 2019 at 4:19 pm

Zachary Kyle KnutsonDear Mathematicians,

This goes to show that even Terrence Tao could be wrong, Hao Huang is a fraud and he copied my work I posted on research gate he is also concurrently wrong… in part, his answer is implied as wrong… Due to his misuse of the mechanisms involved. I would write it up if it didn’t make my stomach turn that somebody would just openly steal my shit or claim money over its false conception…It is not 2^(N-1)+1 it is 2^(n-1)-1 the minus is key and Terrance is probably the only other person on the earth beside the few of you that understands why…

22 August, 2019 at 11:38 pm

bizarre.Highest upvoted comment truly bizarre.

24 August, 2019 at 11:23 pm

bizarre.Self-voting perhaps.

24 August, 2019 at 6:19 am

Zachary Kyle KnutsonThanks to you all lol my faith in humanity is restored…

24 August, 2019 at 6:26 am

Zachary Kyle Knutson#ResearchGate me…

24 August, 2019 at 6:27 am

Zachary Kyle KnutsonBecause People have copied me in physics and engineering aswell…