[This blog post was written jointly by Terry Tao and Will Sawin.]
In the previous blog post, one of us (Terry) implicitly introduced a notion of rank for tensors which is a little different from the usual notion of tensor rank, and which (following BCCGNSU) we will call “slice rank”. This notion of rank could then be used to encode the Croot-Lev-Pach-Ellenberg-Gijswijt argument that uses the polynomial method to control capsets.
Afterwards, several papers have applied the slice rank method to further problems – to control tri-colored sum-free sets in abelian groups (BCCGNSU, KSS) and from there to the triangle removal lemma in vector spaces over finite fields (FL), to control sunflowers (NS), and to bound progression-free sets in -groups (P).
In this post we investigate the notion of slice rank more systematically. In particular, we show how to give lower bounds for the slice rank. In many cases, we can show that the upper bounds on slice rank given in the aforementioned papers are sharp to within a subexponential factor. This still leaves open the possibility of getting a better bound for the original combinatorial problem using the slice rank of some other tensor, but for very long arithmetic progressions (at least eight terms), we show that the slice rank method cannot improve over the trivial bound using any tensor.
It will be convenient to work in a “basis independent” formalism, namely working in the category of abstract finite-dimensional vector spaces over a fixed field . (In the applications to the capset problem one takes to be the finite field of three elements, but most of the discussion here applies to arbitrary fields.) Given such vector spaces , we can form the tensor product , generated by the tensor products with for , subject to the constraint that the tensor product operation is multilinear. For each , we have the smaller tensor products , as well as the tensor product
defined in the obvious fashion. Elements of of the form for some and will be called rank one functions, and the slice rank (or rank for short) of an element of is defined to be the least nonnegative integer such that is a linear combination of rank one functions. If are finite-dimensional, then the rank is always well defined as a non-negative integer (in fact it cannot exceed . It is also clearly subadditive:
For , is when is zero, and otherwise. For , is the usual rank of the -tensor (which can for instance be identified with a linear map from to the dual space ). The usual notion of tensor rank for higher order tensors uses complete tensor products , as the rank one objects, rather than , giving a rank that is greater than or equal to the slice rank studied here.
From basic linear algebra we have the following equivalences:
Lemma 1 Let be finite-dimensional vector spaces over a field , let be an element of , and let be a non-negative integer. Then the following are equivalent:
Proof: The equivalence of (i) and (ii) is clear from definition. To get from (ii) to (iii) one simply takes to be the span of the , and conversely to get from (iii) to (ii) one takes the to be a basis of the and computes by using a basis for the tensor product consisting entirely of functions of the form for various . To pass from (iii) to (iv) one takes to be the annihilator of , and conversely to pass from (iv) to (iii).
One corollary of the formulation (iv), is that the set of tensors of slice rank at most is Zariski closed (if the field is algebraically closed), and so the slice rank itself is a lower semi-continuous function. This is in contrast to the usual tensor rank, which is not necessarily semicontinuous.
Corollary 2 Let be finite-dimensional vector spaces over an algebraically closed field . Let be a nonnegative integer. The set of elements of of slice rank at most is closed in the Zariski topology.
Proof: In view of Lemma 1(i and iv), this set is the union over tuples of integers with of the projection from of the set of tuples with orthogonal to , where is the Grassmanian parameterizing -dimensional subspaces of .
One can check directly that the set of tuples with orthogonal to is Zariski closed in using a set of equations of the form locally on . Hence because the Grassmanian is a complete variety, the projection of this set to is also Zariski closed. So the finite union over tuples of these projections is also Zariski closed.
We also have good behaviour with respect to linear transformations:
Lemma 3 Let be finite-dimensional vector spaces over a field , let be an element of , and for each , let be a linear transformation, with the tensor product of these maps. Then
Furthermore, if the are all injective, then one has equality in (2).
Thus, for instance, the rank of a tensor is intrinsic in the sense that it is unaffected by any enlargements of the spaces .
Proof: The bound (2) is clear from the formulation (ii) of rank in Lemma 1. For equality, apply (2) to the injective , as well as to some arbitrarily chosen left inverses of the .
Computing the rank of a tensor is difficult in general; however, the problem becomes a combinatorial one if one has a suitably sparse representation of that tensor in some basis, where we will measure sparsity by the property of being an antichain.
Proposition 4 Let be finite-dimensional vector spaces over a field . For each , let be a linearly independent set in indexed by some finite set . Let be a subset of .
Let be a tensor of the form
where for each , is a coefficient in . Then one has
where the minimum ranges over all coverings of by sets , and for are the projection maps.
Now suppose that the coefficients are all non-zero, that each of the are equipped with a total ordering , and is the set of maximal elements of , thus there do not exist distinct , such that for all . Then one has
In particular, if is an antichain (i.e. every element is maximal), then equality holds in (4).
Proof: By Lemma 3 (or by enlarging the bases ), we may assume without loss of generality that each of the is spanned by the . By relabeling, we can also assume that each is of the form
with the usual ordering, and by Lemma 3 we may take each to be , with the standard basis.
Let denote the rank of . To show (4), it suffices to show the inequality
for any covering of by . By removing repeated elements we may assume that the are disjoint. For each , the tensor
can (after collecting terms) be written as
for some . Summing and using (1), we conclude the inequality (6).
Now assume that the are all non-zero and that is the set of maximal elements of . To conclude the proposition, it suffices to show that the reverse inequality
holds for some covering . By Lemma 1(iv), there exist subspaces of whose dimension sums to
such that is orthogonal to .
Let . Using Gaussian elimination, one can find a basis of whose representation in the standard dual basis of is in row-echelon form. That is to say, there exist natural numbers
such that for all , is a linear combination of the dual vectors , with the coefficient equal to one.
We now claim that is disjoint from . Suppose for contradiction that this were not the case, thus there exists for each such that
As is the set of maximal elements of , this implies that
for any tuple other than . On the other hand, we know that is a linear combination of , with the coefficient one. We conclude that the tensor product is equal to
plus a linear combination of other tensor products with not in . Taking inner products with (3), we conclude that , contradicting the fact that is orthogonal to . Thus we have disjoint from .
For each , let denote the set of tuples in with not of the form . From the previous discussion we see that the cover , and we clearly have , and hence from (8) we have (7) as claimed.
As an instance of this proposition, we recover the computation of diagonal rank from the previous blog post:
Example 5 Let be finite-dimensional vector spaces over a field for some . Let be a natural number, and for , let be a linearly independent set in . Let be non-zero coefficients in . Then
has rank . Indeed, one applies the proposition with all equal to , with the diagonal in ; this is an antichain if we give one of the the standard ordering, and another of the the opposite ordering (and ordering the remaining arbitrarily). In this case, the are all bijective, and so it is clear that the minimum in (4) is simply .
The combinatorial minimisation problem in the above proposition can be solved asymptotically when working with tensor powers, using the notion of the Shannon entropy of a discrete random variable .
Proposition 6 Let be finite-dimensional vector spaces over a field . For each , let be a linearly independent set in indexed by some finite set . Let be a non-empty subset of .
Let be a tensor of the form (3) for some coefficients . For each natural number , let be the tensor power of copies of , viewed as an element of . Then
as , where is the quantity
and range over the random variables taking values in .
Now suppose that the coefficients are all non-zero and that each of the are equipped with a total ordering . Let be the set of maximal elements of in the product ordering, and let where range over random variables taking values in . Then
as . In particular, if the maximizer in (10) is supported on the maximal elements of (which always holds if is an antichain in the product ordering), then equality holds in (9).
It will suffice to show that
as , where is the projection map. Then the same thing will apply to and . Then applying Proposition 4, using the lexicographical ordering on and noting that, if are the maximal elements of , then are the maximal elements of , we obtain both (9) and (11).
We first prove the lower bound. By compactness (and the continuity properties of entropy), we can find a random variable taking values in such that
Let be a small positive quantity that goes to zero sufficiently slowly with . Let denote the set of all tuples in that are within of being distributed according to the law of , in the sense that for all , one has
By the asymptotic equipartition property, the cardinality of can be computed to be
if goes to zero slowly enough. Similarly one has
and for each , one has
Now let be an arbitrary covering of . By the pigeonhole principle, there exists such that
and hence by (14), (15)
which by (13) implies that
noting that the factor can be absorbed into the error). This gives the lower bound in (12).
Now we prove the upper bound. We can cover by sets of the form for various choices of random variables taking values in . For each such random variable , we can find such that ; we then place all of in . It is then clear that the cover and that
for all , giving the required upper bound.
It is of interest to compute the quantity in (10). We have the following criterion for when a maximiser occurs:
Proposition 7 Let be finite sets, and be non-empty. Let be the quantity in (10). Let be a random variable taking values in , and let denote the essential range of , that is to say the set of tuples such that is non-zero. Then the following are equivalent:
Furthermore, when (i) and (ii) holds, one has
Proof: We first show that (i) implies (ii). The function is concave on . As a consequence, if we define to be the set of tuples such that there exists a random variable taking values in with , then is convex. On the other hand, by (10), is disjoint from the orthant . Thus, by the hyperplane separation theorem, we conclude that there exists a half-space
where are reals that are not all zero, and is another real, which contains on its boundary and in its interior, such that avoids the interior of the half-space. Since is also on the boundary of , we see that the are non-negative, and that whenever .
By construction, the quantity
is maximised when . At this point we could use the method of Lagrange multipliers to obtain the required constraints, but because we have some boundary conditions on the (namely, that the probability that they attain a given element of has to be non-negative) we will work things out by hand. Let be an element of , and an element of . For small enough, we can form a random variable taking values in , whose probability distribution is the same as that for except that the probability of attaining is increased by , and the probability of attaining is decreased by . If there is any for which and , then one can check that
for sufficiently small , contradicting the maximality of ; thus we have whenever . Taylor expansion then gives
for small , where
and similarly for . We conclude that for all and , thus there exists a quantity such that for all , and for all . By construction must be nonnegative. Sampling using the distribution of , one has
almost surely; taking expectations we conclude that
The inner sum is , which equals when is non-zero, giving (17).
Now we show conversely that (ii) implies (i). As noted previously, the function is concave on , with derivative . This gives the inequality
for any (note the right-hand side may be infinite when and ). Let be any random variable taking values in , then on applying the above inequality with and , multiplying by , and summing over and gives
By construction, one has
so to prove that (which would give (i)), it suffices to show that
or equivalently that the quantity
is maximised when . Since
it suffices to show this claim for the quantity
One can view this quantity as
By (ii), this quantity is bounded by , with equality if is equal to (and is in particular ranging in ), giving the claim.
The second half of the proof of Proposition 7 only uses the marginal distributions and the equation(16), not the actual distribution of , so it can also be used to prove an upper bound on when the exact maximizing distribution is not known, given suitable probability distributions in each variable. The logarithm of the probability distribution here plays the role that the weight functions do in BCCGNSU.
Remark 8 Suppose one is in the situation of (i) and (ii) above; assume the nondegeneracy condition that is positive (or equivalently that is positive). We can assign a “degree” to each element by the formula
then every tuple in has total degree at most , and those tuples in have degree exactly . In particular, every tuple in has degree at most , and hence by (17), each such tuple has a -component of degree less than or equal to for some with . On the other hand, we can compute from (19) and the fact that for that . Thus, by asymptotic equipartition, and assuming , the number of “monomials” in of total degree at most is at most ; one can in fact use (19) and (18) to show that this is in fact an equality. This gives a direct way to cover by sets with , which is in the spirit of the Croot-Lev-Pach-Ellenberg-Gijswijt arguments from the previous post.
We can now show that the rank computation for the capset problem is sharp:
Proposition 9 Let denote the space of functions from to . Then the function from to , viewed as an element of , has rank as , where is given by the formula
Proof: In , we have
Thus, if we let be the space of functions from to (with domain variable denoted respectively), and define the basis functions
of indexed by (with the usual ordering), respectively, and set to be the set
then is a linear combination of the with , and all coefficients non-zero. Then we have . We will show that the quantity of (10) agrees with the quantity of (20), and that the optimizing distribution is supported on , so that by Proposition 6 the rank of is .
To compute the quantity at (10), we use the criterion in Proposition 7. We take to be the random variable taking values in that attains each of the values with a probability of , and each of with a probability of ; then each of the attains the values of with probabilities respectively, so in particular is equal to the quantity in (20). If we now set and
we can verify the condition (16) with equality for all , which from (17) gives as desired.
This statement already follows from the result of Kleinberg-Sawin-Speyer, which gives a “tri-colored sum-free set” in of size , as the slice rank of this tensor is an upper bound for the size of a tri-colored sum-free set. If one were to go over the proofs more carefully to evaluate the subexponential factors, this argument would give a stronger lower bound than KSS, as it does not deal with the substantial loss that comes from Behrend’s construction. However, because it actually constructs a set, the KSS result rules out more possible approaches to give an exponential improvement of the upper bound for capsets. The lower bound on slice rank shows that the bound cannot be improved using only the slice rank of this particular tensor, whereas KSS shows that the bound cannot be improved using any method that does not take advantage of the “single-colored” nature of the problem.
We can also show that the slice rank upper bound in a result of Naslund-Sawin is similarly sharp:
Proposition 10 Let denote the space of functions from to . Then the function from , viewed as an element of , has slice rank
Proof: Let and be a basis for the space of functions on , itself indexed by . Choose similar bases for and , with and .
Set . Then is a linear combination of the with , and all coefficients non-zero. Order the usual way so that is an antichain. We will show that the quantity of (10) is , so that applying the last statement of Proposition 6, we conclude that the rank of is ,
Let be the random variable taking values in that attains each of the values with a probability of . Then each of the attains the value with probability and with probability , so
Setting and , we can verify the condition (16) with equality for all , which from (17) gives as desired.
We used a slightly different method in each of the last two results. In the first one, we use the most natural bases for all three vector spaces, and distinguish from its set of maximal elements . In the second one we modify one basis element slightly, with instead of the more obvious choice , which allows us to work with instead of . Because is an antichain, we do not need to distinguish and . Both methods in fact work with either problem, and they are both about equally difficult, but we include both as either might turn out to be substantially more convenient in future work.
Proposition 11 Let be a natural number and let be a finite abelian group. Let be any field. Let denote the space of functions from to .
Let be any -valued function on that is nonzero only when the elements of form a -term arithmetic progression, and is nonzero on every -term constant progression.
Then the slice rank of is .
Proof: We apply Proposition 4, using the standard bases of . Let be the support of . Suppose that we have orderings on such that the constant progressions are maximal elements of and thus all constant progressions lie in . Then for any partition of , can contain at most constant progressions, and as all constant progressions must lie in one of the , we must have . By Proposition 4, this implies that the slice rank of is at least . Since is a tensor, the slice rank is at most , hence exactly .
So it is sufficient to find orderings on such that the constant progressions are maximal element of . We make several simplifying reductions: We may as well assume that consists of all the -term arithmetic progressions, because if the constant progressions are maximal among the set of all progressions then they are maximal among its subset . So we are looking for an ordering in which the constant progressions are maximal among all -term arithmetic progressions. We may as well assume that is cyclic, because if for each cyclic group we have an ordering where constant progressions are maximal, on an arbitrary finite abelian group the lexicographic product of these orderings is an ordering for which the constant progressions are maximal. We may assume , as if we have an -tuple of orderings where constant progressions are maximal, we may add arbitrary orderings and the constant progressions will remain maximal.
So it is sufficient to find orderings on the cyclic group such that the constant progressions are maximal elements of the set of -term progressions in in the -fold product ordering. To do that, let the first, second, third, and fifth orderings be the usual order on and let the fourth, sixth, seventh, and eighth orderings be the reverse of the usual order on .
Then let be a constant progression and for contradiction assume that is a progression greater than in this ordering. We may assume that , because otherwise we may reverse the order of the progression, which has the effect of reversing all eight orderings, and then apply the transformation , which again reverses the eight orderings, bringing us back to the original problem but with .
Take a representative of the residue class in the interval . We will abuse notation and call this . Observe that , and are all contained in the interval modulo . Take a representative of the residue class in the interval . Then is in the interval for some . The distance between any distinct pair of intervals of this type is greater than , but the distance between and is at most , so is in the interval . By the same reasoning, is in the interval . Therefore . But then the distance between and is at most , so by the same reasoning is in the interval . Because is between and , it also lies in the interval . Because is in the interval , and by assumption it is congruent mod to a number in the set greater than or equal to , it must be exactly . Then, remembering that and lie in , we have and , so , hence , thus , which contradicts the assumption that .
In fact, given a -term progressions mod and a constant, we can form a -term binary sequence with a for each step of the progression that is greater than the constant and a for each step that is less. Because a rotation map, viewed as a dynamical system, has zero topological entropy, the number of -term binary sequences that appear grows subexponentially in . Hence there must be, for large enough , at least one sequence that does not appear. In this proof we exploit a sequence that does not appear for .