Quasihomomorphisms from the integers into Hamming metrics

A function $f: \mathbb{Z} \to \mathbb{Q}^n$ is a $c$-quasihomomorphism if the Hamming distance between $f(x+y)$ and $f(x)+f(y)$ is at most $c$ for all $x,y \in \mathbb{Z}$. We show that any $c$-quasihomomorphism has distance at most some constant $C(c)$ to an actual group homomorphism; here $C(c)$ depends only on $c$ and not on $n$ or $f$. This gives a positive answer to a special case of a question posed by Kazhdan and Ziegler.


Introduction
A c-quasihomomorphism from a group G to a group H with a left-invariant metric d is a map f : G → H such that d(f (xy), f (x)f (y)) ≤ c for all x, y in G.A central question in geometric group theory, first raised by Ulam in [Ula60, Chapter 6], is whether there exists an actual homomorphism f : G → H such that d(f (x), f (x)) is at most some constant C for all x.Different versions of this question are of interest: for instance, C may be allowed to depend on c, G, (H, d) but not on f ; or G, (H, d) may be restricted to certain classes and or C is only allowed to depend on c.A well-known example where the answer to this question is negative is the case where G = H = Z with the standard metric.Here, quasihomomorphisms modulo bounded maps are a model of the real numbers [A'C21], and the answer is yes only for those quasihomomorphisms that correspond to integers.
Much literature in this area focusses on quasimorphisms, which are quasihomomorphisms into the real numbers R with the standard metric; we refer to [Kot04] for a brief introduction.In another branch of the research on quasihomomorphisms H is assumed nonabelian, and one of the first positive results on the central question above is Kazhdan's theorem on ε-representations of amenable groups [Kaz82].For more recent results on quasihomomorphisms into nonabelian groups we refer to [FK16] and the references there.
The following instance of the central question was formulated by Kazhdan and Ziegler in their work on approximate cohomology [KZ18].
Question 1.1.Let c ∈ N. Does there exist a constant C = C(c) such that the following holds: For all n ∈ N and all functions f : there exists a matrix g such that Here, G equals Z and H equals C n×n , both with addition, and the metric on H is defined by d(A, B) := rk(A − B).Our main result is an affirmative answer to this question in the special case where all matrices f (x) are assumed to be diagonal.Definition 1.2.Let (Q, +) be an abelian group.For an element v ∈ Q n , the Hamming weight w H (v) is the number of nonzero entries of v.For a pair of elements Remark 1.4.The map diag : C n → C n×n is an isometric embedding from C n with the Hamming metric to C n×n with the rank metric.This connects Definition 1.3 to Question 1.1.
We are ready to state our main result.
Theorem 1.6 (Main Theorem).Let c ∈ N. Then there exists a constant C = C(c) ∈ N such that for all n ∈ N and c-quasihomorphisms f : Z → Q n , we have: Moreover, we can take C = 28c.
Remark 1.7.The coefficient 28 is probably not optimal.However, we certainly have that C(c) ≥ c.Indeed, any map f : Z → Q n for which the only nonzero entries of f (x) are among the first c, is automatically a c-quasimorphism.♦ Corollary 1.8.Theorem 1.6 also holds with Q replaced by any torsion-free abelian group Q, with the same value of C = C(c).
Proof.Suppose, for a contradiction, that we have a c-quasihomomorphism f : Theorem 1.6 shows that for a c-quasihomomorphism f : Z → Q n , the group homomorphism f : Z → Q n defined by f (x) = x • f (1) gives a C-approximation for some constant C ∈ N independent on n.However, f need not be the homomorphism closest to f , as the next example shows.
Example 1.9.Let c = 1 and n ≥ 3. Define f : Z → Q n to be where α x is arbitrary if 5 | x, and α x = 0 otherwise.This is a 1-quasihomomorphism.Note that w H (f (x)−x•f (1)) ≤ 3 where equality is sometimes achieved.However, there also exist 2-approximations of f .For instance, letting v = ( 2 5 , 1 5 , 0, . . ., 0) ∈ Q n , one verifies that In fact, we can show that for every 1-quasihomomorphism f : Z → Q n there exists a 2-approximation, as claimed in [KZ18].The proof will appear in future work.♦ On the other hand, the following shows that the best possible approximation of a given qusimorphism f is at most twice as close as the homomorphism x → x • f (1).
Remark 1.10.Suppose that a map f : Remark 1.11.A result similar to Theorem 1.6 is easily proven in finite characteristic if we allow the constant C to depend on the characteristic.Let K be a field of characteristic p > 0, and let f : Z → K n be a c-quasihomomorphism.Then there exists a constant To see this, we observe that for all u, v ∈ Z with u ≥ 1, we have This follows by repeatedly applying the inequality w here we have used that p • f (1) = 0. We rewrite the latter as We have w 1)c using our observation with u = p, v = k; and also w H (f (r) − rf (1)) ≤ (p − 2)c (in the case r > 0) using our main observation.In total, this gives w H (f (x)−xf (1)) ≤ 2(p−1)c, so we can take C = 2(p − 1)c.♦ The remainder of this paper is organized as follows.In Section 2 we prove an auxiliary result of independent interest: maps from a finite abelian group into a torsion-free group that are almost a homomorphism, are in fact almost zero.Then, in Section 3, we apply this auxiliary result to the component functions of a cquasimorphism Z → Q n to prove the Main Theorem.

Almost homomorphisms are almost zero
Let A be a finite abelian group and let H be a torsion-free abelian group.The only homomorphism A → H is the zero map.The following proposition says that maps that are, in a suitable sense, close to being homomorphisms, are in fact also close to the zero map.
Proposition 2.1.Let a be a positive integer, A an abelian group of order a, H a torsion-free abelian group, q ∈ [0, 1], and f : A → H a map.Suppose that the zero set Z(f has cardinality at most qa.Then the problem set The contraposition of this statement says that if P (f ) is a small fraction of a 2 , so that f can be thought of as an (additive) "almost homomorphism" A → H, then q must be close to 1 so that f is essentially zero.
Proof.Since H is torsion-free, it embeds into the Q-vector space V := Q ⊗ Z H.By basic linear algebra, there exists a Q-linear function ξ : it suffices to prove the proposition for ξ • f instead of f .In other words, we may assume from the beginning that Let λ 1 > λ 2 > . . .> λ k > 0 be the distinct values in f (B), and for each i = 1, . . ., k set as well as so that the left-hand side is not in f (B) and in particular not equal to f (b + c).We have thus found and (b, c) ∈ P (f ).Then and hence b + c ∈ B 1 .But given c, there are at most n 1 values of b with b + c ∈ B 1 .(Note that here we have used that A is a group.)Hence we have at least Similarly, we find at least In total, we have therefore found at least pairs in P (f ); see Figure 1.
Repeating the same argument above with B and n , we find at least n (n + 1)/2 further pairs in P (f ), disjoint from those found above.Since |Z(f )| ≤ qa, we have n + n ≥ a(1 − q).Therefore where the second inequality is the Cauchy-Schwarz inequality Since n + n ≥ a(1 − q), we conclude that Remark 2.2.The lower bound in Proposition 2.1 is sharp.Let a = 2k + 1 ∈ Z, consider A := Z/aZ and define f : A → Z as f (x) :=the representative of x + aZ in {−k, . . ., 0, . . ., k}.We take q = Z(f ) a = 1 2k+1 .Then f (x + y) = f (x) + f (y) if and only if the right-hand side is still inside the interval {−k, . . ., k}, and a straightforward count shows that this is the case for 3k 2 + 3k + 1 pairs (x, y) ∈ A 2 .Hence P (f ) has size k • (k + 1), which equals a−qa 2 • a−qa 2 + 1 .A similar construction for a = 2k yields a problem set of size a 2 4 = k 2 , which equals the ceiling of the lower bound a 2 4 − 1 4 .♦ Below, we will use the following strengthening of Proposition 2.1: Proposition 2.3.Let a, A, H, q and f be as in Proposition 2.1.Furthermore, let p ∈ [0, 1−q 2 ) and let S ⊆ A be a subset of cardinality at most pa.Then the set On left, a graphical proof of the inequality (2.1): the left-hand side is the number of small squares in the shaded region, the right-hand side is the number of squares on or above the main diagonal.
On the right, a proof of the inequality (2.2): the left-hand side is the area of the shaded region, the right-hand side, the area enclosed by the dashed line.
Proof.Keep the notation from the proof of Proposition 2.1.Recall n = |B| and n = |B |.Note that for a fixed b, there can be at most pa choices of c with b + c ∈ S. We then find at least Letting k ≤ k be the largest index for which the second factor is nonnegative, as in the proof of Proposition 2.1, we find that B contributes at least to P S (f ); see Figure 1.Similarly, B contributes at least (n − pa)(n − pa + 1)/2, and these contributions are disjoint.The desired inequality follows as in the proof of Proposition 2.1 but with n, n replaced by n − pa, n − pa.
The key ingredient for the proof of Theorem 1.6 is the following corollary of Proposition 2.3.Here, and in the rest of the paper, we write [a] for the set {1, 2, . . ., a}. where Proof.Let f be the restriction of f to the interval [a], and identify Z/aZ with [a] with the group operation * defined by x * y := x + y (mod a).
Let S = N P (f ), and apply Proposition 2.3 to f .We find that this set is contained in the problem set P (f ).

Proof of the Main Theorem
The main goal of this section is to prove Theorem 1.6.We start with some definitions.
Definition 3.1.Let 1 < a, and f : [2a] → Q.We define the following problem sets of f : }. Furthermore, we recall that the zero set of f is defined as The following proposition says that P 1 (f ), P a (f ), P (f ) cannot be simultaneously small.Proposition 3.2.Let p, q ∈ (0, 1) such that p < 1−q 2 , a ∈ N with 1 < a, and let Proof.Without loss of generality we can assume f (a) = 0 and hence f (1) = 0. Indeed, suppose we have shown the statement for every f with f (a) = 0. Then for any f : [2a] → Q with f (a) = af (1), we take f : [2a] → Q to be f (x) = af (x) − xf (a).Now we observe that f (a) = 0 = a f (1), and that P (f ) = P ( f ), Likewise, at least one of the elements 1, 2, . . ., x 1 − 1 needs to be in P 1 (f ).Thus we have and by assumption we have |N P (f )| = |P a (f )| ≤ pa.Now we can apply Corollary 2.4 to conclude.
We now prove Theorem 1.6.
Proof of the Main Theorem.Consider a c-quasihomomorphism f = (f 1 , . . ., f n ) : Z → Q n .Our goal is to show that for every a ∈ Z we have w H (f (a) − a • f (1)) ≤ C for some constant C depending only on c.We start with the case a > 0.
Write M := w H (f (a)−af (1)).Without loss of generality, we have f 1 (a) = af (1), . . ., f M (a) = af (1).We will show that M ≤ C for some constant C depending on c only.To this end, fix small parameters p, q ∈ (0, 1) (to be optimized over later) and consider the restrictions f i : [2a] → Q of the components of f .By Proposition 3.2, for every i, we have that either Let m 0 be the number of coordinates i such that (iii) holds.We define m 1 and m 2 analogously, for (i) and (ii) respectively.
By counting the number of triples (i, x, y) Because f is a c-quasihomomorphism, the left hand side is at most a 2 c.On the other hand, the right hand side is at least m 0 F (p, q)a 2 , so This completes the proof.
To get the explicit bound 28c from Theorem 1.6, we minimize the function Note that this function is strictly convex for (p, q) ∈ R 2 >0 , so that it has at most one minimum in the positive orthant.We find this by setting the partial derivatives equal to zero and solving for p, q.The minimum is ≈ 27.6817 and attained at (p, q) ≈ (0.1167, 0.16500).