Kaisa Matomäki, Maksym Radziwiłł, and I have just uploaded to the arXiv our paper “Sign patterns of the Liouville and Möbius functions“. This paper is somewhat similar to our previous paper in that it is using the recent breakthrough of Matomäki and Radziwiłł on mean values of multiplicative functions to obtain partial results towards the Chowla conjecture. This conjecture can be phrased, roughly speaking, as follows: if ${k}$ is a fixed natural number and ${n}$ is selected at random from a large interval ${[1,x]}$, then the sign pattern ${(\lambda(n), \lambda(n+1),\dots,\lambda(n+k-1)) \in \{-1,+1\}^k}$ becomes asymptotically equidistributed in ${\{-1,+1\}^k}$ in the limit ${x \rightarrow \infty}$. This remains open for ${k \geq 2}$. In fact even the significantly weaker statement that each of the sign patterns in ${\{-1,+1\}^k}$ is attained infinitely often is open for ${k \geq 4}$. However, in 1986, Hildebrand showed that for ${k \leq 3}$ all sign patterns are indeed attained infinitely often. Our first result is a strengthening of Hildebrand’s, moving a little bit closer to Chowla’s conjecture:

Theorem 1 Let ${k \leq 3}$. Then each of the sign patterns in ${\{-1,+1\}^k}$ is attained by the Liouville function for a set of natural numbers ${n}$ of positive lower density.

Thus for instance one has ${\lambda(n)=\lambda(n+1)=\lambda(n+2)}$ for a set of ${n}$ of positive lower density. The ${k \leq 2}$ case of this theorem already appears in the original paper of Matomäki and Radziwiłł (and the significantly simpler case of the sign patterns ${++}$ and ${--}$ was treated previously by Harman, Pintz, and Wolke).

The basic strategy in all of these arguments is to assume for sake of contradiction that a certain sign pattern occurs extremely rarely, and then exploit the complete multiplicativity of ${\lambda}$ (which implies in particular that ${\lambda(2n) = -\lambda(n)}$, ${\lambda(3n) = -\lambda(n)}$, and ${\lambda(5n) = -\lambda(n)}$ for all ${n}$) together with some combinatorial arguments (vaguely analogous to solving a Sudoku puzzle!) to establish more complex sign patterns for the Liouville function, that are either inconsistent with each other, or with results such as the Matomäki-Radziwiłł result. To illustrate this, let us give some ${k=2}$ examples, arguing a little informally to emphasise the combinatorial aspects of the argument. First suppose that the sign pattern ${(\lambda(n),\lambda(n+1)) = (+1,+1)}$ almost never occurs. The prime number theorem tells us that ${\lambda(n)}$ and ${\lambda(n+1)}$ are each equal to ${+1}$ about half of the time, which by inclusion-exclusion implies that the sign pattern ${(\lambda(n),\lambda(n+1))=(-1,-1)}$ almost never occurs. In other words, we have ${\lambda(n+1) = -\lambda(n)}$ for almost all ${n}$. But from the multiplicativity property ${\lambda(2n)=-\lambda(n)}$ this implies that one should have

$\displaystyle \lambda(2n+2) = -\lambda(2n)$

$\displaystyle \lambda(2n+1) = -\lambda(2n)$

and

$\displaystyle \lambda(2n+2) = -\lambda(2n+1)$

for almost all ${n}$. But the above three statements are contradictory, and the claim follows.

Similarly, if we assume that the sign pattern ${(\lambda(n),\lambda(n+1)) = (+1,-1)}$ almost never occurs, then a similar argument to the above shows that for any fixed ${h}$, one has ${\lambda(n)=\lambda(n+1)=\dots=\lambda(n+h)}$ for almost all ${n}$. But this means that the mean ${\frac{1}{h} \sum_{j=1}^h \lambda(n+j)}$ is abnormally large for most ${n}$, which (for ${h}$ large enough) contradicts the results of Matomäki and Radziwiłł. Here we see that the “enemy” to defeat is the scenario in which ${\lambda}$ only changes sign very rarely, in which case one rarely sees the pattern ${(+1,-1)}$.

It turns out that similar (but more combinatorially intricate) arguments work for sign patterns of length three (but are unlikely to work for most sign patterns of length four or greater). We give here one fragment of such an argument (due to Hildebrand) which hopefully conveys the Sudoku-type flavour of the combinatorics. Suppose for instance that the sign pattern ${(\lambda(n),\lambda(n+1),\lambda(n+2)) = (+1,+1,+1)}$ almost never occurs. Now suppose ${n}$ is a typical number with ${\lambda(15n-1)=\lambda(15n+1)=+1}$. Since we almost never have the sign pattern ${(+1,+1,+1)}$, we must (almost always) then have ${\lambda(15n) = -1}$. By multiplicativity this implies that

$\displaystyle (\lambda(60n-4), \lambda(60n), \lambda(60n+4)) = (+1,-1,+1).$

We claim that this (almost always) forces ${\lambda(60n+5)=-1}$. For if ${\lambda(60n+5)=+1}$, then by the lack of the sign pattern ${(+1,+1,+1)}$, this (almost always) forces ${\lambda(60n+3)=\lambda(60n+6)=-1}$, which by multiplicativity forces ${\lambda(20n+1)=\lambda(20n+2)=+1}$, which by lack of ${(+1,+1,+1)}$ (almost always) forces ${\lambda(20n)=-1}$, which by multiplicativity contradicts ${\lambda(60n)=-1}$. Thus we have ${\lambda(60n+5)=-1}$; a similar argument gives ${\lambda(60n-5)=-1}$ almost always, which by multiplicativity gives ${\lambda(12n-1)=\lambda(12n)=\lambda(12n+1)=+1}$, a contradiction. Thus we almost never have ${\lambda(15n-1)=\lambda(15n+1)=+1}$, which by the inclusion-exclusion argument mentioned previously shows that ${\lambda(15n+1) = - \lambda(15n-1)}$ for almost all ${n}$.

One can continue these Sudoku-type arguments and conclude eventually that ${\lambda(3n-1)=-\lambda(3n+1)=\lambda(3n+2)}$ for almost all ${n}$. To put it another way, if ${\chi_3}$ denotes the non-principal Dirichlet character of modulus ${3}$, then ${\lambda \chi_3}$ is almost always constant away from the multiples of ${3}$. (Conversely, if ${\lambda \chi_3}$ changed sign very rarely outside of the multiples of three, then the sign pattern ${(+1,+1,+1)}$ would never occur.) Fortunately, the main result of Matomäki and Radziwiłł shows that this scenario cannot occur, which establishes that the sign pattern ${(+1,+1,+1)}$ must occur rather frequently. The other sign patterns are handled by variants of these arguments.

Excluding a sign pattern of length three leads to useful implications like “if ${\lambda(n-1)=\lambda(n)=+1}$, then ${\lambda(n+1)=-1}$” which turn out are just barely strong enough to quite rigidly constrain the Liouville function using Sudoku-like arguments. In contrast, excluding a sign pattern of length four only gives rise to implications like “`if ${\lambda(n-2)=\lambda(n-1)=\lambda(n)=+1}$, then ${\lambda(n+1)=-1}$“, and these seem to be much weaker for this purpose (the hypothesis in these implications just isn’t satisfied nearly often enough). So a different idea seems to be needed if one wishes to extend the above theorem to larger values of ${k}$.

Our second theorem gives an analogous result for the Möbius function ${\mu}$ (which takes values in ${\{-1,0,+1\}}$ rather than ${\{-1,1\}}$), but the analysis turns out to be remarkably difficult and we are only able to get up to ${k=2}$:

Theorem 2 Let ${k \leq 2}$. Then each of the sign patterns in ${\{-1,0,+1\}^k}$ is attained by the Möbius function for a set ${n}$ of positive lower density.

It turns out that the prime number theorem and elementary sieve theory can be used to handle the ${k=1}$ case and all the ${k=2}$ cases that involve at least one ${0}$, leaving only the four sign patterns ${(\pm 1, \pm 1)}$ to handle. It is here that the zeroes of the Möbius function cause a significant new obstacle. Suppose for instance that the sign pattern ${(+1, -1)}$ almost never occurs for the Möbius function. The same arguments that were used in the Liouville case then show that ${\mu(n)}$ will be almost always equal to ${\mu(n+1)}$, provided that ${n,n+1}$ are both square-free. One can try to chain this together as before to create a long string ${\mu(n)=\dots=\mu(n+h) \in \{-1,+1\}}$ where the Möbius function is constant, but this cannot work for any ${h}$ larger than three, because the Möbius function vanishes at every multiple of four.

The constraints we assume on the Möbius function can be depicted using a graph on the squarefree natural numbers, in which any two adjacent squarefree natural numbers are connected by an edge. The main difficulty is then that this graph is highly disconnected due to the multiples of four not being squarefree.

To get around this, we need to enlarge the graph. Note from multiplicativity that if ${\mu(n)}$ is almost always equal to ${\mu(n+1)}$ when ${n,n+1}$ are squarefree, then ${\mu(n)}$ is almost always equal to ${\mu(n+p)}$ when ${n,n+p}$ are squarefree and ${n}$ is divisible by ${p}$. We can then form a graph on the squarefree natural numbers by connecting ${n}$ to ${n+p}$ whenever ${n,n+p}$ are squarefree and ${n}$ is divisible by ${p}$. If this graph is “locally connected” in some sense, then ${\mu}$ will be constant on almost all of the squarefree numbers in a large interval, which turns out to be incompatible with the results of Matomäki and Radziwiłł. Because of this, matters are reduced to establishing the connectedness of a certain graph. More precisely, it turns out to be sufficient to establish the following claim:

Theorem 3 For each prime ${p}$, let ${a_p \hbox{ mod } p^2}$ be a residue class chosen uniformly at random. Let ${G}$ be the random graph whose vertices ${V}$ consist of those integers ${n}$ not equal to ${a_p \hbox{ mod } p^2}$ for any ${p}$, and whose edges consist of pairs ${n,n+p}$ in ${V}$ with ${n = a_p \hbox{ mod } p}$. Then with probability ${1}$, the graph ${G}$ is connected.

We were able to show the connectedness of this graph, though it turned out to be remarkably tricky to do so. Roughly speaking (and suppressing a number of technicalities), the main steps in the argument were as follows.

• (Early stage) Pick a large number ${X}$ (in our paper we take ${X}$ to be odd, but I’ll ignore this technicality here). Using a moment method to explore neighbourhoods of a single point in ${V}$, one can show that a vertex ${v}$ in ${V}$ is almost always connected to at least ${\log^{10} X}$ numbers in ${[v,v+X^{1/100}]}$, using relatively short paths of short diameter. (This is the most computationally intensive portion of the argument.)
• (Middle stage) Let ${X'}$ be a typical number in ${[X/40,X/20]}$, and let ${R}$ be a scale somewhere between ${X^{1/40}}$ and ${X'}$. By using paths ${n, n+p_1, n+p_1-p_2, n+p_1-p_2+p_3}$ involving three primes, and using a variant of Vinogradov’s theorem and some routine second moment computations, one can show that with quite high probability, any “good” vertex in ${[v+X'-R, v+X'-0.99R]}$ is connected to a “good” vertex in ${[v+X'-0.01R, v+X-0.0099 R]}$ by paths of length three, where the definition of “good” is somewhat technical but encompasses almost all of the vertices in ${V}$.
• (Late stage) Combining the two previous results together, we can show that most vertices ${v}$ will be connected to a vertex in ${[v+X'-X^{1/40}, v+X']}$ for any ${X'}$ in ${[X/40,X/20]}$. In particular, ${v}$ will be connected to a set of ${\gg X^{9/10}}$ vertices in ${[v,v+X/20]}$. By tracking everything carefully, one can control the length and diameter of the paths used to connect ${v}$ to this set, and one can also control the parity of the elements in this set.
• (Final stage) Now if we have two vertices ${v, w}$ at a distance ${X}$ apart. By the previous item, one can connect ${v}$ to a large set ${A}$ of vertices in ${[v,v+X/20]}$, and one can similarly connect ${w}$ to a large set ${B}$ of vertices in ${[w,w+X/20]}$. Now, by using a Vinogradov-type theorem and second moment calculations again (and ensuring that the elements of ${A}$ and ${B}$ have opposite parity), one can connect many of the vertices in ${A}$ to many of the vertices ${B}$ by paths of length three, which then connects ${v}$ to ${w}$, and gives the claim.

It seems of interest to understand random graphs like ${G}$ further. In particular, the graph ${G'}$ on the integers formed by connecting ${n}$ to ${n+p}$ for all ${n}$ in a randomly selected residue class mod ${p}$ for each prime ${p}$ is particularly interesting (it is to the Liouville function as ${G}$ is to the Möbius function); if one could show some “local expander” properties of this graph ${G'}$, then one would have a chance of modifying the above methods to attack the first unsolved case of the Chowla conjecture, namely that ${\lambda(n)\lambda(n+1)}$ has asymptotic density zero (perhaps working with logarithmic density instead of natural density to avoids some technicalities).