31. The Riemann Hypothesis via the Liouville function

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Lean: Engine/RiemannLiouville.lean. Numbers: \(|L(x)|/\sqrt{x}\) is bounded (the classical Liouville equivalent of RH). Link to the engine: the Liouville sign = (−1)^rank, flipped by deleteFactor.

Where we are

In chapters 30–31 (SeparatingScale, RankDescent, ProductCore) we closed ProductHall by pure arithmetic and shifted the whole burden onto the concrete product core RankNode r — a structure with a sign and r factors, along which a correct rank descent \(r \to r-1 \to \dots \to 1\) runs. The logic of the descent is sound; what remained open were the stitching nodes (extensionality on residual ranks, finiteness of ProdSig against the pigeonhole).

Before continuing to storm these nodes head-on, let us take a step sideways and see onto what exactly our rank apparatus (rank — the "height" of a state, strictly decreasing along descent steps; see the glossary) maps in classical analytic number theory. The answer will turn out to be unexpectedly precise: the dynamics of the sign under rank descent is literally the sign dynamics of the Liouville function, and its balance is equivalent to the Riemann Hypothesis.

The idea of this branch: take a ready-made equivalent of RH

The RiemannBranch line (ch. 28) went straight from "engine \Rightarrow zeros of \zeta" and ran into the analytic bridge EngineBridge, comparable in difficulty to RH itself. Here we proceed differently. It is natural to suppose: instead of building our own bridge to the zeros of the zeta function, take a ready-made arithmetic statement equivalent to RH and decompose it within our theory. Then its truth automatically yields RH — all the analysis is already packed into the known equivalence, and what remains for us is a purely arithmetic goal.

We choose the Liouville equivalence. It is convenient because it speaks of a sign that depends exactly on the number of prime factors — and the number of prime factors is precisely our rank.

Definitions

Let us introduce the Liouville function and its summatory function.

Definition 31.1 (Liouville function). For \(n \ge 1\)

\[ \lambda(n) \;=\; (-1)^{\Omega(n)}, \]

where \(\Omega(n)\) is the number of prime factors of \(n\) counted with multiplicity (for instance \(\Omega(12)=\Omega(2^2\cdot 3)=3\)). In mathlib this is ArithmeticFunction.liouville, and \(\Omega(n)\) is cardFactors n.

Definition 31.2 (summatory function).

\[ L(x) \;=\; \sum_{n=1}^{x} \lambda(n). \]

In Lean: def L (x : ℕ) : ℤ := ∑ n ∈ Finset.Icc 1 x, liouville n.

The function \(\lambda\) is completely multiplicative and takes values \(\pm 1\); \(L(x)\) is an alternating sum, the accumulated imbalance between the numbers with an even and an odd number of prime factors up to \(x\).

Definition 31.3 (LiouvilleBound).

\[ \texttt{LiouvilleBound} \;:\!\!\iff\; \forall\,\varepsilon>0\ \exists\,C>0\ \forall x:\quad |L(x)| \le C\,x^{1/2+\varepsilon}. \]

In Lean this is LiouvilleBound : Prop. In substance: a sum of \(\pm 1\) of length \(x\) behaves like a random walk — growing no faster than \(\sqrt{x}\) (up to \(x^\varepsilon\)), not linearly.

Definition 31.4 (RiemannHypothesis, from mathlib). Following the principle "everything external comes from mathlib", RiemannHypothesis here is the official mathlib formulation (a quantifier over all zeros of riemannZeta, excluding the trivial ones -2(n+1) and s = 1), and not a home-made def. The same goal as in RiemannBranch.

The classical bridge: \(LiouvilleBound \iff RH\)

The supporting fact is a well-known theorem of analytic number theory (mathlib does not yet have it, so we introduce it as an explicit input-bridge — an input, or gate, is an honestly named unproven statement (see the glossary) — rather than postulating RH):

Definition 31.5 (the Liouville bridge \(H_L\), LiouvilleRHBridge).

\[ \texttt{LiouvilleRHBridge} \;:\!\!\iff\; \bigl(\texttt{LiouvilleBound} \iff \texttt{RiemannHypothesis}\bigr). \]

In Lean — LiouvilleRHBridge : Prop.

This is the classical equivalence: the bound \(L(x)=O(x^{1/2+\varepsilon})\) is equivalent to the absence of zeros of \(\zeta\) to the right of the line \(\mathrm{Re}=1/2\). We do not prove it here and do not pass it off as our own — we make the branch conditional on it. This is more honest than postulating RH: what is postulated is merely a known, recognized fact about the connection between two objects, not the thing sought.

Given the bridge, the branch closes by a trivial implication.

Theorem 31.6 (riemann_of_liouville_bound). If the bridge LiouvilleRHBridge is given and LiouvilleBound is proven, then RiemannHypothesis.

\[ \texttt{LiouvilleRHBridge} \;\wedge\; \texttt{LiouvilleBound} \;\Rightarrow\; \texttt{RiemannHypothesis}.\tag{31.1} \]

The Lean proof is a single line, bridge.mp hbound: all the analysis is isolated in the bridge, and the arithmetic (LiouvilleBound) is the only substantive goal.

Thus the branch's only open arithmetic node is LiouvilleBound. It is precisely this node that our rank apparatus must decompose. Let us show why it fits our theory at all.

Why this is our apparatus: \(\lambda = (-1)^{\text{rank}}\)

The key observation is a coincidence of objects. The number of prime factors \(\Omega(n)=\) cardFactors n is exactly our rank RankNode: with us, RankNode r has factors : Fin r → ℕ, so the rank \(r\) is the node's number of factors. From this, instantly:

Theorem 31.7 (liouville_eq_neg_one_pow_rank). For \(n\ne 0\)

\[ \lambda(n) \;=\; (-1)^{\mathrm{cardFactors} n} \;=\; (-1)^{\text{rank}}.\tag{31.2} \]

In Lean this is simply liouville_apply hn: the Liouville sign is the parity of the rank.

So \(L(x)\) is the accumulated sum of the signs \((-1)^{\text{rank}}\) over all nodes up to \(x\). LiouvilleBound says that this sum is small (of order \(\sqrt{x}\)): the signs of our rank are balanced.

The sign flip under deleteFactor: descent = the sign dynamics of Liouville

The second observation ties our descent step to the change of sign. With us, deleteFactor lowers the rank \(r\to r-1\) by removing one role. In terms of numbers this is the passage from \(n=p\cdot m\) to \(m\) (stripping one prime factor). What happens to the sign?

Theorem 31.8 (liouville_flip_of_mul_prime). For a prime \(p\) and \(m\ne 0\)

\[ \lambda(p\cdot m) \;=\; -\,\lambda(m).\tag{31.3} \]

Proof (Lean): \(\mathrm{cardFactors}(p\cdot m)=\mathrm{cardFactors} m + 1\) (via cardFactors_mul and cardFactors_apply_prime), whence \((-1)^{\Omega(m)+1}=-(-1)^{\Omega(m)}\).

The meaning. Our deleteFactor (RankNode (r+1) → RankNode r, stripping one factor while keeping the same sign field) at the level of numbers flips the Liouville sign. Hence product-rank descent is exactly the sign dynamics of \(\lambda\): every rank-descent step turns \(\lambda\) into its opposite. The descent \(r\to r-1\to\dots\to 1\), which in chapters 30–31 drives the collision toward the rank-1 base, in arithmetic language is a sequence of Liouville sign flips along the factorization.

Note. Here two different "descents" meet on one object. Our dynamical descent is a downward walk along a single lineage (\(p\cdot m \to m\)), and it is strictly sign-alternating. The sum \(L(x)\), on the other hand, runs over all \(n\le x\) horizontally. The connection between the "vertical" flip and the "horizontal" balance is precisely the substantive work the rank apparatus must perform: to show that the pairwise flips along lineages force the horizontal sum to cancel down to \(O(\sqrt{x})\).

Numbers: \(|L(x)|/\sqrt{x}\) is bounded

What exactly must be proven is best seen in normalized form. LiouvilleBound is equivalent to the statement that the quantity $\(\frac{|L(x)|}{\sqrt{x}}\)$ remains bounded (up to the correction \(x^{\varepsilon}\)). Numerically \(L(x)\) does indeed behave like a \(\sqrt{x}\)-sized swing: the signs \((-1)^{\text{rank}}\) split almost evenly between even and odd rank, and the accumulated imbalance never leaves the square-root scale. This is exactly the separating scale \(\sqrt{x}\) which in chapter 30 separated the legal layer; here it surfaces as the growth scale of \(L\).

Note (historical caution). The stronger claim \(L(x)\le 0\) for \(x\ge 2\) — Pólya's conjecture — is refuted (the first counterexample near \(x\approx 906{,}150{,}257\)). Therefore our goal is precisely the bound \(O(x^{1/2+\varepsilon})\), not a definite sign of \(L\): the rank balance must be statistical (the sum is small), not pointwise (the sign is fixed). This matters: our descent strictly flips the sign along a lineage, but the horizontal balance is probabilistic, and the two must not be confused.

Honestly: \(LiouvilleBound \equiv RH\) is the parity wall

Now let us say plainly where the wall is, and not pass the reduction off as a proof. We have not proven LiouvilleBound. We have shown only that this equivalent of RH is expressible in our language and that our operation deleteFactor governs exactly the sign whose sum must be bounded.

The essence of the difficulty is parity. The Liouville sign \(\lambda(n)=(-1)^{\Omega(n)}\) is the parity of \(\Omega(n)\), that is, the parity of the rank. The bound \(L(x)=O(x^{1/2+\varepsilon})\) asserts that this parity is distributed almost uniformly (numbers with an even and an odd number of factors occur almost equally often, with a square-root fluctuation).

This is the classical parity wall: control of the parity of \(\Omega\) is a well-known obstruction, inaccessible to sieve methods and, in substance, equivalent to RH itself.

Conclusion. Our rank apparatus supplies vertical information (one flip per one stripped factor), but LiouvilleBound demands a horizontal statistical balance of parities over all \(n\le x\). The passage from the first to the second is exactly the unresolved node.

Conjecture and a plan for closing the LiouvilleBound node

Let us state the open node explicitly.

Conjecture 31.9 (rank balance). The sign dynamics of the descent (Theorem 31.8, liouville_flip_of_mul_prime) combined with our no_infinite_descent (the impossibility of a perpetual engine, ch. 28) entails a statistical balance of rank parities, that is, LiouvilleBound.

The closing plan, step by step, with an honest assessment of each:

1. Pair the lineages into flip-pairs. Every node of rank \(r\), via deleteFactor, produces a parent of rank \(r-1\) with opposite \(\lambda\). The idea: organize the contribution to \(L(x)\) as a sum over pairs \((m,\ p\cdot m)\) in which the two terms cancel. The difficulty: the pairs leave the segment \([1,x]\) asymmetrically (boundary effects of order \(\sqrt{x}\)) — and this is exactly where the square-root scale, and not a linear one, must arise.
2. Tie the flip balance to the impossibility of the engine. If the rank parities were systematically skewed (for instance \(L(x)\gg \sqrt{x}\)), that would give a "free" directed pumping of mass along the descent — the very perpetual engine forbidden by no_infinite_descent (EPMI — the impossibility core for the perpetual engine, see the glossary). This is exactly the logic of the RiemannBranch line, but expressed through the Liouville sign rather than through the height \(H\). The task is to make this implication precise: \(parity skew \Rightarrow Engine\).
3. Fold it into the bound. From the boundedness of the skew — derive \(O(x^{1/2+\varepsilon})\).

Note. Step 2 is the heart of the plan and, at the same time, the boundary of honesty. It repeats the structure of EngineBridge (ch. 28) on a new object: "asymmetry \Rightarrow engine". In all likelihood, its difficulty is comparable to RH itself — and that is precisely what it means to have run into the parity wall rather than gone around it. We record this as a conjecture, not a theorem: Step00 remains a sorry. The progress is real — the node's exact arithmetic face has been found (the Liouville sign = the parity of the rank), along with the exact operation governing it (deleteFactor = flip) — but the balance itself has not been presented.

Conclusion

The Liouville branch gives the shortest honest formulation of RH in our theory: we take the ready-made equivalence \(LiouvilleBound \iff RH\) (the bridge LiouvilleRHBridge), and RH reduces to a single arithmetic goal — the bound \(|L(x)|=O(x^{1/2+\varepsilon})\) (Theorem 31.6, riemann_of_liouville_bound).

This goal fits our apparatus perfectly: \(\lambda=(-1)^{\text{rank}}\) (Theorem 31.7, liouville_eq_neg_one_pow_rank), and our deleteFactor flips the sign (Theorem 31.8, liouville_flip_of_mul_prime), so that product-rank descent is literally the sign dynamics of Liouville.

But LiouvilleBound is not proven: it is the control of the parity of \(\Omega\), the classical parity wall, and the passage from the vertical flip to the horizontal balance remains an open conjecture (the plan runs through no_infinite_descent).

Bridge to the next chapter

So we now have two independent inputs to RH — the analytic EngineBridge (ch. 28, via the zeros of \(\zeta\)) and the arithmetic LiouvilleBound (this chapter, via the parity of the rank) — and both run into one and the same wall, "asymmetry \Rightarrow engine", merely dressed differently.

In the next chapter we return to the concrete product core and examine ProductCore in detail: how RankNode, coreSigOf, and deleteFactor assemble into an honest descent core_step_proved, and exactly where (the stitching with deleteFactor on residual ranks) an input remains — that is, we look at the same parity/extensionality wall no longer through \(\lambda\), but from inside the structure of the core itself.

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