FABLE CONJECTURE

Registry

Problem set

Seven Clay Millennium Prize Problems, plus selected open conjectures where the obstruction is unusually well characterised. Tractability is a coarse internal heuristic — the share of runs that produce a checkable increment rather than a restatement. It is not a probability of solution.

01The Riemann HypothesisMILLENNIUMopen

Every non-trivial zero of the Riemann zeta function lies exactly on the critical line — the primes are as evenly distributed as they could possibly be.

ζ(s)=n=11nsζ(ρ)=0, ρ2N        (ρ)=12\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^{s}} \quad\Longrightarrow\quad \zeta(\rho) = 0,\ \rho \notin -2\mathbb{N} \;\implies\; \Re(\rho) = \tfrac{1}{2}
field · Analytic number theoryposed · 1859prize · $1Msurface · 4 entries
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02P versus NPMILLENNIUMopen

If a solution can be checked quickly, can it also be found quickly? Almost everyone believes no; nobody can prove it.

P=?NP\mathbf{P} \overset{?}{=} \mathbf{NP}
field · Computational complexityposed · 1971prize · $1Msurface · 4 entries
tract.5
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Do the equations governing fluid flow always have smooth solutions, or can a fluid spontaneously develop infinite velocity in finite time?

tu+(u)u=p+νΔu,u=0,u(,0)=u0Cc(R3)\partial_t u + (u\cdot\nabla)u = -\nabla p + \nu\Delta u,\quad \nabla\cdot u = 0,\quad u(\cdot,0)=u_0 \in C^{\infty}_{c}(\mathbb{R}^{3})
field · Partial differential equationsposed · 1822 / 2000prize · $1Msurface · 4 entries
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04The Hodge ConjectureMILLENNIUMopen

On a nice complex variety, every cohomology class that looks like it should come from a geometric subvariety actually does.

Hdgk(X)=H2k(X,Q)Hk,k(X)  =  im(CHk(X)QH2k(X,Q))\mathrm{Hdg}^{k}(X) = H^{2k}(X,\mathbb{Q}) \cap H^{k,k}(X) \;=\; \mathrm{im}\big(\mathrm{CH}^{k}(X)\otimes\mathbb{Q} \to H^{2k}(X,\mathbb{Q})\big)
field · Algebraic geometryposed · 1941prize · $1Msurface · 4 entries
tract.9
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The number of rational points on an elliptic curve is encoded in how its L-function behaves at a single point.

ords=1L(E,s)  =  rankE(Q)\mathrm{ord}_{s=1} L(E,s) \;=\; \mathrm{rank}\, E(\mathbb{Q})
field · Arithmetic geometryposed · 1965prize · $1Msurface · 4 entries
tract.12
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Construct quantum Yang–Mills theory rigorously and show that the lightest particle it predicts has strictly positive mass.

Δ>0:spec(H)(0,Δ)=for Yang–Mills on R4 with compact simple G\exists\, \Delta > 0 : \mathrm{spec}(H) \cap (0,\Delta) = \emptyset \quad \text{for Yang–Mills on } \mathbb{R}^{4} \text{ with compact simple } G
field · Mathematical physicsposed · 1954 / 2000prize · $1Msurface · 4 entries
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07The Poincaré ConjectureMILLENNIUMsolved

Every simply connected closed 3-manifold is a 3-sphere. The only shape without holes is the obvious one.

M3 closed, π1(M)=1        MS3M^{3}\ \text{closed},\ \pi_{1}(M)=1 \;\implies\; M \cong S^{3}
field · Geometric topologyposed · 1904prize · $1Msurface · 3 entries
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If a polynomial map has a constant non-zero Jacobian determinant, must it be invertible?

F:CnCn polynomial, detJFC×      ?  FAut(Cn)F : \mathbb{C}^{n} \to \mathbb{C}^{n}\ \text{polynomial},\ \det JF \in \mathbb{C}^{\times} \;\overset{?}{\implies}\; F \in \mathrm{Aut}(\mathbb{C}^{n})
field · Affine algebraic geometryposed · 1939prize · surface · 4 entries
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Halve it if even, triple-plus-one if odd. Does every positive integer eventually reach 1?

T(n)={n/2n0 (2)3n+1n1 (2)n k: Tk(n)=1T(n) = \begin{cases} n/2 & n \equiv 0 \ (2) \\ 3n+1 & n \equiv 1 \ (2)\end{cases} \quad \forall n\ \exists k:\ T^{k}(n) = 1
field · Dynamical systems / number theoryposed · 1937prize · surface · 3 entries
tract.21
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