nLab 11/8 conjecture

Contents

Idea

The 11/8 conjecture arises from the still not fully answered question, which symmetric unimodular bilinear forms are the intersection forms of oriented closed smooth 4-manifolds.

According to Donaldson's theorem, a definite form is the intersection form of an oriented closed smooth 4-manifolds if and only if it’s either n[1]\oplus n[-1] or n[+1]\oplus n[+1], with both realized by the even simply connected examples #nP 2\#n\mathbb{C}P^2 or #nP 2¯\#n\overline{\mathbb{C}P^2}. According to Serre’s classification theorem, an indefinite odd form is isomorphic to m[1]n[+1]\oplus m[-1]\oplus n[+1], which are all realized by the even simply connected examples #mP 2#mP 2¯\#m\mathbb{C}P^2\#m\overline{\mathbb{C}P^2}. (According to Freedman's theorem, all these simply connected examples are even unique up to homeomorphism.)

Left over is, which indefinite even forms are realized. It comes down to the forms:

±2mE 8nH \pm 2mE_8\oplus nH

according to Serre’s classification theorem with Rokhlin's theorem requiring an even number of definite ±E 8\pm E_8 and indefiniteness requiring n1n\geq 1. Simon Donaldson furthermore showed, that if m0m\neq 0, then n3n\geq 3.

Currently, the strongest result approaching the 11/8 conjecture is Furuta's theorem, which is also called 10/8 theorem.

Statement

The open 11/8 conjecture assumes: Every oriented closed smooth 4-manifold MM with indefinite even intersection form fulfills:

b 2(M)118|σ(M)|. b_2(M) \geq\frac{11}{8}|\sigma(M)|.

Hence, if σ(M)0\sigma(M)\geq 0, then 19b 2 (M)3b 2 +(M)19b_2^-(M)\geq 3b_2^+(M). If σ(M)0\sigma(M)\leq 0, then 19b 2 +(M)3b 2 (M)19b_2^+(M)\geq 3b_2^-(M). Without loss of generality, the 11/8 conjecture can be restricted to one of these cases by reversing orientation with b 2(M¯)=b 2(M)b_2(\overline{M})=b_2 (M) and σ(M¯)=σ(M)\sigma(\overline{M})=-\sigma(M).

Alternatively, for the above structure of the intersection form, the 3/2 conjecture is equivalent to:

n3|m|. n\geq 3|m|.

This follows from:

b 2(M)=16|m|+2n; b_2(M) =16|m|+2n;
|σ(M)|=16|m|. |\sigma(M)| =16|m|.

(Scorpan 05, p. 247)

Both reformulations of the 11/8 conjecture show, that the difference to Furuta's theorem or 10/8 theorem is bigger than the small difference between the fractions makes it seem.

If the 11/8 conjecture holds for two oriented closed smooth 4-manifolds MM and NN with indefinite even intersection form, then it automatically holds for their connected sum when additionally using the triangle inequality:

118|σ(M#N)|=118|σ(M)+118σ(N)|b 2(M)+b 2(N)=b 2(M#N). \frac{11}{8}|\sigma(M\# N)| =\frac{11}{8}|\sigma(M) +\frac{11}{8}\sigma(N)| \leq b_2(M) +b_2(N) =b_2(M\# N).

But that doesn’t mean, that it’s sufficient to prove the 11/8 conjecture just for irreducible (not splitting in a non-trivial connected sum) oriented closed smooth 4-manifolds with indefinite even intersection form. While all other properties reduce back to the splittings, indefiniteness doesn’t. For example, a connected sum of manifolds with a positive and negative definite intersection form has indefinite intersection form. Definite forms are extremely numeruous and therefore lack a simple classification.

Example

Further improvement of the 11/8 conjecture isn’t possible as shown by the simply connected oriented closed K3 surface K3V(z 0 3+z 1 3+z 2 3+z 3 3)P 3K3\coloneqq V(z_0^3+z_1^3+z_2^3+z_3^3)\subset\mathbb{C}P^3 with intersection form:

Q K3=2E 83H. Q_{K3} =-2E_8 \oplus 3H.

It therefore has b 2(K3)=22b_2(K3)=22 and σ(K3)=16\sigma(K3)=16, meaning the 11/8 conjecture is an equality for it. Taking connected sums #mK3\#m K3 including reversing orientation (corresponding to flipping the sign of mm) furthermore produces examples for equalities for every mm\in\mathbb{Z}.

Articles on geometry and topology of 4-manifolds:

References

Last revised on May 16, 2026 at 14:29:03. See the history of this page for a list of all contributions to it.