Dennis Sullivan Course Notes March 18, 2005

Gabriel C. Drummond-Cole

April 15, 2005

Actually, you guys should prove this exercise.

Exercise 1 Let Λ = {λ} be any subset. Let L_{λ} be a straight line passing through λ and not vertical; suppose there exists an ϵ > 0 such that for λ_{1} ≠ λ_{2} then L_{λ_{1}} does not intersect L_{λ_{2}} for −ϵ < x < λ. Then if Λ̄ is the closure of Λ, the system of lines for Λ extends uniquely to a system of lines for Λ̄ with the same ϵ-disjointness property.

All right, so we have these four dimensions, 4i + 2, 4i − 1, 4i, 4i + 1. In the middle dimension we then have F with skew-symmetric ⟨⟩, T with symmetric skew-symmetric ⟨⟩, F with symmetric ⟨⟩, and T with skew symmetric ⟨⟩.

In the last case we have A ⊕ A ⊕ ℤ_{2} or A ⊕ A. In the first case we have that F has even

rank and things pair up canonically as ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ); in the third case we have a symmetric

matrix; in the second we had things like ( \begin{matrix}ϵ_{1} & 1 \\ 1 & ϵ_{2} \end{matrix} ) and in the fourth ( \begin{matrix}t & 1 \\ 1 & t \end{matrix} ) for order

two t mod 2^{n}, and lines.

The realization theorem I think is the following. I’m 100% sure it’s true, and I have 98% of a proof.

For 4i + 2 all are realizable by closed manifolds. All 4i − 1 are realizable as well. This is a little harder. For 4i, this is pretty interesting. There’s a natural partition in 4i − 1; type one means some a_{ii} is odd. The other is all a_{ii} are even. It’s not so easy to write down a matrix of determinant one where all the diagonals are even.

( \begin{matrix}2 & & & 1 \\ & 2 & 1 \\ & 1 & 2 \\ 1 & & 1 & 2 & 1 \\ & & & 1 & 2 & 1 \\ & & & & 1 & 2 & 1 \\ & & & & & 1 & 2 & 1 \\ & & & & & & 1 & 2 \end{matrix} )

Anyone ever taken a determinant of an 8 × 8 matrix?

[I’ve done it on a computer.]

I’ve done a four by four. There’s an eight dimensional manifold related to E_{8} whose boundary is combinatorially equivalent to the sphere, but an exotic seven-sphere. It has this as its intersection matrix.

Show the determinant is one.

So over 4i type two is realizable in all dimensions. Type one is realizable only in dimensions four, eight, or sixteen.

For 4i + 1 let’s call the extra ℤ/2 type I and the lack of it type II. Then type one is realizable only for i = 1 in dimension five. I proved that a realization is diffeomorphic to the manifold SU(3)/SO(3).

The rest later.

I’m not saying, for 4i, that you can get these smoothly; I’m only saying it combinatorially. I’m only saying it for topological manifolds. People are still hammering away at dimension four. The K_{3} surface is like the torus, a very simple object. It’s a simply connected fourmanifold with second Betti number 22. It has matrix two E_{8} and three of the simple planes. There’s a whole school of mathematics that tries to get rid of the plane blocks.

I’ll just give you a construction that gives something in dimension four, but not a manifold. I think I gave the construction. Take a 4i-ball and glue on 2i-handles.

[lost]