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\title{Dennis Sullivan Course Notes\\March 18, 2005}
\author{Gabriel C. Drummond-Cole}
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Actually, you guys should prove this exercise.

\begin{exercise}
Let $\Lambda=\{\lambda\}$ be any subset.  Let $L_\lambda$ be a straight line passing through $\lambda$ and not vertical; suppose there exists an $\epsilon>0$ such that for $\lambda_1\ne \lambda_2$ then $L_{\lambda_1}$ does not intersect $L_{\lambda_2}$ for $-\epsilon<x<\lambda.$  Then if $\bar{\Lambda}$ is the closure of $\Lambda,$ the system of lines for $\Lambda$ extends uniquely to a system of lines for $\bar{\Lambda}$ with the same $\epsilon$-disjointness property.
\end{exercise}

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 $\langle\rangle,\ T$ with symmetric skew-symmetric $\langle\rangle,\ F$ with symmetric $\langle\rangle,$ and $T$ with skew symmetric $\langle\rangle.$

In the last case we have $A\oplus A\oplus \mathbb{Z}_2$ or $A\oplus A.$  In the first case we have that $F$ has even rank and things pair up canonically as $\left(\begin{array}{cc}0&1\\-1&0\end{array}\right);$ in the third case we have a symmetric matrix; in the second we had things like $\left(\begin{array}{cc}\epsilon_1&1\\1&\epsilon_2\end{array}\right)$ and in the fourth $\left(\begin{array}{cc}t&1\\1&t\end{array}\right)$ 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.  
$$
\left(\begin{array}{cccccccc}
2&&&1&&&&\\
&2&1&&&&&\\
&1&2&&&&&\\
1&&1&2&1&&&\\
&&&1&2&1&&\\
&&&&1&2&1&\\
&&&&&1&2&1\\
&&&&&&1&2
\end{array}\right)
$$
Anyone ever taken a determinant of an $8\times 8$ 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 $\mathbb{Z}/2$ 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 four-manifold 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]
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X = a0 + bI +cV

X  = a^2(=) + b^2(=) + ab|| + ab (=o)+ c^2||+ (ac+bc)* + (ac+bc)(=) virt/reg
X 


a^2+b^2+ab(z)+ac+bc=0.

ab+c^2 = 1
(a+b)c=0

b=-a


2a^2 - a^2z = 0, so z=2a

c^2-a^2=1

az+b+c= 1

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