Seiberg-Witten

Section 1 4-dimensional manifolds

We are interested in classifying $4$ -manifolds. Without further constraints on the manifold, this is impossible, given the following theorems:

THM 1.1

There is no algorithm which, given a presentation of a group, whether or not it’s trivial.

THM 1.2

Given a finitely presented group

𝐺

, there exists a smooth

4

-manifold with fundamental group

𝐺

Suppose

𝐺 = ⟨ 𝑔_{1}, \dots, 𝑔_{𝑛} : 𝑟_{1}, \dots, 𝑟_{𝑛} ⟩

. By Seifert-van Kampen theorem, given

4

-manifolds

𝑀

and

𝑁

𝜋_{1} (𝑀 #︎ 𝑁) = 𝜋_{1} (𝑀) * 𝜋_{1} (𝑁)

. Consider

{#︎}_{𝑛} 𝕊^{1} \times 𝕊^{3}

; it has fundamental group

𝑁 = ℤ^{* 𝑛}

. Every relation

𝑟_{𝑖}

then represents a loop

𝛾_{𝑖}

𝑁

. To quotient them out, first find a tubular neighbourhood

𝑈_{𝑖}

𝛾_{𝑖}

for every

𝑖

. It is homeomorphic to

𝕊^{1} \times 𝔻^{3}

, with boundary

𝕊^{1} \times 𝕊^{2}

. Remove

𝑈_{𝑖}

from

𝑁

and in its place we glue back

𝔻^{2} \times 𝕊^{2}

, whose boundary is also

𝕊^{1} \times 𝕊^{2}

. The latter is contractible, so again by Seifert-van Kampen theorem the resulting manifold has fundamental group

𝐺

This argument does not work for dimension $3$ , for the following reason:

THM 1.3

There is no simply-connected

3

-manifold with boundary

𝕊^{1} \times 𝕊^{1}

All compact odd-dimensional manifolds satisfy

𝜒 (𝑀) = \frac{1}{2} 𝜒 (\partial 𝑀)

. Since

\partial 𝑀 = 𝕊^{1} \times 𝕊^{1}

𝜒 (𝑀) = \frac{1}{2} 𝜒 (𝕊^{1} \times 𝕊^{1}) = 0

. An analysis on homology groups shows a contradiction.

Therefore, we put our focus on a smaller subset of $4$ -manifold, simply-connected closed $4$ -manifolds $𝑀$ . Some observations are immediate. First, since $𝜋_{1} (𝑀) = 0$ , we know $𝐻_{1} (𝑀) = 𝐻^{1} (𝑀) = 0$ , and by Poincaré duality $𝐻_{3} (𝑀) = 0$ and $𝐻_{2} (𝑀)$ is torsion-free. Second, again by Poincaré duality, $𝐻_{2} (𝑀) ≅ 𝐻^{2} (𝑀)$ , which is the only interesting homology group.

DEF 1.4 intersection form

The Poincaré pairing map

𝑄 : 𝐻_{2} \times 𝐻_{2} \to ℤ

is called the intersection form. It always has determinant

\pm 1

, and is a perfect symmetric bilinear form.

For example,

$𝑋 = 𝕊^{4}$ , $𝑄 = 0$ ;
$𝑋 = {ℂ P}^{2}$ , $𝑄 = (\begin{matrix} 1 \end{matrix})$ ;
$𝑋 = {ℂ P}^{2}$ ( ${ℂ P}^{2}$ with reversed orientation), $𝑄 = (\begin{matrix} - 1 \end{matrix})$ ;
$𝑋 = 𝕊^{2} \times 𝕊^{2}$ , $𝑄 = (\begin{matrix} 1 \\ 1 \end{matrix})$ ;
$𝑋 = {ℂ P}^{2} #︎ {ℂ P}^{2}$ , $𝑄 = (\begin{matrix} 1 \\ - 1 \end{matrix})$ ;
Both $(𝕊^{2} \times 𝕊^{2}) #︎ {ℂ P}^{2}$ and ${ℂ P}^{2} #︎ 2 {ℂ P}^{2}$ has $𝑄$ congruent (over $ℤ$ ) to $(\begin{matrix} 1 \\ - 1 \\ - 1 \end{matrix})$ .

$𝑄$ is called intersection form for the following reason.

THM 1.5

Suppose

𝑋

is a smooth

4

-manifold. Then every homology class in

𝐻_{2} (𝑋)

is represented by some smoothly embedded

2

-surface, and the Poincaré pairing of two homology classes corresponds to the intersection index of two transverse such surfaces.

There exists an isomorphism from the set of complex line bundles over $𝑋$ to $𝐻^{2} (𝑋; ℤ)$ , taking a complex line bundle to its first Chern class. This is because complex $𝑚$ -dimensional vector bundles $𝐿$ on $𝑋$ correspond to homotopy classes of maps from $𝑋$ to the classification space ${Gr}_{𝑚} (ℂ^{\infty})$ $Gr$ denotes the Grassmanian, and when $𝑚 = 1$ , ${Gr}_{𝑚} (ℂ^{\infty}) = {ℂ P}^{1} = 𝐾 (ℤ, 2)$ . Therefore, complex line bundles corresponds to $⟨ 𝑋, 𝐾 (ℤ, 2) ⟩ ≅ 𝐻^{2} (𝑋; ℤ)$ .

Therefore, for a homology class $𝛼 \in 𝐻^{2} (𝑋; ℤ)$ , consider a complex line bundle $𝐿 \to 𝑋$ representing $𝛼$ . Let $𝑠$ be a generic section $𝑋 \to 𝐿$ transverse to the zero section. In real dimensions, $dim 𝑋 = 4$ , $dim 𝐿 = 6$ , so $𝑠^{- 1} (0)$ has dimension $2$ . This submanifold corresponds to $𝛼$ .

Alternatively, $𝐻^{2} (𝑋; ℤ) ≅ [𝑋, {ℂ P}^{\infty}] ≅ [𝑋, {ℂ P}^{2}]$ , and the pullback of ${ℂ P}^{1} \subseteq {ℂ P}^{2}$ is the desired submanifold.

The question is then how much information does the intersection form carry. In other words, whether it determines homotopy / homeomorphism / diffeomorphism type. On homotopy type the following theorem gives the answer:

THM 1.6 Whitehead’s theorem

Two closed simply-connected

4

-manifolds are homotopy equivalent if and only if their intersection forms are congruent over

ℤ

On homeomorphism type, the answer depends on parity and smoothness. A bilinear form $𝐴 : ℤ^{𝑟} \times ℤ^{𝑟} \to ℤ$ is called even if $𝐴 (𝑎, 𝑎) \equiv 0 (mod 2)$ for all $𝑎$ , and odd otherwise. Parity is an invariant under congruence.

THM 1.7 (Freedman, 1982)

For any unimodular symmetric bilinear form $𝑄$ , there exists a topological simply-connected closed $4$ -manifold $𝑋$ with intersection form $𝑄$ ;
if $𝑄$ is even, then $𝑋$ is unique up to homeomorphism; if $𝑄$ is not even, then there exists exactly $2$ such $𝑋$ , and at most one of them admits a smooth structure.

Therefore, simply-connected closed smooth $4$ -manifolds are determined uniquely by their intersection form up to homeomorphism.

Diffeomorphism type is much more complicated and is core to the study of $4$ -manifolds. In dimension $3$ , every topological manifold admits a unique smooth structure up to diffeomorphism (Moise). This is far from true in dimension $4$ ; for example, $ℝ^{𝑛}$ has a unique smooth structure except for $𝑛 = 4$ , where one gets infinitely many non-diffeomorphic smooth structures on $ℝ^{4}$ (Donaldson et al.). If $𝑋$ is a closed $4$ -manifold, it admits at most countably many smooth structures, and there are closed $4$ -manifolds which admits countably infinitely many smooth structures, for example ${ℂ P}^{2} #︎ 𝑘 {ℂ P}^{2}$ . We don’t even know whether there are $4$ -manifolds with only finitely many smooth structures. For another example, for $𝑛 = 1, 2, 3, 5, 6$ , $𝕊^{𝑛}$ has a unique smooth structure; for $𝑛 = 7$ we have $28$ smooth structures, given by Milnor’s exotic spheres; for $𝑛 = 4$ this is still open.

The next question is naturally the classification of unimodular symmetric bilinear forms over $ℤ$ . Over $ℝ$ , the classification of perfect symmetric bilinear forms is rather simple: they are classified by their signature. Over $ℤ$ things are more complicated and more invariants are involved.

THM 1.8

If $𝑄$ is indefinite and odd, then $𝑄 = 𝑚 \cdot (\begin{matrix} 1 \end{matrix}) \oplus 𝑛 \cdot (\begin{matrix} - 1 \end{matrix})$ ( $𝑚, 𝑛 > 0$ );
if $𝑄$ is indefinite and even, then $𝑄 = 𝑚 \cdot (\begin{matrix} 1 \\ 1 \end{matrix}) \oplus 𝑛 \cdot 𝐸_{8}$ ( $𝑚 > 0$ ), where
$𝐸_{8} = (\begin{matrix} 2 & 0 & - 1 \\ 0 & 2 & 0 & - 1 \\ - 1 & 0 & 2 & - 1 \\ - 1 & - 1 & 2 & - 1 \\ - 1 & 2 & - 1 \\ - 1 & 2 & - 1 \\ - 1 & 2 & - 1 \\ - 1 & 2 \end{matrix}) .$

For $𝑄$ definite, things are much more definite and is not determined by parity, rank and signature alone; for example, $9 \cdot (\begin{matrix} 1 \end{matrix})$ and $𝐸_{8} \oplus (\begin{matrix} 1 \end{matrix})$ are both odd, positive definite and has rank $9$ , but they are not equivalent over $ℤ$ .

When can $𝑄$ become the intersection form of a smooth $4$ -manifold?

THM 1.9 (Roknlin, 1952)

𝑋

is smooth and simply-connected and

𝑄

is even, then

16

divides the signature of

𝑄

Using this we can construct topological manifolds without smooth structures, for example, the topological manifold with intersection form $𝐸_{8}$ . But what about $𝐸_{8} \oplus 𝐸_{8}$ ? It has signature $16$ and is even, but still it doesn’t correspond to any smooth $4$ -manifold.

THM 1.10 Donaldson diagonalization theorem

𝑋

is a closed simply-connected

4

-manifold and if

𝑄

is definite, then it is diagonalizable over

ℤ

. Therefore,

𝑄 = \pm 𝑟 \cdot (\begin{matrix} 1 \end{matrix})

Therefore, although algebraically we failed at classifying definite forms, the geometric constraint solves the problem. Putting everything together,

COR 1.11

The homeomorphism type of a simply-connected closed smooth

4

-manifold

𝑋

is determined uniquely by the signature

𝜎 (𝑄)

, parity of

𝑄

, and

𝜒 (𝑋)

. Equivalently, it’s determined uniquely by the positive and negative index of

𝑄

and parity of

𝑄

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1 Section 1 4-dimensional manifolds

Section 1 4-dimensional manifolds