# 4-manifolds: 1-connected

## 1 Introduction

Any finitely presentable group may occur as the fundamental group of a smooth closed 4-manifold. On the other hand, the class of simply connected (topological or smooth) 4-manifolds still appears to be quite rich, so it appears reasonable to consider the classification of simply connected 4-manifolds in particular.

It appears that the intersection form is the main algebro-topological invariant of simply-connected 4-manifolds.

## 2 Construction and examples, their intersection forms

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### 2.1 First examples


The intersection form of the 4-sphere is the "empty form" of rank 0. The intersection forms of the others are given by

$\displaystyle \begin{array}{ccc} q_{\mathbb{CP}^2} & = & ( \ 1 \ ) , \\ \\ q_{\overline{\mathbb{CP}^2}} & = & ( \, -1 \ ) , \\ \\ q_{S^2 \times S^2} \ & = & \ \begin{pmatrix} \ 0 \ & \ 1 \ \\ 1 & 0 \end{pmatrix} , \\ \\ q_{\mathbb{CP}^2 \# \overline{\mathbb{CP}^2}} \ & = & \ \begin{pmatrix} \ 1 \ & \ 0 \ \\ 0 & -1 \end{pmatrix} . \end{array}$

The manifolds $S^2 \times S^2$$S^2 \times S^2$ and $\mathbb{CP}^2 \# \overline{\mathbb{CP}^2}$$\mathbb{CP}^2 \# \overline{\mathbb{CP}^2}$ both have indefinite intersection forms of same rank and signature, but of different type. Therefore they are not homotopy-equvialent.

### 2.2 Hypersurfaces in CP3

For an integer $d \geq 1$$d \geq 1$ we define a subset $S_d$$S_d$ of $\mathbb{CP}^3$$\mathbb{CP}^3$ by the formula

$\displaystyle S_d = \{ X_0^d + X_1^d + X_2^d + X_3^d = 0 | [X_0:X_1:X_2:X_3] \in \mathbb{CP}^3 \} .$

It is easy to check that in each chart of $\mathbb{CP}^3$$\mathbb{CP}^3$ the $S_d$$S_d$ is cut out transversally by the homogeneous polynomial of degree $d$$d$. Therefore, $X_d$$X_d$ is a submanifold, in fact, an algebraic hypersurface. This is a special case of a complete intersection.

By the Lefschetz hyperplane section theorem the hypersurface $S_d$$S_d$ is simply connected. Its intersection form may be computed as follows: First one computes the Chern classes of $S_d$$S_d$. Evaluating the second Chern class on the fundamental class $[S_d]$$[S_d]$ yields the Euler characteristic and therefore the rank of $H^2(S_d;\mathbb{Z})$$H^2(S_d;\mathbb{Z})$. Likewise, by computing the Pontryagin class and using the Hirzebruch signature theorem, stating that for a closed oriented 4-manifold $X$$X$ one has

$\displaystyle \text{sign}(X) = \frac{1}{3} \langle p_1(TX), [X] \rangle \ ,$

one computes the signature of $S_d$$S_d$. Whether the intersection form is even or odd may be seen from the second Stiefel-Whitney class $w_2(S_d) = c_1(S_d) \ (\text{mod} \ 2)$$w_2(S_d) = c_1(S_d) \ (\text{mod} \ 2)$.

There are three facts that we need to use:

• The normal bundle $\nu_{S_d}$$\nu_{S_d}$ of $S_d$$S_d$ in $\mathbb{CP}^3$$\mathbb{CP}^3$ is given by $H^d$$H^d$, where $H$$H$ is the line bundle dual to the hyperplane $\mathbb{CP}^2 \subseteq \mathbb{CP}^3$$\mathbb{CP}^2 \subseteq \mathbb{CP}^3$. Its first Chern class $h$$h$ generates the cohomology ring of $\mathbb{CP}^3$$\mathbb{CP}^3$.
• The hypersurface $S_d$$S_d$ is Poincaré dual to the class $d h$$d h$, or equivalently $\langle h^2,[S_d] \rangle = d$$\langle h^2,[S_d] \rangle = d$.
• The total Chern class of $S_d$$S_d$ is given by
$\displaystyle c(T\mathbb{CP}^3) = (1+h)^4 .$

We can now apply the Whitney sum formula for the total Chern class to the splitting $T\mathbb{CP}^3 |_{S_d} = TS_d \oplus \nu_{S_d}$$T\mathbb{CP}^3 |_{S_d} = TS_d \oplus \nu_{S_d}$,

$\displaystyle c(T\mathbb{CP}^3 |_{S_d}) = c(TS_d) c(\nu_{S_d}) = c(TS_d) (1+dh)$

which we can invert to obtain the formula

$\displaystyle c(TS_d) = (1+h)^4 (1-dh + d^2 h^2) ,$

and in particular

$\displaystyle c_1(TS_d) = (4-d)h, \; \text{ and } \;\; c_2(TS_d) = (6-4d +d^2) h^2.$

We compute the Euler characteristic $\chi(S_d) = \langle c_2(S_d), [S_d] \rangle = (6-4d+d^2)d$$\chi(S_d) = \langle c_2(S_d), [S_d] \rangle = (6-4d+d^2)d$ by the above mentioned fact. The first Pontryagin class $p_1(TS_d) = -c_2(TS_d \oplus \overline{TS_d}) = (4-d^2) h^2$$p_1(TS_d) = -c_2(TS_d \oplus \overline{TS_d}) = (4-d^2) h^2$ yields the signature $\text{sign}(S_d) = \frac{1}{3} (4-d^2)d$$\text{sign}(S_d) = \frac{1}{3} (4-d^2)d$. We summarise

$\displaystyle b_2(S_d) = (6-4d+d^2)d - 2 \; \text{ and } \;\; \text{sign}(S_d) = \frac{1}{3} (4-d^2)d$

Furthermore $S_d$$S_d$ is spin if and only if $d$$d$ is even. This is because we have

$\displaystyle w_2(TS_d) = d h \ (\text{mod} \ 2) ,$

and because the inclusion $S_d \hookrightarrow \mathbb{CP}^3$$S_d \hookrightarrow \mathbb{CP}^3$ yields an injective restriction map in second cohomology with $\mathbb{Z}/2$$\mathbb{Z}/2$ coefficients because of the hypersection theorem.

Summarising the above discussion, we have

$\displaystyle q_{S_d} = \frac{1}{3}(d^3-6d^2+11d-3)\ (\ 1 \ ) \oplus \frac{1}{3} (2d^3 - 6d^2 +7d -3) \ (\ -1 \ )$

for $d$$d$ odd, and

$\displaystyle q_{S_d} = \frac{1}{24} d(d^2-4)\, (-E_8) \, \oplus \frac{1}{3}(d^3-6d^2 +11d -3) \, H$

A particularly interesting special case is that of $d=4$$d=4$. The surface $S_4$$S_4$ is a K3 surface. It is spin, has signature $-16$$-16$, and has $b_2=22$$b_2=22$. By the classification results of indefinite intersection forms we know that the intersection form of $S_4$$S_4$ is given by

$\displaystyle q_{S_4} = -2 E_8 \oplus 3 H .$

Blowing up the surface $S_4$$S_4$ yields the manifold $S_4 \# \overline{\mathbb{CP}^2}$$S_4 \# \overline{\mathbb{CP}^2}$ which now has an odd intersection form given by

$\displaystyle q_{S_4 \# \overline{\mathbb{CP}^2}} = 3 \ (\ 1\ ) \oplus 20 \ (\ -1\ ),$

the same form as that of the 4-manifold $3 \mathbb{CP}^2 \# 20 \overline{\mathbb{CP}^2}$$3 \mathbb{CP}^2 \# 20 \overline{\mathbb{CP}^2}$. Below we shall see that these two 4-manifolds are homeomorphic by Freedman's classification results, but not diffeomorphic because they have different Seiberg-Witten invariants.

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## 4 Topological classification

By early work of Milnor and Whitehead the following theorem was known since 1958, and is based on a generalised Pontryagin-Thom construction:

Theorem 4.1 (Milnor, Whitehead). Two simply connected 4-manifolds are homotopy equivalent if and only if they have isomorphic intersection forms.

For the classification of topological 4-manifolds Freedman achieved to use surgery theory in order to establish his famous

Theorem 4.2 (Freedman).

• Two simply-connected closed topological 4-manifolds are homeomorphic if and only if they have isomorphic intersection forms and the same Kirby-Siebenmann invariant.
• Given any even unimodular symmetric bilinear form $q$$q$ over $\mathbb{Z}$$\mathbb{Z}$ there is, up to homeomorphism, a unique simply connected topological 4-manifold with intersection form $q$$q$.
• Given any odd unimodular symmetric bilinear form $q$$q$ over $\mathbb{Z}$$\mathbb{Z}$ there are, up to homeomorphism, precisely two simply connected topological 4-manifolds with intersection form $q$$q$. One of them has non-trivial Kirby-Siebenmann invariant and therefore cannot be given a smooth structure.

## 5 Non-existence results for smooth 4-manifolds

Theorem 5.1 (Rohlin). A smooth closed 4-manifold that is spin, and therefore has even intersection form, has its signature divisible by 16.

By Freedman's theorem we know that there are closed simply connected topological 4-manifolds $X$$X$ with even intersection form and $\text{sign}(X)/8 \ \equiv 1 \ (\text{mod} \ 2)$$\text{sign}(X)/8 \ \equiv 1 \ (\text{mod} \ 2)$, as for instance the $E_8$$E_8$ manifold $X_{E_8}$$X_{E_8}$. By Rohlin's theorem these manifolds cannot admit a smooth structure. However, the manifold $X_{E_8} \# X_{E_8}$$X_{E_8} \# X_{E_8}$ might so because its signature is equal to 16.

Using methods from gauge theory Donaldson was able to prove his famous theorem on the intersection form of smooth definite 4-manifolds:

Theorem 5.2 (Donaldson). Let $X$$X$ be a smooth closed 4-manifold with definite intersection form $q_X$$q_X$. Then $q_X$$q_X$ is diagonal, i.e. it is the direct sum of the 1-dimensional forms $( \ 1 \ )$$( \ 1 \ )$ in the positive definite case, and of the forms $( \ -1 \ )$$( \ -1 \ )$ in the negative definite case.

By Donaldson's theorem, the topological 4-manifold $X_{E_8} \# X_{E_8}$$X_{E_8} \# X_{E_8}$ does not admit a smooth structure either, neither does any of the manifolds $X_{E_8} \# X_{E_8} \# \mathbb{CP}^2 \# \dots \# \mathbb{CP}^2$$X_{E_8} \# X_{E_8} \# \mathbb{CP}^2 \# \dots \# \mathbb{CP}^2$ with odd intersection form. In fact, it is an algebraic theorem due to Eichler [Milnor&Husemoller1973] that there is a unique decomposition theorem for definite forms, and so no two of the forms $k E_8 \oplus k (\ 1 \ )$$k E_8 \oplus k (\ 1 \ )$ with different values of $k, l \in \mathbb{N}$$k, l \in \mathbb{N}$ are isomorphic.

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