Failure of the H-Cobordism Theorem in 4 Dimensions

This is a follow-up on Donaldson’s work in gauge theory about the failure of the h-bordism theorem for 4-manifolds. The main example utilizes a K3 surface. Since all K3 surfaces are diffeomorphic by Kodaira’s work, we can use any K3 for our purposes. Recall that $(b^+_2,b^-_2) = (3,19)$ since the intersection form is $3H\oplus 2(-E_8)$.

When studying ASD instantons, sometimes the moduli spaces are 0-dim and this lets us count the moduli space with signs to get a $\mathbb{Z}$-invariant $q$. There is a choice of sign for $q$ but if we fix an orientation $\Omega$ of $H^2_+(X)$, then there is no ambiguity. Normally, there would also be dependence on a lift of some mod 2 class $\alpha$ but since the intersection form of $X$ is even, there is no such dependence. We’ll give more details below.

Proposition: There is a class $\alpha$ and orientation $\Omega$ such that $q(\alpha,\Omega) = 1$.

We’ll sketch the proof below but first, there is a corollary:

Corollary 1: There is no diffeomorphism of $X$ which acts trivially on $H^2(X,\mathbb{Z}/2)$ but reverses the orientation of the positive part of $H^2(X,\mathbb{R})$.

Proof: The naturality of $q$ ensures this (this invariant is an orientation-preserving diffeomorphism invariant). Suppose we had such a diffeomorphism $f:X \to X$. If it reverses overall orientation, then the subspaces $H^+,H^-$ would not be preserved and therefore, not be preserved mod 2. But $f$ acts trivially on $H^2(X,\mathbb{Z}/2)$. So $f$ is orientation-preserving. By naturality, $q(f^*\alpha,f^*\Omega) = q(\alpha,\Omega)$. On the other hand, $f^* \alpha = \alpha$ but $f^* \Omega = -\Omega$ by hypothesis. So $q(\alpha,\omega) = -q(\alpha,\Omega)$ and implies it equals 0, contradicting the proposition above. $\square$

A corollary of this is the following:

Corollary 2: There is a simply-connected 5-dim h-cobordism which is not diffeomorphic to a product.

Proof: Recall that an h-cobordism $W$ between manifolds $X,Y$ is a cobordism such that inclusion of $X$ and $Y$ into $W$ are homotopy equivalences. In particular, the h-cobordism $W$ induces a map $f_W: H^2(X,\mathbb{Z}) \to H^2(Y,\mathbb{Z})$ that preserves the intersection forms of $X$ and $Y$. It turns out that when $X,Y$ are both simply connected, the converse holds: any map $f:H^2(X,\mathbb{Z}) \to H^2(Y,\mathbb{Z})$ which preserves the intersection forms can be realized as a map $f=f_W$ that comes from an h-cobordism $W$.

For us, we’ll take $X=Y$ to be a K3 surface; If we take the map $-1$, then $Q(-\alpha,-\beta) = Q(\alpha,\beta)$ so $-1$ preserves the intersection form and hence, is realized by some h-cobordism $W$. Thus, we have boundary inclusions $i_X,i_Y$ with $i^*_X = -i^*_Y$. If this cobordism $W$ were diffeomorphic to $X \times [0,1]$, then $W$ would be an h-cobordism and hence, we would obtain a diffeomorphism $f:X \to X$ which realizes $-1$ on $H^2(X)$. This map would be trivial on $\mathbb{Z}/2$ but reverse the orientation of $H^+$. This contradicts our Corollary 1. $\square$

Remark: This example is great because it demonstrates that the h-cobordism theorem fails in dim 4. We know that there is some version of it in the topological category by Freedman’s work but it cannot be strengthened to the smooth category. Before we prove the proposition, we’ll give one more corollary.

Corollary 3: There does not exist a connected sum decomposition of a K3 surface as $X = Y \# (S^2 \times S^2)$.

Proof: On $S^2 \times S^2$, we have an antipodal map on each factor and together, we get a diffeomorphism which induces $-1$ on homology. When we have two manifolds $Y,Z$ and diffeomorphisms $f:Y \to Y,g:Z \to Z$, we can take a “connected sum” of $f$ and $g$ to get a diffeomorphism on $Y \#Z$ in the sense that the diffeomorphism induces the map $f^* \oplus g^*$ on cohomology. The construction is easy; just isotope the diffeomorphisms to be the identity map on small balls and then connect sum in that region. So if we had such a connected sum decomposition of $X$, we would have a diffeomorphism which acts as $-1$ on $H = \begin{pmatrix}0 & 1 \ 1 & 0 \end{pmatrix}$ and $+1$ on the complement. This would reverse the orientation of the positive part $H^+$ but act trivially mod 2. Hence, it would contradict Corollary 1. $\square$

Sketch of a proof of the proposition: There is a standard construction of a branched double cover of $\mathbb{CP}^2$ by a K3 with branched locus being a sextic curve. Let $\pi:X \to \mathbb{CP}^2$ denote this branched double cover. Then $E=\pi^*T\mathbb{CP}^2 \to X$ has an associated holomorphic $SO(3,\mathbb{C})$ bundle $P$; it comes from the trace-free endomorphisms of $E$. A computation of the 1st Pontryagin class shows:

$-p_1(P) = 4c_2(E) - c_1(E)^2=2(4c_2(\mathbb{CP}^2) - c_1(\mathbb{CP}^2)^2) = 2(4\cdot 3-9)=6$. The extra 2 comes from the fact that $\pi$ is a branched double cover. So $p_1(P)=-6$. Also, $\alpha = w_2(P) = \pi^* h$ where $h$ generates $H^2(\mathbb{CP}^2)$. Since $\alpha \cdot \alpha = 1$ and $\pi$ is a deg 2 map, then $\alpha^2 \equiv 2 \pmod{4}$.

The reason we want $p_1=-6$ is because we’re going to study ASD $SO(3)$-connections on this bundle and build a moduli space for it. The dimension formula for simply connected 4-manifolds is: $\dim \mathcal{M}_P (g) = -2p_1(P) − 3(1 + b^+_2)$. This formula is quite similar to that of $SU(2)$ bundles; there, we would replace $-2p_1$ with $8c_2$. In our situation with K3, we have $-2(-6)-3(1+3) = 0$. Hence, we’ll be able to count the moduli space and produce an integer invariant $q(\alpha,\Omega)$.

By some hard work which can be seen in Donaldson-Kronheimer Ch. 6, the moduli space of ASD $SO(3,\mathbb{C})$-connections on $P$ can be identified with the moduli space of holomorphic 2-plane bundles topologically equivalent to $E$ which are stable with respect to the ample line bundle $\pi^*\xi$, that is, the tautological line bundle (also called the Hopf bundle) $\xi \to \mathbb{CP}^2$. We need some definitions.

Ampleness means that there is an embedding $X \to \mathbb{CP}^N$ such that the hyperplane class $[X \cap \mathbb{CP}^{n-1}]$ is a positive multiple of $c_1(\pi^* \xi) \in H^2(X)$.

A holomorphic $SL(2,\mathbb{C})$-bundle $E \to X$ ($X$ can be any complex surface with some Kähler metric) is stable if, for each holomorphic line bundle $L \to X$ for which there is a nontrivial holomorphic map $E \to L$, the degree $\deg L > 0$. Donaldson-Kronheimer show that $T\mathbb{CP}^2$ is stable and moreover, that there is only one such holomorphic 2-plane bundle topologically equivalent to $E$ that is stable with respect to $\pi^* \xi$, namely, $E$ itself. Thus, the moduli space has a single point and so we have an invariant $q(\alpha,\Omega) = +1$. $\square$

Example From LeBrun

In this paper, the main subject is that of Einstein manifolds. A Riemannian metric $g$ is said to be Einstein if its Ricci curvature, considered as a function on the unit tangent bundle, is constant. This is equivalent to requiring that the Ricci tensor $r$ of $g$ satisfy $r = \lambda g$ for some real number $\lambda$. In the original context of general relativity, where the metric is taken to be Lorentzian rather than Riemannian, and the equation above is called the Einstein vacuum equation with cosmological constant. In fact, Einstein did briefly consider the possibility that the constant $\lambda$ might be non-zero, but he eventually came to consider this the “biggest blunder of his life.”

Question: For a fixed dimension $n$, if a manifold admits two Einstein metrics with constants $\lambda_1,\lambda_2$, are they of the same sign?

Example: For $n=2,3$, sectional curvature is automatically constant and so the sign of the Einstein constants are predetermined by the fundamental group of the manifold.

For dimension $n=4$, suppose that $X$ and $Y$ are two smooth closed oriented simply-connected non-spin 4-manifolds that have the same Euler characteristic and signature. In light of Donaldson’s thesis and the classification of symmetric bilinear forms over the integers, a classic result of Terry Wall implies that $X$ and $Y$ are then in fact $h$-cobordant, meaning that there is a smooth compact oriented 5-manifold $W$ with $\partial W = \overline{X} \sqcup Y$ such that the inclusions $X,Y \hookrightarrow W$ are homotopy equivalences. But this implies that $X \times X$ and $Y \times Y$ are also $h$-cobordant, via the 9-manifold $(W \times X) \cup_{Y \times X} (Y \times W)$. Since the simply connected-manifold $X \times X$ has dimension $8 > 4$, Smale’s $h$-cobordism theorem implies that $X \times X$ and $Y \times Y$ are diffeomorphic. Similarly, the $h$-cobordism theorem also implies that the $k$-fold Cartesian products $X^k = X \times …\times X$ and $Y^k$ (similarly defined) are diffeomorphic for $k \geq 2$. However, we can find $X,Y$ so that they have different Seiberg-Witten invariants and so the $h$-cobordism theorem fails to hold in dimension 4.

What are these $X,Y$? Rebecca Barlow found a non-singular simply-connected complex surface $B$ with $c^2_1 = 1$ and $h^{2,0}=0$; the surface is non-spin and has Euler characteristic $11$ and signature $-7$. The surface is minimal; i.e. no rational curves with self-intersection $-1$. It does contain four rational curves with self-intersection $-2$; for this reason, it does not have $c_1 < 0$. The Aubin/Yau theorem implies that generic deformations of $B$ all have compatible Kähler-Einstein metrics with $\lambda < 0$. So the underlying smooth 4-manifold of the Barlow surface $B$ carries some Einstein metric $g^-$ with $\lambda < 0$. All Cartesian products $B^k$ have an Einstein product metric $g^- \oplus…\oplus g^-$ with the same negative Einstein constant.

There is also a family of del Pezzo surfaces $S$ with $c_1 > 0$, simply connected, and $c^2_1 = 1, h^{2,0} = 0$; del Pezzo just means Fano which is having an ample anticanonical divisor. So the Euler characteristic and signature are the same as that of $B$. These specific del Pezzo surfaces are diffeomorphic to $\mathbb{CP}^2 \# 8 \overline{\mathbb{CP}}^2$ and Tian/Yau have shown some have compatible Kähler-Einstein metrics with $\lambda > 0$. So the underlying 4-manifold $S$ admits an Einstein metric $g^+$ with $\lambda > 0$.

Using the Cartesian product argument above, we get the following result:

Theorem (Catanese-LeBrun): For each $k \geq 2$, there is a closed simply-connected $4k$-manifold $M$ that admits a pair of Einstein metrics with Einstein constants of opposite sign.

This result has been improved so that there are at least ${k+3 \choose k}$ examples of $4k$-manifolds admitting Einstein metrics with both signs.

Appendix: Constructing a branched double cover $\large X \to \mathbb{CP}^2$

Let’s first give a way of how not to define the double cover. Note that the equation $w^2+x^2+y^2+z^2=0$ defines a quadric $Q \subset \mathbb{CP}^3$. If we choose a point $p \in Q$ and fix a hyperplane $H \cong \mathbb{CP}^2$, then we can draw projective lines from $p$ to $H$ which will intersect $Q$ at one point (for a total of 2 points since $Q$ is degree 2). This gives us an injective map $Q \setminus p \to H$. This allows us to conclude that $Q$ is birational to $\mathbb{CP}^2$ but it may not be isomorphic. Indeed, this particular quadric is isomorphic to $\widetilde{Gr}_2(\mathbb{R}^4)$, the Grassmannian of oriented 2-planes in $\mathbb{R}^4$. This space is actually $S^2 \times S^2$.

So it would be tempting to take the Fermat quartic model of a K3 surface: $F_4 = {w^4+x^4+y^4+z^4=0}$ and define a map $[w:x:y:z] \mapsto [w^2:x^2:y^2:z^2]$ but what this gives is a branched double cover of $F_4 \to Q \cong S^2 \times S^2$.

What we’ll do instead is consider the tautological line bundle $\xi \to \mathbb{CP}^2$ and take a generic section $s$ of $\xi^{\otimes 2k}$. The preimage $s^{-1}(0)$ is a deg $2k$ Riemann surface $B$. Now take the total space of $\pi_k:\xi^{\otimes k} \to \mathbb{CP}^2$ and let the fiber coordinate be $x$. Then consider the subspace $R_k$ cut out by $x^2 = s$. The induced projection map $\pi_k:R_k \to \mathbb{CP}^2$ is a 2-1 cover away from $B$, the branch locus.

The Lefschetz hyperplane theorem tells us that $R_k$ is simply connected. We can compute some other invariants. For example, there is a formula for the Euler characteristic of branched covers: $\chi(R_k) = 2 \chi(\mathbb{CP}^2) - \chi(B)$ (2 is the degree of the cover). Since $B$ is a deg $2k$ curve, the genus-degree formula tells us that the genus of $B$ is $(2k-1)(2k-2)/2 = (k-1)(2k-1)$. So $\chi(R_k) = 2(k-1)(2k-1)+4$.

Next, note that $c_1(R_k) = -c_1(\Lambda^2 T^*_{1,0} R_k)$ which equals $\pi^*c_1(\Lambda^2 T^*_{1,0} \mathbb{CP}^2) + [\pi^{-1}(B)]$ Since $\Lambda^2 T^*_{1,0} \mathbb{CP}^2 \cong \mathcal{O}(-3)$, we can deduce that $c_1(R_k) = (k-3) \pi^*H$ and $c^2_1(R_k) = 2(k-3)^2$.

I’m not sure how Donaldson-Kronheimer deduce this but $b^+ = k^2-3k+3, b^-=3k^2-3k+1$. Also, $\pi^* (k-3)H \cong 0 \pmod{2}$ exactly when $k$ is odd; this is the condition for $R_k$ to be spin. Putting this together, $k=1$: $\chi = 4$, $b^+ = 1,b^- = 1$ and $R_1$ is spin. It cannot be $\mathbb{CP}^2 \# \overline{\mathbb{CP}}^2$ so it must be $S^2 \times S^2$. $k=2$: $\chi = 10$, $b^+ = 1, b^- = 7$, and $R_2$ is not spin. We may conclude that this rational surface is homeomorphic to $\mathbb{CP}^2 \# 7\overline{\mathbb{CP}}^2$. $k=3$: $\chi = 24$, $b^+ = 3, b^- = 19$, and $R_3$ is spin. We see also that $c_1 = 0$. So $R_3$ is a K3 surface.