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## Hypersurfaces, 4-manifolds, and characteristic classes

In this post we’ll compute the (topological) cohomology of smooth projective (complex) hypersurfaces in $\mathbb{CP}^n$. When $n = 3$ the resulting complex surfaces give nice examples of 4-manifolds, and we’ll make use of various facts about 4-manifold topology to try to say more in this case; in particular we’ll be able to compute, in a fairly indirect way, the ring structure on cohomology. This answers a question raised by Akhil Mathew in this blog post.

Our route towards this result will turn out to pass through all of the most common types of characteristic classes: we’ll invoke, in order, Euler classes, Chern classes, Pontryagin classes, Wu classes, and Stiefel-Whitney classes.

Examples in the plane

Recall that a smooth projective hypersurface $X$ of degree $d$ is a projective variety cut out by a single homogeneous polynomial $F(X_0 : ... : X_n) \in \mathbb{C}[X_0, ..., X_n]$ of degree $d$ which is smooth. This is the case if and only if the partial derivatives $\frac{\partial F}{\partial X_i}$ have no zeroes in common with $F$ in $\mathbb{CP}^n$. Such a variety has complex dimension $n - 1$, hence real dimension $2n - 2$.

Example. When $n = 2$ we are considering smooth projective curves in the projective plane $\mathbb{CP}^2$. Examples are given by the Fermat curves

$\displaystyle X_0^d + X_1^d = X_2^d$.

Topologically, these are compact oriented surfaces, and hence their homeomorphism and even diffeomorphism type is completely determined by the rank of their first homology, or equivalently by their genus $g$. The genus-degree formula asserts that the genus of a plane curve of degree $d$ is $g = \frac{(d-1)(d-2)}{2}$.

Subexample. When $d = 1$ or $d = 2$ the genus is $0$, so we just get projective lines $\mathbb{CP}^1$, or topologically we get $2$-spheres $S^2$. When $d = 3$ the genus is $1$, so we get elliptic curves (after choosing identities), or topologically we get tori $T^2 \cong S^1 \times S^1$.

There is a nice heuristic proof of the genus-degree formula (which can be made rigorous; see this MO discussion) which goes as follows. First consider the singular curve of degree $d$ given by $d$ lines in general position, so that every pair of lines intersects exactly once but otherwise there are no intersections. Topologically this gives a collection of $d$ spheres each pairwise intersecting in a point. If we perturb the coefficients of the singular curve, it will become smooth; topologically the $d$ spheres become pairwise connected by tubes. After using $d - 1$ of these tubes to connect the spheres in a line, to obtain a sphere, the remaining ${d \choose 2} - (d - 1) = \frac{(d - 1)(d - 2)}{2}$ tubes each increase the genus of the resulting surface by $1$.

An aside

The following is not necessary for the computation to come but is nevertheless a nice explanation of a particular aspect of how it turns out. Eventually we’ll show that the cohomology of a smooth projective hypersurface depends only on the degree $d$ and the dimension $n$ of the ambient projective space, and this is explained by the fact that an even stronger statement than this holds.

Theorem: The diffeomorphism type of a smooth projective hypersurface of degree $d$ in $\mathbb{CP}^n$ depends only on $d$ and $n$.

Remark. This statement cannot be strengthened to a statement about isomorphism in the holomorphic / algebraic category, as the example of cubic curves in $\mathbb{CP}^2$ already shows.

Rough sketch. The idea is that slightly perturbing the coefficients of a homogeneous polynomial $F$ does not affect the diffeomorphism type of the hypersurface it cuts out, and moreover that the space of homogeneous polynomials defining a smooth hypersurface is the complement of a subvariety (the subvariety of polynomials sharing at least one zero with its partial derivatives), hence has real codimension $2$ and in particular is path-connected, so we can perturb the coefficients of any such polynomial to get any other such polynomial.

Proof. Let $V$ be a complex vector space of dimension $n + 1$, so that we can identify $\mathbb{CP}^n$ with $\mathbb{P}(V)$. A homogeneous polynomial of degree $d$ on $V$ is an element of $S^n(V^{\ast})$, but since we’re only looking at the hypersurface cut out by such a polynomial we can ignore the zero polynomial and scaling, so we are really looking at an element of $\mathbb{P}(S^n(V^{\ast}))$. Now let

$B = \mathbb{P}(S^n(V^{\ast})) \setminus \{ F : \gcd(F, \frac{\partial F}{\partial X_i}) \neq 1 \}$

be the complement in $\mathbb{P}(S^n(V^{\ast}))$ of the singular locus of polynomials having a zero in common with their partial derivatives, and let

$\displaystyle E = \{ (v, F) \in \mathbb{P}(V) \times B : F(v) = 0 \}$.

The space $E$ admits a projection map $\pi : E \to B$ onto the second coordinate $(v, F) \mapsto F$, and the hypersurface cut out by $F$ is precisely the fiber $\pi^{-1}(F)$.

Our goal is to show that the fibers of this map are diffeomorphic by applying Ehresmann’s theorem to it, which tells us that $\pi$ is a locally trivial smooth fibration provided that it is a proper surjective submersion. This implies in particular that the fibers $\pi^{-1}(F)$ are all diffeomorphic if $B$ is path-connected.

We’ll divide up the rest of the proof into the following steps.

Step 1: $B$ is a path-connected smooth manifold. More generally the following is true.

Proposition: Let $V \subseteq \mathbb{CP}^n$ be a Zariski-closed subset. Then $\mathbb{CP}^n \setminus V$ is a path-connected smooth manifold.

Proof. A Zariski-closed subset is in particular closed, so $\mathbb{CP}^n \setminus V$ is an open subset of a smooth manifold and hence a smooth manifold. The key point for path-connectedness is that $V$ has codimension at least $2$, but we can avoid explicitly using this fact as follows. Any two distinct points $p, q \in \mathbb{CP}^n \setminus V$ determine a complex line $\mathbb{CP}^1$ passing through them. The intersection of this complex line with $V$ is finite, since it is a Zariski-closed subset of $\mathbb{CP}^1$ but not the whole thing. Now $\mathbb{CP}^1$ minus a finite set of points is path-connected, so $p, q$ can be connected by a path lying inside $\mathbb{CP}^1$ as desired. $\Box$

It remains to show that the singular locus of polynomials having a zero in common with their partial derivatives is Zariski-closed, but this is a corollary of the existence of the multivariate resultant of the $n + 1$ polynomials $\frac{\partial F}{\partial X_i}, 0 \le i \le n$, which is a polynomial in the coefficients vanishing iff the polynomials have a common zero, together with the identity

$\displaystyle \sum_{i=0}^n X_i \frac{\partial F}{\partial X_i} = d F$

showing that if all of the $\frac{\partial F}{\partial X_i}$ vanish at a point then so does $F$.

Step 2: $E$ is a smooth manifold. To start with, we’ll work locally. On the open subset where $X_0 \neq 0$ and the coefficient of $X_0^d$ in $F$ is also nonzero, $E$ is locally the zero locus of the function

$(x_1, x_2, ..., x_n, f) \mapsto f(x_1, ..., x_n) \in \mathbb{C}$

where $x_i = \frac{X_i}{X_0}$ and $f$ is the dehomogenization $X_0^d F \left( 1 : \frac{X_1}{X_0} : ... : \frac{X_n}{X_0} \right)$ of $F$, scaled so that the constant coefficient (the coefficient of $X_0^d$ in $F$) is $1$. Fixing $f$, the differential of this map in the $x_i$ has coefficients the partial derivatives $\frac{\partial f}{\partial x_i}$, and since by assumption we’ve removed the singular hypersurfaces, at least one of these partial derivatives must be nonzero, so by the regular value theorem the zero locus is locally a smooth manifold. Running this argument with $X_0$ replaced by any $X_i$ and $X_0^d$ replaced by any monomial of degree $d$, we get that $E$ is a smooth manifold as desired.

Step 3: $\pi$ is a submersion. $\pi$ is clearly surjective and proper (since hypersurfaces are compact), so this is the only interesting step remaining. Again working locally and on the open subset where $X_0 \neq 0$ and the coefficient of $X_0^d$ in $F$ is nonzero, $\pi$ locally takes the form

$\displaystyle (x_1, x_2, ..., x_n, f) \mapsto f \in \mathbb{C}^{ {n+d-1 \choose d} - 1 }$

where again $f$ is the dehomogenization scaled to have constant coefficient $1$, and $f(x_1, ..., x_n) = 0$. To show that $\pi$ is surjective on tangent spaces it suffices to show that any infinitesimal deformation in the coefficients of $f$ can be canceled out by a corresponding deformation in the $x_i$ so that the relation $f(x_1, ..., x_n) = 0$ continues to hold (this is what it means to lift a tangent vector from our target to our source). But this is precisely guaranteed by the condition that at least one of the partial derivatives $\frac{\partial f}{\partial x_i}$ is nonzero. Again, running this argument with all of the other coordinates and monomials we get the result. $\Box$

Remark. A simpler version of this argument can be used to give a proof of the fundamental theorem of algebra. The rough sketch here is to argue 1) that the space of polynomials with nonzero discriminant is connected, 2) that the number of roots of a polynomial with nonzero discriminant does not change when you perturb its coefficients, and 3) that establishing the fundamental theorem for polynomials with nonzero discriminant establishes it for all polynomials, since if $p$ is any polynomial then $\frac{p}{\gcd(p, p')}$ has nonzero discriminant, or equivalently is squarefree.

Most of the cohomology

Below all cohomologies are with integer coefficients unless otherwise stated.

Let $X$ be a smooth projective hypersurface of degree $d$ in $\mathbb{CP}^n$. Most of the cohomology of $X$ is determined by the Lefschetz hyperplane theorem, as follows. Thinking again of $\mathbb{CP}^n$ as $\mathbb{P}(V)$ where $\dim V = n+1$, we have a Veronese embedding

$\mathbb{P}(V) \ni v \mapsto v^d \in \mathbb{P}(S^d(V))$

and, essentially by definition, $X$ is the intersection of the image of the Veronese embedding with a hyperplane in $\mathbb{P}(S^d(V))$. The Lefschetz hyperplane theorem then guarantees that the natural map $H^k(\mathbb{CP}^n) \to H^k(X)$ is an isomorphism for $k \le n - 2$ and an injection for $k = n - 1$. Recalling that

$\displaystyle H^k(\mathbb{CP}^n) \cong \begin{cases} \mathbb{Z} & \text{ if } k \le 2n, 2 | k \\ 0 & \text{ otherwise} \end{cases}$

we conclude that $H^k(X)$ is $\mathbb{Z}$ if $k$ is even and $0$ otherwise for all $k \le n - 2$. Moreover, since $X$, by virtue of being a compact complex manifold, is in particular a compact oriented manifold, we can apply Poincaré duality to conclude that the same is true of $H^{2n - 2 - k}(X)$. That is,

$\displaystyle \boxed{ H^k(X) \cong \begin{cases} \mathbb{Z} & \text{ if } k \le n-2, 2 | k \text{ or } k \ge n, 2 | (n - k) \\ ? & \text{ if } k = n - 1 \\ 0 & \text{ otherwise} \end{cases} }$

and so the only remaining question is what the middle cohomology $H^{n-1}(X)$ looks like. So far all we know is that $H^{n-1}(\mathbb{CP}^n)$ injects into it; this is $\mathbb{Z}$ if $n$ is odd but $0$ if $n$ is even.

Reduction to the Euler characteristic

We claim that to compute the middle cohomology of $X$ it suffices to compute its Euler characteristic $\chi(X)$. First, recall that a compact manifold has finitely generated cohomology. It follows that $X$ has a well-defined Euler characteristic. Since we know all of the Betti numbers $b_k = \dim H^k(X, \mathbb{Q})$ except one, computing the Euler characteristic will tell us the remaining Betti number. Explicitly, our computations above give

$\displaystyle \chi(X) = 2 \left\lceil \frac{n-1}{2} \right\rceil + (-1)^{n-1} b_{n-1}(X)$.

However, we still need to rule out the possibility of torsion in the middle cohomology $H^{n-1}(X)$ in order to be confident that knowing the Betti number $b_{n-1}(X)$ is enough. We can do this using the universal coefficient theorem, which gives a short exact sequence

$\displaystyle 0 \to \text{Ext}^1(H_{n-2}(X), \mathbb{Z}) \to H^{n-1}(X) \to \text{Hom}(H_{n-1}(X), \mathbb{Z}) \to 0$.

The group on the right is torsion-free because it is given by homomorphisms into a torsion-free group, and the group on the left is torsion-free because it vanishes: $H_{n-2}(X)$ is free by another part of the Lefschetz hyperplane theorem, hence has no nontrivial extensions. It follows that $H^{n-1}(X)$ is free abelian, so is determined by its rank $b_{n-1}(X)$.

The Euler characteristic via Chern classes

Recall that the Euler characteristic of a compact oriented smooth manifold $X$ can be computed as the evaluation of the Euler class $e(TX)$ of its tangent bundle on the fundamental class $[X]$. (Since the Euler class of a vector bundle can be thought of as Poincaré dual to the zero locus of a generic section, this can be thought of as a restatement of the Poincaré-Hopf theorem.)

On a compact complex manifold, the tangent bundle has a complex structure and hence Chern classes $c_k(TX)$. It is common to refer and to notate these as the Chern classes $c_k(X)$ of $X$ itself. Moreover, the top Chern class $c_n(X)$ is the Euler class. Hence one way to compute the Euler characteristic of a compact complex manifold is to compute its top Chern class, which is the approach we will take: in fact we will compute all Chern classes.

We will first need to compute the Chern classes of $\mathbb{CP}^n$. The key tool is the Euler sequence

$\displaystyle 0 \to \mathbb{C} \to \mathcal{O}(1)^{n+1} \to T \mathbb{CP}^n \to 0$

where $\mathbb{C}$ is the trivial line bundle and $\mathcal{O}(1)$ is the dual of the tautological line bundle whose fiber at a point in $\mathbb{CP}^n$ is the line in $\mathbb{C}^{n+1}$ it represents; equivalently, $\mathcal{O}(1)$ is the line bundle whose holomorphic sections are homogeneous polynomials of degree $1$. Since the total Chern class is multiplicative in exact sequences, we get

$\displaystyle c(T \mathbb{CP}^n \oplus \mathbb{C}) = \sum_{k \ge 0} c_k(T \mathbb{CP}^n) = c(\mathcal{O}(1))^{n+1} = (1 + \alpha)^{n+1}$

where $\alpha$ is a generator of the cohomology ring $H^{\bullet}(\mathbb{CP}^n) \cong \mathbb{Z}[\alpha]/\alpha^{n+1}$. It follows that the Chern classes of $\mathbb{CP}^n$ are given by

$\displaystyle c_k(\mathbb{CP}^n) = {n+1 \choose k} \alpha^k$.

(In particular, the top Chern class is ${n+1 \choose n} \alpha^n$, which when evaluated on the fundamental class gives the Euler characteristic $\chi(\mathbb{CP}^n) = n + 1$ as expected.)

To get from here to the Chern classes of a hypersurface $X$ we need to relate the two tangent bundles, which we do via the short exact sequence

$0 \to TX \to T \mathbb{CP}^n |_X \to N_{\mathbb{CP}^n / X} \to 0$

of vector bundles on $X$, where $N_{\mathbb{CP}^n / X}$ is the normal bundle.

Now, it turns out that the normal bundle is the restriction $\mathcal{O}_X(d)$ to $X$ of the line bundle $(L^{\ast})^{\otimes d} \cong \mathcal{O}(d)$ whose holomorphic sections are homogeneous polynomials of degree $d$; this is essentially the content of the adjunction formula. Roughly speaking this is because $X$ is defined as the zero locus of a nonvanishing section $F$ of $\mathcal{O}(d)$, and the actual map $T \mathbb{CP}^N |_X \to \mathcal{O}_X(d)$ can be thought of as the derivative of this section, although I’m not sure how to make this precise.

In particular, since $c_1(\mathcal{O}(d)) = d c_1(\mathcal{O}(1)) = d \alpha$, the total Chern class of $\mathcal{O}(d)$ is given by $1 + d \alpha$, and hence the total Chern class of $X$ is

$\displaystyle c(X) = \frac{c(T \mathbb{CP}^n | _X)}{c(\mathcal{O}_X(d))} = \frac{(1 + \alpha)^{n+1}}{1 + d \alpha}$

where by abuse of notation $\alpha$ denotes the pullback of our previously chosen generator of $H^{\bullet}(\mathbb{CP}^n)$ to $X$. We can now compute that the top Chern class of $TX$ is

$\displaystyle c_{n-1}(TX) = \sum_{k=0}^{n-1} {n+1 \choose k} (-d)^{n-1-k} \alpha^{n-1}$.

It remains to evaluate $\alpha^{n-1}$ on the fundamental class of $X$. Now, $\alpha^{n-1}$ is Poincaré dual to the intersection of $n - 1$ generic hyperplanes in $\mathbb{CP}^n$, which give a copy of $\mathbb{CP}^1$, and since $X$ is cut out by a hypersurface of degree $d$ intersecting it with a generic line gives $d$ points, so we conclude that $\alpha^{n-1} [X] = d$ and hence that

$\displaystyle \chi(X) = \sum_{k=0}^{n-1} (-1)^{n-1-k} {n+1 \choose k} d^{n-k}$

which gives our desired computation of the rank of the middle cohomology:

$\displaystyle \boxed{ b_{n-1}(X) = \sum_{k=0}^{n-1} (-1)^k {n+1 \choose k} d^{n-k} + (-1)^n 2 \left\lceil \frac{n-1}{2} \right\rceil }$.

Example. Let $n = 2$. As mentioned above, in this case $X$ is topologically a compact oriented surface The Betti numbers of $X$ are $1, b_1, 1$, and

$\displaystyle b_1 = 2g = d^2 - 3d + 2 = (d - 1)(d - 2)$

and we recover the genus-degree formula.

Example. Rewriting the formula above as

$\displaystyle b_{n-1}(X) = \frac{(d - 1)^{n+1} + (-1)^{n+1} ((n + 1) d - 1)}{d} + (-1)^n 2 \left\lceil \frac{n-1}{2} \right\rceil$

makes it more convenient to do some kinds of computations with. In particular, for $d = 1$ we get

$\displaystyle b_{n-1}(X) = \begin{cases} 0 \text{ if } 2 | n \\ 1 \text{ otherwise} \end{cases}$

as expected since in this case $X$ is just $\mathbb{CP}^{n-1}$ and we know its middle cohomology already. For $d = 2$ we get

$\displaystyle b_{n-1}(X) = \begin{cases} 0 \text{ if } 2 | n \\ 2 \text{ otherwise} \end{cases}$

which is a little more interesting; the resulting hypersurfaces, namely the quadric hypersurfaces, are birational to $\mathbb{CP}^{n-1}$ but not necessarily homotopy equivalent. We’ll identify the quadric hypersurface $X$ when $n = 3$ below; when $n = 5$ it turns out to be the Grassmannian $\text{Gr}_{2, 4}$ of complex planes in $\mathbb{C}^4$, with the embedding into $\mathbb{CP}^5$ being given up to projective change of coordinates by the Plücker embedding.

For $d \ge 3$ by inspection the Betti number grows exponentially in $n$.

Complex surfaces as 4-manifolds

Now let $n = 3$. In this case $X$ is topologically a compact oriented 4-manifold. The Betti numbers of $X$ are $1, 0, b_2, 0, 1$, and

$\displaystyle b_2 = d^3 - 4d^2 + 6d - 2 = d(d(d - 4) + 6) - 2$.

Example. When $d = 1$, so that $X$ is $\mathbb{CP}^2$, we get $b_2 = 1$ as expected.

Example. When $d = 2$, so that $X$ is a quadric surface, we get $b_2 = 2$; here $X$ is $\mathbb{CP}^1 \times \mathbb{CP}^1$, so diffeomorphic to $S^2 \times S^2$, with the embedding into $\mathbb{CP}^3$ being given up to projective change of coordinates by the Segre embedding.

Example. When $d = 3$, so that $X$ is a cubic surface, we get $b_2 = 7$.

Example. When $d = 4$, so that $X$ is a quartic surface, we get $b_2 = 22$; in this case $X$ is also a K3 surface.

(When $d \ge 5$, $X$ is a surface of general type.)

For $n \ge 3$, the homotopy group version of the Lefschetz hyperplane theorem implies that the natural map $\pi_1(X) \to \pi_1(\mathbb{CP}^n)$ is an isomorphism; since the latter is trivial, so is the former. Hence as 4-manifolds, our complex surfaces are compact, oriented, and simply connected.

For such a 4-manifold, once we know its cohomology groups the only additional data of the sort that one usually calculates in a first course in algebraic topology is the cup product, which is completely determined by the intersection form

$\langle -, - \rangle : H^2(X) \times H^2(X) \ni (\alpha, \beta) \mapsto (\alpha \cup \beta)[X] \in \mathbb{Z}$

where $[X]$ is the fundamental class in $H_4(X)$. Since $2$ is even, the intersection form is symmetric, so gives $H^2(X)$ the structure of an integral lattice (that is, a free abelian group equipped with a symmetric bilinear $\mathbb{Z}$-valued form), and by Poincaré duality this lattice is unimodular.

Thus identifying invariants of lattices immediately gives us (oriented homotopy) invariants of compact oriented 4-manifolds, and more generally of compact oriented manifolds in dimension $4n$. We’ll focus our attention on three such invariants.

• The rank of a lattice $L$ is its rank as an abelian group; in the case of 4-manifolds this is just the second Betti number $b_2(X)$.
• The signature of a lattice $L$ is the signature of its bilinear form on $L \otimes \mathbb{R}$. More explicitly, by Sylvester’s law of inertia any nondegenerate bilinear form on a real vector space can be diagonalized so that the corresponding quadratic form is

$x_1^2 + ... + x_m^2 - x_{m+1}^2 - ... - x_{m+n}^2$

for two integers $m, n \ge 0$, which can equivalently be described as the number of positive resp. negative eigenvalues of a matrix describing the bilinear form. The signature is then $m - n$; note that the rank is $m + n$, so the signature and the rank together determine the ordered pair $(m, n)$, which is also sometimes called the signature. This gives an invariant of compact oriented manifolds $X$ in dimension $4n$ also called the signature and denoted $\sigma(X)$.

• The parity of a lattice $L$ is defined as follows: if $\langle \beta, \beta \rangle$ is always divisible by $2$, then the lattice is even, and otherwise the lattice is odd. In other words, where the signature comes from looking at $L \otimes \mathbb{R}$, the parity comes from looking at $L \otimes \mathbb{F}_2$.

Remark. In general this is very far from being a complete set of invariants of lattices. In the case that the signature is equal to the rank (so that the lattice is positive definite), the Smith-Minkowski-Siegel mass formula implies that the number of isomorphism classes of lattices grows very rapidly with the rank.

Remark. The signature is a particularly interesting invariant: its definition can be extended to manifolds in dimension not divisible by $4$ by declaring the corresponding signatures to be $0$, and then the signature is a genus, although we won’t use this fact.

The intersection form turns out to be a surprisingly strong invariant. Milnor and Whitehead showed that compact, oriented, simply connected 4-manifolds are determined up to oriented homotopy by their intersection forms as lattices. Freedman showed that every unimodular lattice arises in this way and that the only additional data required to determine such a 4-manifold up to homeomorphism is a class in $H^4(X, \mathbb{Z}_2) \cong \mathbb{Z}_2$ called the Kirby-Siebenmann invariant; moreover,

• if the lattice is even, then there is a unique corresponding 4-manifold up to homeomorphism with Kirby-Siebenmann invariant $\frac{\sigma(X)}{8} \bmod 2$, and
• if the lattice is odd, then there are exactly two corresponding 4-manifolds, one with each possible value of the Kirby-Siebenmann invariant.

The Kirby-Siebenmann invariant vanishes whenever a manifold has a smooth structure, and so in the odd case at least one of the two 4-manifolds does not have a smooth structure.

There are also other obstructions to having a smooth structure involving the intersection form. For example, by the above the $E_8$ lattice occurs as the intersection form of a unique homeomorphism class of compact, orientable, simply connected 4-manifold, the $E_8$ manifold. The $E_8$ lattice is positive definite but not diagonalizable, so by Donaldson’s theorem the $E_8$ manifold does not have a smooth structure.

The computations we’ve done so far don’t tell us what the intersection form is. Fortunately, we’ll be able to compute the intersection form, and hence the cup product structure on cohomology, as follows. First, we can compute the signature using the Hirzebruch signature theorem in terms of Pontryagin classes. Second, if the signature is not equal to plus or minus the rank (so the lattice is indefinite) then the possible lattices have been completely classified. There are only two possibilities if the rank $m + n$ and signature $m - n$ are fixed, depending only on the parity:

• if the lattice is odd, it must be the lattice $I_{m, n}$ of vectors with integer entries in $\mathbb{R}^{m, n}$, the real vector space $\mathbb{R}^{m + n}$ equipped with the symmetric bilinear form of signature $m - n$;
• if the lattice is even, the signature $m - n$ must be divisible by $8$, and the lattice must be the lattice $II_{m, n}$ of vectors in $\mathbb{R}^{m, n}$ whose entries are either all integers or all integers plus $\frac{1}{2}$ and which sum to an even number.

In other words, for indefinite unimodular lattices the rank, signature, and parity form a complete set of invariants. Hence if we compute that the signature $\sigma(X)$ is not equal to the rank $b_2(X)$, the only additional information we need to determine the lattice $H^2(X)$ is its parity. It will turn out that this is determined by whether the second Stiefel-Whitney class $w_2(X)$ vanishes, or equivalently by whether $X$ admits a spin structure.

The signature via Pontryagin classes

Recall that if $V$ is a real vector bundle over a space $X$ then it admits a complexification $V \otimes_{\mathbb{R}} \mathbb{C}$ which is a complex vector bundle, and that the Pontryagin classes of $V$ are characteristic classes defined in terms of the Chern classes of the complexification via

$p_k(V) = (-1)^k c_{2k}(V \otimes_{\mathbb{R}} \mathbb{C}) \in H^{4k}(X)$.

For a compact smooth oriented 4-manifold $X$, the Hirzebruch signature theorem asserts that the signature $\sigma(X)$ is given by

$\displaystyle \sigma(X) = \frac{p_1 [X]}{3}$

where $p_1$ is the first Pontryagin class

$\displaystyle p_1(X) = p_1(TX) = - c_2(TX \otimes \mathbb{C}) \in H^4(X)$

of (the tangent bundle of) $X$ and $[X] \in H_4(X)$ is the fundamental class as usual. In particular, it implies that the first Pontryagin number $p_1 [X]$ is divisible by $3$.

Hence to compute the signature of a hypersurface $X \subset \mathbb{CP}^3$ we need to compute the second Chern class of the complexification of its tangent bundle, regarded as a real vector bundle (whereas above we computed the Chern classes of the tangent bundle, which already had a complex structure). In general we can compute the Chern classes of the complexification of a complex vector bundle in terms of the Chern classes of the original bundle as follows.

Theorem: Let $V$ be a complex vector bundle. Then the complexification $V \otimes_{\mathbb{R}} \mathbb{C}$ of the underlying real vector bundle of $V$ is isomorphic, as a complex vector bundle, to $V \oplus \overline{V}$, where $\overline{V} \cong V^{\ast}$ is the conjugate vector bundle.

Corollary: The Pontryagin classes $p_k(V) = (-1)^k c_{2k}(V \otimes_{\mathbb{R}} \mathbb{C})$ of the underlying real vector bundle of a complex vector bundle $V$ can be computed in terms of its Chern classes via the Whitney sum formula as

$\displaystyle \sum_{k \ge 0} c_k(V \otimes_{\mathbb{R}} \mathbb{C}) = \left( \sum_{k \ge 0} c_k(V) \right) \left( \sum_{k \ge 0} (-1)^k c_k(V) \right)$.

In particular,

$\displaystyle p_1(V) = - c_2(V \otimes_{\mathbb{R}} \mathbb{C}) = c_1(V)^2 - 2c_2(V)$.

Proof. Write

$V \otimes_{\mathbb{R}} \mathbb{C} \cong (V \otimes_{\mathbb{C}} \mathbb{C}) \otimes_{\mathbb{R}} \mathbb{C} \cong V \otimes_{\mathbb{C}} \left( \mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \right)$.

This tells us that to understand the endofunctor $V \mapsto V \otimes_{\mathbb{R}} \mathbb{C}$ on complex vector bundles it suffices to understand $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$ as a $(\mathbb{C}, \mathbb{C})$-bimodule; the left $\mathbb{C}$-module structure tells us how to take the tensor product and the right $\mathbb{C}$-module structure tells us what the complex structure on the tensor product is. The theorem is then equivalent to the claim that, as a bimodule,

$\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \cong \mathbb{C} \oplus \overline{\mathbb{C}}$

where

• $\mathbb{C}$ denotes the identity bimodule, with $\mathbb{C}$ acting on the left and right by left and right multiplication, so that tensoring with this bimodule is the identity endofunctor $V \mapsto V$, and
• $\overline{\mathbb{C}}$ denotes the bimodule where left and right multiplication by $i$ disagree by a sign of $-1$ (more explicitly, we can take the left module structure to be the usual one and the right module structure to be multiplication by the conjugate), so that tensoring with this bimodule is the endofunctor $V \mapsto \overline{V}$.

To see this, we will first think of $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$ as a right $\mathbb{C}$-module with basis $1 \otimes 1, i \otimes 1$, and then we will diagonalize left multiplication by $i$. When we do this we find that on

$1 \otimes 1 - i \otimes i$

left and right multiplication by $i$ agree, whereas on

$1 \otimes 1 + i \otimes i$

left and right multiplication differ by a sign. The left, or equivalently right, $\mathbb{C}$-submodules generated by these vectors gives the desired decomposition. $\Box$

Now let $X$ be a hypersurface of degree $d$ in $\mathbb{CP}^3$. Above we computed the total Chern class $c(X)$ to be

$\displaystyle \sum_{k \ge 0} c_k(X) = \frac{(1 + \alpha)^4}{1 + d \alpha}$

so we compute that

$c_1(X) = (4 - d) \alpha, c_2(X) = (6 - 4d + d^2) \alpha^2$

and hence that

$\displaystyle p_1(X) = (16 - 8d + d^2) \alpha^2 - 2(6 - 4d + d^2) \alpha^2 = (4 - d^2) \alpha^2$

and, using again the fact that $\alpha^2 [X] = d$, we compute the signature of a smooth projective hypersurface of degree $d$ in $\mathbb{CP}^3$ to be

$\displaystyle \boxed{ \sigma(X) = \frac{p_1 [X]}{3} = - \frac{(d - 2) d (d + 2)}{3} }$.

Above the numerator has been written in a form that makes it clear that it is divisible by $3$.

We conclude that for $d \ge 2$ the signature is not equal to plus or minus the rank, and so the intersection form is indefinite in this case, which tells us that to uniquely identify the intersection form we only need to know the parity as we hoped.

Example. When $d = 1$ the signature is $1$. This reflects the fact that the intersection form on $X = \mathbb{CP}^2$ is positive definite, since it is just given by $\langle \alpha, \alpha \rangle = \alpha^2 [X] = 1$.

Example. When $d = 2$ the signature is $0$. This reflects the fact that the intersection form on $X = \mathbb{CP}^1 \times \mathbb{CP}^1$ is indefinite, since by the Kunneth formula $H^2(X)$ is generated by two elements $\alpha \otimes 1, 1 \otimes \alpha$ (where $\alpha$ denotes a generator of $H^2(\mathbb{CP}^1)$) which square to zero but whose cup product is a generator of $H^4(X)$. An explicit diagonalization of the intersection form over $\mathbb{R}$ is given by the basis $\alpha \otimes 1 + 1 \otimes \alpha, \alpha \otimes 1 - 1 \otimes \alpha$.

Example. When $d = 3$ the signature is $-5$. In particular it is not divisible by $8$, so the intersection form is odd and hence must be the lattice $I_{1, 6}$.

Example. When $d = 4$ the signature is $-16$. We’ll see later that in this case the intersection form is even, and hence must be the lattice $II_{3, 19}$.

In general, when $d$ is odd the signature is odd, so the intersection form is odd and hence is uniquely determined. When $d$ is even the signature is divisible by $16$, and in particular is divisible by $8$, so the intersection form could be even or odd.

The parity via Stiefel-Whitney classes

To summarize, the story so far is the following:

• If $X$ is a smooth projective hypersurface in $\mathbb{CP}^3$ of degree $d$, then in particular it is a smooth, compact, oriented, and simply connected 4-manifold.
• For such a manifold, $H^2(X)$ is a free abelian group of finite rank, and $X$ is determined up to homeomorphism by the intersection form on $H^2(X)$, which gives $H^2(X)$ the structure of a unimodular lattice.
• The rank and the signature of $H^2(X)$ are given by

$\displaystyle b_2(X) = d(d(d - 4) + 6) - 2, \sigma(X) = - \frac{(d - 2) d(d + 2)}{3}$

and in particular, for $d \ge 2$, $H^2(X)$ is indefinite.

• By the classification of indefinite unimodular lattices, the only remaining bit of information we need about $H^2(X)$ to completely determine it is its parity. More specifically, if the parity is odd then $H^2(X)$ must be $I_{m, n}$ and if the parity is even then $H^2(X)$ must be $II_{m, n}$, where

$\displaystyle m = \frac{b_2 + \sigma}{2}, n = \frac{b_2 - \sigma}{2}$.

In this section we’ll compute the parity. It will turn out to depend only on $d$, which via Freedman’s work gives an independent confirmation that when $n = 3$ the homeomorphism type of a smooth projective hypersurface of degree $d$ only depends on $d$ (since the Kirby-Siebenmann invariant vanishes when a 4-manifold has a smooth structure).

Let $X$ be a smooth, compact, oriented, simply connected 4-manifold. Since $H_1(X)$ vanishes, we have $H^2(X, \mathbb{F}_2) \cong H^2(X) \otimes \mathbb{F}_2$, and so the parity of $H^2(X)$ is determined by whether or not the map

$\displaystyle H^2(X, \mathbb{F}_2) \ni \beta \mapsto \beta \cup \beta \in H^4(X, \mathbb{F}_2)$

is identically zero. Over $\mathbb{F}_2$ this map is linear; in fact it can be identified with the Steenrod square $\text{Sq}^2$. By Poincaré duality (this step only requires that $X$ is compact, since every compact manifold is orientable over $\mathbb{Z}_2$) there must therefore be a unique cohomology class $\nu_2 \in H^2(X, \mathbb{F}_2)$ such that

$\displaystyle \beta \cup \beta = \text{Sq}^2(\beta) = \nu_2 \cup \beta$.

This class is called the second Wu class, and by definition $\beta \cup \beta = \nu_2 \cup \beta$ vanishes identically iff $\nu_2$ vanishes, so $H^2(X)$ is even iff $\nu_2$ vanishes.

So it remains to compute $\nu_2$. The Wu classes turn out to be closely related to the Stiefel-Whitney classes (of the tangent bundle). More precisely, the total Stiefel-Whitney class is the total Steenrod square of the total Wu class:

$\displaystyle w = w_1 + w_2 + ... = \text{Sq}(v) = \left( \text{Sq}^0 + \text{Sq}^1 + ... \right) \left( \nu_1 + \nu_2 + ... \right)$.

Remark. In particular, the Stiefel-Whitney classes of a compact smooth manifold depend only on its cohomology over $\mathbb{F}_2$ as a module over the Steenrod algebra, which is surprising: a priori the Stiefel-Whitney classes also depend on the additional data of the tangent bundle.

This gives

$\displaystyle w_1 = \nu_1, w_2 = \nu_1^2 + \nu_2$

and hence

$\displaystyle \nu_2 = w_1^2 + w_2$.

Since we assumed that $X$ is oriented, $w_1$ vanishes (although this also follows from the fact that $H^1(X, \mathbb{F}_2)$ vanishes as well), from which it follows that $\nu_2$ vanishes iff the second Stiefel-Whitney class $w_2$ vanishes. Hence we have proven the following.

Theorem: Let $X$ be a compact oriented simply connected 4-manifold. Then $H^2(X)$ is even iff $w_2$ vanishes.

Remark. Even if $X$ is not equipped with a smooth structure, hence is not equipped with a tangent bundle, as long as $X$ is compact we can still define its Stiefel-Whitney classes in terms of its Wu classes, and these will agree with the Stiefel-Whitney classes computed from any smooth structure on $X$. If $X$ is equipped with a smooth structure and is oriented, then the vanishing of $w_2$ is equivalent to $X$ also admitting a spin structure.

Remark. The $E_8$ lattice is even; in fact it is the unique even positive definite unimodular lattice of rank $8$. It follows that if the $E_8$ manifold had a smooth structure, it would also admit a spin structure, and then Rokhlin’s theorem would imply that its signature is divisible by $16$. But its signature is $8$; contradiction. This gives a second proof that the $E_8$ manifold has no smooth structure.

Remark. If $X$ is not simply connected, or more precisely if $H_1(X)$ has $2$-torsion, then it is still true that if $w_2$ vanishes then $H^2(X)$ is even, but the converse need not hold owing to the presence of an additional direct summand $\text{Ext}^1(H_1(X), \mathbb{F}_2)$ in $H^2(X, \mathbb{F}_2)$ coming from universal coefficients.

It remains to compute the second Stiefel-Whitney class. We can in fact compute all Stiefel-Whitney classes of a hypersurface of degree $d$ in $\mathbb{CP}^n$ as follows.

Theorem: Let $V$ be a complex vector bundle. Then the Stiefel-Whitney classes $w_k(V) \in H^k(X, \mathbb{F}_2)$ of the underlying real vector bundle are determined by the Chern classes $c_k(V) \in H^{2k}(X, \mathbb{Z})$ as follows: the odd Stiefel-Whitney classes $w_{2k+1}(V)$ vanish, and the even Stiefel-Whitney classes satisfy

$w_{2k}(V) \equiv c_k(V) \bmod 2$

Proof. We’ll first prove this in the case when $V$ is a line bundle $L$. (This is the only case we need but it’s not much harder to prove the general statement.) In this case we only need to show that $w_1(L)$ vanishes and that $w_2(L) \equiv c_1(L) \bmod 2$.

First, $w_1(L)$ vanishes if and only if $L$ has an orientation. But any complex structure induces an orientation, so this is clear.

To compute $w_2(L)$ we can use the fact that the top Stiefel-Whitney class of an oriented vector bundle is the reduction $\bmod 2$ of its Euler class while the top Chern class of a complex vector bundle is its Euler class, which gives $w_2(L) \equiv c_1(L) \bmod 2$ since they are both the Euler class $\bmod 2$. If we want to avoid the Euler class, we can also argue as follows:

The functor from complex line bundles to real plane bundles is induced, at the level of classifying spaces, by the map

$BU(1) \to BO(2)$

induced by the standard embedding $U(1) \to O(2)$. Since $U(1) \cong SO(2)$ as subgroups of $O(2)$, the map above factors as a composite

$\displaystyle BU(1) \sim BSO(2) \to BO(2)$

where the first map is a homotopy equivalence, showing that the classification of complex line bundles is in fact equivalent to the classification of oriented real plane bundles.

From standard results about characteristic classes we know that on the one hand

$\displaystyle H^{\bullet}(BSO(2), \mathbb{F}_2) \cong \mathbb{F}_2[w_2]$

is a polynomial algebra on the universal second Stiefel-Whitney class $w_2 \in H^2(BSO(2), \mathbb{F}_2)$, while on the other hand

$\displaystyle H^{\bullet}(BU(1), \mathbb{Z}) \cong \mathbb{Z}[c_1]$

is a polynomial algebra on the universal first Chern class $c_1 \in H^2(BU(1), \mathbb{Z})$. In particular, $c_1$ generates $H^2(BU(1), \mathbb{Z})$ while $w_2$ is the unique generator of $H^2(BSO(2), \mathbb{F}_2)$, so the homotopy equivalence $BU(1) \sim BSO(2)$ necessarily identifies the latter with the $\bmod 2$ reduction of the former.

We have the desired result for line bundles. To obtain the result for all bundles we appeal to the splitting principle, which tells us in particular that to prove an equality of characteristic classes it suffices to prove it on a direct sum of line bundles.

So let $L_1, ..., L_n$ be complex line bundles. We now know that the total Stiefel-Whitney class of the underlying real vector bundle of $L_1 \oplus ... \oplus L_n$ can be computed, using the Whitney sum formula, as

$\displaystyle w(L_1 \oplus ... \oplus L_n) = (1 + w_2(L_1)) ... (1 + w_2(L_n))$

since we know that $w_1(L_i)$ vanishes. This implies that all of the odd Stiefel-Whitney classes vanish. Since we also know that $w_2(L_i) \equiv c_1(L_i) \bmod 2$, this tells us that the total Stiefel-Whitney class is

$\displaystyle (1 + c_1(L_1)) ... (1 + c_1(L_n)) \bmod 2$

and this is the $\bmod 2$ reduction of the total Chern class as desired. $\Box$

Now again let $X$ be a hypersurface of degree $d$ in $\mathbb{CP}^3$. Above we computed the first Chern class to be

$\displaystyle c_1(X) = (4 - d) \alpha \in H^2(X)$

where $\alpha$, as before, denotes the pullback of the generator of $H^2(\mathbb{CP}^3)$ to $X$. By the Lefschetz hyperplane theorem, or from the fact that we know $\alpha^2 [X] = d \neq 0$, the cohomology class $\alpha \in H^2(X)$ is nonzero, hence the reduction

$\displaystyle \boxed{ w_2(X) \equiv (4 - d) \alpha \equiv d \alpha \bmod 2 }$

vanishes if and only if $d$ is even. We conclude that the parity of $H^2(X)$ is precisely the parity of $d$. This completes our computation of the cohomology ring of $X$.

Remark. When $d$ is even we also conclude that the hypersurfaces $X$ have a spin structure, and in particular we get an independent confirmation of Rokhlin’s theorem that the signature is divisible by $16$ in this case.

### 5 Responses

1. […] classes also give obstructions to finding complex structures: as we saw earlier, if a real vector bundle has a complex structure then the odd Stiefel-Whitney classes vanish and […]

2. That’s an impressive computation! One complaint, I think: you say that for $d\geq 2$ the signature is nonpositive, and thus the form is indefinite. But this hasn’t ruled out a negative definite form, right? Seems better to just check that $b_2(d)-\sigma(d)$ only has the one integral zero.

• Right, I also need to check that the signature is not equal to minus the rank. Let me fix that.

3. […] « Hypersurfaces, 4-manifolds, and characteristic classes […]

4. Interesting. Sorry I never responded to your comment on my post. It was an extremely hectic end to the semester. I didn’t know the answer.