Let ${\phi: \Omega_{SO} \rightarrow \Lambda}$ be a genus. We might ask when ${\phi}$ satisfies the following multiplicative property:

Property: For any appropriate fiber bundle ${F \rightarrow E \rightarrow B}$ of manifolds, we have

$\displaystyle \phi(E) = \phi(B) \phi(F). \ \ \ \ \ (1)$

When ${B}$ is simply connected, this is true for the signature by an old theorem of Chern, Hirzebruch, and Serre.

A special case of the property (1) is that whenever ${E \rightarrow B}$ is an even-dimensional complex vector bundle, then we have

$\displaystyle \phi(\mathbb{P}(E)) = 0,$

for ${\mathbb{P}(E)}$ the projectivization: this is because ${\mathbb{P}(E) \rightarrow B}$ is a fiber bundle whose fibers are odd-dimensional complex projective spaces, which vanish in the cobordism ring.

Ochanine has given a complete characterization of the genera which satisfy this property.

Theorem 1 (Ochanine) A genus ${\phi}$ annihilates the projectivizations ${\mathbb{P}(E)}$ of even-dimensional complex vector bundles if and only if the associated log series ${g(x) = \sum \frac{\phi(\mathbb{CP}^{2i})}{2i+1} x^{2i+1}}$ is given by an elliptic integral

$\displaystyle g(x) = \int_0^x Q(u)^{-1/2} du,$

for ${Q(u) = 1 - 2\delta u^2 + \epsilon u^4}$ for constants ${\delta, \epsilon}$.

Such genera are called elliptic genera. Observe for instance that in the case ${\epsilon = 1, \delta = 1}$, then

$\displaystyle g(x) = \int_0^x \frac{du}{1 - u^2} = \tanh^{-1}(u),$

so that we get the signature as an example of an elliptic genus (the signature has ${\tanh^{-1}}$ as logarithm, as we saw in the previous post).

I’d like to try to understand the proof of Ochanine’s theorem in the next couple of posts. In this one, I’ll describe the proof that an elliptic genus in fact annihilates projectivizations ${\mathbb{P}(E)}$ of even-dimensional bundles ${E}$.

1. Calculations in the projectivization

Suppose ${\phi}$ is the genera associated to an elliptic integral: that is, suppose the logarithm is given by the formal power series

$\displaystyle g(x) = \int_0^x \frac{1}{\sqrt{1 - 2 \delta u^2 + \epsilon u^4}} du,$

where ${\epsilon, \delta}$ are “generic.”

Let ${M}$ be a manifold, and let ${E \rightarrow M}$ be an even-dimensional complex vector bundle. We would like to show that ${\phi( \mathbb{P}(E)) = 0}$. Let ${\pi: \mathbb{P}(E) \rightarrow M}$ be the natural fiber bundle. First, we note that we have a splitting

$\displaystyle T_{\mathbb{P}(E)} = \pi^* T_M \oplus T',$

where ${T'}$ consists of the vectors “tangent to the fiber.” We have

$\displaystyle T' \oplus \mathbf{1} = \pi^* E \otimes \mathcal{O}(1),$

where ${\mathcal{O}(1)}$ is the dual to the tautological bundle on ${\mathbb{P}(E)}$. This follows as in the case of complex projective space.

Let ${f}$ be the characteristic power series of the genus ${\phi}$, as in the previous post; then

$\displaystyle f(x) = \frac{x}{g^{-1}(x)}$

and ${f}$ defines a stable characteristic class for real vector bundles. Then ${\phi(\mathbb{P}(E)) = \int_{\mathbb{P}(E)} f(T_{\mathbb{P}(E)})}$. This is what we want to compute.

Fortunately, we have a pretty good hand on ${\mathbb{P}(E)}$. In fact,

$\displaystyle H^*(\mathbb{P}(E)) = H^*(M)\left\{1, t, \dots, t^{n-1}\right\}$

if ${\dim E = n}$, and where ${t = c_1(\mathcal{O}(1))}$ is in degree ${2}$. Consequently,

$\displaystyle f(\mathbb{T}_{\mathbb{P}}(E)) = \pi^* f(T_M) f( \pi^* E \otimes \mathcal{O}(1)).$

The fundamental class of ${\mathbb{P}(E)}$ is given by the fundamental class of ${M}$ times ${t^{n-1}}$. So, in other words, we have to show:

Lemma 2 If ${n}$ is even, then the coefficient of ${t^{n-1}}$ in ${f( \pi^* E \otimes \mathcal{O}(1))}$ vanishes.

2. Reduction to a fact about elliptic functions

So, we’ve reduced to a lemma on computing the characteristic class of ${\pi^* E \otimes \mathcal{O}(1)}$. Let ${y_1, \dots, y_n}$ be formal Chern roots of ${E \otimes_{\mathbb{R}} \mathbb{C}}$; then the formal Chern roots for ${\pi^* E \otimes \mathcal{O}(1) \otimes \mathbb{C}}$ are ${y_1 + t, \dots, y_n + t}$ because the Chern classes in ordinary cohomology are additive under the tensor product of line bundles.

In particular, we get

$\displaystyle f( \pi^* E \otimes \mathcal{O}(1)) = \prod_{i = 1}^n f( t + y_i) = \prod_{i = 1}^n \frac{t + y_i}{g^{-1}(t + y_i)}.$

We have to show that there is no coefficient of ${t^{n-1}}$ in this.

Here Ochanine’s clever idea is to use the fact that ${\prod (t + y_i) = 0}$ (this is the relation that ${t}$ satisfies in terms of the Chern classes) and then to use facts about elliptic functions. Namely, let ${u}$ be a variable, and let ${\left\{x_i\right\}}$ be formal variables. We can write

$\displaystyle \prod_{i = 1}^n \frac{u + x_i}{g^{-1}(u +x_i)} = P + Q\prod (u +x_i) ,$

where ${P = P(\left\{y_i\right\}, u)}$ is of degree ${\leq n-1}$ in ${u}$. Substituting ${x_i = y_i}$ in ${H^*(M)}$ and ${u = t}$, we find

$\displaystyle \prod_{i = 1}^n \frac{t + y_i}{g^{-1}(t +y_i)} = P( \left\{y_i\right\}, t) \in H^*(\mathbb{P}(E)).$

So we want to show that the coefficient of ${t^{n-1}}$ in ${P(\left\{y_i\right\}, t)}$ is zero. We will do this by doing so in the universal case, where we can introduce denominators.

Lemma 3 If ${n}$ is even, the coefficient of ${u^{n-1}}$ in ${P(\left\{x_i \right\}, u)}$ is zero.

So the strategy will be to work out what ${P}$ looks like. We can do this in the case of indeterminates, as before. Since ${P}$ is a polynomial of degree ${\leq n-1}$, we can do this using Lagrange interpolation. Namely, we have for each ${j}$,

$\displaystyle P( \left\{x_i\right\} , -x_j) = \prod_{i \neq j} \frac{x_i - x_j}{g^{-1}(x_i - x_j)}.$

Since we have obtained ${n}$ values of ${P}$, we can thus write

$\displaystyle P( \left\{x_i\right\},u) = \sum_j \prod_{i \neq j} \frac{u - x_i}{x_j - x_i}\frac{x_i - x_j}{g^{-1}(x_i - x_j)} = \sum_j \prod_{i \neq j} \frac{u - x_i}{g^{-1}(x_j - x_i)} \ \ \ \ \ (2)$

Note that it is important that we are working in the universal case (with indeterminates) here, because we have introduced denominators. The coefficient of ${u^{n-1}}$ in this expression is given by

$\displaystyle \sum_j \prod_{i \neq j} \frac{1}{g^{-1}(x_j - x_i)} \in \mathbb{Q}((\left\{x_i\right\})).$

3. Some facts about elliptic integrals

So, we need to prove that this is zero. We have used essentially nothing so far; it is here that where we will need the specific facts about elliptic integrals. It follows that we need to prove:

Lemma 4 If ${g(x) }$ is an elliptic integral ${\int_0^x (1 - 2\delta v^2 + \epsilon v^4)^{-1/2 } dv}$, then

$\displaystyle \sum_j \prod_{i \neq j} \frac{1}{g^{-1}(x_i - x_j)} = 0\in \mathbb{Q}((\left\{x_i\right\}))$

.

We will prove this by treating ${g^{-1}}$ as a function and applying to complex numbers ${\left\{x_i\right\}}$.

The functions ${r(x) = \frac{1}{\sqrt{ \wp(x) - e_1}}}$ (for ${\wp}$ the Weierstrass function for an appropriate lattice) can be taken as the inverses to elliptic integrals of the given form, and in particular ${g^{-1}(x) = r(x)}$ for some choice of lattice (at least if $\epsilon, \delta$ are suitably nondegenerate). It follows that we need to prove

$\displaystyle \sum_j \prod_{i \neq j} \frac{1}{r(x_i - x_j)} = 0\in \mathbb{C}.$

The idea is that the function

$\displaystyle \prod_{i=1}^n r(x - x_i)$

is, for ${n}$ even, an honest elliptic function, with simple poles at the ${x_i}$. The function ${r(x)}$ (as a square root of an even function) was not necessarily elliptic; it was only elliptic up to factors of ${\pm 1}$. But since ${n}$ is even, the product of all these is elliptic. Now, for any elliptic function ${p(x)}$, the sum of the residues is zero: this follows by integrating the differential ${p dx}$. But the sum of the residues of ${\prod_{i=1}^n r(x - x_i)}$ is given by the sum in question.

Anyway, this proves one half of Ochaine’s theorem. It turns out that much more is true (namely, one has multiplicative in suitable fiber bundles where the fiber is a spin manifold); this follows from work of Bott and Taubes.