In a previous post, I began discussing a theorem of Ochanine:

Theorem 1 (Ochanine) A genus {\phi: \Omega_{SO} \rightarrow \Lambda} annihilates the projectivization {\mathbb{P}(E)} of every even-dimensional complex bundle {E \rightarrow M} if and only if the logarithm of {\phi} is an elliptic integral

\displaystyle g(x) = \int_0^x (1 - 2\delta u^2 + \epsilon u^4)^{-1/2} du.

In the previous post, we described Ochanine’s proof that a genus whose logarithm is an elliptic integral (a so-called elliptic genus) annihilated any such projectivization. The proof relied on some computations in the projectivization and then some trickery with elliptic functions. The purpose of this post is to prove the converse: a genus with a suitably large kernel comes from an elliptic integral.

1. Preliminaries

The strategy here is going to be to use the images of the Milnor hypersurfaces to say something about the logarithm of the formal group law, and to use the fact that many of the Milnor hypersurfaces are themselves projectivizations. So, we’re going to start with some preliminaries on the Milnor hypersurfaces, most of which will be a review of the previous post.

In the previous post, we defined {H_{i,j} \subset \mathbb{CP}^i \times \mathbb{CP}^j} as the zero locus of a generic section of the complex line bundle {\mathcal{O}(1) \boxtimes \mathcal{O}(1)} on {\mathbb{CP}^i \times \mathbb{CP}^j}. We wrote down an expression in the cobordism ring: namely, we proved the formula

\displaystyle [H_{i,j}] = \sum_{m \leq i, n \leq j} a_{m, n} [\mathbb{CP}^{i-m}][\mathbb{CP}^{j-m}] \in \pi_* MU,

for the {a_{m, n}} the coefficients of the formal group law on {\pi_* MU}. Another way to phrase this is that if we define the power series

\displaystyle h(u, v) = \sum_{i, j} [H_{i,j}] u^i v^j \in \pi_*MU [[u, v]],

and write {f(u, v)} for the formal group law of {\pi_* MU}, then we have the formula

\displaystyle h(u, v) = f(u, v) g'(u) g'(v) \ \ \ \ \ (1)

for {g'(u) = \sum_i [\mathbb{CP}^i] u^i} the derivative of the logarithm.

Note that this formula is valid in the oriented cobordism ring as well (where one can throw out the odd-dimensional projective spaces, as they contribute zero). It persists under any genus {\phi: \Omega_{SO} \rightarrow \Lambda}. That is, if we define analogous power series

\displaystyle h^\phi(u, v) = \sum_{i,j} \phi( [H_{i, j}]) u^i v^j

and {f^\phi(u,v), g^\phi(u)} similarly, then we have

\displaystyle h^{\phi}(u,v) = f^\phi(u, v) g^{\phi \prime}(u) g^{\phi \prime}(v).

2. A basic formula

A key observation that we will need in the future is that the {H_{i,j}} can be obtained as projectivizations. Suppose {i \leq j}. Then the projection map

\displaystyle H_{i,j} \rightarrow \mathbb{CP}^i

is a fiber bundle with fiber {\mathbb{CP}^{j-1}}. In fact, it exhibits {H_{i,j}} as a suitable projectivization of the bundle {\mathcal{O}(1) \oplus \mathbf{1}^{j-i}} over {\mathbb{CP}^i}.

Consequently, we can actually prove a stronger version of the theorem:

Theorem 2 If {\phi: \Omega_{SO} \rightarrow \Lambda} is a genus such that {\phi(H_{3, 2i}) =0} for all {i \geq 2}, then the logarithm of {\phi} is an elliptic integral.

Define the power series

\displaystyle r^\phi(u) = \sum_{i \geq 1} \phi(H_{3, 2i}) u^{2i}.

Then this series is a multiple of {u^2} if and only if {\phi} annihilates the {H_{3, 2i}, i \geq 2}.

In order to prove this result, we will show that, without any hypotheses on the genus,

\displaystyle r^\phi(u) = \frac{1}{6} g^{\phi \prime}(u) \frac{\partial^3 f^\phi(u, 0)}{\partial^3 v},

for any genus. To ease notation, we will drop the {\phi} and prove it as an identity in the oriented (not complex!) cobordism ring. In other words, we will deviate from the previous notation and write the identity as

\displaystyle r(u) = \frac{1}{6} g'(u) \frac{\partial^3 f(u, 0)}{\partial^3 v} \in \Omega_{SO}[[u]] . \ \ \ \ \ (2)

In order to prove this, we will use (1) together with some elementary manipulations. Namely, we have that {r(u)} is the coefficient of {v^3} in {h(u, v)} involving {u}. Now we can write { h(u, v) = f(u, v) g'(u) g'(v) } and the strategy is to Taylor expand {f} in powers of {v}. Write

\displaystyle f(u, v) = u + \frac{\partial f(u, 0) }{\partial v} v + \frac{1}{2}\frac{\partial^2 f(u, 0)}{\partial^2 v} v^2 + \frac{1}{6}\frac{\partial^3 f(u, 0)}{\partial^3 v} v^3 + \dots

where the {\dots} signify terms involving {v^4} and higher (which don’t contribute to {r(u)}). Recalling that one has

\displaystyle g'(u)^{-1} = \frac{\partial f(u, 0)}{\partial v},

we find from the identity {h(u,v ) = g'(u) g'(v) f(u, v)},

Here, again, the {\dots} indicate terms involving {v^4} and higher. Now if we are looking for the coefficient of {v^3} in this, something miraculous happens: the first and third terms are even in {v}, and the second contains no term involving {u}. So, only the last term contributes.

We get for {r(u)}:

\displaystyle r(u) = \frac{1}{6} g'(u) \frac{\partial^3 f(u, 0)}{\partial^3 v},

as desired.

3. Completion of the proof

The goal is now to express {r(u)} completely in terms of the logarithm {g}, so that we can see directly that if {r(u) = u^2}, the logarithm comes from an elliptic integral. In fact, by definition of the logarithm

\displaystyle f(u, v) = g^{-1}(g(u) + g(v)).

The formula {r(u) = \frac{1}{6} g'(u) \frac{\partial^3 f(u, 0)}{\partial^3 v}} together with some computation to get the partial derivatives of {f} give

\displaystyle 6 r(u) = \frac{1}{2} ( b(u)^2)'' - b''(0),

where {b(u) = \frac{1}{g'(u)}}. So, in particular, if {r(u)} is a multiple of {u^2}, we find that {b(u)^2} must be at most quartic, and it is even. So we get

\displaystyle \frac{1}{g'(u)} = \sqrt{ Q(u)}

for a quartic {Q(u)} with only even monomials; this means that {g} comes from an elliptic integral.

I must confess that this proof still feels like magic to me; it’s not at all clear to me what is really going on, and I can’t really tell what to take away from it. Does anyone reading this have any ideas? Is there a “high-concept” explanation for this result?