In a previous post, I began discussing a theorem of Ochanine:
Theorem 1 (Ochanine) A genus annihilates the projectivization of every even-dimensional complex bundle if and only if the logarithm of is an elliptic integral
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.
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 as the zero locus of a generic section of the complex line bundle on . We wrote down an expression in the cobordism ring: namely, we proved the formula
for the the coefficients of the formal group law on . Another way to phrase this is that if we define the power series
and write for the formal group law of , then we have the formula
for 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 . That is, if we define analogous power series
and similarly, then we have
2. A basic formula
A key observation that we will need in the future is that the can be obtained as projectivizations. Suppose . Then the projection map
is a fiber bundle with fiber . In fact, it exhibits as a suitable projectivization of the bundle over .
Consequently, we can actually prove a stronger version of the theorem:
Theorem 2 If is a genus such that for all , then the logarithm of is an elliptic integral.
Define the power series
Then this series is a multiple of if and only if annihilates the .
In order to prove this result, we will show that, without any hypotheses on the genus,
for any genus. To ease notation, we will drop the 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
In order to prove this, we will use (1) together with some elementary manipulations. Namely, we have that is the coefficient of in involving . Now we can write and the strategy is to Taylor expand in powers of . Write
where the signify terms involving and higher (which don’t contribute to ). Recalling that one has
we find from the identity ,
Here, again, the indicate terms involving and higher. Now if we are looking for the coefficient of in this, something miraculous happens: the first and third terms are even in , and the second contains no term involving . So, only the last term contributes.
We get for :
3. Completion of the proof
The goal is now to express completely in terms of the logarithm , so that we can see directly that if , the logarithm comes from an elliptic integral. In fact, by definition of the logarithm
The formula together with some computation to get the partial derivatives of give
where . So, in particular, if is a multiple of , we find that must be at most quartic, and it is even. So we get
for a quartic with only even monomials; this means that 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?