UPDATE added 28.05.2019: An error (detailed in another post) in the paper that this post is about was discovered in Summer 2018, and has since been fixed, but it renders some portion of what is explained in this post either irrelevant or misleading. In particular, everything this post says about the “quadratic unique continuation” lemma should be taken with a grain of salt. I’ve written a pair of more recent posts to try and explain these matters properly.
In the last two posts, I have been describing a machine…
…well, no, not quite like that machine. I was speaking figuratively.
The machine is a stratification theorem, whose purpose is to demystify the transversality properties of multiply covered holomorphic curves. Its user interface consists mainly of a set of “twisted” Cauchy-Riemann operators associated to the normal operator of any multiply covered curve , producing a splitting
as described in the previous post. Operating the machine requires no specialized training beyond the ability to compute the indices of these operators—which are Cauchy-Riemann type operators defined on Sobolev spaces with exponential weight conditions over a punctured surface—plus a certain amount of patience with dimension-counting arguments. If you have this, then the machine gives you a stratification of the moduli space of multiple covers, with chambers in which transversality (or possibly the next best thing) is achieved, separated by smooth walls and further strata whose dimensions can all be computed.
Anyway, that’s what the instruction manual says it does. In this post I want to open up the machine and try to explain why it works.
The statement of the theorem
I am assuming you’ve read the original post in which all of the following notation was explained, but recall in particular that is a moduli space of -fold covered -holomorphic curves with prescribed critical and branching behavior for and respectively, and for and we consider the subset
Theorem D. For generic , is a smooth submanifold of with
where for each of the irreducible representations , the number is defined as the real dimension of .
The proof of this theorem follows the same general outline as most other theorems you’ve seen before that begin with the words “For generic …”. The relevant generic set comes from applying the Sard-Smale theorem to a projection map
where is a suitable Banach manifold of perturbed almost complex structures and is the resulting universal moduli space
The main step is thus to prove that this universal moduli space is a smooth Banach manifold, which will follow from the implicit function theorem after proving that a certain linearized operator is surjective. The latter is, as usual, the hard part, and it resembles more familiar arguments in that it requires a unique continuation result for linear Cauchy-Riemann type equations, but the precise lemma we need is probably different than what you are used to. This so-called “quadratic” unique continuation lemma is the heart of the machine, and in itself it is not very hard to prove, but it is somewhat of a challenge to understand what it means and what it is good for, so one of my main goals for this post will be to explain that.
It should go without saying that when the kernel and cokernel conditions in the definition of are dropped, the resulting universal moduli space
is indeed a smooth Banach manifold; this follows from standard arguments. Thus assuming you’re on board with what I said above about the Sard-Smale theorem, we can agree that the goal is to prove the following:
Main lemma. is a smooth finite-codimensional submanifold with codimension .
I cheated a bit in this statement by replacing with a new object that I have not yet defined. Its precise definition will come in a bit, but for now, suffice it to say that it is an open subset of , defined via an extra condition without which I wouldn’t know how to prove the lemma—but this extra condition will have no impact on the rest of the discussion.
What follows is an extended sketch of the proof, with occasional references to specific results in the paper where the full details are carried out. This may turn out to be somewhat weightier material than you are used to reading on a blog, so maybe it will help to have some appropriately weighty music as accompaniment; I personally recommend Beethoven’s Große Fuge.
Reduction to the case of regular covers
For reasons that were hinted at near the end of the previous post, we can impose the following simplifying assumption without loss of generality:
Assumption (cf. the beginning of Section 3.5). The branched covers for elements are regular (i.e. normal).
To see why this is not a loss of generality, you need to recall how the splitting of is defined in terms of a regular presentation of (see the previous post). Each branched cover of degree with generalized automorphism group has a (canonical up to isomorphism) regular branched cover that factors through it, with degree and . The holomorphic curve then has a splitting
whose summands are the same twisted Cauchy-Riemann operators that appear in the splitting of , but with different multiplicities ; in particular, the regularity of implies via a standard theorem in representation theory that all of the must be nonzero. Since and have exactly the same critical values, their respective neighborhoods in the space of branched covers with fixed branching data can be identified naturally, so that there are canonical identifications and in the neighborhoods of these two curves.
With the preceding understood, the regularity assumption will be in effect from now on. It has the following important consequences:
- acts by biholomorphic maps on the domain , and therefore also on the kernel and cokernel of the normal operator , by reparametrization.
- The multiplicities in the splitting of are all positive.
Notice now that if , then sufficiently nearby elements will belong to if and only if
Indeed, the splittings of these operators vary continuously as moves toward , and each summand is a Fredholm operator, so the dimension of its kernel can jump downward under small perturbations, but not upward. (If you are unfamiliar with this fact, it will follow easily from the discussion of the space of Fredholm operators below.) Since every twisted operator appears with positive multiplicity in the splitting of , any downward jump in will necessarily cause a downward jump in .
How to perturb a normal Cauchy-Riemann operator
To understand the local structure of , we need to understand the effect that moving through has on the normal operators . This is fairly straightforward if we consider only variations in that leave fixed, which will suffice for our purposes. Fix a closed -holomorphic curve , where is somewhere injective, and consider a 1-parameter family with such that along the image of , so remains -holomorphic for all . Its normal bundle also remains unchanged as the parameter moves, but the normal Cauchy-Riemann operator depends on , so let us denote the resulting 1-parameter family of operators by
Now vanishes along the image of , but if is any injective point of , then there is considerable freedom to choose the derivative of near in directions normal to . Choose and write the normal derivative in terms of the tangent-normal splitting as
One can now compute that if is chosen to vanish near all non-injective points of , then the change in the normal Cauchy-Riemann operator is precisely
The point is that we can choose families to make this perturbation more or less anything we want. If we’re working in a symplectic manifold , then of course we’d like to require all the to be compatible with , which gives a nontrivial relation between and , but this does not constrain at all if we are willing to let it determine . The one caveat is that if has , then every perturbation of produced in this way will automatically be -invariant. The important result can thus be stated as follows:
Lemma 1 (see Lemma 6.1). Let denote the open and dense set of injective points of . Then given any smooth -invariant zeroth-order perturbation with support in , there exists a smooth 1-parameter family of compatible almost complex structures , satisfying and matching along the image of for all , such that .
Generic are nowhere integrable
It’s time to fill in the missing detail about the definition of the constrained universal moduli space . For technical reasons that will become clear when we discuss unique continuation at the end of this post, I need to impose an extra open condition on pairs . It amounts to the requirement that must not be integrable on any neighborhood of the image of , though non-integrability as such is not really the point, but is more of a side-effect.
Recall that for any complex vector space , a real subspace is called totally real if . (Sometimes one also requires to have half the dimension of but I am not requiring that here.) Since is a real-linear operator between two complex Banach spaces, one can ask in particular whether its kernel and cokernel (meaning the kernel of its formal adjoint) are totally real. There is an easy criterion for this: recall that if we break up into its complex-linear part and antilinear part , then the latter is a zeroth-order term, meaning a smooth bundle map . Using the standard unique continuation results for linear Cauchy-Riemann operators, one can easily show (cf. Lemma 3.11) that and are guaranteed to be totally real whenever the antilinear bundle map is invertible on some fiber. This is manifestly an open condition, and we shall say that itself is totally real whenever it holds. Notice that if this holds for a given curve , then it automatically also holds for all of its multiple covers .
Definition. Let denote the open subset consisting of pairs for which is totally real.
If is integrable, then is always complex linear and the totally real condition can never be satisfied. Of course, this should not worry us very much, because we are trying to prove a theorem about generic , which cannot be expected to be integrable. It turns out in fact that genericity is enough to guarantee that the totally real condition is always satisfied, and this is why establishing the smoothness of will suffice for proving Theorem D.
Lemma 2 (cf. Lemma 6.2). For generic , every closed -holomorphic curve has the property that is totally real.
The proof of this is not hard once you’ve absorbed the implications of Lemma 1. The point is to show that the universal moduli space of somewhere injective curves that fail to satisfy the totally real condition lives inside a submanifold of arbitrarily large codimension, namely the set of curves such that satisfies an incidence relation with the subvariety of noninvertible linear maps at some chosen point . Having the invertibility condition fail on an open neighborhood of means that the incidence relation can be taken to involve jets of arbitrarily high order, cutting out submanifolds of arbitrarily large codimension.
Walls in the space of Fredholm operators
I would now like to tell you a lovely fact about the space of Fredholm operators that everyone ought to know: for any pair of real Banach spaces and and any integers , the subset
is a smooth finite-codimensional submanifold in the space of bounded linear maps with
One can see it as follows. Given an operator , is evidently Fredholm, so there exist splittings , into closed linear subspaces, with and , thus , and restricts to as a Banach space isomorphism These produce a block decomposition for any bounded linear operator in the form
such that will necessarily be invertible whenever lies in a sufficiently small neighborhood of . We can therefore define a smooth map
whose derivative at is
and is thus manifestly surjective. Now if we also associate to each the linear “coordinate change” on defined in terms of the splitting by
we have , implying
Since is defined on the space with dimension , this implies that for all sufficiently close to , and equality is satisfied if and only if . As a consequence,
and since is surjective, the implicit function theorem gives this zero set the structure of a smooth submanifold with codimension equal to .
Walls in the universal moduli space of multiple covers
We now proceed toward the proof of the main lemma. Given , we observed already that a sufficiently close element will also belong to if and only if , or equivalently if the two cokernels have matching dimensions. Plugging this into the above discussion about Fredholm operators in general, there is a neighborhood of and a smooth map
whose zero set is a neighborhood of in . It is important to notice moreover that in light of the natural action of , can be arranged to have its image in the space of –equivariant linear maps , i.e.
Computing the dimension of the space on the right hand side requires some representation theory: each of and are now representations of , and their decompositions into irreducible representations can be deduced from our splitting of . Schur’s lemma then breaks up the elements of into blocks that always must vanish when they correspond to two non-isomorphic representations, and the result (cf. Equation (3.22)) is
The implicit function theorem will now complete the proof of the main lemma if we can prove that the linearization of at is surjective. Let us write down the derivative of this map in directions of the form that we considered in Lemma 1. Denoting by the zeroth-order perturbation of that results from varying in the direction, the derivative in question gives a linear map
of the form
where denotes the natural linear projection map from the relevant Sobolev space of sections of to .
Lemma 3 (cf. Lemmas 5.4 and 6.4). The operator described above is surjective.
This lemma is the main technical step. To prove it, one can restate the problem in light of Lemma 1 as follows. Let us choose suitable bundle metrics and area forms so that there are well-defined -pairings on spaces of sections, and we can thus identify with the kernel of the formal adjoint of . This makes the -orthogonal complement of , so that the matrix elements of the linear transformation for any given take the form
for and .
So we need to know that for any given -equivariant linear map , we can find a -invariant zeroth-order perturbation , with support away from the non-injective points of the underlying simple curve , such that
for all and .
Notice that if we can find a solution to this problem that is not -invariant, then we can always symmetrize it to produce one that is; this is possible due to the -equivariance of We are therefore free to ignore the -symmetry from now on, and simply look for any zeroth-order perturbation that is supported in a given open set and satisfies the above relation for a given linear map .
This problem does not sound unsolvable when you consider that is required to live in a finite-dimensional vector space, while we are free to choose from a space that is infinite-dimensional. Arguing by contradiction, suppose there is no solution, or equivalently, that contains a nontrivial element orthogonal to every element that can be produced by choices of zeroth-order perturbations . This can be expressed more concretely in terms of the matrix elements with respect to orthonormal bases of and of : if the solution we’re looking for does not exist, then there is a set of real numbers , not all equal to zero, such that
for every in the space of allowed perturbations. This doesn’t sound very likely, but there is one conceivable situation where we would now be in big trouble: to see this, let us write the -product more explicitly as an integral of a real bundle metric , which we can view as a fiberwise linear form on the tensor product of the bundle with itself. The above expression then becomes
The summation in parentheses in this integral is a section of the tensor bundle , defined as a linear combination of products where satisfies a linear Cauchy-Riemann type equation and satisfies its formal adjoint equation. It is not difficult to show (cf. Lemma 5.5) that if this linear combination is nonzero on some open set, then can be chosen with support in that set to ensure that the integral is nonzero, thus giving a contradiction. We know that the and all satisfy unique continuation results, so we can easily find an open set on which all of them are nonvanishing. Is it really possible that a linear combination of this form could nonetheless vanish?
Quadratic unique continuation
Actually yes: if we’re not careful, such a linear combination certainly could vanish. Let’s put the problem in slightly more general terms: assume is a real-linear Cauchy-Riemann type operator on some complex vector bundle over a Riemann surface , with , and denotes its formal adjoint with respect to some chosen -pairing. The question we need to consider is local, so we place no assumptions on the base , i.e. it could be simply a disk.
Question: If and are finite-dimensional subspaces, must the natural map
defined by be injective?
Answer: In general, no. For example take and to be the trivial line bundle over , with and , so their kernels are the spaces of holomorphic and antiholomorphic functions respectively. Now define to be the complex span of the functions and , while is the complex span of and . Then
defines a nontrivial element in the real tensor product of the vector spaces and , but it also defines the zero-section of the bundle .
It turns out that the reason for the horror scenario in the above example is that we allowed the spaces and to be complex. This is where the extra condition in our definition of comes into play, as we will see that the problem goes away if we require and to be totally real.
UPDATE 28.05.2019: The following lemma was the error that is mentioned in the caveat at the top of this post. It is now known to be false, but has been replaced with a similar statement that is true. See this more recent post for details.
Lemma 4 (quadratic unique continuation, cf. Proposition 5.1). For any sets of complex-linearly independent sections and in the above setting, no linear combination
with coefficients satisfying for some vanishes to infinite order at any point.
This is a statement about complex bases and a complex tensor product, but in light of the canonical surjective bundle map and the fact that any real basis of a totally real subspace is also complex-linearly independent, it has the following consequence:
Corollary. If the finite-dimensional subspaces and are both totally real, then the natural map
is injective, and the nontrivial sections in its image do not vanish to infinite order at any point.
You’ll gain some valuable intuition about Lemma 4 if you work out the case where on a trivial vector bundle, so the are all holomorphic functions and the are antiholomorphic functions. The proof in this situation is not very hard: it hinges on the fact that all terms in the Taylor series of are powers of , while those in the Taylor series of are powers of . Putting them together in a complex tensor product thus leaves them “decoupled” so that no nontrivial linear combination can kill them, so long as the original sets of Taylor coefficients are complex-linearly independent. For the general case, one cannot use Taylor series, but it is still useful to note that—as a corollary of the usual unique continuation results (based on the similarity principle or whatever else you prefer)—the first nonvanishing term in the Taylor expansion of each is a power of , and similarly for with powers of . The proof thus proceeds following the same idea as in the (anti-)holomorphic case, but paying specific attention to the first nontrivial term in the Taylor expansion at each step.
A concluding remark on integrability
I’m sure no one will argue with me when I say that integrable complex structures are not “generic,” and indeed, some of the corollaries of Theorem D are known to be false in the integrable case. For instance, Bryan and Pandharipande have found examples in algebraic Calabi-Yau 3-folds for which super-rigidity fails. While I honestly don’t know whether the totally real condition in my definition of is essential, I find it amusing and slightly poetic that the generic almost complex structures provided by Theorem D are guaranteed to be non-integrable. That does not mean of course that results like super-rigidity cannot possibly hold in integrable settings: there are also known cases in which they do, but when it happens, it is for reasons completely unrelated to any of what I’ve been talking about here.
Is the Große Fuge over yet?
Update (19.12.2017): I have edited this post (and will shortly also be updating the paper on the arXiv) to correct a minor error in representation theory. The correction necessitated a change in the definition of the space , so that some dimensions that used to be real are now dimensions over the endomorphism algebra . This change (fortunately) has no adverse impact on the main applications concerning super-rigidity and transversality for multiple covers. Many thanks to Thomas Walpuski and Aleksander Doan for catching the error.