The following is a thing I sometimes say to my students, or anyone else willing to listen to me:

I’m going to try and explain in this post what I mean by that. In particular, I’m going to state a couple of rules for dealing with -spaces that I believe would make symplectic topology a *slightly* happier field if they were generally observed. My students should definitely read this post, because I promise to give them trouble if they submit a thesis that does not follow these rules. There are surely also people who are not my students and ought to read this post, but many of them won’t, and that’s probably not the end of the world.

Probably.

## What on Earth are you talking about, Chris?

I should remind you first what the Floer -space actually is and what it is for. Suppose you want to prove a statement of the following form, which I will refer to from now on simply as “the GTT”:

*Generic transversality theorem** (GTT): For generic , the moduli space is cut out transversely.*

For concreteness, most of my readers will be happy to imagine that is a space of smooth tame/compatible almost complex structures on some symplectic manifold, is a moduli space of J-holomorphic curves, and “cut out transversely” means that every curve in is *Fredholm regular*, implying that is a smooth manifold or (if there are symmetries) orbifold of the “expected” finite dimension derived from a Fredholm index. Results of this kind are also quite standard in other contexts involving nonlinear elliptic PDEs, e.g. in gauge theory. Whichever context you prefer to imagine, I want to assume in general that the set of geometric data on which the PDE is based consists of smooth sections of some fiber bundle over a compact manifold , and is endowed with the -topology. This is the most natural assumption to make in most geometric settings. (Given the name of this blog, I should add that one can also allow to be noncompact if one requires all sections in to match some fixed section outside of a fixed compact subset — this is what one does for defining continuation or cobordism maps in any Floer-type theory. But for this discussion, I will assume is compact just for simplicity.)

Now, up to thorny technical details that we’ll get to in a moment, there is a standard playbook for proving the GTT:

- Define a
*universal moduli space*, and prove via the implicit function theorem that it is a smooth Banach manifold. - Observe that the projection is a smooth Fredholm map, so the Sard-Smale theorem gives a comeager subset such that is cut out transversely for every .

There is an immediate impediment to implementing this strategy if is indeed a space of *smooth* objects with the -topology: is not a Banach manifold, thus the universal moduli space will not be one either if it is defined as described above. In the case I deal with most often, where is a space of smooth almost complex structures on a compact manifold, is at best a Fréchet manifold, not a Banach manifold, just as the space of smooth functions on a compact manifold is a Fréchet space with the -topology, but not a Banach space. Unfortunately, there is no contraction mapping principle in Fréchet spaces, and thus no implicit function theorem (at least not unless one wants to use something overly complicated like Nash-Moser, which I don’t). In short, the naive definition of the universal moduli space does not work; one needs a cleverer trick to define something that is useful.

I’m aware of two such tricks that are popular among symplectic topologists:

**Option 1:**Pick a large integer and replace with its completion in the -topology, which is a Banach manifold. This is the approach taken by e.g. McDuff and Salamon’s book on J-holomorphic curves.**Option 2:**Replace with a smaller space that only contains smooth objects but is endowed with a finer topology making it a smooth Banach manifold. A suitable topology for this purpose was introduced in one of Floer’s original papers on Floer homology, and is thus known as the*Floer -topology*.

The disadvantage of Option 1 is that necessarily contains elements that have only finitely-many derivatives, so the nonlinear operator defining will no longer be smooth in general, and itself becomes (at best) a manifold of class for some large , but not a *smooth* manifold. That is not the end of the world, but it is a major pain in the neck if at every step you have to keep track of how many derivatives of your finitely-differentiable functions you have not yet burned up. I personally prefer to avoid this whenever possible.

Option 2 avoids that problem because Floer’s -space contains only smooth objects. To sketch the idea, suppose first that we are interested in smooth sections of a finite-rank vector bundle over the compact manifold , and we make the usual choices (e.g. a finite covering by coordinate charts and local trivializations) so that the -norm of such a section is well defined for each integer . Choose a sequence

such that ,

and define the -norm of a smooth section by

.

It is easy to check that the space of smooth sections with is a Banach space with respect to this norm, and its obvious inclusion into the Fréchet space is continuous, i.e. -convergence implies -convergence.

For applications to the GTT, one typically fixes a “reference” object and defines the Banach manifold as something along the lines of a -small neighborhood of in . To say this more precisely, let us assume that really is a Fréchet manifold, so every has a tangent space , which is the Fréchet space of -sections of some vector bundle, and one can define an “exponential map”

that sends a neighborhood of 0 homeomorphically to a neighborhood of . (There are typically easy direct ways to define so that it has this local homeomorphism property — one need not think about connections on infinite-dimensional Fréchet manifolds or anything so exotic.) One can then fix a sufficiently -small neighborhood of 0 and define

.

This can be regarded as a smooth Banach manifold in a trivial way: is an open subset of a Banach space with the -norm, and is the inverse of a global chart identifying this subset with . Since only one chart has been defined, there is no need to worry about transition maps.

It is important to understand however that since not every smooth section has finite -norm, does not contain *every* smooth element in , and in fact it does not even contain a -neighborhood of . In this sense, is not a remotely natural space to work with; is much more natural by comparison. However, the obvious inclusion is clearly continuous, and this means that does contain (some, but not all) arbitrarily -small perturbations of the particular element ; moreover, there is nothing special about , as it can be chosen arbitrarily before defining . As we will see below, this makes a good enough space for use in proving the GTT, and — in my opinion at least — proving it this way is typically less painful than dealing with finitely-differentiable moduli spaces.

## The Floer space is even less natural than it looks

I’m planning to say some very positive things about the -topology in a moment, but first, I want to make sure you’re fully aware of its flaws.

Recall that the -norms on sections of the vector bundle are not canonically defined, but since is compact, they are well defined up to equivalence, so the -topology is canonical. Here’s a bit of bad news that may not have occurred to you yet: the -norm is a linear combination of *infinitely many* norms, each of which depends on choices. If we had only finitely many norms to add up, then we could easily prove that the -topology is similarly independent of choices. But if we pick and modify infinitely many of the -norms within their individual equivalence classes, can easily become infinite. In other words, for any given sequence , **the space and its topology are not canonically defined**. They depend on further choices such as local charts and trivializations.

This leads me to the first of the two rules I’d like to propose:

Floer obeyed this rule, though several illustrious people since then have occasionally disregarded it, and some have even ended up with the impression that if one chooses to use in a proof of the GTT, then must also be mentioned in the statement. At the end of the day, though, you want to have a result about generic -small perturbations of your geometric data, not generic perturbations in a much finer and completely unnatural topology that depends on arbitrary choices.

I’m here to tell you that this is possible.

## The Floer space is tricky to get one’s hands on

In the standard proof of the GTT using -spaces, the main thing one needs to know about them is an easy lemma that was proved essentially by Floer — I will state it somewhat informally as follows:

*Bump function lemma** (see Lemma 5.1 in Floer’s paper): If the sequence has sufficiently rapid decay, then contains sections with arbitrarily small support around any given point and arbitrary values at that point.*

This lemma basically says that for a given and any point , one can without loss of generality assume there exist small perturbations of that are “pushed” in any desired direction near but match everywhere else. In practice, this makes a “big enough” space of perturbations to prove the GTT.

In more general applications, however, the bump function lemma does not always suffice, and if you’re not thinking from the right perspective, working with -spaces can then start to seem harder than it actually is. The following is a slight simplification of a question that I recently found myself banging my head against:

**Frustrating Question**: Given a smooth submanifold , a point and a linear map that vanishes on , does there exist a function of class that satisfies and ?

This question arises unavoidably in the approach to equivariant transversality that I’ve recently been trying to promote via my paper on super-rigidity. The answer would be obviously “yes” if we only needed to be smooth, but the ability to make its -norm finite while also choosing it to vanish on the submanifold seems to depend on information about that is not given. In the application I have in mind, is the image of an arbitrary holomorphic curve in a moduli space that is completely unknown, so answering the question seems hopeless. I briefly had the terrible feeling that I was going to have to switch to Option 1 and rewrite large portions of my super-rigidity paper to accommodate finitely-differentiable almost complex structures.

But then I realized that I was thinking about it the wrong way around, and thus learned the second rule:

Let me explain.

## Lack of naturality is not a bug, it’s a feature

There’s one thing about -spaces that you must never, ever forget: just as the topology of depends on plenty of noncanonical choices, the sequence is in itself a noncanonical choice, and you are free to change it. In particular, it can always be useful to make converge to 0 *even faster*. For crying out loud, that’s why it’s called !

Let’s formalize this idea a bit. Denote by the set of all sequences of positive numbers that converge to 0, and define a pre-order on by

.

Intuitively, means that the sequence decays to 0 at least as fast as . One can now define statements of the form “ holds whenever has sufficiently rapid decay” to have the precise meaning, “There exists an such that holds for every “.

**Pre-order L***emma**: Every countable subset of has a lower bound with respect to the pre-order . Moreover, the -spaces of smooth sections of have the following properties:*

*There is a continuous inclusion whenever .**For any countable collection of smooth sections of , one has for all if has sufficiently rapid decay.*

**Proof**: A lower bound for a countable collection of sequences is given by with . The statement about inclusions follows easily from the observation that if and only if there exist constants and such that holds for all . The last statement now follows after observing that any has if for all .

The ability to choose decaying faster than any *countably infinite* collection of choices is a very useful bit of freedom that should not be underestimated. One can use this for instance to make dense in Banach spaces such as , or for , since these are all separable and already contain as a dense subspace.

We can now resolve the question that I was recently banging my head against.

*Answer to the Frustrating Question**: This is the wrong question to ask.*

Indeed, we already know how to construct the desired function if it only needs to be *smooth* instead of belonging to a given -space. The solution is thus simply to construct a smooth function , and *then *choose so that . Now of course, there may actually be infinitely many such -functions we need to find for different choices of the data , however… in all situations I’m familiar with, one can get away with restricting to a *countable* set of such choices, in which case the Pre-order Lemma gives exactly what we need.

To show how this works, let’s work through a slightly novel take on the standard proof of the GTT.

## Proving the Generic Transversality Theorem

To avoid excessive vagueness, we will be concrete now and assume is a closed -manifold, is the space of all smooth almost complex structures on , and is the space of (parametrized) somewhere injective -holomorphic spheres with its natural -topology. Equivalently, is the zero-set of the nonlinear Cauchy-Riemann operator,

,

which we can regard as a smooth section of the Banach space bundle with fibers over the Banach manifold for some . (Note that the smoothness of the section depends on being smooth — this is one of the nice things we’d have to give up if we were following Option 1.) Linearizing at gives rise to a linear Fredholm operator

,

and we call *Fredholm regular* whenever this operator is surjective. This is equivalent to the condition that is a *transverse* intersection of the section with the zero-section of the Banach space bundle . The goal is to find a comeager subset such that for every , the intersection of with the zero-section is everywhere transverse.

**Warm-up**

Let’s first quickly work through what I will call the “fairyland proof” of the theorem — in a fictional world where unicorns are real, sushi grows on trees, and is a smooth Banach manifold, this is how the proof would go.

**Fairyland proof of the GTT**: The universal moduli space is the zero-set of the smooth section

,

where denotes the obvious extension of to a bundle over . Its linearization at some is then a bounded linear operator

,

where we are pretending is a Banach space, and we recall for concreteness that

for some . We claim that is always surjective. Since is Fredholm, a standard exercise in functional analysis implies that has closed image, so we only need to prove that its image is also dense. If it isn’t, then there exists a nontrivial section for that annihilates the image of , implying the two conditions

for all for all .

The first condition means is a weak solution to the Cauchy-Riemann type equation , so by elliptic regularity and the similarity principle, it is smooth and has only isolated zeroes. We can then pick an injective point of at which is nonzero and find a smooth section with support near to make positive, producing a contradiction that proves the claim. The rest of the proof consists of standard applications of big theorems: the surjectivity of implies via the implicit function theorem that is a smooth Banach manifold, and applying the Sard-Smale theorem to the projection gives the desired comeager set of regular values , for which intersects the zero-section transversely.

**Departure from fairyland**

The lazy way of transporting the proof above out of fairyland and into the real world is to replace with the Banach manifold at every step. That produces a correct argument, but it proves a slightly different theorem than the one we really want, one that violates **Rule 1** by providing generic -small perturbations instead of -small perturbations. Floer’s way around this was to settle for a slightly weaker result: since contains arbitrarily -small perturbations of , and one could have chosen *any* element of to call , the argument does provide a *dense* set of almost complex structures in that achieve transversality. “Dense” is a weaker condition than “comeager”, but it is enough for most applications, e.g. it certainly suffices for defining Floer homology. On the other hand, sometimes one would like to intersect the set of regular data with some other generic subset and know that the intersection is still nonempty, in which case dense sets are not good enough, though comeager sets would be.

But there is also a subtler problem: not every step in the fairyland proof has a completely straightforward adaptation for . The trickiest step is where one needs to find an element with support near such that . This was easy in fairyland because finding smooth bump functions with arbitrarily small support is easy; in the real world, this is where one needs to apply Floer’s bump function lemma, which requires to have sufficiently rapid decay. But the application of this lemma is also not so straightforward, for another reason that I haven’t mentioned yet: one of the irritating features of as we’ve defined it is that for any given , it is not so easy to say precisely which sections actually belong to and which do not. The exception is the case , since the construction clearly identifies as the space of elements in that have finite -norm. You may have noticed in any case that by discussing this issue, we are running afoul of **Rule 2**. We are thinking backwards.

**How to prove it without violating the rules**

By this point, most of the necessary ideas for a correct but maximally stress-free proof have been mentioned, they just need to be assembled in the right way. Here we go.

*Step 1**: The universal “epsilon-regular” moduli space.*

*Step 1*

Fix arbitrary and and use these to define the Banach manifold in the usual way. Notice that while this space depends on the choice of sequence , the “reference” almost complex structure belongs to for every choice. One can now define the usual universal moduli space

,

and present it as the zero-set of a smooth section of the obvious extension of to a Banach space bundle over . Let us denote the linearization of this section at by

,

and notice that is just the restriction of the operator in the fairyland proof to a smaller domain. At this point, I find it useful to introduce the following bookkeeping device:

**Definition**: An element is *-regular *if the operator defined by linearizing at is surjective. Similarly, given any smooth almost complex structure , a curve can be called *-regular* if belongs to and the pair is -regular.

Note that a curve can be -regular without being Fredholm regular, as the former is a smoothness condition concerning the neighborhood of (or more accurately of ) in the *universal* moduli space, rather than in . Clearly -regularity is also an open condition, so

is an open subset of . The implicit function theorem then implies that is a smooth Banach manifold, and applying the Sard-Smale theorem in the usual way to the projection gives:

*Lemma* 1:* For every , there exists a comeager subset such that for each , every -regular curve is Fredholm regular.*

Notice that so far, we have not actually done any work, we just applied some standard theorems in a standard way. The lemma is, at this stage, correspondingly free of content: we have not yet shown that the set of -regular curves in is ever nonempty. That is the next task.

*Step 2: **Proving epsilon-regularity for one curve*

*Step 2:*

None of the ideas in our fairyland proof above were fundamentally wrong, they just were not applied in the right context. Salvaging the work done in the main technical step now leads to the following lemma. Note that in this statement, ; since the tangent space for is not easy to describe, we shall avoid thinking about it altogether.

*Lemma 2**: For any given curve , is -regular for all with sufficiently rapid decay.*

**Proof**: The fairyland proof contains a completely valid argument showing that the operator has dense image. Since is a separable Banach space, we can then choose a dense sequence , along with a sequence satisfying for all . By the Pre-order Lemma, all of the smooth sections are of class for with sufficiently rapid decay, hence they are in , and it follows in this case that the image of is also dense. Since it is already known to be closed, the result follows.

*Step 3**: Every space you care about is second countable.*

*Step 3*

The next statement strengthens Lemma 2 via a change in the order of quantifiers.

*Lemma 3**: For all with sufficiently rapid decay, every curve in is -regular.*

**Proof**: As a subset of the separable metrizable space , is also separable, so in particular, it is a second-countable topological space, and thus has the property that every open cover has a countable subcover. Now since -regularity is an open condition, we can apply Lemma 2 and associate to each some and a neighborhood of such that

.

Pick a sequence such that the open sets still cover . The statement then holds if is chosen to be any lower bound for the countable set .

*Step 4**: The Taubes trick*

*Step 4*

Traditionally, the so-called Taubes trick appears in the literature as a method for converting slightly inelegant statements about comeager subsets of or into more elegant statements about comeager subsets of . Some authors don’t bother with it, and thus settle for slightly inelegant statements, especially if they don’t care about violating **Rule 1**. But in this version of the proof of the GTT, the Taubes trick plays a slightly more prominent role.

The Taubes trick depends on the ability to exhaust by a countable collection of compact subsets that depend continuously in some sense on . Let us assume in particular that for each and , a subset can be defined that has the following properties:

- (
**Exhaustion**) . - (
**Compactness**) For any -convergent sequence and any fixed , every sequence has a subsequence -convergent to an element of .

There are various ways to define in general, e.g. by imposing uniform -bounds to force compactness and other conditions to prevent the loss of injective points; such details are tangential to the present discussion, so we will omit them. We can now define

for each as the set of all for which every curve in is Fredholm regular. By the **Exhaustion** property stated above, the countable intersection is then precisely the set that we would like to show is comeager. This follows from the next two lemmas.

*Lemma 4a**: For each , is open.*

**Proof**: If not, then there exists a and a convergent sequence such that for all , meaning there also exists a sequence of curves that are not Fredholm regular. The **Compactness** property then provides a subsequence of convergent to some , and must be Fredholm regular since . That is a contradiction, as Fredholm regularity is an open condition.

Now comes the more interesting part.

*Lemma 4b**: For each , is dense.*

**Proof**: Since the “reference” almost complex structure in the definition of can be chosen arbitrarily, it will suffice to prove that there exists a sequence converging in the -topology to . By Lemma 3, we can choose some with sufficiently rapid decay so that every curve in is -regular. Lemma 1 then provides a comeager subset such that every -regular curve in is also Fredholm regular for . Comeager subsets are dense, so we can choose a sequence that converges in the -topology to , in which case also converges to in . We claim that for each , belongs to for all sufficiently large. If not, then there exists a sequence of curves that are not Fredholm regular, and the **Compactness** property allows us to replace with a subsequence that converges to some . In particular, now converges to in . Since -regularity is an open condition, it follows that is also -regular for all sufficiently large, and is therefore Fredholm regular since , contradicting our assumptions.

## Final word

Here’s what I want to point out about the proof we’ve just completed: every step where one actually has to do some work (rather than just quoting a big theorem) is carried out in the -setting, and I would conjecture that one can get away with this in almost any transversality proof if one approaches it in the right way. One then just needs to know that the collection of *all* the -topologies for all choices of is a good enough “approximation” to the -topology, in a sense that is made precise by the Pre-order Lemma.

If you’re curious to see how one might apply this strategy in more general situations, like the one that motivated the *Frustrating Question* stated above, you’ll find examples in Sections 5.4 and 6 of the newest revision of my super-rigidity paper, which appeared on the arXiv this week.

What one definitely should *not* do is try to understand what it “means” for a function to be of class . This question has no deeply meaningful answer, and one can do considerable harm to one’s own peace-of-mind by thinking about it. I know this from experience.