3.3 Automatic transversality [049T]
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3.3 Automatic transversality
In our later applications, it is not enough to just have a bordism current between Lagrangians constructed from perturbed pseudoholomorphic curves. Two additional conditions are desirable: automatic transversality and positivity condition. These are natural properties of the highly idealized picture of Lotay and Pacini (cf. section 2.9), but may seem rather strong for Floer theorists.
In this section we discuss various sufficient conditions for automatic transversality, which intuitively means that the bordism current is constructed without perturbing the integrable complex structure. This requires that the (extended) linearized Cauchy-Riemann operator is surjective, namely the moduli space is regular. The next section will discuss the positivity condition. Complex integrability and the existence of holomorphic volume form will be assumed throughout. All holomorphic curves are assumed to be nonconstant.
- •
(Automatic transversality) There exist a finite collection of -dimensional smooth moduli spaces of holomorphic curves with respect to the integrable complex structure, constructed from the inputs in , and the bounding cochain data as in section 3.1, such that by taking the weighted sum of the -dimensional universal families of holomorphic curves, we obtain a current with . This bordism current agrees with the bordism currents constructed from generically perturbed almost complex structures, up to the boundary of an -dimensional current.
Morever, considering the boundary of holomophic curves varying in the dimensional regular moduli spaces, we obtain -dimensional universal families, sweeping out the cycle ; we require the evaluation map from these -dimensional universal family to to be immersions, except at the corner points of mapping to the Lagrangian intersections, where the failure of immersion is ‘minimal’ (see below for details). We say that the bordism current consists purely of ‘automatically transverse curves’.
- •
(Automatic transversality, weak version) We can allow certain holomorphic curves arising in -virtual dimensional moduli spaces, which are not automatically transverse, subject to the following requirements on these extra bad curves:
- 1.
When virtual perturbation theory is taken into account, still holds.
- 2.
At any such bad curve , given any first order deformation vector fields, the 1-form restricted to vanishes identically. Intuitively, this means vanishes identically on the Zariski tangent space of the universal family at . As a caveat, these Zariski tangent spaces may be higher dimensional.
- 3.
The boundary evaluation for all such bad holomorphic curves is contained in some subset of of Hausdorff dimension . As such, at almost every point on , only automatically transverse curves pass through it.
- 4.
The Solomon functional can be computed by integrating only on the part of consisting of automatically transverse curves.
- 1.
The automatic transversality assumption should be viewed as a higher dimensional generalization of the fact that on Riemann surfaces, the nontrivial holomorphic polygons are immersions up to the boundary (cf. [69, Section 13 (b)]. The intuition for the weak version is that we sometimes need extra holomorphic curves to maintain , but for questions related to the Solomon functional and the boundary evaluation to the Lagrangians, these extra curves do not contribute.
Index theory preliminary
For a pseudoholomorphic polygon with inputs at and an output at , arranged in clockwise order, the index is , where the degree convention is (63). The index amounts to a Maslov number computation, and an alternative topological description is as follows: take a section of the complex line bundle , which restricts on to a section of the real line bundle . (When several Lagrangians are involved, it is understood that refers to the appropriate Lagrangian on the portion of .) We assume has isolated zeros up to the boundary and the corner (aka. strip like ends). We may also regard as a function on the polygon, by contraction with . Then
| (23) |
Here the order of zeros is computed from winding numbers, and for general sections may take positive and negative values. At a corner where passes from to in the clockwise direction, we can put the tangent spaces into the standard form respecting the complex structure
| (24) |
so if in the complex coordinate of the upper half plane model, the excess vanishing order at the corner is . Formula (23) is equivalent to the standard index formula by a topological version of the Cauchy residue formula.
3.3.1 Automatically transverse cases
Holomorphic strip
We first consider the holomorphic strip case with input and output , and the integrability of the complex structure will be important. The first order deformations of the holomorphic strips are given by holomorphic sections of , which takes boundary value in over , and decays at the corners.
Lemma 3.5.
If are first order deformation vector fields, then either everywhere on , or we must have , and when the equality holds then only vanishes at the ends with excess vanishing order zero.
Proof.
We have a section of given by , which takes boundary value in on . Since are all holomorphic, so must be the function . Assume this function is not identically zero. By holomorphicity, the zeros are isolated. We claim that the order of zeros must be nonnegative everywhere. This is clear for the interior and the boundary points. We analyze the ends of the strip as the origin in the upper half plane model with holomorphic coordinate , putting in the standard form at the corner point. The deformation vector field has the leading asymptote
hence
and the excess vanishing order is nonnegative. By the index formula (23), the index , and when equality is achieved all vanishing orders must be zero. In particular at the corners. ∎
Corollary 3.6.
(Automatic transversality, strip case) Suppose . If are -linearly independent at some point on away from the two corners, then span the space of all first order deformations, the obstruction space vanishes, and the moduli space is smooth at . Morever, the holomorphic strip is an immersion up to the boundary.
Proof.
Since are -linearly independent at a point on , they span at the point, so . By the Lemma above are pointwise complex linearly independent as sections of the holomorphic vector bundle over , so any holomorphic first order deformation can be written as
The functions are holomorphic on up to boundary, and even up to corners due to . Now subtracting a constant linear combination of , we can ensure vanishes at any chosen point on . Then has a zero, so must be identically zero by the above Lemma, whence identically. Similar all , so . This proves that span all first order deformations. Since the index is , and the first order deformation space is -dimensional, we must have vanishing obstruction space.
There is a special deformation vector field from translation. The nonvanishing result then implies that the holomorphic strip is an immersion up to boundary. At the corners, the holomorphic strip is to leading order
By , this translation vector field cannot be at the corner, so for at least one choice of , we have . We say the failure of immersion at the corner is ‘minimal’. ∎
Holomorphic polygon
We now move on to holomorphic polygons with corner points for . The extended linearized Cauchy-Riemann equation (cf. [69, Chapter 9]) involves a vector field decaying at the ends, and representing a tangent vector of the Stasheff associahedron (i.e. the deformation of Riemann surface structure on the domain ), satisfying
where is the complex structure on . Here can be taken to be compactly supported, so near the corners. It immediately follows that
Lemma 3.7.
Given first order deformation vector fields , then the (1,0)-form on is holomorphic.
Remark 3.6.
Adding vector fields on valued in to does not affect as a 1-form on . Thus this 1-form is insensitive to how one represents the Riemann surface structures on the abstract polygon.
We impose that the input at one of the corners maps to an intersection point in , and the other inputs map to degree one self intersections of or . The output maps to .
Proposition 3.8.
(Automatic transversality, polygon case) Suppose are linearly independent first order deformation vector fields. Either vanishes identically as a 1-form on , or we must have , and when the equality holds then only vanishes at corners. In this case, all first order defomation vector fields are spanned by , the holomorphic polygon is an immersion up to the boundary, the obstruction of the extended linearized operator vanishes, and the moduli space is smooth at .
Proof.
By viewing the domain of the polygon as a strip with extra boundary punctures, we produce a holomorphic vector field as the -translation vector field. However, unlike in the strip case, at the degree one self intersection corners does not typically have the required decay to be admitted as a deformation vector field. Indeed, by thinking about such a corner point as the origin in the upper half plane model of with local coordinate , then decays at the corner, but not necessarily itself.
Now is a section of with boundary value on , and is a holomorphic function on . We assume from now on that it is not identically zero. The index of the ordinary Cauchy-Riemann operator is
Invoking (23) this is computable from the vanishing orders of :
The interior and boundary vanishing orders are non-negative. Since the are holomorphic near the corners without correction, the proof of Lemma 3.5 shows that the excess vanishing order at the corner and the corner are both non-negative. At the degree one self intersection corners, the excess vanishing order of is nonnegative by the same previous arguments, so itself has excess vanishing order . Hence namely .
When the equality is achieved, then all bounds are saturated. In particular, can only vanish at the corners, so is an immersion up to boundary. At the corners, the same arguments in Corollary 3.6 shows the failure of immersion is minimal.
If is the deformation vector field corresponding to an arbitrary kernel element of the extended linearized operator, then after subtracting off a constant linear combination of , we may assume is tangent to at any chosen point on . The same argument in Corollary 3.6 shows is tangent to the image of . The immersion property allows us to lift to the domain . There is no room to deform the complex structure of , nor is there any automorphism of , so in fact vanishes identically. This shows that span all first order deformations. But implies that the index of the extended linearized operator is
Thus the cokernel dimension is zero, namely the obstruction space vanishes. Consequently, the moduli space of such holomorphic polygons is smooth. ∎
Remark 3.7.
Using the Floer degree formula (63), the asymptotic behaviour of at the corners can be extracted from the above proof: at the corner point
At the degree one self intersections on or ,
At the degree output ,
Weighted Sobolev space with exponential growth
We now discuss solutions to linearized Cauchy-Riemann equations in weighted Sobolev spaces (cf. [70, section 2]). These spaces agree with their unweighted counterparts along the strip like input ends, but at the strip like output end , a vector field means that lies in . Generally we choose to avoid a discrete set of indicial values. The main point of these weighted Sobolev spaces is that they allow for holomorphic vector fields with prescribed exponential growth along the output end, which is conceptually similar to allowing for meromorphic functions in Riemann surface theory. If we think of the strip like end as the infinity (resp. the origin) in the upper half plane model of , then the natural coodinate is (resp. ), and the exponential growth becomes (resp. .
For larger more vector fields are included in the Sobolev space, and the index increases by one each time crosses an indicial value (counted with multiplicity). In our problem, the indicial values are
where are the characterizing angles at the Lagrangian intersection point at the output end. Then the index for the linearized Cauchy-Riemann operator is
| (25) |
where is the number of input ends. In particular, for holomorphic strips with (resp. ), then the index for is equal to (resp. ). In contrast, the ordinary index (for the case) is zero, and the moduli space obtained by taking -quotient has virtual dimension (resp. zero). There are in fact sufficient conditions to rule out the negative dimension moduli spaces, and constrain the zero dimensional moduli spaces:
Lemma 3.9.
In the holomorphic strip case, assume are in the kernel of the linearized Cauchy-Riemann operator on , such that does not vanish identically. Then . When the equality is achieved, the holomorphic strip is an immersion up to the boundary with minimal vanishing at the corner, and the zero dimensional moduli space is regular.
Proof.
We modify the proof of Lemma 3.5 and Cor. 3.6. We think of the corner as the origin in the upper half plane model. Without loss of generality is the -translation vector field of the holomorphic strip. Then the leading order asymptotic is
and
hence
The excess vanishing order is , where negative order stands for poles. By the index formula (23) for the ordinary linearized Cauchy-Riemann equation, we have
and equality forces to have no interior zero, no boundary zero, minimal zero at , and at . The argument in Cor. 3.6 shows span the real vector space of first order deformations in . In particular, the only first order deformation which decays at is the -translation vector field. Thus the cokernel to the ordinary linearized Cauchy-Riemann operator vanishes, and the moduli space is regular. ∎
A very analogous statement holds in the polygon case, and is left to the reader:
Lemma 3.10.
In the holomorphic polygon case, assume are in the kernel of the extended linearized Cauchy-Riemann operator on , such that does not vanish identically as a 1-form on . Then . When the equality is achieved, the holomorphic polygon is an immersion up to the boundary with minimal vanishing at the corner, and the zero dimensional moduli space of holomorphic polygons is regular at .
A similar statement applies to teardrop curves:
Lemma 3.11.
(Regularity of teardrops) Let be a teardrop curve with a unique output and no input ends. Assume are in the kernel of the linearized Cauchy-Riemann operator on , such that does not vanish identically as a 1-form on . Then . When the equality is achieved, the teardrop curve is an immersion up to the boundary with minimal vanishing at the corner, and the kernel of the ordinary Cauchy-Riemann operator is spanned as a real vector space by the Möbius vector fields on fixing the corner, and the cokernel vanishes.
Proof.
We modify the proof of Lemma 3.9. We think of the corner as the origin in the upper half plane model, and take instead to be the Möbius vector field on . This has one higher order of vanishing:
This leads to
so the excess vanishing order at is . The index of the ordinary linearized Cauchy-Riemann operator is
whence .
When the equality is achieved, then there is no interior or boundary zero, and at the corner, hence the immersion claim. The argument in Cor. 3.6 shows that span the real vector space of first order deformations in . In particular, the only first order deformation which decays at are spanned by and , namely the Möbius generators. Since the index of the ordinary Cauchy-Riemann operator is two, the cokernel must have dimension zero, namely the obstruction vanishes. ∎
The above lemma describes the optimal case for teardrop curves. Such teardrop curves arise in isolated zero dimensional moduli spaces after taking the quotient, and the counting contribution to are depending on the spin structure and the orientation issues.
Structure of linearized Cauchy-Riemann equation
Let be a holomorphic polygon with input ends , and one output end at . The case corresponds to teardrops, and corresponds to strips. We consider the ordinary linearized Cauchy-Riemann operator in weighted Sobolev spaces , to classify the structure of the first order deformation theory. As usual, the complex structure is integrable. Since is elliptic, its cokernel in is finite dimensional, represented by holomorphic 1-forms on , which must have finite order of vanishing at . For large enough , the dual space for imposes an exponential decay condition at , so the cokernel evantually vanishes for . Then the kernel dimension in is equal to the index, computed by (25). For convenience, we use , which avoids the indicial values. Then
| (26) |
It is convenient to view the domain of the holomorphic polygon as the upper half plane with coordinate , with corners on the real line and at infinity.
Lemma 3.12.
If , then for some real coefficient polynomial function on the upper half plane, such that is nonvanishing on , and vanishes minimally at (meaning is indivisible by ).
Proof.
If vanishes at any boundary point on , or if vanishes at beyond minimal order, then (resp. ) is also a first order deformation with the same boundary condition, subject to the growth constraints at infinity. Since the kernel dimension is finite, the divisions can only happen a finite number of times, producing the polynomial . ∎
Let , and let be the maximal number depending on , such that there exist -linearly independent , satisfying
- •
In case is not a corner point, then are -linearly independent vectors,
- •
In case is a corner point, then the nonzero elements in the -span of are vector fields vanishing minimally at .
Lemma 3.13.
Any is of the form for some real coefficient rational functions , nonsingular at .
Proof.
Without loss of generality . Let be any first order deformation. By the maximality of , we can choose real numbers , such that the first order deformation vanishes at zero, so for some first order deformation which is nonzero at the origin. Finite dimensionality means this process can be repeated for only a finite number of times:
We choose the smallest such that are -linearly dependent as vector fields; notice are linearly independent, so . We then get a linear relation
where are polynomials, and . This implies the claim. ∎
We can also apply a Möbius transform to make the output end lie at the origin. The growth condition translates to at zero. Let be maximal, such that there are -linearly independent vector fields , and any nonzero element in the -span of vanishes mimimally at . As a caveat, this does not assume satisfy the growth constraints at infinity to lie in . Minor adaptions give
Lemma 3.14.
Any is of the form for some real coefficient rational functions , nonsingular at .
Corollary 3.15.
The number is independent of the boundary and corner points on .
We view the boundary . By the above lemmas, there is a real algebraic vector bundle of rank over such that provide the basis of local sections. By Grothendieck’s classification of vector bundles,
Proposition 3.16.
for some .
Since the rank is nondecreasing in , it evantually stabilizes for . Since around any given point, the same choice of is valid for all large , the algebraic vector bundle is independent of . The elements of can be interpreted as global sections of . Thus for all large ,
| (27) |
Contrasting with the index formula (26),
Corollary 3.17.
The rank , and the degree .
The structure of provides meromorphic sections which are a basis of local sections on , and have excess vanishing orders at . Consider the function . By construction, it has no boundary zero, and its excess corner vanishing order is . Comparing with the index formula (23),
Since all interior vanishing orders are nonnegative by holomorphicity,
Corollary 3.18.
We have in the interior of .
The significance is that the algebraic vector bundle structure on now extends over the entire . The now provide the basis of local sections for the vector bundle . One upshot is that an algebraic structure arises on from solving the Cauchy-Riemann equation with Lagrangian boundary:
| (28) |
In contrast, the Lagrangians are only assumed to be smooth, not necessarily real analytic.
To analyze obstructions, Serre duality motivates us to consider the dualized cokernel to the ordinary (unweighted, unextended) linearized Cauchy-Riemann operator. A dualized cokernel element is represented by a holomorphic 1-forms in with integrablity, and its factor lies in the annilator of the boundary condition. Equivalently, for all test vector fields ,
where is the pairing of with , and the wedge takes care of the forms on . In the canonical form (28), this dualized cokernel is isomorphic to In particular,
Corollary 3.19.
In the teardrop curve case , the strip case and the triangle case , the vanishing of cokernel is equivalent to for all .
For the deformation of the holomorphic polygons is governed instead by the extended Cauchy-Riemann equation, since the punctured Riemann surface structure on is allowed to vary. The dualized cokernel of the extended Cauchy-Riemann operator, is the subspace of the dualized cokernel of the ordinary Cauchy-Riemann operator, which pairs trivially with for any representing some tangent vector of the Stasheff associahedron.
Hamiltonian deformations and transversality
We now consider the parametrized moduli space of holomorphic curves over the infinite dimensional space of Hamiltonian deformations for the Lagrangian . Infinitesimally around a holomorphic curve , we have a Hamiltonian vector field defined by , viewed as a -valued vector field over . We are interested in whether the Hamiltonian deformation kills the cokernel of the ordinary Cauchy-Riemann operator. This question was first addressed by Oh [64]. The following account follows a similar strategy but differs in details.
Recall the ordinary Cauchy-Riemann operator maps to . The effect of Hamiltonian deformation is to enlarge the domain of the operator, by including the vector fields for all the allowed Hamiltonians . The question is to analyze the pairing of with the dualized cokernel elements.
Proposition 3.20.
Let be a holomorphic disc which is immersed near some point with the boundary injectivity property . Let be a nonzero dualized cokernel element for the ordinary linearized Cauchy-Riemann operator. Then there is a Hamiltonian supported in any prescribed small ball on containing , such that .
Proof.
Since is a holomorphic 1-form valued in , Stokes theorem gives
where stands for the pairing between and . On , we can write for some vector field valued in , and is any local coordinate on . The cokernel element condition implies for any , so must in fact be valued in the Lagrangian subbundle . Thus
We suppose for contradiction, that this pairing vanishes identically for any supported in the prescribed ball.
By the holomorphicity of , its zeros are isolated, so without loss of generality does not vanish in the local portion of where is injective and immersed. Suppose first that is not tangent to the image of . Then we find some local function on a small ball in with and on the local portion of , and another cutoff function with along , supported in a small ball. Taking , then
This contradiction shows is tangent to the image of in the local portion of . We can write for some local function . Then requiring
for any compactly supported local function , implies that is constant in the local portion of . Thus up to multiplying by a nonzero constant, locally
| (29) |
We now produce holomorphic vector fields on . For holomorphic strips or polygons with corners, we select one input end as , and call the output as usual, and represent as a strip with boundary punctures. This perspective provides a natural translation vector field , which have exponential decay along the ends, but may not be near the other ends. Instead, by thinking about the ends as the origin in the upper half plane model, we see
for the characterizing angles at the Lagrangian intersection point. The part of is . Contracting this with the part of yields a 1-form on
which is also holomorphic, with boundary value along
| (30) |
Here since both vectors satisfy the boundary condition. Notably, the boundary condition of is real valued. In the upper half plane model, the Schwartz reflection principle allows us to extend meromorphically over .
At any of the ends, since , we know by holomorphicity , so in the upper half plane model, hence has no pole. At the ends, by the decay of the holomorphic and , we likewise infer that has no pole in the upper half plane model. In conclusion, the extension of over has no pole, so must in fact vanish. However, by (29)(30), on a local portion of
This contradiction proves the Proposition in the case.
Finally, for the teardrop curve case , we replace the holomorphic vector field by the Möbius vector fields vanishing at the corner, and the rest of the arguments are entirely similar. ∎
The upshot is that by the Sard-Smale theorem, provided we can always ensure ‘somewhere boundary injectivity’ for any holomorphic disc in a given moduli space, then generic Hamiltonian perturbation would be able to achieve regularity for the moduli space.
Remark 3.8.
In the exact setting there is no closed holomorphic curve. The failure of ‘somewhere boundary injectivity’ is often associated with multiple cover issues, namely may decompose into several domain components, each of which factorizes through a somewhere boundary injective holomorphic disc (cf. [50] for the case of Lagrangian boundary with no corners).
In the simplest case, if factorizes through another disc, then the corner points would be repeated several times on . This phenomenon does not happen for the curves appearing in the bordism current , which involve only one corner at and one corner at . Nor does this occur for teardrop curves, which have only one corner at a degree two self intersection point. This raises hope that the failure of ‘somewhere boundary injectivity’ may be highly nongeneric, or in certain situations can be ruled out altogether.
Further comments on automatic transversality
We now comment on the gap between what we proved and the (weak version of) automatic transversality that we will later assume.
- 1.
Prop. 3.8, Lemma 3.5 and Cor. 3.6 establish the dichotomy for holomorphic discs arising in virtual dimension moduli spaces, that either is automatically transverse, or vanishes for any first order deformation vectors. This argument does not establish unperturbed regularity for the lower dimensional moduli spaces, so it is not completely clear if complex structure perturbations can be removed in the arguments for in section 2.9.
- 2.
For the bad curves, vanishes identically as a 1-form on , so at any point on the boundary, and the tangent vector to are -linearly dependent. Suppose for the moment that the moduli spaces are regular, then the boundary evaluation to for the bad curves arise in Hausdorff dimension at most . Morever, since the Solomon functional is defined through , and vanishes around the bad curves, smoothness assumptions imply that the bad curves cannot contribute.
When regularity assumptions are dropped, one needs to appeal to virtual techniques, so these conclusions require further justification. One problem is that the standard virtual perturbation techniques based on Kuranishi structures do not necessarily produce virtual cycles inside the original moduli spaces, but only inside their small neighbourhoods. This perturbation step destroys the identical vanishing of , by a small amount corresponding to the size of the perturbation. As one shrinks the size of the perturbations, one needs uniform mass bound on the virtual chains to justify that the integral contribution to from the bad curves actually converges to zero.
- 3.
Alternatively, one can hope to replace Lagrangians by arbitrarily small Hamiltonian perturbations to achieve transversality. This is mostly adequate for our purpose, except that one needs to justify the ‘somewhere boundary injectivity’ property (cf. Remark 3.8).