17.01.2013 Views

PDF (1016 kB)

PDF (1016 kB)

PDF (1016 kB)

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

For the remaining part of the argument leading to C∞ loc compactness of M Λ Υ (x1, x2; y) it is more<br />

convenient to use the original definition of this solutions space and the smooth structure of ΣΛ Υ .<br />

Then the argument is absolutely standard: If by contradiction there is no uniform C1 bound, a<br />

concentration argument (see e.g. [HZ94, Theorem 6.8]) produces a non-constant J-holomorphic<br />

sphere. However, there are no non-constant J-holomorphic spheres on cotangent bundles, because<br />

the symplectic form ω is exact. This contradiction proves the C1 bound. Then the Ck bounds for<br />

arbitrary k follow from elliptic bootstrap, as in [HZ94, section 6.4].<br />

Other solutions spaces, such as the space M Ω Υ<br />

for the traingle products, involve Riemann<br />

surfaces with boundary, and the solutions take value on some conormal subbundle of T ∗ M. In<br />

this case the concentration argument for proving the C 1 bound could produce a non-constant<br />

J-holomorphic disk with boundary on the given conormal subbundle. However, the Liouville oneform<br />

vanishes on conormal subbundles, so such J-holomorphic disks do not exist. Again we find<br />

a contradition, leading to C 1 bounds and - by elliptic bootstrap - to C k bounds for every k.<br />

6.2 Removal of singularities<br />

Removal of singularities results state that isolated singularities of a J-holomorphic map with<br />

bounded energy can be removed (see for instance [MS04, section 4.5]). In Proposition 6.4 below,<br />

we prove a result of this sort for corner singularities. The fact that we are dealing with cotangent<br />

bundles, which can be isometrically embedded into�N , allows to reduce such a statement to the<br />

following easy linear result, where�r is the open disk of radius r in�, andÀ+ is the quarter plane<br />

{Rez > 0, Im z > 0}.<br />

6.3. Lemma. Let V0 and V1 be partially orthogonal linear subspaces ofÊn . Let u : Cl(�1 ∩À+ ) \<br />

{0} →�n be a smooth map such that<br />

for some p > 2, and<br />

u ∈ L p (�1 ∩À+ ,�n ), ∂u ∈ L p (�1 ∩À+ ,�n ),<br />

u(s) ∈ N ∗ V0 ∀ s > 0, u(it) ∈ N ∗ V1 ∀ t > 0.<br />

Then u extends to a continuous map on Cl(�1 ∩À+ ).<br />

Proof. Since V0 and V1 are partially orthogonal, by applying twice the Schwarz reflection argument<br />

of the proof of Lemma 5.6 we can extend u to a continuous map<br />

u :�1 \ {0} →�n ,<br />

which is smooth on�1 \ (Ê∪iÊ), has finite L p norm on�1, and satisfies<br />

∂u ∈ L p (�1).<br />

Since p > 2, the L 2 norm of u on�1 is also finite, and by the conformal change of variables<br />

z = s + it = eζ = eρ+iθ , this norm can be written as<br />

� 0 � 2π<br />

��1<br />

|u(z)| 2 dsdt =<br />

−∞<br />

0<br />

|u(e ρ+iθ )| 2 e 2ρ dθdρ.<br />

The fact that this quantity is finite implies that there is a sequence ρh → −∞ such that, setting<br />

ǫh := e ρh , we have<br />

lim<br />

h→∞ ǫ2 h<br />

� 2π<br />

0<br />

|u(ǫhe iθ )| 2 dθ = lim<br />

h→∞ e2ρh<br />

� 2π<br />

If ϕ ∈ C ∞ c (�1,�N ), an integration by parts using the Gauss formula leads to<br />

=<br />

��1<br />

��ǫ<br />

��1\�ǫ<br />

〈u, ∂ϕ〉dsdt = 〈u, ∂ϕ〉dsdt + 〈u, ∂ϕ〉dsdt<br />

h<br />

h<br />

��ǫ h<br />

〈u, ∂ϕ〉dsdt −<br />

��1\�ǫ h<br />

0<br />

�<br />

〈∂u, ϕ〉dsdt + i<br />

91<br />

|u(e ρh+iθ )| 2 dθ = 0. (158)<br />

∂�ǫ h<br />

〈u, ϕ〉dz.

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!