By Casim Abbas

This ebook presents an advent to symplectic box idea, a brand new and demanding topic that is at the moment being constructed. the place to begin of this thought are compactness effects for holomorphic curves proven within the final decade. the writer offers a scientific advent supplying loads of heritage fabric, a lot of that is scattered in the course of the literature. because the content material grew out of lectures given through the writer, the most target is to supply an access element into symplectic box idea for non-specialists and for graduate scholars. Extensions of convinced compactness effects, that are believed to be actual by way of the experts yet haven't but been released within the literature intimately, replenish the scope of this monograph.

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**Extra info for An Introduction to Compactness Results in Symplectic Field Theory**

**Example text**

39 This example is from [13]. Consider the following two geodesics in H : γk (t) = k + iet , t ∈ R, k = 1, 2. The closed strip S between them is a hyperbolic surface. We have γ1 (R) ∩ γ2 (R) = ∅, but d(γ1 (t), γ2 (t)) → 0 as t → ∞, hence the sets γ1 (R) and γ2 (R) have distance zero. e. we identify γ1 (t) with φ(γ1 (t)) = γ2 (t + log 2). The resulting surface Sˆ is topologically a cylinder, and it carries a metric with sectional curvature −1, but this metric is not complete: Consider the following sequences of points: an = 1 + i2n , bn = φ(an ) = 2 + i2n+1 .

Following similar arguments as before, when we discussed case (i), there is φ ∈ D(π) such that the geodesics γ˜1 and φ(γ˜2 ) intersect transversally. Then the situation is again similar to Fig. 18 and the curves α˜1 , φ(α˜ 2 ) intersect as well. Projecting into S this would imply the contradiction α1 ∩ α2 = ∅. The case where ∂S = ∅ then simply follows from taking the double of S. 65. The statement can be refined as follows: The corresponding geodesics γ1 and γ2 agree as point sets if and only if there are nonzero integers k1 , k2 such that α k1 and α k2 are freely homotopic.

The one on the right has two degenerate boundary components Fig. 11 Building a hexagon with prescribed lengths (b1 ), (b2 ), (b3 ) Assume first that G is a hexagon with three degenerate sides b1 , b2 , b3 . Then there is an isometry φ ∈ I such that φ(b1 ) = 0, φ(b2 ) = 1 and φ(b3 ) = ∞. Since φ transforms geodesics into geodesics and since it preserves angles, φ(G) is again a hexagon, and it looks like the last one in Fig. 8. Let us now assume that one of the sides of G, say b1 , has positive length.