Lorentzian curve straightening and analytic continuation Purdue 8 - PDF document

Lorentzian curve straightening and analytic continuation Purdue 8 April 2002 1

Plan of talk: (1) Poincar´ e → Lorentz (2) Straightening (3) The (2+1)-dimensional case (4) Analytic continuation (5) Hartogs problems (6) Complexified independent variable 2

Starting point: Poincar´ e half-plane dx 2 + dy 2 • metric: y 2 • geodesics: circles ⊥ R ; vertical lines def Let’s look for geometry on G = { geodesics } a 1 a 2 b 1 b 2 Natural invariant: cross-ratio γ = ( a 1 − b 1 )( a 2 − b 2 ) ( a 1 − b 2 )( a 2 − b 1 ) (Label so that | γ | < 1) � � �� 2 1 −√ γ def ∈ R Use ρ = − log 1+ √ γ 3

Hessian of ρ at b j = a j yields 4 da 1 da 2 ( a 2 − a 1 ) 2 , a Lorentz metric! • spacelike geodesics: • timelike geodesics: • lightlike geodesics: 4

Facts about G : • Geodesically complete • a, b, ∈ G ⇒ a, b joined by geodesic (unique if timelike/lightlike) � � 2 � � 4 da 1 da 2 • ρ ( a, b ) = ( a 2 − a 1 ) 2 geod. from a to b top. • G = { unordered pairs in S 1 } ∼ M¨ obius band • G not time-orientable • π 1 ( G ) = Z • Do not get geodesic in each homotopy class (except in spacelike case) • Geodesics don’t separate G • G (or double cover) = 2-dim. deSitter space { geodesics in G } ∼ Poin. half-plane ∪ G ∪ S 1 ∼ real projective plane 5

Let G ′ = G \ { vertical lines } . New coordinates: ( x, y ) = top of semi-circle Get: • metric dx 2 − dy 2 y 2 • geodesics: equilateral hyperbolas ⊥ R ; vertical lines Metric is: • incomplete “as y → ∞ ” • complete “for y bounded” • time-oriented (forward = up) 6

( x 1 , y 1 ) , ( x 2 , y 2 ) joined by • unique timelike geodesic if | x 2 − x 1 | < | y 2 − y 1 | • unique lightlike geodesic if | x 2 − x 1 | = | y 2 − y 1 | • unique spacelike geodesic if | y 2 − y 1 | < | x 2 − x 1 | < y 1 + y 2 • no geodesic if | x 2 − x 1 | ≥ y 1 + y 2 Geodesics do separate G ′ into “half-spaces” “Hahn-Banach Theorem.” Let S ⊂⊂ G ′ . Then the following are equivalent: (A) S is an intersection of half-spaces (B) S contains geodesic joining any two of its points 7

= � of half-spaces containing S def � S Example. Hull of two points. � { ( x 1 , y 1 ) , ( x 2 , y 2 ) } = • geodesic segment if | x 2 − x 1 | < y 1 + y 2 • if | x 2 − x 1 | ≥ y 1 + y 2 8

Next topic: “straightening” of (parameterized) curves I) Smooth curves in R 2 Could call a smooth (endpoint-fixing) homotopy { γ t } a straightening if � ∂ � 2 ∂ ∂tγ t ( s ) = µ ( s, t ) γ t ( s ) , µ ( s, t ) ≥ 0 . ∂s Proposition. Every smooth curve can be straightened to an affine map. 1 Proof. Use µ ( s, t ) = 1 − t . II) Polygonal arcs in R 2 Say that a straightening is a sequence of moves in which two adjacent segments are replaced by a single segment. Of course all polygonal arcs can be straightened to line segments. III) Broken geodesics in Riemannian manifold Same idea, but insist on staying in same homotopy class. All broken geodesics can be straightened if ambient manifold is complete. 9

IV) Broken geodesics in G ′ Clearly cannot straighten γ if there is no geodesic joining the endpoints. What positive results can we obtain? 10

Theorem A. Suppose γ takes values in K compact convex ⊂ G ′ . Then γ can be straightened. Proof. 11

Theorem B. Suppose endpoints of γ can be joined by geodesic (i.e., endpoints have compact hull). Then γ can be straightened. Proof. 12

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Up one dimension: C = { circles/lines in C } C ′ = { genuine circles in C } coordinates on C ′ : ( x, y, r ) metric dx 2 + dy 2 − dr 2 on C ′ extends to r 2 M¨ obius-invariant metric on C Geodesics through (0 , 0 , 1): 14

An important difference from 2-dim.: hull of triangle may not be compact. Insist that each straightening move involve three consecutive vertices with compact hull. Theorem A carries over; Theorem B does not. Combinatorial hypothesis on S = { S ⊂ { v 0 , . . . , v N } : � S compact } will guarantee straightening. 15

Theorem C. Suppose that γ lies in backwards light cone of initial vertex. Then γ can be straightened. Proof. Always eliminate second vertex. 16

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Analytic continuation (finally): Given γ = ( γ X , γ Y , γ R ) : [ α, β ] → C ′ , let Ω γ = { ( z 1 , z 2 ) ∈ C 2 : α < Re z 1 < β, | z 2 − ( γ X + iγ Y ) (Re z 1 ) | < γ R (Re z 1 ) } . Proposition. If γ 1 can be straightened to γ 2 then every holomorphic function on Ω γ 1 can be analytically continued to Ω γ 2 . Question: Is there a converse?   forward timelike   If acceleration at corner is backwards timelike   spacelike   pseudoconcavity   pseudoconvexity get  .  mixture Straightening arguments � analytic continuation theorems. 18

“Hartogs problem” Given: f : ∆ → C cont., f (b∆) ⊂ ∆ Ask: do all functions holo. in nbhd. of � � b∆ × ∆ ∪ graph f have analytic cont. to nbhd. of ∆ × ∆? • (Hartogs) Yes for f ≡ 0, f holo. • (Chirka(-Rosay)) Yes for f : ∆ → ∆ • (Thm C. + ǫ ) Yes for f ( re iθ ) = g ( r ) e ikθ , k ≥ 0 • (Thm B. + ǫ ) Yes for f ( re iθ ) = g ( r ) e − ikθ , g R -valued • (Thm C. + ǫ ) Yes for f ( re iθ ) = g ( r ) e − ikθ , r k | g ( r ) | ≤ 1 • (Alexander-Wermer) No for f ( re iθ ) = g ( r ) e − iθ , g ( r ) = Mr ( r 2 − 1) e ir 2 , M ≥ 26 19

To get more flexible results, complexify independent variable, study maps γ : D planar domain → C ′ ( s,t ) replacing ∆ C ′ t by   − 1 0 0 s,t + 2  ∂γ ∂s × ∂γ ∆ C ′  0 − 1 0 r ∂t 0 0 1 Issues: • Discretization • Need revised ideas for straightening priorities 20

EXTRA **************** If γ is 2-piece broken geodesic then ψ convexity properties of Ω γ depend on “acceleration”: acceleration 21

• backward timelike/lightlike acceleration ⇒ ψ convex • forward timelike/lightlike acceleration ⇒ ψ concave • spacelike acceleration ⇒ mixed 22

Straightening: • ψ convex corner: Ω γ s ⊂ Ω γ g g s • ψ concave corner: Ω γ ⊂ Ω γ s = � Ω γ g s g 23

• mixed corner: Ω γ � � Ω γ s � � Ω γ g g s In some very special cases: � Ω γ = Ω γ ∪ Ω γ s 24