How to recognise the geodesics of a metric connection
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1 Second CoE Talk at the University of Tokyo p. 1/19 How to recognise the geodesics of a metric connection Michael Eastwood Australian National University
2 Second CoE Talk at the University of Tokyo p. 2/19 References Robert Bryant, Michael Eastwood, and Maciej Dunajski, Metrisability of two-dimensional projective structures, Jour. Diff. Geom. 83 (2009) Michael Eastwood and Vladimir Matveev, Metric connections in projective differential geometry, IMA Volumes 144, Springer Verlag 2007, pp Roger Liouville, Sur les invariantes de certaines équations différentiale, Jour. l École Politechnique 59 (1889) 7 76.
3 Second CoE Talk at the University of Tokyo p. 3/19 Metric geodesics metric Levi-Civita connection geodesics unparameterised geodesics What is lost? What do we obtain? Recall that g ab Γ d ab = 1 2 gcd ( a g bc + b g ac c g ab ) gives the Levi Civita connection a φ b = a φ b Γ c ab φ c.
4 Second CoE Talk at the University of Tokyo p. 4/19 Projective equivalence Suppose a is a torsion-free connection. Define ˆ a by ˆ a φ b = a φ b Υ a φ b Υ b φ a. ˆ a is torsion-free. ˆ a has the same unparameterised geodesics as a. Conversely, these two properties force. Definition: a and ˆ a are projectively equivalent. Our questions are now operational: σ : {metrics}? {projective structures}.
5 Example: σ is not injective σ(g ab ) = σ(constant g ab ) but also (1 y 2 )dx 2 + xy dxdy + (1 x 2 )dy 2 (1 x 2 y 2 ) 2 Beltrami on {x 2 + y 2 < 1} has the same unparameterised geodesics as (1 + y 2 )dx 2 xy dxdy + (1 + x 2 )dy 2 (1 + x 2 + y 2 ) 2 Thales or, indeed, as the flat metrics dx 2 + dy 2 or dx 2 dy 2. Second CoE Talk at the University of Tokyo p. 5/19
6 Second CoE Talk at the University of Tokyo p. 6/19 Example: σ is not surjective Writing a φ b = a φ b Γ ab c φ c, Γ ab 1 = x 2 x 0 0 x Γ ab 2 = x 2 0 x x 0 is a metric connection but Γ ab 1 = x x + y x + y y Γ ab 2 = x 0 0 y is not projectively equivalent to a metric connection. Why Not?
7 Second CoE Talk at the University of Tokyo p. 7/19 Another example The connection Γ ab 1 = xy + 2e y + 2ye y + 2x 2 ye y 1 + y + x 2 y x + y 2 x + y 2 x 2(1 + y + x 2 y) Γ ab 2 = y + x 2 y e y 1 + x 2 + 4x + 4xy + 4x 3 y + 4y 2 + 4y 3 + 4y 3 x 2 2(1 + y + x 2 y) e y is projectively equivalent to a metric connection!
8 Second CoE Talk at the University of Tokyo p. 8/19 Naïve dimension count On a surface Jets = Λ 0 + Λ Λ Λ Λ 1 + rank = Therefore k rankj k rankj k+1 (metrics) ditto up to scale rankj k (proj. struc.) deficit
9 Second CoE Talk at the University of Tokyo p. 9/19 Obstructions We expect that J k+1 (metrics up to scale) J k (projective structures) is not surjective for k = 5. Therefore, we expect D(Γ) = polynomial in Γ, Γ,, (5) Γ such that Γ metrisable D(Γ) = 0. For k = 6, expect 5 = #{D(Γ), 1 D(Γ), 2 D(Γ)}+ two more invariants and then 10 < generically this is enough (real-analytic... ).
10 Second CoE Talk at the University of Tokyo p. 10/19 Special connections Ricci curvature ( a b b a )X b = R ab X b is not necessarily symmetric. However, ˆ a φ b = a φ b Υ a φ b Υ b φ a implies ˆR [ab] = R [ab] (n + 1) [a Υ b]. Bianchi [a R bc] = 0. Thus, restrict to R ab = R (ab). Definition Connections with symmetric Ricci special.
11 Second CoE Talk at the University of Tokyo p. 11/19 An overdetermined system Theorem (Liouville,..., Sinjukov, Mikeš,... ) A special connection a is projectively equivalent to a metric connection if and only if the system of PDEs trace-free part of ( a σ bc ) = 0 a σ bc 1 n+1 δ a b d σ cd 1 n+1 δ a c d σ bd for σ ab symmetric has a non-degenerate solution. Proof Check that g ab = det(σ)σ ab works, where det(σ) = ǫ a b ǫ c d σ ac σ bd and a ǫ c d = 0 (OK for special connections).
12 Second CoE Talk at the University of Tokyo p. 12/19 Prolongation Rewrite as a σ bc = δ b a µ c + δ c a µ b. Flat case as warm-up: 0 = ( a b b a )σ bc = n a µ c δ c a b µ b whence But whence a µ b = δ a b ρ. 0 = ( a b b a )µ b = (n 1) a ρ a ρ = 0.
13 Second CoE Talk at the University of Tokyo p. 13/19 Tractor connection Curved case: a σ bc a µ b = δ a b µ c + δ a c µ b = δ a b ρ P ac σ bc + 1 n W ac b dσ cd a ρ = 2P ab µ b + 4 n Y abcσ bc c c R ab d = W ab d + δ c a P bd δ c b P ad Y abc = [a P b]c. Reorganize as tractors: σ bc a σ bc δ b a µ c δ c a µ b µ b a a µ b δ b a ρ + P ac σ bc 1 n W ac b dσ cd ρ a ρ + 2P ab µ b 4 n Y abcσ bc cf. Cartan connection.
14 Second CoE Talk at the University of Tokyo p. 14/19 Surfaces c W ab d = 0 and Cotton-York Y abc is projectively invariant. σ bc a σ bc δ b a µ c δ c a µ b Σ α µ b a a µ b δ b a ρ + P ac σ bc ρ a ρ + 2P ab µ b 2Y abc σ bc Compute curvature (fix a volume form ǫ ab ): σ bc 0 ǫ ab a b µ b = 0 ρ 5Y a µ a + Z ab σ ab where Y a = ǫ bc Y bca and Z ab = (a Y b).
15 Second CoE Talk at the University of Tokyo p. 15/19 Consequences of metrisability If Γ is metrisable, then a Σ α = 0 for some Σ α 0. Now 0 ǫ ab a b Σ α = 0 Ξ α Σ α = 0 where Ξ α Differentiate once more: 5Y a Z ab. 0 = a (Ξ γ Σ γ ) = ( a Ξ γ )Σ γ (dual connection) 5Y a a Σ γ = 5 a Y c + 2Z ac. a Z cd 5P a(c Y d)
16 Second CoE Talk at the University of Tokyo p. 16/19 First obstruction Differentiate again: 0 = ( (a b) Ξ γ )Σ γ. The 6 6 matrix Ξγ, a Ξ γ, (a b) Ξ γ is singular. The 5 th order expression D(Γ) det Ξγ, a Ξ γ, (a b) Ξ γ is a projectively invariant obstruction to metrisability.
17 Second CoE Talk at the University of Tokyo p. 17/19 Example Can compute with MAPLE. Γ ab 1 = x x + y x + y y Γ ab 2 = x 0 0 y is not metrisable because is has non-vanishing primary obstruction:
18 Second CoE Talk at the University of Tokyo p. 18/19 D(Γ) = e 6y2 3x 2 6xy ( x xy y y x x 3 y x 2 y xy x 10 y x 5 y x 5 y x 13 y x y 9 x y 8 x x x 2 y xy x y x 7 y x 2 y x 11 y x 7 y x 5 y x 13 y x 6 y x 4 y x 8 y x 9 y x 9 y x 10 y x 6 y x 6 y x 10 y x 9 y x 5 y x 4 y x 4 y x x y y y 7 x y 5 x x 9 y x 15 y y 10 x x 8 y y 11 x x 7 y x 8 y x 11 y x 3 y x 14 y x 6 y x 11 y x 12 y x 3 y x y x 7 y x 3 y x 12 y x 6 y x 3 y x 3 y x 5 y x 7 y x 5 y x 4 y x 4 y x 2 y y 6 x x 8 y y 11 x y y 10 x )
19 THANK YOU Second CoE Talk at the University of Tokyo p. 19/19
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