Mobius transformations and AdS 3 geometry

Transcription

1 Mobius transformations and AdS 3 geometr Jeffre Danciger September 1, PSL (R + Rτ) and its action on AdS 3 Let B be the real algebra generated b an element τ, with τ = +1. As a vector space B = R + Rτ is two dimensional over R. Note that B is not a field as e.g. There is a conjugation operation that defines a square-norm (1 + τ) (1 τ) = 0. a + bτ a + bτ = a bτ a + bτ = (a + bτ)(a + bτ) = a b. This square-norm comes from the (1, 1) Minkowski product on R (with basis {1, τ}). The space-like elements of B (i.e. square-norm > 0), acting b multiplication on B form a group and can be thought of as the similarities of the Minkowski plane that fix the origin. Note that if a + bτ = 0 then b = ±a and multiplication b a + bτ collapses all of B onto the light-like line spanned b a + bτ. The elements 1+τ and 1 τ are two spanning idempotents which annihilate one another ( 1 ± τ ) = 1 ± τ, and ( 1 + τ ) so B = R R as R algebras via the isomorphism ( ) ( ) 1 + τ 1 τ a + b (a, b). Next, define {( ) a b SL (B) = c d ( ) 1 τ = 0 } : a, b, c, d B, ad bc = 0 1

2 and PSL (B) = SL (B)/{±1, ±τ}. Note that because B = 1+τ R 1 τ R, we have a similar splitting for M (B): ( ) ( ) ( ) 1 + τ 1 + τ 1 + τ A + 1 τ B C + 1 τ D = AC + 1 τ BD and also ( 1 + τ det A + 1 τ ) B = 1 + τ det(a) + 1 τ detb. So SL B = SL R SL R and PSL B = PSL R PSLR = IsomAdS 3 (note that here Im taking AdS to be PSL R, though mabe its more common to define it as SL R). Now, just as PSL C doesn t quite act on C, PSL B doesn t quite act on B. We define a compactification P 1 B as follows (I think this is the same thing as the Lorentz compactification?). {( } P 1 x B = : x α = 0 and α = 0 for α B α = 0 / ) where ( ) x ( ) xλ λ for λ B. With this definition PSL B clearl acts on P 1 B. We can think of this action as the Lorentz Mobius transformations. Lemma 1. P 1 B = P 1 R P 1 R and the isomorphism (given below) identifies the action of PSL B on P 1 B with that of PSL R PSL R on P 1 R P 1 R. Proof. The isomorphism P 1 R P 1 R P 1 B is given b [( )] [( )] [ ( ) ( )] a c 1 + τ a, + 1 τ c b d b d {[( } x Now, P 1 B is a compactification of B = : x B. What new points [( 1)] )] x are added? Well, consider P 1 B with not invertible. There are several cases ( ( x 1 = 0, and so x is invertible. Thus =. ) 0) = a 1+τ and x invertible. So ( ) ( ) x a 1+τ = a 1+τ.

3 Figure 1: P 1 B is gotten b adding a wedge of circles at infinit to B. The Minkowski (1, 1) plane B is drawn in light blue. The red circle (representing limits of assmptoticall light-like paths with slope 1) meets the blue circle (representing limits of assmptoticall light-like paths with slope +1) at. = a 1 τ and x invertible. So = a 1+τ and x = b 1 τ. So ( x ( ) x = a 1 τ and x = b 1+τ. So ( ) ( ) x a 1 τ ) ( ) 1 τ 1 + τ ( ) 1 + τ 1 τ = a 1 τ. = 1 τ 1+τ. = 1+τ 1 τ. So the extra points make up a wedge of two circles as in the figure. A path going to infinit along the light-like axis spanned b 1 + τ converges to 1+τ 1 τ. An affine light-like line of the form r(t) = a + t 1+τ with a 0 converges to a 1+τ as t ±. All space-like and time-like lines converge to = 1 0. A half-space model for AdS 3 In [1] Ahlfors describes a general construction of n-dimensional hperbolic space as a half-space inside a Clifford algebra C n. The isometries in this model are Mobius transformations with coefficients in a sub-algebra C n 1. In the case n = 3, C 3 is the Hamiltonian quaternions, C is the complex numbers and the construction gives the standard upper half-space model of H 3. Following Ahlfors, we construct a half-space model of AdS 3 using generalized Clifford numbers. The isometries in this model will be SL B acting b Mobius transformations. This construction can (surel) be extended to higher anti de Sitter spaces. 3

4 Starting with our algebra B = R + Rτ from the previous section, we add an element j with j = 1 and jτ = τj. This defines a Clifford algebra A that is four dimensional over the reals. In fact, in this low dimensional case A is isomorphic to the algebra of real two-b-two matrices. There are man different copies of C ling inside A, one distinguished cop being R + Rj, just as there are man isomorphic copies of B. There is a conjugation operation on A, z z satisfing 1 = 1, j = j, τ = τ, and (zw) = w z that defines a square-norm z = zz = zz R. Let V = span{1, j, τ} be the subspace of elements of A having degree 1, the so-called vectors of A. The square-norm restricted to V comes from the (, 1) Minkowski product on R,1 (with basis {1, j, τ}). We ve alread seen how SL B acts on the compactification P 1 B of B = R + Rτ. Our goal here is to extend this compactification to a compactification of V in a wa that allows SL B to act naturall. Define P 1 V as follows (again, this is likel a different description of the Lorentz compactification): {( } P 1 x V = A ) : x V, xα = 0 and α = 0 for α A α = 0 / where ( x ) ( ) xλ λ for λ A. Note that when is invertible ( we ) have ( x 1 ) = x/, so that the condition x x 1 x V allows us to think of as the vector x 1 1 in V P 1 V. [( x Lets examine the extra points in P 1 V. Consider P )] 1 V with not invertible (so = 0). There are several cases ( ) ( ) x 1 1. = 0, and so x is invertible. Thus =. 0 ( ) ( ) x a. x is invertible and 0. Then e jθ +τ for some a R and some e jθ = cos θ + j sin θ on the unit circle in R + Rj. ( ) ( ) ( ) x 1 e 3. x is not invertible ( = 0). Then jθ τ e e jθ jθ τ + τ 1 + e jθ. τ These added points are all endpoints of lines in V. Space-like and time-like lines all limit to. Light-like lines limit to the points described in cases and 3. Indeed, an light-like line can be written in orthogonal coordinates as r(t) = bje jθ + a ejθ + τ 4 + t ejθ τ

5 where a, b R. If a 0, then in P 1 V ( ) ( bje r(t) = jθ + a ejθ +τ + t ejθ τ e jθ τ a bj ) + e jθ + τ 1 t ( ) 1+e jθ τ = abj+b t + 1 ejθ bj + 1+ejθ a + 1 ejθ (a bj) e jθ τ a bj t + e jθ +τ ( ) a e jθ +τ as t ± and we see that we are in case. If a = 0 a similar calculation shows that ( ) 1 e r(t) jθ τ e jθ as t ± + τ which is of the form in case 3. This algebraic description of P 1 V is nice because we can easil ( see that ) SL B a b acts on P 1 V (and indeed PSL B acts faithfull). For consider SL c d B ( ) x and with x = e+αj where e R+Rτ and α R. Then x = x = e αj and so ( ( ) ( ) a b x ax + b = c d) cx + d has (ax + b)(cx + d) = (ax + b)(x c + d) = x ac + bd + a(e + αj)d + b(e αj)c = x ac + bd + aed + be c + α(ad bc)j = ( something in R + Rτ ) + αj V. Though P 1 B sits inside P 1 V, P 1 B does not divide P 1 V into two components. In fact points in the upper half of V (i.e. j component > 0) can easil be sent to points in the lower half (i.e. j component < 0) b an element of SL B. The space P 1 V itself is actuall a non-orientable (closed) three manifold. So, breaking from the analog to the hperbolic upper half-space model, we will define our model for AdS 3 as the quotient of P 1 V b and involution fixing P 1 B. Define the algebra homomorphism I on A b I(1) = 1, I(τ) = τ, I(j) = j. Its clear that I is an involution whose fixed set is B. I also defines an involution of P 1 V b [( )] [( )] x I(x) I = I() which has fixed set P 1 B. Its eas to check that the action of PSL B commutes with I, so PSL B acts on the quotient X = P 1 V/I. 5

6 Proposition 1. X = P 1 V/I is a solid torus with boundar P 1 B. Proof. I ll omit this for now. Basicall, drawing a good picture of the added points in P 1 V allows ou to see this. Its kind of a fun exercise. Note that after taking the quotient b I, man of the added points in P 1 V V lie in the interior of X. For example the endpoint of a light-like line l lies in the interior of X if l has non-constant j-component..1 The AdS metric For simplicit, we will work out the metric aspects in the space V B. We let x 1, x, x 3 be real coordinates for V b the formula x = x 1 +x τ +x 3 j. Consider the metric obtained b rescaling the (, 1) Lorentzian metric on V b 1/ x 3 : ds = dx x 3 = dx 1 dx + dx 3. ( ) a b Now consider a transformation g = SL c d B. Assume that x is in a neighborhood where cx + d is invertible so that x 3 g(x) = (ax + b)(cx + d) 1 V. Differentiating the expression g(x)(cx+d) = ax+b and appling it to a tangent vector u V we obtain Dg(x)u(cx + d) = (a g(x)c)u = (ac 1 (cx + d) (ax + b))(cx + d) 1 cu = c 1 (ad bc)(cx + d) 1 cu = (xc + d) 1 u where we ve assumed c invertible (to avoid a more painful calculation). Anwa, we get the formula Dg(x) u = (xc + d) 1 u(cx + d) 1 (1) from which we see that g is conformal (with respect to the Minkowski (, 1) metric on V ) with rate of magnification ( cx + d ). Next, writing g(x) = g 1 + g τ + g 3 j, a quick computation shows x 3 g 3 = cx + d which gives that the metric ds is preserved b g: Dg u g 3 = u x. 3 The curvature of our metric ds is 1 (the calculation should be the same as in the hperbolic case). 6

7 Proposition. X = P 1 V/I with the metric ds is isometric to PSL R with the AdS metric. Proof. We now know its locall isometric to AdS. Show its geodesicall complete (see next section). Then calculate the length of a time-like geodesic.. Geodesics Let P be the (closure of the) plane spanned b {1, j} passing through the origin. The (euclidean) reflection about P is clearl an isometr of our metric ds. It follows that P is a totall geodesic plane in X. Of course, P is the upper half-plane model of H, so in particular the curve γ 0 (t) = e t j is a unit speed space-like geodesic. We translate γ 0 around b SL( B to ) obtain a description of a b all space-like geodesics in this model. Given A = SL c d B, consider γ(t) = Aγ 0 (t) = (ae t j + b)(ce t j + d) 1. For ease of demonstration, we will assume here and in the subsequent computation that c, d are invertible and that t is such that ce t j + d is invertible (this is true for all except possibl one value of t). Then γ(t) = acet + bde t + j c e t + d e t and we note that the endpoints of γ are γ( ) = bd 1 and γ( ) = ac 1. Now, b analog with the hperbolic case we expect γ to be some sort of (euclidean) conic section perpendicular to the boundar with midpoint equal to 1 (ac 1 + bd 1 ). In fact, a calculation shows that γ(t) ac 1 + bd 1 = ( c e t d e t c e t + d e t ) 1 ( cd + 1 c e t + d e t ) j and so we see that γ(t) lies in the affine plane centered at ac 1 +bd 1 and spanned b the directionsj and 1/(cd) = (ac 1 bd 1 )/. Further γ(t) satisfies γ(t) ac 1 + bd 1 = 1 4 c d showing that γ parametrizes an ellipse if c, d have the same sign, or a hperbola if c, d have opposite sign. Note that in the latter case γ is still a smooth path through P 1 V/I though it appears discontinuous when we draw it in V. We have demonstrated the following: Proposition 3 (Space-like geodesics). Let p 1, p B such that the displacement = (p 1 p )/ is not light-like. Let p be the midpoint p = (p 1 + p )/. Then 7

8 the AdS geodesic γ connecting p 1 to p is the conic ling in the affine plane p + span{, j} defined b the equation γ p =. If > 0 then γ is a (euclidean) ellipse, and if < 0 then γ is a hperbola. In either case, γ meets the R + Rτ plane at right angles. Figure : A space-like geodesic is either an ellipse or a hperbola depending on the displacement between its endpoints. In the hperbola case, the geodesic comes out from the boundar along one branch of the hperbola, passes though an added point of P 1 V and then returns to the boundar along the other branch. Time-like geodesics have a similar description, though we won t give a detailed proof: Proposition 4 (Time-like geodesics). Let p, B with < 0. Then, the locus of points γ in the affine plane p + span{, j} satisfing γ p = defines a time-like geodesic. All time-like geodesics are described in this wa. We note that each time-like geodesic closes up after passing through exactl one added point in P 1 V. Proposition 5 (Light-like geodesics). The parametrized light-like geodesics are described b γ(t) = p + 1 t l where p R + Rτ, and l is a lightlike vector in V. Note that in this light-like case γ( ) = γ( ) and that after performing the quotient b I, γ meets itself at an angle at the boundar. This is in contrast with the projective model of AdS in RP 3 in which lightlike geodesics are tangent to the boundar. 8

9 Figure 3: A light-like geodesic. The light cone is drawn (in red) for reference. Proposition 6 (Space-like Planes). Both of the following are space-like planes in this model: 1. A two-sheeted hperboloid in V meeting the boundar B at right angles, described b an equation of the form x p = R where p B. Note that the two components of the hperboloid actuall meet up in P 1 V.. a vertical affine plane of the form p + span{v, j} where p, v R + Rτ with v space-like. 3 AdS ideal tetrahedra Let z 1, z, z 3, z 4 be four points in the ideal boundar P 1 B of AdS. Recalling the isomorphism P 1 B = P 1 R P 1 R, we write each z i as z i = λ i 1 + τ + µ i 1 τ i = 1,, 3, 4 where λ i, µ i P 1 R. If the λ i are distinct and the µ i are distinct, then taking the z i three at a time, we get four space-like hperbolic ideal triangles T 1, T, T 3, T 4. If further (λ 1, λ, λ 3, λ 4 ) and (µ 1, µ, µ 3, µ 4 ) have the same cclic order up to PGL R, then the T i bound an AdS ideal tetrahedron. Ever AdS ideal tetrahedron can be put into standard position, with ideal vertices 0, 1,, z where z satisfies z > 0, 1 z > 0 so that in particular all of the R + Rτ ratios 1 z, 1 z, z 1 z assigned to the edges are defined and space-like. Lets describe this ideal tetrahedron in our half-space model. The three faces containing are vertical 9

10 planes, while the fourth bottom face is a hperboloid x p = R. Writing z = z 1 + z τ, a simple calculation gives p = 1 + z 1 z τ z R = z 1 z z. Note that in the case z 0, our tetrahedron degenerates and all four faces lie in a common plane. Figure 4: The ideal AdS tetrahedra defined b the ideal points 0, 1,, z P 1 B = AdS. Each of the three side faces lies in a(euclidean) plane, while the bottom face lies in a hperboloid. Each face is a totall geodesic ideal triangle isometric to an ideal triangle in H. References [1] Ahlfors, L.V., Mobius tranformations and Clifford numbers, Differential Geometr and Complex Analsis, H.E. Rauch Memorial Volume, Springer- Verlag