Symmetry through Geometry Nalini Joshi @monsoon0 Supported by the Australian Research Council The University of Sydney
Outline ✑ Symmetry ✑ Geometry ✑ Dynamics The University of Sydney
Symmetry Photo by Dmitriy Smaglov/Thinkstock slate.com The University of Sydney 3
Reflection mirror P P perpendicular distance from P to the mirror = perpendicular distance to its reflection.
Translation ( x, y ) ✎ ( a + x, b + y ) ✎ (0 , 0) ( a, b )
An example √ β = ( − 1 , 3) 2 π / 3 (0 , 0) α = (2 , 0) The University of Sydney 6
Place a mirror s α √ β = ( − 1 , 3) α = (2 , 0)
Reflect s α √ √ β = ( − 1 , 3) α + β = (1 , 3) α = (2 , 0)
Reflect again & again s α p p β = ( � 1 , 3) α + β = (1 , 3) s β � α = ( � 2 , 0) α = (2 , 0) p p � ( α + β ) = ( � 1 , � 3) � β = (1 , � 3)
Root system α 2 α 1 This is reflection group called A 2. α 1 and α 2 are “simple” roots
Root system s 1 α 2 α 1 This is reflection group called A 2. α 1 and α 2 are “simple” roots
Root system s 1 α 1 + α 2 α 2 α 1 This is reflection group called A 2. α 1 and α 2 are “simple” roots
Root system s 1 α 1 + α 2 α 2 − α 1 α 1 This is reflection group called A 2. α 1 and α 2 are “simple” roots
Root system s 1 α 1 + α 2 α 2 s 2 − α 1 α 1 This is reflection group called A 2. α 1 and α 2 are “simple” roots
Root system s 1 α 1 + α 2 α 2 s 2 − α 1 α 1 − α 2 This is reflection group called A 2. α 1 and α 2 are “simple” roots
Root system s 1 α 1 + α 2 α 2 s 2 − α 1 α 1 − α 2 − α 1 − α 2 This is reflection group called A 2. α 1 and α 2 are “simple” roots
Root system s 1 α 1 + α 2 α 2 s 2 − α 1 α 1 − α 2 − α 1 − α 2 This is reflection group called A 2. α 1 and α 2 are “simple” roots
Reflection groups • Roots: α 1 , α 2 , . . . , α n w i ( α j ) = α j − 2 ( α i , α j ) • Reflections: ( α i , α i ) α i α i • Co-roots: α i = 2 ˇ ( α i , α i ) • Weights: h 1 , h 2 , . . . , h n ( h i , ˇ α j ) = δ ij
longest root s 1 α 1 + α 2 α 2 h 2 s 2 h 1 − α 1 α 1 − α 2 − α 1 − α 2 The University of Sydney 12
Translation by longest route A 2(1)
Dynamics on the lattice Translation 1 T 1 4 0 4 0 2 a 2 = 0 a 1 = 0 a 0 = 0
On the lattice ✑ Define to be s 0 , s 1 , s 2 reflections across each edge a 2 = 0 a a 1 = 0 0 = 0 equilateral triangle
On the lattice ✑ Define to be s 0 , s 1 , s 2 reflections across each edge a 2 = 0 a a 1 = 0 0 = 0 equilateral triangle s 0 ( a 0 , a 1 , a 2 ) = ( − a 0 , a 1 + a 0 , a 2 + a 0 )
Translations as reflections 1 1 T 1 4 0 4 0 0 2 2 a 2 = 0 2 1 a 1 = 0 a 0 = 0 s 1 ( 4 ) 0
Translations as reflections 1 1 0 T 1 4 0 4 s 2 ( s 1 ( 4 )) 0 0 2 2 2 1 a 2 = 0 2 1 = 0 a 0 = 0 1 a 0
Translations as reflections 1 1 1 T 1 π 4 0 4 π ( s 2 ( s 1 ( 4 ))) 0 0 2 2 0 2 a 2 = 0 2 1 0 a 0 = = 1 0 a 0
Translations So we have T 1 ( a 0 ) = π s 2 s 1 ( a 0 ) = π s 2 ( a 0 + a 1 ) = π ( a 0 + a 1 + 2 a 2 ) = a 1 + a 2 + 2 a 0 = a 0 + k ⇒ T 1 ( a 0 ) = a 0 + k, T 1 ( a 1 ) = a 1 − k, T 1 ( a 2 ) = a 2
Translations again Using T 1 ( a 0 ) = a 0 + 1 , T 1 ( a 1 ) = a 1 − 1 , T 1 ( a 2 ) = a 2 Define u n = T n 1 ( f 1 ) , v n = T n 1 ( f 0 )
Translations again Using T 1 ( a 0 ) = a 0 + 1 , T 1 ( a 1 ) = a 1 − 1 , T 1 ( a 2 ) = a 2 Define u n = T n 1 ( f 1 ) , v n = T n 1 ( f 0 ) ( = t − v n − a 0 + n ⇒ u n + u n +1 v n = t − u n + a 1 − n v n + v n − 1 u n
Translations again Using T 1 ( a 0 ) = a 0 + 1 , T 1 ( a 1 ) = a 1 − 1 , T 1 ( a 2 ) = a 2 Define u n = T n 1 ( f 1 ) , v n = T n 1 ( f 0 ) ( = t − v n − a 0 + n ⇒ u n + u n +1 v n = t − u n + a 1 − n v n + v n − 1 u n This is a discrete Painlevé equation.
How Columbus Discovered America. London: Bancroft & Co. 1961, by Vojt ě ch Kuba š ta https://library.bowdoin.edu/arch/exhibits/popup/practitioners.shtml
Dynamics Gridlock in India http://thedailynewnation.com/news/57912/a-total-traffic- chaos-at-shahbagh-intersection-in-city-as-vehicles-got-stuck-up-to-make-way- through-the-massive-gridlock-this-photo-was-taken-on-saturday.html The University of Sydney
Motion follows curves ✑ In initial value space • ✑ But continuation of the flow fails at singularities of curves and at base points where families of curves all intersect. The University of Sydney
Motion of plane pendulum ⇢ ˙ u ( t ) = v ( t ) u ( t ) � � v ( t ) = − sin ˙ u ( t ) The University of Sydney
Phase Space H ( u, v ) = v 2 � � 2 − cos u ( t ) ✑ Level curves of H are curves through initial values. ✑ Phase space ⟺ space of initial values.
Another View f ( t ) = e iu ( t ) 8 ˙ f 2 ¨ f 2 − 1 f − 1 � � f = < 2 ⇒ ⇣ ⌘ ˙ f 2 2 f + 1 f + 1 E = : f 2
Transformed phase plane 6 ✑ Phase curves are given by y 2 = − x ( x − E + ) ( x − E − ) 4 p E ± = E ± E 2 − 1 2 ✑ This is a pencil of cubic curves. ✑ All curves go through (0,0). 0 ✑ Such a point is called a base point of the pencil. 2 4 6 6 4 2 0 2 4 6
Canonical example ✑ Weierstrass cubic curves y 2 = 4 x 3 − g 2 x − g 3 where is fixed and is g 2 g 3 free. ✑ Phase space is no longer compact. In homogeneous coordinates in the curves CP 2 dlmf.nist.gov are w v 2 = 4 u 3 − g 2 uw 2 − g 3 w 3 ✑ Base point: (0, 1, 0) The University of Sydney
Base point 6 4 2 0 2 4 6 Continuation of the flow is not defined through a base point. 6 4 2 0 2 4 6 The University of Sydney 27
Resolving a base point From JJ Duistermaat, Springer Verlag, 2010 The University of Sydney
An example y 2 = x 3 The curve has a singularity at (0,0), which can be resolved. The University of Sydney
Resolution 1 ( ( e 0 x 1 = x x = x 1 ⇔ = y y 1 y = x 1 y 1 x C (1) The University of Sydney
Resolution 1 ( ( e 0 x 1 = x x = x 1 ⇔ = y y 1 y = x 1 y 1 x C (1) The University of Sydney
Resolution 1 ( ( e 0 x 1 = x x = x 1 ⇔ = y y 1 y = x 1 y 1 x f = y 2 − x 3 2 y 1 2 − x 3 = x 1 1 C (1) 2 � 2 − x 1 � = x 1 y 1 The University of Sydney
Resolution 1 ( ( e 0 x 1 = x x = x 1 ⇔ = y y 1 y = x 1 y 1 x f = y 2 − x 3 2 y 1 2 − x 3 = x 1 1 C (1) 2 � 2 − x 1 � = x 1 y 1 e 0 = { x 1 = 0 } , f (1) = y 1 2 − x 1 The University of Sydney
Resolution 2 e (1) ( ( = x 1 0 x 2 x 1 = x 2 y 2 • y 1 ⇔ C (2) y 1 = y 2 y 2 = y 1 f (1) ( x 1 , y 1 ) = y 2 2 − x 2 y 2 • e 1 � � = y 2 y 2 − x 2 • e 1 = { y 2 = 0 } , f (2) = y 2 − x 2 The University of Sydney
Resolution 3 ( ( x 3 = x 2 x 2 = x 3 • ⇔ = y 2 y 3 y 2 = x 3 y 3 x 2 C (3) e (2) • f (2) ( x 2 , y 2 ) = x 3 y 3 − x 3 0 e 2 � � = x 3 y 3 − 1 e (1) 1 • e 2 = { x 3 = 0 } , f (3) = y 3 − 1 The University of Sydney
Intersection theory ✑ Each line has a self- intersection number. C (3) -2 ✑ Each blow up reduces e (2) the self-intersection 0 number by 1 . -1 e 2 ✑ The lines of self- e (1) intersection -2 play a -2 1 special role. The University of Sydney
Intersection theory ✑ Each line has a self- intersection number. C (3) -2 ✑ Each blow up reduces e (2) the self-intersection 0 number by 1 . -1 e 2 ✑ The lines of self- e (1) intersection -2 play a -2 1 special role. The University of Sydney
DuVal Correspondence C (3) ✑ The curves with self- e (2) 0 intersection -2 e 2 correspond to nodes of a Dynkin diagram. e (1) ✑ This is DuVal (or 1 Mackay) correspondence. The University of Sydney
DuVal Correspondence C (3) ✑ The curves with self- e (2) 0 intersection -2 e 2 correspond to nodes of a Dynkin diagram. e (1) ✑ This is DuVal (or 1 Mackay) correspondence. The University of Sydney
DuVal Correspondence C (3) ✑ The curves with self- e (2) 0 intersection -2 e 2 correspond to nodes of a Dynkin diagram. e (1) ✑ This is DuVal (or 1 Mackay) correspondence. ⇓ A 2 The University of Sydney
This A 2 is the reflection group we saw earlier.
Painlevé and discrete Painlevé equations ✑ Equations with initial value space that is compactified and regularised after 9 P 2 blow-ups in . ✑ Sakai classified all such equations. ✑ They each have a symmetry group obtained as an orthogonal complement of the resolved initial value space inside the Picard lattice. The University of Sydney
Sakai described all such equations. The University of Sydney 37
Sakai’s Description II (1) Ell: A 0 (1) 0 A 7 (1) (1) (1) (1) (1) (1) (1) (1) (1) A A A A A A A A A Mul: 0 1 2 3 4 5 6 7 8 (1) (1) (1) (1) (1) (1) (1) (1) A A A D D D D D Add: 0 1 2 4 5 6 7 8 (1) (1) (1) E E E 6 7 8 Initial-value spaces of Painlevé equations Sakai 2001 Rains 2016
Recommend
More recommend
