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finding a symplectic recursion fixed point

  • To: mathgroup at smc.vnet.net
  • Subject: [mg112019] finding a symplectic recursion fixed point
  • From: Roger Bagula <roger.bagula at gmail.com>
  • Date: Tue, 24 Aug 2010 06:16:27 -0400 (EDT)

The solution of a sp(2) symplectic group like
a0={{a,b},
      {c,-a}}
For a Migdal-Kadanoff like recursion to get a fixed point isn't easy.
m'=Exp[a0].Exp[a0]->m=Exp[a0']
Here is Mathematica code that gets a 3d surface from the recursion:
T = Exp[{{a, b}, {c, -a}}]
T1 = Exp[{{a1, b1}, {c1, -a1}}]
T10 = {a1, b1, c1}
T2 = T.T
B = T2 - T1
ca = Flatten[Table[B[[i, j]] == 0, {i, 2}, {j, 2}]]
Solve[Delete[ca, 1], T10]
Solve[Delete[ca, 4], T10]
A = Table[a /. Solve[Log[Exp[2
           *a] + Exp[b + c]] == -Log[Exp[-2
            *a] + Exp[b + c]], a][[k]], {k, 4}];
B = Table[
   FullSimplify[
    ExpandAll[{Log[Exp[-a + b] + Exp[a + b]],
       Log[Exp[-a + c] + Exp[a + c]], Log[Exp[2*a] + Exp[b + c]]} /.
      a -> A[[k]]]], {k, 4}];
g = ParametricPlot3D[Abs[B[[2]]], {b, -Pi, Pi}, {c, -Pi, Pi},
  Boxed -> False, Axes -> False,
  ColorFunction -> Function[{x, y, z}, Hue[z]]]
Show[g, ViewPoint -> {Pi, Pi/2, 2}]

Can anyone figure out how to isolate the fixed points?


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