On 2017-10-02 13:08, rcs@xmission.com wrote:

This seems worth passing along to the Math-Fun crowd. Read the bottom half first. My apologies to Propp & Speyer. -- Rich

----- Forwarded message from speyer@UMICH.EDU ----- Date: Mon, 2 Oct 2017 12:34:10 -0400 From: David Speyer <speyer@UMICH.EDU> Reply-To: The Robbins Forum <ROBBINS@LISTSERV.UML.EDU> Subject: Re: [ROBBINS] A 7-cycle To: ROBBINS@LISTSERV.UML.EDU

This provides no insight but, yes, the composition has order 7. Define the seven cones

1: 0 < x,y 2: 0 < -x < y 3: 0 < y < -x 4: x<y<0 5: y<x<0 6: 0 < x < -y 7: 0 < y < -x

Then one can check that h and v are linear on each cone, with image another cone. Specifically, h(k) = 4-k mod 7 and v(k) = -k mod 7. So hv permutes the cones by k+4 mod 7 and (hv)^7 is the identity on the set of cones; it is then easy to check that it is the identity on each cone.

I'm curious about the tropicalization question -- seven isn't a number which tends to come out of cluster algebras.

READ THIS FIRST

On Mon, Oct 2, 2017 at 12:20 PM, James Propp <jamespropp@gmail.com> wrote:

...

Apologies for the double posting to the DAC listserv and Robbins listserv; both seem equally appropriate.

If we define piecewise-linear continuous maps h and v from R^2 to itself with h((x,y)) = (-x-max(y,0),y) and v((x,y)) = (x,-y-max(x,0)), then the composition of h and v appears to be of order 7.

Is this true?

Usually the way I prove a PL identity like this is by finding a birational identity that tropicalizes to it (and having Mathematica do the heavy lifting), but in this case that strategy failed; I tried the obvious de-tropicalization H((x,y)) = (1/(x (1 + y)), y) and V((x,y)) = (x, 1/(y (1 + x))) but the composition of these maps is not of order 7.

Is the composition of h and v the tropicalization of some birational map of order 7?

(If you're wondering where h and v come from: they're the horizontal and vertical fliber-flipping maps for the polygon bounded by the lines x=-1, x=1, y=-1, y=1, and x+y=1; the composition of h and v leaves the origin fixed and cyclically permutes the 7 lattice points on the boundary. If the term "fiber-flipping" is unfamiliar, see the bottom of page 3 and the top of page 4 of https://arxiv.org/pdf/1404.3455.pdf, where the fiber-flipping maps are called piecewise-linear toggles.)

Jim Propp

Mathematica comes bizarrely close to proving order 7 symbolically, in ℂ rather than ℝ² : h = I Im@# - Re@# - Max[0, Im@#] & v = Re@# - I (Im@# + Max[0, Re@#]) & The 1st iterate: In[29]:= hv = Function @@ {FullSimplify[Composition[h, v]@#]} Out[29]= -I (Im[#1] - I Max[0, -Im[#1] - Max[0, Re[#1]]] + Max[0, Re[#1]]) - Re[#1] & The 2nd iterate: In[7]:= Function[x, Evaluate[FullSimplify[Composition[%, %]@x]]] // tim During evaluation of In[7]:= 1.486366,2 Out[7]= Function[x, I Im[x] + Max[0, -Im[x] - Max[0, Re[x]]] - Max[0, Im[x] - Max[0, -Max[0, -Im[x] - Max[0, Re[x]]] - Re[x]] + Max[0, Re[x]]] - I Max[0, -Max[0, -Im[x] - Max[0, Re[x]]] - Re[x]] + I Max[0, Re[x]] + Re[x]] The 4th iterate: In[8]:= f4=Function[y, Evaluate[FullSimplify[Composition[%, %]@y]]] // tim During evaluation of In[8]:= 8034.496115,2 Out[8]= Function[y, Piecewise[ {{I y - Re[y], y == 0 || (Re[y] > 0 && ((Im[y] < 0 && Im[y] + Re[y] != 0) || Im[y] > 0))}, {(-1 + I) Re[y], (Im[y] == Re[y] && Re[y] < 0) || (y == Re[y] && Re[y] > 0)}, {I y, (Re[y] < 0 && ((Im[y] + Re[y] < 0 && Im[y] >= 0) || (Im[y] < 0 && Im[y] != Re[y]) || (Im[y] > 0 && Im[y] + Re[y] > 0))) || (Re[y] == 0 && y != 0)}, {y + 2 I Re[y], Im[y] + Re[y] == 0 && Im[y] > 0}}, (1 + I) y - (2 - I) Re[y]]] The 3rd iterate: f3 = Function[{x}, Evaluate[FullSimplify[Composition[hv,hv,hv]@x]]] Function[{x}, Piecewise[ {{Im[x], (Re[x] == 0 && Im[x] < 0) || x == 0}, {-I x, Im[x] < 0 && ((Re[x] < 0 && Im[x] != Re[x]) || (Re[x] > 0 && Im[x] + Re[x] != 0))}, {(1 - I) Im[x], Im[x] > 0 && Re[x] == 0}, {(1 - I) Re[x], Im[x] == Re[x] && Im[x] < 0}, {(-1 - I) x + Re[x], Im[x] >= 0 && ((Re[x] < 0 && Im[x] + Re[x] != 0) || Re[x] > 0)}, {-Re[x], Re[x] < 0 && Im[x] + Re[x] == 0}}, (-1 - I) Re[x]]]; And for the grand finale 7th iterate: In[14]:= Function[z, Evaluate[FullSimplify[Composition[f4, f3]@z]]] // tim During evaluation of In[14]:= 46.694108,2 Out[14]= Function[z, Piecewise[ {{z, z == 0 || z == 0 || ((1 + I) z == Re[z] && Re[z] < 0 && Im[z] + Re[z] != 0 && Im[z] >= 0) || ((1 + I) z == Re[z] && Im[z] >= 0 && Re[z] > 0) || (Im[z] + Re[z] < 0 && Im[z] > 0) || (Re[z] < 0 && Im[z] + Re[z] > 0) || (Re[z] > 0 && Im[z] > 0) || (Im[z] < 0 && Re[z] != 0 && Im[z] + Re[z] != 0 && ((Re[z] < 0 && (-z + (1 + I) Im[z] < 0 || -z + (1 + I) Im[z] > 0) && Im[z] != Re[z]) || (Re[z] > 0 && z + (1 - I) Im[z] != 0)))}, {I Im[z], Re[z] == 0 && z != 0}, {(1 - I) Re[z], Im[z] + Re[z] == 0 && (Im[z] < 0 || Re[z] < 0)}, {(2 - I) Re[z], (z == Re[z] && (z == 0 || Re[z] == 0 || ((1 + I) z == Re[z] && (Re[z] >= 0 || Im[z] + Re[z] != 0)))) || (Im[z] >= 0 && Re[z] != 0 && (((1 + I) z == Re[z] && (Im[z] + Re[z] != 0 || Re[z] >= 0)) || (Im[z] > 0 && Im[z] + Re[z] != 0)))}, {(1 + I) Re[z], Re[z] < 0 && (Im[z] == Re[z] || (z - (1 + I) Re[z] == 0 && -z + (1 + I) Im[z] >= 0 && Im[z] < 0 && -z + (1 + I) Im[z] <= 0))}, {(1 - I) z + I Re[z], (z == Re[z] && (1 + I) z != Re[z] && ((Re[z] < 0 && Im[z] + Re[z] != 0) || Re[z] > 0)) || (Im[z] < 0 && ((z + (1 - I) Im[z] != 0 && Im[z] + Re[z] != 0 && Re[z] > 0) || (Re[z] < 0 && Im[z] != Re[z] && (-z + (1 + I) Im[z] < 0 || -z + (1 + I) Im[z] > 0)) || Re[z] == 0))}, {(-1 - 3 I) z, (Im[z] == Re[z] || z - (1 + I) Re[z] == 0) && Im[z] < 0 && Re[z] < 0}, {0, Re[z] == 0 && Im[z] < 0}, {(1 - I) z - (2 - I) Im[z], Im[z] < 0 && ((Re[z] < 0 && Im[z] != Re[z]) || (Re[z] > 0 && Im[z] + Re[z] != 0))}, {z - I Re[z], Im[z] >= 0 && ((Re[z] < 0 && Im[z] + Re[z] != 0) || Re[z]

0)}}, (2 - 3 I) z]]

The 4th iterate took 2 hrs, This 7th took 46 secs. Notice the first branch is z, the identity! Is that crazy condition an unsimplified tautology? Anything but! Out[22]= z == 0 || z == 0 || ((1 + I) z == Re[z] && Re[z] < 0 && Im[z] + Re[z] != 0 && Im[z] >= 0) || ((1 + I) z == Re[z] && Im[z] >= 0 && Re[z] > 0) || (Im[z] + Re[z] < 0 && Im[z] > 0) || (Re[z] < 0 && Im[z] + Re[z] > 0) || (Re[z] > 0 && Im[z] > 0) || (Im[z] < 0 && Re[z] != 0 && Im[z] + Re[z] != 0 && ((Re[z] < 0 && (-z + (1 + I) Im[z] < 0 || -z + (1 + I) Im[z] > 0) && Im[z] != Re[z]) || (Re[z] > 0 && z + (1 - I) Im[z] != 0))) Out[24]= {{z -> 1}, {z -> 1 + I}, {z -> 2 I}, {z -> -2 + 2 I}, {z -> -4}, {z -> -4 - 4 I}, {z -> -8 I}, {z -> 8 - 8 I}, {z -> 16}} In[25]:= %22 /. % Out[25]= {False, True, False, False, False, False, False, False, False} So Mathematica blew it and the 7th iterate isn't really the identity -- NOT!! In[17]:= (4/5 - 3 I/5)^Range[-5, 5] Out[17]= {-(3116/3125) - (237 I)/3125, -(527/625) + (336 I)/625, -(44/125) + (117 I)/125, 7/25 + (24 I)/25, 4/5 + (3 I)/5, 1, 4/5 - (3 I)/5,7/25 - (24 I)/25, -(44/125) - (117 I)/125, -(527/625) - (336 I)/625, -(3116/3125) + (237 I)/3125} In[19]:= %14 /@ %17 Out[19]= {-(3116/3125) - (237 I)/3125, -(527/625) + (336 I)/625, -(44/125) + (117 I)/125, 7/25 + (24 I)/25, 4/5 + (3 I)/5, 1, 4/5 - (3 I)/5,7/25 - (24 I)/25, -(44/125) - (117 I)/125, -(527/625) - (336 I)/625, -(3116/3125) + (237 I)/3125} It's just a bizarrely unsimplified identity. Maybe someday FullSimplifyInTheSky will say so. Something for WRI's To Do List. --rwg