Homology of rotatopes

Discussion of shapes with curves and holes in various dimensions.

Homology of rotatopes

Postby PWrong » Sun Nov 22, 2009 9:40 am

Using Mayer-Vietoris, I've managed to find a recursive equation for homology groups of extrusions. So if you know all the homology groups of all the frames of some shape A, then this formula will tell you the homology groups of all the frames of A1 = A x I.

I write FkA for the k-frame of A. I haven't checked it yet, so there could be a mistake somewhere, but this is what I've got. Let X = A1 = A extruded. Then

H0 FkX = H0 Fk-1A

H1 FkX = 2H1 FkA + 2H0 Fk-1A - 2Z

Hq FkX = 2Hq FkA + Hq-1 Fk-1A
for all q > 1

EDIT: fixed Hq

This combined with an initial condition: H0 F0 {.} = Z, where {.} is a point,
gives us the homology groups for all the cubes.
Since we also know the homology groups for hyperspheres, we can work out any rotatope with one sphere and a bunch of lines, e.g. 311 or 5111.
Working out the rest might be tricky because of the way Mayer-Vietoris works (which I'll try to explain eventually).

I'll do up a table after I check the equation.
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Re: Homology of rotatopes

Postby PWrong » Mon Nov 23, 2009 12:10 am

I've made a mistake somewhere. According to this formula,
H0 F1 II = H0 F0 I = 2
but H0 of a wireframe square must be 1.

I think I missed a couple of other things, so I'll derive the equation again.
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Re: Homology of rotatopes

Postby PWrong » Mon Nov 23, 2009 12:40 am

It seems I have a problem when the frame number is equal to the minimum or maximum frame of X. This is because the shape is then a union of itself with the empty set. This is a problem for Mayer Vietoris, which I'll explain soon in another thread.
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Re: Homology of rotatopes

Postby PWrong » Mon Nov 23, 2009 4:17 am

For the minimum frame, we have
Hq Fmin X = 2 Hq Fmin A

For the maximum frame,
Hq Fmax X = Hq Fmax-1 A

For other frames, the equations in my first post should still work.
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Re: Homology of rotatopes

Postby PWrong » Mon Nov 23, 2009 5:05 am

Turns out I made a mistake for H0 of frames between the min and max.

H0 X simply equals Z except when X is a minimal frame.
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Re: Homology of rotatopes

Postby PWrong » Sat Nov 28, 2009 1:54 am

My computer died so I'm stuck posting on my iPhone for a while. So that might make equations and long posts difficult.

I've got a method for working out the homology of any rotatope in the minimal frame. I can also do larger frames if the shape is a product of any two spheres. So I can do 54 in any frame, and 433 only in the minimal frame. The simplest shape I can't work out is the 4-frame 222. Problem is it's naturally a union of three cells, so it's hard to cut up into 2 simple pieces. I have one approach that kind of works, but i have to guess what the maps are. I'm not sure how I would extend it to larger shapes.
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Re: Homology of rotatopes

Postby PWrong » Mon Nov 30, 2009 11:22 am

Here are the rules for extrusions, all in one post.

For minframe < k < maxframe

H0 FkX = Z

H1 FkX = 2H1 FkA + 2H0 Fk-1A - 2Z

Hq FkX = 2Hq FkA + Hq-1 Fk-1A
for all q > 1

For the minimum frame:
Hq Fmin X = 2 Hq Fmin A

For the maximum frame,
Hq Fmax X = Hq Fmax-1 A
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Re: Homology of rotatopes

Postby PWrong » Mon Nov 30, 2009 4:19 pm

I just realised I've been missing a simple rule all this time. I can do cartesian products of circles with anything now :D. All of the following applies only to minimal frames.

Hq 2xY = HqA ⊕ Hq-1A

Proof:
Let X = 2xY for any shape Y. We can treat the circle like any other sphere and cut it into the northern part and the southern part.
Let A = S1N x Y which is homotopic to Y. So it has the same homology groups as Y
Let B = S1S x Y which is homotopic to Y. Again, same homology groups as Y.
Then AnB is S0 x Y. All the homology groups are doubled.

Now each bit of the Mayer-Vietoris sequence looks like:

Hq+1X --> HqAnB --> HqA⊕HqB --> HqX

Hq+1X --> 2HqY --> 2HqY --> HqX

The map in the middle sends (a,b) to (a+b, -a-b).
It has an image isomorphic to HqY, the complement of which is HqY.
It has a kernal isomorphic to HqY.

So we have, for all q:
Hq+1X -->> HqY
HqY >--> HqX

and therefore:
HqY >--> HqX --> Hq-1Y

so HqX = HqA ⊕ Hq-1A as required.


I just have to check this is consistent with my previous rules, and then I can do more complicated rotatopes and we can start making a proper table. I'll hopefully do this over the next few days.
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Re: Homology of rotatopes

Postby PWrong » Wed Dec 02, 2009 2:52 am

This rule contradicts shapes I already worked out :(

According to this rule,
H222 = 1,3,3,1

But using the other method (cutting out a cube) it's H222 = 1,3,0,1
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Re: Homology of rotatopes

Postby PWrong » Wed Dec 02, 2009 3:42 am

According to wikipedia, 1,3,3,1 is right :D.

The k-th homology group of an n-torus is a free abelian group of rank n choose k.


The other method is apparently valid, so I must be doing something wrong. Maybe when you cut a cube out of 222, you're left with a wedge sum of 3 circles and 3 spheres.
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Re: Homology of rotatopes

Postby Keiji » Wed Dec 02, 2009 7:06 am

Ugh... it'd be nice if the homology groups didn't have to keep changing!

I take it an n-ocylinder is a row of Pascal's triangle, then?
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Re: Homology of rotatopes

Postby PWrong » Wed Dec 02, 2009 7:17 am

Sorry, these things are hard to calculate :(.

I take it an n-ocylinder is a row of Pascal's triangle, then?

Yep. And many other shapes look like Pascal's triangle with a different starting point.

The triangle for the sequence 3, 32, 322, 3222, ... looks like this:

1,0,1
1,1,1,1
1,2,2,2,1
1,3,4,4,3,1
1,4,7,8,7,4,1
1,5,11,15,15,11,5,1
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Re: Homology of rotatopes

Postby PWrong » Wed Dec 02, 2009 7:22 am

So we can now easily work out shapes of the form a-sphere x b-sphere x arbitrarily many circles.

The next shape to beat is 333.
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Re: Homology of rotatopes

Postby Keiji » Wed Dec 02, 2009 7:54 am

PWrong wrote:And many other shapes look like Pascal's triangle with a different starting point.


Yeah, I just worked that out too. The hole-sequence generating part of my program now looks like this:

Code: Select all
def numsum(a):
   r = 0
   for c in a:
      r += int(c)
   return r

def elemult(s, p):
   r = []
   for item in s:
      r.append(item*p)
   return r

def pascal(n, r = None):
   if (r == None): r = []
   i = len(r)
   end = r[i-1]
   while (i < n):
      j = i-1
      while (j > 0):
         r[j] += r[j-1]
         j -= 1
      r.append(end)
      i += 1
   return r

def mfhs(r):
   t = ''
   ones = 0
   twos = 0
   HSlen = numsum(r)-len(r)+1
   for c in r:
      if (c == '1'):
         ones += 1
      elif (c == '2'):
         twos += 1
      else:
         t += c
   if (t == ''):
      r = [1]
   elif (len(t) == 1):
      r = [1] + [0]*(int(t)-2) + [1]
   elif (len(t) == 2):
      r = [1] + [0]*(numsum(t)-3) + [1]
      r[int(t[1])-1] += 1
      r[len(r)-int(t[1])] += 1
   else:
      return None
   if (twos): r = pascal(HSlen, r)
   return elemult(r, 1<<ones)
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Re: Homology of rotatopes

Postby PWrong » Wed Dec 02, 2009 8:10 am

Hey, that looks awesome :D.

Also, I worked out 333 with the hemisphere approach. It's [1,0,3,0,3,0,1]
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Re: Homology of rotatopes

Postby PWrong » Thu Dec 03, 2009 12:39 am

We have two options for 433. We can cut the 4 in half or a 3.

X = 433
A = 33, B = 33,
[1, 0, 2, 0, 1]
AnB = 333
[1, 0, 3, 0, 3, 0, 1]

This option has maps from 3Z to 4Z and from 4Z to 2Z. I have no idea what these maps would be.


X = 433
A = 43, B = 43,
[1, 0, 1, 1, 0, 1]
AnB = 432
[1, 1, 1, 2, 1, 1, 1]

This is easier, I can guess what most of these maps look like. I'm not sure about the map from 2Z --> 2Z though. Usually it would be (a,b) |--> (a+b, -a-b), but since a and b come from different places (one from the 3-hole in a sphere, the other from the 4-hole in a glome), maybe they don't both count. It could be (a,b) |--> (b,-b).
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Re: Homology of rotatopes

Postby PWrong » Thu Dec 03, 2009 12:56 am

What I'll do is a compromise between 433 alone and A3 for general A.

I'll work out Sa x Sb x S2, with the assumption that 1 < a < b-1 so there isn't too much overlap. Then I'll see if the resulting formula works when you take away the assumption (it should do).

X = Sa x Sb x S2
A = Sa x Sb
B = Sa x Sb
AnB = Sa x Sb x S1

So we have a map
Sa x Sb x S1 --> 2 Sa x Sb

h0 +h1 + ha + ha+1 + hb + hb+1 + ha+b + ha+b+1 --> h0 +ha + hb + ha+b

Assuming a and b are sufficiently far apart, all the maps are either Z --> 0 or Z --> 2Z.
We can replace these with ...-->> Z --> 0 >-->... and ...-->>0 --> Z >-->... respectively.

So we have four bits of Mayer-Vietoris sequence.

... 0 >--> H1X -->> Z --> 0 >--> H1X -->> 0 --> Z >--> H0X -->> 0

... 0 >--> Ha+2X -->> Z --> 0 >--> Ha+1X -->> 0 --> Z >--> HaX -->> 0 ...

... 0 >--> Hb+2X -->> Z --> 0 >--> Hb+1X -->> 0 --> Z >--> HbX -->> 0 ...

... 0 >--> Ha+b+2X -->> Z --> 0 >--> Ha+b+1X -->> 0 --> Z >--> Ha+bX -->> 0 ...

So the solution is
h0 +h2 + ha + ha+2 + hb + hb+2 + ha+b + ha+b+2.
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Re: Homology of rotatopes

Postby PWrong » Thu Dec 03, 2009 1:53 am

I've done 433 assuming that the map I was unsure about is (a,b) |--> (a+b, -a-b).

H433 = [1, 0, 2, 1, 1, 2, 0, 1], which is exactly what the formula gives us. So that suggests the formula works even when the sphere's are similar.

The formula is also consistent with 333 :D.
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Re: Homology of rotatopes

Postby Keiji » Thu Dec 03, 2009 10:50 am

I'm seeing a symmetry in the hole-sequences here.

Is it true that all hole-sequences for rotatopes are indeed symmetric? (I presume so.)
It's obviously not true that all hole-sequences for unions or wedge products are.
Are there any other supersets of rotatopes which preserve the symmetry?
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Re: Homology of rotatopes

Postby PWrong » Thu Dec 03, 2009 10:56 am

Well all the toratopes are symmetric, since they're homeorphic to a rotatope. The symmetry follows from my new conjecture. Also the n-torus (G2) is always symmetric.
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Re: Homology of rotatopes

Postby Keiji » Thu Dec 03, 2009 11:04 am

Well, I wasn't counting toratopes as a superset in this case as all toratopes as they're all homeomorphic anyway. I meant supersets that include hole-sequences which do not correspond to any rotatope.

Basically, it would be nice to say `X is homeomorphic to a rotatope <=> HX is symmetric`
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Re: Homology of rotatopes

Postby PWrong » Fri Dec 04, 2009 1:56 am

Klein bottles and projective plains are asymmetric, but the n-torus is [1,n,1], so that's a counterexample. Maybe the connected sum of shapes with symmetric homology also has symmetric homology.

The symmetry might relate to the fundamental polygon somehow.

http://en.wikipedia.org/wiki/Fundamental_polygon
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Re: Homology of rotatopes

Postby Keiji » Fri Dec 04, 2009 9:34 am

PWrong wrote:Klein bottles and projective plains are asymmetric, but the n-torus is [1,n,1], so that's a counterexample.

The symmetry might relate to the fundamental polygon somehow.


So maybe non-orientable surfaces are asymmetric, and orientable surfaces are symmetric?

Also, what are the hole-sequences for the Möbius strip, the Klein bottle and the real projective plane?

Maybe the connected sum of shapes with symmetric homology also has symmetric homology.


What's the connected sum? I imagine it's not the same as the wedge sum as that one definitely introduces asymmetry.
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Re: Homology of rotatopes

Postby PWrong » Fri Dec 04, 2009 10:28 am

So maybe non-orientable surfaces are asymmetric, and orientable surfaces are symmetric?

What about wedge sums and unions?


Also, what are the hole-sequences for the Möbius strip, the Klein bottle and the real projective plane?

The Mobius strip can deform into a circle, so it's [Z,Z].

Klein bottle is [Z, Z+Z2] where Z2 is the cyclic group with 2 elements.

Projective plane is [Z, Z2].


What's the connected sum? I imagine it's not the same as the wedge sum as that one definitely introduces asymmetry.

Take a disk out of each shape, and glue them together at the boundary.
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Re: Homology of rotatopes

Postby Keiji » Fri Dec 04, 2009 11:00 am

PWrong wrote:The Mobius strip can deform into a circle, so it's [Z,Z].


Using that argument, the Klein bottle can be deformed into a torus.

Take a disk out of each shape, and glue them together at the boundary.


Isn't that the same as the wedge sum?
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Re: Homology of rotatopes

Postby PWrong » Fri Dec 04, 2009 11:13 am

Wedge sum connects the shapes at a point, connected sum connects at a disk (or some appropriate k-ball).

I'm sure you can't deform a Klein bottle into a torus or duocylinder. But a möbius strip can retract into the circle in the middle.
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Re: Homology of rotatopes

Postby Keiji » Fri Dec 04, 2009 12:51 pm

Hmm, I see.

So where do you get the hole-sequences from then?
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Re: Homology of rotatopes

Postby PWrong » Sat Dec 05, 2009 12:56 am

The wikipedia article on Mayer-Vietoris explains one way to get the Klein bottle. I learned another way, where you can calculate the torus, Klein bottle and projective plane all in one go. They have the same sequence, but different maps. I'm trying to remember how it went but it's not working :\.
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Re: Homology of rotatopes

Postby PWrong » Wed Dec 09, 2009 2:02 am

Basically you cut out a square from the shape and call that A. Then B is what's left, which can be retracted to S^1vS^1. AnB can be deformed to a circle. When you do Mayer-Vietoris the maps depend on what shape you start with. I never understood why the maps weren't all zero though.
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