[sage-support] Re: computing the cup-product in the cohomology of a simplicial complex
Hi John! Thanks for your detailed answer! I will try to figure this out and get back to you! Felix -- To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to sage-support+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-support URL: http://www.sagemath.org
[sage-support] Re: computing the cup-product in the cohomology of a simplicial complex
Hello again! I have followed your instructions and come up with the following: X = simplicial_complexes.Torus() C = X.chain_complex(cochain=True) print C._chomp_repr_() H = C.homology(generators=True) gen1 = H[1][1][0] gen2 = H[1][1][1] d1 = C.differential()[1] This works very well so far. In particular d1*gen1 gives the zero vector as it should. Now: How can I get the bijection between the indices of the vectors gen1 and gen2 and the corresponding 1-faces in X? C._chomp_repr_() gives the boundary matrices in the form: dimension 1 boundary a1 = - 1 * a3 - 1 * a10 boundary a2 = - 1 * a2 - 1 * a5 boundary a3 = + 1 * a4 + 1 * a9 boundary a4 = + 1 * a6 + 1 * a12 boundary a5 = + 1 * a2 + 1 * a11 boundary a6 = - 1 * a4 - 1 * a11 boundary a7 = + 1 * a4 + 1 * a8 boundary a8 = + 1 * a5 + 1 * a14 boundary a9 = - 1 * a9 + 1 * a10 boundary a10 = + 1 * a1 + 1 * a6 boundary a11 = - 1 * a6 - 1 * a8 boundary a12 = + 1 * a5 + 1 * a7 boundary a13 = + 1 * a8 + 1 * a14 boundary a14 = + 1 * a7 + 1 * a9 ... But I don't see any information what face, say, a1 corresponds to. Am I missing something? Thank you very much for your help! Felix -- To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to sage-support+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-support URL: http://www.sagemath.org
[sage-support] Re: computing the cup-product in the cohomology of a simplicial complex
On Sunday, December 4, 2011 7:47:13 PM UTC-8, Felix Breuer wrote: > > Hello again! > > I have followed your instructions and come up with the following: > > X = simplicial_complexes.Torus() > C = X.chain_complex(cochain=True) > print C._chomp_repr_() > H = C.homology(generators=True) > gen1 = H[1][1][0] > gen2 = H[1][1][1] > d1 = C.differential()[1] > > This works very well so far. In particular d1*gen1 gives the zero vector > as it should. Now: How can I get the bijection between the indices of the > vectors gen1 and gen2 and the corresponding 1-faces in X? C._chomp_repr_() > gives the boundary matrices in the form: > > dimension 1 >boundary a1 = - 1 * a3 - 1 * a10 >boundary a2 = - 1 * a2 - 1 * a5 >boundary a3 = + 1 * a4 + 1 * a9 >boundary a4 = + 1 * a6 + 1 * a12 >boundary a5 = + 1 * a2 + 1 * a11 >boundary a6 = - 1 * a4 - 1 * a11 >boundary a7 = + 1 * a4 + 1 * a8 >boundary a8 = + 1 * a5 + 1 * a14 >boundary a9 = - 1 * a9 + 1 * a10 >boundary a10 = + 1 * a1 + 1 * a6 >boundary a11 = - 1 * a6 - 1 * a8 >boundary a12 = + 1 * a5 + 1 * a7 >boundary a13 = + 1 * a8 + 1 * a14 >boundary a14 = + 1 * a7 + 1 * a9 ... > > > But I don't see any information what face, say, a1 corresponds to. Am I > missing something? > You should be able to do sage: X.n_faces(1) to get the list of 1-simplices. a_i should be dual to the ith element in that list. -- John -- To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to sage-support+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-support URL: http://www.sagemath.org
[sage-support] why is this contour integral wrong?
I keep wondering whether Sage is making a mistake, or I'm not
understanding complex analysis. I'm a little afraid to learn the answer.
:)
Take f(z) = (z-I)*(z-1)^2/(z-(-1/2-I/3)). It's analytic everywhere
except at -1/2-I/3, where it has a simple pole. So, if I integrate over
a circle centered at 0 of radius, say, 2, the answer should be
2*pi*I*(residue of f at -1/2 - I/3),
which is pi*(181/27 + 19*I/36). However, when I try to do the contour
integral, I get:
sage: integrate(f(2*exp(I*t)) * 2*I*exp(I*t), (t, 0, 2*pi))
0
even though the contour encloses the pole. It works if I center the
circle around the pole:
sage: integrate(f(-1/2-I/3 + exp(I*t)) * I*exp(I*t), (t, 0, 2*pi))
(19/36*I + 181/27)*pi
and also if I integrate over the square with vertices 1+i, 1-i, -1-i,
-1+i. What's wrong with the circle at the origin?
Note that Mathematica gets this right, although you need to ask for full
simplification: with f[z_] = (z-I)*(z-1)^2/(z-(-1/2-I/3)), you get
In[5]:= Integrate[f[2*Exp[I*t]] * 2*I*Exp[I*t], {t, 0, 2*Pi}]//FullSimplify
181 19 I
Out[5]= (--- + ) Pi
27 36
Any ideas?
Dan
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--- Dan Drake
- http://mathsci.kaist.ac.kr/~drake
---
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[sage-support] Re: why is this contour integral wrong?
On Dec 5, 5:31 am, Dan Drake wrote:
> I keep wondering whether Sage is making a mistake, or I'm not
> understanding complex analysis. I'm a little afraid to learn the answer.
> :)
>
> Take f(z) = (z-I)*(z-1)^2/(z-(-1/2-I/3)). It's analytic everywhere
> except at -1/2-I/3, where it has a simple pole. So, if I integrate over
> a circle centered at 0 of radius, say, 2, the answer should be
>
> 2*pi*I*(residue of f at -1/2 - I/3),
>
> which is pi*(181/27 + 19*I/36). However, when I try to do the contour
> integral, I get:
>
> sage: integrate(f(2*exp(I*t)) * 2*I*exp(I*t), (t, 0, 2*pi))
> 0
>
> even though the contour encloses the pole. It works if I center the
> circle around the pole:
>
> sage: integrate(f(-1/2-I/3 + exp(I*t)) * I*exp(I*t), (t, 0, 2*pi))
> (19/36*I + 181/27)*pi
>
> and also if I integrate over the square with vertices 1+i, 1-i, -1-i,
> -1+i. What's wrong with the circle at the origin?
>
> Note that Mathematica gets this right, although you need to ask for full
> simplification: with f[z_] = (z-I)*(z-1)^2/(z-(-1/2-I/3)), you get
>
> In[5]:= Integrate[f[2*Exp[I*t]] * 2*I*Exp[I*t], {t, 0, 2*Pi}]//FullSimplify
>
> 181 19 I
> Out[5]= (--- + ) Pi
> 27 36
>
> Any ideas?
>
> Dan
>
> --
> --- Dan Drake
> - http://mathsci.kaist.ac.kr/~drake
> ---
>
> signature.asc
> < 1KViewDownload
Note that pari has a good approximate solution:
sage: pari('f(z) = (z-I)*(z-1)^2/(z-(-1/2-I/3))')
(z)->(z-I)*(z-1)^2/(z-(-1/2-I/3))
sage: 2*n(pi)*I*pari('intcirc(z=0,2,f(z))')
21.0603063073982 + 1.65806278939461*I
sage: CC(pi*(181/27 + 19*I/36))
21.0603063073982 + 1.65806278939461*I
Andrzej Chrzeszczyk
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