Ask Your Question
0

Connection Forms not Anti-Symmetric

asked 2023-12-31 17:46:37 +0100

idontgetoutmuch gravatar image

M = Manifold(2, 'M', r'\mathcal{M}')

c_xy.<x,y> = M.chart('x:(-1,1) y:(-1,1)', coord_restrictions=lambda x,y: x^2+y^2<1)

g = M.riemannian_metric('g')

g[0,0], g[1,1] = 4/(1 - x^2 - y^2)^2, 4/(1 - x^2 - y^2)^2

e1 = M.vector_field((1 - x^2 - y^2) / 2, 0)

e2 = M.vector_field(0, (1 - x^2 - y^2) / 2)

e = M.vector_frame('e', (e1, e2), non_coordinate_basis=True)

nabla = g.connection()

omega = nabla.connection_form

omega(0,0).display(e), omega(0,1).display(e)

(nabla_g connection 1-form (0,0) = x e^0 + y e^1, nabla_g connection 1-form (0,1) = y e^0 - x e^1)

omega(1,0).display(e), omega(0,1).display(e)

(nabla_g connection 1-form (1,0) = -y e^0 + x e^1, nabla_g connection 1-form (0,1) = y e^0 - x e^1)

edit retag flag offensive close merge delete

Comments

1

What anti-symmetry do you expect and why?

rburing gravatar imagerburing ( 2024-01-01 09:27:16 +0100 )edit
1

\omega_{ij} = -\omega{ji} - the connection forms for the Levi-Civita connection are well known to be anti-symmetric so the diagonal elements should be 0. Anyhow see my answer.

idontgetoutmuch gravatar imageidontgetoutmuch ( 2024-01-03 10:30:47 +0100 )edit

1 Answer

Sort by » oldest newest most voted
0

answered 2024-01-03 10:28:39 +0100

idontgetoutmuch gravatar image

It seems one should add the frame as an argument to the connection form as well as to the request for display.

omega(0,0,e).display(e), omega(0,1,e).display(e)

(nabla_g connection 1-form (0,0) = 0, nabla_g connection 1-form (0,1) = ye^0 - x e^1)

omega(1,0,e).display(e), omega(1,1,e).display(e)

(nabla_g connection 1-form (1,0) = -ye^0 + x e^1, nabla_g connection 1-form (1,1) = 0)

edit flag offensive delete link more

Comments

1

Indeed, if the frame argument is not provided, the connection 1-forms are computed for the manifold's default frame, see the documentation: https://doc.sagemath.org/html/en/refe...

The antisymmetry you referring to holds only for an orthonormal frame, which is the case of e, but not of the manifold's default frame, the latter being the coordinate frame (d/dx, d/dy).

eric_g gravatar imageeric_g ( 2024-01-03 15:15:39 +0100 )edit

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Question Tools

1 follower

Stats

Asked: 2023-12-31 17:46:37 +0100

Seen: 366 times

Last updated: Jan 03