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lavinia

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## Main Question or Discussion Point

I am interested to know how to realize this abstract surface as a subset of Euclidean space.

The surface as a point set is the 2 dimenional Euclidean plane minus the origin.

the metric is given by declaring the following 2 vector fields to be an orthonormal frame:

e[itex]_{1}[/itex] = x[itex]\partial[/itex]x - y[itex]\partial[/itex]y

e[itex]_{2}[/itex] = x[itex]^{2}[/itex][itex]\partial[/itex]x + e[itex]^{-xy}[/itex][itex]\partial[/itex]y

I think that this surface has constant negative curvature equal to -1 but I do not think it is a Poincare disk minus a point because the origin is infinitely far away from any point on the x or y axes.

BTW: Here is the curvature calculation but I am not 100% sure of it.

the Lie bracket, [e[itex]_{1}[/itex],e[itex]_{2}[/itex]], equals e[itex]_{2}[/itex] so the covariant derivative can be defined using the following formulas:

[itex]\nabla_{e_{1}}[/itex]e[itex]_{1}[/itex] = 0

[itex]\nabla_{e_{1}}[/itex]e[itex]_{2}[/itex] = 0

[itex]\nabla_{e_{2}}[/itex]e[itex]_{1}[/itex] = -e[itex]_{2}[/itex]

[itex]\nabla_{e_{2}}[/itex]e[itex]_{2}[/itex] = e[itex]_{1}[/itex]

Extend these formulas to arbitrary vector fields by linearity and the Leibniz rule.

The connection 1 form is ω(e[itex]_{2}[/itex]) = -1, ω(e[itex]_{1}[/itex]) = 0.

Since this connection is torsion free the exterior derivative of ω easily computes and is the volume element of the metric. Therefore the Gauss curvature is -1.

It may be helpful to note that the rectangular hyperpolas y = k/x are geodesics as are the rays from the origin along the axes.

The surface as a point set is the 2 dimenional Euclidean plane minus the origin.

the metric is given by declaring the following 2 vector fields to be an orthonormal frame:

e[itex]_{1}[/itex] = x[itex]\partial[/itex]x - y[itex]\partial[/itex]y

e[itex]_{2}[/itex] = x[itex]^{2}[/itex][itex]\partial[/itex]x + e[itex]^{-xy}[/itex][itex]\partial[/itex]y

I think that this surface has constant negative curvature equal to -1 but I do not think it is a Poincare disk minus a point because the origin is infinitely far away from any point on the x or y axes.

BTW: Here is the curvature calculation but I am not 100% sure of it.

the Lie bracket, [e[itex]_{1}[/itex],e[itex]_{2}[/itex]], equals e[itex]_{2}[/itex] so the covariant derivative can be defined using the following formulas:

[itex]\nabla_{e_{1}}[/itex]e[itex]_{1}[/itex] = 0

[itex]\nabla_{e_{1}}[/itex]e[itex]_{2}[/itex] = 0

[itex]\nabla_{e_{2}}[/itex]e[itex]_{1}[/itex] = -e[itex]_{2}[/itex]

[itex]\nabla_{e_{2}}[/itex]e[itex]_{2}[/itex] = e[itex]_{1}[/itex]

Extend these formulas to arbitrary vector fields by linearity and the Leibniz rule.

The connection 1 form is ω(e[itex]_{2}[/itex]) = -1, ω(e[itex]_{1}[/itex]) = 0.

Since this connection is torsion free the exterior derivative of ω easily computes and is the volume element of the metric. Therefore the Gauss curvature is -1.

It may be helpful to note that the rectangular hyperpolas y = k/x are geodesics as are the rays from the origin along the axes.

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