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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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