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This article defines and describes hyperbolic triangles whose vertices are allowed to lie outside the hyperbolic plane and its boundary. We call these triangles generalized hyperbolic triangles because their representation as polygons can have more than three sides.

By "outside the plane," we mean that from the perspective of the hyperboloid model of the hyperbolic plane, the point corresponds to a vector in Lorentz space with real as opposed to imaginary norm. We explain the meaning of this further below.

When viewed as polygons, generalized hyperbolic triangles are, roughly speaking, the polygons one can get by replacing one or more vertices of a normal hyperbolic triangle with an extra side and two right-angled vertices. In addition to normal triangles with three sides, the polygons can be — quadrilaterals with at least two right angles, pentagons with at least four right angles, and hexagons with all right angles. The figure below shows the five possible cases. (There are also some self-intersecting cases, which we don't currently show.)

This figure shows the five possible shapes of a generalized hyperbolic triangle.

Definition

Following a notation similar to [Th, p. 64], let \(\mathbf{E}^{2,1}\) denote the three-dimensional Lorentz space (aka Minkowski space) with indefinite metric \(ds^2 = dx_1^2 + dx_2^2 - dx_3^2\) and associated quadratic form \(Q(x) = x_1^2 + x_2^2 - x_3^2\). The set of vectors \(v\in \mathbf{E}^{2,1}\) with \(Q(v)=-1\) is a 2-sheeted hyperboloid. Let \(H\) denote the upper sheet (i.e. \(x_3>0\)) of this hyperboloid. The surface \(H\) is the hyperboloid in the hyperboloid model of the hyperbolic plane.

Define a generalized hyperbolic triangle to be a triple \((v_1, v_2, v_3)\) of vectors in \(\mathbf{E}^{2,1}\) that are linearly independent. Requiring linear independence ensures that the triangles aren't degenerate (e.g. all points lying on the same line). In the next section, we describe how to interpret a generalized hyperbolic triangle as a polygon in the plane. We also call the polygons that can result generalized hyperbolic triangles.

Interpretation as polygons

Before describing how to interpret generalized hyperbolic triangles as polygons, we first describe how to interpret a single vector in \(\mathbf{E}^{2,1}\) in terms of the hyperbolic plane. See [Th, pp. 76-82] for a more detailed treatment of some of this material.

Thus, consider any non-zero vector \(v\in \mathbf{E}^{2,1}\). Then \(Q(v)\) is either positive, negative, or zero. The related quantity \(\sqrt{Q(v)}\) is known as the length, or norm, of \(v\) and is either real, imaginary, or zero, respectively. If \(Q(v)<0\) (imaginary length \(v\)), then the 1-dimensional subspace spanned by \(v\) intersects \(H\) in a single point, and \(v\) can be interpreted as that point in \(H\). If \(Q(v)>0\) (real length \(v\)), the 1-dimensional subspace spanned by \(v\) doesn't intersect \(H\), and we are in the case of \(v\) lying outside the plane. In this case, consider the 2-dimensional plane \(v^\perp\) in \(\mathbf{E}^{2,1}\) orthogonal to \(v\). The intersection of \(v^\perp\) with \(H\) forms a geodesic in \(H\). We will use this geodesic below when addressing triangles. Finally, if \(Q(v)=0\) (zero length \(v\)), then \(v\) corresponds to an ideal point, or point on the boundary of \(H\).

Consider now a triple \((v_1, v_2, v_3)\) of vectors in \(\mathbf{E}^{2,1}\). To interpret this triple as a polygon, apply the interpretation above for each \(v_i\) separately. Namely, if \(Q(v)\leq 0\), the vector corresponds to a point in \(H\) or on the boundary of \(H\). If \(Q(v)>0\), the vector corresponds to a geodesic in \(H\).

Finally, connect the resulting three points and/or geodesics to form a polygon, as follows. To connect a point with an adjacent point, use the unique geodesic segment connecting the two points. To connect a geodesic with an adjacent geodesic, if they don't already intersect, use the unique geodesic segment that intersects both geodesics perpendicularly. To connect a geodesic with an adjacent point, if they don't already intersect, use the unique geodesic segment from the point to the geodesic that intersects the geodesic perpendicularly.

This figure shows the construction of the polygon corresponding to a generalized hyperbolic triangle \((v_1, v_2, v_3)\) when \(v_1\) and \(v_2\) have imaginary length and \(v_3\) has real length (i.e. one vertex lies outside the hyperbolic plane). The resulting polygon is a quadrilateral with two right angles.

The figure above illustrates the interpretation described above for a generalized hyperbolic triangle \((v_1, v_2, v_3)\) with \(v_1\) and \(v_2\) of imaginary length and \(v_3\) of real length.

References

  • [Th] William P. Thurston, Three-dimensional Geometry and Topology, Vol. 1, edited by Silvio Levy, Princeton Mathematical Series, 35, 1997.

This article defines and describes hyperbolic triangles whose vertices are allowed to lie outside the hyperbolic plane and its boundary. We call these triangles generalized hyperbolic triangles because their representation as polygons can have more than three sides.

By "outside the plane," we mean that from the perspective of the hyperboloid model of the hyperbolic plane, the point corresponds to a vector in Lorentz space with real as opposed to imaginary norm. We explain the meaning of this further below.

When viewed as polygons, generalized hyperbolic triangles are, roughly speaking, the polygons one can get by replacing one or more vertices of a normal hyperbolic triangle with an extra side and two right-angled vertices. In addition to normal triangles with three sides, the polygons can be — quadrilaterals with at least two right angles, pentagons with at least four right angles, and hexagons with all right angles. The figure below shows the five possible cases. (There are also some self-intersecting cases, which we don't currently show.)

This figure shows the five possible cases for the shape associated to a generalized hyperbolic triangle.

Definition

Following a notation similar to [Th, p. 64], let \(\mathbf{E}^{2,1}\) denote the three-dimensional Lorentz space (aka Minkowski space) with indefinite metric \(ds^2 = dx_1^2 + dx_2^2 - dx_3^2\) and associated quadratic form \(Q(x) = x_1^2 + x_2^2 - x_3^2\). The set of vectors \(v\in \mathbf{E}^{2,1}\) with \(Q(v)=-1\) is a 2-sheeted hyperboloid. Let \(H\) denote the upper sheet (i.e. \(x_3>0\)) of this hyperboloid. The surface \(H\) is the hyperboloid in the hyperboloid model of the hyperbolic plane.

Define a generalized hyperbolic triangle to be a triple \((v_1, v_2, v_3)\) of vectors in \(\mathbf{E}^{2,1}\) that are linearly independent. Requiring linear independence ensures that the triangles aren't degenerate (e.g. all points lying on the same line). In the next section, we describe how to interpret a generalized hyperbolic triangle as a polygon in the plane. We also call the polygons that can result generalized hyperbolic triangles.

Interpretation as polygons

Before describing how to interpret generalized hyperbolic triangles as polygons, we first describe how to interpret a single vector in \(\mathbf{E}^{2,1}\) in terms of the hyperbolic plane. See [Th, pp. 76-82] for a more detailed treatment of some of this material.

Thus, consider any non-zero vector \(v\in \mathbf{E}^{2,1}\). Then \(Q(v)\) is either positive, negative, or zero. The related quantity \(\sqrt{Q(v)}\) is known as the length, or norm, of \(v\) and is either real, imaginary, or zero, respectively. If \(Q(v)<0\) (imaginary length \(v\)), then the 1-dimensional subspace spanned by \(v\) intersects \(H\) in a single point, and \(v\) can be interpreted as that point in \(H\). If \(Q(v)>0\) (real length \(v\)), the 1-dimensional subspace spanned by \(v\) doesn't intersect \(H\), and we are in the case of \(v\) lying outside the plane. In this case, consider the 2-dimensional plane \(v^\perp\) in \(\mathbf{E}^{2,1}\) orthogonal to \(v\). The intersection of \(v^\perp\) with \(H\) forms a geodesic in \(H\). We will use this geodesic below when addressing triangles. Finally, if \(Q(v)=0\) (zero length \(v\)), then \(v\) corresponds to an ideal point, or point on the boundary of \(H\).

Consider now a triple \((v_1, v_2, v_3)\) of vectors in \(\mathbf{E}^{2,1}\). To interpret this triple as a polygon, apply the interpretation above for each \(v_i\) separately. Namely, if \(Q(v)\leq 0\), the vector corresponds to a point in \(H\) or on the boundary of \(H\). If \(Q(v)>0\), the vector corresponds to a geodesic in \(H\).

Finally, connect the resulting three points and/or geodesics to form a polygon, as follows. To connect a point with an adjacent point, use the unique geodesic segment connecting the two points. To connect a geodesic with an adjacent geodesic, if they don't already intersect, use the unique geodesic segment that intersects both geodesics perpendicularly. To connect a geodesic with an adjacent point, if they don't already intersect, use the unique geodesic segment from the point to the geodesic that intersects the geodesic perpendicularly.

This figure shows the construction of the polygon corresponding to a generalized hyperbolic triangle \((v_1, v_2, v_3)\) when \(v_1\) and \(v_2\) have imaginary length and \(v_3\) has real length (i.e. one vertex lies outside the hyperbolic plane). The resulting polygon is a quadrilateral with two right angles.

The figure above illustrates the interpretation described above for a generalized hyperbolic triangle \((v_1, v_2, v_3)\) with \(v_1\) and \(v_2\) of imaginary length and \(v_3\) of real length.

References

  • [Th] William P. Thurston, Three-dimensional Geometry and Topology, Vol. 1, edited by Silvio Levy, Princeton Mathematical Series, 35, 1997.