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Proving a version of the Ultraparallel theorem
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I have been asked to prove a version of the Ultraparallel theorem.
Let $l_1$ and $l_2$ be hyperbolic lines with normals $n_1$ and $n_2$. Show that there exists a unique hyperbolic line $l_3$ orthogonal to $l_1$ and $l_2$ if and only if $|langle,n_1,n_2rangle| gt1$.
My progress so far:
"$Rightarrow$" Assume that $l_3$ exists. Suppose that $|langle,n_1,n_2rangle| leq 1$. Then two cases can occur.
Case 1: $|langle,n_1,n_2rangle| lt 1$
Then $l_1$ and $l_2$ intersect. Since $l_3$ intersects the other lines orthogonally we get a hyperbolic triangle with two right angles. This is a contradiction since the angle sum of a hyperbolic triangle may never exceed $pi$.
Case 2: $|langle,n_1,n_2rangle|= 1$
In this case $l_1$ and $l_2$ do not intersect but are parallel. Hence we get a triangle with two right angles where one vertex is a point at infinity. I know that in the Beltrami-Klein-Model this construction corresponds to a triangle where one vertex lies on the unit circle. The problem is that I don't see how to construct a contradiction.
"$Leftarrow$"
Consider the Beltrami-Klein-Model.
Since $|langle,n_1,n_2rangle| gt1$ $l_1$ and $l_2$ intersect outside of the unit disk. Denote the point of intersection by A. Consider the two tangents to the unit circle passing through A. Denote the points where these tangents touch the circle by B and C. Now consider the line passing through B and C. This should be the perpendicular line. The problem here is that I do not see how to formally prove that this line corresponds to a line that intersects $l_1$ and $l_2$ orthogonally in the hyperboloid model. Essentially I need to show that $|langle,n_1,n_3rangle|= 0$ and $|langle,n_1,n_3rangle|= 0$ where $n_3$ is the normal of the constructed line in the hyperboloid model. But I do not see how to do this.
hyperbolic-geometry
asked 2 days ago
Polymorph
965
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up vote
1
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I have been asked to prove a version of the Ultraparallel theorem.
Let $l_1$ and $l_2$ be hyperbolic lines with normals $n_1$ and $n_2$. Show that there exists a unique hyperbolic line $l_3$ orthogonal to $l_1$ and $l_2$ if and only if $|langle,n_1,n_2rangle| gt1$.
My progress so far:
"$Rightarrow$" Assume that $l_3$ exists. Suppose that $|langle,n_1,n_2rangle| leq 1$. Then two cases can occur.
Case 1: $|langle,n_1,n_2rangle| lt 1$
Then $l_1$ and $l_2$ intersect. Since $l_3$ intersects the other lines orthogonally we get a hyperbolic triangle with two right angles. This is a contradiction since the angle sum of a hyperbolic triangle may never exceed $pi$.
Case 2: $|langle,n_1,n_2rangle|= 1$
In this case $l_1$ and $l_2$ do not intersect but are parallel. Hence we get a triangle with two right angles where one vertex is a point at infinity. I know that in the Beltrami-Klein-Model this construction corresponds to a triangle where one vertex lies on the unit circle. The problem is that I don't see how to construct a contradiction.
"$Leftarrow$"
Consider the Beltrami-Klein-Model.
Since $|langle,n_1,n_2rangle| gt1$ $l_1$ and $l_2$ intersect outside of the unit disk. Denote the point of intersection by A. Consider the two tangents to the unit circle passing through A. Denote the points where these tangents touch the circle by B and C. Now consider the line passing through B and C. This should be the perpendicular line. The problem here is that I do not see how to formally prove that this line corresponds to a line that intersects $l_1$ and $l_2$ orthogonally in the hyperboloid model. Essentially I need to show that $|langle,n_1,n_3rangle|= 0$ and $|langle,n_1,n_3rangle|= 0$ where $n_3$ is the normal of the constructed line in the hyperboloid model. But I do not see how to do this.
hyperbolic-geometry
asked 2 days ago
Polymorph
965
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have been asked to prove a version of the Ultraparallel theorem.
Let $l_1$ and $l_2$ be hyperbolic lines with normals $n_1$ and $n_2$. Show that there exists a unique hyperbolic line $l_3$ orthogonal to $l_1$ and $l_2$ if and only if $|langle,n_1,n_2rangle| gt1$.
My progress so far:
"$Rightarrow$" Assume that $l_3$ exists. Suppose that $|langle,n_1,n_2rangle| leq 1$. Then two cases can occur.
Case 1: $|langle,n_1,n_2rangle| lt 1$
Then $l_1$ and $l_2$ intersect. Since $l_3$ intersects the other lines orthogonally we get a hyperbolic triangle with two right angles. This is a contradiction since the angle sum of a hyperbolic triangle may never exceed $pi$.
Case 2: $|langle,n_1,n_2rangle|= 1$
In this case $l_1$ and $l_2$ do not intersect but are parallel. Hence we get a triangle with two right angles where one vertex is a point at infinity. I know that in the Beltrami-Klein-Model this construction corresponds to a triangle where one vertex lies on the unit circle. The problem is that I don't see how to construct a contradiction.
"$Leftarrow$"
Consider the Beltrami-Klein-Model.
Since $|langle,n_1,n_2rangle| gt1$ $l_1$ and $l_2$ intersect outside of the unit disk. Denote the point of intersection by A. Consider the two tangents to the unit circle passing through A. Denote the points where these tangents touch the circle by B and C. Now consider the line passing through B and C. This should be the perpendicular line. The problem here is that I do not see how to formally prove that this line corresponds to a line that intersects $l_1$ and $l_2$ orthogonally in the hyperboloid model. Essentially I need to show that $|langle,n_1,n_3rangle|= 0$ and $|langle,n_1,n_3rangle|= 0$ where $n_3$ is the normal of the constructed line in the hyperboloid model. But I do not see how to do this.
hyperbolic-geometry
asked 2 days ago
Polymorph
965
I have been asked to prove a version of the Ultraparallel theorem.
Let $l_1$ and $l_2$ be hyperbolic lines with normals $n_1$ and $n_2$. Show that there exists a unique hyperbolic line $l_3$ orthogonal to $l_1$ and $l_2$ if and only if $|langle,n_1,n_2rangle| gt1$.
My progress so far:
"$Rightarrow$" Assume that $l_3$ exists. Suppose that $|langle,n_1,n_2rangle| leq 1$. Then two cases can occur.
Case 1: $|langle,n_1,n_2rangle| lt 1$
Then $l_1$ and $l_2$ intersect. Since $l_3$ intersects the other lines orthogonally we get a hyperbolic triangle with two right angles. This is a contradiction since the angle sum of a hyperbolic triangle may never exceed $pi$.
Case 2: $|langle,n_1,n_2rangle|= 1$
In this case $l_1$ and $l_2$ do not intersect but are parallel. Hence we get a triangle with two right angles where one vertex is a point at infinity. I know that in the Beltrami-Klein-Model this construction corresponds to a triangle where one vertex lies on the unit circle. The problem is that I don't see how to construct a contradiction.
"$Leftarrow$"
Consider the Beltrami-Klein-Model.
Since $|langle,n_1,n_2rangle| gt1$ $l_1$ and $l_2$ intersect outside of the unit disk. Denote the point of intersection by A. Consider the two tangents to the unit circle passing through A. Denote the points where these tangents touch the circle by B and C. Now consider the line passing through B and C. This should be the perpendicular line. The problem here is that I do not see how to formally prove that this line corresponds to a line that intersects $l_1$ and $l_2$ orthogonally in the hyperboloid model. Essentially I need to show that $|langle,n_1,n_3rangle|= 0$ and $|langle,n_1,n_3rangle|= 0$ where $n_3$ is the normal of the constructed line in the hyperboloid model. But I do not see how to do this.
hyperbolic-geometry
hyperbolic-geometry
asked 2 days ago
Polymorph
965
asked 2 days ago
Polymorph
965
asked 2 days ago
Polymorph
965
asked 2 days ago
Polymorph
965
asked 2 days ago
Polymorph
965
965
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