Example 2a

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On Poincaré upper half plane, we can give an assignment [math]\displaystyle{ A = -\frac{x}{y} }[/math] following the flow equation and right-hand rule from additive axis to multiplicative axis. [math]\displaystyle{ A }[/math] is also an eigenvector of Laplacian with eigenvalue of [math]\displaystyle{ 2 }[/math].

The assignment

The assignment is given as

[math]\displaystyle{ A = - \frac{x}{y} }[/math]

 

in below geometry settings.

Geometry settings

For the upper half plane

[math]\displaystyle{ \{\mathcal{H}: (x, y) | y \gt 0 \} }[/math]

 

equipped with an inner product

[math]\displaystyle{ \mathbf{a} \cdot \mathbf{b} = \begin{bmatrix} a_x & a_y \end{bmatrix} \begin{bmatrix} \frac{1}{y^2} & 0 \\ 0 & \frac{1}{y^2} \end{bmatrix} \begin{bmatrix} b_x \\ b_y \end{bmatrix} }[/math]

 

and the metrics

[math]\displaystyle{ ds^2 = \frac{dx^2 + dy^2}{y^2} }[/math]

 

We have the Gauss curvature of [math]\displaystyle{ -1 }[/math] and the Laplacian is [1]

[math]\displaystyle{ \Delta = - y^2 (\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}) }[/math]

 


Proof

The above [math]\displaystyle{ A }[/math] is an arithmetic expression space.

[math]\displaystyle{ da = d(-\frac{x}{y}) = \frac{xdy - ydx}{y^2} = -\frac{dx + ady}{y} }[/math]

 

and

[math]\displaystyle{ ds = \frac{\sqrt{dx^2 + dy^2}}{y} }[/math]

 

then

[math]\displaystyle{ \frac{da}{ds} = - \frac{dx + ady}{y} \frac{y}{\sqrt{dx^2 + dy^2}} = - \frac{dx + ady}{\sqrt{dx^2 + dy^2}} }[/math]

 

Considering the local coordinate is given by [math]\displaystyle{ (-1, 0) }[/math][math]\displaystyle{ (0, -1) }[/math] under the right-hand rule, we have

[math]\displaystyle{ \cos \theta = \frac{\begin{bmatrix} dx & dy \end{bmatrix} \begin{bmatrix} \frac{1}{y^2} & 0 \\ 0 & \frac{1}{y^2} \end{bmatrix} \begin{bmatrix} -1 \\ 0 \end{bmatrix}}{\sqrt{\begin{bmatrix} dx & dy \end{bmatrix} \begin{bmatrix} \frac{1}{y^2} & 0 \\ 0 & \frac{1}{y^2} \end{bmatrix} \begin{bmatrix} dx \\ dy \end{bmatrix}}\sqrt{\begin{bmatrix} -1 & 0 \end{bmatrix} \begin{bmatrix} \frac{1}{y^2} & 0 \\ 0 & \frac{1}{y^2} \end{bmatrix} \begin{bmatrix} -1 \\ 0 \end{bmatrix}}} }[/math]

 

hence

[math]\displaystyle{ \cos \theta = \frac{-dx}{\sqrt{dx^2 + dy^2}} }[/math]

 

and similarly

[math]\displaystyle{ \sin \theta = \frac{-dy}{\sqrt{dx^2 + dy^2}} }[/math]

 

then

[math]\displaystyle{ \frac{da}{ds} = \cos \theta + a \sin \theta }[/math]

 

As eigenvector of Laplacian

We can verify [math]\displaystyle{ A }[/math] is an eigenvalue of the Laplacian

[math]\displaystyle{ \Delta A = - y^2 (\frac{\partial^2}{\partial x^2} A + \frac{\partial^2}{\partial y^2} A) = y^2 (\frac{1}{\partial y} (\frac{1}{\partial y} \frac{x}{y})) = 2 A }[/math]

 

References

  1. Negro, Giusepp. "Laplacian on Poincaré upper half plane", Mathematics Stack Exchange. (2019).