Derivative of (Norm of) Möbius Addition w.r.t. Curvature

Question

Does anyone know a resource showing the formula for the derivative of the (norm) of the Möbius addition?

This is the Möbius addition: $$ x \oplus_cy=\frac{\overbrace{(1+2c\langle x,y\rangle+c||y||_2^2)}^{A(c)}x + \overbrace{(1-c||x||_2^2)}^{B(c)}y}{\underbrace{ 1+2c\langle x,y\rangle + c^2||x||^2_2||y||^2_2}_{D(c)}} $$

What is

$$ \frac{\partial}{\partial c}(x\oplus_c y) $$

or

$$ \frac{\partial}{\partial c}( ||x\oplus_c y||_2) $$ ?

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So far what I've found is:

$$ \frac{\partial}{\partial c } (x\oplus_c y) = \frac{A'(c)x+B'(c)y+D'(c)(x\oplus_c y)}{D(c)} $$

How can this expression be interpreted in gyrovector space notation?

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The expression $\sqrt{c}||(-x)\oplus_c y||_2$ represents the geodesic length between $x$ and $y$.

Now what would be its derivative w.r.t. $c$? The change in geodesic length between $x$ and $y$ depending on the curvature $c$, right?

Is that expressible in some way in the gyrovector notation?

Answer 1

Using a slightly modified version of your addition (but equivalent): $$x \oplus_K y = \frac{(1-2K\langle x, y\rangle - K||y||^2)x + (1+K||x||^2)y}{1-2K\langle x, y \rangle + K^2 ||x||^2 ||y||^2}$$

After doing symbolic differentiation using Mathematica: $$\frac{\partial}{\partial K} (x \oplus_K y) = \frac{ (2\langle x, y \rangle + ||x||^2 - 2K||x||^2||y||^2-K^2||x||^4||y||^2)}{(1-2K\langle x, y \rangle + K^2 ||x||^2 ||y||^2)^2}y + \frac{(-||y||^2-2K||x||^2||y||^2 + 2K^2 \langle x, y \rangle ||x||^2||y||^2 +K^2||x||^2||y||^4)}{(1-2K\langle x, y \rangle + K^2 ||x||^2 ||y||^2)^2}x.$$ Unfortunately, I don't really see a way to simplify it much. With some more notation, we can get $$\frac{\partial}{\partial K} (x \oplus_K y) = \frac{ (2\langle x, y \rangle + ||x||^2 - 2\alpha-K||x||^2\alpha)}{(1-2K\langle x, y \rangle + K\alpha)^2}y + \frac{(-||y||^2-2\alpha + 2K \langle x, y \rangle \alpha +K||y||^2\alpha)}{(1-2K\langle x, y \rangle + K\alpha)^2}x,$$ where $\alpha = K||x||^2||y||^2$, but that doesn't really help much.

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Original Mathematica code:

kplus[x_,y_,K_]:=((1-2K*Dot[x,y]-K*Norm[y]^2)*x+(1+K*Norm[x]^2)*y)/(1-2K*Dot[x,y]+K^2*Norm[x]^2*Norm[y]^2);
Expand[Simplify[D[kplus[x,y,K],K]]]

and the original result:

(-x Norm[y]^2 - K^2 y Norm[x]^4 Norm[y]^2 + 2 x.y (y + K^2 x Norm[x]^2 Norm[y]^2) + Norm[x]^2 (y - 2 K (x + y) Norm[y]^2 + K^2 x Norm[y]^4)) / (1 - 2 K x.y + K^2 Norm[x]^2 Norm[y]^2)^2