# Equalities and inequalities for quadrilaterals in hyperbolic space

In euclidean space any quadrilateral satisfies equalities and inequalities

$$a^2 + b^2 + c^2 + d^2 = p^2 + q^2 + 4x^2$$ $$a^2 + b^2 + c^2 + d^2 \ge p^2 + q^2$$

where $a,b,c,d$ are the side lenghts, $p,q$ the lenghts of the diagonals and $x$ is the distance between the midpoints of the diagonals.

Are there any similar formulae in hyperbolic space?

The reason I am asking is the follwing: Any CAT($\kappa$) space X satisfies a 4-point condition which informally states that for any quadrilateral $Q$ in the space X, we can find a quadrilateral $\bar{Q}$ in the model space $M^2_\kappa$ with the same side lenghts and whose diagonals are equal or larger than the diagonals in $Q$.

In the case $\kappa = 0$ this falls back to euclidean geometry and we can use the well known formulas. But in the case that $\kappa = -1$, I fail to see a convenient way to check the conditions.

We observe that the quadrilateral result is actually one about tetrahedra. A Euclidean quadrilateral with diagonals $p$ and $q$ is the projection of a tetrahedron with opposite edges $p$ and $q$ into a plane parallel to those edges. If $h$ is the corresponding perpendicular distance from edge $p$ to edge $q$, then the quadrilateral edges $a$, $b$, $c$, $d$ are projections of edges of length $a^\prime$, $b^\prime$, $c^\prime$, $d^\prime$ such that $$(s^\prime)^2 = s^2 + h^2 \qquad \text{for}\quad s = a, b, c, d$$ Moreover, the midpoint segment of length $x$ in the quadrilateral is the projection of the segment of length $x^\prime$ joining the midpoints of tetrahedral edges $p$ and $q$, with $(x^\prime)^2 = x^2 + h^2$. In the OP's quadrilateral relation, we see that we can add $h^2$ terms to obtain

$$(a^\prime)^2 + (b^\prime)^2 + (c^\prime)^2 + (d^\prime)^2 \;=\; p^2 + q^2 + 4 (x^\prime)^2$$

We'll show this hyperbolic counterpart of the tetrahedral result (suppressing "$\prime$"s in the tetrahedral edge lengths).

For a hyperbolic tetrahedron with opposite edges of length $p$ and $q$, remaining edges of length $a$, $b$, $c$, $d$, and distance $x$ between the midpoints of segments $p$ and $q$, \begin{align} \cosh a + \cosh b + \cosh c + \cosh d &\;=\; 4 \cosh\frac{p}{2}\;\cosh\frac{q}{2}\;\cosh x \tag{\star} \\[12pt] \cosh a + \cosh b + \cosh c + \cosh d &\;\geq\; 4 \cosh\frac{p}{2}\;\cosh\frac{q}{2} \tag{\star\star} \end{align}

Getting here is a bit of a symbolic slog, which I'll outline below. Note, however, that $(\star)$ is consistent with the Euclidean equality when applied to "infinitesimal" hyperbolic quadrilaterals. We see this by expanding the cosines as power series and ignoring terms of degree greater than 2:

$$\sum_{s=a,b,c,d}\left( 1 + \frac{s^2}{2!} + \cdots \right) \;=\; 4 \left( 1+\frac{(p/2)^2}{2!} +\cdots \right)\left(1+\frac{(q/2)^2}{2!}+\cdots\right)\left(1 + \frac{x^2}{2!}+\cdots\right)$$

\begin{align} \implies\qquad 4 + \frac{a^2}{2} + \frac{b^2}{2} + \frac{c^2}{2} + \frac{d^2}{2} &\;\;\approx\;\; 4 \left( 1+\frac{p^2}{8}+\frac{q^2}{8}+\frac{x^2}{2} \right) \\[10pt] \implies\qquad a^2 + b^2 + c^2 + d^2 &\;\;\approx\;\; p^2 + q^2 + 4x^2 \end{align}

Proof of $(\star)$. It's straightforward, if tedious, to proceed via coordinatized hyperbolic space, where a point $P$ has coordinates $(x_P, y_P, z_P)$ such that

• $z_P$ is the signed length of the perpendicular from $P$ to point $P_{xy}$ in the $xy$-plane
• $y_P$ is the signed length of the perpendicular from $P_{xy}$ to point $P_x$ on the $x$-axis
• $x_P$ is the signed distance from $P_x$ to the origin.

We'll need this:

The Distance Formula. The distance from $A(x_a,y_a,z_a)$ to $B(x_b,y_b,z_b)$ is given by $$\cosh|AB| \;\;=\;\; \ddot{z_a}\ddot{z_b}\;\left(\; \ddot{y_a}\ddot{y_b} \;\left(\; \ddot{x_a} \ddot{x_b} - \overline{x_a}\,\overline{x_b}\;\right) - \overline{y_a}\,\overline{y_b} \;\right) - \overline{z_a}\,\overline{z_b}$$

where $\ddot{x} := \cosh x$ and $\overline{x} := \sinh x$ (and, later, $\ddot\theta := \cos \theta$ and $\overline{\theta} := \sin\theta$).

A fun fact about hyperbolic tetrahedra is that, for each pair of opposite edges, there exists a line perpendicular to the lines containing those edges. Let our tetrahedron have opposite edges $P_1P_2$ and $Q_1Q_2$, with $P_0$ and $Q_0$ the points where the mutual perpendicular meets these (extended) edges. Situating our tetrahedron such that $P_0Q_0$ (of length $h$) aligns with the $x$-axis, edge $P_1 P_2$ aligns with the $y$-axis, and opposite edge $Q_1 Q_2$ makes angle $\theta$ with the $xy$-plane, we have these coordinates: \begin{align} P_0 &:= (0,0,0) \qquad \,P_1 := (0,p_1,\,0\;) \qquad \,P_2 := (0,-p_2,\phantom{-}0\,) \qquad \text{where}\quad p_i := |P_0P_i| \\ Q_0 &:= (h,0,0) \qquad Q_1 := (h,y_1,z_1) \qquad Q_2 := (h,-y_2,-z_2) \qquad\text{where}\quad q_i := |Q_0 Q_i|\end{align} $$\text{and}\quad y_i := \operatorname{atanh}(\;\tanh q_i \cos\theta\;)\qquad z_i = \operatorname{asinh}(\;\sinh q_i\sin\theta\;)$$

With these coordinates, the Distance Formula provides these expressions for the lengths of the tetrahedron's edges: $$\ddot{a} := |P_1Q_1| = \ddot{p_1}\ddot{q_1}\ddot{h}-\overline{p_1}\,\overline{q_1}\ddot{\theta} \qquad\qquad \ddot{b} := |Q_1 P_2| = \ddot{p_2}\ddot{q_1}\ddot{h} + \overline{p_2}\,\overline{q_1}\ddot{\theta}$$ $$\ddot{c} := |P_2Q_2| = \ddot{p_2}\ddot{q_2}\ddot{h}-\overline{p_2}\,\overline{q_2}\ddot{\theta} \qquad\qquad \ddot{d} := |Q_2 P_1| = \ddot{p_1}\ddot{q_2}\ddot{h} + \overline{p_1}\,\overline{q_2}\ddot{\theta}$$ so that, defining $m := (p_1-p_2)/2$ and $n := (q_1-q_2)/2$, \begin{align} \ddot{a}+\ddot{b}+\ddot{c}+\ddot{d} &= (\ddot{p_1}+\ddot{p_2})(\ddot{q_1}+\ddot{q_2})\ddot{h}-(\overline{p_1}-\overline{p_2})(\overline{q_1}-\overline{q_2})\cos\theta \\[6pt] &= 4\,\cosh\frac{p}{2}\;\cosh\frac{q}{2}\;\left(\ddot{m} \ddot{n}\;\ddot{h}-\overline{m}\,\overline{n}\;\ddot{\theta}\right) \end{align}

It remains, then, to show that $$\cosh|MN| = \ddot{m}\ddot{n}\;\ddot{h}-\overline{m}\,\overline{n}\;\ddot{\theta}$$ where $M$ and $N$ are respective midpoints of $P_1P_2$ and $Q_1Q_2$. We can verify this relation via the Distance Formula applied to points with coordinates $$M = \left(\;0, m, 0\;\right) \qquad N = \left(\;h, y_N, z_N\;\right)$$ with $$y_N := \operatorname{atanh}\left(\;\tanh n \,\cos\theta\;\right)\qquad z_N = \operatorname{asinh}\left(\;\sinh n \,\sin\theta\;\right)$$ thusly: \begin{align} \cosh|MN| &= 1\cdot\ddot{z_N}\;\left(\; \ddot{m}\ddot{y_N} \;\left(\; 1\cdot \ddot{h} - 0\cdot\overline{h}\;\right) - \overline{m}\,\overline{y_N} \;\right) - 0\cdot\overline{z_M} \\ &= \ddot{z_N}\;\left(\; \ddot{m}\ddot{y_N}\ddot{h} - \overline{m}\,\overline{y_N} \;\right)\\ &= \ddot{z_N}\ddot{y_N}\;\left(\; \ddot{m}\ddot{h} - \overline{m}\,\tanh y_N \;\right) \\ &= \ddot{n} \;\left(\;\ddot{m}\ddot{h} - \overline{m} \tanh n \ddot{\theta} \;\right) \\ &= \ddot{m}\ddot{n}\ddot{h} - \overline{m}\,\overline{n} \ddot{\theta} \quad\square \end{align}

Caveat. I transcribed various formulas from the Coordinates section of my note "Hedronometric Formulas for Hyperbolic Tetrahedra". In doing so, I changed various variable names and made different signing choices, which may have led to typographical errors in the above.