1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.