2.1 Hyperbolic geometry

A background reference for complex hyperbolic geometry is [Reference GoldmanGo]. We will use the ball model of complex hyperbolic space $\mathbb {CH}^{n}.$ In that model, the manifold for $\mathbb {CH}^{n}$ is the unit ball, $\mathbb {B} _{n}\subset \mathbb {C}^{n},$ and the geometry is the Poincare Bergman geometry determined by the transitive group of biholomorphic automorphisms of the ball, $\operatorname *{Aut}\mathbb {B}_{n}$ . For each $\alpha \in \mathbb {B}_{n}$ , there is a unique involution $\phi _{a}\in \operatorname *{Aut}\mathbb {B}_{n}$ , which satisfies $\phi _{a}(a)=0.$ The group $\operatorname *{Aut}\mathbb {B}_{n}$ is generated by those involutions together with the unitary maps. We will say two sets Z, $W\subset \mathbb {CH}^{n}$ are congruent, $Z\sim W,$ if there is a $\phi \in \operatorname *{Aut}\mathbb {B}_{n}$ with $\phi (Z)=W.$ If the sets are ordered then, absent other comment, we suppose that $\phi $ respects the ordering. Congruence is an equivalence relation, and we are particularly interested in congruence equivalence classes.

The pseudohyperbolic metric $\delta $ on $\mathbb {CH}^{n}$ is defined by $\forall \alpha ,\beta \in \mathbb {CH}^{n}$

(2.1) $$ \begin{align} \delta(\alpha,\beta)=\left\vert \phi_{\alpha}(\beta)\right\vert =\left\vert \phi_{\beta}(\alpha)\right\vert. \end{align} $$

Equivalently, $\delta $ is the distance on $\mathbb {B}_{n}$ , which satisfies $\delta (0,z)=\left \vert z\right \vert $ for $z\in \mathbb {B}_{n}$ and is $\operatorname *{Aut}\mathbb {B}_{n}$ invariant [Reference Duren and WeirDW]. $\delta $ is not a length metric; the length metric it generates is the Poincare Bergman metric for the ball normalized to agree with the Euclidean metric to second order at the origin. For $n=1$ , the formula is $\delta (\alpha ,\beta )=\left \vert \frac {\alpha -\beta }{1-\bar {\beta }\alpha }\right \vert .$ The formula for general n is given in (2.9). $\delta $ satisfies a strengthened version of the triangle inequality, [Reference Duren and WeirDW]; for $x,y,z\in \mathbb {B}_{n}$ ,

(2.2) $$ \begin{align} \frac{\left\vert \delta(x,z)-\delta(z,y)\right\vert }{1-\delta(x,z)\delta (z,y)}\leq\delta(x,y)\leq\frac{\delta(x,z)+\delta(z,y)}{1+\delta (x,z)\delta(z,y)}. \end{align} $$

The space $\mathbb {CH}^{n}$ contains various totally geodesic submanifolds of interest here. These include the classical geodesics which are totally geodesic copies of $\mathbb {RH}^{1},$ and also include totally geodesic copies of $\mathbb {CH}^{1},$ sometimes called complex geodesics. Every pair of points is contained in a unique classical geodesic which is in turn contained in a unique complex geodesic. There are also totally geodesic copies of the real hyperbolic plane $\mathbb {RH}^{2}$ inside $\mathbb {CH} ^{n}.$ In particular, for ${\mathbb {R}^{2}=\left \{ \left ( x,y,0,...,0\right ) :x,y\in \mathbb {R}\right \} \subset \mathbb {C}^{n}}$ , consider the two-dimensional real ball, $RB_{2}=\mathbb {R}^{2}\cap \mathbb {CH}^{n}.$ That intersection is a totally geodesically embedded submanifold whose induced geometry is that of the Beltrami–Klein model of $\mathbb {RH}^{2}$ with constant curvature $-1/4.$ All of the automorphic images of $RB_{2}$ are also totally geodesically embedded submanifolds of real dimension 2, and they, together with the complex geodesics, are the only such.

The geometry of $RB_{2}$ is not conformal with the Euclidean geometry of the containing $\mathbb {R}^{2}.$ However, the two geometries are conformally equivalent at the origin, in particular angles with vertex at the origin have the same size in both geometries. There is a useful picture of $RB_{2}$ in [Reference GoldmanGo, p. 83], and there is a discussion of the geometry of that model (although the version with curvature $-1)$ as well as the more familiar Poincaré disk model in Appendices B and C of [Reference JansonJ].

If G is a complex geodesic and $x\in \mathbb {CH}^{n}$ , we define the metric projection of x onto G, $P_{G}x,$ to be that point in G which is closest to x in the pseudohyperbolic metric.

For $z,w\in \mathbb {B}_{r}$ , we define the kernel function k by

(2.3) $$ \begin{align} k_{z}(w)=k(w,z)=\frac{1}{1-\left\langle \left\langle w,z\right\rangle \right\rangle }. \end{align} $$

(We write $\left \langle \left \langle \cdot ,\cdot \right \rangle \right \rangle $ for the inner product on $\mathbb {C}^{n}$ to distinguish from the inner products on the general Hilbert spaces we consider.)

There is a fundamental identity which describes the interaction of the involutive automorphisms with the kernel functions [Reference RudinRu]: for any $y,z,w\in \mathbb {B}_{r}$ ,

(2.4) $$ \begin{align} k(\phi_{y}(z),\phi_{y}(w))=\frac{k\left( y,y\right) ^{1/2}}{k(z,y)} \frac{k\left( y,y\right) ^{1/2}}{k(y,w)}k(z,w). \end{align} $$

By a triangle in $\mathbb {CH}^{n}$ , we mean an ordered set of three distinct points called vertices, or those vertices together with the sides, the geodesic segments connecting the vertices. We allow the degenerate case of three points in a single geodesic. The length of a side is the $\delta $ distance between the corresponding vertices. Triangles in $\mathbb {CH}^{n}$ do not necessarily have “faces”; three points in $\mathbb {CH}^{n}$ are not generally contained in a totally geodesic submanifold of real dimension 2.

Similarly, a tetrahedron in $\mathbb {CH}^{n}$ is an ordered set of four vertices. Its four subsets of three vertices are its triangular faces, even though they may not be faces in the geometric sense.

2.2 Hilbert spaces with reproducing kernels

Our background references for Hilbert spaces are [Reference Agler and PickAM, Reference Arcozzi, Rochberg, Sawyer and WickARSW2, Reference QuigginPR, Reference RochbergRo].

Except for the spaces $DA_{r}$ discussed this section, all the Hilbert spaces in this paper are finite-dimensional.

An n-dimensional RKHS is an n-dimensional Hilbert space H together with a distinguished basis of vectors called reproducing kernels, $RK(H)=\left \{ h_{i}\right \} _{i=1}^{n}.$ For any $v\in H$ , we write $\hat {v}$ for its normalized version; $\hat {v}=v/\left \Vert v\right \Vert .$ For $h_{i},h_{j}\in RK(H)$ , we write $h_{ij}$ for their inner product and $\widehat {h_{ij}}$ for the inner product of their normalizations:

$$\begin{align*}h_{ij}=\left\langle h_{i},h_{j}\right\rangle ,\ \ \ \widehat{h_{ij} }=\left\langle \widehat{h_{i}},\widehat{h_{j}}\right\rangle =\left\langle \frac{h_{i}}{\left\Vert h_{i}\right\Vert },\frac{h_{j}}{\left\Vert h_{j}\right\Vert }\right\rangle. \end{align*}$$

The Gram matrix of H is the matrix $\operatorname *{Gr}(H)=\left ( h_{ij}\right ) _{i,j=1}^{n}.$

Any v in H is regarded as a function on the index set of $RK(H)$ by setting ${v(i)=\left \langle v,h_{i}\right \rangle} $ . If m is a function on $RK(H)$ , then the associated multiplier operator $M_{m}$ acting on H is the linear operator which satisfies $M_{m}v(i)=m(i)v(i).$

A regular subspace of H is a subspace J spanned by a subset S of $RK(H)$ and regarded as an RKHS by setting $RK(J)=S.$

We now recall the Drury–Arveson spaces, $DA_{r}$ ; some basic references are [Reference Agler and PickAM, Reference ArvesonAr, Reference Shalit and AlpaySh]. With $k_{z}(\cdot )$ the functions from (2.3), $DA_{r}$ is the infinite-dimensional RKHS with kernel functions $\left \{ k_{z}:z\in \mathbb {B}_{r}=\mathbb {CH}^{r}\right \} $ and inner product given by $\left \langle k_{s},k_{t}\right \rangle =k_{s}(t).$ $DA_{1}$ is the classical Hardy space $H^{2}$ of the unit circle, a space which can also be described as the space of holomorphic functions on the unit disk with square summable sequences of Taylor coefficients. For $r>1$ , $DA_{r}$ is a proper subspace of the Hardy space of the sphere. It is in the scale of Besov–Sobolev spaces, which also contains the Hardy space of the sphere and the Bergman space of the ball [Reference Arcozzi, Rochberg, Sawyer and WickARSW2].

We are particularly interested in finite-dimensional regular subspaces of $DA_{r}$ . For $Z=\left \{ z_{j}\right \} _{j=1}^{n}\subset \mathbb {CH}^{r}$ , let $DA_{r}(Z)$ be the regular subspace of $DA_{r}$ spanned by the kernel functions $\left \{ k_{z_{i}}\right \} _{i=1}^{n}.$ We abbreviate them by $\left \{ k_{i}\right \} $ and set $k_{ij}=\left \langle k_{i},k_{j} \right \rangle .$

If $r^{\prime }>r$ , there are natural inclusions of $DA_{r}$ , $\mathbb {C}^{r}$ , $\mathbb {B}_{r}$ , and $\mathbb {CH}^{r}$ into the corresponding objects with $r^{\prime }.$ These inclusions interact in harmless ways with the constructions we are using. For instance, an inclusion of $\mathbb {B}_{r}$ into $\mathbb {B}_{r^{\prime }}$ takes $Z\subset \mathbb {B}_{r}$ to a $Z^{\prime }\subset \mathbb {B}_{r^{\prime }}.$ There is then an obvious natural map of $DA_{r}(Z)$ onto $DA_{r^{\prime }}(Z^{\prime })$ which preserves all the structure of interest here. Going forward, we will suppose r is sufficiently large, identify such pairs of sets and spaces, and drop the subscript $r;$ thus, $DA(Z).$

2.2.1 Rescaling

Rescaling is a fundamental equivalence relation between RKHSs. Given a finite-dimensional RKHS, $G,H$ with $RK(H)=\left \{ h_{\alpha }\right \} _{\alpha \in A}$ , $RK(G)=\left \{ g_{\beta }\right \} _{\beta \in B}$ , we say G is a rescaling of H and write $G\sim H$ if there is a one-to-one map $\theta $ of A onto B and a nonvanishing complex valued function $\gamma $ defined on A such that for all $\alpha _{1},\alpha _{2}$ in $A,$

(2.5) $$ \begin{align} \left\langle h_{\alpha_{1}},h_{\alpha_{2}}\right\rangle =\left\langle \gamma(\alpha_{1})g_{\theta(\alpha_{1})},\gamma(\alpha_{2})g_{\theta (\alpha_{2})}\right\rangle =\gamma(\alpha_{1})\overline{\gamma(\alpha_{2} )}\left\langle g_{\theta(\alpha_{1})},g_{\theta(\alpha_{2})}\right\rangle. \end{align} $$

For instance, the linear map which sends each kernel function $h_{\alpha }$ to its normalized version, $\widehat {h_{\alpha }}=h_{\alpha }/\left \Vert h_{\alpha }\right \Vert $ is a rescaling. If A and B are ordered, then, unless we specify otherwise, we suppose that $\theta $ respects the ordering.

A basic class of rescalings in our discussion are those induced on spaces $DA(Z)$ by ball automorphisms. It is a consequence of (2.4) that given any $Z\subset \mathbb {B}_{r}$ and any $y\in \mathbb {B}$ , the automorphism $\phi _{y}$ induces a rescaling of $DA(Z)$ to $DA(\phi _{y}(Z)).$

If $G_{H}$ is a regular subspace of G and $H\sim G_{H}$ , we will write $H\leadsto G$ or $H\leadsto G_{H}\subset G$ and will say that H has a rescaling into $G.$

2.3 Invariants

Given $H\in \mathcal {RK}$ , we now define several numerical functions of tuples of elements of $RK(H)$ the set of reproducing kernels of $H.$ The definitions involve algebraic combinations of Gram matrix entries which by (2.5) are seen to be unchanged when H is replaced by a rescaling of $H.$ Thus, the functions are rescaling invariants, well defined on the rescaling equivalence classes.

These functionals can also be regarded as defined for tuples of points in $\mathbb {CH}^{n}$ as follows. Given a functional F defined on, for instance, pairs of kernel functions, and given $x,y\in \mathbb {CH}^{n}$ select by $X\subset \mathbb {CH}^{n}$ with $x,y\in X$ and set $F(x,y)=F(k_{x},k_{y})$ where $k_{x}$ and $k_{y}$ are the kernel functions in $RK(DA(X)).$ It is straightforward to check that this value is independent of the choice of X (and we will not mention X again) and that if F acting on Hilbert spaces is rescaling invariant, then the “same” functional acting on $\mathbb {CH}^{n}$ is automorphism invariant. We will use the same names and notation for the functionals acting on an $RK(H)$ and for the induced functionals acting on points in $\mathbb {CH}^{n}.$

2.3.1 Distance

For $H\in \mathcal {RK}$ and $h_{1}$ , $h_{2}\in RK(H)$ , we define $\delta _{H}$ by

(2.6) $$ \begin{align} \delta_{H}(i,j)=\delta_{H}(h_{i},h_{j})=\sqrt{1-\frac{\left\vert h_{ij}\right\vert ^{2}}{h_{ii}h_{jj}}}=\sqrt{1-|\widehat{h_{ij}}|^{2}}. \end{align} $$

This function is a metric on H ([Reference Agler and PickAM, Lemma 9.9], [Reference Arcozzi, Rochberg, Sawyer and WickARSW]) and is clearly invariant under rescaling. Using the scheme of the previous paragraphs, we can extend this definition to a functional, call it $\delta ^{\prime }$ for the moment, acting on pairs of points in $\mathbb {CH}^{n}.$ One of the main links between the Hilbert space theory and hyperbolic geometry is that $\delta ^{\prime }$ equals the pseudohyperbolic distance, $\delta ,$ between points, already defined in (2.1). To see this, use (2.3) and rewrite (2.4) as, for $w,x,y,z\in \mathbb {CH}^{n},$

(2.7) $$ \begin{align} 1-\frac{k(w,y)k(y,z)}{k(y,y)k(w,z)}=\left\langle \left\langle \phi_{y} (z),\phi_{y}(w)\right\rangle \right\rangle. \end{align} $$

Taking $z=w$ , we have

(2.8) $$ \begin{align} \delta^{2}(y,w)=\left\vert \phi_{y}(w)\right\vert ^{2}=1-\frac{\left\vert k(y,w)\right\vert ^{2}}{k(y,y)k(w,w)}=\delta^{\prime2}(y,w). \end{align} $$

The first equality in (2.8) is the definition of $\delta $ , the second is a special case of (2.7), and the third is the definition of $\delta ^{\prime 2}.$ Thus, $\delta =\delta ^{\prime }.$ Alternatively, note that for any $z\in \mathbb {B}_{n}$ , we have $\delta (0,z)=\left \vert z\right \vert =\delta ^{\prime }(0,z)$ and both $\delta $ and $\acute {\delta }^{\prime }$ are invariant under automorphisms; hence, the two metrics are equal. We now drop the notation $\delta ^{\prime }$ and simply write $\delta $ for the functional defined on RKHS and for the pseudohyperbolic metric. Using (2.7), we can express $\delta $ in terms of coordinates; for $y,w\in \mathbb {B}_{n}$ ,

(2.9) $$ \begin{align} \delta^{2}(y,w)=1-\frac{(1-\left\langle \left\langle y,y\right\rangle \right\rangle )(1-\left\langle \left\langle w,w\right\rangle \right\rangle )}{\left\vert 1-\left\langle \left\langle y,w\right\rangle \right\rangle \right\vert ^{2}}. \end{align} $$

2.3.2 Angular invariant

For $H\in \mathcal {RK}$ , we define the angular invariant $\alpha $ by, for $k_{1}$ , $k_{2}$ , $k_{3}\in RK(H),$

(2.10) $$ \begin{align} \alpha(1,2,3)=\alpha(k_{1},k_{2},k_{3})=-\arg\left\langle k_{1},k_{2} \right\rangle \left\langle k_{2},k_{3}\right\rangle \left\langle k_{3} ,k_{1}\right\rangle =-\arg k_{12}k_{23}k_{31}. \end{align} $$

(In general situations, care is needed in selecting a branch of $\arg .$ However, for $k\in RK(DA(Z))$ , $\operatorname {Re}k>0$ and that lets us avoid problems.) Notice that $\alpha $ satisfies a cocycle identity; if $k_{4}$ is a fourth kernel function, then

(2.11) $$ \begin{align} \alpha(1,2,3)-\alpha(2,3,4)+\alpha(3,4,1)-\alpha(4,1,2)=0. \end{align} $$

We discuss some of the geometry associated with $\alpha $ in Section 4.4. More about this invariant is in [Reference Burger and IzzoBIM, Reference ClercC, Reference Clerc and ØrstedCO, Reference GoldmanGo, Reference Hangan and MasalaHM, Reference MonodM, Reference RochbergRo].

2.3.3 $\mathbf {kos}$

For $H\in \mathcal {RK}$ , we define $\operatorname *{kos}$ , a functional of triples of kernel functions. For $k_{1}$ , $k_{2}$ , $k_{3}\in RK(H)$ , $k_{1}\neq k_{2}$ , $k_{3},$ set

(2.12) $$ \begin{align} \operatorname*{kos}\nolimits_{k_{1}}(k_{2},k_{3})=\operatorname*{kos} \nolimits_{1}(2,3)=\frac{1}{\delta_{12}\delta_{13}}\left( 1-\frac {k_{21}k_{13}}{k_{11}k_{23}}\right) , \end{align} $$

and note the symmetry $\operatorname *{kos}\nolimits _{1}(2,3)=\overline {\operatorname *{kos}\nolimits _{1}(3,2)}$ .

If $H=DA(X)$ for some $X\subset \mathbb {CH}^{n}$ , then $\operatorname *{kos}$ is related to the geometry of $X.$ Recall that if $x\in \mathbb {CH}^{n}$ , then $\phi _{x}$ is the ball involution which interchanges x and $0.$ We can use (2.7) and obtain

(2.13) $$ \begin{align} \operatorname*{kos}\nolimits_{x}(y,z)=\frac{1}{\delta(x,y)\delta (x,z)}\left\langle \left\langle \varphi_{x}\left( y\right) ,\varphi _{x}\left( z\right) \right\rangle \right\rangle. \end{align} $$

In particular, if $\xi $ is at the origin, $\xi =\mathit {o}$ , then $\phi _{\xi }$ is the identity, for any $w \ \phi _{\xi }(w)=w,$ and $\delta (\xi ,w)=\delta (o,w)=\left \vert w\right \vert .$ In that case,

(2.14) $$ \begin{align} \operatorname*{kos}\nolimits_{\mathit{o}}(y,z)=\left\langle \left\langle \frac{y}{\left\vert y\right\vert },\frac{z}{\left\vert z\right\vert }\right\rangle \right\rangle =\left\langle \left\langle \widehat{y} ,\widehat{z}\right\rangle \right\rangle. \end{align} $$

(Although $\operatorname *{kos}$ is invariant under automorphisms of $\mathbb {B}_{n}$ , this formula is an inhomogenous representation and is not invariant.)

If the vectors y and z were in $\mathbb {R}^{n}$ , this would be the inner product of unit vectors in $\mathbb {R}^{n}$ and would equal the cosine of the angle between the segments $\mathit {o}y$ and $\mathit {o}z$ . That is the source of the name $\operatorname *{kos}.$

If $\dim H=n$ , then for $1\leq s\leq n$ , we define the $(n-1)\times (n-1)$ matrices

(2.15) $$ \begin{align} \!\!\!\operatorname*{KOS}(H,s) & =\operatorname*{KOS}(\operatorname*{Gr} (H),s)=\left( \operatorname*{kos}\nolimits_{s}(i,j)\right) _{\substack{i,j=1\\i,j\neq s}}^{n},\qquad\ \ \end{align} $$

(2.16) $$ \begin{align} \operatorname*{MQ}(H,s) & =\left( \delta_{si}\delta_{sj}\operatorname*{kos} \nolimits_{s}(i,j)\right) _{\substack{i,j=1\\i,j\neq s}}^{n}=\left( 1-\frac{k_{is}k_{sj}}{k_{ij}k_{ss}}\right) _{\substack{i,j=1\\i,j\neq s}}^{n}. \end{align} $$

We also write $\operatorname *{KOS}(X,s)$ for $\operatorname *{KOS}(DA(X),s).$

2.4 Matrix notation

For an $n\times n$ matrix A, we write $A\succ 0$ if A is positive definite and $A\succcurlyeq 0$ if it is positive semidefinite. We say B is a principal submatrix of A if it is obtained from A by removing certain rows and also the corresponding columns. Note that if $H\in $ $\mathcal {RK}$ , then J is a regular subspace of H if and only if $\operatorname *{Gr}(J)$ is a principal submatrix of $\operatorname *{Gr}(H)$ . We denote the set of all principal submatrices of A by $\mathcal {PS} {\large (A).}$ If $B\in \mathcal {PS}{\large (A)}$ and the indices of the rows and columns retained in B are an initial segment, indices j with $1\leq j\leq k$ for some $k\leq n,$ then B is said to be a leading principal submatrix. The determinants of those matrices are called principal minors and leading principal minors, respectively.

3.1 Assembly and coherence

We might know that each of several spaces $\left \{ J_{i}\right \} $ is equivalent under rescaling to a subspace $H_{i}$ of some $H; \ J_{i}\leadsto H_{i}\subset H.$ In that case, there are coherence conditions connecting the $\left \{ J_{i}\right \} $ ; if $H_{i}\cap H_{j}=H_{ij}$ , then the corresponding subspaces, $J_{i(j)}\subset J_{i}$ which corresponds to $H_{ij}$ and $J_{j(i)} \ \subset \ J_{j}$ which also corresponds to $H_{ij}$ , must be rescalings of each other. This condition is nontrivial if $\dim H_{ij}>1.$ If these conditions are met, then the $\left \{ J_{i}\right \} $ are said to be a coherent set of spaces and we will write $\left \{ J_{i}\right \} \rightrightarrows \,H$ .

Here is how this works in the context of Question 2. For a four-dimensional $H\in \mathcal {CPP}$ with $RK(H)=\left \{ h_{i}\right \} _{i=1}^{4}$ , denote the four regular three-dimensional subspaces $\left \{ H_{i}\right \} _{i=1}^{4}$ by

$$\begin{align*}RK(H_{i})=\left\{ h_{r}:1\leq r\leq4,\text{ }r\neq i\right\}. \end{align*}$$

Question 2 asks for necessary and sufficient conditions on four three-dimensional spaces $\left \{ J_{i}\right \} _{i=1}^{4}\subset \mathcal {RK}$ to insure that there is an $H\in \mathcal {CPP}$ with for $1\leq i\leq 4,$ $H_{i}\sim J_{i}.$ If there are such rescalings, then we suppose the indices on the reproducing kernels $j_{ir}$ of $J_{i}$ have been chosen so that for each $i,r$ , the rescaling of $H_{i}$ and $J_{i}$ matches the kernel function $h_{r}$ of $H_{i}$ with the kernel function $j_{ir}$ of $J_{i}.$

If $\left \{ J_{i}\right \} $ is a coherent set of spaces, then the space $J_{1},$ with kernel functions $\left \{ j_{12},j_{13},j_{14}\right \} ,$ is a rescaling of $H_{1}$ which has kernel functions $\left \{ h_{2},h_{3} ,h_{4}\right \} $ with $j_{1s}$ pairing with $h_{s};$ similarly for $J_{3}$ with kernels $\left \{ j_{31},j_{32},j_{34}\right \} $ and $H_{3}$ with kernels $\left \{ h_{1},h_{2},h_{4}\right \} .$ Thus, subspace $J_{1,24}$ of $J_{1}$ with kernel functions $\left \{ j_{12},j_{14}\right \} $ and the subspace $J_{3,24}$ of $J_{3}$ with kernel functions $\left \{ j_{32} ,j_{34}\right \} $ are both rescalings of the subspace of $H_{24}$ with kernel functions $\left \{ h_{2},h_{4}\right \} ,$ and hence $J_{1,24}\sim J_{3,24}.$ The other coherence conditions are all of this form; given $p<q,r<s$ distinct elements of $\left \{ 1,2,3,4\right \} $ , we must have $J_{p,rs}\sim J_{q,rs}.$ These conditions are necessary for there to be rescalings of $\left \{ J_{i}\right \} $ into $H.$ These conditions are the Hilbert space analogs of the matching side length conditions necessary to assemble triangles into a tetrahedron.

Notice that the conditions we obtained just now, $J_{p,rs}\sim J_{q,rs},$ did not involve the space H and in fact we can have a coherent set of spaces $\left \{ J_{i}\right \} $ without having an $H.$ It suffices to have a consistent set of coherence conditions, statements that certain subspaces of the various $\left \{ J_{i}\right \} $ are rescalings of each other. The consistency requirement is that the set of specified rescalings must be compatible with the fact that rescaling is an equivalence relation which interacts coherently with passage to subspaces. That is automatic if $\left \{ J_{i}\right \} \rightrightarrows \,H$ but is also possible otherwise. In that case, we still call the $\left \{ J_{i}\right \} $ a coherent set of spaces and we will write $\left \{ J_{i}\right \} \rightrightarrows \,??.$

Thus, the statement $\left \{ J_{i}\right \} \rightrightarrows \,??$ is the statement that there is a description (perhaps only implicit) of a type of target space H and an assembly scheme for constructing such an $H.$ The statement $\left \{ J_{i}\right \} \rightrightarrows H$ is that there is such an $H.$

Given $\left \{ J_{i}\right \} \rightrightarrows H$ , we would like to obtain information about H from the spaces $\left \{ J_{i}\right \} $ together with the coherence data $\left \{ J_{i}\right \} \rightrightarrows \,??.$ We know from Theorem 3.2 that the values of $\delta _{ij}$ and $\operatorname *{kos}_{i}(j,k)$ for H describe H up to rescaling and would like to compute them from the $\left \{ J_{i}\right \} .$ Given $\left \{ J_{i}\right \} \rightrightarrows \,??$ , we can construct the imputed value of $\delta _{H}(a,b)$ by selecting any $J_{s}$ whose image under the rescaling $J_{s}\leadsto H$ contains the kernel functions $h_{a}$ and $h_{b}$ of $RK(H).$ If $j_{ra}$ and $j_{rb}$ are the elements of $RK(J_{s})$ which correspond to $h_{a}$ and $h_{b}$ under that rescaling map, then $\delta _{J_{r}}(j_{ra},j_{rb})$ is our candidate value for $\delta _{H}(a,b).$ Note that this is defined even if there is no H; however, if there is an H, then the rescaling $J_{s}\leadsto H$ insures that this value is $\delta _{H}(a,b).$ Also, if there is another possible choice for $J_{s}$ , then the coherence conditions insure that choice will produce the same value of $\delta $ . If there is no such $J_{s}$ , then we have no candidate for the value $\delta _{H}(a,b)$ and that value would be a free variable in our analysis; the free variable $K_{42}$ in Corollary 6.3 is an example. The procedure for constructing our candidate values for $\operatorname *{kos}_{a}(b,c)$ is the same except that we need to select a $J_{s}$ whose image would contain the three elements of $RK(H)$ with indices $a,b,$ and $c.$

The general pattern is that given $\left \{ J_{i}\right \} \rightrightarrows \,??$ , we can use the coherence data and compute (some of) the entries $\operatorname *{kos}_{a}(b,c)$ that the matrix $\operatorname *{KOS}(H,a)$ will have if there is an H with $\left \{ J_{i}\right \} \rightrightarrows \,H.$ (We are supposing that it has been specified which kernel functions (if any) of the spaces $J_{i}$ are to be associated with the distinguished kernel $k_{a}$ in $H.)$ We denote the partially defined matrix computed this way by $\operatorname *{KOS}(\left \{ J_{i}\right \} ,a).$ If $\left \{ J_{i}\right \} \rightrightarrows \,H$ , then $\operatorname *{KOS}(\left \{ J_{i}\right \} ,a)=\operatorname *{KOS}(H,a).$ In Section 5.1, we give necessary and sufficient conditions on a matrix for it to be $\operatorname *{KOS}(H,a)$ for an $H\in \mathcal {CPP}$ . Those conditions applied to $\operatorname *{KOS} (\left \{ J_{i}\right \} ,a)$ give necessary and sufficient conditions for the assembly question $\left \{ J_{i}\right \} \rightrightarrows \,??$ to have a positive solution.

By Corollary 3.3, statements about spaces, subspaces, rescalings, values of invariants, and coherence are equivalent to statements about sets in $\mathbb {CH}^{m},$ subsets, congruences, values of invariants, and an appropriate notion of coherence. We will use the same language and notation in both contexts. For instance, given sets $\left \{ Y_{i}\right \} $ in $\mathbb {CH}^{n}$ , we will write $\left \{ Y_{i}\right \} \rightrightarrows X$ if there are specified congruences of each $Y_{i}$ into some X describing which points from the $\left \{ Y_{i}\right \} $ are to be mapped to which points of $X.$ Those specifications force congruences of various subsets of the various $Y_{i}$ that are analogs to the rescaling statements for the subspaces $J_{p,rs}$ in the previous paragraphs. We write $\left \{ Y_{i}\right \} \rightrightarrows \,??$ if we know the $\left \{ Y_{i}\right \} $ satisfy those congruences even if we do not know that there is an $X.$

We will pass freely between these ideas for RKHS and for sets in $\mathbb {CH}^{n}.$

3.2 Equivalence of the two questions

Using Theorems 3.1 and 3.2, we can see that the two questions in the introduction are equivalent. Those theorems give us the following facts:

1. (1) Given a set of triangles $\left \{ Y_{i}\right \} _{i=1}^{4}$ in $\mathbb {CH}^{n}$ , there are three-dimensional $\left \{ J_{i}\right \} _{i=1}^{4}\subset \mathcal {CPP}$ such that

   (3.7) $$ \begin{align} J_{i}\sim DA(Y_{i}),\text{ }i=1,...,4. \end{align} $$

   In the other direction, given three-dimensional $\left \{ J_{i}\right \} _{i=1}^{4}\subset \mathcal {CPP}$ , there are $\left \{ Y_{i}\right \} _{i=1}^{4}$ such that (3.7) holds. In either case, (3.7) continues to hold if the $\left \{ J_{i}\right \} $ are replaced by rescalings $\left \{ J_{i} ^{\prime }\right \} $ or if the $\left \{ Y_{i}\right \} $ are replaced by sets congruent to the $\left \{ Y_{i}^{\prime }\right \} $ .

2. (2) Given a tetrahedron X in $\mathbb {CH}^{n}$ , there is a four-dimensional $H\in \mathcal {CPP}$ such that

   (3.8) $$ \begin{align} H\sim DA(X). \end{align} $$

   In the other direction, given $H\in \mathcal {CPP}$ , there is an X such that (3.8) holds. In either case, (3.8) continues to hold if H is replaced by a rescaling $H^{\prime }$ or X is replaced by a congruent $X^{\prime }.$

Hence, we have the following proposition.

In the proposition, the assumption that (3.7) holds insures the $J_{i} \in \mathcal {CPP}$ . Even with that condition, having $H\in \mathcal {RK}$ and $\left \{ J_{i}\right \} \rightrightarrows H$ is not enough to insure that $H\in \mathcal {CPP}$ . This is shown by, for instance, Quiggin’s example which is discussed in Section 6.3.

4.1 Evaluating $Kos$

Because $\operatorname *{kos}$ is an automorphism invariant, we can study it for a general triple by first using an automorphism to place our triple in the configuration $\Gamma $ described in (3.1): $\left \{ x_{1},x_{2} ,x_{3}\right \} =\left \{ \left ( 0,0\right ) ,\left ( a,0\right ) ,\left ( x,b\right ) \right \} $ . In that case, computing using (2.14) gives

(4.1) $$ \begin{align} \operatorname*{kos}\nolimits_{1}(2,3)=\left\langle \left\langle \frac {(a,0)}{\left\Vert (a,0)\right\Vert },\frac{\left( x,b\right) }{\left\Vert \left( x,b\right) \right\Vert }\right\rangle \right\rangle =\frac{a\bar{x} }{\left\Vert x_{2}\right\Vert \left\Vert x_{3}\right\Vert }=\frac{\bar{x} }{\left\Vert x_{3}\right\Vert }=\frac{\left\vert \bar{x}\right\vert }{\left\Vert x_{3}\right\Vert }\frac{\bar{x}}{\left\vert \bar{x}\right\vert }. \end{align} $$

To evaluate $\operatorname *{kos}\nolimits _{1}(2,3)$ for a general triple $\left \{ x_{1},x_{2},x_{3}\right \} $ , we want a description that is invariant under automorphism of hyperbolic space and which gives the value in (4.1) for the particular triple $\left \{ \left ( 0,0\right ) ,\left ( a,0\right ) ,\left ( x,b\right ) \right \} .$ Given the general triple, let $G(1,2)$ be the complex geodesic which contains $x_{1}$ and $x_{2} $ . Recall that $P_{G(1,2)}$ is the metric projection onto $G(1,2).$

We will use the notion of angle between two geodesic segments in $G(1,2).$ Two geodesic segments in $\mathbb {CH}^{n}$ form a bivalent vertex; however, there is not a natural notion of “vertex angle” which determines the shape of the vertex, rather the geometry of the vertex is determined by the complex number $\operatorname *{kos},$ this is discussed in Section 4.3. If, however, the two geodesics lie in the same complex geodesic, then there is a conformal map of that complex geodesic to the Poincare disk. In that case, we take the angle between the two segments to be the angle between their images in the Poincare geometry of the disk (which is in fact the same as the Euclidean angle).

Here is the invariant statement we want.

Proposition 4.1 Set $P_{G(1,2)}x_{3}=y.$ Writing $\operatorname *{kos} \nolimits _{x_{1}}(x_{2},x_{3})=re^{i\theta }$ , $r>0$ , $-\pi \leq \theta <\pi $ and $\operatorname *{angle}$ for the hyperbolic angle, we have

(4.2) $$ \begin{align} \!r & =\frac{\delta(x_{1},y)}{\delta(x_{1},x_{3})},\qquad\ \ \end{align} $$

(4.3) $$ \begin{align} \theta & =\operatorname*{angle}(x_{1}x_{2},x_{1}y). \end{align} $$

Given points $\left \{ x_{1},x_{2},x_{3}\right \} $ , we are using $\operatorname *{kos}$ to describe the geometry of the intersection of the geodesics $x_{1}x_{2}$ and $x_{1}x_{3}.$ Other parameters can be used for the same purpose, for instance, in [Reference GoldmanGo, p. 88], Goldman uses angles, $\phi ,\theta $ which satisfy ${\cos \phi =\left \vert \operatorname *{kos} _{1}(2,3)\right \vert }$ and $\cos \theta =\operatorname {Re}\operatorname *{kos} _{1}(2,3)$ .

If the triangle is not in general position, we can say more about the relationship between the values of $\operatorname *{kos}$ at the vertices and the shape of the triangle. Given a triangle $T=\left \{ x_{1},x_{2} ,x_{3}\right \} \subset \ \mathbb {CH}^{2}$ , we use an automorphism to suppose that $x_{1}$ is at the origin. Regard $\mathbb {CH}^{2}$ as $\mathbb {B}_{2}$ inside $\mathbb {C}^{2},$ denote by V the real linear span of the points of $T,$ and set $W=V\cap \mathbb {B}_{2}.$ It may be that W is a totally geodesic submanifold of $\mathbb {CH}^{2}.$ In that case, with the origin in $W,$ there are three possibilities. First, W may be a classical geodesic in which case it will be a line segment through the origin. Second, W may be a complex geodesic which contains the origin. In that case, we can suppose it is the unit disk in the plane of the first coordinate and hence it is the classical Poincaré disk with constant negative curvature $-1$ . The final possibility is that W is a totally real totally geodesic disk and hence, after an automorphism, we can suppose it is ${RB_{2}=\left \{ \left ( r,s\right ) \in \mathbb {B}_{2}:r,s\in \mathbb {R}\right \},}$ the Beltrami-Klein disk of constant curvature $-1/4.$

In the first case, noting (4.2) and (4.3), we see that $\operatorname *{kos}\nolimits _{1}(2,3)=\pm 1.$ The value $-1$ occurs when $x_{1}$ separates the other two points, the value $+1$ when it does not. In the second case, $x_{3}$ is in $G(1,2)$ and so $P_{G(1,2)}x_{3}=x_{3}.$ From (4.2), we see that $\left \vert \operatorname *{kos}\nolimits _{x_{1} }(x_{2},x_{3})\right \vert =1$ and then from (4.3) that $\operatorname *{kos}\nolimits _{1}(2,3)=e^{-i\gamma }$ where $\gamma $ is the Euclidean angle between the segments $x_{1}x_{2}$ and $x_{1}x_{3}.$ Thus, $\gamma $ is the Euclidean and also the hyperbolic angle at vertex $x_{1}.$ Similarly, the values of $\operatorname *{kos}$ at the other two vertices give the other angles. The congruence class of a triangle in a plane of constant negative curvature is determined by its angles and hence in this case also by the three values of $\operatorname *{kos}$ . Finally, in the third case, the triangle is in a totally real vector space and we are in the situation discussed in Proposition 6.7. By automorphism invariance, we may suppose $x_{1}$ is at the origin. From (2.14), we see that $\operatorname *{kos}\nolimits _{x_{1}}(x_{2},x_{3})=\cos \theta $ where $\theta $ is the Euclidean angle of the vertex at the origin of the triangle with vertices $\left \{ x_{i}\right \} _{i=1}^{3}.$ However, the intersection of the real plane with those vertices with $\mathbb {CH}^{2}$ is the Beltrami–Klein model of $\mathbb {RH}^{2}$ which, at the origin, is conformal with the Euclidean geometry. Hence, $\theta $ is also the angle at $x_{1}$ of the hyperbolic triangle with vertices $\left \{ x_{i}\right \} _{i=1}^{3}$ sitting in a copy of $\mathbb {RH}^{2}.$ Similarly for the other two vertices. Hence, again, we know the three angles of a triangle in a plane of constant negative curvature and hence know its congruence class. Finally, looking backward, we see that the first case is the second and third cases holding simultaneously.

In each of these cases, the argument can be reversed. If $\operatorname *{kos} \nolimits _{1}(2,3)=\pm 1$ , then W is a line through the origin, if $\left \vert \operatorname *{kos}\nolimits _{x_{1}}(x_{2},x_{3})\right \vert =1$ , then $P_{G(1,2)}x_{3}=x_{3}$ , and hence the triangle lies in a complex geodesic, and finally, again noting Proposition 6.6, if $\operatorname *{kos} \nolimits _{x_{1}}(x_{2},x_{3})$ is real, then after automorphism, the triangle is in the position described.

4.2 $Kos$ and other invariants

In Theorem 3.1, we recalled the result of Brehm [Reference BrehmBr] that equality of the data sets $S^{\prime }$ is a congruence criterion for triangles in $\mathbb {CH}^{n}$ . That criterion and variations are often used in describing the geometry of finite sets in $\mathbb {CH}^{n}$ [Reference Brehm and Et-TaouiBE, Reference Cunha and GusevskiiCG, Reference Hakim and SandlerHS, Reference GusevskiiG, Reference RochbergRo]. Here, we mainly use the congruence criterion given by the dataset $S^{\prime \prime }$ which is, as we noted in Remark 1, an analog of the classical side-angle-side congruence criterion for Euclidean triangles.

It is straightforward to pass between $S^{\prime }$ and $S^{\prime \prime }.$ The parameters are invariant, so we can assume that we are in the model case and the triangle is $\Gamma =\left \{ x_{1},x_{2},x_{3}\right \} =\left \{ \left ( 0,0\right ) ,\left ( a,0\right ) ,\left ( x,b\right ) \right \} .$ First, suppose we have the data $S^{\prime \prime },$ that is, $a=\delta (x_{1},x_{2}),$ ${\omega =\delta (x_{1},x_{3}),}$ and $\operatorname *{kos}\nolimits _{1}(2,3)$ are known. Using those values and (2.14), we also know $a\bar {x}$ . Hence, using (2.9), we can find the third side length using

(4.4) $$ \begin{align} \delta^{2}(x_{2},x_{3})=1-\frac{(1-\left\Vert x_{2}\right\Vert ^{2} )(1-\left\Vert x_{3}\right\Vert ^{2})}{\left\vert 1-\left\langle \left\langle x_{2},x_{3}\right\rangle \right\rangle \right\vert ^{2}}=1-\frac{(1-\alpha ^{2})(1-\omega^{2})}{\left\vert 1-\alpha\bar{x}\right\vert ^{2}}. \end{align} $$

To finish determining $S^{\prime }$ , we need to find $\beta ,$ the angular invariant. For this triangle, $\beta =\arg (1-a\bar {x})$ , and, as we just mentioned, $a\bar {x}$ is known; hence, $\beta $ is known.

In the other direction, to go from $S^{\prime }$ to $S^{\prime \prime }$ , we need to find $\operatorname *{kos}_{1}(2,3).$ In this case, we know the three side lengths, and hence, noting (4.4), we know $\left \vert 1-\alpha \bar {x}\right \vert .$ We also know the angular invariant $\beta =\arg (1-a\bar {x}).$ Combining these two, we know $a\bar {x}.$ With that information, and using (4.1), we can find $\operatorname *{kos}_{1}(2,3).$

4.3 $Kos$ and the geometry of vertices

For a bivalent vertex in $\mathbb {R}^{n}$ , the cosine of the vertex angle carries all the intrinsic geometric data about the vertex. The geometry of a trivalent vertex is determined by the geometry of the three component bivalent vertices. The values of $\operatorname *{kos}$ provide similar information for vertices in $\mathbb {CH}^{n}.$

To set the stage, we first consider sets in $\mathbb {R}^{n}$ . By a vertex $\mathbf {V}$ in $\mathbb {R}^{n}$ , we mean a collection of two or more line segments, rays, with a common starting point the vertex point, or vertex, V. We call $\mathbf {V}$ bivalent if it has two rays, trivalent if there are three, multivalent in general. We suppose the rays of a bivalent vertex are ordered. (That will be important in the complex case.) The two rays of a bivalent vertex span an affine plane and form an angle in that plane, the vertex angle. We say two vertices are (Euclidean) congruent if there is a Euclidean isometry placing the second vertex point at the same position as the first and so that the initial segments of the rays of the second vertex coincide with the initial segments of the rays of the first.

If three Euclidean triangles have pairs of matching side lengths, then they can be put together as faces of a tetrahedron if and only if the three dihedral angles that would be joined at a vertex can, in fact, be joined to form a trivalent vertex. (The simple classical condition for this is in Corollary 6.9.) In particular, there is no constraint on the side lengths beyond the matching condition. Here is a precise statement.

Suppose we are given $\mathfrak {S}$ , a set of three triangles $\left \{ T_{i}\right \} _{i=1}^{3}$ each contained in some $\mathbb {R}^{n}.$ Suppose for $i=1,2,3$ that $T_{i}$ has vertices $\left \{ x_{i1},x_{ia},x_{ib} \right \} ,$ that the Euclidean length of the side $x_{i1}x_{ia}$ is $l_{ia}$ and similarly for $l_{ib}$ , and that $\mathbf {W}_{i}$ is the bivalent vertex in $T_{i}$ with vertex point $x_{i1}$ and rays ordered similarly to the sides of the triangle; $x_{i1}x_{ia}$ first, $x_{i1}x_{ib}$ second. It may or may not be possible to join these three triangles as faces of a tetrahedron in some $\mathbb {R}^{n}.$ That is there may be or may not be a tetrahedron $\left \{ y_{1},y_{2},y_{3},y_{4}\right \} $ in some $\mathbb {R}^{n}$ with the triangular face $\left \{ y_{1},y_{2},y_{3}\right \} $ congruent to $T_{1},$ $\left \{ y_{1},y_{3},y_{4}\right \} $ congruent to $T_{2},$ and $\left \{ y_{1},y_{2},y_{4}\right \} $ congruent to the triangle ${\widetilde {T_{3}}=\left \{ x_{31},x_{3b},x_{3a}\right \} }$ . (The triangle $\widetilde {T_{3}}$ has the same vertices as $T_{3}$ but in a different order.) Certainly, a necessary condition for building a tetrahedron is that certain side lengths match.

Definition 4.3 The set of three triangles $\mathfrak {S}$ is said to be a matched set if there are numbers $\left \{ L_{i}\right \} _{i=1}^{3}$ such that

(4.5) $$ \begin{align} l_{1b}=l_{2a},=L_{1},\text{ }l_{2b}=l_{3a}=L_{2},\text{ }l_{3b}=l_{1a}=L_{3}. \end{align} $$

We now prove analogs of the two previous propositions for $\mathbb {CH}^{n}.$ By a vertex $\mathbf {V}$ in $\mathbb {CH}^{n}$ , we mean a collection of two or more geodesic segments, rays, $R_{i},$ with a common starting point the vertex point, $V.$ Again, a vertex may be bivalent, trivalent, or multivalent. We say two vertices are congruent if there is an automorphism placing the second vertex point on the first and so that initial segments of the two sets of rays overlap.

The geometry of a bivalent vertex in $\mathbb {CH}^{n}$ can be described by two real numbers or one complex number. We will use the complex quantity $\operatorname *{kos}\left ( \mathbf {V}\right ) $ which we define to be $\operatorname *{kos}_{V}(x_{1},x_{2})$ where V is the vertex point and $x_{i}$ is chosen on the ray $R_{i}$ , $i=1,2.$ The formula (2.14) shows that this value does not depend on those choices. Thus, if $\left \{ x_{1},x_{2},x_{3}\right \} $ is a triangle and $\mathbf {V}$ the bivalent vertex with vertex point $x_{1}$ , then ${\operatorname *{kos}\left ( \mathbf {V}\right ) =\operatorname *{kos}_{1}(2,3)}$ .

Now, suppose we are given $\mathfrak {S,}$ a set of three triangles each in $\mathbb {CH}^{n}$ . We continue the earlier notation and terminology, but now we use the pseudohyperbolic side lengths $\delta (\cdot ,\cdot )$ . With that change, the analog of Proposition 4.4 holds.

In particular, note that in both the real and complex cases, the congruence class is completely described by the shapes of three faces and the geometry of the vertex where they meet. There need not be any mention of the geometry of the fourth face.

The analogous congruence result for multivalent vertices in $\mathbb {CH}^{n}$ is given in Corollary 5.3.

4.4 $Kos$ and area

There is not a natural notion of area for a general triangle T in $\mathbb {CH}^{m}$ ; however, there is a related invariant, the symplectic area of $T,$ obtained by integrating the symplectic form of $\mathbb {CH}^{n}$ over a real two manifold bounded by the sides of $T.$

However, if T sits inside a complex geodesic $A\subset \mathbb {CH}^{m}$ , then we can define and compute its area as follows. After an automorphism, we can suppose T is inside a copy of $\mathbb {CH}^{1}$ inside $\mathbb {CH}^{n}.$ That copy of $\mathbb {CH}^{1}$ is isometric to the classical Poincare disk of curvature $-1.$ We define the area of T, $\operatorname *{Area}(T),$ to br the area of that copy of T in the Poincare disk. This definition is an automorphism invariant, and it can be shown that $\operatorname *{Area}(T)$ equals both the symplectic area of T and is also twice the angular invariant, $\alpha ,$ of T (defined in Section 2.3) [Reference GoldmanGo, Reference Hangan and MasalaHM].

We can also evaluate $\operatorname *{Area}(T)$ using $\operatorname *{kos}.$

Proposition 4.7 If $T=\left \{ x_{1},x_{2},x_{3}\right \} $ is a triangle in $\mathbb {CH}^{n}$ which sits inside a complex geodesic, then

(4.6) $$ \begin{align} \pi-\arg\left( \operatorname*{kos}\nolimits_{1}(2,3)\operatorname*{kos} \nolimits_{2}(3,1)\operatorname*{kos}\nolimits_{3}(1,2)\right) =2\alpha (x_{1},x_{2},x_{3})=\operatorname*{Area}(T). \end{align} $$

Proof After an automorphism of $\mathbb {CH}^{n}$ , we suppose that T is in the unit disk of $\mathbb {C}$ which we identify with $\mathbb {CH}^{1}.$ From (2.13), we see that $\operatorname *{kos}_{1}(2,3)$ is a positive multiple of

$$\begin{align*}\kappa_{123}=\left\langle \left\langle \phi_{x_{1}}(x_{2}),\phi_{x_{1}} (x_{3})\right\rangle \right\rangle. \end{align*}$$

Hence, in computing the left-hand side, $LHS$ , of (4.6), we can replace $\operatorname *{kos}(2,3)$ with $\kappa _{123}$ and similarly for other indices. The conformal involutions of the disk are given by Blaschke factors. Hence, noting that for $a,b$ in the disk $\left \langle \left \langle a,b\right \rangle \right \rangle =a\bar {b}$ , we find that $\kappa _{123} =\frac {x_{1}-x_{2}}{1-x_{1}\overline {x_{2}}}\overline {\frac {x_{1}-x_{3} }{1-x_{1}\overline {x_{3}}}}.$ Thus,

$$\begin{align*}LHS=\pi-\arg\frac{-\Pi\left\vert x_{i}-x_{j}\right\vert ^{2}}{\Pi\left( 1-x_{i}\overline{x_{j}}\right) ^{2}} \end{align*}$$

with both products over the index pairs $\left \{ \left ( 1,2\right ) ,\left ( 2,3\right ) ,\left ( 3,1\right ) \right \} .$ The positive factor $\Pi \left \vert x_{i}-x_{j}\right \vert ^{2}$ does not affect the value of $\arg $ , and hence we continue with

$$\begin{align*}LHS=\pi-\left( \pi+2\arg\Pi k(x_{i},x_{j}\right) )=2\alpha, \end{align*}$$

the last equality by (2.10).

To finish, we need to know that $2\alpha = \ \operatorname *{Area}(T).$ That is in [Reference GoldmanGo]. Alternatively, going back to (4.6), let $\gamma _{i}$ be the angle at vertex $x_{i}$ . Because $T\subset \mathbb {CH}^{1}$ , we see from Proposition 4.1 that $\operatorname *{kos}_{1}(2,3)=e^{i\gamma _{1}}$ where $\gamma _{1}$ is the angle at $x_{1}$ of the triangle $T.$ Similarly for the other indices. Using that, we see that $LHS$ in (4.6) equals $\pi -(\gamma _{1}+ \gamma _{2}+\gamma _{3})$ , which equals $\operatorname *{Area} (T)$ by the classical result based on the Gauss–Bonnet theorem $.$

In contrast to T, consider now a triangle R that sits inside a totally real submanifold M. (If M exists, it can be taken to have real dimension $2$ .) The previous discussion does not apply, and, in fact, the symplectic area of R is $0.$ However, we have the following observation. We can suppose M is two-dimensional, and using a conformal automorphism, we can move M to $RB_{2}=\left \{ \left ( x,y,0,...,0\right ) \in \mathbb {B}_{n}:x,y\in \mathbb {R}\right \} $ inside $\mathbb {CH}^{n}=\mathbb {B}_{n}.$ That subspace of complex hyperbolic space is, isometrically, a copy of the Beltrami–Klein model of $\mathbb {RH}^{2}$ which has constant curvature $-1/4.$ As such, it carries a natural area measure, and we let $\mathit {Area}(R)$ denote the area of R as a triangle in that space. This quantity is also invariant under conformal automorphisms of $\mathbb {CH}^{n}.$

Proposition 4.8 Suppose $R=\left \{ x_{1},x_{2},x_{3}\right \} $ sits in a totally real totally geodesic submanifold of $\mathbb {CH}^{n}.$ Then

$$\begin{align*}4(\pi-(\cos^{-1}\operatorname*{kos}\nolimits_{1}(2,3)+\cos^{-1} \operatorname*{kos}\nolimits_{3}(1,2)+\cos^{-1}\operatorname*{kos} \nolimits_{2}(3,1)))=\mathit{Area}(R). \end{align*}$$

One proposition gives a result involving a sum of values of $\arg \operatorname *{kos}$ and the other a sum of values of $\cos ^{-1} \operatorname *{kos}.$ It would be interesting to have a general result which unifies the two.

5.1 Describing sets by their triangles

In Theorem 3.2, we saw that the congruence class of a finite $X\subset \mathbb {CH}^{n}$ is determined by the congruence classes of those of its subtriangles which share a specified designated vertex. From Theorem 3.1, we know that the congruence class of each of those triangles can be described using side lengths and the angular invariant. Those parameters have been used in describing the congruence class of finite $X\subset \mathbb {CH}^{n}$ [Reference Brehm and Et-TaouiBE, Reference Cunha and GusevskiiCG, Reference GusevskiiG, Reference Hakim and SandlerHS, Reference RochbergRo] and have also been used for more general geometric questions, [Reference ClercC, CG]. When those parameters are restricted to sets in real hyperbolic space, the angular invariant trivializes leading to descriptions of polyhedra in $\mathbb {RH}^{n}$ in terms of edge lengths.

Here, instead of side lengths and the angular invariant, we use side lengths and $\operatorname *{kos}$ to describe the constituent triangles of $X.$ This alternative description emphasizes a different type of geometric data. For instance, when restricted to sets in real hyperbolic space, it produces a description of polyhedra using side lengths and vertex angles. The parameters only have natural geometric constraints, and their values interact well with passage to subsets. That lets us give explicit answers to the questions in the introduction.

Suppose we are given $X=\left \{ x_{i}\right \} _{i=1}^{n+1} \ \subset \mathbb {CH}^{n}$ . If we connect $x_{1}$ to each of the other $x_{i}$ by a geodesic $\gamma _{1i}$ , then the point $x_{1}$ will be the vertex point of an n-valent vertex $\mathbf {V}$ which is composed of the bivalent vertices $\mathbf {V}_{ij}$ , $2\leq i,j\leq n+1$ having $\gamma _{1i}$ as a first ray and $\gamma _{1j}$ as a second. We can describe X using the distances between $x_{1}$ and the other $x_{j}$ and the numbers $K_{ij} =\operatorname *{kos}(\mathbf {V}_{ij})$ . Specifically, recalling the definition (2.15), we define the n-vector $\rho (X)$ and the $n\times n$ matrix $\mathcal {M}(X)$ by

(5.1) $$ \begin{align} \,\rho(X) & =\left( \delta(x_{1},x_{2}),...,\delta(x_{1},x_{n+1})\right) ,\text{ and}\qquad\qquad\qquad \end{align} $$

(5.2) $$ \begin{align} \mathcal{M}(X) & =\left( K_{ij}\right) _{i,j=2}^{n+1}=\left( \operatorname*{kos}\left( \mathbf{V}_{ij}\right) \right) _{i,j=2} ^{n+1}=\operatorname*{KOS}(DA(X),1). \end{align} $$

The functionals that give the entries of the vector $\rho $ and the matrix $\mathcal {M}$ are defined for any $H\in \mathcal {RK}$ , and hence those same definitions can be used to produce $\rho (H)$ and $\mathcal {M}(H)$ .

The following result parametrizes congruence classes of finite sets in $\mathbb {CH}^{n}$ as well as rescaling equivalence classes of $\mathcal {RK}$ with the CPP.

Thus the congruence class of a set X of k ( $=n+1)$ points in $\mathbb {CH}^{m}$ , $m\geq k-1,$ is determined by the $k-1$ real numbers in $\rho (X)$ together with the $(k-1)(k-2)/2$ complex numbers which specify, $\mathcal {M}(X)$ a positive semidefinite matrix with ones on the diagonal. Together these give $(k-1)^{2}$ real parameters, as expected. We should emphasize that $\rho (X)$ and $\mathcal {M}(X)$ only depend on the congruence class of ordered set X. The choice of which element is placed at the origin by an automorphism substantially affects all the values in $\rho (X)$ and $\mathcal {M}(X)$ . Once that point is specified, the remaining ordering only affects the ordering of the entries in $\rho (X)$ and $\mathcal {M}(X).$

Details of $\mathcal {M}(X)$ contain information about the geometry of $X.$ From the previous theorem and proof, we have the following corollary.

Here are the details of $\rho (X)$ and $\mathcal {M}(X)$ in some simple cases. Suppose $n+1=3$ . From Theorem 3.1, we know that the congruence class of the triangle $X=\{x_{1},x_{2},x_{3}\}$ is described by the set $S^{\prime \prime }=(\delta _{12},\delta _{13},\operatorname *{kos}_{1}(2,3)).$ In the notation of the previous theorem,

$$\begin{align*}\rho(X)=\left( \delta_{12},\delta_{13}\right) ,\text{ }\mathcal{M}(X)= \begin{pmatrix} 1 & \operatorname*{kos}_{1}(2,3)\\ \operatorname*{kos}_{1}(3,2) & 1 \end{pmatrix}. \end{align*}$$

In this case, the condition $\mathcal {M}(X)\succcurlyeq 0$ is equivalent to $\left \vert \operatorname *{kos}_{1}(2,3)\right \vert \leq 1$ , which is the condition (3.5) in Theorem 3.1.

The case of $n+1=4$ points is discussed in Section 6.

For $X\subset \mathbb {CH}^{1}$ , we use the complex coordinate of $\mathbb {C} ^{1}$ and write $X=\left \{ r_{s}e^{i\theta _{s}}\right \} _{s=1}^{n+1}$ with $r_{1}=0.$ Using the computations in Section 4.1, we see that

$$\begin{align*}\rho(X)=\left( r_{2},...,r_{n+1}\right) ,\text{ }\mathcal{M}(X)=\left( \exp i\left( \theta_{s}-\theta_{t}\right) \right) _{s,t=2}^{n+1}. \end{align*}$$

The automorphisms of $\mathbb {CH}^{1}$ which fix the base point are rotations. They do not change the entries in $\mathcal {M}(X)$ or the congruence class of $X.$ On the other hand, complex conjugation, which is not in $\operatorname *{Aut}\mathbb {B}_{1},$ can change the matrix entries and the congruence class.

For $Y \ \subset RB_{2},$ the Beltrami–Klein model of $\mathbb {RH}^{2},$ we use the polar coordinates of the containing $\mathbb {R}^{2}.$ We have $Y=\left \{ \left ( r_{s},\theta _{s}\right ) \right \} _{s=1}^{n+1}$ with $r_{1}=0.$ Now, using Section 4.1 gives

$$\begin{align*}\rho(Y)=\left( r_{2},...,r_{n+1}\right) ,\text{ }\mathcal{M}(Y)=\left( \cos\left( \theta_{s}-\theta_{t}\right) \right) _{s,t=2}^{n+1}. \end{align*}$$

In this case, the map $(r,\theta )\rightarrow (r,-\theta ),$ which looks like complex conjugation, is the restriction of an element of $\operatorname *{Aut} \mathbb {B}_{2}$ to $RB_{2},$ namely, the map $\left ( z,w\right ) \rightarrow (z,-w).$ That map changes the sign of the $\theta ^{\prime }s$ , but that does not change $\mathcal {M}(Y)$ or the congruence class of $Y.$

The previous theorem also gives the extension to n-valent vertices of the results in Section 4.3 for bivalent and trivalent vertices. Suppose $\mathbf {V}$ is an n-valent vertex in $\mathbb {CH}^{n}$ with vertex point $x_{1}$ and rays $\left \{ \gamma _{1i}\right \} _{i=2}^{n+1}.$ We are only interested in congruence classes, and hence we suppose $x_{1}$ is at the origin. For $i=2,...,n+1$ , select a point $x_{i}$ on $\gamma _{1i}$ and set $X=\left \{ x_{i}\right \} _{i=1}^{n+1}.$ From the previous theorem, we know the congruence class of X is determined by $\mathcal {M}(X)$ and $\rho (X).$ Looking at that proof, we see that knowing $\mathcal {M}(X)$ is equivalent to knowing the congruence class of the projected set $\widehat {X}.$ From the definitions, we see that knowing $\widehat {X}$ is equivalent to knowing $\mathbf {V.}$ Hence, defining $\mathcal {M}(\mathbf {V})$ to be $\mathcal {M}(X)$ , we have the following corollary of the previous theorem.

5.2 Comparison with the McCullough–Quiggin theorem

Theorem 5.1 has some similarity to the McCullough–Quiggin theorem. Here is brief informal discussion of that relation; more details are in Chapters 7 and 8 of [Reference Agler and PickAM] and the Historical Notes to those chapters.

Recall that we are only considering finite-dimensional spaces.

The first observation is that our “definition of convenience,” that H has the CPP if and only if it is a rescaling of a space $DA(X),$ is actually inconvenient in this context. We need to distinguish between that definition and the actual complete Pick property (ACPP) defined in terms of extension properties of certain matrix multipliers on $H.$ That property is defined and discussed in Chapter 5 of [Reference Agler and PickAM].

Given H and $1\leq s\leq \dim H$ , we can form the matrices $\operatorname *{KOS}(H,s)$ as described in (2.15).

Consider now the following four conditions:

1. (1) H has the ACPP.

2. (2) H has the CPP.

3. (3) For each s, $1\leq s\leq \dim H$ , $\operatorname *{KOS}(H,s)\succcurlyeq 0.$

4. (4) For some s, $1\leq s\leq \dim H$ , $\operatorname *{KOS}(H,s)\succcurlyeq 0.$

We saw that if $H=DA(X)$ , then $\operatorname *{KOS}(H,s)$ is a Gram matrix; hence, statement 2 implies statements 3 and 4.

The result of McCullough and Quiggin, Theorem 7.6 in [Reference Agler and PickAM], is that statements 1 and 3 are equivalent. (Their result actually uses an equivalent formulation based on matrices $\operatorname *{MQ}(H,s)$ defined at (2.16). Their result also involves matrices of all sizes, but in this finite-dimensional case, it suffices to only consider the matrices of maximum size.) It is a result of Agler and McCarthy, Theorem 8.2 in [Reference Agler and PickAM], that conditions 1 and 2 are equivalent. Combining these two results, we see that 3 implies 2.

On the other hand, by Theorem 5.1, if condition 4 holds, then 2 holds; H is a rescaling of some $DA(X).$ Considering the details of the proof of Theorem 5.1, this is a simpler path to that conclusion than the path through condition 1. Also, it only requires information about a single matrix, condition 4 rather than 3. However, this path says nothing about the relationship between conditions 1 and 2, which is one of the centerpieces of the theory of spaces with the ACPP.

5.3 Assembly questions

We now return to the assembly and coherence question we discussed in Section 3.1. We will consider questions of congruence involving subsets of sets in $\mathbb {CH}^{n};$ however, recall, as we commented earlier, that these are equivalent to questions about rescaling of regular subspaces of spaces in $\mathcal {CPP}$ .

The generalized version of Question 1 from the Introduction is: given a finite collection of finite sets $\left \{ Y_{i}\right \} $ in some $\mathbb {CH}^{n}$ , is there an X in some $\mathbb {CH}^{m}$ which contains congruent copies of the various $Y_{i}?$ We may also impose requirements on the overlap of the copies of the $\left \{ Y_{i}\right \} $ inside of $X.$ In Section 3.1, we saw that meeting such overlap conditions may require congruence of certain subsets of the various $Y_{i}.$ If all those congruence constraints are satisfied, we write $\left \{ Y_{i}\right \} \rightrightarrows ??$ and if there is such an X, we write $\left \{ Y_{i}\right \} \rightrightarrows \ X.$

For $X\subset \mathbb {CH}^{n}$ with distinguished base point $x_{1}$ , recall that we write $\operatorname *{KOS}(X,1)$ as a shorthand for $\operatorname *{KOS}(DA(X),1)$ . For each i, let $1_{i}$ be that point in $Y_{i}$ which is mapped to the base point $x_{1}$ of X under the assumed congruences, with $1_{i}$ an arbitrary point of Y $_{1}$ if that definition is unfilled.

If $\left \{ Y_{i}\right \} \rightrightarrows \ X$ , then, as we noted in Section 3.1, we can use data from the matrices $\operatorname *{KOS} (Y_{i},1_{i})$ to compute some of the entries $\operatorname *{KOS}(X,1)$ and those computations can be done even without knowing if there is an $X.$ The computations produce a (possibly only partially filled) matrix, which we denote $\operatorname *{KOS}(\left \{ Y_{i}\right \} ,1),$

We now study the possibility that there is an X by comparing $\operatorname *{KOS}(\left \{ Y_{i}\right \} ,1)$ with the properties which we know from Theorem 5.1 that $\operatorname *{KOS}(X,1)$ must have. More specifically, from our previous analysis culminating in Theorem 5.1, we know that the congruence class of X is described by a vector of side lengths, $\rho (X),$ and the matrix $\mathcal {M(}X)=\operatorname *{KOS}(X,1).$ We extract information from $\operatorname *{KOS}(X,1)$ by working with the principal submatrices. From the definitions, we have the following lemma.

The following classical result will let us relate the fact that $\mathcal {M(} X\mathcal {)}\succcurlyeq 0$ to properties of its principal submatrices $\mathcal {PS}(\mathcal {M(}X\mathcal {))}$ .

Now, we look at three special cases, the case where the $\left \{ Y_{i}\right \} $ are three point sets whose congruent images fill an $n+1$ point set, the question of specifying the geometry of the n point subsets of a set of $n+1$ points, and after those general questions, we consider specific ad hoc variation chosen to show how these ideas work in a more complicated situation.

In the next section, we use these ideas for a systematic study of the question in the introduction of assembling four triangles into a tetrahedron.

5.3.1 Variation 1

We know from Theorem 3.2 that the congruence class of a set X is determined by the congruence classes of the triangles in X which contain a specified base point. We now ask if a given set of triangles can be congruent to those faces of some $X.$ We want to know if we can map triangles $\left \{ Y_{i}\right \} _{i=1}^{n}$ in $\mathbb {CH}^{n},$

$$\begin{align*} Y_{j} & =\left\{ y_{j1},y_{j,j+1},y_{j,j+2}\right\} ,\text{ } \ j<n,\\ Y_{n} & =\left\{ y_{n1},y_{n,n+1},y_{n2}\right\} , \end{align*}$$

into a set $X=\left \{ x_{1},...,x_{n+1}\right \} $ with each $y_{jk}$ is mapped to $x_{k}.$ That is, each image has its first vertex at $x_{1}$ and the images fill X with each segment $x_{1}x_{t}$ in X covered twice. The coherence conditions are that two triangle sides that cover the same $x_{1}x_{t}$ must be the same length.

Recall that the matrix $\operatorname *{KOS}(\left \{ Y_{i}\right \} ,1)$ is defined in (2.15).

Theorem 5.6 Given the coherence data $\left \{ Y_{i}\right \} \rightrightarrows \,??$ just described, the following are equivalent:

1. (1) $\exists X$ , $\left \{ Y_{i}\right \} \rightrightarrows X.$

2. (2) $\operatorname *{KOS}(\left \{ Y_{i}\right \} ,1)\succcurlyeq 0$ .

3. (3) $\forall A\in \mathcal {PS(}\operatorname *{KOS}(\left \{ Y_{i}\right \} ,1))$ , $\det A\geq 0.$

4. (4) $\forall S\subset \left \{ 1,...,n\right \}$ , $1\in S$ , $\det \operatorname *{KOS}(\left \{ Y_{i}\right \} _{y\in S},1)\geq 0.$

Proof This is a direct consequence of Theorem 5.1 and Lemmas 5.4 and 5.5.

Taking note of Corollary 3.3, we also have the same result for spaces $J_{i}\in \mathcal {CPP}$ and the question of moving from $\left \{ J_{i}\right \} \rightrightarrows \,??$ to $\left \{ J_{i}\right \} \rightrightarrows H$ with $H\in \mathcal {CPP}$ .

For both $\left \{ Y_{i}\right \} $ and $\left \{ J_{i}\right \} $ , the condition $\det A\geq 0$ is automatic for the A that are $1\times 1$ principal submatrices and the $2\times 2$ case is insured by Theorem 3.1 $.$ For A that are $3\times 3$ , the situation is more complicated. For instance, if we suppose $\left \{ J_{i}\right \} \subset \mathcal {RK}$ but not necessarily in $\mathcal {CPP}$ , then $\det A\geq 0$ is not automatic; see the comment after Theorem 6.4.

The case $n=3$ of this result is the tetrahedron assembly question of the introduction. We discuss it in more detail in the next section.

5.3.2 Variation 2

Suppose we are given $\left \{ Y_{i}\right \} _{i=1}^{n},$ sets of size n in $\mathbb {CH}^{n}$ . Write $Y_{i}=\left \{ y_{ij}:1\leq j\leq n+1,\text { }j\neq i+1\right \} .$ We impose the coherence conditions $\left \{ Y_{i}\right \} \rightrightarrows \,??$ that would hold if there were a set

$$\begin{align*}X=\left\{ x_{1},...,x_{n+1}\right\} \end{align*}$$

and congruences $Y_{i}\leadsto X$ which mapped the points $y_{is}$ to $x_{s},$ for all $s\neq i.$ Informally, we are trying to specify the congruence type of all the subsets of size n inside a set of size $n+1.$ This coherence condition $\left \{ Y_{i}\right \} \rightrightarrows \,??$ requires strong interrelations between the $Y_{i}$ . Given $Y_{r}$ and $Y_{s}$ , there are $Y_{rs}\subset Y_{r}$ and $Y_{sr}\subset Y_{s}$ both of size $n-1$ with $Y_{rs}\sim Y_{sr}$ . Also, note that every $A\in \mathcal {PS(} \operatorname *{KOS}(\left \{ Y_{i}\right \} ,1))$ which is not maximal, $A\neq \operatorname *{KOS}(\left \{ Y_{i}\right \} ,1),$ satisfies $A\in \mathcal {PS(}\operatorname *{KOS}(Y_{r},1))$ for some individual $Y_{r}.$ We know $\operatorname *{KOS}(Y_{r},1)\succcurlyeq 0$ , and hence, by Sylvester’s criterion, $\det A\geq 0.$ In sum, the only ${A\in \mathcal {PS(} \operatorname *{KOS}(\left \{ Y_{i}\right \} ,1))}$ for which we do not know $\det A\geq 0$ is the matrix $\operatorname *{KOS}(\left \{ Y_{i}\right \} ,1)$ itself. This discussion, together with Theorem 5.1, and the previous two lemmas, complete the proof of the following.

Theorem 5.7 Given $\left \{ Y_{i}\right \} \rightrightarrows \,??$ , there is an X so that $\left \{ Y_{i}\right \} \rightrightarrows X$ if and only if $\det \operatorname *{KOS}(\left \{ Y_{i}\right \} ,1)\geq 0$ .

6.1 Question 1

We are given $\left \{ T_{i}\right \} _{1}^{4}$ a set of four triangles in $\mathbb {CH}^{n}$ which satisfy the coherence conditions for assembly into a tetrahedron, $\left \{ T_{i}\right \} _{i=2}^{4}\rightrightarrows ??,$ and we want to know if they can, in fact, be assembled into a tetrahedron $X\subset \mathbb {CH}^{n}$ , $\left \{ T_{i}\right \} _{i=2}^{4} \rightrightarrows X.$ That can be done if and only if the parameter values imputed to X from the details of the $\left \{ T_{i}\right \} $ and the coherence conditions describe a possible tetrahedron $.$ We begin by reviewing those parameters.

Any tetrahedron in $\mathbb {CH}^{n}$ is congruent to one of the forms $X=\left \{ 0,x_{2},x_{3},x_{4}\right \} \subset \mathbb {CH}^{2}.$ Here, we identify $\mathbb {CH}^{2}$ with the unit ball in $\mathbb {C}^{2}$ and also regard the $\left \{ x_{i}\right \} $ as points in that space. Let $\mathbf {V}_{ij}$ be the bivalent vertex $x_{i}0x_{j}.$ From Theorem 5.1, we see that we must have each $\left \vert x_{i}\right \vert <1$ and that the matrix

$$\begin{align*}\mathcal{M}=\mathcal{M}(X)=\left( \operatorname*{kos}\left( \mathbf{V} _{ij}\right) \right) _{i,j=2}^{4}=\left( \left\langle \left\langle \widehat{x_{i}},\widehat{x_{j}}\right\rangle \right\rangle \right) _{i,j=2}^{4} \end{align*}$$

must be positive semidefinite. We will make that last condition more explicit.

Lemma 6.1 Given $\left \{ a_{i}\right \} _{i=1}^{3} \subset \mathbb {C}$ with each $\left \vert a_{i}\right \vert \leq 1,$ set

$$\begin{align*}\mathcal{N}= \begin{pmatrix} 1 & a_{1} & a_{2}\\ \overline{a_{1}} & 1 & a_{3}\\ \overline{a_{2}} & \overline{a_{3}} & 1 \end{pmatrix}. \end{align*}$$

The following are equivalent:

1. (1) $0\preccurlyeq \mathcal {N}$ .

2. (2) $0\leq \det \mathcal {N}.$

3. (3) $0\leq 1+2\operatorname {Re}a_{1}\overline {a_{2}}a_{3}-\left \vert a_{1}\right \vert ^{2}-\left \vert a_{2}\right \vert ^{2}-\left \vert a_{3}\right \vert ^{2}.$

4. (4) $\left \vert a_{1}\overline {a_{2}}-a_{3}\right \vert ^{2}\leq (1-\left \vert a_{1}\right \vert ^{2})(1-\left \vert a_{2}\right \vert ^{2}).$

Proof By Sylvester’s criterion, the first condition implies the second. The second and third conditions are equivalent by definition. That the third and fourth are equivalent can be seen by expanding both sides of the fourth statement giving

$$\begin{align*}|a_{1}\overline{a_{2}}|^{2}-2\operatorname{Re}a_{1}\overline{a_{2}} a_{3}+\left\vert a_{3}\right\vert ^{2}\leq1-\left\vert a_{1}\right\vert ^{2}-\left\vert a_{2}\right\vert ^{2}+\left\vert a_{1}a_{2}\right\vert ^{2}. \end{align*}$$

Cancellation and rearrangement shows that is equivalent to the third statement.

To go in the other direction, we again use Sylvester’s criterion and show the second statement implies the first. That states that the first statement is a consequence of the nonnegativity of the seven principal minors. Three are the determinants of the ${1\times 1}$ matrices given by the diagonal entries and they are positive. The determinants of the three $2\times 2$ submatrices are accounted for by the assumption on the size of the $a_{i}.$ Finally, the positivity of $\det \mathcal {N}$ is the second statement.

We want to know if a set of four triangles in $\mathbb {CH}^{n}$ , $\left \{ T_{i}\right \} _{1}^{4}$ which satisfy $\left \{ T_{i}\right \} _{i=1} ^{4}\rightrightarrows\ ??$ , might satisfy $\left \{ T_{i}\right \} _{i=1} ^{4}\rightrightarrows X$ for some tetrahedron $X=\left \{ 0,x_{2},x_{3} ,x_{4}\right \} .$ By congruence invariance, we may suppose that for $i=2,3,4$ , the triangle $T_{i}$ has vertices $\left \{ 0,x_{i1},x_{i2}\right \} $ for various $x_{ij}$ of length at most one. The coherence requirements on the set of triangles insure that $\left \vert x_{22}\right \vert =\left \vert x_{31}\right \vert $ , $\left \vert x_{32}\right \vert =\left \vert x_{41} \right \vert $ , and $\left \vert x_{42}\right \vert =\left \vert x_{21}\right \vert .$ If the assembly of the triangles into an X is possible, these three quantities will be the side lengths $\left \vert x_{2}\right \vert ,$ $\left \vert x_{3}\right \vert ,$ and $\left \vert x_{4}\right \vert ,$ of X, and those lengths, the values of the $\rho $ of Theorem 5.1, can be any numbers between zero and one. Hence, it only remains to check if the values of $\operatorname *{kos}\left ( \mathbf {V}_{ij}\right ) $ imputed from the triangles lead to a matrix $\mathcal {M}$ with the required properties. If assemble is possible, then the vertex $\mathbf {V}_{23}$ of X will be (congruent to) the vertex of $T_{2}$ at the origin, similarly for $\mathbf {V}_{34}$ and $T_{3}$ , $\mathbf {V}_{42}$ and $T_{4}.$ Thus, $\mathcal {M(}\left \{ T_{i}\right \} \mathcal {)}$ , the imputed value of $\mathcal {M(}X\mathcal {)}$ , is given by

(6.1) $$ \begin{align} \mathcal{M(}\left\{ T_{i}\right\} \mathcal{)}= \begin{pmatrix} 1 & \operatorname*{kos}(T_{2}) & \overline{\operatorname*{kos}(T_{4})}\\ \overline{\operatorname*{kos}(T_{2})} & 1 & \operatorname*{kos}(T_{3})\\ \operatorname*{kos}(T_{4}) & \overline{\operatorname*{kos}(T_{3})} & 1 \end{pmatrix} = \begin{pmatrix} 1 & K_{23} & K_{24}\\ K_{32} & 1 & K_{34}\\ K_{42} & K_{43} & 1 \end{pmatrix}, \end{align} $$

where the last equality defines the matrix $\left ( K_{ij}\right ) .$

We have collected all of the pieces to answer Question 1.

The condition (6.2) was obtained by specializing general results. However, in this low-dimensional case, we could have worked directly with the coordinates of $X=\Delta $ as described in (3.2). In that case, the $K_{ij}$ can be computed using the points $\{\widehat {x_{i}}\}_{i=2}^{4}$ on the unit sphere with coordinates

(6.3) $$ \begin{align} & \widehat{x}_{2}=(1,0,0),\text{ }\widehat{x}_{3}=(\xi,\beta,0),\text{ }\widehat{x}_{4}=(\eta,\zeta,\gamma);\nonumber\\ & \beta,\gamma\geq0;\text{ }\xi,\eta,\zeta\in\mathbb{C}; \text{ }\nonumber\\ & \left\vert \xi\right\vert ^{2}+\beta^{2}=\left\vert \eta\right\vert ^{2}+\left\vert \zeta\right\vert ^{2}+\gamma^{2}=1. \end{align} $$

For $2\leq i,j\leq 4$ , we have $K_{ii}=1$ , $K_{ij}=\overline {K_{ji}}.$ The rest of the story is given by

$$\begin{align*}K_{23}=\bar{\xi},\text{ } \ K_{24}=\bar{\eta},\text{ } \ K_{34}=\xi\bar{\eta }+\beta\bar{\zeta}. \end{align*}$$

We also have

$$\begin{align*}\left\vert \beta\right\vert ^{2}=1-\left\vert K_{23}\right\vert ^{2};\text{ }\left\vert \bar{\zeta}\right\vert ^{2}=1-\left\vert K_{24}\right\vert ^{2}-\gamma^{2}. \end{align*}$$

Hence, noting that for all $i,j$ , $\left \vert K_{ij}\right \vert \leq 1,$ we must have

(6.4) $$ \begin{align} \left\vert K_{34}-\overline{K_{23}}K_{24}\right\vert ^{2}\leq(1-\left\vert K_{23}\right\vert ^{2})(1-\left\vert K_{24}\right\vert ^{2}), \end{align} $$

which is (6.2). In the other direction, it is not hard to start from $\left \{ K_{ij}\right \} $ , which satisfy these conditions and find coordinates of points on the sphere which generate these data.

Our path to answering Question 1 had many digressions, and it is perhaps worthwhile to summarize the essential steps. First, if there is a tetrahedron, then the interpretation of $\operatorname *{kos}$ in (2.14) together with the definition $\mathcal {M=M}(\left \{ T_{i}\right \} )$ insures that $\mathcal {M}$ is a Gram matrix and hence is positive definite. In the other direction, given that $\mathcal {M}$ is positive definite, it must be the Gram matrix of a three-point set on the sphere. Reversing the construction of that three-point set determines three points in the ball which, together with the origin, are the vertices of our candidate tetrahedron. To finish, we need to show that this candidate tetrahedron has faces in the correct congruence classes. For the faces that meet at the origin, this holds by the complex analog of the side-angle-side criterion for congruence of triangles, Remark 1. Proposition 4.6 then shows that that data determine the congruence class of the fourth face and the proof of that same proposition shows that that same congruence class is forced by the data of the first three triangles and the assumed coherence conditions.

Here is a variation on these ideas. Given two triangles, $T_{2}$ and $T_{3},$ with one pair of matching side lengths, is it possible to find a third triangle $T_{4}$ so that $\left \{ T_{i}\right \} _{2}^{4}\rightrightarrows X$ for some tetrahedron $X?$ Again the question is whether the imputed parameters for X are an allowable set. If $T_{2}$ and $T_{3}$ are given and have a pair of matching side lengths, then all of the data for a putative $\rho (X)$ are specified and, as required, each value is between zero and one. The value of $\operatorname *{kos}$ at the distinguished vertex of $T_{2}$ will give the value of $K_{23}$ for the potential matrix $\mathcal {M}$ , similarly for $T_{3}$ and $K_{24}.$ However, the data from $T_{2}$ and $T_{3}$ are not enough to compute $K_{42}$ , which would be the value of $\operatorname *{kos}$ at its distinguished vertex of $T_{4}$ . Hence, the congruence class of $T_{4}$ is indeterminate. To go forward, we set $K_{42}=z$ , and using z, we complete the matrix $\mathcal {M(}\left \{ T_{2},T_{3}\right \} )$ to a matrix $\mathcal {M}$ which has no missing values. We then apply the previous theorem to that matrix.