# Solid Angles Beyond Dimension Three

There is a theorem by Ribando (Measuring Solid Angles Beyond Dimension Three, Discrete Comput Geom 36:479–487 (2006)) https://link.springer.com/content/pdf/10.1007/s00454-006-1253-4.pdf?pdf=button:

Let $$\Omega \subseteq \Bbb{R}^n$$ be a solid-angle spanned by unit vectors $$\lbrace v_1 , \dots , v_n \rbrace$$, let $$V$$ be the matrix whose ith column is $$v_i$$ , and let $$\alpha _{ij} = v_i \cdot v_j$$ as above. Let $$T_{\alpha}$$ be the following infinite multivariable Taylor series: $$T_{\alpha} = \dfrac{det \ V}{(4 \pi )^{n/2}} \sum _{a \in \Bbb{N}^{{n \choose 2}}} \left[ \dfrac{(-2)^{\sum _{i < j} a_{ij}}}{ \Pi _{i The series $$T_{\alpha}$$ agrees with the normalized measure of solid-angle $$\Omega$$ whenever $$T_{\alpha}$$ converges.

My question is how someone can find $$a$$ values and what this part means exactly $$\sum _{a \in \Bbb{N}^{{n \choose 2}}}$$ ? Is it finally an infinite sum over all possible $$a \in \Bbb{N}^{{n \choose 2}}$$? thanks

• Here's my read of it. Note first the multi-index notation $\alpha^a:=\prod_{i<j}\alpha_{ij}^{a_{ij}}$. In analogy with the symmetric $\alpha_{ij}$ which are trivial on-diagonal viz. $\alpha_{ii}=1$, the matrices $a$ included in this sum will only be allowed e.g. $i<j$ degrees of freedom, so the set thereof is denoted $\Bbb N^{\binom{n}{2}}$. But this is analogous to $\Bbb N^{n\times n}$, not to any set of the form $\Bbb N^k$.
– J.G.
Commented Dec 28, 2022 at 16:29
• @ J.G Thanks for your response. Yes, I know that $a$ is not from any size, so for example, for $n = 4$, the size os the set is 6. But still, I don't know how we can determine each element of $a$. Commented Dec 28, 2022 at 17:01

Let's consider a simpler expression to gain some intuition.

$$e^{x+y} = \left(e^x\right)\left( e^y \right)\approx \left(\sum_{k =0}^{N-1} \frac{x^k}{k!} \right)\left(\sum_{l =0}^{N-1} \frac{y^l}{l!} \right)$$

This contains terms $$x^ky^l$$ with various combinations of $$k$$ and $$l$$ and the summation is over those terms. In this case you get $$N^2$$ terms. In the limit $$N \to \infty$$, the pairs $$k,l$$ will be all possible values in $$\mathbb{N}^2$$ and the sum will equal the exponential.

With your application you get products of upper or lower triangular terms $$\alpha_{ik}$$. For instance when $$\alpha$$ is 4 by 4, then you get sums of products of $${4 \choose 2} = 6$$ terms like $$\boldsymbol{\alpha}^{\boldsymbol{a}} = \alpha_{12}^{a_{12}} \alpha_{13}^{a_{13}} \alpha_{14}^{a_{14}} \alpha_{23}^{a_{23}} \alpha_{24}^{a_{24}} \alpha_{34}^{a_{34}}$$

The $$a_{ik}$$ will range from $$0$$ to $$(N-1)$$ and there will be $$N^{n \choose 2}$$ combinations. (and in the limit you get all values in the space $$\mathbb{N}^{n \choose 2}$$)

In this answer you see a code that computes this for the first few terms. You can use that code to get a better idea to see which terms are being computed. The code computes eventually products like

prod1 = prod(factorial(aj))
term_i = sapply(1:dim, FUN = function(i) {gamma( 0.5 * (sum(Maj[i,-i])+1) )})
prod2 = prod(term_i)
prod3 = prod(alpha^aj)
pow = (-2)^sum(aj)
pow/prod1*prod2*prod3


And it does this for $$N^{n \choose 2}$$ different values of the vector $$a_j$$ (which contains the powers).

In the case of $$N = 3$$ and $$n=4$$ then the $$a_j$$ will look like the $$3^6 = 729$$ rows of the matrix below

        [,1] [,2] [,3] [,4] [,5] [,6]
[1,]    0    0    0    0    0    0
[2,]    1    0    0    0    0    0
[3,]    2    0    0    0    0    0
[4,]    0    1    0    0    0    0
[5,]    1    1    0    0    0    0
[6,]    2    1    0    0    0    0
[7,]    0    2    0    0    0    0
[8,]    1    2    0    0    0    0
[9,]    2    2    0    0    0    0
[10,]    0    0    1    0    0    0
[11,]    1    0    1    0    0    0
[12,]    2    0    1    0    0    0
...
[729,]    2    2    2    2    2    2

• Many thanks for your response. So these $a_{ik}$s are not certain and definite values, they can be anything between $0$ to $N$. But how you can determine $N$ as small as possible to guarantee convergence? Commented Dec 29, 2022 at 11:56
• I would 't call it 'not certain and definite' (but maybe this is more semantic, I guess we understand it as the same). The values are fixed like 0,1,2,3, etc. The values $a_{ik}$ are the indices in a summation, just like used in the single dimensional Taylor series. Instead of a single sum $\sum_{a=0}^N$ you get multiple sums $\sum_{a_{12}=0}^N \sum_{a_{13}=0}^N \sum_{a_{23}=0}^N$. Commented Dec 29, 2022 at 12:04
• To determine convergence rate and how far to increase $N$ I would keep increasing $N$ untill the change becomes below a certain limit. I am not sure whether that is close to the generally accepted method. If you want to make a lot of use of this formula then probably you can make use of smarter ways to determine convergence or size of the error. Commented Dec 29, 2022 at 12:11
• Okay I see. I was wondering if we could determine $N$ as a function of other known parameters with a certain error order for convergence, then we could claim that we have an approximated closed-form probability for positive orthant for any arbitrary covariance and dimension. Commented Dec 29, 2022 at 12:21
• @ Sextus Empiricus: based on the paper, if Matrix $M(1,-|\alpha_{i,j}|)$ is positive semi-definite then the probability always converge. Do you think if Matrix $A$ or actually the covariance matrix is PSD, then M is always PSD, and convergence is guaranteed? Commented Jan 2, 2023 at 12:51