Solid Angles Beyond Dimension Three

Question

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: $$T_{\alpha} = \dfrac{det \ V}{(4 \pi )^{n/2}} \sum _{a \in \Bbb{N}^{{n \choose 2}}} \left[ \dfrac{(-2)^{\sum _{i < j} a_{ij}}}{ \Pi _{i<j} a_{ij}!} \Pi _{i} \Gamma \left( \dfrac{1 + \sum _{m \neq i} a_{im}}{2} \right) \right] \alpha^{a}$$ 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

Answer 1

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.

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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