the t in spherical t-design I would like to understand exactly what the "t" in a spherical t-design means.
I'm facing the following questions so I can understand the concept:

*

*does "t" represent some kind of order?  And if so,

*what does that order mean?  for example:

*What's the difference between a t = 3 vs. t = 5 design?

*what does the value of t signify/mean regarding the physical description of the points on a sphere?

*Am I correct in assuming that for designs to distribute nodes on the surface of a sphere I'm interested in spherical t-designs on $S^2$

*Are the polynomials that the math descriptions give (refer to link provided above) the spherical harmonics in my case?

For context, I'm designing a layout for an Ambisonic decoding system based on spherical geometry. It is discussed that using spherical t-designs for locating nodes on the sphere are best for maintaining energy across the sphere.  I understand this.  Where I get lost is that there of lots of different t-designs that have been developed, and trying to determine which would be best for the job.
Thank you for assisting me understand just what the math behind t-design means in a practical sense.  :)
 A: As taken from your link, here is the definition.
Definition: A set of $N$ points is called a spherical $t$-design if the integral of any polynomial of degree at most $t$ over the sphere is equal to the average value of the polynomial over the set of $N$ points.
Let me fill in some of the gaps in this definition. They are (mostly) interested in the unit sphere in 3D space $$\mathbb{S}^2=\{(x,y,z)\in\mathbb{R}^3\mid x^2+y^2+z^2=1\}\subset\mathbb{R}^3,$$
(although some other sources, if you search the literature, rescale the sphere so that it has surface area $1$). Now we consider all polynomials on $\mathbb{R}^3$ with degree at most $t$. For example, if $t=3$ these are all functions of the form $$P(x,y,z)=a+b_1x+b_2y+b_3z+c_1xy+c_2yz+c_3zx+d_1x^2+d_2y^2+d_3z^2$$$$+e_1x^2y+e_2y^2z+e_3z^2x+e_4xy^2+e_5yz^2+e_6zx^2+fxyz,$$
where $a, b_i,c_i,d_i,e_i,f\in\mathbb{R}$ are constants. Given such a polynomial, we can compute its average value over the surface of the sphere by integrating $$A(P)=\frac{1}{\textrm{area}(\mathbb{S}^2)}\int_{\mathbb{S}^2}P(x,y,z)\;\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z.$$
Computing $A$ directly might be difficult, so we might instead want to approximate $A$ by just computing it at finitely many points $X=\{(x_1,y_1,z_1),\dots,(x_N,y_N,z_N)\}\subset\mathbb{S}^2$: $$A(P)\simeq\frac{1}{N}\sum_{i=1}^NP(x_i,y_i,z_i)$$.
Now, the definition says that $X$ is a spherical $t$-design if for every polynomial with degree at most $t$ then this approximation exactly equals $A(P)$. So we have two variables: $N$ and $t$ and they work against each other. The larger $N$ is the more points there are and so the better we can make the approximation - conversely, the larger $t$ is the more complicated the polynomials $P$ can be, and so the harder they become to approximate. That is why in the list at the link the value of $N$ increase as $t$ increases.
The way this might be useful in practice is as follows. Suppose you have some collection $\mathfrak{F}$ of functions (not necessarily polynomials) on $\mathbb{S}^2$ and you want to efficiently approximate $A(f)$ for any $f\in \mathfrak{F}$. Suppose further that you  do not mind approximating the functions in $\mathfrak{F}$ by polynomials of some bounded degree $t$. Then for any $f\in\mathfrak{F}$ there is some $P(x,y,z)$ such that $A(f)\simeq A(P)$ and you just have to compute $A(P)$. With a spherical $t$-design $X$ you can compute this efficiently and accurately.
So to answer your questions:

*

*$t$ represents the maximum "complexity" of the functions (polynomials) whose average value over the sphere we want to compute.

*Specifically, the degree is the maximum sum $k_x+k_y+k_z$ where $cx^{k_x}y^{k_y}z^{k_z}$ is a term in the polynomial for $c\in\mathbb{R}\backslash\{0\}$.

*Every $t=5$ design is a $t=3$ design since if a design can be used to compute $A(P)$ for $P$ of degree at most $5$, then it can certainly be used for polynomials of degree at most $3$. The reverse is not true, a polynomial of degree $5$ may be too complicated for $A(P)$ to be computed with set of points only designed to deal with polynomials of degree $3$.

*It's difficult, I think, to say something concrete, but in general the higher $t$ is, the more points there will be, and so the more evenly spaced out they will be. If you look at the library of 3D points at that link then for $t=1$ and $N=2$, $X$ is two opposite points, but for $t=2$ and $N=4$, $X$ is the set of vertices of an inscribed tetrahedron.

*Yes, this is the setting I've focussed on, but everything generalises immediately to unit spheres in higher dimensions.

*I'm by no means certain, but I suspect the answer is no.

