2

In some scientific article (see bottom of page one) it is written that the dimension of the space of $3$-variate polynomials, $\mathbb{P}^3_n(S^2)$, of degree at most $n$, restricted on the $2$-sphere, $S^2$, is $(n+1)^2$.

If we just consider the set of $3$ variate polynomials of degree at most $n$ on $\mathbb{R}^3$, then it is obvious that the dimension is $\binom{n+3}{3}$. But I don't really understand the restricted case on the $2$-sphere.

hardmath
  • 37,015
PLP
  • 51
  • Perhaps the identity $x^2+y^2+z^2=1$ is applied to "simplify" such polynomials. Knowing the citation of that article might clear up the context. – hardmath Jan 24 '18 at 14:09
  • Indeed if we quotient the commutative ring $\mathbb{P}^3_n$ of real $3$-variate polynomials over the ideal generated by $x^2+y^2+z^2-1$ and count the number of generator we get to $(n+1)^2$. – PLP Jan 24 '18 at 14:31
  • A minor quibble: $\mathbb{P}^3_n$ is a real vector space, but not a ring because of the degree bound. – hardmath Jan 24 '18 at 15:34
  • Can't we define $\mathbb{P}^3_n$ itself as the quotient of $\mathbb{R}[x,y,z]$ by the ideal generated by $x^4-1,y^4-1,z^4-1,x^3y-1,x^2y^2-1,\cdots$ ? – PLP Jan 24 '18 at 15:52
  • Possibly, but I'd like to use the article you asked about to tell us if that is the context. Generally the dimension of a vector space is more crisply defined than the "number of generators" of a ring. – hardmath Jan 24 '18 at 16:05
  • 1
    Here it is, in the introduction, at the bottom of page one:

    http://www.math.unipd.it/~marcov/pdf/optisphere.pdf

     – PLP Jan 24 '18 at 16:09

1 Answers1

2

In the notation of the paper linked above, polynomials in $d$ variables of degree at most $n$ form a (real) vector space $\mathbb{P}^d_n$, and their "restrictions" to a compact set or manifold $S$ is $\mathbb{P}^d_n(S)$.

The context of the paper makes clear that it is concerned with polynomials as functions to be evaluated, i.e. computing approximations to "the global minimum of arbitrary $n$-th degree polynomials on the sphere". Hence the restriction of polynomials to $S$ has the sense of evaluating these polynomials at points of $S$.

In this context we can calculate and compare the dimensions of the two vector spaces, unrestricted functions $\mathbb{P}^d_n$ on $\mathbb{R}^d$ and restricted functions $\mathbb{P}^d_n(S)$ on $S$ when $S$ is the unit $d$-sphere:

$$ S = \{(x_1,x_2,\ldots,x_d) \mid x_1^2 + x_2^2 + \ldots + x_d^2 = 1 \} $$

In the unrestricted case there is a basis of functions consisting of monomials of degree at most $n$:

$$ 1,x_1,x_2,\ldots,x_d,x_1^2,x_1x_2,\ldots,x_ix_j,\ldots,x_d^2,x_1^3, \ldots,x_d^n $$

Counting these gives $\dim(\mathbb{P}^d_n)= \binom{n+d}{d}$. This can be proved by induction, and the interested Reader may enjoy formulating a stars-and-bars approach to counting them.

In the case of restriction to the unit sphere, two polynomial functions are equal if and only if their evaluations at all points of the unit sphere are equal. Note that given:

$$ x_1^2 = 1 - x_2^2 - \ldots - x_d^2 $$

a polynomial $p(x_1,x_2,\ldots,x_d)$ can always be rewritten to eliminate all terms with more than a single power of $x_1$:

$$ p(x_1,x_2,\ldots,x_d) = q(x_2,\ldots,x_d) + x_1\; r(x_2,\ldots,x_d) $$

It is important to note that this rewriting, replacing instances of $x_1^2$ with the right-hand quadratic in $x_2,\ldots,x_d$, does not create terms of higher degree than those being replaced.

Therefore a basis for $\mathbb{P}^d_n(S)$ consists of monomials of degree $n$ or less in $x_2,\ldots,x_d$ combined with monomials of the form $x_1$ times monomials of degree $n-1$ or less in $x_2,\ldots,x_d$. Counted together (they are disjoint and independent) we have:

$$ \begin{align*} \dim(\mathbb{P}^d_n(S)) &= \binom{n+d-1}{d-1} + \binom{n-1+d-1}{d-1} \\ &= \left[\frac{n+d-1}{n} + 1 \right] \binom{n+d-2}{d-1} \end{align*} $$

As noted in the linked paper, for $d=3$ (trivariate polynomials) this gives:

$$ \dim(\mathbb{P}^3_n(S)) = \frac{2n+2}{n} \binom{n+1}{2} = (n+1)^2 $$

which is less (for $n\gt 1$) than:

$$ \dim(\mathbb{P}^3_n) = \binom{n+3}{3} = \frac{(n+3)(n+2)(n+1)}{6} $$

hardmath
  • 37,015