Let $\mathbb{S}=\{x\in\mathbb{R}^n|x\_1^2+\ldots +x\_n^2=1\}$ be the unit sphere in $\mathbb{R}^n$, $\mathbb{C}[x]=\mathbb{C}[x\_1,\ldots ,x\_n]$ the complex-valued polynomial functions on $\mathbb{R}^n$, and $\Delta=\partial\_1^2+\ldots +\partial\_n^2$ the Laplacian. Let $H\_n=\ker(\Delta)$ be the space of harmonic polynomials. Consider the restriction to the sphere


$$\rho:\mathbb{C}[x\_1,\ldots ,x\_n]\to\mathbb{C}[x\_1,\ldots ,x\_n]/(x\_1^2+\ldots +x\_n^2-1)$$


and the image $\bar{H}\_n=\rho(H\_n)\subseteq\mathbb{C}[x\_1,\ldots ,x\_n]/(x\_1^2+\ldots +x\_n^2-1)$.


Now, there is the curious fact that $\bar{H}\_n$ is in fact the whole of $\mathbb{C}[x\_1,\ldots ,x\_n]/(x\_1^2+\ldots +x\_n^2-1)$. In other words, for every $f\in\mathbb{C}[x\_1,\ldots ,x\_n]$, the polynomial restriction $f|\_{\mathbb{S}}$ on the sphere has a harmonic representative $h\in H\_n$ such that $f|\_{\mathbb{S}}=h|\_{\mathbb{S}}$.


This follows from a $\sum(\mathrm{radial})\cdot (\mathrm{harmonic})$ decomposition


**Theorem.** Let $f$ be a polynomial. Then $f(x)=\sum\_i f\_i(x\_1^2+\ldots +x\_n^2)\cdot h\_i(x)$ where $h\_i\in H\_n$ and $f\_i\in\mathbb{C}[t]$.


Indeed: $f|\_\mathbb S =(\sum\_i f\_i(x\_1^2+\ldots+x\_n^2) \cdot h\_i)|\_\mathbb S=\sum\_i f\_i(1)\cdot h\_i|\_\mathbb S = (\sum\_i f\_i(1)h\_i)|\_\mathbb S$ and now $h:=\sum\_i f\_i(1)h\_i$ is harmonic.


In particular, harmonic restrictions on the sphere form an algebra (equal to the algebra of all polynomial restrictions). From this, I think, one can prove the spherical harmonics are a Hilbert basis on the sphere by using Stone-Weierstrass (and density of continuous functions in $L^2$) without proving that they form a basis of eigenfunctions of the Laplace-Beltrami operator $\Delta\_\mathbb{S}$ on $\mathbb S$ (which by the way they do, and it's kinda interesting in itself).



> 
> **Q.** Is this stuff part of some theory, or just a coincidence for the usual Laplacian?
> 
> 
> 


For example, what if $D$ is a differential operator with symbol $\sigma\_D(x)$ and $X\_D\subseteq\mathbb{R}^n$ the variety $\{x|\sigma\_D(x)=1\}$. Is there sometimes a polynomial decomposition


$$f=\sum\_i f\_i(\sigma\_D(x))\cdot h\_i(x)$$


with $Dh\_i=0$? Or an "intrinsic" operator $P$ on $X\_D$, related to $D$ in a similar way as $\Delta\_\mathbb{S}$ is related to $\Delta$, such that the homogeneous $h\in\ker D$ are eigenfunctions of $P$?



Harmonic polynomials on the sphere
I view this as a concatenation of two facts:




---


**Fact 1:** Let $k$ be a field, let $I$ be an ideal of $k[x\_1, \ldots, x\_n]$ and let $J$ be associated graded ideal of $I$, meaning that a degree $d$ homogenous polynomial $g(x)$ is in $J$ if and only if there is an element of $I$ of the form $g(x) + (\text{terms of degree $< d$})$. Let $\{ p\_a \}$ be a set of homogenous polynomials which maps to a basis of $k[x]/J$. Then $\{ p\_a \}$ also maps to a basis of $k[x]/I$.
The proof of Fact 1 is just an upper triangularity argument.


Apply Fact 1 with $I$ the ideal $\langle \sum x\_i^2 -1 \rangle$ and $J$ the ideal $\langle \sum x\_i^2 \rangle$. So this reduces us to the question of why harmonic polynomials are a basis for $\mathbb{R}[x]/\langle \sum x\_i^2 \rangle$.




---


Turn $\mathbb{R}[x\_1, x\_2, \ldots, x\_n]$ into a module over itself by 
$$g(x\_1, \ldots, x\_n) \ast f(x\_1, \ldots, x\_n) := g(\tfrac{\partial}{\partial x\_1}, \ldots, \tfrac{\partial}{\partial x\_n}){\big(} f(x\_1, \ldots, x\_n) {\big)}$$
To be clear, this is a differential operator, encoded by $g$, acting on the polynomial $f$.
This action induces an inner product on the degree $d$ homogenous polynomials by $\langle g,f \rangle := g \ast f$, since $g \ast f$ is a degree $0$ polynomial and can be thought of an element of $\mathbb{R}$.


Let $J \subseteq \mathbb{R}[x\_1, \ldots, x\_n]$ be any graded ideal of $\mathbb{R}[x\_1, \ldots, x\_n]$. We write $J = \bigoplus J\_d$ for its composition into graded pieces. Let $J^{\perp} = \bigoplus J\_d^{\perp}$, where the orthogonal complement is taken with the above inner product on $\mathbb{R}[x\_1, \ldots, x\_n]\_d$. By generalities of inner products, the map $J^{\perp} \hookrightarrow R[x\_1, \ldots, x\_n] \twoheadrightarrow R[x\_1,\ldots, x\_n]/J$ is an isomorphism.


**Fact 2** We have $$J^{\perp} = \{ f : g \ast f=0 \ \forall g \in J \}.$$ Moreover, if $g\_1$, $g\_2$, ..., $g\_N$ is a list of generators for $J$, it is sufficient to impose that $g\_i \ast f=0$ for each generator $g\_i$.


Thus, the set of polynomials with $\sum (\tfrac{\partial}{\partial x\_i})^2 f =0$ is a basis for $\mathbb{R}[x]/(\sum x\_i^2)$.




---


I learned this by reading Mark Haiman's papers on "diagonal harmonics", and trying to figure out why he was calling them harmonic functions. I'll try to find some better references.


