Signed factors of harmonic polynomials

Let ${\rm Harm}_n^d$ be the space of real harmonic polynomials in $n$ variables, homogeneous of degree $d$. If $P\in{\rm Harm}_n^d$, then $$\left(\frac{\partial^2}{\partial x_1^2}+\cdots+\frac{\partial^2}{\partial x_n^2}\right)P=0.$$ A harmonic polynomial is not necessarily irreducible in ${\mathbb R}[X_1,\ldots,X_n]$. For instance every non-zero $P\in{\rm Harm}_2^4$ splits as the product of two quadratic forms; it turns out that none of them is positive definite. Besides, it is not two difficult to show that if $X_1^2+\cdots+X_n^2$ divides a harmonic polynomial $P$, then $P=0$. These observations lead me to me following question:

Is it possible that a non-zero harmonic polynomial (say homogeneous) factorizes $P=QR$ in ${\mathbb R}[X_1,\ldots,X_n]$, with the factor $Q$ being non-constant and positive definite (i.e. $Q(x)>0$ for every $x\ne0$) ?

I incline toward a negative answer, of course.

• Veterans of the 2005 Putnam exam may dispute your assessment of the difficulty of the result that 0 is the only harmonic polynomial divisible by $X_1^2 + \cdots + X_n^2$... This was that year's Problem B-5, and was the hardest on the exam, solved by only five of the top 200 scorers. See the Monthly article on that Putnam exam by Klosinski, Alexanderson, and Larson (Oct.2006 = Vol.**113** #8, pages 733-743). Jun 28 '11 at 17:46
• @Noam. Here is a short and easy proof. If $P\neq0$, write $P=|X|^{2k}Q(X)$, with $Q$ not divisible by $|X|^2$. The degree of $Q$ is denoted $m$. Then $$\Delta P=2k(2k+n-2+2m)|X|^{2(k-1)}Q+|X|^{2k}\Delta Q.$$ Since $\Delta P=0$, this shows that $|X|^2$ divides $Q$, a contradiction. Jun 29 '11 at 7:55
• Yes, that's nice, thanks; but still not all that easy — one must somehow come up with the formula for $\Delta |X|^{2k} Q$ (and under the constraints of a Putnam exam!). Jul 8 '11 at 2:44
• @Noam. This is not a problem. Just use $\Delta(fg)=f\Delta g+g\Delta f+2\nabla f\cdot\nabla g$. Then $\Delta |X|^\alpha$ and $\nabla|X|^\alpha$ are easy by using spherical coordinates. At last $X\cdot\nabla Q=mQ$ is the Euler's identity for homogeneous functions. Jul 8 '11 at 6:48

For $n=2$ the answer is negative for reasons so obvious that it may be considered a coincidence: a harmonic polynomial of degree $n$ is, in trigonometric form, a combination of $r^n\cos(n\phi)$ and $r^n\sin(n\phi)$, and so it always has $n$ distinct roots: $$a\cos(n\phi)+b\sin(n\phi)=0$$ means $$\tan(n\phi)=-\frac{a}{b}.$$ For $n=3$ it looks plausible too, since Legendre polynomials appear in explicit formulas, and maybeone can play with the known fact on their roots.
S. Kharlamov pointed out to me that it is a consequence of the diagonalization of the Laplacian $\Delta_S$ over the unit sphere. Its eigenvalues are the integers $\lambda_d=d(d+n-2)$, and the corresponding eigenspace $E_d$ is given by the trace over $S$ of the harmonic polynomials of degree $d$. Finally, the space of traces of polynomials of degree $\le d$ is the sum of the $E_{d-2k}$ for $k=0,\ldots,[d/2]$. This implies that ${\rm Harm}_n^d$ is orthogonal to ${\rm Hom}_n^{d-2k}$, the space of homogeneous polynomials of degree $d-2k$, whenever $k=1,\cdots,[d/2]$. The orthogonality refers to the $L^2$-scalar product over $S$: $$(f,g):=\int_Sf(x)g(x)d\sigma(x).$$
Suppose now that $P$ is harmonic of degree $d$ and it splits as $QR$, with $Q$ positive definite and non constant. Then $Q$ has degree $2k$ for some $k\ge1$. We have seen that $(P,R)=0$, which means $$\int_SQ(x)R(x)^2d\sigma(x)=0.$$ Because $Q$ has a constant sign, this implies $QR^2\equiv0$, that is $P=0$.