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.

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    $\begingroup$ 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). $\endgroup$ Jun 28, 2011 at 17:46
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    $\begingroup$ @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. $\endgroup$ Jun 29, 2011 at 7:55
  • $\begingroup$ 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!). $\endgroup$ Jul 8, 2011 at 2:44
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    $\begingroup$ @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. $\endgroup$ Jul 8, 2011 at 6:48

3 Answers 3


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 $P=0$.

Of course, instead of integrating over the unit sphere, one may integrate against the standard Gaussian measure. A classical trick in several topics, including random matrices theory.


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.

In general, it seems that the answer is negative too but it is not clear at all what sort of statement to attempt to prove. Positive definiteness of polynomials is a subtle thing (cf. Hilbert's 17th problem etc.)...

  • $\begingroup$ +1. I forgot to mention that I had this case solved too. $\endgroup$ Jun 13, 2011 at 6:40

It seems that Dennis Serre has already addressed the question. I just want to point out that the result is a special case of a theorem of Brelot and Choquet.

Suppose that $h$ is a (not necessarily homogeneous) harmonic polynomial of degree $m>1$ such that $h = h_1 \cdot h_2$ with $h_1, h_2$ each having degree $<m$. Then $h_1, h_2$ must change sign.

See Exercise 17.4 (and the notes at the end of that chapter) in the book ``Linear Holomorphic PDE'' by Khavinson and Lundberg MR3821527.


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