# Basis for the space of Harmonic homogeneous polynomial in N variables.

Hello,

Does someone know an explicit basis of the space of harmonic homogeneous polynomial in N variables.

When $N=3$, if I'm not mistaking Legendre polynomial allow to write an explicit basis.

Is there a known explicit basis when $N > 3$ ?

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I really want a basis for harmonic homogeneous polynomials, by harmonic I mean in the kernel of the euclidean laplacian. And, such a polynomial restrict to the sphere give a eigenvector for the spherical laplacian, right ? But, I don't get your "If the former, apply the laplacian on monomials to see what happens and why I think that you may not really want to know about harmonic homogeneous polynomials." Do you mean that such a basis will be awfully too complicated ? –  Ludo Marquis Oct 20 '11 at 12:45
I think I misunderstood the question. My comment does not apply and I will delete it. Sorry. –  José Figueroa-O'Farrill Oct 20 '11 at 15:55
The Gegenbauer Polynomials are the generalization of associated Legendre Polynomials to higher dimensions: mathworld.wolfram.com/GegenbauerPolynomial.html –  Yoav Kallus Aug 3 '13 at 4:25

Let $K$ denote the Kelvin transform, and let $|\alpha|:=\sum_{j=1}^n\alpha_j$ denote the weight of the multi-index $\alpha\in\mathbb{N}^n$. Then, an explicit base for the space of homogeneous harmonic polynomials in $n$ variables and degreee $m$, $\mathcal H^m:=\mathcal H^m[x_1,\dots ,x_n]$, is $$\big\{ K\big(\partial^\alpha\|x\|^{2-n}\big) : |\alpha|=m, \alpha_n\leq 1 \big\}.$$ Indeed, an easy combinatorial computation shows it has the right cardinality $$\operatorname{card}\{\alpha\in\mathbb{N}^n : |\alpha|=m, \alpha_n\leq 1 \}= {n+m-1\choose n-1} - {n+m-3\choose n-1} = \operatorname{dim}\mathcal H^m,$$ (the latter dimension being already known from linear algebra considerations on the operator $\Delta$). On the other hand, one can verify that it spans the same linear space as the analogous set without the constraint on $\alpha_n$: $$\{ K\big(\partial^\alpha\|x\|^{2-n}\big): |\alpha|=m\},$$ which is the whole space $\mathcal H^m$. So it is a base.