Timeline for Spherical Harmonics
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 3, 2017 at 10:06 | comment | added | Bazin | @Venkataramana Thank you very much. | |
| Oct 30, 2017 at 15:04 | comment | added | Venkataramana | @Bazin, sorry, I forgot to complete the argument. Let $H_m$ be the space oh Harmonic homogeneous polynomials of degree $m$; we have a surjection $sym ^m (V^*)\rightarrow H_m$ with kernel consisting of polynomials divisible by $Q$. Thus, $$sym ^m(V^*)=H_m\oplus Qsym ^{m-2}(V^*).$$ By induction, every element of $sym ^m(V^*$ is a sum of polynomials of the form $q^if_i$ where $f_i$ is harmonic of degree $m-2i$. | |
| Oct 30, 2017 at 14:40 | comment | added | Venkataramana | For a reference to this, you may look at Goodman and Wallach book. | |
| Oct 30, 2017 at 14:38 | comment | added | Venkataramana | @Bazin, this identity, while simple and algebraic, is a little involved to explain. BY Zariski density, we may deal with $V=\mathbb C ^n$ instead of ${\mathbb R }^n$. Consider the quadric $Z$ defined by $Q=0$ in the projective space ${\mathbb P}({\mathbb C}^n)$. The $m$-th symmetric power $sym ^m (V^*)$ is irreducible for $GL_n(\mathbb C)$, and restricts non-trivially to the quadric $Z$; the quadric $Z=G/P$ where $G=SO(n)$ and $P$ a suitable parabolic. Then (by the Borel-Weil theorem), the restriction of $sym ^m (V^*)$ to $Z$ is irreducible. This is the space of Harmonic degree $m$ polys . | |
| Oct 29, 2017 at 15:55 | comment | added | Bazin | @Venkataramana Thanks for your comment: How do you get that identity? | |
| Oct 13, 2017 at 11:59 | comment | added | Bazin | Thanks for your comment: How do you get that identity? | |
| Oct 11, 2017 at 14:34 | comment | added | Venkataramana | The space $P$ of all polynomials on ${\mathbb R} ^n$ can be written as a direct sum $$P=\sum _{m=0}^{\infty} Q^m H$$ where $H$ is the space of harmonic polynomials and $Q=\sum x_i^2$ is the standard quadratic form on $\mathbb R ^n$. It follows, upon restriction to the unit sphere, that all polynomials on $S^{n-1}$ are sums of harmonic polynomials. Then by Stone Weierstrass, Harmonic polynomials are dense in the space of continuous functions on the sphere. | |
| Oct 11, 2017 at 13:11 | history | answered | Alexandre Eremenko | CC BY-SA 3.0 |