# What is the simplest chiral spherical harmonic?

What is the simplest (lowest order) real spherical harmonic that cannot be rotated into its own mirror image? I would like to make an illustration of a simple smooth chiral object, but cannot find a good example.

Edit: I should have clarified that I need a linear combination of spherical harmonics, not just one of them.

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I'm not sure if I understand chirality correctly, but maybe you could take a linear combination of basic harmonics composed with incommensurate rotations? – anon Aug 26 '11 at 9:58
Apologies for my incorrect answer; you're quite right that linear combinations of $l=1$ harmonics can be rotated into their mirror images. Adding an $l=0$ function won't help, so that shows you'll need at least $l=2$. An arbitrary linear combination of the nine functions up to $l=2$ is chiral (even three $l=1$ and one $l=2$ should suffice), so that raises the question how exactly you define "simplest", since "lowest order" doesn't distinguish between the various possibilities including $l=2$ functions. – joriki Aug 26 '11 at 14:10
Thanks for your comment. I don't think arbirary combinations are chiral, for exampel you can make the functions independent of z and then they are not chiral. But maybe most functions are indeed chiral. For example we can take the harmonic polynomial $x+y+z + xy$, can you prove that this is chiral? I am also interested in how to prove these things. – uekstrom Aug 26 '11 at 14:41
I've updated and undeleted my answer. About "arbitrary combinations": By "arbitrary" I meant something like "in general position". As it happens, your example is in the set of measure zero of achiral combinations :-) – joriki Aug 29 '11 at 9:20

Here's a way to determine rigorously whether a given linear combination of spherical harmonics is chiral. The functions with different values of $l$ don't get mixed under rotations. Thus, a combination is achiral if and only if there's a single rotation that rotates the mirror images of all $l$ components into their originals. So you can look at the $l$ values individually and see whether there's any intersection among the sets of rotations.
For your example, $x+y+z+xy$, the $l=1$ component $x+y+z$ is a rotated version of $Y_{10}$ oriented towards $(1,1,1)$. The rotations that rotate the mirror image of this into the original are rotations through $\pi$ about an axis perpendicular to $(1,1,1)$. (Subsequently rotating about $(1,1,1)$ doesn't add anything; it just corresponds to a different choice of axis for the first rotation.)
The $l=2$ component $xy$ is invariant under inversion, so we need to find the rotations that leave it invariant. Its angular dependence is $\sin^2\theta\sin\phi\cos\phi$, and the only rotations that leave the $\theta$ dependence invariant are ones with axes in the $x$-$y$ plane or orthogonal to it (since the poles have to end up at the poles). There's one axis in the $x$-$y$ plane that's orthogonal to $(1,1,1)$, namely $(1,-1,0)$. Coincidentally, rotating through $\pi$ about that axis corresponds to $x\to-y$, $y\to-x$, $z\to-z$, and thus works for $x+y+z+xy$. However, this is only one of two axes in the $x$-$y$ plane for which rotating through $\pi$ leaves $xy$ invariant, since rotating the rotation axis about the $z$ axis by $\alpha$ amounts to rotating the result about the $z$ axis by $2\alpha$, and that only leaves $xy$ invariant if $2\alpha$ is a multiple of $\pi$. So you can pick almost any other combination of the $l=1$ functions, as long as the associated vector isn't orthogonal to either the $z$ axis or the axes $(1,-1,0)$ and $(1,1,0)$, and add $xy$ to get a chiral combination.