MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a set of $n$ real unit vectors, in 3-dimensional space.

(It is a follow up of Sets of vectors related by a rotation.)

Is there a construction providing a complete set of independent*) invariants under SO(3)?

*) I'm the most interested in independent invariants, but a solution without this assumption is of mine interest as well.

I had two initial ideas, but I run in some problems:

  • construct Grammian matrix of vectors; however, then the matrix itself depend on a particular permutation of vectors, making it impossible to use as if two set of vectors are equivalent; also, determinant of the matrix is not enough as for $n=3$ it gives only 2 out of 3 invariants,
  • calculate moments, i.e. $M_{ijk} = \langle x^i y^j z^k \rangle$; then for $i+j+k=1$ there is length squared, for $i+j+k=2$ characteristic polynomial of the inertia matrix, but I don't know how to go further.
share|cite|improve this question
This seems related to some work in mechanics on separating the "internal" degrees of systems of $n$-particles from those arising from translations and rotations. One motivation of this is the "falling cat problem": how does a cat falling upside down from a tree manage to reorient itself to fall on its feet, given that it cannot create torques by clawing the air? I don't know enough to post a real answer to your question but there is a review here: (there also seem to be some copies findable by google if you don't have access) – j.c. Nov 14 '12 at 16:46
@Piotr: I'm not sure what you mean by 'complete' and 'independent'. You should be aware that the ring $R_n$ of polynomial functions on $n$-tuples of points in $S^2$ that are invariant under permutation of the points and rotations in $S^2$ is not (except for small values of $n$) a free polynomial ring on $2n{-}3$ generators, though it contains subrings that are (which would probably satisfy your notion of 'independent' but not 'complete'). The full $R_n$ (which would probably satisfy your notion of 'complete') is usually a nontrivial quotient of a free polynomial ring on more generators. – Robert Bryant Nov 15 '12 at 13:24
@Robert: 'complete' - such that two sets of vectors are equivalent iff all invariants coincide; 'independent' - no invariant can be written as a function of other invariants; I know that especially the second thing is nontrivial, but I can live without that. – Piotr Migdal Nov 15 '12 at 16:22
@Piotr: The tag 'vector' is not at all useful here (or probably anywhere else), though maybe 'lie-groups' would be. You are asking a question about invariant theory related to compact Lie group actions. As Robert Bryant points out, this kind of problem arose in classical invariant theory. But it's certainly not trivial to answer in full detail. I guess that's why the subject died out a few times and had to be re-invented. – Jim Humphreys Nov 15 '12 at 23:26

The following construction reduces your problem it to a classical and well-studied problem in invariant theory. First, I claim that there is a natural way to interpret an $n$-tuple of points in $S^2$ as a point of $\mathbb{CP}^n$ and vice-versa. This depends on interpreting $S^2$ as $\mathbb{CP}^1$ and the classical fact that the symmetric product of $n$ copies of $\mathbb{CP}^1$ is naturally regarded as $\mathbb{CP}^n$. Moreover, this can be done in an $\mathrm{SO}(3)$-equivariant way, so that your problem becomes that of finding invariants for an action of $\mathrm{SO}(3)$ on $\mathbb{CP}^n$, a very classical problem.

To see this, remember that $\mathrm{SU}(2)$, which is the double cover of $\mathrm{SO}(3)$ acts on $\mathbb{C}^2$ in the obvious way and that this action is transitive on the $1$-dimensional subspaces $L\subset\mathbb{C}^2$, the set of which is naturally $S^2=\mathbb{CP}^1$. Moreover, the induced action on $S^2$ is that of $\mathrm{SO(3)}=\mathrm{SU}(2)/\{\pm I_2\}$. Thus, nonzero vectors in $\mathbb{C}^2$ up to nonzero complex multiples correspond to points of $S^2$, and, given any $n$-tuple of unit vectors $u_i\in S^2$ for $1\le i\le n$, they can be represented by an $n$-tuple of lines in $\mathbb{C}^2$ of the form $u_i = [v_i] = \mathbb{C}\cdot v_i\subset \mathbb{C}^2$. Now consider the symmetric product $$ v_1v_2\cdots v_n\in \mathsf{S}^n(\mathbb{C}^2) \simeq \mathbb{C}^{n+1} $$ The line spanned by this product, $[v_1v_2\cdots v_n]\in \mathbb{CP}^n$ is well-defined, independent of the choice of the $v_i$ to represent the $u_i$. Conversely, since any nonzero complex polynomial in two variables that is homogeneous of degree $n$ can be factored into linear factors uniquely up to complex multiples and permutations, it follows that any element $p\in \mathbb{CP}^n$ can be constructed this way and, moreover, the $n$-tuple of points $u_i\in S^2 = \mathbb{CP}^1$ that gives rise to $p$ is uniquely determined up to permutation.

Now, what about the action of $\mathrm{SO}(3)$? Since this is induced by the action of $\mathrm{SU}(2)$ on $\mathbb{C}^2$, and that action extends in the usual way to the symmetric power $\mathsf{S}^2(\mathbb{C}^2)=\mathbb{C}^{n+1}$, on which it is irreducible (as a complex representation), it follows that this construction is $\mathrm{SO}(3)$-equivariant.

Thus, you are reduced to finding a 'complete' (in your sense) and 'independent' (in your sense) set of invariants for the action of $\mathrm{SO}(3)$ on $\mathbb{CP}^n$ that is induced by the irreducible representation of $\mathrm{SU}(2)$ on $\mathbb{C}^{n+1}$. This is, of course, a very classical problem, about which an enormous amount has been known since the 19th century. In particular, the Clebsch-Gordan formulae can be used to give one a procedure for generating all of the polynomial invariants, but they inevitably have complicated relations among them as soon as $n$ gets bigger than $3$ or $4$.

share|cite|improve this answer
Thanks! However, then the problem goes back to my very original one (I went the other way around, hoping that it actually simplifies the problem). – Piotr Migdal Nov 16 '12 at 1:22
@Piotr: Perhaps you could say why you thought that your formulation could circumvent classical invariant theory. (If I had known that you already knew the classical problem, I wouldn't have spent so much space explaining it.) Also, perhaps you could say what you find unsatisfactory about the algorithm for constructing the ring of invariants using CIT and the Clebsch-Gordan formulae. It's true that explicit answers for each $n$ require further work than CIT, but the answers are known to be complicated as $n$ increases, and no reformulation will do away with that. – Robert Bryant Nov 16 '12 at 13:00
@Robert The story is that I'm working on quantum information. There a relation of permutation-symmetric state for n qubits (n two-dimensional complex vectors) to n points on 2d real sphere is called "Majorana's stellar representation" (it is basically: sym. subspace for d=2 -> polynomial -> factorization). (I should have said what I know, excuse me for that.) Such relation is specific to d=2 case; so I thought that maybe in this specific case there is an easier way to calculate invariants (clearly, there is at least an easy way to find some invariants). – Piotr Migdal Mar 18 '13 at 13:42
@Piotr: There is an explicit, easily implemented algorithm that generates the space of homogeneous polynomial invariants of any given degree. (The dimension of this finite-dimensional vector space can be found without much difficulty; there is a generating function which is the Poincaré-Hilbert polynomial of the full ring that can be computed fairly explicitly as a rational function). However, determining the algebraic relations among these invariants is nontrivial when $n$ is sufficiently large, and I'm not aware of any simple way of doing it. – Robert Bryant Mar 18 '13 at 14:54
@Piotr: degree bounds are not easy either. For binary form CIT there is an old one by Jordan see: I don't know if it has been improved recently. – Abdelmalek Abdesselam Mar 18 '13 at 20:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.