Let $S$ be a spin representation of the Euclidean spin group $Spin(d)$ and let ${\mathbb R}^d$ be Euclidean $d$-space with $Spin(d)$ action on it in the canonical way, via the 2:1 cover to $SO(d)$. (I am being careful here: A spin rep.'', notthe spin rep.'')

Is there, for all $d$, an onto quadratic $Spin(d)$-equivariant map $S \to {\mathbb R}^d$? If so, is there a `universal' ($d$-independent) construction of this map?

MOTIVATION: For $d=2, 3$ I know these maps. They are famous in celestial mechanics and yield the
standard regularizations of the Kepler problem, or, what is the same, of binary collisions in the classical N-body problem. They turn Kepler for negative energies into a harmonic oscillator.

Case $d=2$. I take $Spin(2)$ to also be $S^1$, but wrapped `twice' around $S^1 = SO(2)$. $S = {\mathbb C}$. The quadratic map is $w \to w^2$. This is the Levi-Civita regularization.

Case $d= 3$. This is the standard Hopf map ${\mathbb C}^2 \to {\mathbb R}^3$, or if you prefer, from the quaternions ${\mathbb H }$ to ${\mathbb R}^3$, sending $q$ to $q k \bar q$. The astronomers call this Kuustanheimo-Steifel regularization.

WHERE I'VE LOOKED SO FAR: I tried to make sense out of Deligne's discussion on spinors in the AMS two-volume set from some Princeton year on string theory from a decade or so ago. I understand that over the complexes, there is either exactly one or exactly two spin representations, depending on the parity of $d$. So even there , we don't get a `universal' d-dependent map. Over the reals things decompose in a rather complicated dimension dependent way (mod 8 probably) and there is no clear choice. I also looked in Reese Harvey's book which I find too baroque and signature depend to penetrate.

Case $d=4$. Here I am not sure. But I know $Spin(4) = SU(2) \times SU(2)$ which I can think of as two copies of the unit quaternions, each acting on ``its own'' ${\mathbb H}$. I guess in this case I better take $S = {\mathbb H} \times {\mathbb H}$ Then I get the desired quadratic map ${\mathbb H} \times {\mathbb H} \to {\mathbb H} = {\mathbb R}^4$ as $(q_1, q_2) \mapsto q_1 \bar q_2$.


1 Answer 1


The answer is no.

For $Spin(5)\simeq USp(4)$, the spin representation of $Spin(5)$ is the defining representation $\mathbf{4}$ of $USp(4)$. The tensor product $\mathbf{4}\otimes \mathbf{4}$ decomposes as $$\mathbf{1}\oplus\mathbf{5}\oplus\mathbf{10}.$$ The vector representation of $SO(5)$ is this $\mathbf{5}$, but it's in the antisymmetric part of this tensor product decomposition: $$\wedge^2\mathbf{4} = \mathbf{1}\oplus\mathbf{5}$$ The $\mathbf{1}$ is there, because of the definition of $USp(4)$.

So, there's no quadratic $Spin(5)$-equivariant map $\mathbf{4}\to\mathbf{5}$.

  • $\begingroup$ Thank you very much Yuji. Could you give me a reference for what $USp(4)$ means and for the isomorphism you began with? -Richard $\endgroup$ Jul 19, 2011 at 4:27
  • $\begingroup$ @Richard Montgomery: Sometimes it is useful to denote $Sp(n)$ as $Usp(2n)$, for any integer $n$. The isomorphism between $Spin(5)$ and $Sp(2)$ is reviewed in section 2 of the following paper by Salamon, calvino.polito.it/~salamon/T/tour.pdf. $\endgroup$
    – agt
    Jul 23, 2011 at 15:56

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