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.'', not`

the 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 $(q_1, q_2) \mapsto q_1 \bar q_2$.

This post imported from StackExchange MathOverflow at 2015-02-20 16:59 (UTC), posted by SE-user Richard Montgomery