TITLE: Spinor space to Euc. vector space: does there exist a universal bilinear map?
QUESTION [7 upvotes]: 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  $(q_1, q_2)  \mapsto q_1 \bar q_2$.

REPLY [9 votes]: 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 $\mathbf{4}\to\mathbf{5}$.