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**Division Algebras,
Lie Algebras, Lie Groups and Spinors**

Let A be a pure
imaginary quaternion as on the previous page, so U = exp(A) is an element

of SU(2) = 3-sphere. Suppose this SU(2) acts on some space, **M**.
Well, does U act on **M** from

the left or right? It can do either, and it matters. **M**
has a copy of **Q** in it, and it's this copy

that receives the action of U.

Given A = A^{k}q_{k},
sum k=1,2,3,

define A_{L} = A^{k}q_{Lk}, and U_{L} = exp(A_{L});

define A_{R} = A^{k}q_{Rk}, and U_{R} = exp(A_{R}).

So U_{L}[**M**]
= U**M**, and U_{R}[**M**]
= **M**U, and both of these are SU(2)
actions, but they're distinct

SU(2)'s, and the left action SU(2) commutes with the right action SU(2).

By the way, using
U_{L} and U_{R} we can construct a copy of SO(3), the automorphism
group of

**Q**. In particular, if X is in **Q**, then U_{L}U^{-1}_{R}[X]
= UXU^{-1} leaves the real part of X alone and

performs an SO(3) rotation on the imaginary 3-dimensional part (the reader should
see

this action is obviously a **Q** automorphism).

**Q** is also
a Clifford algebra, and an integral part of Clifford algebra theory. However,
Clifford

algebras also act on some space (which are called spinor spaces), so we should
once again

specify the direction of action. Let CL(p,q) be the Clifford algebra of the
real pseudo-orthogonal

space with signature p(+), q(-) (see the
book). Then **Q**_{L} is isomorphic to CL(0,2), a 1-vector

basis being {q_{L1}, q_{L2}}, and the sole 2-vector basis element:
q_{L3} = q_{L1}q_{L2}. Likewise **Q**_{R}
is

isomorphic to CL(0,2).

What if we allow
elements of both **Q**_{L} and **Q**_{R}? We'll denote
by **Q**_{A} the algebra of combined

left/right actions of **Q** on itself. This algebra is isomorphic to **R**(4)
(hence also to CL(3,1) and

CL(2,2)). But this is a path down which I haven't the patience at present to
trod. Time for

octonions.

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