From: Urs Schreiber (Urs.Schreiber@uni-essen.de)
 Subject: Lie derivative of spinors
 Newsgroups: sci.physics.research
 Date: 2003-02-21 23:01:04 PST
 

Since in another thread the formula L_v = {i_v,d} for the Lie
derivative on differential forms is being discussed, I was
reminded of what

Alfred Einstead (whopkins@csd.uwm.edu) wrote a while ago:

>baez@galaxy.ucr.edu (John Baez) wrote:
>> The basic idea is this: there's no god-given way to just take a smooth
>> manifold and define spinor fields on this manifold!  To define spinor
>> fields you first need to pick a "spin structure".
>
>... which gets to a related question I've been meaning to ask for
>a while: what's the Lie derivative L_X psi of a spinor field psi
>by a vector X?

Years before there was a related question on this group, which
apparently remained unanswered (but certainly not unanswerable
;-):

From: Andrea Giulio Livotto (livotto@maths.ox.ac.uk)
 Subject: Lie derivative on spinors
 Newsgroups: sci.physics.research
 Date: 1997/07/31

>
>Does anybody know how to derive ( or some bibliography on) the following
>expression for the Lie derivative on spinors?
>
> L_{K} = i (  K_{j} D_{j} + (1/8) [\Gamma^{j}, \Gamma^{l}] (D_{j}K_{l})  )
>
>where: * \Gamma^{j} are the generators of the Clifford algebra;
>       * K is a Killing vector on the base manifold, with components
>         K_{j};
>       * the summation convention applies (summing over indices repeated
>         twice).
>
>This expression appears in a paper of E. Witten, Fermion Quantum Numbers
>in Kaluza-Klein theories (1983).

So the short answer is: "In general there is no Lie derivative
on spinors, but along a Killing vector field there is."

This was of course already implicit in Arkadiusz Jadczyk's
answer to the first question:

Arkadiusz Jadczyk (ark@cassiopaea.org) wrote:

>On Thu, 19 Dec 2002 02:49:10 GMT, whopkins@csd.uwm.edu (Alfred Einstead)
>wrote:
>> The basic idea is this: there's no god-given way to just take a smooth
>> manifold and define spinor fields on this manifold!  To define spinor
>> fields you first need to pick a "spin structure".
>
>... which gets to a related question I've been meaning to ask for
>a while: what's the Lie derivative L_X psi of a spinor field psi
>by a vector X?
>
>
>The problem with spinor Lie derivative comes from the fact that
>if X is not a Killing vector field, then transport of an orthonormal
>frame at x(t) back to x(0) along the flow of X is not an orthonormal
>frame.
[...]

Since Arkadiusz Jadczyk did not mention the formula, I here
take the opportunity to derive it and relate it to the formula
L_v = {i_v,d} mentioned above. (I think I know what I am
doing, but in case I screw up there will certainly be somebody
to correct me.)

So let me start with differential forms. This is maybe not the
most immediate approach to the question, but I think it
provides some interesting insight.

On differential forms, we have, as discussed currently in the
thread "Gravity from Weitzenboeck", the form creators and
annihilators a*^n and a^n. Their canonical anticommutation
relation {a*^n,a^m} = g^nm is isomorphic to the two
anticommuting Clifford algebras generated by

 y^n = a*^n + a^n

and

 Y^n = i(a*^n - a^n)

which satisfy

 {y^n,y^m} = 2 g^nm

 {Y^n,Y^m} = 2 g^nm

 {y^n,Y^m} = 0 .

As is well known, a differential form may be thought of as the
product of two spinors. To see this, it is sufficient to note
that the "symbol map", sends any element C of the above
Clifford algebra to the differential form C|0>, where |0> is
the constant unit 0-form, e.g.

 y^[n y^m] |0> = a*^[n a*^m]|0> = dx^[n /\ dx^m],

and that, for instance,

  Y^[n Y^m] y^[k y^l] |0>
= y^[k y^l] Y^[n Y^m] |0>
= y^[k y^l] y^[m y^n] |0>,

(here the square brackets are clumsy ASCII for antisymmetrized
indices)
i.e. that the Y^n may be thought of as acting from the
*right*, while the y^n act from the left on the Clifford
element that corresponds to a differential form (or the other
way round, if desired).

Now let's check how the Lie derivative on forms looks like
when we think of forms as products of spinors:

First of all we have

 L_v
 = {d,i_v}
 = {a*^n nabla_n , v_n a^n}
 = v^n nabla_n + (nabla_n v_m)a*^n a^m .

As for the notation I refer to the thread "Gravity from
Weitzenboeck". From the discussion there one in particular
finds that

nabla_n
= partial_n +       omega_nab a*^a a^b
= partial_n + (1/4) omega_nab (y^a y^b + Y^a Y^b) ,

where in the last line I have rewritten the form creators and
annihilators in terms of Clifford generators. The y and Y are
"decoupled" in this expression, and a little reflection shows
that this nabla_n is in fact the sum of the two spinor
covariant derivatives nabla^S_n on the two spinor bundles
whose product makes up the form bundle:

 nabla^S_n = partial_n + (1/4) omega_nab y^a y^b  .

This is one way to see that a covariant derivative makes sense
for both, forms and for spinors. That the same is not true in
general for the Lie derivative, however, can be seen by
similarly replacing creators/annihilators by Clifford
generators in the above expression for the Lie derivative
along v:

L_v
= v^n nabla_n + (nabla_n v_m)a*^n a^m
= v^n nabla_n + (nabla_n v_m)(1/4)(y^n - i Y^n)(y^m + i Y^m) .

As is obvious, on the right there will in general be a term
"foo" that couples "left" and "right" Clifford generators:

 foo = (nabla_n v_m)(i/4)(y^n Y^m + y^m Y^n) .

These are the terms that prevent us from interpreting L_v as
acting on the two spinor bundles (the left and the right one)
seperately. Hence they prevent us from splitting L_v into two
parts that act on the left and the right spinor bundle
seperatly, as was well possible for the covariant derivative.

Everything would be fine, however, if foo = 0. Noting that the
combination of Clifford generators on the right of foo is
symmetric in the indices n and m, we see that foo vanishes iff

 nabla_n v_m + nabla_m v_n = 0 ,

i.e.

 foo = 0  <=>   v is Killing !

In that case the above formula for L_v reads
 

 L_v = v^n nabla_n + (nabla_n v_m)(1/4)(y^n y^m + Y^n Y^m)
                                        (iff v is Killing) .

Note that now the term  nabla_n v_m  is antisymmetric in n,m.
To emphasize this one can equivalently write

L_v
= v^n nabla_n + (nabla_n v_m)(1/8)([y^n, y^m] + [Y^n, Y^m])
                                         (iff v is Killing).

It is now obvious that this is the sum of two analogous
operators L^S_v that induce the Lie derivative along the
Killing vector v on each of the two spinor bundles:

 L^S_v = v^n nabla^S_n + (nabla_n v_m)(1/8)[y^n, y^m] .

This is the formula given by Andrea Giulio Livotto above,
citing a paper by Witten.

Another place where I have seen this fomula mentioned is on p.
195 of

R. W. Tucker, A Clifford Calculus for Physical Field
Theories,  in
J. Chisholm and A. Common (eds.), Clifford Algebras and their
Applications in Mathematical Physics (1986), D. Reidel
Publishing, pp. 177-199

But probably it is given in most relevant books, and I just
have not realized it so far.
 

--
Urs.Schreiber@uni-essen.de