Discussion:
Dirac operators on Lorentzian manifolds
Kasper J. Larsen
2005-04-20 08:42:15 UTC
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Hi all,

I'm a bit confused. Can one define a Dirac operator D on a Lorentzian
manifold in the same way as one defines D on a Riemannian manifold? If
yes, how do the definitions differ? (I am already acquainted with the
Riemannian manifold definition.)

Moreover, assuming that a Lorentzian Dirac operator D can be defined,
what is its physical significance? If you substitute this D for the
"flat space-time" D in the ordinary Dirac equation, do you get a
"curved space-time" equation describing spin-œ-particles?

Kasper Jens Larsen
Urs Schreiber
2005-04-20 09:46:38 UTC
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Post by Kasper J. Larsen
Can one define a Dirac operator D on a Lorentzian
manifold in the same way as one defines D on a Riemannian manifold?
Yes.
Post by Kasper J. Larsen
If yes, how do the definitions differ? (I am already acquainted with the
Riemannian manifold definition.)
The definition in terms of Clifford algebra is the same in both cases. The
only thing that changes is

i) the signature of the Clifford algebra, which leads to different matrix
representations of the algebra

ii) the definition of the scalar product on spinors

So in every case the Dirac operator comes from applying a connection
composed with the "symbold map", which sends forms to Clifford generators.

So locally in components it always reads

D = \gamma^\mu \nabla_\mu .

Which connection nabla precisely you want to use depends on a couple of
details. For Dirac fermions in 4 dimensions you would just use the "spin
connection" which is essentially nothing but the Levi-Civita connection on
the Clifford bundle.

If you are unfamiliar with this have a look for instance at the discussion
in chapter 19.5 of the book

Th. Frankel
"The Geometry of Physics".

or, if you like lots of indices, section 12.5 (pp. 365) in the book

S. Weinberg
"Gravitation and Cosmology".

For a more mathematical discussion see for instance chaper 3 in the book

N. Berline, E. Getzler & M. Vergne
"Heat Kernels and Dirac Operators"

If you are interested in the field theory of fermions on curved spaces you
have to know about Weyl fermions and how they are treated in the physics
literature in terms of "2-component spinors". This is discussed in great
detail for instance in chaper 13 of

Robert Wald
"General Relativity"
Post by Kasper J. Larsen
Moreover, assuming that a Lorentzian Dirac operator D can be defined,
what is its physical significance? If you substitute this D for the
"flat space-time" D in the ordinary Dirac equation, do you get a
"curved space-time" equation describing spin-œ-particles?
Yes, sure. For instance you might want to know how fermions behave in black
hole backgrounds.

See for instance the paper

C. Doran, A. Lasenby, S. Dolan, I. Hinder,
Fermion absorption cross section of a Schwarzschild black hole
http://arxiv.org/abs/gr-qc/0503019

for a nice application of the single-particle Dirac equation for a curved
Lorentzian background.

These authors from Cambridge, incidentally, are proponents of a certain
school of thought that has a very pedagogical approach to Clifford algebras
and spinors. They call that "Geometric Algebra". It's the same formalism
that everybody else uses, just expressed with a different emphasis on
different aspects. Have a look at their website

http://www.mrao.cam.ac.uk/~clifford/index.html .

There is much, much more on spinors in curved space available. Maybe have a
look at the list of references in this thesis:

Fermion Quantum Field Theory in Black-hole Spacetimes

I hope these references suffice for a start.
Jack Tremarco
2005-04-20 17:25:19 UTC
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Post by Kasper J. Larsen
Can one define a Dirac operator D on a Lorentzian
manifold in the same way as one defines D on a Riemannian manifold?
Yes.
This is true if you ignore non-compactness issues, which can be quite
serious. In generic time-dependent backgrounds the honest answer is
closer to a "no", at least if you demand mathematical rigor.

[Moderator's note: It depends on what sort of rigor you want to have.
The spinors (and the Dirac equation) on a curved manifold is defined
with the help of the vielbein (tetrad) that allows one to treat
the curved space locally in exactly the same way as the Euclidean
or Minkowski space. It's easy to define the Dirac operator. LM]
Urs Schreiber
2005-04-20 18:12:40 UTC
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Post by Jack Tremarco
Post by Kasper J. Larsen
Can one define a Dirac operator D on a Lorentzian
manifold in the same way as one defines D on a Riemannian manifold?
Yes.
This is true if you ignore non-compactness issues, which can be quite
serious. In generic time-dependent backgrounds the honest answer is
closer to a "no", at least if you demand mathematical rigor.
The only global condition that you have is the obvious one, that your
manifold admits spinors at all, globally, which means that it admits a spin
bundle which means that it is spin. This same condition is there for
Riemannian signature. So the answer is indeed Yes.

It would be strange if otherwise, given that we do live in a Lorentzian
spacetime and we do observe spinors.

Note that the question "Can one define a Dirac operator D on a spin bundle
over M" is different from the question "Can we make sense of quantum field
theory of fermions on M." QFT on curved spaces is hard.
Thomas Mautsch
2005-04-23 10:11:10 UTC
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Post by Urs Schreiber
Post by Jack Tremarco
Post by Kasper J. Larsen
Can one define a Dirac operator D on a Lorentzian
manifold in the same way as one defines D on a Riemannian manifold?
Yes.
This is true if you ignore non-compactness issues, which can be quite
serious. In generic time-dependent backgrounds the honest answer is
closer to a "no", at least if you demand mathematical rigor.
Jack, could you be a little bit more specific about what you mean here?
Post by Urs Schreiber
The only global condition that you have is the obvious one, that your
manifold admits spinors at all, globally, which means that it admits a spin
bundle which means that it is spin. This same condition is there for
Riemannian signature. So the answer is indeed Yes.
I am sorry, Urs, but this statement looks like word-juggling to me. -
If you don't specify how you define these terms,
how can we believe you that the whole construction is so "obvious"?

I would agree with Jack, that there are certain difficulties
with the definitions of spin structures, spinors, and Dirac operators
in pseudo-Riemannian spaces.

I had to look this up in the book:

"Spin-Strukturen und Dirac-Operatoren
ueber pseudoriemannschen Mannigfaltigkeiten"
by Helga Baum (Teubner-Verlag, Leipzig, 1981),

The results of this book have also appeared in English
without proofs in:

* Dlubek, Helga
Spinor-structures and Dirac-operators on pseudo-Riemannian manifolds.
Proceedings of the Conference on Differential Geometry and
its Applications (Nove Mesto na Morave, 1980), pp. 17--23,
Univ. Karlova, Prague, 1982.
* Baum, Helga
Spinor structures and Dirac operators on pseudo-Riemannian
manifolds. Bull. Polish Acad. Sci. Math. 33 (1985), no. 3-4, 165--171

Here are some facts from the book above:

First of all, the existence of pseudo-Riemannian structures
on manifolds is topologically restricting. -
For the existence of a Lorentz structure on a manifold M,
there has to exist a nowhere-vanishing line field on M.
Similar for pseudo-Riemannian metrics
of index (k,n-k) on an n-dimensional manifold,
there has to exist a splitting of the tangent bundle TM
into the direct sum
of a k-dimensional subbundle V and an (n-k)-dimensional subbundle W.
The first Stiefel-Whitney classes w_1 of these bundles,
which describe their orientability,
are invariants of the underlying pseudo-Riemannian structure.

Now, a metric of index (k,n-k) on M
gives us the bundle of orthonormal frames over M,
a principal bundle with structure group O(k,n-k).

The construction of Clifford algebra and Pin group Pin(k,n-k),
which is a natural double cover of O(k,n-k),
is very much like in the Riemannian case,
and there is plenty of literature about it, like, e.g.:

* Lawson, Michelsohn
Spin geometry. Princeton University Press, 1989.
Chapter I.
* Harvey, Spinors and calibrations. Academic Press, Boston, MA, 1990.

A spin structure over the manifold M can be defined as reduction
of the principal O(k,n-k)-bundle of orthonormal frames
to a principal Pin(k,n-k)-bundle
which is compatible with the natural projection from Pin to O.

Already here appears the first slight complication:
While in Riemannian geometry the condition on a manifold
to carry a spin structure is that the second Stiefel-Whitney class
w_2(TM) be zero, in non-Riemannian geometry
the condition is connected to the splitting of M in space- and time-like
directions, TM = V + W, and it becomes:

w_2(TM) = w_1(V) u w_1(W).

This looks like a first reason to restrict oneself to
manifolds that carry space- and time-orientations,
and we have not yet started to define the spin group...

O.K., given a spin structure,
we can define the associated spinor bundle,
lift the Levi-Civita connection
to a compatible connection on the spinor bundle,
and define the Dirac operator.

Problem is, that all we can do in this setting is "geometric"
(we can e.g. define parallel spinors and harmonic spinors,
spinors annihilated by the Dirac operator)
but nothing "analytic", because for analytical considerations
we need the Dirac operator to act on a (complex!) *Hilbert space*. ...
Post by Urs Schreiber
It would be strange if otherwise, given that we do live in a Lorentzian
spacetime and we do observe spinors.
Note that the question "Can one define a Dirac operator D on a spin bundle
over M" is different from the question "Can we make sense of quantum field
theory of fermions on M." QFT on curved spaces is hard.
Even to do simple spectral theory for the Dirac operator,
we need a Hilbert space for it to act on.

Now Harvey goes in his book a great length
to prove that there exists a natural inner product on (real) spinors
that is compatible with the action of the Clifford algebra,
so that the Dirac operator would be *formally* self-adjoint
with respect to this product.
*BUT* this product may take values in R, C, or H,
and it may be symmetric or anti-symmetric,
and after complexification of the spinor bundle
(which we *must* inevitably do
to get a *complex* Hilbert space of spinor sections),
the inner product we obtain will - in the non-Riemannian case -
not be positive definite.

Now, in the later chapters of Helga Baum's book
there is a description for the definition of a "natural"
positive-definite hermitean product on the complex spinor bundle,
but to construct this product,
one has to fix a time- and a space-orientation on M,
i.e. one has to break the non-compact symmetry group
to the compact group SO(k)xSO(n-k).

But even then, the Dirac operator will *not* be essentially self-adjoint
in the Hilbert space defined by this scalar product,
only its "real and imaginary part" will be essentially self-adjoint;
and this makes that the spectrum of the Dirac operator
will not only consist of pure eigenvalues. -
There will also be essential spectrum and even rest spectrum.

Well, a Dirac operator with rest spectrum might
from the point of view of physics sound like nonsense,
but who can say, where we went wrong??
Urs Schreiber
2005-04-24 18:37:47 UTC
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Post by Thomas Mautsch
Post by Urs Schreiber
Post by Jack Tremarco
Post by Kasper J. Larsen
Can one define a Dirac operator D on a Lorentzian
manifold in the same way as one defines D on a Riemannian manifold?
Yes.
This is true if you ignore non-compactness issues, which can be quite
serious. In generic time-dependent backgrounds the honest answer is
closer to a "no", at least if you demand mathematical rigor.
Jack, could you be a little bit more specific about what you mean here?
Post by Urs Schreiber
The only global condition that you have is the obvious one, that your
manifold admits spinors at all, globally, which means that it admits a
spin bundle which means that it is spin. This same condition is there for
Riemannian signature. So the answer is indeed Yes.
I am sorry, Urs, but this statement looks like word-juggling to me. - If
you don't specify how you define these terms, how can we believe you that
the whole construction is so "obvious"?
My apologies if what I wrote looked like word-juggling. I am not sure why it
did to you, but that was certainly not my intention. I also don't see a
disagreement between what I wrote and what you write. But let's see. We'll
sort that out.

First of all let's distinguish between various flavors of the question that
is being discussed. Originally there was the question under what conditions
we can have a Dirac operator on a Lorentzian manifold and how it is defined.

Due to certain applications that people have in mind, this question will
immediately make them think of the question, when we can have a Dirac
operator and in addition have this operator satisfy some desired list of
properties.

I replied to the first question that you construct a Dirac operator on a
Lorentzian manifold in precisely the same way as you do on a Riemannian
manifold. In order for that to be meaningful we need to have a manifold
which admits a spin bundle. The Dirac operator acts on spinors (sections of
a spin bundle) and hence it is "obvious" (and please apologize if you find
this usage of the word obvious too vague) that we need the condition that
there are spinors in the first place on our manifold. If that is the case,
we can define the Dirac operator.

You don't seem to disagree with this stament. What you do point out is what
the conditions of having

1) Lorenttian manifold
2) with spin structure

mean in detail.

I would like to point out that once we have this we can write down
expressions like

\bar\psi D \psi

for \psi a spinor field (a section of the spin bundle) and construct an
action principle for this. For instance we could consider the field theory
describing those spinors coupled to general relativity, which would read

\int vol ( R + \bar \psi D\psi ) .

Or we could write down an action for some flavor of supergravity, for
instance. People do this all the time.

You point out that we will in general not have the additional properties
that there is a Hilbert space of spinors with a positive semi-definite inner
product (=scalar product) with respect to which the Dirac operator is

These problems are related to the fact that relativistic wave equations are
hard and under certain conditions impossible to be interpretable as
describing single particle dynamics.

For illustration purposes, let me mention some aspects of this for the case
where the Dirac operator under consideration is that on the exterior bundle,
i.e. the Dirac-Kaehler operator

D = d + del

where d is the exterior derivative for our manifold and del = \pm *d* is its
(formal, possibly) adjoint with respect to the Hodge inner product

<a|b> = \int a /\ * b

on differential p-forms a,b.

This case features most of the issues which we discussed by avoiding some
less essential technicalities.

First of all, this Dirac operator exists on every Riemannian or
pseudo-Riemannian manifold, no matter what.

We also always have an inner product <|>, which in the Lorentzian case is
indefinite. In order to really define it we need to restrict to sections of
the exterior bundle which are square integrable. Depending on the context
that one is dealing with this can be done in several ways.

Next one might ask under which conditions we can get from D a 'spatial'
Dirac operator D_s together with a true Hilbert space (H,[,]) for sections
of the exterior bundle restricted to some spatial hyperslice of our
manifold, such that the original
Dirac equation D psi = 0 becomes a Schroedinger-Dirac equation

i d_t psi = D_s psi

and hence can be (more or less) interpreted as a 1-particle theory.

Once necessary condition for this to work is that our spacetime is
stationary, i.e. that it has a timelike Killing vector.

Calling this Killing vector v we may be able to use the Clifford element
c(v) as an operator that turns our indefinite inner product space into a
Krein space and construct a true Hilbert space for D_s from it.

And so on and forth.

Agreed?
Volker Braun
2005-04-26 15:47:35 UTC
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I just want to point out one detail. Lets take the usual 9+1 (or 3+1 for
historical reasons :-) dimensions, and split the tangent bundle TM=V+W
into time and space directions. If the "time-direction" line bundle W is
not trivial, then we are in such deep troubles that closed timelike curves
seem harmless in comparison. So for any physical spacetime, w_1(W) better
be zero.
Post by Thomas Mautsch
While in Riemannian geometry the condition on a manifold
to carry a spin structure is that the second Stiefel-Whitney class
w_2(TM) be zero, in non-Riemannian geometry
the condition is connected to the splitting of M in space- and time-like
w_2(TM) = w_1(V) u w_1(W).
And for any physically sensible M, that just reduces to w_2(TM)=0.
Volker
Robert C. Helling
2005-04-27 07:44:54 UTC
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Post by Volker Braun
If the "time-direction" line bundle W is
not trivial, then we are in such deep troubles that closed timelike curves
seem harmless in comparison. So for any physical spacetime, w_1(W) better
be zero.
Why? Could you please expand this? What can you say about solution spaces
of you favourite wave equations?

Robert
--
.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oO
Robert C. Helling School of Science and Engineering
International University Bremen
print "Just another Phone: +49 421-200 3574
stupid .sig\n"; http://www.aei-potsdam.mpg.de/~helling
Volker Braun
2005-04-27 15:14:01 UTC
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I was not thinking of anything this complicated, certainly I did not
construct any solutions to the Dirac equation on such a manifold.

But if w_1(W) <> 0, then there is a closed loop in your manifold along
which the time-forward-direction flips. So how would you go about defining
past, future, and asymptotic states.

For example (ignoring spinors), a Moebius strip (S^1 x R^1)/Z_2. You take
local coordinates (t,x) where x runs along the circle and t is
perpendicular to it, pointing in the noncompact direction. The metric is
diag(-1,+1) such that x is space and t is time. Please define scattering
amplitudes :-)

Volker
Post by Robert C. Helling
If the "time-direction" line bundle W is not trivial, then we are in
such deep troubles that closed timelike curves seem harmless in
comparison. So for any physical spacetime, w_1(W) better be zero.
Why? Could you please expand this? What can you say about solution
spaces of you favourite wave equations?
Robert C. Helling
2005-04-28 12:15:57 UTC
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Post by Volker Braun
I was not thinking of anything this complicated, certainly I did not
construct any solutions to the Dirac equation on such a manifold.
But if w_1(W) <> 0, then there is a closed loop in your manifold along
which the time-forward-direction flips. So how would you go about defining
past, future, and asymptotic states.
As you know, microscopic physics is invariant under time reversal
(well, most of it) so I don't see any immerdiate problems here.
Post by Volker Braun
For example (ignoring spinors), a Moebius strip (S^1 x R^1)/Z_2. You take
local coordinates (t,x) where x runs along the circle and t is
perpendicular to it, pointing in the noncompact direction. The metric is
diag(-1,+1) such that x is space and t is time. Please define scattering
amplitudes :-)
But is that more complicated than scattering in other spaces that are
not asymptotically flat (or AdS or something similar with good
asymptotic regions)?

Robert
--
.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oO
Robert C. Helling School of Science and Engineering
International University Bremen
print "Just another Phone: +49 421-200 3574
stupid .sig\n"; http://www.aei-potsdam.mpg.de/~helling
Volker Braun
2005-04-28 19:23:28 UTC
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Post by Robert C. Helling
For example (ignoring spinors), a Moebius strip (S^1 x R^1)/Z_2 [...]
But is that more complicated than scattering in other spaces that are
not asymptotically flat
Well, the Moebius strip is completely flat. The time-slices are
compact, but you could easily extend that by adding more flat,
noncompact space-direction(s). I don't see any problem with finding
asymptotic regions. Apart from being unable to decide whether they are
past or future, of course.

Volker
r***@yahoo.ca
2005-05-02 19:50:29 UTC
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Post by Jack Tremarco
This is true if you ignore non-compactness issues, which can be quite
serious. In generic time-dependent backgrounds the honest answer is
closer to a "no", at least if you demand mathematical rigor.
I have an idea about how to generalize the Dirac equation to
4-dimensional curved space-times which have a property P such that all
harmonic charts which verify P are related by Lorentz tranformations.

In one harmonic chart which verifies P, set w: T(M) -> R^4 as,

w = sqrt( eta g )

With the help of vector fields,

w^a(v_b) = d_ab

the Dirac equation can be generalized to,

( gamma^a(i v_a - A(v_a)) - m ) Psi = 0

The current might be,

j = Psi^h gamma^0 gamma^a Psi eta_ab w^b

which has *d*j = 0.

mihai cartoaje
mihai cartoaje
2005-05-29 20:03:55 UTC
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Post by r***@yahoo.ca
j = Psi^h gamma^0 gamma^a Psi eta_ab w^b
which has *d*j = 0.
This is not always true from the construction.
Actually,

j = Psi^h gamma^0 gamma^a Psi eta_ab w^b
= j^b w_b

has the divergence *d*j = j^b w^a(v_a, v_b).

FAIK, it might work to find a coordinate
system in which w = sqrt( eta g ) has the
property v_a . w^a = 0. At first order,
a solution exists iff R = 0 + O(h^2).