Grassmann numbers and other questions

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wchogg

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May 28, 2002, 9:18:18 PM5/28/02
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This is probably a silly question, but I just came across Grassmann
numbers in Peskin and Schroeder's Introduction to Quantum Field Theory
(for various reasons I need to get through most of this book on my own
over this summer), and while they describe what Grassmann numbers are,
they don't really say what they *are*.

My guess is that they're just vectors in a Clifford algebra, and that
when they talk about a Grassmann field they mean a map from R^n to
said algebra. I'm also not sure what algebra they're using. Since
it's in the context of Dirac fields, I'm assuming it's the one spanned
by the gamma matrices.

And one other question on this topic. Why are they called Grassmann
*numbers*? It seems odd to me to call them numbers.

Also, a question on the Dirac equation. If I, for some reason, wanted
to try a semi-classical meshing of gr and qed, then how would I change
the Dirac operator to reflect this? I'm pretty sure I'd have to
change the definitions of the gamma matrices, for example, but I'm not
really sure. Also, is there any reason why I would want to do this?

Thanks,
C. Hogg

Aaron Bergman

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May 29, 2002, 10:51:48 PM5/29/02
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In article <5e464d68.02052...@posting.google.com>,
wcho...@hotmail.com (wchogg) wrote:

> This is probably a silly question, but I just came across Grassmann
> numbers in Peskin and Schroeder's Introduction to Quantum Field Theory
> (for various reasons I need to get through most of this book on my own
> over this summer), and while they describe what Grassmann numbers are,
> they don't really say what they *are*.

They're Grassman numbers. They're not anything else. It's a perfectly
fine algebra.

> My guess is that they're just vectors in a Clifford algebra,

No. Elements in a Clifford algebra square to 1. Elements in a Grassman
algebra square to zero,

[...]


>
> Also, a question on the Dirac equation. If I, for some reason, wanted
> to try a semi-classical meshing of gr and qed, then how would I change
> the Dirac operator to reflect this? I'm pretty sure I'd have to
> change the definitions of the gamma matrices, for example, but I'm not
> really sure.

The short and not quite wrong answer is to pick an orthonormal basis.
Contract that gamma matrices with this. Proceed as usual.

For a better answer, you can check out math.DG/0005239.

> Also, is there any reason why I would want to do this?

Fermions on a curved spacetime are worth studying.

Aaron
--
Aaron Bergman
<http://www.princeton.edu/~abergman/>

gvhooydo

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May 29, 2002, 10:54:49 PM5/29/02
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That there is an error in the Dirac equation is known ever since the Lamb
shift was measured 1947 (see the Lamb Retherford paper in Phys Rev of
1950).
The Dirac equation is linked with the existence of (charged) antiparticles
(fermions) only. The question is to verify if there is a hidden symmetry
within neutral antiparticles (bosons). This is why the CERN-study of
antihydrogen is eagerly followed by the physics community (see the
nucl-ex/0202020 paper from the ATHENA collaboration at the archive).
If there already were an overlooked mirror symmetry in natural or
conventional neutral boson atom H, not covered by the Dirac equation, this
must show in the observed H line spectrum, available for many a decade, in
particular in the Lyman ns1/2-singlet series.
If such a mirror of left-right symmetry really existed in atom H, then a
Mexican hat shaped energy curve must be hidden in the observed terms for
this famous H-series.
Now, this is exactly what is observed (see CERN-EXT-2002-045, available
from the preprint server of CERN at http://cds.cern.ch or at CPS:
physchem/0205004).
It appears that some of the observed natural states of atomic hydrogen
must be classified as 'antihydrogenic', which would not only weigh heavily
on the Dirac equation itself, on STR but also on the final validation of
QED.

wchogg

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May 30, 2002, 3:55:30 PM5/30/02
to

Aaron Bergman <aber...@princeton.edu> wrote in message news:<abergman-0EA5D1...@news.bellatlantic.net>...

> In article <5e464d68.02052...@posting.google.com>,
> wcho...@hotmail.com (wchogg) wrote:
>
> > This is probably a silly question, but I just came across Grassmann
> > numbers in Peskin and Schroeder's Introduction to Quantum Field Theory
> > (for various reasons I need to get through most of this book on my own
> > over this summer), and while they describe what Grassmann numbers are,
> > they don't really say what they *are*.
>
> They're Grassman numbers. They're not anything else. It's a perfectly
> fine algebra.
>
> > My guess is that they're just vectors in a Clifford algebra,
>
> No. Elements in a Clifford algebra square to 1. Elements in a Grassman
> algebra square to zero.

Oops, sorry that should have been rather obvious. In fact, now that
I'm looking a bit more closely at the book and at the reference you
gave me it seems to make sense. I'm getting the impression that when
I've heard people say Grassmann algebra, that they're just talking
about an exterior algebra on a vector space (or at least that's a
guess considering that apparently Grassmann developed the notion of
exterior algebra), or is there more to it than that? Also, then
wouldn't the Grassmann numbers just be elements in the exterior
algebra generated by taking C as a 1-dimensional complex vector space?
Please feel free to tell me if I'm completely wrong at this point,
which is a *distinct* possibility. I'm just trying to think if this
algebra is just a special case of something I've seen before, rather
than something completely new. Of course, maybe it is just something
new and different...

Also, how do we know we can do calculus over this algebra? This
probably has an obvious answer, but I'm not seeing it at the moment.

> [...]
> >
> > Also, a question on the Dirac equation. If I, for some reason, wanted
> > to try a semi-classical meshing of gr and qed, then how would I change
> > the Dirac operator to reflect this? I'm pretty sure I'd have to
> > change the definitions of the gamma matrices, for example, but I'm not
> > really sure.
>
> The short and not quite wrong answer is to pick an orthonormal basis.
> Contract that gamma matrices with this. Proceed as usual.
>
> For a better answer, you can check out math.DG/0005239.
>
> > Also, is there any reason why I would want to do this?
>
> Fermions on a curved spacetime are worth studying.
>
> Aaron

Thanks for the reference, it looks really interesting. I'll probably
be reading it over the next few days when I'm not so tired.

Thanks again,

C. Hogg

Greg Weeks

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May 31, 2002, 5:44:53 PM5/31/02
to
wchogg (wcho...@hotmail.com) wrote:

: And one other question on this topic. Why are they called Grassmann


: *numbers*? It seems odd to me to call them numbers.

It *is* odd to call them numbers, in physics anyway.

In QFT, Grassman elements are used as *formal variables* used to create
formal functions that can be formally differentiated and integrated. But
the formal functions aren't functions. There is no domain, no range, and
no mapping. But if there *were*, the elements of their domain and range
would be anticommuting "numbers".

Greg


Aaron Bergman

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May 31, 2002, 7:37:30 PM5/31/02
to

> Aaron Bergman <aber...@princeton.edu> wrote in message
> news:<abergman-0EA5D1...@news.bellatlantic.net>...

> > In article <5e464d68.02052...@posting.google.com>,
> > wcho...@hotmail.com (wchogg) wrote:

> > > This is probably a silly question, but I just came across Grassmann
> > > numbers in Peskin and Schroeder's Introduction to Quantum Field Theory
> > > (for various reasons I need to get through most of this book on my own
> > > over this summer), and while they describe what Grassmann numbers are,
> > > they don't really say what they *are*.

> > They're Grassman numbers. They're not anything else. It's a perfectly
> > fine algebra.

> > > My guess is that they're just vectors in a Clifford algebra,

> > No. Elements in a Clifford algebra square to 1. Elements in a Grassman
> > algebra square to zero.

> Oops, sorry that should have been rather obvious. In fact, now that
> I'm looking a bit more closely at the book and at the reference you
> gave me it seems to make sense.

Shankar's book on quantum mechanics also has a bit on Grassman
variables. I think Freund's book on supersymmetry does also.

> I'm getting the impression that when
> I've heard people say Grassmann algebra, that they're just talking
> about an exterior algebra on a vector space (or at least that's a
> guess considering that apparently Grassmann developed the notion of
> exterior algebra), or is there more to it than that?

Pretty much. This identification between differential forms and fermions
is very important when talking about things like supersymmetric index
theorems.

[...]

> Also, how do we know we can do calculus over this algebra? This
> probably has an obvious answer, but I'm not seeing it at the moment.

You just define it. It's called the Berezin integral. Shankar talks
about it. The rules are pretty simple.

Daniel Doro Ferrante

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Jun 1, 2002, 1:59:33 PM6/1/02
to
On Thu, 30 May 2002, wchogg wrote:

> I'm getting the impression that when
> I've heard people say Grassmann algebra, that they're just talking
> about an exterior algebra on a vector space (or at least that's a
> guess considering that apparently Grassmann developed the notion of
> exterior algebra), or is there more to it than that? Also, then
> wouldn't the Grassmann numbers just be elements in the exterior
> algebra generated by taking C as a 1-dimensional complex vector space?

There is a little bit of nice info on this in: "Quantum Field Theory of
Point Particles and Strings", by Brian Hatfield. He "compares" spinor
fields with Grassmann numbers... i just don't remember in which page...

> Also, how do we know we can do calculus over this algebra? This
> probably has an obvious answer, but I'm not seeing it at the moment.

For this one, you can try: "Geometry, Spinors, and Applications"
(Springer-Praxis Books in Mathematics), by Donal J. Hurley, Michel A.
Vandyck.

--
Daniel
,-----------------------------------------------------------------------------.
> Daniel Doro Ferrante | www.het.brown.edu www.fma.if.usp.br <
> dani...@het.brown.edu | <
> Linux Counter #34445 | Suaviter in modo, fortiter in re. <
> | Se non e vero, e ben trovato. <
`-----------------------------------------------------------------------------'

Urs Schreiber

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Jun 3, 2002, 2:33:20 PM6/3/02
to
"Aaron Bergman" <aber...@princeton.edu> schrieb im Newsbeitrag
news:abergman-E4B0A0...@news.bellatlantic.net...

[...]

> Pretty much. This identification between differential forms and fermions
> is very important when talking about things like supersymmetric index
> theorems.
>
> [...]
>
> > Also, how do we know we can do calculus over this algebra? This
> > probably has an obvious answer, but I'm not seeing it at the moment.
>
> You just define it. It's called the Berezin integral. Shankar talks
> about it. The rules are pretty simple.

A nice thing is that these rules become even simpler, or at least more
transparent, when one sticks to thinking in terms of differential forms.
Namely then the Grassmann-Berezin "Fourier transformation" [1]

f -> f~ = int_Berezin f exp(-c*^n c_n) dc*

(I don't care for signs and other details here) is just the Hodge duality

f -> *f

operation (if you take care to get your signs right) and the Grassmann
inner product defined from this is just the local inner product of forms,

<f~,g>_Berezin = *(f/\*g)_(projected on 0-form sector) .

The combined ordinary and Grassmann integration then gives the
global Hodge inner product on forms (if you provide for the necessary
sqrt(g) factor)

int sqrt(g) <f~,g> dx = int f/\*g

and one sees that the rules of Berezin integration are there
to say that only top forms give a nonvanishing contribution when integrated
over the manifold.

The fact that Berezin integration is the same as Grassmann differentiation
finds its counterpart in the fact that

*(f/\*g) = [g]^\dag [f] |0>

where on the right hand side |0> is the form vacuum and [f] the form
creation (aka form multiplication) operator so that [f]|0> = f
and [g]^\dag is the adjoint of [g], which is a form annihilation (aka
Grassmann derivative) operator. (Eric Forgy has a beautiful way to
explain this and related facts, maybe he wants to provide more details.
On the other hand, you can also find more in the archives.)

The (N=1) superspace over bosonic base space M is just Lambda(M),
the exterior bundle over M.

The usual statement "susy exchanges bosons and fermions" translates
to the fact that to every coordinate x there is a differential dx and they
are related (very loosely speaking) by

[d, x] = dx

{del, dx} = -@x = -i p_x ,

where d (del) are the exterior (co-) derivatives which play the role
of susy generators.

I believe the exterior calculus point of view is a very fruitful one in
supersymmetry. I like to think of it this way:
Whenever you encounter a fermionic quantum field,
switch to the Schroedinger representation, make a mode decomposition,
and the mode amplitudes of the Fermi field will be Grassmann numbers.
Add the superpartner bosonic field, make a mode decomposition,
its amplitudes will be ordinary numbers that parameterize the bosonic
configuration space M. Thanks to supersymmetry the fermionic amplitudes
will be the differential forms over that space. The Schroedinger rep.
Hamiltonian will be the (deformed) Laplace-Beltrami operator and
the susy generators deformed exterior derivatives on Lambda(M)
I believe this should be true in general. I can show it for the system
Klein-Gordon + Dirac field and I am currently trying to show that
and how exactly (the "that"-part is pretty simple) it works in
canonical N=1, D=4 quantum supergravity.

It was (and is) E. Witten who showed and stressed the tight relations
between geometry and susy quantum physics. It started out with

Witten:1982,
author = {E. Witten},
title = {Supersymmetry and Morse Theory},
journal = {J. Diff. Geom.},
year = {1982},
volume = {17},
pages = {661-692}

and

Witten:1982b,
author = {E. Witten},
title = {Constraints on Supersymmetry Breaking},
journal = {Nucl. Phys. B},
year = {1982},
volume = {202},
pages = {253-316}

(see last section on the sigma-model),

which is nicely reviewed in

Pete:1999,
author = {G. Pete},
title = {{M}orse theory},
year = {1999},
howpublished = {{\tt http://www.math.u-szeged.hu/gpete/morse.ps}}

and reviewed and extended in the fascinating

FroehlichGrandjeanRecknagel:1996,
author = {J. Fr{\"o}hlich and O. Grandjean and A. Recknagel},
title = {Supersymmetric Quantum theory and (non-commutative) differential
geometry},
journal = {\tt hep-th/9612205},
year = {1996}

and

FroehlichGrandjeanRecknagel:1997,
author = {J. Fr{\"o}hlich and O. Grandjean and A. Recknagel},
title = {Supersymmetric Quantum theory, non-commutative geometry, and
gravitation},
journal = {\tt hep-th/9706132},
year = {1997} .

Of course there are also texts such as

Hull:1999,
author = {C. Hull},
title = {The geometry of supersymmetric quantum mechanics},
journal = {\tt hep-th/9910028},
year = {1999}

MichelsonStrominger:2000,
author = {J. Michelson and A. Strominger},
title = {The geometry of (super) conformal quantum mechanics},
journal = {\tt hep-th/9907191},
year = {2000},

but beware that these are often mostly concerned with N=1 susy QM,
which *cannot* be (naturally) described with differential forms but
which does call for a Clifford algebra instead of a Grassmann algebra.
The precise relation

N=1 SQM : N=2 SQM :: Clifford : Grassmann :: spin geom. : Riem. geom.

is detailed in the (first part of the) J. Froehlich papers cited above.

I gather that nowadays the interest has shifted from sigma-models and
SQM to homological field theories, see e.g.

Park:1999,
author = {J. Park},
title = {Cohomological Field Theories with K{\"a}hler Structure},
journal = {\tt hep-th/9910209},
year = {1999}

but I still do not know much about these. The point is that here one
always considers generalized deRahm operators on *Riemannian*
manifolds, where lots of nice relations hold and everything is
pretty much determined by topology alone. Due to my occupation
with sugra I am inclined to think (unless I am fundamentally confused,
which is always possible) that one should also study such things
on semi-Riemannian manifolds, but I am not aware of any work in this
direction. On semi-Riemannian manifolds, instead of toplogical field
theories one finds something formally similar to source-free
electromagnetism, in fact a generalization thereof. It's not topological
but still rather pretty.


-- more references:

[1] Some more useful information on superanalysis can be found in the
introductions to

Rogers:1992,
author = {A. Rogers},
title = {Stochastic calculus in superspace {I}: {S}upersymmetric
Hamiltonians},
journal = {J. Phys. A},
year = {1992},
volume = {25},
pages = {447-468}

and

@article{
Rogers:2002,
author = {A. Rogers},
title = {Supersymmetry and {B}rownian motion on supermanifolds},
journal = {\tt quant-ph/0201006},
year = {2002}
}

The Grassmann inner product and Fourier transformation in the
context of field theory are nicely explained in section 2.4 of

FaddeevSlavnov:1980,
author = {L. Faddeev and A. Slavnov},
title = {Gauge Fields, Introduction to Quantum Theory},
publisher = {Benjamin/Cummings},
year = {1980} .

Of course there is also the standard reference on superanalysis

Berezin:1987,
author = {F. Berezin},
title = {Introduction to superanalysis},
publisher = {D. Reidel},
year = {1987} .


--
Urs.Sc...@uni-essen.de

Alejandro Rivero

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Jun 4, 2002, 1:05:46 AM6/4/02
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"Aaron Bergman" <aber...@princeton.edu> wrote in message
news:abergman-E4B0A0...@news.bellatlantic.net


> > > They're Grassman numbers. They're not anything else. It's a perfectly
> > > fine algebra.

Actually, Grassman intended them to be a sort of fundation
for infinitesimal calculus. Centuries ago.

> Pretty much. This identification between differential forms and fermions
> is very important when talking about things like supersymmetric index
> theorems.

Aaron, I like this explicit statement "identification between
differential forms and fermions". It not accurate, at least it
is very stimulating. I wonder, have you heard of read it directly
in the literature. I did a search for it time ago without results.
(I mean, for the statement, no for the identification)

Alejandro


--
Posted via Mailgate.ORG Server - http://www.Mailgate.ORG

Aaron Bergman

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Jun 4, 2002, 1:49:43 AM6/4/02
to

In article
<99501694e054a5b8a9a...@mygate.mailgate.org>,
"Alejandro Rivero" <riv...@sol.unizar.es> wrote:

> > Pretty much. This identification between differential forms and fermions
> > is very important when talking about things like supersymmetric index
> > theorems.
>
> Aaron, I like this explicit statement "identification between
> differential forms and fermions". It not accurate, at least it
> is very stimulating. I wonder, have you heard of read it directly
> in the literature. I did a search for it time ago without results.
> (I mean, for the statement, no for the identification)

I'm sorry. I'm not sure what you're asking for here. If you want a more
explicit statement, I'll quote directly from hep-th/9411210:

Let us introduce the supermanifold M whose odd coordinates are gnerated
from the fibers of T^*M. In local coordinates we may write (x^i,psi^i).
A function on M is the same thing as a differential form on M ... the
correspondence simply given by psi^i <-> dx^i,

Is that better?

John Baez

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Jun 5, 2002, 2:16:19 AM6/5/02
to
In article <5e464d68.02052...@posting.google.com>,
wchogg <wcho...@hotmail.com> wrote:

>This is probably a silly question, but I just came across Grassmann
>numbers in Peskin and Schroeder's Introduction to Quantum Field Theory
>(for various reasons I need to get through most of this book on my own
>over this summer), and while they describe what Grassmann numbers are,
>they don't really say what they *are*.

If you need to know what they *are* you must secretly be a mathematician;
a physicists would be content to calculate with them. :-)

>My guess is that they're just vectors in a Clifford algebra,

No, they're vectors in an exterior algebra. A "Grassmann algebra" is
another name for an exterior algebra, since Grassmann invented exterior
algebras.

I can see why you're making the mistake you're making, but don't
forget: while "gamma matrices" are really generators of a Clifford
algebra, "spinor fields" are best thought of as elements of an
exterior algebra.

>And one other question on this topic. Why are they called Grassmann
>*numbers*? It seems odd to me to call them numbers.

Heh... this is just a continuation of a physics tradition started
by Dirac, who spoke of "c-numbers" and "q-numbers" - classical numbers
and quantum numbers - instead of complex numbers and operators on
Hilbert spaces. I don't know any mathematicians who call them
Grassmann "numbers", but lots of physicists do.

And the physicists may not be completely wrong on this. After
all, if you want to get really comfortable with elements of some
algebra, it may help to think of them as "numbers". We did this
with complex numbers, and Hamilton probably tried to do this with
quaternions, and some people refer to elements of Clifford algebras
as "hypercomplex numbers".

>Also, a question on the Dirac equation. If I, for some reason, wanted
>to try a semi-classical meshing of gr and qed, then how would I change
>the Dirac operator to reflect this? I'm pretty sure I'd have to
>change the definitions of the gamma matrices, for example, but I'm not
>really sure.

I think you'd want to use the tetrad formalism, to make the Dirac
operator have a well-defined meaning even when the metric was a
quantum field. At least that's what most of us do!


Alejandro Rivero

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Jun 5, 2002, 4:16:58 PM6/5/02
to
"Aaron Bergman" <aber...@princeton.edu> wrote in message
news:abergman-8F2887...@news.bellatlantic.net

> <99501694e054a5b8a9a...@mygate.mailgate.org>,
> "Alejandro Rivero" <riv...@sol.unizar.es> wrote:
> > > Pretty much. This identification between differential forms and fermions
> > > is very important when talking about things like supersymmetric index
> > > theorems.
> >
> > Aaron, I like this explicit statement "identification between

> > differential forms and fermions". If not accurate, at least it
> > is very stimulating.

> I'm sorry. I'm not sure what you're asking for here. If you want a more

> explicit statement, I'll quote directly from hep-th/9411210:
>
> Let us introduce the supermanifold M whose odd coordinates are gnerated
> from the fibers of T^*M. In local coordinates we may write (x^i,psi^i).
> A function on M is the same thing as a differential form on M ... the
> correspondence simply given by psi^i <-> dx^i,
>
> Is that better?

A lot better. Just notational correction:

... supermanifold $\hat M$ ...
A function on $\hat M$ is the same thing as a differential form on M

But what for the other term of the syllogism? You told "identification
with fermions"? Are fermions just functions in superspace? Or are
fermions just the fiber of T^*M?

Yours,

Alejandro

Alejandro Rivero

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Jun 5, 2002, 4:22:26 PM6/5/02
to
"Urs Schreiber" <Urs.Sc...@uni-essen.de> wrote in message
news:adadhv$jk4$1...@rs04.hrz.uni-essen.de
> "Aaron Bergman" <aber...@princeton.edu> schrieb

> > Pretty much. This identification between differential forms and fermions
> > is very important when talking about things like supersymmetric index
> > theorems.
> The usual statement "susy exchanges bosons and fermions" translates
> to the fact that to every coordinate x there is a differential dx and they
> are related (very loosely speaking) by
> [d, x] = dx
> {del, dx} = -@x = -i p_x ,
> where d (del) are the exterior (co-) derivatives which play the role
> of susy generators.

Er... just brainstorming myself, then... I could understand this
as standard lore when x is the infinite-dimensional space of
field operators. But if we go back to 0+1, ie to quantum
mechanics, then the operator x is just the coordinate of
the particle, isn't it? And this is supossed to be a
bosonic supersymmetric partner?

But wait, I could accept this. If could be telling us that we
have always been looking at the SUSY partners of the fermions
and not seeing them: because they show to us just as the
coordinates of the particle!

Alejandro Rivero

Urs Schreiber

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Jun 5, 2002, 5:00:39 PM6/5/02
to
"Alejandro Rivero" <riv...@sol.unizar.es> schrieb im Newsbeitrag
news:99501694e054a5b8a9a...@mygate.mailgate.org...

[...]

> Aaron, I like this explicit statement "identification between
> differential forms and fermions". It not accurate, at least it
> is very stimulating. I wonder, have you heard of read it directly
> in the literature. I did a search for it time ago without results.
> (I mean, for the statement, no for the identification)

According to

author = {S. Cordes and G. Moore and S. Ramgoolam},
title = {Lectures on {2D} {Y}ang-{M}ills Theory, Equivariant Cohomology
and Topological Field Theories},
journal = {\tt hep-th/9411210},
year = {1994}

(see section 11.9.1 on page 110) it is a "basic tautology" (and it's true: An
antisymmetric tensor is an antisymmetric tensor is an antisymmetric tensor.).
In

author = {M. Claudson and M. B. Halpern},
title = {Supersymmetric ground state wave functions},


journal = {Nucl. Phys. B},

year = {1984},
volume = {250},
pages = {689-715}

you'll find a detailed discussion of the N=2 non-linear sigma model where it is
shown explicitly that and how the connection to exterior calculus emerges.
There is a lot of literature on this, see for instance

author = {A. Dolgallo and K. Ilinski},
title = {Generalized supersymmetric quantum mechanics on {R}iemannian
surfaces with meromorphic superpotentials},
journal = {J. Math. Phys.},
year = {1994},
volume = {35},
number = {5},
pages = {2074}

for the same for N=4 SQM, and, of course, my favorite reference

author = {J. Fr{\"o}hlich and O. Grandjean and A. Recknagel},
title = {Supersymmetric Quantum theory and (non-commutative) differential
geometry},
journal = {\tt hep-th/9612205},
year = {1996}

on "Susy is geometry." Note the "non-commutative" part.

More towards field theory: The identification

fermionic Fock space <-> form bundle

is emphasized in

author = {O. Rudolph},
title = {Super {H}ilbert Spaces},
journal = {\tt math-ph/9910047},
year = {2000}

and

author = {J. Kupsch and O. Smolyanov},
title = {Functional Representation for Fock Superalgebras},
journal = {\tt hep-th/9708069},
year = {1997}

I think it is correct to say that the "basic tautology" was first recognized in

author = {E. Witten},
title = {Supersymmetry and Morse Theory},
journal = {J. Diff. Geom.},
year = {1982},
volume = {17},
pages = {661-692}

author = {E. Witten},


title = {Constraints on Supersymmetry Breaking},
journal = {Nucl. Phys. B},
year = {1982},
volume = {202},

pages = {253-316} .

BTW: It is not only "A fermion is a form.", but also "A ghost is a form." (at
least ghosts of bosons are). The connection

exterior calculus <-> BRST theory

becomes most striking in the BRST/coBRST framework. See

author = {{J. W. van} Holten},
title = {Aspects of BRST Quantization},
journal = {\tt hep-th/0201124},
year = {2002}

author = {M. Spiegelglas},
title = {{Q_BRST} Cohomology},


journal = {Nucl. Phys. B},

year = {1986},
volume = {283},
pages = {205-216}

author = {G. F\"ul\"op and R. Marnelius},
title = {Gauge fixing and co{BRST}},


journal = {Nucl. Phys. B},

year = {1995},
volume = {456},
pages = {443-472}

author = {G. F\"ul\"op },
title = {{BRST--co-BRST} quantization of reparametrization-invariant
theories},
journal = {Int. J. Theor. Phys.},
year = {1997},
volume = {13},
pages = {3629-3647}

author = {W. Kalau and {J. W. van Holten}},
title = {{BRST} cohomology and {BRST} gauge fixing},


journal = {Nucl. Phys. B},

year = {1991},
volume = {361},
pages = {233-252}

for the close formal relation between BRST cohomology and de Rahm
cohomology.

It is tantalizing how similar BRST and SUSY are. Is it deep or is it
trivial?

--
Urs.Sc...@uni-essen.de

Aaron Bergman

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Jun 6, 2002, 5:00:18 PM6/6/02
to
In article
<370cbaf15948c45a15c...@mygate.mailgate.org>,
"Alejandro Rivero" <riv...@sol.unizar.es> wrote:

> But what for the other term of the syllogism? You told "identification
> with fermions"? Are fermions just functions in superspace? Or are
> fermions just the fiber of T^*M?

N=1 Superfields can be identified with differential forms. Within a
superfield, there are fermionic and bosonic fields.

Urs Schreiber

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Jun 6, 2002, 5:04:18 PM6/6/02
to
"Alejandro Rivero" <riv...@sol.unizar.es> schrieb im Newsbeitrag
news:5f3f24d18c1828abe35...@mygate.mailgate.org...

> "Urs Schreiber" <Urs.Sc...@uni-essen.de> wrote in message
news:adadhv$jk4$1...@rs04.hrz.uni-essen.de

>> The usual statement "susy exchanges bosons and fermions" translates


>> to the fact that to every coordinate x there is a differential dx and they
>> are related (very loosely speaking) by
>> [d, x] = dx
>> {del, dx} = -@x = -i p_x ,
>> where d (del) are the exterior (co-) derivatives which play the role
>> of susy generators.

> Er... just brainstorming myself, then... I could understand this
> as standard lore when x is the infinite-dimensional space of
> field operators. But if we go back to 0+1, ie to quantum
> mechanics, then the operator x is just the coordinate of
> the particle, isn't it? And this is supossed to be a
> bosonic supersymmetric partner?
>
> But wait, I could accept this. If could be telling us that we
> have always been looking at the SUSY partners of the fermions
> and not seeing them: because they show to us just as the
> coordinates of the particle!

Probably Aaron Bergman or someone else can give a more helpful comment on this
then I can, but I'll try to do my best:

First, yes, in the 1+0 dim non-linear susy sigma-model the "coordinate fields"
are certainly the superpartners of the Grassmann fields, which, as has been
discussed, can be interpreted as forms, or "Grassmann coordinates" for that
matter.

Second, somebody told me and I once briefly looked at a paper which showed,
IIRC, that when reformulating ordinary QED in the so-called "worldline"
formulation, then the worldline action, which is essentially the worldline
action of a Dirac particle, turns out to be supersymmetric with respect to
exchange of the particle's coordinates with its spin degrees of freedom, which
are Grassmann valued.

So this makes clear that one should carefully distinguish two ways in which
"superpartners" arise: On the one hand side the ordinary Dirac particle can be
shown to be described by a theory which has some formal supersymmetry within
itself. On the other hand this is not the susy that people mean when they are
looking for superpartners at accelerators.

Similar examples where some sort of susy algebra appears in non-susy theories
is ordinary classical electromagnetism and single Dirac particle theory. I find
it helpful to recognize the susy algebra in these theories, but they are rarely
mentioned in the context of susy.

And then, there is a way in which the "coordinates are bosonic
superpartners"-point-of-view coincides with the ordinary "bosonic fields are
bosonic superpartners"-point-of-view, namely in the Schroedinger representation
of field theory. I have mentioned before in this thread that I think I
understand how this works for simple complex scalar+spinor field theory (and
perhaps this is obvious to everyone else), but let me stick to supergravity,
which I am actually more familiar with: In the Schroedinger representation of
canonical quantum supergravity the spatial graviton field mode amplitudes b^n
parameterize the bosonic configuration space. Hence they are naturally
interpreted as coordinates on configuration space: b^n = x^n. Ordinary quantum
gravity can be regarded as ordinary point particle quantum mechanics in this
config space. Now, when the gravitino field is added we also have the
amplitudes f^n of the gravitino modes, which upon quantization become the usual
anticommuting fermionic Fock space operators. These are naturally interpreted
as differential forms over configuration space: f^n = dx^n. Let @b^n be the
functional derivative with respect to the n-th graviton mode (the n-th
coordinate on config space) and f^n the (Grassmann) functional multiplication
operator with respect to the n-th gravitino mode, a form over config space.
Then the primed quantum supersymmetry generators of canonical supergravity read
schematically (I am suppressing some details, obviously)

S = f^n (@b^n + V_n) .

for some functional V of the graviton field.
Does this look familiar?

We can identify the f^n with forms, so

f^n @b^n =: d,

where d is the exterior derivative on config space. We can ask this d: "What is
the superpartner of the graviton b^n?". And it answers:

[d, b^n] = f^n,

which is S-ish for: "It's the gravitino, the gravitino!!". Of course this is
just saying

[d, x^n] = dx^n

on config space.

Since you're into NCG, I should say that yet another slight reformulation is
fruitful: The generator S has an adjoint, bar S given by

bar S = f#^n ( @b^n + ... ) ,

where f#^n is a linear combination of the functional gravitino mode derivatives
so that

{f#^n, f^m} = G^nm ,

where G_nm is the metric on config space (which is essentially defined by this
relation). We may want to switch to the linear combination

D = S + bar S
= (f^n - f#^n)( @b^n + ... )

A little reflection shows that

y^n = f^n - f#^n

is the usual Clifford algebra constructed from fermionic creators/annihilators
and

D = y^n @b^n + ...

is a Dirac operator on config space. Asking D for superpartners gives or course

[D, b^n] = y^n + ... .

So maybe in this sense every pair of superpartners is always of the form
(x,dx) - in configuration space.

--
Urs.Sc...@uni-essen.de

Urs Schreiber

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Jun 6, 2002, 5:04:31 PM6/6/02
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"Alejandro Rivero" <riv...@sol.unizar.es> schrieb im Newsbeitrag
news:370cbaf15948c45a15c...@mygate.mailgate.org...

[...]

> But what for the other term of the syllogism? You told "identification
> with fermions"? Are fermions just functions in superspace? Or are
> fermions just the fiber of T^*M?

A form is a superfield on T^*M, coordinates on M being coordinates and
Grassmann coordinates being 1-forms.

It's best to think of the fermions in terms of fermionic Fock space elements.
So here is the vacuum:

|0>,

a zero form. Create a fermion in some mode

c^n|0> = dx^n,

a one form. Create another fermion

c^m c^n |0> = dx^m /\ dx^n,

a two form. Create a bunch of fermions in wild superposition:

f_n1 n2 n3 ... c^n1 c^n2 c^n3 ... |0>,

a (possibly inhomogeneous) element of the form bundle - a superfield. In this
geometric representation fermions like to be represented by creation and
annihilation operators, but bosons like to be represented as multiplication and
differentiation operators, so that the susy generator reads

S = c^n @_n + ...

Only for certain potentials, most notably the oscillator potential, is it
natural to introduce the usual bosonic creators and annihilators

a^n = (a @_n + b x^n)

so that for instance

S = c^n (a^n + a^dag n) + ... .

Formulated this way people sometimes refer to the "bosonic oscillator", the
"fermionic oscillator" and the "susy oscillator", e.g. in

author = {H. Kalka and G. Soff},
title = {Supersymmetrie},
publisher = {{T}eubner {S}tudienb{\"u}cher},
year = {1997} ,

but I do not find that very helpful. The differential geometry point of view is
*way* more general and powerful. To apply it to field theory, I think one has
to switch to the Schroedinger representation and work over M = configuration
space. I give some details on this in another post.

--
Urs.Sc...@uni-essen.de

Alejandro Rivero

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Jun 8, 2002, 5:18:25 PM6/8/02
to

"Aaron Bergman" <aber...@princeton.edu> wrote in message
news:abergman-FA0A52...@news.bellatlantic.net

> <370cbaf15948c45a15c...@mygate.mailgate.org>,
> "Alejandro Rivero" <riv...@sol.unizar.es> wrote:
> > But what for the other term of the syllogism? You told "identification
> > with fermions"? Are fermions just functions in superspace? Or are
> > fermions just the fiber of T^*M?
> N=1 Superfields can be identified with differential forms. Within a
> superfield, there are fermionic and bosonic fields.

OK,

So fermions are not a separate object of the formalism.

Next question could be: Which is the interpretation of
SUSY breaking from the "differential form" point of view. If
we break the symmetry, some degrees of freedom go to the boson,
some degrees of freedom go to the fermion.

In the cotangent bundle, the differential forms should equaly
divide in two mathematical objects. Which are them? Are them
objects known by mathematicians, or just some arbitrary
factorization of the bundle?

Alejandro Rivero

John Baez

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Jun 8, 2002, 8:18:32 PM6/8/02
to

In article <10226860...@cswreg.cos.agilent.com>,
Greg Weeks <we...@vitus.scs.agilent.com> wrote:

>wchogg (wcho...@hotmail.com) wrote:

>: And one other question on this topic. Why are they called Grassmann
>: *numbers*? It seems odd to me to call them numbers.

>It *is* odd to call them numbers, in physics anyway.

They are, after all, odd! Odd elements of a supercommutative algebra,
that is. :-)

It's mainly when you take supersymmetry superseriously that you
start wanting to build it into mathematics at a very deep level,
talking about supermanifolds, supervector spaces, and sometimes
even "supernumbers". Yes, I'm pretty sure I've seen that term
somewhere - it means the same thing as "Grassmann numbers".


Aaron Bergman

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Jun 10, 2002, 2:23:05 PM6/10/02
to
In article
<c44b82c8b4e74e1ea5c...@mygate.mailgate.org>,
"Alejandro Rivero" <riv...@sol.unizar.es> wrote:

> > N=1 Superfields can be identified with differential forms. Within a

> > superfield, there are fermionic and bosonic fields.

> OK,
>
> So fermions are not a separate object of the formalism.
>
> Next question could be: Which is the interpretation of
> SUSY breaking from the "differential form" point of view.

The exact same as it is with superfields. There is no difference at all.

Urs Schreiber

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Jun 10, 2002, 5:05:21 PM6/10/02
to

"Alejandro Rivero" <riv...@sol.unizar.es> schrieb im Newsbeitrag
news:c44b82c8b4e74e1ea5c...@mygate.mailgate.org...

>
> "Aaron Bergman" <aber...@princeton.edu> wrote in message

[...]

> > N=1 Superfields can be identified with differential forms. Within a
> > superfield, there are fermionic and bosonic fields.
>
> OK,
>
> So fermions are not a separate object of the formalism.

That depends on what you do with your superfield.

> Next question could be: Which is the interpretation of
> SUSY breaking from the "differential form" point of view. If
> we break the symmetry, some degrees of freedom go to the boson,
> some degrees of freedom go to the fermion.
>

> In the cotangent bundle, the differential forms should equally


> divide in two mathematical objects. Which are them? Are them
> objects known by mathematicians, or just some arbitrary
> factorization of the bundle?

I am not sure this is the right question to ask. Forms/superfields are part of
the kinematics. Susy breaking is part of the dynamics. It does not affect the
way you formulate your theory.


--
Urs.Sc...@uni-essen.de

Alfred Einstead

unread,
Jun 11, 2002, 6:53:49 PM6/11/02
to
Aaron Bergman <aber...@princeton.edu> wrote:
> > My guess is that they're just vectors in a Clifford algebra,
> No. Elements in a Clifford algebra square to 1. Elements in a Grassman
> algebra square to zero,

The vector elements of a Clifford algebra square to the inner
product
a*a = <a,a>

which includes, as a special case, the 0 inner product <a,a> = 0.

A Grassman algebra is a Clifford algebra whose vector elements
have 0 square.

Alejandro Rivero

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Jun 12, 2002, 9:24:19 PM6/12/02
to
"Aaron Bergman" <aber...@princeton.edu> wrote in message
news:abergman-850D6F...@news.bellatlantic.net

>
> > > N=1 Superfields can be identified with differential forms. Within a
> > > superfield, there are fermionic and bosonic fields.
>
Last weekend in the library xeroxing the bibliography.

I have found the original definition of superspace very clear,
directly in Salam Strathdee NPB 76, p. 477. One gets
space-time coordinates, then one adds Grassman variables
at will. Initially we are not restricted to have the same
number of Grassman and Space-Time degrees of freedom.

But in most interesting cases the number coincides, and then
we can do the substitution between grassman variables
\theta and differentials dx. In other cases one proceeds
ad hoc via dimensional reduction technique, it seems.

The main source of confusion, to me, is that grassmann variables
relate to the supercharge, but have nothing with the fermions
of the theory, i.e. with the representation. Fermions and
bosons are pieces of the superfield. The superfield can
be seen as a differential form, ie an element of \Omega^*.
As explained by Urs, the supercharge operator acts in
\Omega^* very like (d+ *d*); the differential forms are
bosons or fermions depending of their degree.

I believe this answers my question about SUSY breaking: it
divides the differential complex \Omega^* in its
components \Omega^0, \Omega^1,...,\Omega^n; or at least
in the even and odd graded pieces.

The superfield can be scalar or vector-valued. This paves
the way to Witten's analysis of the non-linear sigma model,
NPB 202, p 253-316, section 11. There the superfield takes
values over the coordinates of a target manifold, so that
one of the bosonic fields ranges again over this set of
coordinates. It is to be noticed that there are not
evident restrictions in the dimension of the target
manifold at this level. I am not sure if we should
like to have just N=1 or if we should to take as many
superchargues as the dimension of the target.

BTW, the whole mechanism of differential forms remembers me
of the Dirac-Kahler equation, where the fermionic degrees
of freedom increase by doubling and then they are scattered
in different pieces of the whole differential form. Has
somebody taken a look to this?

yours,

Alejandro Rivero

Marginal note
While reading Witten's 1982, the feeling is very like attending
Lucas' new trilogy. One sees him as the new Feymann, the
heritage of american theoretists, the promise...
If he is mirroring the career path of Anakin Skywalker, will
we come to see the return of the Jedi?

Alfred Einstead

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Jun 13, 2002, 11:26:51 PM6/13/02
to
we...@vitus.scs.agilent.com (Greg Weeks) wrote:
> In QFT, Grassman elements are used as *formal variables* used to create
> formal functions that can be formally differentiated and integrated.

The use of Grassman numbers has nothing, per se, to do with QFT. It's
the *classical* Dirac field psi (which is sometimes called a "semiclassical"
or "1st quantized" field), which is a Grassmann field. The only
additional element that QFT brings into the picture is that it
quantizes this field (or "2nd quantizes" it), by turning psi from
a Grassman c-number into a Grassman q-number, to coin a phrase.

Urs Schreiber

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Jun 18, 2002, 11:34:59 PM6/18/02
to
"Alejandro Rivero" <riv...@sol.unizar.es> schrieb im Newsbeitrag
news:351d9e49cec98b54694...@mygate.mailgate.org...

[...]

> I believe this answers my question about SUSY breaking: it
> divides the differential complex \Omega^* in its
> components \Omega^0, \Omega^1,...,\Omega^n; or at least
> in the even and odd graded pieces.

I am not sure what you really mean here. A system is supersymmetric if its
Hamiltonian commutes with the supercharges. A state is supersymmetric if it is
annihilated by the supercharges. "SUSY breaking" refers to supersymmetric
systems which are found in a state (i.e. have a gound state) which is not
annihilated by all supersymmetry generators. Often this can be discussed in
each fermion number sector (each \Omega^p component) separately, whether susy
is broken or not.

> The superfield can be scalar or vector-valued. This paves
> the way to Witten's analysis of the non-linear sigma model,
> NPB 202, p 253-316, section 11. There the superfield takes
> values over the coordinates of a target manifold, so that
> one of the bosonic fields ranges again over this set of
> coordinates. It is to be noticed that there are not
> evident restrictions in the dimension of the target
> manifold at this level. I am not sure if we should
> like to have just N=1 or if we should to take as many

> supercharges as the dimension of the target.

The way to have higher N-extended susy here is by means of Kaehler structures
on the target space which allows the supercharges to split as

d = @ + bar @

in holomorphic and anti-holomorphic parts.

> BTW, the whole mechanism of differential forms remembers me
> of the Dirac-Kahler equation, where the fermionic degrees
> of freedom increase by doubling and then they are scattered
> in different pieces of the whole differential form. Has
> somebody taken a look to this?

Yes, it is intimately related. I have thought a bit about this in
http://www-stud.uni-essen.de/~sb0264/sqm.ps.gz .


> Marginal note
> While reading Witten's 1982, the feeling is very like attending
> Lucas' new trilogy. One sees him as the new Feymann, the
> heritage of american theoretists, the promise...
> If he is mirroring the career path of Anakin Skywalker, will
> we come to see the return of the Jedi?

I think he is rather mirroring the career path of George Lucas ;-) .

--
Urs.Sc...@uni-essen.de

Dr Tim

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Jun 19, 2002, 1:51:15 PM6/19/02
to
What "is" a Grassman number?

Grassman numbers are used as homogenous coordinates in projective
geometry.
For example, a point in three dimensional space can be represented by
a 4-dimensional Grassman 1-form. The line that passes through two
points A and B is represented by the product of their Grassman forms.
And the plane that passes through a line and a point not on it is
represented by the product of the Grassman forms of the line and the
point.

A point lies on a line, or a point on a plane, or a line on a plane if
the product of their Grassman forms is zero.

You can use the conjugate product (if that's what it's called) to find
the point where a line passes through a plane.

I have actually used Grassman numbers in this way to do calculations
for 3-D graphics.

So you can say that a 4-dimensional Grassman 1-form "is" a point in
3-dimensional space, a 2-form "is" a line and a 3-form "is" a plane.

Doug B Sweetser

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Jun 19, 2002, 11:38:32 PM6/19/02
to undisclosed-recipients., news.std.com
Hello:

John Baez wrote [a number of weeks ago, and this post I misplaced,
oops]:

>And the physicists may not be completely wrong on this. After
>all, if you want to get really comfortable with elements of some
>algebra, it may help to think of them as "numbers". We did this
>with complex numbers, and Hamilton probably tried to do this with
>quaternions, and some people refer to elements of Clifford algebras
>as "hypercomplex numbers".

Both physicists and mathematicians agree that the real numbers and
complex numbers are, well, numbers. So a question is what properties
do these two share? The technical way to say it is they are both
topological algebraic fields. Let's break that down. The elements of
an algebraic field can be added, subtracted, multiplied, and divided,
except by zero. Numbers form a group under the addition and
subtraction operator. By group it means there is an identity element,
zero for addition, each member has an inverse, the negative of a
number for the addition operator, and the associative law holds. If
zero is omitted, the real and complex numbers form a group under the
multiplication and division operators. Both real and complex numbers
have a non-zero norm between two numbers (unless the numbers are
identical). A non-zero norm is required for the division operator.
One can define division as the conjugate over the norm for both real
and complex numbers. The conjugate of a real number is the real
number, and the norm is the square of a number, so division is what we
learned in grade school. The nice thing about this longer definition
is that it works without modification for the complex numbers.

Up to an isomorphism, the only other finite-dimensional associative
topological algebraic field is the quaternions. It would be fair on
technical grounds to call quaternions numbers. Culturally though, it
is not so easy. In physics, quaternions play a bit part in 3D
rotations. The unit quaternions, SU(2), are part of the standard
model. De Leo figured out how to directly represent the Lorentz group
using real quaternions only in 1996. It is much easier to use a
Clifford algebra or complex-valued quaternions (not a division
algebra) to write out the Lorentz group. The Maxwell equations can be
written compactly using Clifford algebras as threads on this
newsgroups have shown. I have been told on several occasions that the
Maxwell equations cannot be written using real quaternions - it is even
in print - although I have done it for the simple reason that
quaternions are an algebraic field. If you cannot find a way from
here to there, you just have not looked hard enough.

Real numbers, complex numbers, and quaternions are all topological
algebraic fields as well as being Clifford algebras. All the other
Clifford algebras are not topological algebraic fields. One should
not expect this mixed collection to have the same power as the real
and complex numbers (and perhaps someday the quaternions, but very
little money is riding on them, certainly less public dollars than the
Clifford algebras).

Complex numbers are now everywhere, but the battle for them to gain
acceptance was nasty. I have not studied either the topic in detail,
but I have heard that the fights over quaternions were every bit as
unpleasant. The quaternion camp did not win. By the way, Gauss
invented quaternions first. Rodriguez developed them independently to
do rotations. Hamilton's story has the greatest tragic flair, so is
usually the one cited. Most books that teach about scalars and
vectors fail to mention quaternions. The claim is that scalars and
vectors are more general. This always baffles me. What is THE most
useful thing in all of math? Can one argue that it is the real
numbers? Real numbers show up more often than anything else, and by
that simple form of measure, real numbers are at the least more
widespread than any other tool. From this perspective, second place
goes to the complex numbers. They are certainly critical for doing
quantum mechanics. Third place does not go to quaternions.

This reminds me of what happens to the word "science". Science has
been so successful, it has been tacked onto things that have little
to do with the core of science, from the science diet to Christian
Science. Numbers can be added, subtract, multiplied, and divided and
one still ends up with a number. There is a notion of distance
between any two numbers, which is never negative, and only zero if
the two numbers are identical. Since Clifford algebras do no share
all these properties, it is my opinion that is misleading to use the
word "number" in that context, even if it makes them more attractive.
At least my position that quaternions should be considered numbers is
grounded in technical details.


doug <swee...@theworld.com>
quaternions.com

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