On Noether's Second (BV, Part VI) | The n-Category Café

October 31, 2007

On Noether’s Second (BV, Part VI)

Posted by Urs Schreiber

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One aim of

R. Fulp, T. Lada, J. Stasheff
Noether’s Variational Theorem II and the BV formalism
math/0204079

was to

[…] restore […] an emphasis [on] the relevance of Noether’s theorem in […] the BV approach

Namely it is Noether’s second theorem (see page 6 of the above article) for Lagrangian theories which is reincarnated equivalently in the BV statement that

the space of ghosts is canonically isomorphic to that of anti-ghosts.

Meaning that

For every Noether identity there is a symmetry. And vice versa.

In terms of the little toy example (which is not that toy-ish, actually, rather skeletalized, I think), which I talked about last time (see also parts I, II, III, IV), this means that in our little complex

$\array{ 0 &\to& \mathrm{ker}(d S(\cdot)) &\hookrightarrow& \Gamma(T X) &\stackrel{d S(\cdot)}{\to}& C^\infty(X) &\stackrel{\rho}{\to}& C^{\infty}(X) \otimes g &\to& 0 \\ \\ && anti-ghosts && anti-fields && fields && ghosts \\ \\ && Tate && Koszul && && Chevalley-Eilenberg \\ \\ && Noether identities && && && symmetries \\ \\ && deg -2 && deg -1 && deg 0 && deg 1 }$ which is induced entirely from a smooth function $S : X \to \mathbb{R}$ on a manifold $X$ , we have a canonical isomorphism between the first and the last term $\mathrm{ker}(d S (\cdot)) \simeq C^\infty \otimes g \,.$ And this canonical isomorphism is, I think, Noether’s second theorem in this context.

And I’ll claim: this is here nothing but a special case of Cartan’s magic formula (or whatever you call that).

For that to make sense, I’ll first need to say mor precisely how the $g$ in $C^\infty(X) \otimes g$ is defined in the first place.

$MathML-enabled post (click for more details).$

Instead of plunging into the jet space gymnastics performed by Fulp, Lada and Stasheff, I shall here follow the discussion of

P.O. Kazinski, S.L. Lyakhovich, A.A. Sharapov
Lagrange structure and quantization
hep-th/0506093

which pretends throughout that the space of fields is a smooth, finite dimensional manifold. The relation we are after is their equation (10) on p. 7 (take notice of the paragraph right beneath that!).

All I want to do here is actually point out the proof of that simple statement, using slightly more invariant language than used there.

Definition (Equations of motion, Symmetries, Noether identities) The equation of motion induced by $S$ is $d S = 0 \,.$ A local symmetry of $S$ is a vector field $v \in \Gamma(T X)$ preserving the equations of motion $L_{\epsilon v} (d S) \; = 0$ ( $L_v$ denotes the Lie derivative) for all $\epsilon \in C^\infty(X)$ . A Noether identity is a vector field $v$ preserving the function $S$ itself $L_v S = 0 \,.$

It seems too trivial to point it out here, but I shall do anyway: people usually think of a Noether identity as a vector field $v$ such that $d S (v) = 0 \,.$

Then if you write the components of $d S$ as $E_a$ in some basis, a Noether identity becomes $v^a E_a = 0 \,.$ In that form you may more easily recognize it in the existing component-ridden literature.

Now

Simple proposition. The space of symmetries is canonically isomorphic to that of Noether identities.

If $L_v d S = 0$ then $L_{\epsilon v} d S = \epsilon L_v d S + d \epsilon \wedge v(S)$ . For this to vanish for all $\epsilon$ we need $v(S)$ to vanish.

Definition (ghosts and anti-ghosts) Suppose that the space of symmetries is generated, over $C^\infty(X)$ , from a subspace $g \subset \Gamma(T X)$ on which the bracket of vector fields closes. Then $g$ is the Lie algebra of symmetries, and we identify the space of symmetries with $C^\infty(X) \otimes g \,.$ BV practitioners will call the elements of $g$ the ghosts.

By the above theorem this induces a similar decomposition of the space of Noether identites. The Noether identites corresponding to the elements in $g$ BV practicioners will call the anti-ghosts.

(Again, keep in mind that the full BV formalism is designed to handle much more general cases than that. In particular, its whole raison d’être is the case where the symmetries do not form a Lie algebra on the nose. But that shall not concern us right now.)

Posted at October 31, 2007 11:42 PM UTC

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17 Comments & 3 Trackbacks

Re: On Noether’s Second (BV, Part VI)

Sorry to correct you but from v being a symmetry of dS does not follow that fv is also a symmetry of dS for any smooth function f. Or maybe i understood the equation L_v dS=0 wrong.

Posted by: Anonymous Coward on November 1, 2007 8:52 PM | Permalink | Reply to this

Re: On Noether’s Second (BV, Part VI)

$MathML-enabled post (click for more details).$

Sorry to correct you

Well, thanks!

I didn’t put it right. I should have said:

$v$ is a local symmetry if $L_{f v} d S = 0$ for all $f \in C^\infty(X)$ . Since $L_{f v} d S = f L_v d S + d f \wedge d S(v)$ this implies that $L_v S = 0$ .

That’s what I should have said, I think. Fixed now. Thanks again.

Posted by: Urs Schreiber on November 2, 2007 3:15 PM | Permalink | Reply to this

Re: On Noether’s Second (BV, Part VI)

From the original post of part VI

And I’ll claim: this is here nothing but a special case of Cartan’s magic formula (or whatever you call that).

For that to make sense, I’ll first need to say mor precisely how the g in C^\INFTY(X)\OTIMES g is defined in the first place.

I LOOK FORWARD TO ELUCIDATION!

Instead of plunging into the jet space gymnastics performed by Fulp, Lada and Stasheff,

BECAUSE THAT’S WHERE LAGRANGIAN’S HAVE THEIR DOMAIN!

AS TO
A Noether identity is a vector field - NO WAY, THOUGH HER THEOREM MAY SAY IT CORRESPONDS TO ONE

AN IDENTITY IS AN EQUATION THAT IS AUTOMATICALLY SATISFIED
CF. SYZYGIES
OR URS’ DISPLAYED FORMULA JUST BEFORE HIS SIMPLE PROPOSITION

WHICH ASSUMES THE SPACE AND THE IDENTITIES ARE DEFINED OVER SOME ALGEBRA
AS HE SAYS IN HIS DEFINITION OF GHOSTS AND ANTIGHOSTS

I BELIEVE THE ELEMENTS OF g^* AND NOT g ARE THE GHOSTS, CF GENERATORS OF THE CHEVALLEY EILENBERG COCHAIN COMPLEX

AS TO
where symmetries do not form a Lie algebra
THAT’S PHYSSPEAK FOR THE CASE IN WHICH THE SPACE OF GENERATORS OVER C^\INFTY(M) DOES NOT CLOSE UNDER THE BRACKET BUT OF COURSE THE BRACKET OF SYMMS IS A SYMM

Posted by: jim stasheff on November 8, 2007 4:23 PM | Permalink | Reply to this

Re: On Noether’s Second (BV, Part VI)

$MathML-enabled post (click for more details).$

I’ ll first need to say mor precisely how the $g$ in $C^\infty(X) \otimes g$ is defined in the first place.

I LOOK FORWARD TO ELUCIDATION!

Okay, let me recall it.

First of all, as you remark below, I have a consistent typo throughout: all my $g$ s need to be read as $g^*$ s, unfortunately.

Then, recall that I was looking at the toy example where the “action” is not a functional on a space of maps, but a mere function on a compact manifold.

(Physically speaking, this is like assuming that we are looking at the action of 0-particles aka (-1)-branes: their “worldvolume” is a single point in target space, hence the action is a mere function on target space.)

With that setup, I said:

- a symmetry is a vector field $v$ such that $\epsilon v$ preserves the equations of motion $\epsilon v( d S) = 0$ for all $\epsilon \in C^\infty(X)$

- a Noether identity is a vector field which preserves the action itself

$v(S) = 0 \,.$

To compare this to the usual definitions, note two things

- a “variational symmetry” which preserves the Lagrangian up to a divergence leaves the equations of motion unchanged.

- the toy setup I discuss follows from th full Jet-space discusssion by truncating that at 0th order, i.e. by setting all derivatives to 0 (since for the 0-particle, there is no differentiation along its worldvolume).

So you see in this case, that the definition of symmetry and Noether identity actually does coincide with the one you consider in your paper. I think.

Instead of plunging into the jet space gymnastics performed by Fulp, Lada and Stasheff,

BECAUSE THAT’S WHERE LAGRANGIAN’S HAVE THEIR DOMAIN!

Yes, I know. Sorry for expressing myself badly: I didn’t mean to say that it was somehow unnecessary to consider jet space. What I tried to do was to see if we can gain some insight into the general structure by restricting attention to a case where the jet space machinery completely disappears. Just to see if that helps to get a clearer picture of the structure that remains.

For the 0-particle, the action is just a function (instead of a functional) and I found it enjoyable to discuss the BV formalism and Noether identities etc in this toy case. I found it helped me get insight into the general problem to observe that in this special case, we are really dealing with the simple exact sequence that I discuss above:

$0 \to \mathrm{ker}(d S(\cdot)) \to \Gamma(T X) \stackrel{d S(\cdot)}{\to} C^\infty(X) \to C^\infty(X) \otimes g^* \to 0$

A Noether identity is a vector field -

NO WAY

Well, let’s see. The $R^{a I} D_I \epsilon$ in your paper reduce to $R^{a} \epsilon$ in the 0-particle case which I am talking about. These are components of vector fields on configuration space.

I can make a comment on what happens to this statement as we allow nontrivial $D_I \epsilon$ , but before I do so I’d like to hear your response to the above.

Posted by: Urs Schreiber on November 11, 2007 3:10 PM | Permalink | Reply to this

Re: On Noether’s Second (BV, Part VI)

That’s fine except a vector (field) is not its set of components! but I came to realize that that is an identification physicists especially mean. I even met a particle physicist who couldn’t rememeber his youth when a vector was an arrow! On the other hand, consider Rainich’s advice in his book on relativity.

Posted by: jim stasheff on November 11, 2007 8:26 PM | Permalink | Reply to this

Re: On Noether’s Second (BV, Part VI)

$MathML-enabled post (click for more details).$

That’s fine except a vector (field) is not its set of components!

Let me see.

Maybe I expressed myself badly:

I tried to say that what appears as a collection of component maps in your paper – and in Noether’s old paper as well as in most of the existing physics literature, for that matter – can be identified as components of a vector field on configuration space.

With due attention to the subtleties of the jet space setting, I think it is right to say that

a Noether identity IS a vector field on configuration space preserving the action functional.

This is the way Kazinski-Lyakhovich-Sharapov define it in their paper #, even though one needs to be quite aware of the fact that they completely ignore all the subtleties of infinite-dimensional function spaces and instead behave as if everything took place on a finite dimensional manifold – the way I did, too, in my toy example setup discussed here.

Most of the subtlety comes from the situation where the quantities $R^{a I}$ – in the notation of your paper with Fulp and Lada # – are nontrivial for nontrivial $I$ . But it seems to me, as a slogan, the above statement still remains true.

I should try to find the time to make that more precise.

Posted by: Urs Schreiber on November 12, 2007 6:41 PM | Permalink | Reply to this

Re: On Noether’s Second (BV, Part VI)

Identified as components - fine
IS ok as a slogan but not much more

Posted by: jim stasheff on November 13, 2007 2:17 AM | Permalink | Reply to this

Re: On Noether’s Second (BV, Part VI)

$MathML-enabled post (click for more details).$

A while ago Jim stasheff wrote:

AS TO

where symmetries do not form a Lie algebra

THAT’S PHYSSPEAK FOR THE CASE IN WHICH THE SPACE OF GENERATORS OVER $C^\infty(M)$ DOES NOT CLOSE UNDER THE BRACKET BUT OF COURSE THE BRACKET OF SYMMS IS A SYMM

I might have a confusion about this point:

clearly, the bracket of two symmetries is always a symmetry.

But it’s not always true that the space of symmetries is of the form $C^\infty(X) \otimes g$ for $g$ a finite-dimensional Lie-algebra with an action on $C^\infty(X)$ . Right?

I was thinking that this is the extra condition one might or might not have satisfied in a given example.

Posted by: Urs Schreiber on November 14, 2007 9:21 PM | Permalink | Reply to this

Re: On Noether’s Second (BV, Part VI)

You are thinking of X as the principal bundle? and the symms as ?
If X were a trivial bundle, then
C^\infty(X) \otimes g would make sense.
Cf. ghosts as (duals of) a *generating set* of the full symm algebra

Posted by: jim stasheff on November 15, 2007 1:41 PM | Permalink | Reply to this

Re: On Noether’s Second (BV, Part VI)

$MathML-enabled post (click for more details).$

You are thinking of $X$ as the principal bundle?

I wasn’t. But let’s do that.

So, say $X \to X/G$ is a principal $G$ -bundle with $g = Lie(G)$ .

Then the

- vertical vector fields $\Gamma_v(T X)$ would be the symmetries

- the left-invariant vertical vector fields would be the generators $\Gamma_{v,li}(T X) \simeq g$ .

Every vertical vector field would be a tensor product of smooth functions on $X$ with left-invariant vertical vector fields.

$\Gamma_v(T X) \simeq C^\infty(X) \otimes \Gamma_{v, li}(T X)$

Right?

Posted by: Urs Schreiber on November 15, 2007 4:47 PM | Permalink | Reply to this

Re: On Noether’s Second (BV, Part VI)

$MathML-enabled post (click for more details).$

I wrote:

Right?

Luckily I can discuss such things with Danny now.

Of course that was not quite right. It’s true only for trivial $G$ -bundles. In general instead of $C^\infty(X) \otimes g$ we have sections of the adjoint bundle.

Posted by: Urs Schreiber on November 15, 2007 5:08 PM | Permalink | Reply to this

Re: On Noether’s Second (BV, Part VI)

$MathML-enabled post (click for more details).$

I wrote:

Of course that was not quite right

Let me try to clean the mess I have cause here up a little.

So, we have a space of

$local symmetries \subset \Gamma(T X)$

which are those vector fields leaving a function $S \in C^\infty(X)$ invariant.

We say we have a Lie algebra of local symmeties if

$local symmetries \simeq C^\infty(X) \otimes g$

for $g$ some Lie algebra.

Do we agree on that so far?

Then, let’s assume we have the special case that a group $G$ integrating $g$ acts freely on $X$ , such that $p : X \to X/G$ is a principal $G$ -bundle and such that our invariant function

$S = p^* s$

is the pullback of a function $s \in C^\infty(X/G)$ on the quotient.

Then, I suppose, we have

$local symmetries = vertical vector fields on X \to X/G$

and moreover

$(vertical vector fields on X \to X/G) \simeq C^\infty(X) \otimes g \,.$

Posted by: Urs Schreiber on November 16, 2007 1:03 PM | Permalink | Reply to this

Re: On Noether’s Second (BV, Part VI)

I agree only if local means
X = G x X/G

Posted by: jim stasheff on November 17, 2007 12:42 PM | Permalink | Reply to this

Re: On Noether’s Second (BV, Part VI)

$MathML-enabled post (click for more details).$

I agree only if local means [trivial bundle]

Wait, what would you otherwise disagree with?

Do we agree that vertical vector fields on the total space is functions on the total space tensor the Lie algebra?

Posted by: Urs Schreiber on November 18, 2007 12:43 PM | Permalink | Reply to this

Re: On Noether’s Second (BV, Part VI)

My confusion
Remind me how you are seeing this
There is a map g –> Gamma(TX)
given by differentiating the principal action
then you cna multiply by anything in C^\infty(X)?

Posted by: jim stasheff on November 18, 2007 7:55 PM | Permalink | Reply to this

Re: On Noether’s Second (BV, Part VI)

Indeed the component ridden version (Urs is disparaging component versions!!!) is what is usually meant.

N.B. `the space of’ needs to be specified with care.

Kosmann-Schwarzbach has a marvelous book about Noether and her Varitional Theorems I and II with due attention to Noether’s Second Vartiational Thm.

Posted by: jim stasheff on November 2, 2007 1:17 PM | Permalink | Reply to this

Re: On Noether’s Second (BV, Part VI)

$MathML-enabled post (click for more details).$

Urs is disparaging component versions!!!

Not in general. What I don’t like is to have only the component version.

I want

a) the intrisic version to know what’s going on

b) the component version for computations.

Kosmann-Schwarzbach has a marvelous book about Noether and her Varitional Theorems I and II with due attention to Noether’s Second Vartiational Thm.

Thanks, I’ll see if I can look at that. Finally met her in Sheffield, by the way. Last time she gave a talk in Hamburg I was travelling and hence missed her.

Posted by: Urs Schreiber on November 2, 2007 3:20 PM | Permalink | Reply to this

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October 31, 2007