## October 30, 2007

### BV for Dummies (Part V)

#### Posted by Urs Schreiber

On my way back from Oxford, I am spending a night in a hotel close to Manchester airport to get my plane tomorrow morning. Luckily they have a public terminal here. This allows me to talk a little about BV formalism.

John began his last course on Quantization and Cohomology by focusing a bit of attention on a seemingly boring special case: that of statics instead of dynamics.

Here I’ll do something similar for the BV formalism (Part I, II, III, IV):

I use a 4-term complex of vector spaces to study the simple situation of a compact manifold $M$ equipped with a smooth real-valued function $S : X \to \mathbb{R}$ which you may think of as a Lagrangian depending only on the fields (not on their derivatives) which are elements of $M$. That complex of vector spaces will extract for us the nature of the critical points of $L$.

If you like, read this in parallel with Jim Stasheff’s hep-th/9712157 from which it follows by truncating the jet space completely down to its 0th component.

On the other hand, if you follow in thoughts the point of view adopted with considerable success by Lyakhovich and Sharapov, who get quite far with thinking of field theory Lagrangians as functions on a finite-dimensional space, you may regard, I guess, the following also as a picture of aspects of the full BV machinery.

In any case, the little pedagogical exercise here is mainly supposed to make explicit the simple nature of the complex we are dealing with, which is essentially

$0 \to \mathrm{ker}(d S(\cdot)) \hookrightarrow \Gamma(T X) \stackrel{d S(\cdot)}{\to} C^\infty(X) \to C^\infty(X) \otimes g^* \to 0 \,,$ where $g$ denotes the Lie algebra of symmetries $\rho : g \to \Gamma(T X)$ $d S (\rho(\cdot)) = 0$ of our function.

And probably I won’t be able to refrain from making some comments on the higher categorical interpretation of what is going on.

Of course for a general BV situation this $g$ may be a higher Lie algebra (a Lie $n$-algebra) and its action may be weak and all that, but that shall be ignored here for the time being.

Instead, we just focus on understanding the critical points of our function $S$ on $X$. These are the zeros of the differential $d S \,.$ They form the space $\{x \in X | d S(x) = 0\}$ called the shell.

If that were it, we’d be done. The point is that we want more: what we really want are the connected components of the space of critical points. $\pi_0(\{x \in X | d S(x) = 0\}) \,.$ Namely if there is, for instance, a whole line of critical points, we say that all points on this line are isomorphic configurations. What we are interested in, then, are the isomorphism classes of critical points.

A function on such isomorphism classes would be a ‘gauge invariant observable’. These are one of the main objects of interest.

So we need to do two things:

a) first identify the functions on the space of critical points among all functions on $X$

b) then identify the gauge invariant functions among all functions on critical points.

We want to tackle a) by quotienting the space of all functions by those vanishing on the critical points. One assumes ‘sufficient regularity’ of $S$, such that the algebra of functions vanishing on the space of critical points is generated from all functions of the form $d S(v)$ for $v$ any vector field on $X$. Think of this as picking out all the different ‘components’ of the ‘covector’ $d S$.

We can say this elegantly as follows: forming the 2-term complex of vector spaces $BV_a = (0 \to \Gamma(T X) \stackrel{d S(\cdot)}{\to} C^\infty(X) \to 0 )$ which we think of as being concentrated in degree -1 and 0, with the differential being of degree +1, we find that its cohomology in degree 0 is precisely the space of functions on the critical points:

$H^0(BV_a) = C^\infty(X)/\mathrm{im}(d S(\cdot)) \,.$

We know that we can think of this little complex as a Baez-Crans type 2-vector space (a $\mathrm{disc}(\mathbb{R})$ module category) – and we should in fact do that, I belive. Because then passing to the cohomology is not a trick pulled out of thin air – but something that just is: 2-vector spaces are equivalent to their cohomology. Meaning that any 2-vector space of the form

$0 \to V_{-1} \stackrel{f}{\to} V_0 \to 0$ is equivalent to $0 \to \mathrm{ker}(f) \stackrel{0}{\to} \mathrm{coker}(f) \to 0 \,.$ And for vector spaces we simply have $\mathrm{coker}(f) = V_0/\mathrm{im}(f) \,.$

Of course it doesn’t matter whether you think of this happeneing in the 2-category of 2-vector spaces or, more traditionally, in the 2-category $2\mathrm{Term}$, of 2-term chain complexes (which is equivalent to $2\mathrm{Vect}$, even as a symmetric monoidal 2-category) – but you need to take care of the 2-morphisms (otherwise you’ll end up talking about “quasi-isomorphisms”, and who wants a quasi 1-thing if we can have an honest 2-thing?).

So we would like to say: whether or not we pass to the cohomology explicitly, i.e. whether or not we pass to the skeleton of our 2-vector space, it is in any case equivalent to the (1-)vector space of functions on the critical points.

That would be what we want. But we are not quite there yet.

Trouble is: we would want the cohomology in degree -1 to vanish, for our intended statement to be true. If that vanished, our 2-vector space would indeed be equivalent to the 1-vector space of functions on the critical points.

But $H^{-1}(BV_a)$ does in general, not vanish. It is equal to $H^{-1}(BV_a) = \mathrm{ker}(d S(\cdot)) \,.$ This is equal to the space of all those vector fields on $X$, which have the property that flowing along them leaves our function $S$ invariant.

This is the space of Noether relations. These vector fields detect to which degree the different ‘equations of motion’ given by the components of the 1-form $d S$ are not independent.

Notice the subtle but crucial difference between Noether relations and symmetries, conceptually:

Noether relations are those vector fields along which our functions $S$ is invariant.

Symmetries are those vector fields along which the equations of motion are invariant, on shell.

In our case these two concepts actually coincide. But for more general setups they need not. See page number 8 (below equation (10)) of the article by Kazinski, Lyakhovich & Sharapov for a very clear-sighted discussion.

(They are interested also in the case where we have a system defined by its equations of motion for which no Lagrangian description exists. Like self-dual $(2k)$-form theories. By considering this generalization, they are lifting accidental degeneracies of concepts, which gives them a clearer view.)

In any case, it follows that, in general, our 2-vector space of functions in not yet equivalent to the 1-vector space of functions on the critical points of $S$.

Instead, we have an extension $(\mathrm{ker}(d S(\cdot)) \to 0) \;\;\;\; \stackrel{t}{\hookrightarrow} \;\;\;\; ( \Gamma(T X) \stackrel{d S(\cdot)}{\to} C^\infty(X)) \;\;\;\;\to\;\;\;\; (0 \to C^\infty(X)/\mathrm{im}(d S(\cdot)))$ of the space we are interested in by the 2-vector space which has a trivial space of objects and the Noether identities of our function $S$ as its space of morphisms.

We would like to quotient the 2-vector space that we have by that of Noether identites. So we should just take the cokernel of the inclusion $t$. But let’s be sophisticated about: let’s take this cokernel not in the world of 2-vector spaces, but in the world of 3-vector spaces. In other words, let us take the weak cokernel (aka homotopy quotient) of $t$.

The result is, simple enough, the 3-vector space which is given by the 3-term chain complex $BV_b := \mathrm{wcoker}(t) = ( 0 \to \mathrm{ker}(d S(\cdot)) \hookrightarrow \Gamma(T X) \stackrel{d S(\cdot)}{\to} C^\infty(X) \to 0 ) \,.$

By its very construction, this 3-vector space now is equivalent to the 1-vector space of functions on the critical points. The additional part in degree -2 kills the cohomology in degree -1.

Of course the construction the way I presented here is highly revisionistic (even when compared to Jim Stasheff’s already revisionistic desciption). People have plenty of other names for this beast.

Mathematicians know this construction as the Koszul-Tate resolution, while physicists call it the antifield-antighost construction. Jim Stasheff emphasized this point in The (secret?) homological algebra of the BV approach.

Here is the table that assigns the names to the symbols

$\array{ \mathrm{ker}(d S(\cdot)) &\hookrightarrow& \Gamma(T X) &\stackrel{d S(\cdot)}{\to}& C^\infty(X) \\ \\ Tate && Koszul \\ \\ antighosts && antifields && fields \\ \\ deg = -2 && deg = -1 && deg = 0 } \,.$

Whichever way you like to look at it, we obtain a 3-vector space which is a puffed-up version of the 1-vector space of functions on critical points that we are looking for.

And I should maybe admit that this space of critical points is called the shell in physics (since in the case that $S$ is the action functional of a relativistic particle, this space is a standard hyperbola in Minkowski space, of all vectors whose Minkowski norm is a given value called the mass. This hyperbola is traditionally called the mass shell, for not so un-obvious reasons.)

So we obtain a 3-vector space which is a puffed-up version of the 1-vector space of functions on the shell.

All right. That solves part a). Next there is part b): we want to restrict to those functions that are gauge invariant, i.e those that are defined on shell and are invariant under the flow along the vector fields sitting in $\mathrm{ker}(d S(\cdot))$.

We know from BV, Part IV that the Lie algebroid which is obtained by differentiation from the action Lie groupoid which is associated to the action of the group of symmetries of $S$ (the subgroup of the group of diffeomorphisms of $X$ leaving $S$ invariant) is, dually, the Chevalley-Eilenberg algebra of the corresponding Lie algebra with values in the module of functions on $X$.

Somehow we want to be acting with that action Lie algebroid on our 3-vector space of on-shell functions and then pass to the subspace of functions acted on trivially.

I still don’t quite know the nice abstract $n$-categorical way to say this. It seems there should be one, related to a general concept of actions of Lie $n$-groups, but right now I feel I am still missing something.

So next I’ll need to fall back to what everybody is used to do in this BV business anyway: we simply go ahead proposing constructions and check in the end that these constructions are justified by the results they produce for us.

After that disclaimer, let’s agree that it is kind of obvious that if we want to incorporate the action of the symmetry algebra into our context, we simply add the vector space underlying the corresponding Chevalley-Eilenberg action Lie algebroid in degree 1, to form the 4-term chain complex:

$BV = ( 0 \to \mathrm{ker}(d S(\cdot)) \hookrightarrow \Gamma(T X) \stackrel{d S(\cdot)}{\to} C^\infty(X) \stackrel{\rho}{\to} C^\infty(X) \otimes g^* \to 0 ) \,. )$

Here $g$ is the Lie algebra of the group of symmetries, such that the space of vector fields in the image of $\rho : g \to \Gamma(T X)$ completed as a $C^\infty(X)$-module yields the space of all vector fields fixing the function $S$. We may think of $\rho$ as an element in $\rho \in \Gamma(T X) \otimes g^* \,.$ Then the map denoted $C^\infty(X) \stackrel{\rho}{\to} C^\infty(X) \otimes g^*$ consists of first tensoring with $\rho$ and then postcomposing with the obvious action $\Gamma(T X ) \otimes C^\infty(X) \to C^\infty(X) \,.$

You might wonder if I am now thinking about this 4-term complex as a 4-vector space. But I do not. Rather, I think we are seeing some notion of action Lie 3-algebroid which is slightly more general than usually considered:

Recall that ordinary Lie algebroids are the same as Lie-Rinehart pairs (these are pairs consisting of a Lie algebra and an associative algebra, both modules over each other in a consistent way that models the archetypical obvious Lie-Rinehart pair $(C^\infty(C),\Gamma(T X))$). If you try to categorify that, you clearly expect to find pairs consisting of Lie $n$-algebras acting – not on 1-algebras but on $n$-algebras.

It seems that we are seeing here something of this sort. But, as I said, I am not entirely sure yet what the best abstract way yo say this is. (On the other hand, you can see that the kind of Lie $n$-algebroid that I am talking about here, which is more like categorified Lie-Rinehart pairs than like that underlying the usual notion of NPQ-manifolds, is related to the construction of Lie algebras of inner automorphism groups of Lie $n$-groupoids that I talked about in The $G$ and the $B$ (and yesterday with Prof. Nigel Hitchin). Maybe more on that later, when the time is ripe.)

In any case, we end up looking at this 4-term complex

$\array{ \mathrm{ker}(d S(\cdot)) &\hookrightarrow& \Gamma(T X) &\stackrel{d S(\cdot)}{\to}& C^\infty(X) &\stackrel{\rho}{\to}& C^\infty(X) \otimes g^* \\ \\ Tate && Koszul && && Chevalley-Eilenberg \\ \\ antighosts && antifields && fields && ghosts \\ \\ deg -2 && deg -1 && deg 0 && deg 1 } \,.$

Its cohomology in degree 0 is now that of on-shell gauge-invariant functions.

Posted at October 30, 2007 6:17 PM UTC

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### Re: BV for Dummies (Part V)

When extending the BV formalism to contain not only fields, antifields and ghosts, but also their conjugate momenta, I ran into a subtlety: the BV formalism does not work even for the harmonic oscillator, unless you introduce extra antifields.

Let us do this explicitly in Fourier space. A basis for X are the modes φ(k). We want to implement the Euler-Lagrange equations

E(k) = (k2 - m2) φ(k) = 0.

Hence we introduce at degree -1 the antifields φ*(k). The differential acts as

δ φ(k) = 0
δ φ *(k) = E(k).

The nice thing about this is that different k decouple, so we can consider each k separately. For k2 != m2, the cohomology vanishes: φ(k) is both closed and exact, and φ*(k) is neither. However, for k = +-m, we have e.g.

δ φ(m) = δ φ *(m) = 0.

Therefore the cohomology is generated by

φ(m), φ(-m), φ*(m), φ*(-m)

This gives the right result for H0(δ), but the higher cohomology groups do not vanish, as they should. It is easy to fix this, of course; just add two new generators at degree -2, θ(m) and θ(-m), such that

d θ(k) = φ*(k), for k = +- m.

But generators at degree -2 seems like a gauge symmetry.

The phenomenon is not particular to the harmonic oscillator, as a simple counting argument shows. There are two physical dofs which solve the EL equations, and hence the BV cohomology must have two dofs. But it consists of infinitely many φ(k) and infinitely many φ *(k), which count negative because they are fermions, and in this case infinity - infinity = 0. To fix the balance, we need two bosonic θ’s, as is seen explicitly for the harmonic oscillator.

### Re: BV for Dummies (Part V)

I am not a physicist, but I think that what Thomas says sounds right.

Also, there are different approaches to BV quantization, and (for the time being) I happen to believe that what is needed is a dg wheeled PROP as shown in this paper.

Posted by: Charlie Stromeyer Jr on October 31, 2007 12:57 PM | Permalink | Reply to this

### Re: BV for Dummies (Part V)

BV formalism does not work

It seems to me that the apparent paradox which you seem to find is a result of using a discontinuous object where something smooth is required.

To slightly simplify your example, consider on $X = \mathbb{R}$ the function $S : X \to \mathbb{R}$ $x \mapsto \frac{1}{2} x^2 \,.$

Its “equations of motion” are $d S = 0 \Leftrightarrow x = 0 \,.$

This is a single isolated critical point.

So we build the corresponding Koszul complex

$\array{ \Gamma(T \mathbb{R}) &\stackrel{d S(\cdot)}{\to}& C^\infty(\mathbb{R}) \\ \\ anti-fields && fields }$

You were making the point that this map has a nontrivial kernel. But that is only true if we’d allow something like distributional vector fields on $\mathbb{R}$. Namely, the apparent kernel of this map are the “vector fields” that are zero everywhere, except at $x = 0$.

But these are not really elements of $\Gamma(T \mathbb{R})$, which are supposed to be at least differentiable. So the kernel is in fact trivial, as it should be.

That’s because there are no continuous symmetries here leaving the “action” invariant.

This changes for instance as soon as we include one more “static harmonic oscillator” and consider for instance

$S : X \to \mathbb{R}^2$ $x \mapsto \frac{1}{2} (x^2 + y^2) \,.$

Now there is still, naively, a single isolated point $d S = 0 \Leftrightarrow (x,y) = 0$ but now it is in fact infinitely degenerate, as all the (smooth!) vector fields in $\mathrm{ker}(d S(\cdot) ) = \left\{ v \in \Gamma(T \mathbb{R}^2), v(x,y) = f(x,y) \left( \array{ y\\ -x } \right) | f \in \C^\infty(\mathbb{R}^2) \right\}$ act trivially on the function $S$ (namely by rotating around the origin).

This kernel is indeed isomorphic to the corresponding Chevalley-Eilenberg space of ghosts, which is $C^\infty(X) \otimes \mathrm{Lie}(so(2)) \,.$

Posted by: Urs Schreiber on October 31, 2007 9:55 PM | Permalink | Reply to this

### Re: BV for Dummies (Part V)

It seems to me that the apparent paradox which you seem to find is a result of using a discontinuous object where something smooth is required.

It is still worthwhile to point out that if you relax this smoothness requirement (whose necessity I dispute), BV only works if you treat solutions and gauge symmetries in the same way.

I follow the prescription in HT, chapter 16 (or maybe 17), with E = dS/dφ. You can do the same analysis for the free field in d dimensions; E(k) vanishes for every k on mass-shell, so you must add extra antifields to kill the unwanted cohomology for all k on the (d-1)-dimensional manifold k2 = m2.

The difference is that a genuine gauge symmetry is labelled by a d-dimensional manifold whereas the solutions depend on a (d-1)-dimensional manifold. But you don’t see this distinction in the abbreviated formalism of HT.

What could be simpler than the harmonic oscillator?

### Re: BV for Dummies (Part V)

What could be simpler than the harmonic oscillator?

The static harmonic oscillator.

Equivalently, the dynamical 0-particle (aka (-1)-brane) in a square potential.

Posted by: Urs Schreiber on November 1, 2007 11:40 AM | Permalink | Reply to this

### Second thought

On second thought, your claim that continuity has something to do with it is incorrect. Recall that the second-order antifields were introduced to fix the problem with unwanted cohomology. However, in the absense of the θ’s, the unwanted cohomology generated by φ*(+-m) has exactly the same continuity properties as the wanted cohomology generated by φ(+-m). Hence continuity is irrelevant, unless you also claim that the usual solutions to the harmonic oscillator are ruled out because they are discontinuous.

### Re: Second thought

Hence continuity is irrelevant, unless you also claim that the usual solutions to the harmonic oscillator are ruled out because they are discontinuous.

We want to rerstrict to the shell without talking about discontinuous functions on the total space. So instead we talk about all smooth functions on the total space and weakly quotient out by those that vanish on shell (namely we don’t quotient out explicitly but instead pass to the weak quotient, which is the complex with the quotient as its cohomology).

Gauge symmetries are smooth vector fields (or at least sufficiently often differentiable ones) on this setup.

It’s things like that which are hard to appreciate in components, which is why I went through the trouble of spelling it all out in invariant terms.

The gauge symmetry you imagine is the vector field

$k \mapsto \delta(k,m)\mathbf{\varphi}^*(k)$

which is not a continuous function. It would describe a gauge transformation that flows no point of config space to anywhere else. That’s the paradox you are seeing. And it is due to the fact that there is indeed no gauge orbit to flow along in your example.

Posted by: Urs Schreiber on November 1, 2007 11:48 AM | Permalink | Reply to this

### Re: Second thought

We want to rerstrict to the shell without talking about discontinuous functions on the total space. So instead we talk about all smooth functions on the total space and weakly quotient out by those that vanish on shell (namely we don’t quotient out explicitly but instead pass to the weak quotient, which is the complex with the quotient as its cohomology).

But I don’t introduce any discontinuous functions (the θ’s) until after the spurious cohomology has already been established.

Gauge symmetries are smooth vector fields (or at least sufficiently often differentiable ones) on this setup.

Sure. Of course I don’t claim that the harmonic oscillator has any gauge symmetries, only that you get spurious cohomology at nonzero degree unless you introduce second-order antifields.

It’s things like that which are hard to appreciate in components, which is why I went through the trouble of spelling it all out in invariant terms.

It would be a bit disturbing if the BV formalism did not work in components, no? Especially since much of it is done in very sparse component formalism, where one does not even distinguish between discrete and continuous indices.

### Re: Second thought

But I don’t introduce any discontinuous functions (the $\theta$’s) until after the spurious cohomology has already been established.

The antifield

$\delta(k,\pm m)\mathbf{\varphi}^*(k)$

is not in the domain that I would admit. For $X$ the config space, this should be something in $\Gamma(T X)$, with $\Gamma$ understood as being smooth (or at least sufficiently differentiable) sections.

Posted by: Urs Schreiber on November 1, 2007 4:38 PM | Permalink | Reply to this

### Re: Second thought

You should reread my original post again. The prescription in HT is this: introduce φ(k) and φ*(k) for all k in R. The cohomology of δ acting on these fields is

H*(δ) = ker δ/im δ = C(φ(m), φ(-m), φ*(m), φ*(-m))

rather than the desired

H*(δ) = C(φ(m), φ(-m)).

I do not introduce any delta function in k for the antifields. It comes out directly from the cohomology, but I start with antifields for all k in R, just as I start with fields for all k in R.

### Re: Second thought

You should reread my original post again.

Somehow I have the feeling the two of us, once again, don’t get anywhere here.

Instead of replying again, I think I’ll wait if maybe somebody else’s comment helps to break this deadlock.

Posted by: Urs Schreiber on November 1, 2007 6:32 PM | Permalink | Reply to this

### Re: Second thought

Maybe things become clearer if we compactify time on a circle with radius N/m, N integer. This does not change anything substantial, except that k becomes a discrete variable, delta functions become Kronecker deltas, and continuity ceases to be relevant.

### Re: Second thought

and continuity ceases to be relevant.

I don’t know. I’d think BRST and BV is all about continuous (smooth, even) symmetries. The BRST operator is essentially the differential on the Lie group of symmetries. It’s not clear to me immediately how to generalize it away from the Lie setup.

If your setup has no continuous symmetries, then there is nothing for BRST-BV to do. As far as I understand.

Posted by: Urs Schreiber on November 2, 2007 3:27 PM | Permalink | Reply to this

### Re: Second thought

If your setup has no continuous symmetries, then there is nothing for BRST-BV to do.

Hamiltonian BRST: nothing. BV: restrict from arbitrary fields to those that solve the equations of motion.

I view BV, and resolutions in general, as a trick to describe some complicated space in terms of simpler spaces. One can even do it for finite-dimensional spaces, which applies to the harmonic oscillator with time on a finite lattice with periodic boundary conditions; differentials are replaced by finite differences. The formulas are identical to those in my first comment.

Let u be an n-dimensional vector and A an n x n matrix. Assume that we are interested in the space of solutions to the equation

Au = 0.

Nontrivial solutions only exist if A is singular; assume that there are p solutions, i.e. rank A = n-p. Introduce fermionic n-dimensional vectors u* and define the differential by

δ u = 0
δ u* = Au

We can only invert A on an (n-p)-dimensional supspace, so p u’s are not exact. Vectors of the form Au span an (n-p)-dimensional subspace, so p u*’s are closed. H0(δ) is the desired space of u’s, but the cohomology groups at nonzero degree do not vanish, because some u*’s survive.

Counting: we start with +n u’s and -n u*’s, totally zero dofs with sign. The cohomology is generated by +p u’s and -p u*’s, still totally zero dofs. To kill the u*’s, we need to introduce +p second-order antifields θ.

Everything is of course exactly as in my first comment, except that we deal with finite-dimensional vectors and matrices.

### Re: Second thought

Resolutions of course work in every context. But you were wondering why in your example the cohomology in degree -1 of the resoltution did not coincide with what you expected from the symmetries. The reason was that for that to happen, you need to impose contiunity or smoothness.

Posted by: Urs Schreiber on November 2, 2007 6:22 PM | Permalink | Reply to this

### Re: Second thought

Um, I was not wondering anything, I pointed out a subtlety. It is true that the extra antifields are not smooth in the sense that they are much fewer than the ones connected to gauge symmetries: in d dimensions, the modes are labelled by a (d-1)-dimensional manifold rather than a d-dimensional one. However, I think it’s better to have a complex that works even without extra assumptions about smoothness. Also, the error is small since H0 is correct, and that’s what we really care about.

The problem arose when I wanted to apply the BV formalism to do canonical quantization in a covariant way, which requires an honest Poisson bracket and not just an antibracket. Then the problem becomes acute, because the antifield momenta have degree +1 and expressions like φ*(m)π*(m) contribute to H0. Fixed by adding the extra antifields and their momenta, however at the price of losing manifest covariance to a degree.

### Re: Second thought

Finite dim needs a modifier. Sometimes we have an algebra over a ring, e.g. C^\infty M,
with a fin dim vector space of gnerators, although the algebra is inf dim and the structure constants may be structure functions

Posted by: jim stasheff on November 9, 2007 1:40 PM | Permalink | Reply to this

### Re: Second thought

I don’t understand this comment. Surely the space of functions over a finite-dimensional space is almost always infinite-dimensional, e.g. M = real numbers.

Anyway, in my original comment, I exactly follow eqn (17.4) of Henneaux and Teitelboim for the harmonic oscillator. The nonzero cohomology groups are

H0 = { F }
H-1 = { F1 φ*(m) + F2 φ*(-m) }
H-2 = { G φ*(m) φ*(-m) }

where F, F1, F2, G are smooth functions of φ(m) and φ(-m) only. Clearly H-1 and H-2 are as smooth as H0, so smoothness has nothing to do with the extra cohomology.

It would be surprising if nobody has checked that BV (at least the version of BV described in HT chapter 17) works out right for the harmonic oscillator, but actually this seems to be the case. Perhaps this is what happens if you spend too much time thinking about the big picture without checking that the details work out in concrete examples.

### Re: Second thought

Thomas,
Thanks for following through. I agree about the big picture versus significant examples, or as Frank Adams said: there are machine builders and machine operators.
I wasn’t following your earlier observations,
so could you explain why extra cohomology is a problem OR in what sense is it extra?
Can discuss by e-mail before telling the whole cafe if you prefer.

Posted by: jim stasheff on November 10, 2007 1:55 PM | Permalink | Reply to this

### Re: Second thought

Well, I only check my job email, which I cannot access until Monday, so I prefer to answer here.

By extra cohomology, I mean all cohomology groups except H0; we only have a resolution of a space V if H0 = V and all other Hn = 0. You know better than me why acyclity is desirable; suffice it to say that Henneaux and Teitelboim claim, in Theorem 17.2, that all cohomology at nonzero degree vanishes.

For notation and definition of the BV complex for the harmonic oscillator, I refer to my first comment in this thread.

### Re: Second thought

Thanks
That’s what I suspected
sound like they didn’t go far enough
or apply the Stalinist? principle:
when you encounter an obstrcution, kill it
or more technically if the Koszul complex is
not a resoluiton
proceed to Koszul-Tate
only be sure then to adjoin the dual generators on the symmetry side.

Posted by: jim stasheff on November 11, 2007 12:26 PM | Permalink | Reply to this

### Re: Second thought

only be sure then to adjoin the dual generators on the symmetry side.

I might have a problem with understanding which point Thomas Larsson is making, but it seemed to me that in his original comment he was wondering why to the antifields he found no ghosts seemed to correspond.

I still think it is due to the fact that his antifields were introduced in error, an error induced by thinking entirely in components.

Consider configuration space $X$ to be the real line, $X = \mathbb{R}$ and take the “action” to be the simple function $S : X \to \mathbb{R}$ $S : x \mapsto x^2 \,.$

This has a single isolated extremum $(d S)_x = 0 \Leftrightarrow x = 0 \,.$

If instead we had taken config space to be $X = \mathbb{R}^2$ and $S : (x,y) \mapsto x^2$ then this extremum would have been infinitely degenerate, as there would have been a smooth vector field $v = \frac{\partial}{\partial y}$ preserving the action, its equations of motion and the space of extrema.

In this situation BV formalism would have told us that the invariance of the action correspond to the symmetry of the equations of motion, and both to that vector field.

But in the case at hand, there is no gauge orbit running through the “shell”. The shell is a single point. There is nothing in the kernel of $\Gamma(T X) \stackrel{d S }{\to} C^\infty(C) \,.$

The analog of the thing in the kernel that Thomas Larsson is considering would be the “vector field”

$v = \delta_{x=0} \frac{\partial}{\partial x}$

Certainly admitting this as an element in $\Gamma(T X)$ leads to all kinds of unexpected pathologies. But there is no reason to include it. In the present example, the kernel of $\Gamma(T X) \stackrel{d S }{\to} C^\infty(C)$ is simply 0-dimensional, as is hence the space of antighosts, and hence the space of ghosts. Just as it should be.

Please let me know if it is clear, or not, how this example coresponds to the oscillator example that Thomas Larsson is describing. If it is not clear, I can spell out the analogous discussion for the harmonic oscillator.

Posted by: Urs Schreiber on November 11, 2007 3:29 PM | Permalink | Reply to this

### Re: Second thought

The way I look at it, we have symmetries before restricting on shell, not just on shell symms. In the case of structure *functions*, they don’t commute, i.e. the
sum of the CE and koszul differentials no longer squares to 0 so we add antighosts.

Posted by: jim stasheff on November 11, 2007 8:30 PM | Permalink | Reply to this

### Re: Second thought

The way I look at it, we have symmetries before restricting on shell, not just on shell symms. In the case of structure functions, they don’t commute, i.e. the sum of the CE and koszul differentials no longer squares to 0 so we add antighosts.

Okay.

I notice that the literature on BRST-BV roughly splits in two parts:

- in one part the symmetries are taken as the starting point and the antighosts are introduced as a fix

- in the other part the Koszul-Tate complex is taken as the starting point, and the symmetries then as a derived concept.

I am equally happy with both these points of view.

But what I am really looking for is a nice unified description in terms of reps of Lie $n$-algebras on $n$-vector spaces.

I have come to the suspicion, that the point of view that a Lie $n$-algebroid is, dually, a non-negatively graded dg-algebra is not the full stroy.

The case of a Lie 1-algebroid shows what is going on: the Lie algebra acts on the degree-0 part, which is an ordinary vector space.

Hence as we pass to actions of Lie $n$-algebras, they should be acting on $n$-vector spaces, situated in degree $0, -1, \cdots -(n-1)$.

It becomes obvious when we think of Lie alghebroids as Lie-Rinehart pairs $(A,B)$. Categorifying these, we expect both $A$ and $B$ to increase in categorical dimension. But assuming non-negatively graded dg-algebras is like assuming $A$ raises in dimension, but not $B$.

I might have to learn about homotopy Lie-Rinehart pairs…

Posted by: Urs Schreiber on November 12, 2007 9:27 AM | Permalink | Reply to this

### Re: Second thought

In the Hamiltonian situation, going by ghost degree (= - antighost degree)
the Chevballey-Eilenberg part is non-negatively graded and the Koszul-Tate part is non-positively graded.
Categorify that!
Analaogous remarks for the Lagrangian = BV case where the pairing is of degree -1.

One way out: Regard CE cochain algebra as Hom(CE chain alg, )
and the chain alg is then also non-pos graded
so flip signs on all the gradings??

Posted by: jim stasheff on November 12, 2007 2:06 PM | Permalink | Reply to this

### Re: Second thought

I might have a problem with understanding which point Thomas Larsson is making, but it seemed to me that in his original comment he was wondering why to the antifields he found no ghosts seemed to correspond.

My simple point was that in order for BV to yield a resolution of the phase space, we must add two bosons. That’s all I’m saying. A triviality really, except that HT explicitly say otherwise.

You could introduce ghosts corresponding to these extra antifields if you want to. But you don’t want to, because then the cohomology vanishes completely, in particular H0 = 0. This is in agreement with the counting argument. Since passage to cohomology preserves the net surplus of bosons, we must start with two more bosons than fermions. If we also add ghosts, the net surplus is again zero.

### Re: Second thought

My simple point was that in order for BV to yield a resolution of the phase space, we must add two bosons.

And I keep saying: no, not if you take all fields to be properly well behaved.

Hopefully we can move away from this deadlock eventually.

Posted by: Urs Schreiber on November 12, 2007 11:42 AM | Permalink | Reply to this

### Re: Second thought

OK, one last try. Forget about the θ’s. Then the cohomology is
generated by

φ(m), φ(-m), φ*(m), φ*(-m).

Do we agree on this?

The desired cohomology is generated only by

φ(m), φ(-m).

Do we agree on this?

So you claim that φ(m), φ(-m) are well behaved, but
φ*(m), φ*(-m) are ill behaved.

Why this difference???

### Re: Second thought

Do we agree on this?

That’s the point we are disagreeing about.

Why this difference???

Because in one case we have the quotient of a space of smooth functions by another space of smooth functions. This space contains elements which may usefully be thought of as “functions supported over a single point”, although technically they are really smooth functions representing a class of smooth functions that all have the same value at that point.

In the other case, the delta-like entity would have to be an element of a space of smooth functions itself. But it is not. Hence it does not exist.

Consider again the smooth manifold $X = \mathbb{R}$ and a smooth function $S : x \mapsto x^2$ on $X$.

Then the cohomology of $\Gamma(T X ) \stackrel{d S(\cdot)}{\to} C^\infty(X) \to 0$ in the middle part is the quotient space

“smooth functions on $X$ modulo smooth functions on $X$ which vanish at $x = 0$”.

This quotient space may heuristically be thought of as being spanned by a delta-distribution supported at $x = 0$. But it does not hurt to remember that that’s not quite what it really is, technically. But anyway, it does contain the objects that you think it should contain.

The next question is: what is $\mathrm{ker}(d S (\cdot))$? Even before we think about the cohomology of $\mathrm{ker}(d S(\cdot)) \hookrightarrow \Gamma(T X ) \stackrel{d S(\cdot)}{\to} C^\infty(X) \to 0$

in the second term, we need to know what this $\mathrm{ker}(d S(\cdot))$ is. And if we assume vector fields to be smooth (or at least continuous), then $\mathrm{ker}(d S(\cdot)) = 0$ is the 0-vector space. It contains no element. No well-behaved non-vanishing vector field acts trivially on the function $S$.

This is the reason that there are no antighosts here.

Posted by: Urs Schreiber on November 12, 2007 3:21 PM | Permalink | Reply to this

### Re: Second thought

OK, I see your point now. Thanks.

However, I don’t like the corollaries that follow from the continuity assumption:
1. You cannot treat the harmonic oscillator by making a Fourier transformation.
2. The harmonic oscillator on the line is not a limiting case of the harmonic oscillator on the circle.
3. There are strictly fewer antifields than fields, but one nevertheless index them by the same label.

OTOH, I don’t see why we can’t relax the continuity assumption. After all, BV is just a trick to describe the space we are interested in, and if there is some other complex with the same cohomology, it should work just as well. Without the assumption of continuity, one must introduce the extra antifields.

### Re: Second thought

OK, I see your point now. Thanks.

All right, thanks.

You cannot treat the harmonic oscillator by making a Fourier transformation.

Hm, now why is that? :-)

I think you can. I think that has been the case we have been discussing. You switch to Fourier space. There the “shell” is given by two points (momenta $\pm m$). Hence the cohomology of physical fields is the quotient space

“smooth functions on $\mathbb{R}$ (now thought of as the space of frequencies) modulo smooth functions vanishing at $k = \pm m$”.

Again, no antighosts exist (as long as we keep ignoring time translation, that is) because no nontrivial smooth vector field on $\mathbb{R}$ preserves the action.

There are strictly fewer antifields than fields, but one nevertheless index them by the same label.

Hm, I am not sure what this means.

I don’t see why we can’t relax the continuity assumption

If it all were just about Koszul-Tate, we could greatly relax the assumptions. The Koszul-Tate resolution alone can be discussed in the context of rather general rings.

But the point of BV is to merge the Koszul-Tate complex with the Chevalley-Eilenberg complex. One wants to see the gauge symmetries act on the space of fields and on the shell.

Here one might be able to drop smoothness, possibly. I don’t know. But as we drop continuity, it becomes hard to tell what “gauge symmetry” is supposed to mean.

In any case, the point of BV is to deal with quotients of large spaces by Lie groups. If not a Lie group is acting but something discrete, BV seems to be the wrong tool to look at. I think.

Posted by: Urs Schreiber on November 12, 2007 4:50 PM | Permalink | Reply to this

### Re: Second thought

Here one might be able to drop smoothness, possibly. I don’t know. But as we drop continuity, it becomes hard to tell what “gauge symmetry” is supposed to mean.

In the case of free electromagnetism, one gets two extra antifields when k2 = 0, corresponding to the two transverse massless photons which solve the equations of motion. The BV complex can be read off from (8.15) of hep-th/0701164, if you put α, β and γ to zero.

Anyway, should not BV work even if the time dimension is compact and energy discrete? Continuity is irrelevant then.

### Re: Second thought

Thanks for the reference to your paper. I no longer subscribe to hep-th. Suggest you cross list to QA.

Posted by: jim stasheff on November 13, 2007 2:36 AM | Permalink | Reply to this

### Re: Second thought

I thought phase space referred to the Hamiltonian version, hence BFV not BV??
In BV, the antifields are there to kill the EL equations, leaving the algebra of fcns on their soln space.

Posted by: jim stasheff on November 12, 2007 2:00 PM | Permalink | Reply to this

### Re: Second thought

HT call it the “covariant phase space”. An orbit which solve the EL equations is completely determined by its position and velocity, or momentum, at time t=0. Hence the space of such orbits can be identified with phase space.

### Re: Second thought

Oh, abus de langage!
in that case, I’ll give in

Posted by: jim stasheff on November 13, 2007 2:28 AM | Permalink | Reply to this

### Re: Second thought

cf. Noether II

Posted by: jim stasheff on November 12, 2007 2:13 PM | Permalink | Reply to this

### Re: BV for Dummies (Part V)

In re:
See page number 8 (below equation (10)) of the article by Kazinski, Lyakhovich, Sharapov for a very clear-sighted discussion.

It’s p.7 in the version now at the arXiv
but the comments below (10) are indeed enlightening. They call attention to the
pairing/duality which is the essence of Noether’s II and explains the graded bracket

The misleading identification of identities AS vector fields is clarified indicially by calling attention to the COEFFICIENTS being the same.

Some other comments before (10) I find
not helpful at all in clarifying what’s going on.

Posted by: jim stasheff on November 8, 2007 4:53 PM | Permalink | Reply to this

### Re: BV for Dummies (Part V)

This is really a response to Urs’ original posting
but I’m blog illiterate.

in re: Lie n-algs acting on n-algs
something like this appears in

Publications results for “Author=(kjeseth)”
MR1854643 (2002i:17030) Kjeseth, Lars A homotopy Lie-Rinehart resolution and classical BRST cohomology. Homology Homotopy Appl. 3 (2001), no. 1, 165–192 (electronic). (Reviewer: Daniel K. Nakano) 17B55
PDF Doc Del Clipboard Journal Article
MR1854642 (2002i:17029) Kjeseth, Lars Homotopy Rinehart cohomology of homotopy Lie-Rinehart pairs. Homology Homotopy Appl. 3 (2001), no. 1, 139–163 (electronic). (Reviewer: Daniel K. Nakano) 17B55

second order anti-fields??

second order in the same sense as e.g. ghosts of ghosts?

Posted by: jim stasheff on November 8, 2007 4:39 PM | Permalink | Reply to this

### Re: BV for Dummies (Part V)

Jim Stasheff pointed me to

Kjeseth, Lars A homotopy Lie-Rinehart resolution and classical BRST cohomology. Homology Homotopy Appl. 3 (2001), no. 1, 165–192 (electronic). (Reviewer: Daniel K. Nakano) 17B55

Great, thanks.

I am currently looking at HOMOTOPY RINEHART COHOMOLOGY OF HOMOTOPY LIE-RINEHART PAIRS by the same author.

This looks quite like what I was looking for. Will have to absorb this a little further, though.

Do you know if the author ever manages or even tries to merge an $L_\infty$ algebra together with the module it acts on into one single differential structure, generated in both positive as well as negative degree?

I would be really surprised if that didn’t turn out to be the simple answer.

Posted by: Urs Schreiber on November 12, 2007 10:35 PM | Permalink | Reply to this

### Re: BV for Dummies (Part V)

In fact, Lars is doing the opposite
Given one single differential structure
he tries to untangle it in terms of an L_\infty alg acting on a moduel up to higher homtopy

In the regular case, he does get an L_\infty-algebra L and an L_\ifnty algebra on L\oplus M but NOT split as one would’ve expected.

Posted by: jim stasheff on November 13, 2007 2:21 AM | Permalink | Reply to this

### Re: BV for Dummies (Part V)

What does “Lie–Rinehart” mean?

Posted by: John Baez on November 12, 2007 11:02 PM | Permalink | Reply to this

### Re: BV for Dummies (Part V)

Lie-Rinehart has several other names.
It consists of a pair (L,A) of a Lie alg and an assoc alg such that each is a module over the other
and satisfying one or two additional conditions.

Posted by: jim stasheff on November 13, 2007 2:42 AM | Permalink | Reply to this

### Re: BV for Dummies (Part V)

I’m not sure but I thought Lie-Rinehart
was an alternative name for a Lie algebroid?

Posted by: David Ben-Zvi on November 13, 2007 6:24 AM | Permalink | Reply to this

### Re: BV for Dummies (Part V)

I thought Lie-Rinehart was an alternative name for a Lie algebroid?

The definition of a Lie-Rinehart pair is essentially equivalent, while different and somewhat more general, to the standard definition of a Lie algebroid.

(Lie-Rinehart is a little more general, at least as compared to the standard textbook definition of Lie algebroid, since it directly admits noncommutative base spaces.)

Lie-Rinehart pair. A Lie-Rinehart pair $(A,L)$ is a Lie algebra $L$ and an associative algebra $A$, such that $L$ is a module over $A$ and $A$ is a Lie module over $L$, with the two actions being mutually compatible in the obvious way.

This “obvious way” is directly deduced by looking at the archetypical Lie-Rinehart pair $(C^\infty(X),\Gamma(T X))$ where the associative algebra is that of smooth functions on some manifold $X$, and where the Lie algebra is that of vector fields on $X$.

I am not sure yet I have found the best online reference, but googling for “Lie-Rinehart” turns up a bunch of references, for instance

Johannes Huebschmann, Lie-Rinehart algebras, descent, and quantization

In the special case that the associative algebra $A$ in the Lie-Rinehart pair $(A,L)$ is that of smooth functions on some manifold $X$ $A = C^\infty(X)$ the Lie-Rinehart pair is equivalent to a Lie algebroid structure on $X$: the Lie algebra $L := \Gamma(E)$ is to be identified with the space of sections of the vector bundle defining the Lie algebroid

$\array{ E &&\stackrel{\rho}{\to}&& T X \\ & \searrow && \swarrow \\ && X }$

where the Lie bracket is induced by the bracket on sections of $E$ provided by the Lie algebroid structure.

While the definitions are equivalent (as far as I understand, Lie-Rinehart is much older but Lie-algebroids became much more popular after they had been introduced) they are different as definitions, clearly. That matters as one tries to categorify.

For one, the definition of a Lie-Rinehart pair admits a smooth elegant way, to my mind, to obtain a Lie-Rinehart pair $(A_C,L_C)$ from a Lie groupoid $C$:

the Lie algebra $L_C$ is the Lie algebra of the Lie group of inner automorphisms of $C$. (See p. 9 of Tangent Categories). The action of that on functions on $\mathrm{Obj}(C)$ is induced from the canonical projection $C \to \mathrm{Obj}(C)\times \mathrm{Obj}(C)$ of any Lie groupoid onto the codiscrete groupoid over its space of objects.

The reason I have the feeling we should be looking at Lie-algebroids in terms of Lie-Rinehart pairs is:

I have the suspicion we all keep missing something with the standard definition of Lie $n$-algebroids (in dual terms of non-negatively graded dg-manifolds): this definition is like categorifying the Lie algebra $L$ in the Lie-Rinehart pair $(A,L)$, but not the associative algebra $A$ it acts on.

I bet the missing negative degrees in the non-negatively graded dg-manifolds can be interpreted as coming from the categorification of the associative algebra part $A$ of a categorified Lie-Rinehart pair.

Posted by: Urs Schreiber on November 13, 2007 9:27 AM | Permalink | Reply to this

### Re: BV for Dummies (Part V)

Lie-Rinehart and Lie algebroid have in common the example given, but Lie-Rinehart is pure unanchored algebra or at least no smoothness involved, hence? more easily categorified?

Posted by: jim stasheff on November 13, 2007 2:11 PM | Permalink | Reply to this

### Re: BV for Dummies (Part V)

Lie-Rinehart is pure unanchored algebra or at least no smoothness involved, hence? more easily categorified?

Yes, that’s what I am thinking.

Personally, I find it important to realize: the Lie bracket on the sections of the Lie algebroid is actually the Lie bracket of the group of inner automorphisms of the correspoinding Lie groupoid.

That fact is captured by Lie-Rinehart, but not as manifestly by the vector-bundle-with-anchor-and-brackets definition.

The difference might seem negligible, but I think it is worthwhile to make it.

Zoran Skoda will arrive in a few minutes, and he, Danny and I are in the process of thinking about categorifying Lie algebroids and Lie-Rinehart pairs.

By the way: in Sheffield I had heard about Kyrill Mackenzie’s double Lie algebroids (which I was ignorant of before). From just the talk I heard I had the strong impression that these are pretty much categories internal to Lie algebroids. But if so, it wasn’t known. Do you know?

Posted by: Urs Schreiber on November 13, 2007 2:21 PM | Permalink | Reply to this

### Re: BV for Dummies (Part V)

Sorry to be notationally challenged except in the symbolism I gre up with. I should have paid more attention here. Indeed, in degree -2 i would expect anti ghosts with
conjugate’ ghosts in degree 1. Doesn’t the harmonic oscillator have symms?

Posted by: jim stasheff on November 11, 2007 8:40 PM | Permalink | Reply to this

### Re: BV for Dummies (Part V)

Doesn’t the harmonic oscillator have symms?

Sure, (time)translation symmetry. But in order to see this, we need to pass to jet space, properly, since this comes from $\delta x = x'$ or equivalently, in the notation of equation (9) of your paper, from $Q^a_E := u_1^a \,.$

But the discussion above did not take the $u^a_1$ into account.

Posted by: Urs Schreiber on November 12, 2007 9:14 AM | Permalink | Reply to this

### Re: BV for Dummies (Part V)

When we pass to cohomology, bosonic and fermionic degrees of freedom cancel in pairs. The net bosonic surplus therefore remains the same. Since we eventually want to end up with two bosons and no fermions, we must start with a net bosonic surplus of 2. Hence there are two unmatched bosons.

Thus we must choose between acyclicity and a non-degenerate antibracket. We cannot have both.

### Re: BV for Dummies (Part V)

The bosons are fields and the fermions are anti-fields
so the differential applied to the fermions kills not the bosons
but the EoM of the bosons.

Does that help?

The acyclicity is only for the K-T part
The total cohomology of the BV complex
is supposed to look right in dim 0
not everywhere

cf. in the Ham case, the rest of the cohomology picks up the foliation cohomology
the de Rham but only with rewspect to diff forms along the leaves but with global coeff

Posted by: jim stasheff on November 12, 2007 2:11 PM | Permalink | Reply to this

### Re: BV for Dummies (Part V)

The acyclicity is only for the K-T part
The total cohomology of the BV complex
is supposed to look right in dim 0
not everywhere

Did you really mean that? The BV complex is not supposed to be acyclic?

Since the harmonic oscillator has no gauge symmetries, BV = KT, no?

Read the post On Noether's Second (BV, Part VI)
Weblog: The n-Category Café
Excerpt: On Noether's second theorem and ghost/antighost pairing.
Tracked: November 1, 2007 12:39 AM
Read the post Modules for Lie infinity-Algebras
Weblog: The n-Category Café
Excerpt: On modules for Lie infinity algebras in general and the definition given by Lars Kjeseth in particular.
Tracked: November 13, 2007 9:14 PM
Read the post Something like Lie-Rinehart infinity-pairs and the BV-complex (BV, part VII)
Weblog: The n-Category Café
Excerpt: Notes on something like Lie infty-algebroids in the light of the BV complex.
Tracked: November 20, 2007 8:10 PM
Read the post On BV Quantization, Part VIII
Weblog: The n-Category Café
Excerpt: Towards understading BV by computing the charged n-particle internal to Z-categories, secretly following AKSZ.
Tracked: November 29, 2007 10:21 PM
Weblog: The n-Category Café
Excerpt: On adjunctions between spaces and n-groupoids induced by homming out of the fundamental n-groupoid.
Tracked: May 6, 2008 6:44 PM
Read the post Frobenius algebras and the BV formalism
Weblog: The n-Category Café
Excerpt: Bruce Bartlett is looking at the latest article by Cattaneo and Mnev on BV-quantization of Chern-Simons theory.
Tracked: November 14, 2008 1:24 PM

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