Notes on BRST II: Lie Algebra Cohomology, Physicist’s Version

My initial plan was to have the second part of these notes be about gauge symmetry and the problems physicists have encountered in handling it, but as I started writing it quickly became apparent that explaining this in any detail would take me into various issues that are quite interesting, but far afield from what I want to get to. So, I hope to get back to this at some point, but for now will just assume that most of my readers know what gauge symmetry is, and that the rest just need to know that:

  • The gauge group is an infinite dimensional Lie group. Locally (on space-time), it looks like a group of maps into a finite dimensional Lie group.
  • The conventional assumption is that physics is invariant under the gauge group, so the gauge group and its Lie algebra should act trivially on physical states.
  • The actual situation is quite a bit more complicated than this, but for now we’ll focus on the simplest version of the mathematical problem that comes up here, and see how the BRST formalism deals with it. This posting will begin explaining one part of this story, starting with the simplest version of BRST cohomology, in a language familiar to physicists. The next posting will deal with Lie algebra cohomology in a more general mathematical context and work out some examples. For more about the material in this posting, see, for instance, Green, Schwarz and Witten, volume I, section 3.2.1, where they go on to apply this to the Virasoro algebra, or these lecture notes from Jose O’Figueroa-Farrill .

    Physicists always begin by choosing a basis, in this case a basis X_i of \mathfrak g satisfying [X_i,X_j]=f_{ij}^kX_k, where f_{ij}^k are called the structure constants of \mathfrak g. A representation (\pi,V) is then a set of linear operators K_i=\pi (X_i) on V satisfying [K_i,K_j]=f_{ij}^kK_k. Let \alpha^i be a basis of the dual space \mathfrak g^*, dual to the basis X_i.

    Now, extend V to \mathcal =V\otimes \Lambda^* (\mathfrak g^*), where \Lambda^* (\mathfrak g^*) is the exterior algebra on \mathfrak g^*. On this space, define the “ghost” operator c^i to be wedge-product with \alpha^i, and “anti-ghost” operator b_i to be contraction (interior product) with X_i. These operators satisfy “fermionic” anti-commutation relations

    \{c^i,c^j\}=\{b_i,b_j\}=0,\ \ \{c^i,b_j\}=\delta^i_j

    and one can get all vectors in \mathcal H from linear combinations of decomposable elements of \mathcal H (those given by repeated application of the c^i to the “vacuum vector” V\otimes \mathbf 1).

    The ghost number operator N=c^ib_i on \mathcal H has eigenvectors the decomposable elements, with integer eigenvalues from 0 to dim \mathfrak g, given by the number of ghost operators needed to produce the eigenvector from a vacuum vector.

    The BRST operator is given by

    Q=c^iK_i -\frac{1}{2}f_{ij}^kc^ic^jb_k

    which increases the ghost number by one, and has the crucial property of Q^2=0 (this comes from the fact that the f_{ij}^k satisfy the Jacobi identity). The BRST cohomology is given by considering the space ker\ Q of elements \chi of \mathcal H that are “BRST-closed”, i.e. satisfy Q\chi=0, and identifying two such elements if they are “BRST-exact”, i.e. differ by Q\lambda for some \lambda. So BRST cohomology is defined by

    H^*_Q(V)=\frac{ker\ Q}{im\ Q}|_{V\otimes\Lambda^*(\mathfrak g^*)

    with H^j_Q(V) the component of the BRST cohomology of ghost number j.

    A vector \chi=v\otimes\mathbf 1 of ghost number zero satisfies Q\chi =0 iff and only if K_iv=0 for all i, so we can identify H^0_Q(V) with the space V^\mathfrak g of \mathfrak g – invariant vectors in V.

    The essence of the BRST method is to replace the problem of finding the invariant subspace V^{\mathfrak g} of a representation V by the problem of finding the degree zero BRST cohomology H^0_Q(V).

    There are two different ways of putting an inner product on \Lambda^*(\mathfrak g*) and thus getting an inner product on \mathcal H ((\pi,V) is assumed to be unitary, so preserves a given inner product on V).

  • Given \omega_1,\omega_2\in \Lambda^*(\mathfrak g*), one can define

    < \omega_1,\omega_2> = \int \omega_1\omega_2\equiv coeff.\ of\  \alpha_1\wedge\cdots\wedge\alpha_{dim\ \mathfrak g}\  in\  \omega_1\wedge\omega_2

    (this uses the “fermionic” or “Berezin” integral \int, although I have not properly dealt with signs here. ).
    This inner product is indefinite, but it makes the BRST operator Q and ghost-operator c^i self-adjoint.

  • Use an inner product on \mathfrak g, e.g. the Killing form for a semi-simple Lie algebra, to identify 	\mathfrak g and \mathfrak g^*. This gives a Hodge operator *_{Hodge} on \Lambda^*(\mathfrak g*) that takes \Lambda^i(\mathfrak g*) to \Lambda^{dim\ \mathfrak g -i}(\mathfrak g*), and one can define

    < \omega_1,\omega_2> = \int_G \omega_1\wedge *_{Hodge} \omega_2

    (Note, here the integral sign is not Berezin integration, but the usual integration of differential forms over a compact manifold, in this case G)

    With this inner product Q and c^i are not self-adjoint on \mathcal H. To get something self-adjoint, one can consider the operator Q + Q^\dagger where Q^\dagger is the adjoint of Q, but this operator does not have a definite ghost-number.

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    15 Responses to Notes on BRST II: Lie Algebra Cohomology, Physicist’s Version

    1. MathPhys says:

      Peter,

      Can you make your BRST notes available as pdf files? That would make them easier to print.

    2. anon. says:

      MathPhys, use the free ‘Cute PDF’ print engine: http://www.cutepdf.com/Products/CutePDF/writer.asp

      It installs as a printer, so you click print, choose Cute PDF as the printer, then it asks you where you want to save the PDF. I’ve just done it for this blog post, and it works perfectly (including all the mathematics).

    3. Peter Woit says:

      As I go along, I’ll put the material from the blog posts into a single document, which will be available as a pdf, and get updated as I go along. I’ll post the link for that here soon.

    4. MathPhys says:

      anon.

      I tried CutePDF and it was very useful. Thanks.

    5. Hendrik says:

      Dear Peter,
      I know this is the physicist’s version, but as it seems so easy, let me point at some underlying technical difficulties, (mostly arising from the fact that the Lie algebra of the gauge group is infinite dimensional).

      1) “The gauge group is an infinite dimensional Lie group. Locally (on space-time), it looks like a group of maps into a finite dimensional Lie group.”
      Well, its topology is important, and there are actually quite a lot of choices (e.g. in terms of how much differentiability is needed). Usually it is given the k-Sobolev topology (w.r.t. a volume form on the base manifold M), but sometimes the smooth maps of compact support are taken too, or even Schwartz groups. It has an interesting structure;- see the thesis of Christoph Wockel on his website.

      2) “Physicists always begin by choosing a basis, in this case a basis X_i of the Lie algebra g…”
      Since the Lie algebra g of the gauge group G is infinite dimensional, the existence (and meaning) of a basis is not clear. If you mean that every element Y in the space has a unique expansion in terms of a convergent sum of multiples of the X_i (which is what your next expression suggests), then that means the set of X_i is a Schauder basis, and there are deep theorems in functional analysis stating that not every separable Banach space has a Schauder basis. So this is a point where the choice of topology on the gauge group G is important. In the case that you choose a Sobolev topology, then there are Schauder bases, but it is not clear that there are if you choose the smooth topology.

      3) “….satisfying [X_i,X_j]=f_{ij}^kX_k, where f_{ij}^k are called the structure constants of g.”
      If this is an infinite sum (over k), then a topology must be stated to get convergence of the series.

      4) “A representation (\pi,V) is then a set of linear operators K_i=\pi (X_i) on V …..”
      The type of representation appearing in physics, is where the K_i are unbounded operators on Hilbert spaces, and these cannot have full domains. There is always a dense invariant subspace of the Hilbert space (also in the domain of your observables).

      5) “Let \alpha^i be a basis of the dual space g*, dual to the basis X_i.”
      Here g* is a topological (not algebraic) dual I assume. As it is infinite dimensional, your assumption of a dual basis severely limits the type of space which g can be. It will work for Hilbert spaces, but very few other infinite dimensional topological spaces (if any).

      6) “extend V to V \otimes \Lambda (g*)…”
      Probably the space H below is the label for this. Topologies are not clear. I presume if g* is a Hilbert space, you will want \Lambda (g*) to be the antisymmetric Fock space, in which case you need to take completion of the space. Also, \otimes for (infinite dimensional) topological spaces involves choices (e.g. of cross-norm, if they are Banach spaces).

      7) “..ghost number operator N=c^ib_i on H has eigenvectors the decomposable elements, with integer eigenvalues from 0 to dim(g),…”
      The ghost number operator N=c^ib_i is written as an infinite sum;- convergence and basis dependence?

      8 ) “BRST operator is given by Q = c^iK_i -(1/2)f_{ij}^kc^ic^jb_k”
      Apart from the question on infinite sums (convergence?) in this expression, this is the formula of the BFV-BRST formalism which does not produce the right constrained theory (without ad-hoc adjustments), when you carry out your BRST prescription below the formula. The Q-operator which produces the correct one, is from the Kugo-Ojima approach where the field aspect has to be taken into account, and you have an integral over the space-time variables rather than a discrete sum.

      9) “…uses the “fermionic” or “Berezin” integral”
      How do you define a Berezin integral when you have infinitely many fermions? In the supersymmetric Taylor expansion, you need a “top term” which will not exist if there are infinitely many fermions.

      10) “Killing form”….“Hodge operator”
      Both of these are problematic in infinite dimensions.

      Of course, most of these problems do not arise when the dimension of g is finite.

    6. Thomas Larsson says:

      Hendrik,

      Perhaps we should refrain from rushing ahead. I am sure that Peter will eventually arrive at various problems associated with infinite dimensionality, but starting with the finite-dimensional case makes sense to me.

      Having said that, I will rush ahead and emphasize that the type of infinities matter, already on the Lie algebra level. For Lie algebras that are linearly infinite-dimensional, i.e. possess a Z-grading by finite-dimensional vector spaces, and positive and negative degrees are isomorphic, things are not too bad; you just have to replace loop algebras by their central extensions, the affine Kac-Moody algebras. But for algebras of gauge transformations in d dimensions, you have a natural Z^d-grading rather than a Z-grading, and then things start to become interesting…

    7. Peter Woit says:

      Hendrik,

      Thanks for the useful explanation of the many places where infinite-dimensionality can cause trouble. I should have made explicit that I’m just sticking to finite dimensions for now.

      At Thomas mentions, for d=1 gauge groups there is some understanding of what to do, with the algebraic part of the story going under the name “semi-infinite cohomology”, which I hope to get to at some point. For d larger than one, I’m not convinced that anyone knows what’s even the right way to formulate the problem. One reason for investigating different approaches in finite dimensions is that they may suggest a more promising way to deal with the situation in higher dimensions.

    8. Pingback: A semana nos arXivs… « Ars Physica

    9. Hendrik says:

      Dear Peter and Thomas:
      OK, if your lie algebra g and the representation spaces V (in your next BRST posting) are finite dimensional, then the functional analysis issues all disappear, and there is not much I can say. Then the theory at this point can only cover global gauge theories (though the observable algebra will still force your representation spaces to be infinite dimensional).

      Point (8) above still remains, in that this approach will lead to multiple copies of the physical space, and give problems with positive definiteness when you add the usual inner products.

      A last comment, is that the BRST-algorithm which you summarized for the constraint space is not yet complete, it also needs a (algebraic) part for the constrained observables. Is that coming later?

    10. Peter Woit says:

      Hendrik,

      I’m not quite sure what you mean in your question. Part of the story is the Kostant-Sternberg formalism for BRST, where the Lie algebra cohomology idea is put together with a resolution of functions on the constrained surface, in a package using Clifford algebra techniques. I’ll write about that, including about the problems with that. For the trivial case it doesn’t work…

      Another thing I plan to get to is work by various mathematicians about how “quantization and reduction commute” in certain contexts, and how to interpret this a la BRST. There are still some things about that though that I’m confused about.

    11. Hendrik says:

      Dear Peter,
      To clarify;- the BRST algorithm, like any constraint enforcement algorithm, constructs not just the constrained representation space (V^g in your notation), it also gives a method for finding the constrained observables on V^g from the original given algebra of observables. In particular this piece of the BRST algorithm specifies the constrained observables through “operator cohomology” by

      O_BRST := (Ker d)/(Ran d)

      where d(A) is the supercommutatior of Q with A. It also shows how O_BRST is represented on the physical space V^g.

      We found that this prescription produced an algebra O_BRST which is too large, i.e. it properly contains the physical algebra of a straightforward Dirac constraining.

      I’ve also looked at the “quantization and reduction commute” papers, and look forward to see your take on those. As you know, there are huge problems with general quantization schemes;- there are always choices involved, and limitations of the algebras which can be quantized. Moreover, some quantum constraints (which can be generators of gauge groups) are purely quantum, i.e. with no classical counterpart, i.e. if you do a “classical limit” they become trivial.

    12. Peter, for those of us who are not steeped in this subject, could you clarify a few points that you zip over (assuming that they are obvious):

      First- At the start of your discussion, you refer to extending V to
      V \tensor /\(g*) where /\*(g*) …
      Is there a typo here and you actually mean that you are extending V to
      V \tensor /\*(g*)
      ?
      I could imagine some sort of weirdness that involves not just an exterior algebra of forms, but an exterior algebra of multi-vector valued forms, and that’s what you mean by the notation, but it seems easier to assume a typo.

      Second- when you get to discussing inner products, what do mean by the integral notation. You and commenters agree that, at this stage of the game we are discussing a finite-dimensional space (that presumably will become a fiber at some point). Presumably we are not (yet) discussing fields of forms and vectors. So what does this integral mean — is it simply meant to be some sort of notational convenience (but what convenience is it buying?)
      Beyond that, there’s still so much elided that I don’t get.
      – Is the second inequality in the Berezin inner product a result (this integral, defined however the hell it is defined, happens to equal this cofficient) or is it a definition?
      – As for the second definition, should there not be another hodge dual outside the integral to convert the result from a (dim-g) form to a 0-form? Under normal (field) circumstances, the integral would take us from a dim-g form to a scalar, but in this context I’ve no idea what the integral means.
      – Is there a good reason why, in both these integrals, we don’t refer to the integrand as \omega_1 \wedge \omega_2 rather than the rather less obvious \omega_1 \omega_2. As far as I can tell in both cases (making allowance for the fact that I’ve no idea what the integral actually does) the result seems to be dependent on the omega1-omega2 ordering.

    13. Peter Woit says:

      Maynard,

      Thanks for the questions. You’re right about the typo, will fix.

      The part about inner products was written in a very confusing way, I’ll try and fix it a bit. One problem is that I was using the integral signs in two ways, one for the Berezin integral, the other for the fundamental class of the compact Lie group, i.e. the standard integral over differential forms. I’ll rewrite this to make it clearer.

      In the second, definite, way of defining the inner product, all sign problems can be buried in the definition of the Hodge dual. For the Berezin case, I don’t have the signs straight, but that’s not the inner product I want to use anyway.

    14. ramón says:

      Can I get the pdf files on BRST cohomology?

    15. Peter Woit says:

      ramon,

      There’s a pdf file at

      http://www.math.columbia.edu/~woit/notesonbrst.pdf

      I’ll be updating that with a new version soon.

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