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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeSep 20th 2011
    • (edited Sep 21st 2011)

    The BV-BRST complex of some local Lagrangian LL over some manifold Σ\Sigma famously comes equipped with a canonical Poisson bracket of grade +1: the antibracket.

    The whole point of this construction is that it serves as a tool for computing another bracket that one is interested in: the ordinary grade-0 bracket induced from the symplectic form that is obtained by descending the canonical presymplectic form on covariant phase space (as discussed there) to the quotient by the symmetries.

    Recently I was beginning to entertain myself with the idea that there should be a more immediate relation between these two brackets than I see considered in the usual literature. I was beginning to think that it should be possible to regard a Poisson n-algebra as a “homotopy” between two Poisson (n1)(n-1)-algebras. (Does that sound reasonable to anyone?)

    So I started toying around with some formulas to see if I can substantiate my guess. I am not as sophisticated yet with these computations as Frédéric, who is reading here, and likely I am making some silly mistakes. But taken together this is are two good reasons to chat about my thoughts here!

    So for the purposes of this post here, I’ll be a bit naive, following the physics literature, and say things like:

    Let, locally, {Φ a(x)}\{\Phi^a(x)\} be a collection of generators for the fields and ghosts…

    as well as

    … and let {Φ¯ a(x)}\{\bar \Phi_a(x)\} be the corresponding collection of generators for the antifields and antighosts.

    The point is that I want to write down the following suggestive symbols:

    In this basis the symplectic form of which the antibracket is the Poisson bracket is locally

    ω BV:= ΣdΦ¯ a(x)dΦ a(x). \omega_{BV} := \int_\Sigma \mathbf{d} \bar \Phi^a(x) \wedge \mathbf{d} \Phi_a(x) \,.

    Here this expression is meant to be an object in the (Kähler-)differential forms on our BV-complex. If we think of the BV-complex as the Chevalley-Eilenberg algebra of a derived L-infinity algebroid, then this expression lives in the corresponding Weil algebra. As such, the situation is very similar to that of symplectic Lie n-algebroids.

    Contrary to that case, however, the 2-form here is not closed under the given differential – here: the BV-differential – and what I am after is precisely the question:

    What is the image of this 2-form under the BV-differential?

    Notice that ω BV\omega_{BV} is of “grade” -1 and that the BV-differential will send it therefore to a 2-form of grade 0, hence an ordinary 2-form. I am about to claim that this ordinary 2-form is

    ω inω out, \omega_{in} - \omega_{out} \,,

    where ω in\omega_{in} is the canonical presymplectic form on covariant phase space for the incoming boundary of Σ\Sigma, and ω out\omega_{out} that for the outgoing boundary. So I am about to be claiming that

    ω out=ω in+d BVω BV. \omega_{out} = \omega_{in} + d_{BV} \omega_{BV} \,.

    If true, this would formalize the idea that I mentioned at the beginning: ω BV\omega_{BV} would be exhibited as a homotopy between the two choices for the canonical presymplectic form.

    In order to check this, I now appeal to the notation and results of the old but seminal article

    Gregg Zuckerman, Action principles and global geometry (pdf).

    For the remainder of this message here I will assume that you either are familiar with this, or that you now spend five minutes with glancing over the first 10 pages, which is all I’ll need.

    I think to make sense of my ω BV\omega_{BV} I want to identify the symbol d\mathbf{d} there with the variational derivative that Zuckerman writes \partial on p. 4

    d= \mathbf{d} = \partial

    For the action of the BV-differential I observe that by definition

    d BV ΣΦ¯ a(x)dΦ a(x) = Σ(d BVΦ¯ a(x))dΦ a(x) =δ ΣLδΦ a(x) ELdΦ a(x) = ΣE adΦ a = ΣE, \begin{aligned} d_{BV} \int_\Sigma \bar \Phi_a(x) \wedge \mathbf{d} \Phi^a(x) &= \int_\Sigma (d_{BV} \bar \Phi_a(x)) \wedge \mathbf{d} \Phi^a(x) \\ & = \frac{\delta \int_\Sigma L}{\delta \Phi^a(x)}_{EL} \wedge \mathbf{d} \Phi^a(x) \\ & = \int_\Sigma E_a \wedge \mathbf{d} \Phi^a \\ & = \int_\Sigma E \,, \end{aligned}

    where EE is Zuckerman’s Euler-Lagrange form from page 9. With the equations there this is equivalently (still writing “d\mathbf{d}” for Zuckerman’s “\partial”)

    = ΣdL ΣDM, \cdots = \int_\Sigma \mathbf{d} L - \int_\Sigma D M \,,

    where now on the right we have the boundary contribution MM that induces the canonical presymplectic form that I am after: this is defined as

    ω in:= inΣdM \omega_{in} := \int_{\partial_{in} \Sigma} \mathbf{d} M

    and similarly for ω out\omega_{out}.

    Collecting all this together I get finally

    d BVω BV =d BVdΦ¯ adΦ a =d(d BVΦ¯ adΦ a) =d( ΣdL ΣDM) =dd ΣL ΣDdM. \begin{aligned} d_{BV} \omega_{BV} & = d_{BV} \mathbf{d} \bar \Phi_a \wedge \mathbf{d} \Phi^a \\ & = - \mathbf{d} \left( d_{BV} \bar \Phi_a \wedge \mathbf{d}\Phi^a \right) \\ & = - \mathbf{d}\left( \int_\Sigma \mathbf{d} L - \int_\Sigma D M\right) \\ & = - \mathbf{d}\mathbf{d} \int_\Sigma L - \int_\Sigma D \mathbf{d} M \,. \end{aligned}

    Here the first term vanishes, due to d 2=0\mathbf{d}^2 = 0. To the second term the Stokes theorem applies, as on p. 10 of Zuckerman’s article. Therefore we are left with

    = ΣdM = outΣdM inΣdM =ω outω in. \begin{aligned} \cdots &= - \int_{\partial \Sigma} \mathbf{d} M \\ & = \int_{\partial_{out} \Sigma} \mathbf{d}M - \int_{\partial_{in} \Sigma} \mathbf{d}M \\ & = \omega_{out} - \omega_{in} \,. \end{aligned}

    QED.

    Anyone who made it so far, please give me a sanity check.

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeSep 21st 2011

    Igor Khavkine kindly points out by private email that all this boils down to the following simple statement:

    let in the variational bicomplex, following Zuckerman’s notation:

    • LΩ dimX,0(ConfigSpace)L \in \Omega^{dim X, 0}(ConfigSpace) be the local Lagrangian;

    • MΩ dimX1,1(ConfigSpace)M \in \Omega^{dim X-1, 1}(ConfigSpace) be the presymplectic potential density;

    • UΩ dimX,2(ConfigSpace)U \in \Omega^{dim X, 2}(ConfigSpace) be the presymplectic density;

    and finally

    • AΩ dimX,2(BVSpace)A \in \Omega^{dim X, 2}(BVSpace) be the density of the symplectic form corresponding to the anti-bracket;

    Then the statement is simply that.

    d BVA=DU. d_{BV} \, A = D U \,.

    Simple as it is, has this been considered anywhere before?

    • CommentRowNumber3.
    • CommentAuthorUrs
    • CommentTimeOct 5th 2011

    I have added a discussion along the above lines to phase space in a new section via BV.

    Since this still needs attention, maybe I should better put this on my personal web. For the moment I just left a cautionary remark. I think I’ll move it tomorrow to a more suitable place.