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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeJul 26th 2018

    added pointer to

    • R. Catenacci, P.A. Grassi, S. Noja, Superstring Field Theory, Superforms and Supergeometry (arXiv:1807.09563)

    diff, v20, current

    • CommentRowNumber2.
    • CommentAuthorDavid_Corfield
    • CommentTimeJul 26th 2018
    • (edited Jul 26th 2018)

    the differential forms on a supermanifold are characterized by two numbers: the form degree and the picture number,

    Presumably the picture number is a number assigned to a picture, something changed by a picture changing operator.

    Notice that in the supergeometric setting the de Rham cohomology depends in general on two numbers, ii and jj above, and indeed the de Rham complex in supergeometry is not really a complex, but a bicomplex instead. The first number, denoted as ii, refers to the actual degree of the forms and for j=0j=0 it is indeed just the analog of the usual degree of a differential form. The second one, denoted with jj, is instead a supersymmetric novelty, indeed it refers to the so-called picture number associated to a form in supergeometry (arXiv:1708.02820)

    It seems to count the number of terms of a certain kind in the form. And the ’picture’ language comes from physics, as in

    the “picture number” gives precisely the number of applications of ghosts needed until one annihilates the ground state. (comment)

    • CommentRowNumber3.
    • CommentAuthorDavid_Corfield
    • CommentTimeJul 26th 2018

    May as well add a sentence on the above about picture numbers.

    diff, v21, current

    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTimeJul 26th 2018
    • (edited Jul 26th 2018)

    Hm, this sounds strange. The plain de Rham complex on a supermanifolds is ×(/2)\mathbb{N} \times (\mathbb{Z}/2)-graded, the first being form degree, the second being superdegree (number of odd variables in a monomial, mod 2). This is basic, but see for instance Deligne-Freed 99.

    In the article “picture number” comes in by, in addition, requiring a concept of integral super forms, needed for integration. (equations 5.9 to 5.12).

    My understanding was that the geometric meaning of the infamous picture changing operators appearing in the quantization of the NSR string has been clarified in Belopolsky 97 and advertized in Witten 12. But I have not actually studied that, it’s just what I got out of the talk that Witten gave about it at StringMath2012 in Bonn.

    • CommentRowNumber5.
    • CommentAuthorUrs
    • CommentTimeJul 26th 2018

    I have made it read as follows:

    If a choice of integral top-forms is made, needed for a notion of integration over supermanifolds, then there is an additional grading by “picture number” (Belopolsky 97b, Witten 12), see (Catenacci-Grassi-Noja 18 (5.8) to (5.12)).

    diff, v22, current

    • CommentRowNumber6.
    • CommentAuthorDavid_Corfield
    • CommentTimeJul 26th 2018

    Ah, ok.

    Idle thought: do we see a differential hexagon appear for this supergeometric case?

    I have a vague memory that we agreed there should be a hexagon for the super-modalities. Ah yes, a fracture square here.

    • CommentRowNumber7.
    • CommentAuthorUrs
    • CommentTimeJul 26th 2018
    • (edited Jul 26th 2018)

    At the moment I have no insight into any potential abstract/synthetic formulation of integration over supermanifolds. It might well deserve to have such, because it is notoriously a thorny subject, but I don’t know.

    Notice though that the use of this in string theory is entirely in the path integration formulation of the quantization of the NSR superstring, namely in the integration over the moduli space of super-Riemann structures on the worldsheet, which is the residual path integral for the 2-d worldsheet super gravity after super-diffeomorphism gauge fixing. This is one example of a path integral quantization right at the boundary between toy examples (1d QFT = quantum mechanics) and ill-defined examples (3\geq 3d interacting QFT ). Whether this boundary position means that it should be counted among those cases where the path integral has an actual right of existence, or those cases where it should be replaced by something else that is not just a fantasy, I don’t know.

    • CommentRowNumber8.
    • CommentAuthorUrs
    • CommentTimeJul 27th 2018
    • (edited Jul 27th 2018)

    spelled out an actual definition (here) of super differential forms, first on super Cartesian spaces, then from there on any super formal smooth set.

    Also expanded the table of signs appearing in the graded-commutativity:

    A\phantom{A}sign ruleA\phantom{A} A\phantom{A}Deligne’sA\phantom{A} Bernstein’sA\phantom{A}
    A\phantom{A}x ax b=x^{a} \; x^{b} = A\phantom{A}+x bx a+ x^{b} \; x^{a}A\phantom{A} A\phantom{A}+x bx a+ x^{b} \; x^{a}A\phantom{A}
    A\phantom{A}x aθ α=x^a \;\theta^\alpha =A\phantom{A} A\phantom{A}+θ αx a+ \theta^\alpha \; x^aA\phantom{A} A\phantom{A}+θ αx a+ \theta^\alpha \; x^aA\phantom{A}
    A\phantom{A}θ αθ β=\theta^{\alpha} \; \theta^{\beta} =A\phantom{A} A\phantom{A}θ βθ α- \theta^{\beta} \; \theta^{\alpha}A\phantom{A} A\phantom{A}θ βθ α - \theta^{\beta} \; \theta^{\alpha}A\phantom{A}
    A\phantom{A}x a(dx a)=x^{a} (\mathbf{d}x^{a}) =A\phantom{A} A\phantom{A}+(dx b)x a+ (\mathbf{d}x^{b}) x^{a}A\phantom{A} A\phantom{A}+(dx b)x a+ (\mathbf{d}x^{b}) x^{a}A\phantom{A}
    A\phantom{A}θ α(dx a)=\theta^\alpha (\mathbf{d}x^a) =A\phantom{A} A\phantom{A}+(dx a)θ α+ (\mathbf{d}x^a) \theta^\alphaA\phantom{A} A\phantom{A}(dx a)θ α{\color{blue}{-}} (\mathbf{d}x^a) \theta^\alphaA\phantom{A}
    A\phantom{A}θ α(dθ β)=\theta^{\alpha} (\mathbf{d}\theta^{\beta}) = A\phantom{A} A\phantom{A}(dθ β)θ α- (\mathbf{d}\theta^{\beta}) \theta^{\alpha}A\phantom{A} A\phantom{A}+(dθ β)θ α{\color{blue}{+}} (\mathbf{d}\theta^{\beta}) \theta^{\alpha}A\phantom{A}
    A\phantom{A}(dx a)(dx b)= (\mathbf{d}x^{a}) (\mathbf{d} x^{b}) =A\phantom{A} A\phantom{A}(dx b)(dx a)- (\mathbf{d} x^{b}) (\mathbf{d} x^{a})A\phantom{A} A\phantom{A}(dx b)(dx a) - (\mathbf{d} x^{b}) (\mathbf{d} x^{a})A\phantom{A}
    A\phantom{A}(dx a)(dθ α)= (\mathbf{d}x^a) (\mathbf{d} \theta^{\alpha}) =A\phantom{A} A\phantom{A}(dθ α)(dx a) - (\mathbf{d}\theta^{\alpha}) (\mathbf{d} x^a) A\phantom{A} A\phantom{A}+(dθ α)(dx a) {\color{blue}{+}} (\mathbf{d}\theta^{\alpha}) (\mathbf{d} x^a) A\phantom{A}
    A\phantom{A}(dθ α)(dθ β)=(\mathbf{d}\theta^{\alpha}) (\mathbf{d} \theta^{\beta}) = A\phantom{A}+(dθ β)(dθ α) + (\mathbf{d}\theta^{\beta}) (\mathbf{d} \theta^{\alpha})A\phantom{A} A\phantom{A}+(dθ β)(dθ α) + (\mathbf{d}\theta^{\beta}) (\mathbf{d} \theta^{\alpha})A\phantom{A}

    diff, v23, current

    • CommentRowNumber9.
    • CommentAuthorLuigi
    • CommentTimeJan 10th 2020
    • (edited Jan 10th 2020)

    Hello, I would have a quite trivial question on notation.

    What do we exactly mean by the name “bosonic differential form”?

    Do we mean just an even differential form, e.g. R:=R abdx adx b+R αβdθ αdθ βR := R_{ab} dx^a \wedge dx^b + R_{\alpha\beta} d\theta^\alpha \wedge d\theta^\beta ?

    Or do we mean a “horizontal” form, e.g. R:=R abdx adx bR := R_{ab} dx^a \wedge dx^b ?

    Or do we mean a form which can be expressed by using only bosonic vielbeins, e.g. R:=R abe ae bR := R_{ab} e^a \wedge e^b ?

    Thanks in advance!

    • CommentRowNumber10.
    • CommentAuthorUrs
    • CommentTimeJan 10th 2020

    I am using it for the latter.

    This matches the use of bosonic modality: With X\overset{\rightsquigarrow}{X} the bosonic aspect of the supermanifold XX, hence its body of the supermanifold, a bosonic form on XX is one pulled back from X\overset{\rightsquigarrow}{X}.

    • CommentRowNumber11.
    • CommentAuthorLuigi
    • CommentTimeJan 13th 2020
    • (edited Jan 13th 2020)

    Thank you for your reply. I am still not sure I understood. If B=B μν(x)dx μdx νB=B_{\mu\nu}(x) dx^\mu \wedge dx^\nu is a B-field on the X\overset{\rightsquigarrow}{X}, we can pull it back to XX to construct

    H=dB μν(x)dx μdx ν+i(ψ¯Γ aΓ 10ψ)e aH = dB_{\mu\nu}(x) \wedge dx^\mu \wedge dx^\nu + i(\bar{\psi}\wedge\Gamma_a\Gamma_{10}\psi)\wedge e^a

    where the vielbein satisfies de a=ψ¯Γ aψde^a = \bar{\psi}\Gamma^a\psi.

    Is this statement correct? Does this mean B μν(x)dx μdx νB_{\mu\nu}(x)dx^\mu \wedge dx^\nu on X\overset{\rightsquigarrow}{X} does not depend on the θ\theta-coordinates?

    Thanks

    • CommentRowNumber12.
    • CommentAuthorUrs
    • CommentTimeJan 13th 2020

    So the pullback itself is just B μνdx μdx νB_{\mu \nu} d x^\mu \wedge d x^\nu (the bosonic modality knows about super-geometry but not about super-symmetry, it just forgets the odd-graded coordinates θ\theta not caring about which forms are super left-invariant).

    Maybe you could say again what exactly your question is. Your original question seemed to be purely about terminological convention. I think the convention I am promoting makes good sense, but in the end it’s just a choice of terminology, without mathematical content.

    • CommentRowNumber13.
    • CommentAuthorLuigi
    • CommentTimeJan 13th 2020

    Of course, I was just trying to clumsily apply the definition to some physical statement to see if I understood, but I definitely need to read carefully the papers first

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