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    • CommentRowNumber1.
    • CommentAuthorMirco Richter
    • CommentTimeFeb 25th 2012
    • (edited Feb 25th 2012)

    In, at

    “… the differential respects the brackets: for all xg 0 x \in g_0 and hg 1h \in g_1 we have

    δ[x,h]=[x,δh] \delta [x,h]=[x,\delta h]


    is wrong. The equation should be:

    δα(x,h)=[x,δh]\delta \alpha(x,h) = [x,\delta h]

    Since I don’t know if I have the right to change an nLab entry,I post this here as an suggestion.

    • CommentRowNumber2.
    • CommentAuthorDavid_Corfield
    • CommentTimeFeb 25th 2012

    Change away. Everyone has the right. For a typo, no need to comment here. For a substantial change, leave a note at the forum.

    • CommentRowNumber3.
    • CommentAuthorMirco Richter
    • CommentTimeFeb 25th 2012
    • CommentRowNumber4.
    • CommentAuthorjim_stasheff
    • CommentTimeFeb 26th 2012

    the differential respects the brackets: for all
    x∈g 0 and
    h∈g 1 we have



    is wrong. The equation should be:


    yes and no
    1. it is common to denote $\alpha$ as [ , ]
    2. in the context of respecting the brackets, it might be better to write

    δ[x,h]= [δx,h]\pm [x,δh]
    and then remark for h in ...
    • CommentRowNumber5.
    • CommentAuthorMirco Richter
    • CommentTimeFeb 28th 2012
    I agree on that, because the last equation is closer to the many bracket stuff of $L_\infty$-algebras.

    But then the definition in the nLab entry must be changed (or another definition must be added), because in the entry as it is there is no defined bracket $[.,.]$ for elements of degree zero and one.
    • CommentRowNumber6.
    • CommentAuthorUrs
    • CommentTimeFeb 28th 2012

    there is no

    Now there is.

    • CommentRowNumber7.
    • CommentAuthorMirco Richter
    • CommentTimeFeb 28th 2012
    • (edited Feb 28th 2012)