\nRecall that a Q-manifold<\/a> is a supermanifold \\(M\\) equipped with a homological vector field, that is an odd vector field \\(Q\\) such that \\([Q,Q]=2 Q^{2}=0\\).<\/p>\n
The algebra of functions \\(C^{\\infty}(M)\\) is thus a supercommutative differential algebra<\/a>. The product of two functions is again a function and importantly this is supercommutative:<\/p>\n \\(f g = (-1)^{\\widetilde{f} \\widetilde{g} } g f\\),<\/p>\n where I denote the Grassmann parity of a homogeneous function by the “tilde”. Just as on a classical manifold, the product of two functions is associative:<\/p>\n \\(f(gh) = (fg)h\\),<\/p>\n for all functions f,g,h. Also important is that the product itself does not carry any Grassmann parity, thus<\/p>\n \\(\\widetilde{fg} = \\widetilde{f} + \\widetilde{g}\\).<\/p>\n The prime example here is to take \\(M = \\Pi TN\\) where \\(N\\) is a classical manifold. The functions \\(C^{\\infty}(\\Pi TM)\\) are identified with (pseudo)differential forms and the homological vector field is just the de Rham differential.<\/p>\n The derived product<\/strong><\/p>\n Definition<\/strong> Let \\((M,Q)\\) be a Q-manifold. Then the derived product on \\(C^{\\infty}(M)\\) is defined as<\/p>\n \\(f \\ast g = (-1)^{\\widetilde{f}+1} Q(f)g\\).<\/p>\n Some comments are in order.<\/p>\n Remark<\/strong> One can consider a derived product on any differential algebra, one does not need the supercommutivity, but as I am interested in supergeometry the definition here suits well.<\/p>\n Right away the derived product seems a strange beast, it carries Grassmann parity itself. Being noncommutative is also interesting, but we are used to noncommutative forms of multiplication.<\/p>\n Theorem<\/strong> The derived product is associative.<\/em><\/p>\n Proof<\/strong> Explicitly Now consider<\/p>\n \\((f \\ast g)\\ast h = (-1)^{\\widetilde{f} + \\widetilde{g}}Q ((-1)^{\\widetilde{f}+1}Q(f)g )h\\).<\/p>\n Using the fact that \\(Q\\) is homological, that is odd and squares to zero we get<\/p>\n \\((f \\ast g)\\ast h =(-1)^{\\widetilde{f} + \\widetilde{g}} Q(f)Q(g) h\\).<\/p>\n QED<\/em><\/p>\n Theorem<\/strong> \\(f \\ast g = (-1)^{(\\widetilde{f}+1)(\\widetilde{g}+1)} g \\ast f + (-1)^{\\widetilde{f}+1}Q(fg)\\).<\/p>\n Proof<\/strong> Left as an exercise for the readers.<\/p>\n The above theorem shows very explicitly that the derived product is not supercommutative. It is however antisymmetric if and only if \\(Q(fg)=0\\).<\/p>\n Where has this come from?<\/strong> J.L. Loday. Dialgebras. In Dialgebras and related operads<\/em>, 7-66, Lecture Notes in Math.<\/em>, 1763<\/strong>, Springer, Berlin 2001.<\/p>\n Concluding remarks<\/strong> I am sure there is lots more to say about derived products, but this is something I am only just beginning to explore. Watch this space…<\/p>\n","protected":false},"excerpt":{"rendered":" This went “under my radar” for a while, I am not sure why, but anyway… Q-manifolds Recall that a Q-manifold is a supermanifold \\(M\\) equipped with a homological vector field, that is an odd vector field \\(Q\\) such that \\([Q,Q]=2 Q^{2}=0\\). The algebra of functions \\(C^{\\infty}(M)\\) is thus a supercommutative differential algebra. The product of … Continue reading The derived product on a Q-manifold<\/span> \n
\n\\(f \\ast (g \\ast h) = (-1)^{\\widetilde{f} + \\widetilde{g}} Q(f)Q(g)h\\).<\/p>\n
\nAs far as I am aware, the notion of a derived product can be found in the work of Loday;<\/p>\n![\"loday\"](\"http:\/\/smf.emath.fr\/files\/loday.jpg\")
\nJean-Louis Loday,
\n (12 January 1946 \u2013 6 June 2012)<\/em><\/p>\n
\nWe have a kind of noncommutative deformation of the algebra of functions on a Q-manifold via the derived product. This “new product” carries Grassman parity and is associative, it is closer to the standard notion of a product than say a Lie algebra.<\/p>\n