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Sep 16, 2018 at 19:30 answer added Deane Yang timeline score: 2
Sep 16, 2018 at 16:49 comment added Deane Yang According to my calculation, this is not a tensor. And, if you choose a connection on the tangent bundle and replace the last two terms in the formula above by the appropriate term, you get minus the Hessian of $e$, whose definition requires both the connection on $E$ and the connection on the tangent bundle.
Sep 14, 2018 at 17:37 comment added Deane Yang I haven't worked this out carefully, but I think what Willie wrote extends to the vector bundle case. Namely $E(e)$ is some combination of the Hessian of $e$ and the covariant derivative of $e$. I'll work out the details, if I find the time.
Sep 14, 2018 at 13:20 comment added Willie Wong If I am not mistaken, if your vector bundle is the tangent bundle $TM$ and $D$ the Levi-Civita connection, then your endormophism $e$ is a section of $T^{1,1}M$ and your $E(e)(X,Y)$ simplifies to $- \nabla^2_{X,Y} e$. In particular, $T(\mathrm{Ric}) = -\triangle_g \mathrm{Ric}$.
Sep 14, 2018 at 13:14 comment added Willie Wong Ah, do you mean , as in Liviu's comment, that $E(e)$ is a covariant two-tensor with values in the endomorphisms of your vector bundle? (Just to double check if I understand your correctly.) // Also, if you are taking traces, isn't $T(e)$ just a field of endormophisms? (You call it a new "tensor", so I am wondering if I am missing something.)
Sep 14, 2018 at 13:06 comment added Willie Wong Hi, the [reference-request] tag is normally used for asking for references to known results/concepts. I don't think it applies here, so I removed it. // I don't quite understand your definition for $E(e)$. Given a section $W$ of your vector bundle, do you intend $D_X eD_Y W$ to mean $D_X(e(D_YW))$?
Sep 14, 2018 at 13:02 comment added Liviu Nicolaescu Perhaps you should write $E(e)(X,Y)$ in the left-hand side of the first eqya
Sep 14, 2018 at 13:00 history edited Willie Wong
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Sep 14, 2018 at 12:09 history asked A.Balan CC BY-SA 4.0