If we have a vector fiber bundle with a connection $D_X=D(X)$ and an endomorphism $e$; we can then define a new tensor by the following formula: $$E(e)=D_X e D_Y + D_Y e D_X  e D_Y D_X  D_X D_Y e  e D_Z + D_Z e$$ with $Z=(X,Y)$ the Lie brackets of vector fields. If $e$ is parallel, then $E(e)=0$.The antisymmetric part is reduced to the curvature tensor, but we can define with the symmetric part a new tensor: $$ T(e)= 2D(e_i) e D(e_i)  e D(e_i) D(e_i)  D(e_i) D(e_i) e$$ with the Einstein's convention. For example, we can take $e=Ric$ the Ricci curvature and deduce a new symmetric tensor. When is this tensor $E(Ric)$ an automorphism and can we define new interesting differential equations?

$\begingroup$ Perhaps you should write $E(e)(X,Y)$ in the lefthand side of the first eqya $\endgroup$ – Liviu Nicolaescu Sep 14 '18 at 13:02

$\begingroup$ Hi, the [referencerequest] 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 $D_X(e(D_YW))$? $\endgroup$ – Willie Wong Sep 14 '18 at 13:06

$\begingroup$ Ah, do you mean , as in Liviu's comment, that $E(e)$ is a covariant twotensor 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.) $\endgroup$ – Willie Wong Sep 14 '18 at 13:14

$\begingroup$ If I am not mistaken, if your vector bundle is the tangent bundle $TM$ and $D$ the LeviCivita 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, $T(\mathrm{Ric}) = \triangle_g \mathrm{Ric}$. $\endgroup$ – Willie Wong Sep 14 '18 at 13:20

1$\begingroup$ 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. $\endgroup$ – Deane Yang Sep 16 '18 at 16:49
Let $B$ be a vector bundle over a manifold $M$, $D$ be a connection on $B$, and $\nabla$ a torsionfree connection on the tangent bundle $T_*$. Given a section $f$ of $\mathrm{End}(B) = B\otimes B^*$ and a section $v$ of $B$, let $\langle f, v\rangle$ denote the section of $B$ obtained by evaluating $f$ at $v$.
Given a section $e$ of $\mathrm{End}(B)$ and vector fields $X$, $Y$, and $Z=[X,Y]$, let $E$ denote the section of $\mathrm{End}(B)$, where for any section $v$ of $B$, \begin{align*} \langle E, v\rangle &= D_X\langle e,D_Yv\rangle + D_Y\langle e,D_Xv\rangle  \langle e,D_Y(D_Xv)\rangle  D_X(D_Y\langle e, v\rangle)\\ &\quad \langle e, D_Zv\rangle + D_Z\langle e, v\rangle\\ &= \langle D_Xe, D_Yv\rangle + \langle e, D_X(D_Yv)\rangle + \langle D_Ye, D_Xv\rangle + \langle e, D_Y(D_Xv)\rangle\\ &\quad \langle e, D_Y(D_Xv)\rangle  D_X\langle D_Ye, v\rangle D_X\langle e,D_Yv\rangle\\ &\quad \langle e,D_Zv\rangle + D_Z\langle e, v\rangle\\ &= \langle D_Xe, D_Yv\rangle + \langle e, D_X(D_Yv)\rangle + \langle D_Ye, D_Xv\rangle\\ &\quad  \langle D_X(D_Ye), v\rangle  \langle D_Ye, D_Xv\rangle  \langle D_Xe, D_Yv\rangle  \langle e,D_X(D_Yv)\rangle\\ &\quad \langle e, D_Zv\rangle + D_Z\langle e, v\rangle\\ &= \langle D_X(D_Ye),v\rangle  \langle e, D_Zv\rangle + D_Z\langle e,v\rangle\\ &= \langle D^2_{XY}e, v\rangle  \langle D_{\nabla_XY}e, v\rangle  \langle e, D_Zv\rangle + \langle D_Ze, v\rangle + \langle e, D_Zv\rangle\\ &= \langle D^2_{XY}e, v\rangle  \langle D_{\nabla_XY}e, v\rangle + \langle D_Ze, v\rangle\\ &= \langle D^2_{XY}e  D_{\nabla_YX}e, v\rangle. \end{align*} Therefore, $$ E = D^2_{XY}e  D_{\nabla_YX}e $$