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?

  • Perhaps you should write $E(e)(X,Y)$ in the left-hand side of the first eqya – Liviu Nicolaescu Sep 14 at 13:02
  • 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 $D_X(e(D_YW))$? – Willie Wong Sep 14 at 13:06
  • 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.) – Willie Wong Sep 14 at 13:14
  • 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, $T(\mathrm{Ric}) = -\triangle_g \mathrm{Ric}$. – Willie Wong Sep 14 at 13:20
  • 1
    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. – Deane Yang Sep 16 at 16:49

Let $B$ be a vector bundle over a manifold $M$, $D$ be a connection on $B$, and $\nabla$ a torsion-free 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, $$ E = D^2_{XY}e - D_{\nabla_YX}e $$

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.