If we have a connection $\nabla$ different than the Levi-Civita connection, and for a Riemannian metric $g$ and $\nabla$ this relation is valid: \begin{align} Zg(X,Y)=g(\nabla_{Z}X,Y)+g(X,\nabla^{*}_{Z}Y), \end{align} $\nabla^{*}$ is the dual connection of $\nabla$. Does it make any sense to define a second order derivative as: \begin{align} \nabla\nabla^{*}_{X,Y}s=\nabla_{X}(\nabla^{*}_{Y}s)-\nabla_{\nabla^{*}_{X}Y}s? \end{align} Thank you!

  • $\begingroup$ What do you mean by dual connection? $\endgroup$ Jun 13, 2015 at 4:15
  • $\begingroup$ It is just called the dual connection(with respect to g) if it satisfies the first equation I wrote. $\endgroup$ Jun 13, 2015 at 16:29
  • $\begingroup$ Oh right. I guess it depends what you mean by "makes sense". Unless I'm making a silly mistake it's $C^\infty$-linear in $X$ and $Y$ (though the same could be said for such a construction using any two affine connections on the tangent bundle - the duality condition doesn't come in to it). $\endgroup$ Jun 13, 2015 at 16:54
  • $\begingroup$ Yes, it is $C^{\infty}$ linear in $X$ and $Y$. I just wonder would it the definition above be correct or not? $\endgroup$ Jun 13, 2015 at 17:39

1 Answer 1


If we extend $\nabla$ and $\nabla^*$ to derivations of the tensor algebra via the Leibniz rule, then actually

$$ (\nabla \nabla^* s)(X,Y) = (\nabla_X\nabla^*s)(Y)=\nabla_X(\nabla^*s(Y)) - \nabla^* s(\nabla_XY) = \nabla_X (\nabla^*_Y s) - \nabla^*_{\nabla_X Y} s. $$


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