I could not find a definition of second covariant derivative of vector field (sort of a second affine connection), so I tried to formulate it on my own. Can you check if it is correct?

A second covariant derivative on a differentiable manifold $M$ is a mapping

$\nabla\nabla: \mathfrak{X}(M)\times\mathfrak{X}(M)\times\mathfrak{X}(M)\rightarrow\mathfrak{X}(M)$

$(X,Y,Z) \mapsto\nabla_Y\nabla_XZ$

which satisfies following properties:

  • $\nabla_{fY+gV}\nabla_XZ=f\nabla_Y\nabla_XZ+g\nabla_V\nabla_XZ=\require{cancel} \color{red}{\cancel{\nabla_Y\nabla_{fX+gV}Z}}$

  • $\color{red}{\nabla_Y\nabla_{fX+gV}Z=f\nabla_Y\nabla_XZ+Y(f)\nabla_XZ+g\nabla_Y\nabla_VZ+Y(g)\nabla_VZ}$

  • $\nabla_Y \nabla_X(Z+V)=\nabla_Y\nabla_XZ+\nabla_Y\nabla_XV$

  • $\nabla_Y\nabla_X(fZ)=\nabla_Y\left(f\nabla_XZ+(Xf)Z\right)=f\nabla_Y\nabla_XZ+(Yf)\nabla_XZ+(Xf)\nabla_YZ+(Y(Xf))Z$,

    for $X,Y,Z,V\in \mathfrak{X}(M)$, $f,g\in C^\infty(M)$ and $Xf=X^i\frac{\partial f}{\partial x_i}$ is directional derivative.

Third properties is taken from do Carmo (p.90), but I have doubts about first one. I will be gratefull for any suggestions.

Edit: I post my calculations:

$\nabla_{fY+gV}(\nabla_XZ)=f\nabla_Y\nabla_XZ+g\nabla_V\nabla_XZ$ $\color{red}{\nabla_Y\nabla_{fX+gV}Z=\nabla_Y(f\nabla_XZ+g\nabla_VZ)=f\nabla_Y\nabla_XZ+Y(f)\nabla_XZ+g\nabla_Y\nabla_VZ+Y(g)\nabla_VZ}$ $\nabla_Y \nabla_X(Z+V)=\nabla_Y(\nabla_XZ+\nabla_XV)=\nabla_Y\nabla_XZ+\nabla_Y\nabla_XV$


1 Answer 1


You seem to be misunderstanding something about the covariant derivative. It is not true that $\nabla_X(fY) = f\nabla_XY$, instead $\nabla_X(fY) = X(f)Y + f\nabla_XY$. You make this mistake in the second equality of the first dot point and the second equality of the second line of your edit. Everything else looks correct.

  • $\begingroup$ Sure, my mistake. Thank you for pointing it out. $\endgroup$
    – Ender
    Commented Oct 2, 2017 at 12:27

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