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The Levi-civita connection on $\mathbb R^n$ corresponds to the usual directional derivative. In this sense I expect the following to hold: $$ \left(\nabla_{\partial_i}\partial_j\right)f=\partial_{ij}f. $$ On the other hand: $\nabla_{\partial_i}\partial_j=\Gamma_{ij}^k\partial_k=0$. These results are contradictory (of course) and I'd be happy to get some clarification.

John
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2 Answers2

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You've got your parentheses in the wrong place. $\big(\nabla_{\partial_i}\partial_j\big)$ is a vector field (in this case the zero vector field), and $\big(\nabla_{\partial_i}\partial_j\big)f$ represents the derivative of $f$ in the direction of this vector field (in this case zero).

On the other hand, if you want the second covariant derivative of $f$, you need $\nabla_{\partial_i}\big(\partial_j f)$, which in Euclidean space is just $\partial_i\partial_j f$.

EDIT: Actually, what I wrote above is not quite right. Because $\partial_j f$ is just a function, its "covariant" derivative $\nabla_{\partial_i}\big(\partial_j f)$ is just $\partial_i\partial_j f$, regardless of what connection we're using. The second covariant derivative of $f$ is the $2$-tensor field $\nabla^2 f$, whose $ij$-component is $$ \nabla^2 _{ij} f = \partial_i\partial_j f - \Gamma_{ij}^k \partial_k f, $$ which in Euclidean space is just $\partial_i\partial_j f$.

Jack Lee
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  • I find it unfortunate that $\nabla_i\nabla_j f$ is often used to mean $(\nabla\nabla f)_{ij}$ instead of $\nabla_i(\nabla_jf)=\partial_i\partial_j f$, as intuition would suggest (to me, at least).

    On the other hand, using abstract indices we do have $\nabla_a\nabla_b f=\nabla_a(\nabla_b f)=(\nabla\nabla f)_{ab}$.

     – Jackozee Hakkiuz May 15 '21 at 22:59
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For the covariant derivative of $\mathbb R^n$, you have $\nabla_XY=dY.X$ where $dY$ is the differential of $Y$. This implies that $\nabla_{\partial_i}\partial_j=0$.

Jackozee Hakkiuz
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Tsemo Aristide
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