All my notation follows Chrusciel's book "Elements of General Relativity". For a vector field $Y$ and connection $\nabla$, we define $$\nabla_a Y^b:=dx^b(\nabla_a Y)$$ where $\nabla_a:=\nabla_{\partial_a}$. So in English, this would be taking the b-th component of the vector field $\nabla_a Y$, NOT taking the covariant derivative of the coefficient function $Y^b$ in the direction of $a$. Apologies if this is standard, but I haven't encountered such notation before. Continuing, we also define Christoffel symbols as $$\nabla_a \partial_b:=\Gamma^c_{ab}\partial_c$$ My question concerns the following lines in proof of Proposition 1.4.1 on page 29:

\begin{align*} \nabla_a\nabla_b X^d&=\partial_a(\nabla_b X^d)+ \Gamma^d_{ae}\nabla_bX^c - \Gamma^e_{ab}\nabla_e X^d\\ &=\partial_a\partial_bX^d+\partial_a\Gamma^d_{be}X^e+\Gamma^d_{be}\partial_aX^e+\Gamma^d_{ac}\partial_bX^c+\Gamma^d_{ac}\Gamma^c_{be}X^e-\Gamma_{ab}^e\nabla_eX^d \end{align*}

I don't understand the first equality. Let me try computing in a way more familar to me. If we take $X=X^i\partial_i$, and define coefficeints $Y^i$ by $\nabla_b X=Y^i\partial_i$, we compute \begin{align*} \nabla_a\nabla_b X&=\partial_aY^i\partial_i+Y^j\nabla_a\partial_j\\ &=\partial_a Y^i\partial_i+Y^j\Gamma^i_{aj}\partial_i\\ &=(\partial_a Y^i+Y^j\Gamma^i_{aj})\partial_i \end{align*} A similar computation determines $Y^i=\partial_b X^i+X^k\Gamma^i_{bk}$, so we subsitute to get \begin{align*} \nabla_a\nabla_bX&=\left[\partial_a(\partial_bX^i+X^k\Gamma^i_{bk})+(\partial_bX^j+X^k\Gamma^j_{bk})\Gamma^i_{aj}\right]\partial_i\\ &=\left[\partial_a\partial_bX^i+\partial_aX^k\Gamma^i_{bk}+X^k\partial_a\Gamma^i_{bk}+\partial_bX^j\Gamma^i_{aj}+X^k\Gamma^j_{bk}\Gamma^i_{aj}\right]\partial_i \end{align*}

Note that the two answers do NOT match! In Chrusciel's computation, there's an extra number of terms, namely $$-\Gamma^e_{ab}\nabla_e X^d,$$ but the rest of the terms match up. Could someone help me find the computation in my mistake, explain where these extra negative terms come from, or explain the first equality in Chrusciel's proof?


1 Answer 1


Your mistake is a subtle one, and is related to the notational ambiguity you mention at the beginning of your question. In abstract index notation, the expression $X^b$ is to be understood to mean the vector field $X$, with the superscript $b$ not being a numerical label (i.e. the $b^{th}$ component) but rather as a formal symbol to remind us that $X$ has one "slot" to fill (i.e. it is a function of a single covector).

Therefore, in the expression $\nabla_a\nabla_b X^d$, we should identify $X^d$ as a $(1,0)$-tensor, $\nabla_b X^d$ as a $(1,1)$-tensor, and $\nabla_a \nabla_b X^d$ as a $(1,2)$-tensor. In other words, $\nabla_a \nabla_b X^d \equiv \nabla \nabla X$ in coordinate-free notation.

The action of $\nabla$ on a $(1,1)$-tensor $T$ is given in component form$^\dagger$ by $$(\nabla T)^i_{jk} = \partial_j T^i_{\ \ k} + \Gamma^i_{j\ell}T^\ell_{\ \ k} \color{red}{- \Gamma^\ell_{jk} T^i_{\ \ \ell}}$$

which is the origin of the mysterious extra term.

$^\dagger$Note that this is not abstract index notation - the indices in this final expression take numerical values, and the quantities being labeled are the components of tensors (and the Christoffel symbols).

  • $\begingroup$ I see, thank you for making me aware of this new notation! In that case, it would be wrong to define $\nabla_a Y^b$ as I have above, but rather it is just $\nabla_a Y$ in more familiar notation. It also looks like $\nabla_a$ isn't $\nabla_{\partial_a}$, but rather it's just $\nabla$, as the variance is increasing. Finally, just to check, you mean to say $\nabla_a\nabla_b X^d$ is a $(1,2)$-tensor correct? $\endgroup$
    – user344261
    Aug 28, 2022 at 13:18
  • $\begingroup$ @ZackFox Yes. The seemingly intentional ambiguity between index notations can be very frustrating, unfortunately - it arises because there's really no good alternative. My rule of thumb is that whenever $\nabla$ is involved, I evaluate the derivatives on the tensor fields first and then add the indices after (so e.g. $\nabla_a T^b_{\ \ c} \equiv (\nabla T)^b_{\ \ ac}$. And yes, thank you for the correction on that typo. $\endgroup$
    – J. Murray
    Aug 28, 2022 at 13:52
  • $\begingroup$ What exactly then are the "Christoffel symbols" you write in your last formula? My understanding is that Christoffel symbols are always coefficients of $\nabla_{\partial_i}\partial_j$, but your formula is a basis-independent formula. Do these "Christoffel symbols" become normal Christoffel symbols when evaluated in a basis? Surely we aren't treating $\Gamma^i_{jl}$ as a (1,2) tensor! $\endgroup$
    – user344261
    Aug 28, 2022 at 14:08
  • $\begingroup$ @ZackFox My last expression refers to the components in a particular basis, not to the tensors themselves. The fact that I'm giving expressions for components means in particular that I am not employing abstract index notation - in which $X^d$ is a vector field - but rather referring to the $d^{th}$ component of the vector field $X$. $\endgroup$
    – J. Murray
    Aug 28, 2022 at 14:32
  • $\begingroup$ I see, thanks for all the help! $\endgroup$
    – user344261
    Aug 28, 2022 at 14:52

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