I am learning GR and the notation is killing me here. So my understanding is, the comma notation is used to indicate a derivative, such as:

\begin{equation} V^{\alpha}_{\;\;,\gamma}=\partial_{\gamma}V_{\alpha} \end{equation}

and a semicolon is used to represent a covariant derivative, such as:

\begin{equation} V^{\alpha}_{\;\;;\gamma}= \partial_{\gamma}V^{\alpha}+\Gamma_{\gamma\mu}^{\alpha}V^{\mu} = V^{\alpha}_{\;\;,\gamma}+\Gamma_{\gamma\mu}^{\alpha}V^{\mu} = \nabla_{\gamma}V^{\alpha} \end{equation}

However! In problem 7.7 in "The Problem Book of Relativity and Gravitation" they write (for the metric tensor g):

\begin{equation} g_{\alpha \beta , \gamma} = \nabla_{\gamma}(\mathbf{e}_{\alpha}\cdot \mathbf{e}_{\beta}) = \Gamma^{\mu}_{\alpha\gamma}\mathbf{e}_{\mu}\cdot\mathbf{e}_{\beta}+\Gamma^{\mu}_{\beta \gamma}\mathbf{e}_{\mu}\cdot\mathbf{e}_{\alpha} \end{equation} Christoffel symbols?! How? I thought these only popped up when taking the COVARIANT derivative. Then later on, they write:

\begin{equation} A^{\alpha}_{\;;\alpha} = A^{\alpha}_{\;,\alpha}+\Gamma^{\alpha}_{\;\beta\alpha}A^{\beta} \end{equation}

Which makes sense given my definition above, but does not make sense with the notation used in the first example. Am I missing something? Is it just a typo??

  • 2
    $\begingroup$ I agree that this is confusing. When acting on a scalar, the covariant derivative and partial derivative are exactly the same. So, since $\vec{e}_\alpha \cdot \vec{e}_\beta$ is a scalar, $\partial_\gamma (\vec{e}_\alpha \cdot \vec{e}_\beta) = \nabla_\gamma (\vec{e}_\alpha \cdot \vec{e}_\beta)$. However, if we want to expand this with the product rule, we'd better use the covariant derivative to do so, because partial derivatives of vectors make no sense. $\endgroup$
    – gj255
    Commented Sep 22, 2016 at 23:47
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    $\begingroup$ Using the comma for ordinary derivative and a semi-colon for the covariant derivative ( or the other way around) is not very clear. I stick to what I think is the usual convention, for ordinary partial derivatives I use the same squiggle as usual , and for covariant derivatives I use D or nabla. ${\displaystyle D_{t}T=\nabla _{{\dot {\gamma }}(t)}T.}$ $\endgroup$
    – user108787
    Commented Sep 23, 2016 at 0:11
  • $\begingroup$ Thanks for the comments! I think I get it now. I really do not like the comma/semicolon notation at ALL. But my professor (and this problem book) seem to like using it, so I better try to get used to it. :\ $\endgroup$
    – user41178
    Commented Sep 23, 2016 at 1:26
  • $\begingroup$ Hey, @user41178! The expression which you are confused about is not a definition of partial derivative, but a nontrivial relation between partial derivatives of the metric and Christoffels. The answer by joshphysics explains how to derive this relation. $\endgroup$ Commented Sep 23, 2016 at 6:13

2 Answers 2


All formulas you shown above are using abstract index notation except the third formula which is fully expressed is a basis. For a vector field, you can write for example $$V = V^\mu e_\mu\;,$$ where $V^\mu$ is a scalar while $e_\mu$ is a vector basis. This is a sort of confusion because in abstract index notation we view $V^\mu$ as a vector field.

When we take the covariant derivative, it reads $$\nabla_\mu V=\nabla_\mu (V^\nu e_\nu) = \nabla_\mu (V^\nu) e_\nu + V^\nu \nabla_\mu ( e_\nu)$$ \begin{eqnarray} &=& \partial_\mu (V^\nu) e_\nu + V^\nu \Gamma_\mu{}^\lambda{}_\nu e_\lambda\;,\\ &=&\big( \partial_\mu V^\nu +\Gamma_\mu{}^\nu{}_\lambda V^\lambda\big)e_\nu \end{eqnarray} If we define $$\nabla_\mu V =: (\nabla_\mu V^\nu) e_\nu $$ we will have the relation in the abstract index notation $$\nabla_\mu V^\nu = \partial_\mu V^\nu +\Gamma_\mu{}^\nu{}_\lambda V^\lambda\;.$$ (More general, you can start with $\nabla V$ and then define $\nabla V =:(\nabla_\mu V^\nu) e^\mu \otimes e_\nu$ ) Next, the metric $g$, it is (o,2) tensor so it has two slots for inserting 2 vectors if we insert the basis into these slots we will get a component of the metric tensor which is a scalar field $$g(e_\mu, e_\nu)=g_{\mu\nu}$$

($g= g_{\alpha\beta} e^\alpha \otimes e^\beta,\;g(e_\mu, e_\nu) =g_{\alpha\beta} e^\alpha(e_\mu) \otimes e^\beta(e_\nu) =g_{\alpha\beta}\delta^\alpha_\mu \delta^\beta_\nu= g_{\mu\nu} $)

It is also usually define that $\eta(A,B):= A\cdot B$, $\eta$ is a Minkowskian metric $A\cdot B$ is a scalar so invariants under coordinate transformations $$A\cdot B =\eta(A,B) \equiv \eta_{IJ} A^I B^J$$ $$= g(A,B) \equiv g_{\mu\nu} A^\mu B^\nu$$ where $A^I = e^I_\mu A^\mu$ for some scalar $e^I_\mu$ (a vierbein), and you can easily prove that $g_{\mu\nu} = \eta_{IJ} e^I_\mu e^J_\nu$.

So now we have $$e_\mu \cdot e_\nu =\eta_{IJ}e^I_\mu e^J_\nu= g_{\mu\nu}$$

In this step, we can view $\eta_{IJ},g_{\mu\nu}$ as the scalar fields $e^I_\mu$ as a vector field \begin{eqnarray} \nabla_\gamma g_{\alpha\beta} (= \partial _\gamma g_{\alpha\beta})&=& \nabla_\gamma (e_\alpha\cdot e_\beta) \\ &=&\eta(\nabla_\gamma e_\alpha,e_\beta) +\eta(e_\alpha,\nabla_\gamma e_\beta) \equiv \eta_{IJ}\nabla_\gamma(e^I_\alpha) e^J_\beta + \eta_{IJ}e^I_\alpha \nabla_\gamma(e^J_\beta) \\ &=&\eta(\Gamma_\gamma{}^\rho{}_\alpha e_\rho,e_\beta) +\eta(e_\alpha,\Gamma_\gamma{}^\sigma{}_\beta e_\sigma) \equiv \eta_{IJ}\Gamma_\gamma{}^\rho{}_\alpha e^I_\rho e^J_\beta + \eta_{IJ}e^I_\alpha \Gamma_\gamma{}^\sigma{}_\beta e^J_\sigma \\ &=&\Gamma_\gamma{}^\rho{}_\alpha \eta( e_\rho,e_\beta) + \Gamma_\gamma{}^\sigma{}_\beta\eta(e_\alpha, e_\sigma) \equiv \Gamma_\gamma{}^\rho{}_\alpha \eta_{IJ} e^I_\rho e^J_\beta + \Gamma_\gamma{}^\sigma{}_\beta \eta_{IJ}e^I_\alpha e^J_\sigma \\ &=& \Gamma_\gamma{}^\rho{}_\alpha e_\rho \cdot e_\beta + \Gamma_\gamma{}^\sigma{}_\beta e_\alpha \cdot e_\sigma \equiv \Gamma_\gamma{}^\rho{}_\alpha g_{\rho \beta} + \Gamma_\gamma{}^\sigma{}_\beta g_{\alpha \sigma} \end{eqnarray} Note: Not fully detailed as much as possible but may be helpful for you.


This may not be the answer you're looking for because I provide no intuition herein, but from a computational perspective, it's not so hard to see why the derivative of the metric involves Christoffel symbols.

The affine connection commonly used in general relativity is chosen to be both torsion free and metric compatible. The second condition means that the covariant derivative of the metric vanishes. $$ \nabla_\gamma g_{\alpha\beta} = 0. $$ These two conditions uniquely specify the connection which is called the Levi-Civita connection. One can show that the associated covariant derivative of an arbitrary 2-tensor satisfies (see, for example, Carroll's GR, section 3.2): $$ \nabla_\gamma T_{\alpha\beta} = \partial_\gamma T_{\alpha\beta} - \Gamma_{\gamma\alpha}^\mu T_{\mu\beta} - \Gamma_{\gamma\beta}^\mu T_{\alpha\mu} $$ Plugging in the metric, noting that the left hand side vanishes, and rearranging gives the desired result.

  • $\begingroup$ Thanks for the comment. I understand WHY the Christoffel symbols are there, my confusion is from the fact that based on what the comma notation means, I would not expect them (Sorry, I was unclear about that) $\endgroup$
    – user41178
    Commented Sep 23, 2016 at 1:28
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    $\begingroup$ @user41178 No apology necessary in the least first of all. I must say, however, that I'm still note entirely sure what you're asking. The partial derivative and covariant derivative can each be thought of as a mapping that takes a tensor of a certain rank, and increases the rank by 1, and that's precisely what the comma and semicolon notation reflect. It's crucial, however, that what the semicolon means (namely what the covariant derivative does) varies based on the rank of tensor one starts with. It's not clear to me how other notations make any of this more transparent to be honest. $\endgroup$ Commented Sep 23, 2016 at 2:43
  • $\begingroup$ @joshphysics: the partial derivative of a tensor is not a tensor of one higher rank. This is only true for the covariant derivative. $\endgroup$
    – gj255
    Commented Sep 23, 2016 at 16:37
  • $\begingroup$ @gj255 Indeed thanks for correcting that error. In any case, the comment was meant to address the notation. I should have said something like "an object with one more free index." $\endgroup$ Commented Sep 23, 2016 at 17:09

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