A couple of times when I've tried to prove symmetries of various tensors (for learning), I've ended up with the expression below, and the fact that either a) I made mistake, or b) the expression is symmetric with respect to switching k and l.

$$ \frac{\partial g_{ij}}{\partial x^k} \frac{\partial g^{ij}}{\partial x^l} $$

Where $g_{..}$ and $g^{..}$ are the covariant and contravariant metric tensor respectively, and $x^.$ is the coordinate.

Is the expression symmetric wrt switching $k$ and $l$? If so, is it possible to prove this using only indicial notation?

up vote 11 down vote accepted

Since the product rule tells us $0 = \partial( g g^{-1} ) = (\partial g) g^{-1} + g (\partial g^{-1})$, we have a formula for the derivative of the inverse metric:

$$ \partial_l g^{ij} = -g^{ia} g^{jb} \partial_l g_{ab}.$$

Substituting this in to your expression we get

$$ -g^{ia} g^{jb} \partial_l g_{ab} \partial_k g_{ij}.$$

If we swap the dummy indices $a \leftrightarrow i$, $b \leftrightarrow j$ then this is equal to

$$ -g^{ai} g^{jb} \partial_l g_{ij} \partial_k g_{ab};$$

so it's symmetric in $k$ and $l$.

  • Nicely done — I suspected there was a simpler way to do it than the approach I used. – Michael Seifert Jun 30 '16 at 15:39
  • Anthony, you win. Did not expect the answer to be this simple! @Michael, your answer was also very insightful and fun to read, thanks a lot. On a side note, as a first-time poster in math.stackexchange, I'm shocked at how quickly this got two answers. – Simplex Jun 30 '16 at 17:25

It does appear to be symmetric, though the proof I came up with requires the introduction of a covariant derivative operator. There may be another proof out there that doesn't require quite so much heavy machinery.

Let $\nabla_k$ be a torsion-free derivative operator defined such that $\nabla_k g_{ij} = 0$. By the general properties of derivative operators, we know that there will exist a tensor $C^i {}_{jk}$ such that the coordinate derivative and our new derivative of a covariant rank-2 tensor are related by $$ \nabla_k g_{ij} = \partial_k g_{ij} - C^m {}_{ki} g_{mj} - C^m {}_{kj} g_{im} $$ and similarly, the derivative of a contravariant rank-2 tensor are given by $$ \nabla_k g^{ij} = \partial_k g^{ij} + C^i {}_{km} g^{mj} + C^j {}_{km} g^{im}. $$ Since by definition $\nabla_k g_{ij} = 0$ and $\nabla_l g^{ij} = 0$, the quantity in your question becomes $$ \partial_k g_{ij} \partial_l g^{ij} = (C^m {}_{ki} g_{mj} + C^m {}_{kj} g_{im})(-C^i {}_{ln} g^{nj} - C^j {}_{ln} g^{in}) = - 2 C^m {}_{ki} C^i {}_{lm} - 2 C^m {}_k {}^j C_{mlj}. $$ Both terms in this expression are manifestly symmetric under the exchange of $k$ and $l$, and so the expression $\partial_k g_{ij} \partial_l g^{ij}$ is symmetric as well.

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.