# Why isn't $g^{\mu \nu}\partial_{\nu}$ equal to $\partial^{\mu}$ in General Relativity?

The question is pretty short. I have been told that unlike for flat spacetime, in curved spacetime we cannot write

$$g^{\mu \nu}\partial_{\nu}f=\partial^{\mu}f$$

With $$f$$ an scalar function, but I don't really get why.

• What is the definition of $\partial^\mu$?
– MBN
Oct 6, 2020 at 11:05
• I thought it was $g^{\mu \nu} \partial_{\nu}$ but apparently it isn't. Oct 6, 2020 at 11:09
• What else would it be? Oct 6, 2020 at 15:29

Well I'll answer but I would be interested to know if my answer is wrong. I thought we defined $$\partial^a$$ such that $$\partial^a f \equiv g^{a\mu} \partial_\mu f .$$ The thing to be careful about is then the fact that $$\partial_\mu (g^{a\mu} f) = (\partial_\mu g^{a\mu}) f + g^{a\mu} \partial_\mu f$$ so usually $$\partial_\mu (g^{a\mu} f) \ne \partial^a f .$$ It also follows that you cannot assume that $$\partial^a$$ operators commute, nor even that $$\partial^\mu \partial_\mu$$ is equal to $$\partial_\mu \partial^\mu$$ unless you make some other definition of $$\partial^a$$.
• So, if I didn't misunderstand, we might actually define $\partial^{\alpha} f \equiv g^{\alpha \mu} \partial_{\mu} f$ but by doing so we would lose some interesting properties of our usual derivative $\partial_{\alpha}$ and therefore it is not a convenient way of proceeding. Oct 6, 2020 at 11:35
• Yes that is my understanding. But there is nothing wrong with the definition; it makes reasonable sense so is arguably no more inconvenient than it has to be. If in doubt you can always avoid $\partial^a$ completely in your own work ... but you still need to know what others mean by it. Oct 6, 2020 at 12:22
For a scalar function; $$\nabla_{\mu}f=\partial_{\mu}f$$, its a definition, we know $$\nabla_{\mu}f$$ is a $$(0,1)$$ tensor so $$g^{\mu\nu}\nabla_{\mu}f=\nabla^{\nu}f$$ but there doesn't exist any similar definition where we identify contravariant version of $$\nabla^{\mu}f$$ with $$\partial^{\mu}f$$. So ultimately it boils down to the definition of $$\partial^{\mu}$$.
The definition $$\partial^{\mu} = g^{\mu \nu} \, \partial_{\nu} \tag{1}$$ doesn't have more sense than $$x_{\mu} = g_{\mu \nu} \, x^{\nu}. \tag{2}$$ Using (1) may lead to many troubles that you want to avoid at all cost. Don't use it!
However, you may use the same for the covariant derivative: $$\nabla^{\mu} = g^{\mu \nu} \, \nabla_{\nu}, \tag{3}$$ which may be usefull since $$\nabla_{\lambda} \, g^{\mu \nu} = 0$$.