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I understand that the gradient $\partial_i$ is covariant. Let f be a function of 3 variables So I can write the total differential as $$ df=\partial_1fdx^1+\partial_2fdx^2+\partial_3fdx^3 = \partial_kfdx^k, $$ summing correctly over the same lower and upper index. But when I write $ \displaystyle\partial_i=\frac{\partial}{\partial x^i} $ (and this is how it is written in all books) and want to write the total differential, I have to write $$ df= \frac{\partial f}{\partial x^1}dx^1+\frac{\partial f}{\partial x^2}dx^2+\frac{\partial f}{\partial x^3}dx^3=\frac{\partial f}{\partial x^k}dx^k $$ and so I am summing over the same upper indices.

My question: Why does an index get raised when it is moved to the denominator?

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  • $\begingroup$ Isn't this asking "why is the gradient covariant?", which you claim to understand in the first sentence? $\endgroup$ – fqq Jun 9 at 17:28
  • $\begingroup$ No, it is not. I want to know why the index gets raised in the denominator $\endgroup$ – Fuzzy Jun 9 at 17:31
  • $\begingroup$ I think the logic flows in the opposite sense: $f(x^1, x^2, x^3, x^4)$ is a function of contravariant components. When we expand it as a differential, the invariance of $df$ requires that the gradient is covariance, and hense the index must be lowered. $\endgroup$ – Vadim Jun 9 at 17:34
  • $\begingroup$ raised compared to what? The gradient is normally defined as the derivative wrt the components $x^i$, which in the usual notation for partial derivatives is denoted $\partial/\partial x^i$. $\endgroup$ – fqq Jun 9 at 17:35
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    $\begingroup$ I might have a glue. If I write $\partial/\partial x^i$ then $x^i$ is the name of the variable, and this expression is not index notation. If I write $\partial_idx^i$ it is index notation and Einstein's summation applies. If I write $\displaystyle\frac{\partial}{\partial x^i}dx^i$ it should not be summed, because Einstein's summation does not apply, but it is done anyway $\endgroup$ – Fuzzy Jun 9 at 18:08
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$\partial_i$ is defined as $\dfrac{\partial}{\partial x^i}$. Therefore the index does not get raised when brought to the denominator, rather it is notation for the derivative justified rigorously:

In Cartesian coordinates the partial derivative transforms as a covariant vector $$\frac{\partial}{\partial \bar{x}^i}=\frac{\partial x^j}{\partial \bar{x}^i}\frac{\partial}{\partial x^j}$$

Covariant vectors are written with lowered indices.

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A raised index in the denominator counts as a lowered index (because it is in the denominator)

See "Introduction to Smooth Manifolds" by J. Lee. I believe it is in both editions but in the second edition see Chapter 3 on tangent vectors and look at the text below Eq. 3.2:

$$D_v |_a f = v^i \frac{\partial f}{\partial x^i}(a)$$ (Here we are using the summation convention as usual, so the expression on the right-hand side is understood to be summed over $i=1, \ldots, n$. This sum is consistent with our index convention if we stipulate that an upper index "in the denominator" is to be regarded as a lower index.)

Here $D_v|_a$ is the derivative of function $f$ in the direction of vector $v = v^i e_i|_a$ at point $a$ where $v^i$ are the components of $v$ with respect to the basis $e_i|_a$, where the basis $e_i|_a$ is defined at point $a$.

Lee goes on to introduce the concept of derivations and how they can be used to generally define tangent vectors on manifolds. It turns out that the map:

$$ \frac{\partial}{\partial x^i}\big|_a $$

defined by

$$ \frac{\partial}{\partial x^i}\big|_a\left(f\right) = \frac{\partial f}{\partial x^i}(a) $$

is a derivation and can be thought of as a tangent vector. Since we agreed above that something like $\frac{\partial}{\partial x^i}$ is to be thought of as a `lower index thing' we can equate it with another lower index thing and define

$$ \partial_i = \frac{\partial}{\partial x^i} $$

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  • $\begingroup$ Thanks, but I don't remember that any of my beginners relativity books say that explicitely. Does it follow from $x_i=g_{ij}x^j$ or similar? Or, alternatively, can you name a book (lecture text) $\endgroup$ – Fuzzy Jun 9 at 17:36
  • $\begingroup$ fqq, can you be a bit more helpful? If it helps I can claim that I might not have understood what I claimed I did $\endgroup$ – Fuzzy Jun 9 at 17:46
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    $\begingroup$ I've edited to add a reference and some extra information. $\endgroup$ – jgerber Jun 9 at 18:33

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