# Index notation and total differential

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?

• Isn't this asking "why is the gradient covariant?", which you claim to understand in the first sentence?
– fqq
Commented Jun 9, 2020 at 17:28
• No, it is not. I want to know why the index gets raised in the denominator Commented Jun 9, 2020 at 17:31
• 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. Commented Jun 9, 2020 at 17:34
• 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 $\partial/\partial x^i$.
– fqq
Commented Jun 9, 2020 at 17:35
• 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 Commented Jun 9, 2020 at 18:08

$$\partial_i$$ 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.

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

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

• 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) Commented Jun 9, 2020 at 17:36
• 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 Commented Jun 9, 2020 at 17:46
• I've edited to add a reference and some extra information. Commented Jun 9, 2020 at 18:33
• @Jagerber48 Hi I'm a bit confused, are you sure this statement at the start of your answer is correct: "A raised index in the denominator counts as a lowered index (because it is in the denominator)"? If true then say Hamilton's equations will be $\dot q_i=\frac{\partial H}{\partial p^i }$ and $\dot p_i=-\frac{\partial H}{\partial q^i }$ but everywhere I've seen them they look like $\dot q_i=\frac{\partial H}{\partial p_i }$ and $\dot p_i=-\frac{\partial H}{\partial q_i }$ like here Commented Nov 8, 2023 at 16:16
• @Electra yes, a raised index in the denominator counts as a lowered index when using differential geometry covariant/contravariant notation. The reference you linked looks like it is just using $i$ as a standard index not to indicate covariance or contravariance. You'll notice there are NO upper indices on that whole webpage. Commented Nov 8, 2023 at 17:25