In Lee's Introduction to Smooth Manifolds he writes

$$ \omega = \omega_i dx^i$$

where $\omega$ is a differential form. See for example page 293.

What does $\omega_i dx^i$ stand for?

According to my understanding we should have

$$ \omega = f(x) dx^1 \wedge \dots \wedge dx^n$$

for $f: U \to \mathbb R $ a smooth function of $n$ variables. But this doesn't explain why $\omega_i$ has an index...

Or could it be that the notation means

$\omega_i dx^i = \sum_i \omega_i dx^1 \wedge \dots \wedge dx^n$?


2 Answers 2


The text here uses the Einstein summation convention, defined below: In any local coframe on a manifold, here we'll use $(dx^i)$, defined, say, on $U \subseteq M$, we can write (the restriction to $U$ of) a $1$-form $\omega$ as a linear combination of coframe elements, namely, as $$\omega = \omega_1 \,dx^1 + \cdots + \omega_n \,dx^n = \sum_{i = 1}^n \omega_i \,dx^i.$$

Linear combinations of frame and coframe elements are ubiquitous in differential geometry---after all, we are generally interested in properties that don't depend on a choice of frame, and so expressions that involve only some frame or coframe elements generically do depend on such a choice---and as you encounter more involved tensorial expressions, you'll find expressions that involve many nested sums. So, we often suppress the sigma notation for any paired index, and the resulting notation (where one implicitly sums over all values of a paired index) is the Einstein summation convention; it's good practice, as Lee does, to do this only when a pair includes one lower index and one upper index (you'll see some literature, especially in physics, ignore this restriction).

  • $\begingroup$ So the version of the lemma in the book is only about $1$-forms? $\endgroup$
    – a student
    Jan 26, 2015 at 10:41
  • 1
    $\begingroup$ I don't have my copy handy, what is the text of the lemma? One can play the same game with tensors of higher rank, e.g., we can write (the restriction of) any covariant $2$-tensor $\psi$ in a local coordinate coframe $(dx^i)$ as $\psi_{ij} dx^i \otimes dx^j$. $\endgroup$ Jan 26, 2015 at 10:50
  • $\begingroup$ Sorry I thought I had linked to it in the question. You can find the lemma here... well, that is, if you can access it. $\endgroup$
    – a student
    Jan 26, 2015 at 11:08
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    $\begingroup$ Lemma 11.49? Yes, it applies just to covector fields, which is what (as I recall) Lee calls $1$-forms before he introduces forms. $\endgroup$ Jan 26, 2015 at 11:27

In a coordinate chart $(U, (x^1, \dots, x^n))$ on an $n$-dimensional manifold, every one-form $\omega \in \Omega^1(U)$ can be written as $\omega = \omega_1 dx^1 + \dots + \omega_n dx^n$ where $\omega_1, \dots, \omega_n : U \to \mathbb{R}$ are smooth functions. So we can write $\omega$ as

$$\omega = \sum_{i=1}^n\omega_i dx^i.$$

Using Einstein notation, we can supress the sigma and write $\omega_i dx^i$ to denote the same sum so that $\omega = \omega_i dx^i$. In general, when using Einstein notation, if an index appears both lowered and raised within a term (in this case $i$), it denotes that the sum is taken over all possible values of that index (in this case $1, \dots, n$).

The $\omega$ you wrote down in your example is an $n$-form, not a one-form. In a coordinate chart $(U, (x^1, \dots, x^n))$ on an $n$-dimensional manifold, it is true that every $\omega \in \Omega^n(U)$ can be written in the form $\omega = f dx^1\wedge\dots\wedge dx^n$ where $f : U \to \mathbb{R}$ is a smooth function.


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