Vectors on a manifold are defined as directional derivatives evaluated at a point and one-forms (in the associated cotangent space) are defined as maps from vectors to real numbers. However, why aren't functions on the manifold themselves one-forms? Clearly functions also take vectors to real numbers. I haven't seen any book talk about this.

In addition since any linear map from a vectors to a real numbers on a tangent space can be expressed as a linear combination of one-forms, what does this mean for functions? It doesn't make sense for functions to be expressed as linear combinations of one-forms because they are 0-forms. Here's my understanding of how functions relate to one-forms.

Given a coordinate chart $\Phi:M \rightarrow \mathbb{R}^n$ on a n-dimensional manifold $M$, we can think of the coordinates as functions $x^\alpha:M \rightarrow \mathbb{R}$. Now, vectors in a tangent space at some point $p \in M$ can be expressed as $V=V^\beta \partial_\beta|_p$. The coordinate one-form $dx^\alpha$ is defined by it's action on a vector:

$$dx^\alpha (V^\beta \partial_\beta) = V^\beta dx^\alpha (\partial_\beta) \stackrel{?}{=} V^\beta \partial_\beta (x^\alpha)|_p = V^\beta \delta^\alpha _\beta = V^\alpha$$

Although this is standard, I think about the one-form $dx^\alpha$ as placing the coordinate function in the directional derivative $\partial_\beta$. Clearly, the coordinate function only changes when the directional derivative is along the direction of $x^\beta$. Hence, we get the kronecker delta. However, I do not know if this is the right way of think about one-forms (hence the question mark on the equal sign). But if this is right, the one-form is just the coordinate function $x^\alpha$.

So what's the difference between a one-form and a function?

I seem to be misunderstanding something but I am not sure what. Any help is appreciated.

  • $\begingroup$ Your last element in the formula should be $V^\alpha$ $\endgroup$
    – GReyes
    May 20, 2021 at 18:15
  • 2
    $\begingroup$ "why aren't functions on the manifold themselves one-forms? Clearly functions also take vectors to real numbers." What do you mean here by "function"? Not every function takes vectors to real numbers. A function defined on a manifold $M$, for example, takes as input a point on $M$, not a vector. A differential $k$-form on $M$ is a function that assigns to each point $p \in M$ an alternating $k$-tensor on the tangent space to $M$ at $p$. $\endgroup$
    – littleO
    May 20, 2021 at 18:16
  • $\begingroup$ Thank you, corrected it. $\endgroup$
    – Chandrahas
    May 20, 2021 at 18:17
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    $\begingroup$ What do you mean by "vector"? Do you mean a tangent vector to a manifold $M$ at a point $p \in M$? If $M$ is a smooth manifold and $f: M \to \mathbb R$, for example, then you can't take a tangent vector to $M$ and plug it into $f$. $\endgroup$
    – littleO
    May 20, 2021 at 18:23
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    $\begingroup$ The object you are referring to is called the exterior derivative $df$ of the function $f$. It's a one form defined by $df(X_p)=X_p\cdot f$ for any $X_p\in T_pM$. This is, however, a different object as the function $f$, since $f$, by definition, maps points of the manifold to real numbers. You can compare this to the case of a function $g:\mathbb{R}^n\to \mathbb{R}$ and its derivative $dg:\mathbb{R}^n\to L(\mathbb{R}^n,\mathbb{R})$, where $L(\mathbb{R}^n,\mathbb{R})$ denotes the linear maps from $\mathbb{R}^n$ to $\mathbb{R}$. $\endgroup$
    – Jake28
    May 20, 2021 at 18:45

1 Answer 1


Let $C^\infty M$ be the space of smooth functions $M\to\mathbb{R}$, and let $\mathfrak{X}^*M$ denote the space of covector fields (also known as 1-forms). These spaces are not isomorphic, but it is possible to associate each function $f\in C^\infty M$ a corresponding $1$-form, which we refer to as the differential* of $f$, denoted $df$, exactly as you describe: $$ df(X):=X(f) $$ Where $X$ is a vector field, the left side is interpreted as the dual pairing of vector fields and 1-forms, and the right side as the defining action of vector fields on functions. Even given this association, one should not think of smooth functions and 1-forms as the same, since different functions may have the same differential, and not all $1$-forms can be written as the differential of a function. Put another way, if we think of the differential as a map $d:C^\infty M\to\mathfrak{X}^*M$, then this map is neither injective nor surjective.

*Also sometimes called the exterior derivative, or just the derivative. There are also other notions of derivative which are equivalent when acting on smooth functions.


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