# Why aren't differential forms just functions?

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.

• Your last element in the formula should be $V^\alpha$ May 20, 2021 at 18:15
• "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 $p$. May 20, 2021 at 18:16
• Thank you, corrected it. May 20, 2021 at 18:17
• 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 $f$. May 20, 2021 at 18:23
• 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 $\mathbb{R}$. May 20, 2021 at 18:45

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.