1-Forms and the Dual Basis

Enrichment — beyond 18.02

This note gives the precise meaning of the symbols $dx$, $dy$, $df$ that appear in Derivatives, Differentials, and the Chain Rule. At the 18.02 level, $dx$ is treated as a “small increment.” Here we show that $dx$ is a function --- a linear map that takes a vector and returns a number. This viewpoint comes from differential geometry (Spivak, Calculus on Manifolds, Ch. 4; Hubbard & Hubbard, Vector Calculus, Linear Algebra, and Differential Forms, §6.1) and is not required for chain rule computations or arc-length parametrization.

Vectors and their components

In ${\mathbb{R}}^2$ with the standard basis ${\mathbf{e}}_1=(1,0)$ and ${\mathbf{e}}_2=(0,1)$, any vector $\mathbf{v}$ can be written as

$$\mathbf{v}=v_1{\mathbf{e}}_1+v_2{\mathbf{e}}_2.$$

Here ${\mathbf{e}}_1,{\mathbf{e}}_2$ are the basis elements (vectors) and $v_1,v_2$ are the coefficients (scalars). For example, $\mathbf{v}=(4,1)=4{\mathbf{e}}_1+1{\mathbf{e}}_2$.

1-forms: linear maps on vectors

A 1-form is a linear function that takes a vector as input and returns a scalar:

$$\omega :\{\text{vectors}\}\to \mathbb{R},\omega (a\mathbf{v}+b\mathbf{w})=a\omega (\mathbf{v})+b\omega (\mathbf{w}).$$

“Linear” means the 1-form distributes over vector addition and respects scalar multiplication, just like a dot product with a fixed vector would.

Vectors vs. 1-forms

Vectors represent directions --- they are the arrows you draw. 1-forms represent linear measurements of directions --- they are machines that eat a vector and output a number. The 1-forms at a point form a vector space of their own, called the dual space.

The dual basis: $dx$ and $dy$ are functions

Given the vector basis $\{{\mathbf{e}}_1,{\mathbf{e}}_2\}$, the dual basis $\{dx,dy\}$ is defined by the rule: each dual basis element extracts the corresponding component of whatever vector you feed it.

$$dx({\mathbf{e}}_1)=1,dx({\mathbf{e}}_2)=0,$$ $$dy({\mathbf{e}}_1)=0,dy({\mathbf{e}}_2)=1.$$

In compact notation using the Kronecker delta (${\delta }_{ij}=1$ if $i=j$, $0$ otherwise):

$$d{x}_i({\mathbf{e}}_j)={\delta }_{ij}.$$

In words: $dx$ picks out the $x$-component and ignores the $y$-component; $dy$ does the reverse.

Evaluating on an arbitrary vector

For $\mathbf{v}=(4,1)=4{\mathbf{e}}_1+1{\mathbf{e}}_2$, linearity gives:

$$dx(\mathbf{v})=4\cdot dx({\mathbf{e}}_1)+1\cdot dx({\mathbf{e}}_2)=4\cdot 1+1\cdot 0=4,$$ $$dy(\mathbf{v})=4\cdot dy({\mathbf{e}}_1)+1\cdot dy({\mathbf{e}}_2)=4\cdot 0+1\cdot 1=1.$$

No surprise: $dx$ extracts the first component, $dy$ extracts the second. But the key insight is that this extraction is itself a linear function on vectors, not just a notational convenience.

The parallel between vectors and 1-forms

Vectors and 1-forms have the same algebraic structure --- both are expressed as coefficients times basis elements:

	Basis elements	Coefficients	Expansion
Vector $\mathbf{v}$	${\mathbf{e}}_1,{\mathbf{e}}_2$ (directions)	$v_1,v_2$ (scalars)	$\mathbf{v}=v_1{\mathbf{e}}_1+v_2{\mathbf{e}}_2$
1-form $df$	$dx,dy$ (component extractors)	$\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}$ (scalars)	$df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$

This answers the original question: in the formula $df={\sum }_i\frac{\partial f}{\partial x_i}d{x}_i$, the $d{x}_i$ are the basis elements (1-forms) and the partial derivatives $\frac{\partial f}{\partial x_i}$ are the coefficients (scalars). This is exactly analogous to how ${\mathbf{e}}_j$ is the basis and $v_j$ is the coefficient in $\mathbf{v}={\sum }_j{v}_j{\mathbf{e}}_j$.

The differential $df$ as a 1-form

The differential operator $d$ takes a smooth function $f:{\mathbb{R}}^n\to \mathbb{R}$ and produces a 1-form:

$$d:C^1({\mathbb{R}}^n)\to \{\text{1-forms}\},df=\sum _{i=1}^n\frac{\partial f}{\partial x_i}d{x}_i.$$

The output $df$ is not a number --- it is a linear map waiting to be fed a vector.

Worked example

Let $f(x,y)=3x+2y$. Then:

$$df=3dx+2dy.$$

Evaluate on $\mathbf{v}=(4,1)$:

$$df(\mathbf{v})=3dx(\mathbf{v})+2dy(\mathbf{v})=3\cdot 4+2\cdot 1=14.$$

Connection to the gradient and directional derivatives

Compare with the dot product $\nabla f\cdot \mathbf{v}$, where $\nabla f=(3,2)$:

$$\nabla f\cdot \mathbf{v}=(3,2)\cdot (4,1)=12+2=14.$$

This is the same number. The general relationship is:

$$df(\mathbf{v})=\nabla f\cdot \mathbf{v}\text{for all }\mathbf{v}.$$

When $\mathbf{v}$ is a unit vector $\mathbf{u}$, this equals the directional derivative $D_{\mathbf{u}}f$. When $\mathbf{v}$ is not unit-length, $df(\mathbf{v})$ gives the rate of change scaled by $\Vert \mathbf{v}\Vert$ --- the linear approximation of how much $f$ changes when you displace by $\mathbf{v}$.

Why both $df$ and $\nabla f$ exist

The gradient $\nabla f$ is a vector (it lives in the same space as $\mathbf{v}$). The differential $df$ is a 1-form (it lives in the dual space). The dot product is the bridge between them: $df(\mathbf{v})=\nabla f\cdot \mathbf{v}$.

In ${\mathbb{R}}^n$ with the standard Euclidean inner product, this bridge is so natural that the distinction seems pedantic. But it matters because:

• The gradient $\nabla f$ requires an inner product (the dot product) to define. It is the vector that “represents” $df$ via the dot product.
• The differential $df$ does not require an inner product. It is defined purely from the function $f$ and the limiting process of differentiation.

In curved spaces or non-Euclidean settings (which arise in general relativity, fluid mechanics on curved surfaces, and abstract differential geometry), there may be no natural inner product, and only $df$ --- not $\nabla f$ --- is available. This is why differential geometers work with 1-forms rather than gradients. For 18.02, the distinction is not needed, but it explains why the notation $df$ exists as something separate from $\nabla f$.

Exact differentials

A 1-form $\omega =Pdx+Qdy$ is called exact if there exists a function $\Phi :{\mathbb{R}}^n\to \mathbb{R}$ (called a potential function) such that

$$\omega =d\Phi .$$

This means $P=\frac{\partial \Phi }{\partial x}$ and $Q=\frac{\partial \Phi }{\partial y}$.

Example. The 1-form $\omega =xdx+ydy+zdz$ is exact because

$$\omega =d\left(\frac{1}{2}(x^2+y^2+z^2)\right)=\frac{1}{2}d(r^2)=rdr,$$

where $r=\sqrt{x^2+y^2+z^2}$. The potential function is $\Phi =\frac{1}{2}(x^2+y^2+z^2)$.

Why exactness matters. If $\omega =d\Phi$ is exact, then the line integral of $\omega$ along any curve from $A$ to $B$ depends only on the endpoints:

$${\int }_C\omega =\Phi (B)-\Phi (A).$$

The path does not matter. This is the multivariable generalization of the Fundamental Theorem of Calculus and is the starting point for conservative vector fields (a topic in later 18.02 material).

See also

• Derivatives, Differentials, and the Chain Rule --- the 18.02-level treatment of derivatives and differentials as linear approximations
• Directional Derivatives and the Gradient --- directional derivatives and the gradient vector, which $df$ generalizes
• Arc-Length Parametrization --- uses the chain rule from the derivatives note; the unit tangent vector $\mathbf{T}=d\tilde{\mathbf{r}}/ds$ is a concrete example of differentiation with respect to arc length