Vector-Valued Functions and Manifolds

A vector-valued function $f:\mathbb{R}^n\to\mathbb{R}^m$ assigns to each input $\mathbf{x}=(x_1,\dots,x_n)$ an output with $m$ components $f(\mathbf{x})=(f_1(\mathbf{x}),\dots,f_m(\mathbf{x}))$. Each component $f_i$ is an ordinary scalar function: studying $f$ means studying its components together.

Depending on $n$ and $m$ the geometric meaning changes:

• $n=1$ (curve): $\mathbf{r}(t)$ describes a trajectory — seen in the Curves chapter.
• $n=m$ (transformation): $f:\mathbb{R}^2\to\mathbb{R}^2$ "deforms" the plane. This is the case of coordinate changes (polar, spherical) and conformal maps.
• $m=n$ seen as an arrow (vector field): each point carries a vector — fluid velocity, electric field.
• $\mathbb{R}^2\to\mathbb{R}^3$ (parametric surface): $\mathbf{r}(u,v)$ draws a surface in space.

The key idea of differential calculus stays the same: near a point, $f$ is almost linear. That linear map is the Jacobian matrix.

The differential of $f:\mathbb{R}^n\to\mathbb{R}^m$ is represented by the Jacobian matrix $J_f$, of size $m\times n$, gathering all partial derivatives:

The $i$-th row is the gradient of component $f_i$; the $j$-th column tells how all components react to a change in $x_j$.

It is the best linear approximation. Near $\mathbf{x}_0$,

the generalisation of $f(x_0+h)\approx f(x_0)+f'(x_0)h$: the Jacobian plays the role of the derivative.

The Jacobian determinant (when $m=n$) has a precise geometric meaning: $\det J_f$ is the local volume-scaling factor. A region of volume $dV$ is mapped by $f$ to one of volume $|\det J_f|\,dV$. This is exactly the factor appearing in the change of variables for multiple integrals (e.g. the $r$ in polar coordinates is $|\det J|$).

Composing two vector functions $g:\mathbb{R}^p\to\mathbb{R}^n$ and $f:\mathbb{R}^n\to\mathbb{R}^m$, the Jacobian of the composite is the product of the Jacobians:

It is the matrix version of $(f\circ g)'=f'(g)\,g'$: the scalar chain rule generalises exactly, with number multiplication replaced by row-by-column product. The dimensions match: $(m\times n)\cdot(n\times p)=m\times p$.

Why order matters. Matrix product is not commutative, so the order $J_f\cdot J_g$ (apply $g$ first, then $f$) is essential. All differentiation formulas for composite multivariable functions — including the implicit and inverse ones in the next sections — follow from this rule.

When is a transformation $f:\mathbb{R}^n\to\mathbb{R}^n$ invertible, at least locally? The answer lies in the Jacobian determinant.

Inverse function theorem. If $f$ is $C^1$ and $\det J_f(\mathbf{x}_0)\neq 0$, then $f$ is locally invertible around $\mathbf{x}_0$: there is a neighbourhood on which $f$ is a bijection with $C^1$ inverse. Moreover the Jacobian of the inverse is the inverse of the Jacobian:

Intuition. If the best linear approximation $J_f$ is invertible (does not collapse volumes to zero, $\det\neq0$), then so is $f$ near that point. If instead $\det J_f=0$, the transformation "collapses" a direction locally and cannot be inverted.

Mind: only local. The theorem does not guarantee global invertibility: $f(x,y)=(e^x\cos y,e^x\sin y)$ has $\det J_f\neq0$ everywhere but is not globally injective (it is periodic in $y$).

This generalises Dini's theorem to systems of equations. Given $F:\mathbb{R}^{n+m}\to\mathbb{R}^m$ and the system $F(\mathbf{x},\mathbf{y})=\mathbf{0}$ ($n$ free variables $\mathbf{x}$, $m$ to solve for $\mathbf{y}$), we ask whether we can solve for $\mathbf{y}$ as a function of $\mathbf{x}$.

Condition. If the Jacobian submatrix with respect to $\mathbf{y}$ is invertible, $\det(\partial F/\partial\mathbf{y})\neq 0$, then locally there exists $\mathbf{y}=\mathbf{g}(\mathbf{x})$ with $F(\mathbf{x},\mathbf{g}(\mathbf{x}))=\mathbf{0}$, and

Where the formula comes from. Differentiating $F(\mathbf{x},\mathbf{g}(\mathbf{x}))=\mathbf{0}$ via the chain rule: $\dfrac{\partial F}{\partial\mathbf{x}}+\dfrac{\partial F}{\partial\mathbf{y}}J_g=0$, then isolate $J_g$. It is the matrix analogue of $\varphi'=-F_x/F_y$.

Geometrically: the set $F=\mathbf{0}$ is a manifold (curve, surface...) that, where the condition holds, is locally the graph of a function.

A parametric surface $\mathbf{r}(u,v)=(x(u,v),y(u,v),z(u,v))$ is the two-dimensional analogue of a curve: two parameters instead of one. The coordinate curves (with $u$ or $v$ fixed) have tangents $\mathbf{r}_u$ and $\mathbf{r}_v$.

Normal vector. The cross product of the two tangents is perpendicular to the surface:

The surface is regular where $\mathbf{n}\neq\mathbf{0}$ (the two tangents are not parallel). The direction of $\mathbf{n}$ defines the orientation (the "side" of the surface), crucial for flux.

Area element and total area. The magnitude $|\mathbf{n}|=|\mathbf{r}_u\times\mathbf{r}_v|$ is the area of the parallelogram spanned by the tangents: it is the area element $dS=|\mathbf{r}_u\times\mathbf{r}_v|\,du\,dv$. The surface area is obtained by integration:

It is the two-dimensional analogue of $L=\int|\mathbf{r}'|\,dt$: there an "infinitesimal ruler", here an "infinitesimal area tile".

To optimise $f(\mathbf{x})$ under a constraint $g(\mathbf{x})=0$, at the optima the gradients are parallel:

The geometric reason (seen in Calculus in Rⁿ): at the optimum the level surface of $f$ is tangent to the constraint, so their gradients are aligned.

Several constraints. With $k$ constraints $g_1=\cdots=g_k=0$, the optimum is where $\nabla f$ is a linear combination of the constraint gradients:

That is, $\nabla f$ lies in the span of the $\nabla g_i$ (the normal space to the constraint manifold). One obtains a system in $\mathbf{x}$ and the multipliers $\lambda_i$. The condition requires the $\nabla g_i$ to be independent (constraint qualification), i.e. the constraint Jacobian has full rank.

Coordinate changes are invertible vector functions $T:\mathbb{R}^n\to\mathbb{R}^n$. The Jacobian determinant measures how much $T$ "stretches" areas/volumes, and this is exactly the correction factor in multiple integrals:

Polar coordinates $x=r\cos\theta$, $y=r\sin\theta$: the Jacobian is $\begin{pmatrix}\cos\theta&-r\sin\theta\\\sin\theta&r\cos\theta\end{pmatrix}$, with $\det J=r$. That is why $dx\,dy=r\,dr\,d\theta$: a thin ring at radius $r$ has area proportional to $r$.

Spherical coordinates $x=\rho\sin\phi\cos\theta$, etc.: $|\det J|=\rho^2\sin\phi$, hence $dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$. Choosing the right coordinates — adapted to the symmetry of the problem — is often the key to making a multiple integral computable.