MAT257Y1/Unit 11
Unit 11 · Analysis II

Tensors & Differential Forms

A $k$-form is the thing you can integrate over a $k$-dimensional piece of space. That is the whole motivation, and everything technical here — alternating multilinear maps, the wedge product, the $dx_{i_1}\wedge\dots\wedge dx_{i_k}$ basis — exists because signed volume is antisymmetric and forms are built to be exactly as antisymmetric as volume is.

By the end you can
Why it matters & where it comes from
Why it exists

Vector fields, line integrals, surface integrals and flux are four notations for two ideas, and in $\mathbb{R}^3$ the coincidence that a plane has a normal line hides the difference. Forms separate them, and the reward is that grad, curl, div and the three classical integral theorems collapse into one operator and one theorem.

In the real world

Maxwell's equations are two equations in the language of forms. Symplectic forms carry all of Hamiltonian mechanics; work and flux are $1$- and $2$-forms; thermodynamic relations are statements about which forms are exact.

Story & open questions

Grassmann invented the exterior algebra in 1844 and was ignored for decades. Élie Cartan turned it into the calculus of differential forms around 1900, and de Rham's theorem in 1931 showed the forms on a space see its topology — the beginning of a subject the last two units only touch.

Multilinear and alternating tensors

Definition

A $k$-tensor on $V=\mathbb{R}^n$ is a map $T:V^k\to\mathbb{R}$ linear in each slot with the others fixed. It is alternating if swapping any two arguments flips the sign: $T(\dots,v,\dots,w,\dots)=-T(\dots,w,\dots,v,\dots)$. Write $\Lambda^k(V)$ for the alternating $k$-tensors.

Theorem Immediate consequences
An alternating $T$ vanishes whenever two arguments are equal — swap them and get $T=-T$.
Hence $T$ vanishes on any linearly dependent list.
Hence $\Lambda^k(V)=0$ for $k>n$: in $\mathbb{R}^n$ any $k>n$ vectors are dependent.
Theorem Dimension

$\dim\Lambda^k(\mathbb{R}^n)=\binom{n}{k}$, with basis the $dx_{i_1}\wedge\dots\wedge dx_{i_k}$ over increasing index tuples $i_1<\dots<i_k$.

An alternating tensor is determined by its values on basis tuples, and antisymmetry means only the increasing tuples carry independent information. For $n=3$ the dimensions are $1,3,3,1$ — and the middle two both being $3$ is the coincidence that lets $\mathbb{R}^3$ get away with vector calculus for so long.

Example Determinant as a top form

$\det:(\mathbb{R}^n)^n\to\mathbb{R}$ is alternating and $n$-linear in its columns. Since $\dim\Lambda^n(\mathbb{R}^n)=1$, every alternating $n$-tensor is a scalar multiple of it. That one-dimensionality is the real content of "the determinant is the unique normalised signed volume".

The wedge product

Definition

For $\omega\in\Lambda^k$ and $\eta\in\Lambda^l$, the wedge $\omega\wedge\eta\in\Lambda^{k+l}$ is the alternation of the tensor product, normalised so that

$dx_{i_1}\wedge\dots\wedge dx_{i_k}(v_1,\dots,v_k) =\det\big[(v_j)_{i_m}\big],$

the determinant of the $k\times k$ matrix of the chosen coordinates of the chosen vectors. Different books insert different constants; only consistency matters.

Theorem Algebra of the wedge
Bilinear over $\mathbb{R}$, and associative: $(\omega\wedge\eta)\wedge\zeta=\omega\wedge(\eta\wedge\zeta)$.
Graded commutative: $\omega\wedge\eta=(-1)^{kl}\eta\wedge\omega$.
Consequently $\omega\wedge\omega=0$ whenever $k$ is odd — in particular $dx\wedge dx=0$ — but not in general for even $k$.
Example A wedge in $\mathbb{R}^3$

With $\omega=dx+2dy$ and $\eta=3dy-dz$:

$\omega\wedge\eta=3\,dx\wedge dy-dx\wedge dz +6\,dy\wedge dy-2\,dy\wedge dz =3\,dx\wedge dy-dx\wedge dz-2\,dy\wedge dz.$

The $dy\wedge dy$ term dies. Expand mechanically, then delete repeats and sort indices, picking up a sign for each transposition.

Example Even degree need not square to zero

In $\mathbb{R}^4$, $\omega=dx_1\wedge dx_2+dx_3\wedge dx_4$ gives $\omega\wedge\omega=2\,dx_1\wedge dx_2\wedge dx_3\wedge dx_4\ne0$. This $\omega$ is the standard symplectic form and the non-vanishing of its powers is why phase-space volume is preserved.

Differential forms and pullback

Definition

A differential $k$-form on an open $U\subseteq\mathbb{R}^n$ is a map assigning to each $x\in U$ an element $\omega(x)\in\Lambda^k$. In coordinates,

$\omega=\sum_{i_1<\dots<i_k} a_{i_1\dots i_k}(x)\,dx_{i_1}\wedge\dots\wedge dx_{i_k},$

and $\omega$ is smooth when its coefficient functions are. A $0$-form is just a function.

Definition Pullback

For a smooth $g:U\to V$ and a $k$-form $\omega$ on $V$, the pullback $g^{*}\omega$ is the $k$-form on $U$ given by

$(g^{*}\omega)(x)(v_1,\dots,v_k) =\omega(g(x))\big(Dg(x)v_1,\dots,Dg(x)v_k\big).$

Push the vectors forward with the derivative, then evaluate. Forms pull back along maps while vector fields do not push forward — the main practical reason forms are the better object.

Theorem Pullback is algebraic
$g^{*}(\omega\wedge\eta)=g^{*}\omega\wedge g^{*}\eta$.
$g^{*}(df)=d(f\circ g)$, so on functions pullback is just composition.
$(h\circ g)^{*}=g^{*}\circ h^{*}$: pullback reverses order.

In practice this means: to pull back, substitute $y_i=g_i(x)$ everywhere and expand $dy_i=\sum_j(\partial g_i/\partial x_j)\,dx_j$ using the wedge rules.

Theorem Top forms and the determinant

For $g:U\to\mathbb{R}^n$ with $U\subseteq\mathbb{R}^n$,

$g^{*}(dy_1\wedge\dots\wedge dy_n) =\det Jg(x)\;dx_1\wedge\dots\wedge dx_n.$

The change of variables factor was a determinant all along, and here it appears with its sign. Unit 10 needed an absolute value only because unoriented volume had nowhere to record orientation.

Example Pulling back to polar coordinates

With $x=r\cos\theta$, $y=r\sin\theta$: $dx=\cos\theta\,dr-r\sin\theta\,d\theta$ and $dy=\sin\theta\,dr+r\cos\theta\,d\theta$. Wedging,

$dx\wedge dy=(\cos^2\theta+\sin^2\theta)\,r\, dr\wedge d\theta=r\,dr\wedge d\theta.$

Three lines of algebra reproduce the polar area element, no geometric picture required.

Forms are integrands, not functions. A $k$-form eats $k$ vectors and returns the signed $k$-volume they span, weighted by position. That is precisely the data you need to integrate over a $k$-dimensional surface, which is why the definition looks the way it does.
Common pitfalls

• Forgetting antisymmetry when sorting indices. $dy\wedge dx=-dx\wedge dy$, always.
• Writing $\omega\wedge\omega=0$ for even-degree forms. It holds for odd degree only.
• Mixing normalisation conventions for the wedge between sources. Pick one and check your text.
• Treating a $2$-form in $\mathbb{R}^3$ as a vector field without saying which identification you used. It works in $\mathbb{R}^3$ and nowhere else.
• Trying to push a form forward. Forms pull back; only diffeomorphisms let you go the other way, via the inverse.

Practice

Original problems written for this guide in the style of Spivak and Munkres exercises. Write a full solution before opening the answer.

Practice 1 · Dimension count

What is $\dim\Lambda^2(\mathbb{R}^4)$, and list a basis.

Solution
$\binom42=6$, with basis $dx_1\wedge dx_2$, $dx_1\wedge dx_3$, $dx_1\wedge dx_4$, $dx_2\wedge dx_3$, $dx_2\wedge dx_4$, $dx_3\wedge dx_4$.
Practice 2 · A wedge

Compute $(dx+dy)\wedge(dy+dz)$ in $\mathbb{R}^3$.

Solution
$dx\wedge dy+dx\wedge dz+dy\wedge dy+dy\wedge dz =dx\wedge dy+dx\wedge dz+dy\wedge dz$.
Practice 3 · Evaluating a $2$-form

Evaluate $dx\wedge dy$ on $v=(1,2,0)$ and $w=(3,1,0)$.

Solution
The determinant of $\begin{smallmatrix}1&3\\2&1\end{smallmatrix}$, which is $-5$. That is the signed area of the parallelogram they span in the $xy$-plane.
Practice 4 · Pullback

Pull back $x\,dy$ under $g(u,v)=(u^2,uv)$.

Solution
$x=u^2$ and $dy=v\,du+u\,dv$, so $g^{*}(x\,dy)=u^2v\,du+u^3\,dv$.
Practice 5 · Dependence detector

Show $v_1\wedge\dots\wedge v_k=0$ exactly when the $v_i$ are dependent.

Solution
If dependent, expand one vector in the others; every resulting term repeats a factor and dies. If independent, extend to a basis; the wedge is then a nonzero multiple of a basis element of $\Lambda^k$.
Practice 6 · Graded commutativity

For a $1$-form $\alpha$ and a $2$-form $\beta$, relate $\alpha\wedge\beta$ and $\beta\wedge\alpha$.

Solution
$(-1)^{1\cdot2}=1$, so they are equal. Odd against odd is the only case that produces a sign flip.
Practice 7 · Volume form of a sphere

Pull $dx\wedge dy\wedge dz$ back to spherical coordinates.

Solution
The determinant of the spherical Jacobian is $\rho^2\sin\varphi$, so the pullback is $\rho^2\sin\varphi\,d\rho\wedge d\varphi\wedge d\theta$. Same factor as Unit 10, now derived algebraically.
Practice 8 · Forms versus vector fields

Why is the identification of $2$-forms with vector fields special to $\mathbb{R}^3$?

Solution
It needs $\binom{n}{2}=n$, which holds only for $n=3$. In $\mathbb{R}^4$ there are six independent $2$-forms and four coordinate directions, so no such identification exists.

Expanded topic checklist

Course coverage. This unit is part of the complete MAT257Y1 topic map. It should cover the following items before you move on:

Textbook/source alignment

Spivak's chapter on integration on chains develops tensors and forms compactly; Munkres gives a longer treatment of the alternating algebra first. Conventions for the wedge constant differ between texts — match the syllabus.

Core theorem map

Antisymmetry gives the dimension count, which makes top forms one-dimensional, which makes the determinant appear under pullback, which is where the change of variables factor comes from. Unit 12 adds the derivative of a form, and Unit 14 integrates it.

Study routine. Compute wedge products by hand until the sign bookkeeping is automatic. Almost every lost mark in the last third of the course is a sign, and there is no way to get fast at it except repetition.

Deep practice extensions

Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.

Extension 1 · Why alternating

Why insist on antisymmetry?

Guide
Because signed volume is antisymmetric: swapping two edges of a parallelepiped reverses its orientation.
Extension 2 · Top degree

Why is $\Lambda^n(\mathbb{R}^n)$ one-dimensional?

Guide
$\binom nn=1$. Every alternating $n$-tensor is a multiple of the determinant.
Extension 3 · Direction of transport

Do forms push forward or pull back?

Guide
Pull back, along any smooth map. Pushing forward needs an inverse.
Extension 4 · Where the Jacobian comes from

Why does $\det Jg$ appear when pulling back a top form?

Guide
Because the top exterior power is one-dimensional, so $Dg$ acts on it by a single scalar, and that scalar is by definition the determinant.