A tangent vector differentiates a function. A covector takes a tangent vector and returns a number. Keeping the two roles separate makes coordinate formulas predictable.
4.1. Dual vectors
Definition 4.1. Cotangent space and differential of a function
The cotangent space \(T_p^*M=(T_pM)^*\) is the space of linear maps \(T_pM\to\mathbb R\). Its elements are covectors. For smooth \(f:M\to\mathbb R\), the covector \(df_p\) is \(df_p(v)=v(f)\). This identifies the differential from chapter 3 with a covector through the coordinate basis of \(T_{f(p)}\mathbb R\).
Proposition 4.2. Coordinate covectors
The covectors \(dx^i|_p\) form the basis dual to \(\partial_i|_p\): \(dx^i(\partial_j)=\delta^i_j\). Thus \(df=\sum_i(\partial_i f)dx^i\). Under \(y=y(x)\), \(dy^a=\sum_i(\partial_i y^a)dx^i\).
Proof.
Each equation follows by applying both covectors to every coordinate basis vector. Equality on a basis implies equality by linearity.
□Example. At \((1,2)\), \(d(x^2+y^2)=2dx+4dy\). Its value on \(3\partial_x-\partial_y\) is \(2\), as in chapter 3. For \(y=2x\), \(dy=2dx\), and a fixed covector \(a\,dx\) has component \(a/2\) in the \(dy\) basis.
Definition 4.3. Tensor and tensor field
On a finite-dimensional real vector space \(V\), a tensor of type \((r,s)\) is a multilinear map
\[T:(V^*)^r\times V^s\longrightarrow\mathbb R.\]Multilinear means linear in each argument when the others are fixed. A \((0,0)\) tensor is a scalar. In finite dimension a \((1,0)\) tensor is a vector: \(v\) acts on \(\alpha\) by \(\alpha(v)\). To see that this accounts for every functional \(L\) on \(V^*\), take a basis \(e_i\) and its dual \(\epsilon^i\); then \(v=\sum_iL(\epsilon^i)e_i\) has \(L(\alpha)=\alpha(v)\).
A tensor field assigns a tensor on \(T_pM\) to every point \(p\). It is smooth when its components, obtained by inserting coordinate vectors and covectors in the slots, are smooth functions in each chart. A covariant 2-tensor has type \((0,2)\), so is a bilinear form on tangent vectors.
Multilinearity is the condition that lets basis values determine the whole object. A tensor field itself does not depend on the basis.
Proposition 4.4. Transformation and smoothness
For a covariant 2-tensor \(A\), write \(A_{ij}=A(\partial_{x^i},\partial_{x^j})\). If \(J^i_a=\partial x^i/\partial y^a\), then
\[A'_{ab}=\sum_{i,j}J^i_a J^j_b A_{ij}.\]The old components on the right are evaluated at \(x\circ y^{-1}(y)\). Smoothness can be checked on any covering smooth atlas. For a general tensor the same rule applies to each slot, using the vector or covector basis change appropriate to that slot.
Proof.
Insert \(\partial_{y^a}=\sum_iJ^i_a\partial_{x^i}\) into each slot and expand by bilinearity. For any number of slots, expand by multilinearity instead. All coefficients are derivatives of smooth transitions or inverse transitions, so are smooth. Products, sums, and composition preserve smoothness by Proposition 1.5. This proves atlas independence.
□Example. For the Euclidean bilinear form \(A=dx\otimes dx\) on the line and \(y=2x\), its component is \(A'_{yy}=1/4\). Here \((\alpha\otimes\beta)(v,w)=\alpha(v)\beta(w)\) defines the displayed tensor product. The scalar \(A(v,v)\) is unchanged: \(v_y=2v_x\), so \((1/4)(2v_x)^2=v_x^2\).
Definition 4.5. Pullback
For \(F:M\to N\) smooth and a covariant \(s\)-tensor field \(A\) on \(N\), its pullback is the field on \(M\) defined by
\[(F^*A)_p(v_1,\ldots,v_s)=A_{F(p)}(dF_pv_1,\ldots,dF_pv_s).\]For a function, \(F^*f=f\circ F\).
Worked example for Definition 4.5
Counterexample: Pulling back a metric can destroy nondegeneracy
Proof.
Linearity of \(dF\) makes the result multilinear. Its components are sums of components of \(A\circ F\) times Jacobian entries, hence smooth. The chain rule gives \((G\circ F)^*=F^*G^*\) and \(F^*(df)=d(f\circ F)\) by direct evaluation on vectors.
□Example. For \(F(t)=(t,t^2)\), \(F^*(dx\otimes dx+dy\otimes dy)=(1+4t^2)dt\otimes dt\). This follows from \(dF(\partial_t)=\partial_x+2t\partial_y\).
4.2. Alternating tensors
Definition 4.6. Differential forms and wedge product
An alternating \(k\)-form on \(V\) is a \(k\)-linear real-valued map that changes sign when two slots are exchanged. A differential \(k\)-form on \(M\) is a smooth field of such maps on tangent spaces. A 0-form is a smooth function. The wedge product of a \(k\)-form \(\alpha\) and an \(\ell\)-form \(\beta\) is
\[(\alpha\wedge\beta)(v_1,\ldots,v_{k+\ell})=\frac1{k!\ell!}\sum_{\sigma\in S_{k+\ell}}\operatorname{sgn}(\sigma)\alpha(v_{\sigma(1)},\ldots,v_{\sigma(k)})\beta(v_{\sigma(k+1)},\ldots,v_{\sigma(k+\ell)}).\]\(S_j\) is the set of permutations of \(j\) labels; \(\operatorname{sgn}(\sigma)\) is the determinant of the corresponding permutation matrix. In particular \((\alpha\wedge\beta)(v,w)=\alpha(v)\beta(w)-\alpha(w)\beta(v)\) for covectors.
Proposition 4.7. Components of a form
Every \(k\)-form has a unique coordinate expansion
\[\omega=\sum_{i_1<\cdots<i_k}\omega_{i_1\cdots i_k}\,dx^{i_1}\wedge\cdots\wedge dx^{i_k},\qquad \omega_{i_1\cdots i_k}=\omega(\partial_{i_1},\ldots,\partial_{i_k}).\]In particular a 2-form is a skew-symmetric covariant 2-tensor, and forms of degree greater than the dimension vanish.
Proof.
Multilinearity expands evaluation on arbitrary vectors in terms of evaluations on basis vectors. A repeated basis vector gives zero because swapping its two occurrences negates the value. All remaining ordered tuples are permutations of increasing tuples. The wedge of distinct coordinate covectors evaluates on those tuples as the corresponding determinant: the wedge formula enumerates precisely its signed permutations, with its factorial divisor cancelling repetitions within the two groups at each product. It is one on the matching increasing tuple and zero on all others. This proves the expansion and uniqueness. There is no distinct tuple longer than the basis.
□Example. On \(\mathbb R^2\), \((dx\wedge dy)((a,b),(c,d))=ad-bc\), the signed area of the two vectors. On Minkowski coordinates \((x^0,x^1,x^2,x^3)\), the electromagnetic field can be written
\[F=-\sum_{i=1}^3\frac{E_i}{c}\,dx^0\wedge dx^i+B_1dx^2\wedge dx^3+B_2dx^3\wedge dx^1+B_3dx^1\wedge dx^2.\]Evaluating on coordinate pairs gives exactly \(F_{0i}=-E_i/c\) and \(F_{ij}=\epsilon_{ijk}B_k\) from the electromagnetic two-form definition. Here \(c>0\) is the model’s speed constant and \(\epsilon_{ijk}\) is zero for repeated indices, +1 for even permutations of 123, and −1 for odd ones. This defines a field; Maxwell’s equations are a separate postulate.
Definition 4.8. Exterior derivative
In a chart define the exterior derivative of a smooth form by
\[d\omega=\sum_{i_1<\cdots<i_k}\sum_j\partial_j\omega_{i_1\cdots i_k}\,dx^j\wedge dx^{i_1}\wedge\cdots\wedge dx^{i_k}.\]For functions this is the differential already defined. A form is closed if \(d\omega=0\), and exact if \(\omega=d\eta\) for some form \(\eta\).
Why this definition is coordinate independent, and why d² = 0
Proof.
The coordinate formula gives \(d(\alpha\wedge\beta)=d\alpha\wedge\beta+(-1)^k\alpha\wedge d\beta\) when \(\alpha\) has degree \(k\). Indeed expand into increasing coordinate wedges, apply the scalar product rule to each pair of coefficients, and move the new derivative covector past the \(k\) factors of \(\alpha\) for the second term. Terms with repeated covectors vanish, and exchanging adjacent covectors changes sign; both follow from the alternating sum defining the wedge.
For a scalar \(f\), \(d(df)=\sum_{i,j}\partial_j\partial_i f\,dx^j\wedge dx^i=0\): paired terms cancel by equality of mixed partials, and diagonal terms vanish. Thus \(d(dy^a)=0\) for every smooth coordinate function expressed in an \(x\)-chart. Expand a form as \(\sum_I a_I(y)dy^{i_1}\wedge\cdots\wedge dy^{i_k}\). Applying the \(x\)-formula and the just-proved product rule differentiates only \(a_I(y)\). The chain rule gives its intrinsic differential, so the resulting expression is exactly the \(y\)-formula. This proves coordinate independence; see the intrinsic exterior derivative.
Finally apply the coordinate formula twice to any form. Each second coefficient derivative is symmetric in its derivative indices and each wedge is antisymmetric in them, so terms cancel in pairs. Hence \(d^2=0\), and every exact form is closed; see the square-zero theorem and closed and exact forms. This does not assert that every closed form is globally exact.
□Example. For \(A=x\,dy\) on \(\mathbb R^2\), \(dA=dx\wedge dy\), and \(d(dA)=0\). Thus a smooth potential automatically gives a closed field. Existence of a potential is an additional question.
4.3. Orientation and volume
Definition 4.9. Orientation and volume form
For a positive-dimensional real vector space, two ordered bases have the same orientation if the change-of-basis determinant is positive. An orientation is one of the two resulting equivalence classes. An oriented smooth manifold has a covering atlas whose transition determinants are positive. A volume form is a nowhere-zero differential form of top degree; it agrees with an orientation if it is positive on its positive bases.
Proof.
Determinant multiplicativity and \(\det(B^{-1})=1/\det B\) prove that the positive-determinant relation is an equivalence relation. Every invertible real matrix has positive or negative determinant; changing one basis vector’s sign interchanges the two classes. A top-degree form evaluates on a changed basis as its original value times the change determinant, by its alternating multilinear expansion. Thus positivity on one positive basis is positivity on every such basis.
□Example. On oriented \(\mathbb R^n\), \(dx^1\wedge\cdots\wedge dx^n\) is the standard volume form. In a positively oriented Lorentz-orthonormal basis, \(dx^0\wedge dx^1\wedge dx^2\wedge dx^3\) is unchanged by another such basis: if \(L^T\eta L=\eta\), determinants give \((\det L)^2=1\), and orientation selects \(\det L=1\). This proves the frame independence of Minkowski volume. The Lorentz form \(\eta\) is defined in chapter 5.
Worked examples
Each example checks the hypotheses and the calculation for the linked statements. The links in the statements return here.
Worked example 1. A covector and its dual-coordinate calculation
Applies to: Definition 4.1, Proposition 4.2.
At \((1,2)\), let \(f(x,y)=xy\). Then \(df=2dx+dy\), so \(df(3\partial_x-\partial_y)=5\). This agrees with the derivation calculation and proves linearity in the tangent vector. With \(u=2x,v=y\), \(du=2dx,dv=dy\) and \(\partial_u=\tfrac12\partial_x\); hence \(du(\partial_u)=1\), \(du(\partial_v)=0\). The same covector is \(du+dv\) at this point, and the vector is \(6\partial_u-\partial_v\), giving again \(6-1=5\). This checks the differential, dual basis, and covector change law together.
Worked example 2. A bilinear tensor and a pullback
Applies to: Definition 4.3, Proposition 4.4, Definition 4.5.
On \(\mathbb R^2\), let \(A=dx\otimes dx+dy\otimes dy\), where \(A(v,w)=v_xw_x+v_yw_y\). It is linear in either slot separately, hence a smooth \((0,2)\)-tensor. Under \(u=2x,v=y\) its matrix is \(\operatorname{diag}(1/4,1)=J^TIJ\), with \(J=\operatorname{diag}(1/2,1)\). For vectors having old coordinates \((a,b),(c,d)\), the new evaluation is \((1/4)(2a)(2c)+bd=ac+bd\), verifying the transformation law. For \(F(t)=(t,t^2)\), the pullback is \((1+4t^2)dt\otimes dt\), since \(dF_t(a\partial_t)=(a,2ta)\). Inserting two such images gives \((1+4t^2)ab\), exactly the definition of \(F^*A\).
Worked example 3. Exterior differentiation and its signs
Applies to: Definition 4.6, Proposition 4.7, Definition 4.8.
On \(\mathbb R^3\), put \(\alpha=x\,dy\). Then \(d\alpha=dx\wedge dy\), and \((dx\wedge dy)((1,2,0),(3,4,0))=4-6=-2\). Reversing the vectors gives 2; repeating a vector gives zero. In its increasing-coordinate expansion the only nonzero coefficient is \(\omega_{12}=1\), as evaluation on \(\partial_x,\partial_y\) confirms. The resulting 2-form is exact and closed: \(d(dx\wedge dy)=0\). Every 4-form on this 3-dimensional space vanishes because four basis choices force repetition in an alternating expression.
The graded product rule can be checked on \(\alpha\wedge z=xz\,dy\): its derivative is \(z\,dx\wedge dy+x\,dz\wedge dy=d\alpha\,z-\alpha\wedge dz\). Also \(d(d(xy))=d(y\,dx+x\,dy)=dy\wedge dx+dx\wedge dy=0\). Under \(u=2x,v=y\), \(\alpha=(u/2)dv\) and \(d\alpha=\tfrac12du\wedge dv=dx\wedge dy\), checking coordinate independence as well as the unboxed square-zero result.
Worked example 4. Orientation and a volume form
Applies to: Definition 4.9.
On \(\mathbb R^2\), \(\Omega=dx\wedge dy\) never vanishes. For an ordered basis \(a=(a_1,a_2),b=(b_1,b_2)\), \(\Omega(a,b)=a_1b_2-a_2b_1\), so it is positive exactly for positive determinant. The chart \((u,v)=(2x,y)\) has positive transition determinant 2 and \(\Omega=\tfrac12du\wedge dv\), giving the same orientation. Swapping coordinates gives determinant \(-1\) and changes the sign. This explicitly connects the basis-equivalence definition, the atlas definition, and a compatible nowhere-zero top form.
Counterexamples and limits of the hypotheses
Each construction below refutes the explicitly stated stronger claim. It does not refute the related theorem with its original hypotheses.
Counterexample 1. Closed does not imply globally exact
Related statements: Definition 4.8.
Claim refuted. Every closed differential form is exact on any domain.
Proof.
On \(U=\mathbb R^2\setminus\{0\}\), let \(\omega=-y\,dx/(x^2+y^2)+x\,dy/(x^2+y^2)\). Its coefficients are smooth on this open set. Writing them as \(P,Q\), \(d\omega=(Q_x-P_y)dx\wedge dy\). Direct differentiation gives \(Q_x=(y^2-x^2)/(x^2+y^2)^2=P_y\), hence \(d\omega=0\). On \(\gamma(t)=(\cos t,\sin t)\), the pullback is \(\gamma^*\omega=(\sin^2t+\cos^2t)dt=dt\), so its integral is \(2\pi\). If \(\omega=df\) for a single-valued smooth \(f\), the same integral would be \(f(\gamma(2\pi))-f(\gamma(0))=0\), contradiction. Local coordinate definitions do not remove the global hole in \(U\).
Counterexample 2. Pulling back a metric can destroy nondegeneracy
Related statements: Definition 4.5.
Claim refuted. The pullback of a Riemannian metric by any smooth map is a Riemannian metric.
Proof.
Let \(F:\mathbb R^2\to\mathbb R\), \(F(x,y)=x\), with target metric \(dt\otimes dt\). Then \(dF(a,b)=a\) and \(F^*g((a,b),(c,d))=ac\), so the pullback is \(dx\otimes dx\). It kills the nonzero vector \(\partial_y\) against every vector and is degenerate, although \(F\) is smooth and onto. The missing hypothesis for a positive-definite target is injectivity of \(dF\): a kernel vector always has zero pullback length, whereas if \(dF(v)\ne0\) for every nonzero \(v\), positive target length gives \(F^*g(v,v)>0\). This proves both the failure and the precise pointwise repair.
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