On a curved space, subtracting points need not make sense. Differentiating a function along a curve does. We use that operation to define a tangent vector.
3.1. Vectors as differentiation at a point
Definition 3.1. Smooth germ
Two smooth real functions defined on neighbourhoods of \(p\in M\) have the same germ at \(p\) if they agree on some neighbourhood of \(p\). A germ is an equivalence class under this relation. Write \(C^\infty_p(M)\) for the germs. Addition and multiplication use representatives on their common domain.
Worked example for Definition 3.1
Counterexample: Equal velocities do not determine curves or function germs
Proof.
Agreement near \(p\) is reflexive and symmetric. Intersecting two neighbourhoods proves transitivity. If representatives are changed, the sums and products still agree on the intersection of the neighbourhoods of agreement. Thus these operations are well defined.
□Example. The function \(x\mapsto x^2\) on \(\mathbb R\) and its restriction to \((-1,1)\) have the same germ at zero. The functions zero and \(x^2\) have the same value there but different germs.
Definition 3.2. Derivation and tangent space
A derivation at \(p\) is a real linear map \(v:C^\infty_p(M)\to\mathbb R\) obeying the product rule
\[v(fh)=f(p)v(h)+h(p)v(f).\]The tangent space \(T_pM\) is the vector space of these maps, with pointwise addition and scalar multiplication. We write \(v(f)\) for its action on a germ or any representative. In a chart \(x\), define the coordinate derivation
\[\partial_i|_p(f)=\partial_i(f\circ x^{-1})(x(p)).\]The product rule captures first-order differentiation. Using germs ensures that a vector at \(p\) depends only on what happens arbitrarily near \(p\). The Euclidean product rule proves that \(\partial_i|_p\) is a derivation; equality on a neighbourhood makes it independent of the germ representative. Linear combinations of derivations satisfy the same identity, so \(T_pM\) really is a vector space.
Proposition 3.3. Coordinate derivations form a basis
On an \(n\)-manifold, every \(v\in T_pM\) has the unique expression
\[v=\sum_{i=1}^n v^i\partial_i|_p,\qquad v^i=v(x^i).\]In particular \(\dim T_pM=n\).
Proof.
First \(v(1)=v(1\cdot1)=2v(1)\), so constants are killed. Put \(a=x(p)\). The smooth remainder formula in chapter 1 gives, as germs,
\[f=f(p)+\sum_i(x^i-a_i)H_i(x-a),\qquad H_i(0)=\partial_i(f\circ x^{-1})(a).\]Apply \(v\). In each product the factor \(x^i-a_i\) vanishes at \(p\), leaving \(v(f)=\sum_i v(x^i)\partial_i|_p(f)\). Finally \(\partial_i|_p(x^j)=\delta_i^j\), where \(\delta_i^j=1\) for \(i=j\) and zero otherwise, proves uniqueness and linear independence.
□Example. At \((1,2)\in\mathbb R^2\), the vector \(v=3\partial_x-\partial_y\) sends \(x^2+y^2\) to \(3(2)-1(4)=2\). A vector is determined by its effect on the two coordinate functions, but acts on every smooth function.
Definition 3.4. Velocity of a curve
A smooth curve is a smooth map \(\gamma:I\to M\), with \(I\subset\mathbb R\) an open interval. Its velocity at \(t\) is
\[\dot\gamma(t)(f)=\frac{d}{ds}\bigg|_{s=t}f(\gamma(s)).\]A curve is regular if its velocity never vanishes. For a \(C^1\) curve the same formula defines its velocity.
Worked example for Definition 3.4
Counterexample: Equal velocities do not determine curves or function germs
Proposition 3.5. Every tangent vector is a velocity
Velocities are derivations. In a chart, \(\dot\gamma(t)=\sum_i (x^i\circ\gamma)'(t)\partial_i|_{\gamma(t)}\). Conversely every tangent vector is the velocity of a smooth curve at some parameter value.
Worked example for Proposition 3.5
Counterexample: Equal velocities do not determine curves or function germs
Proof.
The one-variable sum and product rules prove the derivation laws. The basis theorem identifies the components by applying the velocity to \(x^i\). Given components \(v^i\) at \(p\), the curve \(\gamma(t)=x^{-1}(x(p)+t(v^1,\ldots,v^n))\) is defined for sufficiently small \(|t|\), since the chart image is open; its velocity at zero has exactly those components.
□Example. For \(\gamma(t)=(\cos t,\sin t)\) on the circle, \(\dot\gamma(0)\) has ambient components \((0,1)\). Intrinsically it acts on a circle function by differentiating that function along the circle. The ambient components describe the differential of the inclusion, not a second definition of the intrinsic vector.
3.2. Differentiating a map
Definition 3.6. Differential
For a smooth map \(F:M\to N\), its differential at \(p\), also called its pushforward at \(p\), is
\[dF_p:T_pM\to T_{F(p)}N,\qquad (dF_pv)(h)=v(h\circ F).\]Proposition 3.7. Coordinate formula and chain rule
The differential is well defined and linear. In coordinates it is the Jacobian of \(y\circ F\circ x^{-1}\). Moreover \(d(G\circ F)_p=dG_{F(p)}dF_p\) and \(dF_{\gamma(t)}\dot\gamma(t)=(F\circ\gamma)'(t)\).
Proof.
Continuity of \(F\) pulls neighbourhoods of \(F(p)\) back to neighbourhoods of \(p\), so composition is well defined on germs. The derivation product rule and linearity pass through composition. Applying \(dF_pv\) to each coordinate \(y^a\) gives \(\sum_i v^i\partial_i(y^a\circ F\circ x^{-1})(x(p))\). For the chain rule, evaluate both sides on \(h\): both are \(v(h\circ G\circ F)\). The velocity identity is the same equality with \(v=\dot\gamma(t)\).
□Example. The map \(F(x,y)=(x^2,y)\) sends \(3\partial_x-\partial_y\) at \((1,2)\) to \(6\partial_{y^1}-\partial_{y^2}\) at \((1,2)\). A diffeomorphism has invertible differential because the differential of each identity is the identity.
Proposition 3.8. Coordinate change for a vector
If \(y=y\circ x^{-1}\) denotes a transition, then
\[\partial_{x^i}|_p=\sum_a\frac{\partial y^a}{\partial x^i}\bigg|_p\partial_{y^a}|_p,\qquad v_y^a=\sum_i\frac{\partial y^a}{\partial x^i}\bigg|_p v_x^i.\]Every transition derivative is invertible.
Proof.
Apply the Euclidean chain rule to \(f\circ x^{-1}=(f\circ y^{-1})\circ(y\circ x^{-1})\). This identifies the two derivations on every germ. Summing against \(v_x^i\) proves the component formula. Differentiating the two inverse transition identities proves invertibility.
□Example. For \(y=2x\) on the line, \(\partial_x=2\partial_y\). Thus \(3\partial_x=6\partial_y\). The vector is the same; its component changes.
3.3. Vectors varying with the point
Definition 3.9. Tangent bundle and vector field
The tangent bundle is the disjoint union \(TM=\coprod_{p\in M}T_pM\), with projection \(\pi(p,v)=p\). A chart on \(M\) induces coordinates \((p,v)\mapsto(x(p),v_x)\) on \(\pi^{-1}(U)\). Declare these to be smooth charts. A smooth vector field is a smooth map \(X:M\to TM\) with \(\pi\circ X=\operatorname{id}_M\); such a map is called a section. Equivalently, \(X=\sum_i X^i\partial_i\) has smooth components on each chart.
Proof.
The bundle transitions are \((x,v)\mapsto(y(x),Dy_xv)\), smooth with smooth inverses. They specify a topology by taking coordinate-open sets as a basis; their agreement on overlaps makes this well defined. Distinct base points are separated using the Hausdorff base; distinct vectors over one point are separated in a common bundle chart. A second-countable space is Lindelöf: for each basis member contained in a cover member, choose one such cover member; the chosen countable family covers. Hence \(M\) has a countable chart subcover, and the countable Euclidean bases of its bundle charts give a countable basis for \(TM\). Finally the coordinate representative of a section is \(x\mapsto(x,X^1(x),\ldots,X^n(x))\), proving the equivalence.
□Example. On \(\mathbb R^2\), \(X=-y\partial_x+x\partial_y\) is smooth and is tangent to every circle centred at the origin: \(X(x^2+y^2)=0\). Its restriction is a nonzero vector field on the unit circle.
Proposition 3.10. Tangent vectors to a regular constraint
For the level set in Proposition 2.7, the differential of its inclusion identifies \(T_pQ\) with \(\ker DG_p\).
Proof.
Since \(G\) is constant on \(Q\), its differential kills every included tangent vector. In the slice coordinates constructed in chapter 2, the inclusion has derivative \(u\mapsto(u,0)\), so is injective. Both its image and \(\ker DG_p\) have dimension \(m-r\), by the basis theorem and rank–nullity, so containment is equality.
□Example. On \(S^1\), the constraint gives \(T_{(X,Y)}S^1=\{(h,k):Xh+Yk=0\}\) under the inclusion. At \((1,0)\) this is the vertical line, agreeing with the curve example.
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 derivation uses the germ, not just the value
Applies to: Definition 3.1, Definition 3.2, Proposition 3.3.
At \(p=(1,2)\) in \(\mathbb R^2\), define \(v(f)=3\partial_x f(p)-\partial_y f(p)\). Functions agreeing near \(p\) have the same derivatives there, so this is well defined on germs. Linearity and \(\partial_i(fh)=f\partial_i h+h\partial_i f\) give the derivation identity explicitly. Its components are \(v(x)=3,v(y)=-1\), and \(v(xy)=3\cdot2-1\cdot1=5\). The constant 2 and the function \(y\) have equal values at \(p\) but different germs, detected by \(v(2)=0,v(y)=-1\). The basis formula says precisely that knowing 3 and \(-1\) recovers all these actions; linear independence is checked by applying a proposed zero combination to \(x\) and \(y\).
Worked example 2. A curve and its pushed-forward velocity
Applies to: Definition 3.4, Proposition 3.5, Definition 3.6, Proposition 3.7.
Let \(\gamma(t)=(1+3t,2-t)\), \(t\in\mathbb R\). It is smooth and regular, and \((f\circ\gamma)'(0)=3f_x(p)-f_y(p)\), so it realizes the preceding tangent vector. For \(F(x,y)=(x^2,y)\), \(dF_p(v)=(6,-1)\) at \(F(p)=(1,2)\). Direct differentiation of \(F\circ\gamma=((1+3t)^2,2-t)\) gives the same velocity. With \(G(u,w)=uw\), \(dG_{F(p)}(6,-1)=12-1=11\), also the derivative at zero of \((1+3t)^2(2-t)\). This checks the derivation definition of pushforward, its Jacobian formula, and both chain rules with all base points specified.
Worked example 3. Changing vector components
Applies to: Proposition 3.8.
Take global coordinates \(u=2x,v=y+x\) on the plane, inverse \(x=u/2,y=v-u/2\). The Jacobian \(\begin{pmatrix}2&0\\1&1\end{pmatrix}\) has determinant 2. Therefore \(\partial_x=2\partial_u+\partial_v\), \(\partial_y=\partial_v\), and \(3\partial_x-\partial_y=6\partial_u+2\partial_v\). Testing on \(u,v\) returns 6 and 2 directly. The inverse Jacobian recovers 3 and \(-1\), illustrating invertible coordinate changes rather than a change of the vector itself.
Worked example 4. A vector field tangent to a circle
Applies to: Definition 3.9, Proposition 3.10.
On \(S^1\subset\mathbb R^2\) let \(X_{(x,y)}=(-y,x)\). The constraint derivative is \(DG_{(x,y)}(a,b)=2xa+2yb\), which kills \(X\). At every circle point its kernel is exactly the line spanned by \((-y,x)\), because the row \((x,y)\) is nonzero. The curve \(s\mapsto(x\cos s-y\sin s,x\sin s+y\cos s)\) lies on the circle and has velocity \((-y,x)\) at zero; scalar multiples follow by rescaling \(s\). Thus the included tangent space is exactly this kernel, not just a subset. In a local angle coordinate \(X=\partial_\theta\), so its bundle components are smooth and \(\pi\circ X=\operatorname{id}\). This verifies the section and smooth-field requirements of the tangent-bundle definition.
Further proof details
Smoothness can be checked in the explicit stereographic charts of chapter 2 as well. Write ambient circle coordinates as \((X,Y)\). For \(u=X/(1-Y)\), the field \((-Y,X)\) has component
\[X_{\mathrm{field}}(u)=\frac{-Y(1-Y)+X^2}{(1-Y)^2}=\frac1{1-Y}=\frac{1+u^2}{2}.\]For the other chart \(v=X/(1+Y)\), its component is \((-Y(1+Y)-X^2)/(1+Y)^2=-(1+v^2)/2\). Both are smooth and nonzero. The induced bundle charts therefore give a smooth section on a covering atlas, establishing the asserted field property without assuming a global angle coordinate.
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. Equal velocities do not determine curves or function germs
Related statements: Definition 3.1, Definition 3.4, Proposition 3.5.
Claim refuted. A tangent vector records an entire local curve, or a germ is determined by its value and first derivatives.
Proof.
In \(\mathbb R^2\), the curves \(\gamma(t)=(t,t^2)\) and \(\delta(t)=(t,0)\) both pass through the origin and have velocity \((1,0)\) there. For any smooth \(h\), the chain rule gives \((h\circ\gamma)'(0)=h_x(0,0)=(h\circ\delta)'(0)\), so they define the same derivation. Yet for every nonzero \(t\), their second coordinates differ; the curves agree on no neighborhood of the parameter zero. Their accelerations are \((0,2)\) and \((0,0)\), which the common tangent does not retain.
Likewise the real functions zero and \(x^2\) have equal value at zero, and every derivation \(v\) there kills \(x^2\): Leibniz gives \(v(x^2)=2x(0)v(x)=0\). They nevertheless have different germs, since \(x^2\) is nonzero at every nearby nonzero point. Germs retain more local information than tangent vectors can detect.
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