What does it mean to differentiate on a space without preferred coordinates? We first need a derivative that survives a change of coordinates. This chapter builds it in Euclidean space. Chapter 2 then defines the spaces on which we will use it.
This series supplies the differential geometry used by the opening pages of Alphabeta Math’s physics library, checked on October 15, 2026. We assume basic topology, Euclidean metric spaces, limits, sequences, single-variable differentiation, and basic linear algebra. We introduce multivariable calculus, manifolds, tangent vectors, tensors, affine and Lorentzian geometry, and connections. The last chapter gives the exact reading routes. No field equation follows from these mathematical definitions.
We write every coordinate sum explicitly. An upper or lower index labels a component; it is not an exponent unless a power is stated. In the physics library, the Einstein summation convention often omits the sum sign for an index repeated once upstairs and once downstairs. Abstract indices instead indicate tensor slots without choosing a coordinate basis. The concrete transformation laws below give the coordinate meaning in either notation.
Seven-chapter reading order
1.1. A linear approximation
Throughout, \(\mathbb R^n\) has its Euclidean norm \(|\cdot|\). An open set contains a ball about each of its points. All Euclidean domains below are open unless stated otherwise. For a linear map \(A:\mathbb R^m\to\mathbb R^n\), put \(\|A\|=\sup_{|v|=1}|Av|\). This number is finite: writing the rows of \(A\) as \(a_i\), Cauchy–Schwarz gives \(|Av|^2\le\sum_i|a_i|^2|v|^2\). Consequently \(|Av|\le\|A\||v|\).
Definition 1.1. Total derivative
A map \(f:U\subset\mathbb R^m\to\mathbb R^n\) is differentiable at \(a\in U\) if a linear map \(Df_a\) satisfies
\[f(a+h)=f(a)+Df_a(h)+r(h),\qquad \lim_{h\to0}\frac{|r(h)|}{|h|}=0.\]The last condition is abbreviated \(r(h)=o(|h|)\). A map is differentiable if this holds at every point. A partial derivative \(\partial_i f(a)\) is the single-variable derivative at zero of \(t\mapsto f(a+te_i)\), where \(e_i\) is the \(i\)-th standard basis vector.
Worked example for Definition 1.1
Counterexample: A total derivative need not vary continuously
Linearity makes the approximation a map between displacement spaces. The error condition requires it to work in every direction at once.
Proposition 1.2. Uniqueness and partial derivatives
The total derivative is unique, differentiability implies continuity, and \(Df_a(e_i)=\partial_i f(a)\). Thus the derivative matrix, called the Jacobian matrix, has entries \((Df_a)_{ji}=\partial_i f^j(a)\), where \(f^j\) is the \(j\)-th component.
Proof.
If \(A,B\) satisfy the definition, substitute \(h=tv\) and divide by \(t\ne0\). The errors tend to zero, so \((A-B)v=0\) for every \(v\). The expansion and the bound for \(A\) imply \(f(a+h)\to f(a)\). Substituting \(h=te_i\) identifies each partial derivative.
□Example. For \(f(x,y)=x^2+y^2\), \(Df_{(a,b)}(h,k)=2ah+2bk\). The remainder is \(h^2+k^2\), whose quotient by \(\sqrt{h^2+k^2}\) tends to zero. For \(f(x,y)=xy\), the derivative is \(bh+ak\); the remainder obeys \(|hk|\le(h^2+k^2)/2\).
Definition 1.3. The classes Cᵏ
\(C^0(U,\mathbb R^n)\) means continuous maps. Recursively, \(f\) is \(C^{k+1}\) if it is differentiable and the matrix-valued function \(a\mapsto Df_a\) is \(C^k\), entry by entry. A map is smooth, or \(C^\infty\), if it is \(C^k\) for every nonnegative integer \(k\). Here \(k\) is always an integer unless we write \(\infty\).
Continuity of the derivative ensures that nearby linear approximations vary continuously. Higher regularity repeats that requirement.
Proposition 1.4. Continuous partial derivatives suffice
If all first partial derivatives exist and are continuous on \(U\), then \(f\) is \(C^1\). Recursively, \(C^k\) is equivalent to existence and continuity of every iterated partial derivative of order at most \(k\).
Worked example for Proposition 1.4
Counterexample: A total derivative need not vary continuously
Proof.
For one scalar component, telescope \(f(a+h)-f(a)\) by changing one coordinate at a time. The one-variable mean value theorem writes the \(i\)-th difference as \(h_i\partial_i f(\xi_i)\), with \(\xi_i\to a\). After subtracting \(\sum_i h_i\partial_i f(a)\), the absolute remainder is at most \(\sum_i|h_i|\max_i|\partial_i f(\xi_i)-\partial_i f(a)|=o(|h|)\), since \(\sum_i|h_i|\le\sqrt m|h|\). Apply this to each component. The recursive claim follows by applying the first-order result to the entries of \(Df\).
□Example. Polynomials are smooth: differentiating a monomial lowers an exponent and gives another polynomial, and sums and products of continuous functions are continuous. The function \(t|t|\) is \(C^1\) but not \(C^2\): its first derivative is \(2|t|\), which is continuous but not differentiable at zero. Existence of partial derivatives alone is insufficient. Define \(f(x,y)=xy/(x^2+y^2)\) off zero and \(f(0,0)=0\). Both partials at zero are zero, but \(f(t,t)=1/2\), so \(f\) is not even continuous there.
Proposition 1.5. The chain rule and algebra rules
If \(f\) is differentiable at \(a\) and \(g\) at \(f(a)\), then \(D(g\circ f)_a=Dg_{f(a)}Df_a\). Sums differentiate termwise; for scalar functions, \(D(fg)_a=f(a)Dg_a+g(a)Df_a\). Sums, products, and composable composites of \(C^k\) maps are \(C^k\).
Proof.
Write \(f(a+h)-f(a)=Ah+r(h)=k\), where \(r=o(|h|)\); then \(|k|=O(|h|)\), meaning \(|k|\le C|h|\) for some fixed \(C\) near zero. Substitution into \(g(f(a)+k)=g(f(a))+Bk+s(k)\) gives remainder \(Br(h)+s(k)=o(|h|)\). For products, multiply the two expansions: the cross terms beyond the displayed derivative are \(o(|h|)\). The sum rule follows by addition. Continuity proves the \(k=0\) closure statements; these derivative formulas and induction prove all higher ones. If a scalar \(f\) never vanishes, its reciprocal is \(C^k\) whenever \(f\) is: the one-variable reciprocal rule and the chain rule give \(D(1/f)=-f^{-2}Df\), and induction proves the higher orders. This justifies the rational functions used below.
□Example. If \(f(t)=(t,t^2)\) and \(g(x,y)=xy\), then \(Dg_{f(t)}Df_t=(t^2,t)(1,2t)^T=3t^2\), the derivative of \(t^3\).
Definition 1.6. Diffeomorphism
A \(C^k\) diffeomorphism, for \(k\ge1\), is a bijection between open sets for which both the map and its inverse are \(C^k\). Without a superscript, “diffeomorphism” means smooth.
Example. The map \(x\mapsto x^3\) is a smooth bijection \(\mathbb R\to\mathbb R\), but is not a diffeomorphism: the difference quotient of its inverse at zero is \(t^{-2/3}\), which has no finite limit. A translation \(x\mapsto x+b\) is a diffeomorphism with inverse \(y\mapsto y-b\).
1.2. Two calculus tools used in the proofs
The next facts let us prove the tangent-space basis theorem and define proper time without assuming integration in advance.
Integration, the fundamental theorem, and a smooth remainder formula
For continuous \(h:[a,b]\to\mathbb R\), a partition is a finite increasing list \(a=t_0<\cdots<t_N=b\). Its mesh is \(\max_i(t_i-t_{i-1})\). A tagged sum is \(\sum_i h(\xi_i)(t_i-t_{i-1})\), with \(\xi_i\in[t_{i-1},t_i]\). The integral \(\int_a^b h(t)\,dt\) is the limit of all tagged sums as the mesh tends to zero; reversing endpoints changes its sign. Vector integrals are componentwise.
A sequence \((z_j)\) is Cauchy if for every \(\varepsilon>0\), eventually \(|z_j-z_k|<\varepsilon\) for all pairs \(j,k\). Euclidean space is complete: every Cauchy sequence converges. Indeed a Cauchy sequence is bounded; repeated bisection of a box containing it produces a convergent subsequence, using completeness of the real numbers in each coordinate. For a Cauchy sequence with a subsequence tending to \(z\), the triangle inequality \(|z_j-z|\le|z_j-z_{k_\ell}|+|z_{k_\ell}-z|\) proves that the whole sequence tends to \(z\). See Cauchy sequences.
Existence. A continuous function on a closed bounded interval is uniformly continuous. One proof: if uniform continuity fails, choose \(x_j,y_j\) with \(|x_j-y_j|\to0\) but a fixed positive gap in function values. A convergent subsequence of \(x_j\), supplied by repeated bisection of the interval, forces both sequences to the same limit and contradicts continuity. On a sufficiently fine partition the oscillation on each piece is at most \(\varepsilon\), so upper and lower sums differ by at most \(\varepsilon(b-a)\). Taking common refinements makes all sufficiently fine tagged sums a Cauchy family, hence gives a unique limit in \(\mathbb R\). The same sums prove linearity, additivity in the interval, and \(|\int h|\le(b-a)\sup|h|\).
Fundamental theorem. If \(H(t)=\int_a^t h(s)\,ds\), then
\[\left|\frac{H(t+u)-H(t)}u-h(t)\right|\le\sup_{|s-t|\le|u|}|h(s)-h(t)|\longrightarrow0.\]Thus \(H'=h\). If \(F'=h\), the mean value theorem makes \(F-H\) constant, proving \(\int_a^b F'=F(b)-F(a)\). A substitution \(t=\phi(s)\) follows by applying the chain rule to a primitive: \(\int h(\phi(s))\phi'(s)\,ds\) has primitive \(H\circ\phi\). See the integral and primitives.
Smooth remainder formula. If \(f\) is smooth near \(a\), restrict to a ball about \(a\). The fundamental theorem along its straight segments gives
\[f(a+h)-f(a)=\sum_i h_i H_i(h),\qquad H_i(h)=\int_0^1\partial_i f(a+th)\,dt,\quad H_i(0)=\partial_i f(a).\]Each \(H_i\) is smooth. To justify differentiation under this integral, a difference quotient of the integrand is the average of its next partial derivative over a shrinking segment, by the fundamental theorem. On a compact smaller ball times \([0,1]\), that partial derivative is uniformly continuous, by the same subsequence argument in finitely many coordinates. The quotients therefore converge uniformly; the integral bound above permits passage to the limit. Repeat for every derivative.
Why mixed partials commute
For a scalar \(C^2\) function, take the rectangular increment
\[R(h,k)=f(a+he_i+ke_j)-f(a+he_i)-f(a+ke_j)+f(a).\]Applying the fundamental theorem in the two coordinate directions gives \(R(h,k)=\int_0^h\int_0^k\partial_j\partial_i f(a+se_i+te_j)\,dt\,ds\). Reversing the order in which we form the differences gives the same rectangle using \(\partial_i\partial_j f\). After division by \(hk\), continuity makes the two expressions tend respectively to these mixed partials at \(a\). They are equal. For example, both mixed partials of \(x^2y\) are \(2x\). This is Clairaut’s theorem.
Worked examples
Each example checks the hypotheses and the calculation for the linked statements. The links in the statements return here.
Worked example 1. The derivative with its remainder bound
Applies to: Definition 1.1, Proposition 1.2, Definition 1.3, Proposition 1.4.
On the open plane let \(f(x,y)=xy\). At \((a,b)\), \(f(a+h,b+k)-f(a,b)=bh+ak+hk\). Thus the proposed derivative is the linear map \((h,k)\mapsto bh+ak\), and \(|hk|/\sqrt{h^2+k^2}\le\sqrt{h^2+k^2}/2\to0\), checking total differentiability. Its values on the standard basis are \(b,a\), exactly the partials. Any other derivative differs by a linear map \(L\) with \(L(tv)/t\to0\); this is the constant \(L(v)\), so \(L=0\). The expansion also gives continuity. The Jacobian \((y,x)\) has continuous partials of every order (second derivatives are constants and higher ones zero), so \(f\) is \(C^k\) for every \(k\), verifying the recursive and partial-derivative descriptions of smoothness.
Worked example 2. A coordinate change and its inverse
Applies to: Proposition 1.5, Definition 1.6.
Let \(F(x,y)=(x,y+x^2)\) on \(\mathbb R^2\). Its inverse is \(G(u,v)=(u,v-u^2)\), so both are polynomial and smooth. Their Jacobians are \(DF=\begin{pmatrix}1&0\\2x&1\end{pmatrix}\), \(DG=\begin{pmatrix}1&0\\-2u&1\end{pmatrix}\). At \(F(x,y)\) their product is \(I\), as the chain rule for \(G\circ F\) requires. For the scalar product \(p(x,y)=xy\), the product rule gives \(Dp=y\,dx+x\,dy\); the sum \(p+x^2\) has derivative \((y+2x)dx+xdy\). These formulas explicitly check the algebra rules as well as a global diffeomorphism.
Worked example 3. The calculus tools on a polynomial
For \(h(t)=2t\) on \([0,1]\), right-endpoint uniform tagged sums equal \(N^{-2}\sum_{j=1}^N2j=1+1/N\to1\). Any tags change this by at most \(2/N\), since the oscillation on an interval is at most \(2/N\); hence the integral, independent of tags, is 1. Its primitive \(H(t)=t^2\) has \(H'=h\), checking the fundamental theorem. Substitution \(t=s^2\) gives \(\int_0^1 2s^2(2s)ds=1\). The sequence \(1+1/N\) is Cauchy because late differences are at most \(1/N+1/M\), and its limit is real.
For \(f(x,y)=x^2y\) at \((0,0)\), the smooth remainder formula has \(H_1(h,k)=\int_0^1 2t^2hk\,dt=2hk/3\), \(H_2(h,k)=\int_0^1t^2h^2\,dt=h^2/3\). Then \(hH_1+kH_2=h^2k=f(h,k)-f(0,0)\), and both mixed partials are \(2x\). Finally the derivative \(A(h,k)=2h+3k\) has operator norm \(\sqrt{13}\) by Cauchy–Schwarz, attained at \((2,3)/\sqrt{13}\). These give explicit instances of the unboxed integration, remainder, mixed-partial, and norm tools used later.
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. A total derivative need not vary continuously
Related statements: Definition 1.1, Proposition 1.4.
Claim refuted. Differentiable at every point means C¹; continuity of the first partials is necessary for differentiability.
Proof.
On \(\mathbb R\), set \(f(0)=0\) and \(f(x)=x^2\sin(1/x)\) for \(x\ne0\). The difference quotient at zero is \(x\sin(1/x)\), of absolute value at most \(|x|\), so \(f'(0)=0\). Off zero the ordinary rules give \(f'(x)=2x\sin(1/x)-\cos(1/x)\). Along \(x_n=(2\pi n)^{-1}\), these derivatives equal \(-1\); along \(y_n=((2n+1)\pi)^{-1}\), they equal 1. Both sequences tend to zero, so the derivative is discontinuous there. The function is differentiable everywhere but is not \(C^1\). The sufficient condition in Proposition 1.4 is stronger than differentiability, not a necessary condition at a point.
Comments