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At a current approximation $p$, Taylor's theorem replaces $\Phi (p+h)=0$ by the affine equation $\Phi (p)+D\Phi (p)h=0$. Solving this equation gives the Newton iteration in a Banach spaceThe inverse is a bounded linear inverse by the bounded inverse theorem. The displayed operator is defined where the derivative is invertible, in particular on the specified open domain; the hypotheses do not supply a globally defined Newton operator on all of $\mathcal{X}$. If $\Phi (\bar{p})=0$ with $\bar{p}$ in that domain, then $N(\bar{p})=\bar{p}$.

For quadratic convergence, choose a sufficiently small neighbourhood of $\bar{p}$. Continuity of the inverse derivative and of $D^2\Phi$ gives constants $M,L$ there such thatPut $e=p-\bar{p}$ and $B(p)=[D\Phi (p){]}^{-1}$. The quadratic Newton error bound in a Banach space follows from the integral remainder:soTake a closed radius-$r$ ball lying in the neighbourhood with $Cr<1$. This inequality maps the ball into itself and makes errors tend to zero. More explicitly, for $e_n=\Vert p_n-\bar{p}\Vert$, induction gives $C{e}_{n+j}\le (C{e}_n{)}^{2^j}$. Hence every sufficiently close initial point converges to $\bar{p}$ at least quadratically; special maps can converge still faster.

For a fixed-point equation, apply the same method to $F(p)=\Phi (p)-p$. Wherever $D\Phi (p)-I$ is invertible, the exact adapted map isEvery fixed point of $\Phi$ is fixed by $A$, and conversely on this domain. This additional invertibility condition is distinct from invertibility of $D\Phi$; for example, $D\Phi =I$ is invertible but $D\Phi -I=0$ is not.

For certification, a frozen Newton correction for a fixed-point equation is often more useful. Choose a bounded, injective approximate inverse $J$ of $D\Phi (p_0)-I$ and setAgain $A_J(p)=p$ if and only if $\Phi (p)=p$. A well-chosen $J$ makes the corrected operator contract even if the original operator has an expanding direction. The subsequent contraction mapping argument requires the stated derivative and residual bounds; it does not assume that arbitrary inverse approximations provide them.

The Lanford contraction proof of the Feigenbaum fixed point replaces an unstable renormalization operator by a contracting Newton correction. Use a coefficient Banach space for normalized even maps, represented by $p(x)=1-x^{2}h(x^2)$ with an absolutely summable coefficient norm on $h$. Its complex domain and a small ball are chosen so that every map is admissible and the renormalization composition is analytic there.

A high-accuracy polynomial $p_0$ gives an approximate fixed point. Use a bounded invertible approximate inverse $J$ for $D\mathcal{T}(p_0)-I$, and defineRigorous estimates certifyThe finite coefficient computations use interval arithmetic with controlled rounding; analytic estimates bound the infinite tail. Thus the certification applies to the full function space, rather than merely a truncated polynomial system.

Part (b) gives a unique fixed point of $A$ in the certified ball. Invertibility of $J$ makes it a unique fixed point $g$ of $\mathcal{T}$ there. The ball also preserves the quadratic critical maximum and nontrivial shape, excluding the constant formal normalized solution. Consequently an actual locally unique analytic Feigenbaum fixed point exists, with the error bound $\Vert g-p_0\Vert \le \epsilon /(1-\kappa )$. The numerical approximation locates it; rigorous residual, derivative and tail bounds establish its existence and uniqueness. This conclusion is local, not a claim of global uniqueness among all analytic solutions.

Write $N_{i,d}$ for the partition-of-unity-normalized degree-$d$ B-spline, supported on $[t_i,t_{i+d+1}]$. Assume $d\ge 1$. First take distinct knots; endpoint repetitions will be interpreted as confluent limits. For fixed parameter $t$, let $f_t(\tau )=(\tau -t{)}_{+}^d$, and formwhere the brackets denote a divided difference in the knot variable $\tau$. This representation follows from the defining properties, rather than assuming a derivative recurrence. If $t<t_i$, $f_t$ is a degree-$d$ polynomial in all its knot arguments and its order-$(d+1)$ divided difference is zero. If $t>t_{i+d+1}$, every truncated-power term is zero. Between knots it is a degree-$d$ polynomial with $d-1$ continuous derivatives. Those support and smoothness conditions determine a one-dimensional spline space: there are $d+1$ spans, the continuity conditions leave $2d+1$ coefficients, and vanishing through order $d-1$ at both support ends imposes $2d$ conditions.

The remaining scalar is fixed by the prescribed integral. Put $a=t_i$, $b=t_{i+d+1}$. Integrating the truncated power first givesThe order-$(d+1)$ divided difference of this polynomial is its leading coefficient $1/(d+1)$, so the proposed $N_{i,d}$ has integral $(b-a)/(d+1)$, exactly the specified normalization.

Now differentiate with respect to $t$ and apply the final recursion for a divided difference:In terms of normalized lower-degree B-splines, this is the B-spline differentiation formulaA zero denominator corresponds to a collapsed-support basis term and contributes zero. Repeated knots follow by coalescing distinct knots; at knots where the classical derivative is discontinuous, read the formula on each open span or one-sided. The stated maximal interior continuity corresponds to simple interior knots, while the clamped endpoint repetitions do not affect the interior argument.

Let the control-point abscissae be the Greville abscissaeCollect the coefficient of $N_{i,d-1}$ after differentiating:The difference of the two consecutive knot averages telescopes toso every noncollapsed coefficient is one. The lower-degree basis on the endpoint-trimmed knot vector has partition of unity; equivalently the outer lower-degree basis functions on the original vector have collapsed support. Therefore $X^{\prime }(t)=1$ on the parameter interval. With $d+1$ repeated left endpoint knots at $a$, the clamped basis has $N_{0,d}(a)=1$, all other values zero, and ${\xi }_0=a$. Thus $X(a)=a$ fixes the constant of integration. Continuity and endpoint limits now giveThis is linear precision at Greville abscissae. It requires neither equal knot spacing nor a restriction on the other control-point coordinates; those determine the remaining coordinates of the parametric curve.

The basic parametric surface interrogation enquiries provide the parameter domain and patch adjacency; evaluation of $S(u,v)$; first derivatives $S_u,S_v$; and, for curvature or error-controlled stepping, second derivatives $S_{uu},S_{uv},S_{vv}$. Useful search enquiries also supply conservative bounding volumes over parameter rectangles, subdivision or restriction of a patch, and regularity/singularity information. A local inverse or closest-point enquiry can help seed a search but does not replace global component isolation.

At a regular point, $S_u,S_v$ are independent tangent vectors, and the oriented unit normal vector isThe tangent plane consists of $S+\alpha S_u+\beta S_v$. If the cross product is zero, this chart does not define a normal; use a regular alternative chart or treat a true singularity separately.

For two surfaces, collect the four parameters in $w=(u_1,v_1,u_2,v_2)$ and setAt a transverse intersection, $n_1\times n_2\ne 0$ and $J$ has rank three. Hence the implicit function theorem makes $F=0$ locally a curve. Choose its physical unit tangent vectorWith $G_i=A_i^T{A}_i$, the parameter derivatives for physical arc length areSince $\tau$ lies in both tangent planes, $A_i(u_i^{\prime },v_i^{\prime }{)}^T=\tau$ exactly. Thus these two pairs form a vector $d$ in the kernel of $J$.

A practical transversal intersection of two parametric surfaces algorithm is:

with Newton method. The final equation intersects the curve with the plane through $X_p$ perpendicular to the predicted tangent. Its Jacobian row is $[{\tau }^T{A}_1,0,0]$; together with $J$ it is nonsingular at a sufficiently nearby transverse point. This removes the otherwise free along-curve parameter in the three coincidence equations.

A finite tolerance supplies a polygonal or fitted-curve approximation; regularity and certified bounds determine its accuracy. Parallel normals, singular charts, branch contacts and coincident patches require separate handling: their intersection may be a singular curve, an isolated point or a two-dimensional region. The transverse marching equations must not be used to claim a unique curve in those cases.

The cubic spline is even, so the symmetry of the normalized B-splines gives $N_{j,4}(-x)=N_{6-j,4}(x)$. It therefore suffices to extract three coefficients. The two polynomial pieces join with matching values and first two derivatives at zero, so the function belongs to the stated cubic spline space. At zero its right-hand derivatives areUse the supplied De Boor–Fix spline coefficient functional with the unnormalized knot polynomial ${\psi }_j(x)={\prod }_{r=j+1}^{j+3}(t_r-x)$. In this sign convention the coefficient isFor $j=1$, ${\psi }_1=(-1-x{)}^3$. Taking the right-hand limit at $-1$ kills the first three terms and gives $c_1=f(-1)=1$. For the next two coefficients take the right-hand limit at zero. Their knot polynomials and derivatives there areConsequently $c_2=(-27+6)/6=-7/2$ and $c_3=(27+6)/6=11/2$. Although the third derivative jumps at zero, its multiplier ${\psi }_j(0)$ vanishes for these two functionals; taking either one-sided limit gives the same coefficients. Symmetry gives $c_4=c_2$ and $c_5=c_1$. ThusFor a normalized B-spline basis, define the synthesis linear map $Tc={\sum }_j{c}_j{N}_{j,4}$ from the maximum coefficient norm into the spline space with its supremum norm. Nonnegativity and partition of unity give $\Vert T\Vert =1$, including equality on the all-ones vector. The coefficient condition number of a normalized B-spline basis isOn $[0,1]$, $f^{\prime }(x)=(27/2)x(3x-2)$. The only interior extremum is at $2/3$, where $f=-1$; at zero and one, $f=1$. Reflection therefore gives $\Vert f{\Vert }_{\infty }=1$. Its largest absolute coefficient is $11/2$, proving

Fix a clamped nondecreasing spline knot sequence $t_1,\dots ,t_{n+k}$ on $[a,b]$, with $t_1=\cdots =t_k=a$ and $t_{n+1}=\cdots =t_{n+k}=b$. For continuous splines assume interior multiplicities at most $k-1$, and assume $t_i<t_{i+k}$. An order-$k$ spline has degree at most $k-1$ on each nonempty knot interval; at a knot of multiplicity $r$ it has $k-1-r$ continuous derivatives. Its dimension is $n$, and a normalized B-spline basis $N_1,\dots ,N_n$ has local support, nonnegative values and sum one on $[a,b]$.

These properties are visible in the Cox-de Boor recurrence. Starting with interval indicators at order one, setwith a term of zero denominator defined to be zero. On each contributing support the weights are nonnegative, and summing the recurrence makes the coefficients of each lower-order B-spline add to one. This gives partition of unity and $\mathrm{supp}{N}_{i,k}=[t_i,t_{i+k}]$, with endpoint values defined by limits.

At distinct ordered sites $x_1<\cdots <x_n$, interpolation is the linear systemIt has a unique solution for every data vector exactly when the B-spline collocation matrix is invertible. The Schoenberg–Whitney theorem characterizes this by $N_i(x_i)>0$ for every $i$; away from clamped endpoints this is $t_i<x_i<t_{i+k}$. At the clamped endpoints the first and last B-splines have value one, so the endpoint version uses this positive-value formulation rather than the strict inequalities. Both ordering and the support conditions matter. A spline space with interior knots is generally not a Chebyshev system on the entire interval: a locally supported B-spline vanishes on an interval. Thus arbitrary distinct nodes need not permit interpolation. Merely specifying all data sites as knots also need not fix the remaining freedom; for example, cubic interpolation with simple interior knots at $m-2$ data sites has dimension $m+2$, and two extra boundary conditions are needed to interpolate $m$ values uniquely.

The positivity governing this existence theorem is total nonnegativity of B-spline collocation matrices: every square minor with increasing row and column indices is nonnegative. In spline terminology this is often called total positivity, but it permits zero minors caused by disjoint supports. It is not the stronger assertion that every minor is strictly positive. A useful proof uses knot insertion. Each refinement step changes coefficients bywith the unchanged endpoint pieces. The refinement matrix is nonnegative bidiagonal, and its minors are nonnegative: any nonzero determinant has its allowed diagonal matching and is a product of nonnegative entries. Products preserve this property by the Cauchy–Binet formula. Insert each sampling site to full multiplicity, so evaluation at that site becomes selection of an ordered refined coefficient. The collocation matrix is then an ordered row submatrix of the refinement product, proving total nonnegativity. The strict support criterion is the strict part of this argument: an ordered path through the consecutive support intervals exists exactly when every diagonal B-spline value is positive, giving the Schoenberg–Whitney theorem determinant criterion.

If $A$ is invertible, its determinant is positive and the adjugate identity yieldsThis checkerboard inverse of a totally nonnegative matrix is the central stability fact. With alternating data $e_i=(-1{)}^i$, the absolute row sums satisfyFor general data, the cardinal splines are ${\ell }_i(x)={\sum }_j{N}_j(x)(A^{-1}{)}_{ji}$. The spline interpolation operator and its exact Lebesgue constant of interpolation areThe equality for the operator norm follows by extending the signs at distinct sites to a continuous unit-norm data function, as in Solution 2. That solution also shows the last inequality can be very strict. Stability concerns the range function as well as its coefficients.

An optimal interpolation set must be specified relative to an objective. For the normalized B-spline coefficients, every admissible sampling set satisfiessince $c(s)=A_{\mathbf{x}}^{-1}(s(x_i))$ and sampling cannot increase the supremum norm. If a unit-norm spline $s_{\ast }$ has $n$ ordered alternating extrema $s_{\ast }(x_i)=(-1{)}^i$ at admissible sites, the checkerboard identity givesTaking the largest row shows $\Vert A_{\mathbf{x}}^{-1}{\Vert }_{\infty }\le \kappa (\mathcal{S})$, so equality holds. This proves the optimal spline coefficient interpolation sites property and the Chebyshev spline coefficient and dual norm equality. In particular, for the cubic space of Solution 1 the alternating extrema $-1,-2/3,0,2/3,1$ satisfy the support conditions. Its coefficients have maximum magnitude $11/2$, so these sites are optimal for coefficient recovery andThe argument proves optimality whenever this admissible equioscillating spline is available; it does not identify extrema of an arbitrary spline as optimal sites, or transfer coefficient optimality automatically to the Lebesgue constant of interpolation.

For the intrinsic interpolation objective, minimize ${\Lambda }_{\mathbf{x}}$ itself. There is a minimizer for a fixed finite-dimensional continuous spline space. Indeed the sites lie in a compact ordered simplex. On admissible sets, inverse entries and hence ${\Lambda }_{\mathbf{x}}$ vary continuously. As a collocation matrix tends to a singular one, its inverse norm diverges. The uniform-norm stability of a B-spline basis gives the complementary boundchoose sample signs attaining a largest inverse row, extend them to a continuous unit-norm function, and use $\Vert Tc{\Vert }_{\infty }\ge \Vert c{\Vert }_{\infty }/\kappa (\mathcal{S})$. Thus a minimizing sequence with bounded Lebesgue constants of interpolation cannot approach a singular site set. A convergent subsequence has an admissible limit attaining the minimum.

Fekete interpolation sites for a continuous function space offer another useful, constructive criterion: maximize the absolute evaluation determinant. A nonzero maximum exists by compactness and linear independence. Replacing one row by evaluation at $x$ multiplies the determinant by the corresponding cardinal function, so maximality gives $|{\ell }_i(x)|\le 1$, and therefore ${\Lambda }_{\mathbf{x}}\le n$. These sites are determinant-optimal, with a proven interpolation bound; one should not claim without further argument that they minimize ${\Lambda }_{\mathbf{x}}$. Greville abscissae provide inexpensive admissible sites for the usual continuous clamped spline spaces of order at least two. Admissibility alone does not give the best stability bound.

Finally, the coefficient condition number of a normalized B-spline basis is bounded by a constant depending only on order. To see the mechanism, for each coefficient choose a longest nonempty knot cell in that B-spline's support. If its length is $h$ and the support width is $W$, then $W/h\le k$. On that cell the spline is a degree-at-most-$k-1$ polynomial; repeated Markov polynomial inequalities bound its $r$th derivative by $C_k{h}^{-r}\Vert s{\Vert }_{\infty }$. In the De Boor–Fix spline coefficient functional, the accompanying derivative of the knot polynomial is bounded by $C_k{W}^r$. Each term is consequently at most $C_k(W/h{)}^r\Vert s{\Vert }_{\infty }$, yielding $\Vert c(s){\Vert }_{\infty }\le C_k\Vert s{\Vert }_{\infty }$ independently of the number and spacing of knots. This is basis stability; it does not bound $\Vert A_{\mathbf{x}}^{-1}\Vert$ at arbitrarily poor sampling sites.

For approximation, reproduction gives the Lebesgue inequality for approximation projectorsAs the maximum knot gap tends to zero with fixed order, spline quasi-interpolation gives an approximant with error at most $\omega (f,kh)$, so the infimum tends to zero. A sequence of uniformly bounded interpolation linear operators therefore converges to every continuous target. Local support and a stable choice of sites make spline interpolation effective; mesh refinement without control of the interpolation operator norm is not by itself a convergence proof.

Use Minkowski spacetime with signature $(+,-,\dots ,-)$ and $\hslash =1$, with a vacuum time-ordering prescription. At a regulator, define the normalized generating functionalThe normalized vacuum correlation functions of a time-ordered product areThe connected generating functional isThe logarithm selects connected source diagrams, or equivalently cumulants. With the assumed vanishing one-point function, derivatives of $Z=e^{iW}$ give the centered four-point cumulant decomposition:In detail, four differentiations either act on one connected block or form one of three pair partitions. Contributions with blocks of size one vanish. If $W_{ij}$ and $W_{1234}$ denote source derivatives at zero, the same relation is $G_{1234}=i{W}_{1234}-W_{12}{W}_{34}-W_{13}{W}_{24}-W_{14}{W}_{23}$, since $C_2=-i{W}_2$ and $C_4=i{W}_4$.

Let $\phi (x)=\delta W/\delta J(x)$ be the source-dependent mean field. On a locally invertible source-to-field branch, define the quantum effective action by the Minkowski Legendre transformIts variation is $\delta \Gamma =-\int J\delta \phi$, so $\delta \Gamma /\delta \phi =-J$. Differentiating the two inverse source-field maps yields the Minkowski inverse-Hessian relation for an effective action:Thus ${\Gamma }^{(2)}=-(W^{(2)}{)}^{-1}$. The inverse is understood at the regulator and on a nonsingular fluctuation sector; it is not an assertion that every source-field map is globally invertible.

For a free real scalar field, integrate by parts to writewhere repeated spacetime variables are integrated. Specify the propagator convention explicitly:Completing the Gaussian path integral around $\varphi =-A_{ϵ}^{-1}J$ cancels the source-independent determinant between numerator and denominator. ThereforeThe positive imaginary term in $A_{ϵ}$ damps the oscillatory Gaussian. Together with vacuum projection at the time boundaries, it selects the Feynman poles and the time-ordered, in-out correlation function rather than a retarded inverse. It can also be obtained by continuation from the Euclidean vacuum integral. A denominator convention without the numerator $i$ redistributes the displayed factors of $i$.

Finally $\phi =-A_{ϵ}^{-1}J$, hence $J=-A_{ϵ}\phi$. Substitution into the Legendre transform gives the vacuum-normalized effective action of a free scalar field:All connected functions beyond order two vanish and all proper vertices beyond the classical quadratic kernel vanish. Vacuum normalization fixes the additive constant; without it, the quantum effective action equals the classical action up to a field-independent determinant contribution.

Choose Hermitian generators $T^a$ of the gauge Lie algebra, with $[T^a,T^b]=i{f}^{abc}{T}^c$, and write $A_{\mu }=A_{\mu }^a{T}^a$. For the usual compact gauge algebra choose $\mathrm{tr}(T^a{T}^b)={\delta }^{ab}/2$. More generally use a nondegenerate invariant symmetric bilinear form on the algebra in place of this trace contraction; a completely arbitrary component metric would not give gauge invariance.

A matter vector is useful to fix conventions: with ${\psi }^{\prime }=U(x)\psi$ and $D_{\mu }={\partial }_{\mu }-ig{A}_{\mu }$, require $D_{\mu }^{\prime }{\psi }^{\prime }=U{D}_{\mu }\psi$. Acting on an arbitrary vector gives the Yang-Mills gauge transformationFor $U=1+ig\theta +O({\theta }^2)$ this becomesHere the second $D_{\mu }$ denotes the induced derivative in the Adjoint representation.

Define the curvature by $[D_{\mu },D_{\nu }]=-ig{F}_{\mu \nu }$. Expanding the commutator derivesSince commutators of the transformed derivatives conjugate, $F_{\mu \nu }^{\prime }=U{F}_{\mu \nu }{U}^{-1}$. Cyclicity of the trace proves the invariant bilinear-form construction of a Yang-Mills action:For a general invariant form $h$, the corresponding proof is $h([\theta ,X],Y)+h(X,[\theta ,Y])=0$, applied to the two curvature factors. No matter field is required in this pure-gauge Lagrangian.

Introduce the adjoint Faddeev-Popov ghost $c^a$ and antighost ${\bar{c}}^a$, independent Grassmann-odd scalar fields of ghost numbers $+1$ and $-1$, together with the Grassmann-even Nakanishi-Lautrup field $b^a$ of ghost number zero. The bar labels the independent antighost, not an additional propagating complex-conjugate matter field. Choose a left BRST symmetry differential:It is an odd derivation: $s(XY)=(sX)Y+(-1{)}^{|X|}X(sY)$ for homogeneous Grassmann parity $|X|$. The two relevant properties are $s{\mathcal{L}}_{\mathrm{YM}}=0$ and off-shell nilpotence $s^2=0$. The first is gauge invariance with parameter replaced by the ghost, and the second follows from the Lie algebra's graded Jacobi identities. ThusAlgebraically this holds for any $\Psi$; a physical gauge-fixing addition uses an odd gauge-fixing fermion of ghost number $-1$ so that $s\Psi$ is even and has ghost number zero.

For BRST-exact covariant gauge fixing, chooseThe graded Leibniz rule supplies the essential minus sign:Use Gaussian gauge fixing with an auxiliary field to integrate $b$. Pointwise,Translation of the regulated oscillatory Gaussian leaves only a field-independent normalization. The remaining Lagrangian is thereforeThis is covariant Feynman gauge, with gauge parameter one. Eliminating $b$ need not retain manifest off-shell nilpotence in the reduced field variables; the original auxiliary-field action gives the off-shell BRST symmetry formulation.

Finally, the ungauge-fixed quadratic momentum kernel isGauge directions are null vectors, so there is no inverse defining a vector propagator, and the functional integral also counts an infinite gauge orbit volume. The added gauge-fixing term cancels the longitudinal part, leaving $Q_{\mu \nu }^{ab}=-{\delta }^{ab}{p}^2{\eta }_{\mu \nu }$. With the vacuum prescription, the propagators areThe ghost action represents the Faddeev-Popov determinant and supplies ghost-vector vertices; closed ghost loops have the fermionic minus sign. Expanding the cubic and quartic Yang-Mills terms and this ghost interaction now defines consistent perturbative Feynman rules. Physical mass-shell poles remain, whereas the gauge degeneracy has been removed. Possible global gauge-fixing obstructions and boundary zero modes are separate from the perturbative inverse around the trivial vacuum.

An affine connection on the tangent bundle of a smooth manifold assigns a vector field ${\nabla }_{X}Y$ to two vector fields, is $\mathbb{R}$-bilinear, and satisfiesThus its first slot is tensorial and its second obeys the Leibniz rule. Its action on covectors is defined by differentiating the pairing, and this extends it to tensor fields. A torsion-free connection has ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]$.

Assume that $g$ is a smooth nondegenerate metric tensor. Metric compatibility and a vanishing torsion tensor imply the Koszul formula,The right-hand side is determined by $g$ and the Lie bracket of vector fields. Nondegeneracy determines ${\nabla }_{X}Y$ uniquely; conversely this formula defines an affine connection with both required properties, proving existence as well as uniqueness. In a coordinate frame the brackets vanish. Combining the three differentiated metric tensor identities givesThis is the Levi-Civita connection. Positive definiteness is unnecessary: the argument also works for a pseudo-Riemannian metric.

For the other affine connection define $S(X,Y)={\bar{\nabla }}_{X}Y-{\nabla }_{X}Y$. The derivative of $f$ cancels between the two Leibniz rules, so $S$ is $C^{\infty }$-linear in both slots. The difference of affine connections is a tensor: here $S$ is a section of $TM\otimes T^{\ast }M\otimes T^{\ast }M$ with components ${\bar{\Gamma }}^a{}_{bc}-{\Gamma }^a{}_{bc}$. Both affine connections are torsion-free, so $S(X,Y)=S(Y,X)$.

Agreement of geodesics means agreement of their unparametrized curves; different affine parameters are allowed. For every nonzero tangent vector $v$, existence of a geodesic with that initial velocity implies that $S(v,v)$ is parallel to $v$. The symmetric bilinear diagonal-parallel lemma now determines $S$. Choose a basis $e_i$ and write $S(e_i,e_i)={\alpha }_i{e}_i$. Applying the parallelism condition to $e_i+e_j$ and $e_i-e_j$ givesWith $V(e_i)={\alpha }_i/2$, bilinearity yieldsThis is projective equivalence of affine connections. It also covers dimension one. For two Levi-Civita connections, the projective covector from metric volume densities gives the explicit answerIndeed ${\Gamma }^a{}_{ac}={\partial }_c\log \sqrt{|\det g|}$. The determinant ratio is a scalar, so this expression is a genuine covector, although either determinant alone is coordinate-dependent. This determinant expression requires nondegeneracy of $\bar{g}$; the difference-tensor and earlier trace formula require only the two torsion-free affine connections.

Conversely, the displayed tensor gives ${\bar{\nabla }}_{\dot{\gamma }}\dot{\gamma }=2V(\dot{\gamma })\dot{\gamma }$ along an affinely parametrized $\nabla$-geodesic. Choosing a new parameter $t(\tau )$ satisfying $t^{\prime \prime }/t^{\prime }=2V(\dot{\gamma })$ removes that tangential acceleration. This geodesic reparametrization under projective equivalence proves sufficiency. If agreement were required with the very same affine parameter, polarization would instead force $S=0$ and $V=0$.

Use $c=1$, background Minkowski spacetime metric tensor ${\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$, and $g_{ab}={\eta }_{ab}+h_{ab}$ with small metric perturbation. All indices and derivatives in this linear approximation use $\eta$. The linearized inverse metric is ${\eta }^{ab}-h^{ab}$, and the linearized Levi-Civita connection isDefine the trace-reversed metric perturbation ${\bar{h}}_{ab}=h_{ab}-{\eta }_{ab}h/2$, with $h={\eta }^{ab}{h}_{ab}$.

An infinitesimal coordinate change $x^{\prime a}=x^a+{\xi }^a$ gives the linearized coordinate gauge transformationThe metric perturbation therefore has redundant components. The gauge invariance of the linearized Riemann tensor follows because its change contains third derivatives of $\xi$ cancelling by commutation of partial derivatives. Conversely, flat linearized metric perturbations are locally pure gauge on a contractible patch. Thus linearized curvature, rather than a particular coordinate value of $h$, detects the physical disturbance. Gauge freedom can also be restricted by global topology or boundary conditions.

Keeping the paper's curvature sign convention, its linearized Ricci tensor and scalar areThese are the negatives of the alternative frequently used curvature convention. Put $v_b={\partial }^a{\bar{h}}_{ab}$. The Linearized Einstein equations in vacuum areUnder a gauge change $v_b\mapsto v_b-\square {\xi }_b$, so solving $\square {\xi }_b=v_b$ imposes the Lorenz gauge in linearized gravity. The field equations then become $\square {\bar{h}}_{ab}=0$. The remaining residual gauge symmetry of linearized gravity has $\square {\xi }_b=0$; Lorenz gauge does not exhaust the coordinate freedom.