Connections Levi Civita and Parallel Transport — Examples

Prescribe $\omega =0$ in the constant frame. By [F1] the resulting derivative is precisely $du$ componentwise. For a scalar $f$, $X(f{u}^a)=X(f)u^a+fX(u^a)$ proves the section Leibniz identity, and $(fX)(u^a)=fX(u^a)$ proves direction-linearity. Since the components of $e_a$ are constant, every ${\mathrm{\nabla }}_X{e}_a$ is zero.

Applying the formula twice to a section gives the components of ${\mathrm{\nabla }}_X{\mathrm{\nabla }}_{Y}s-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_{X}s-{\mathrm{\nabla }}_{[X,Y]}s$ as $X(Y{u}^a)-Y(X{u}^a)-[X,Y]u^a=0$, by the defining commutator of vector fields. This is the stated direct meaning of flatness. For rank zero the formula is the unique zero operator; rank one gives $\mathrm{\nabla }u=du$. An empty base or a zero-dimensional base causes no exception. The product frame is given, so no choice of trivialization for an arbitrary bundle is involved.

Apply [F1] with the one-by-one matrix $a$. The formula is real-linear, and $d(fu)+a(fu)=dfu+f(du+au)$ explicitly verifies the one-form Leibniz identity. Applying it to $u=1$ gives $\mathrm{\nabla }e=ae$, hence exactly the asserted coefficient.

For $a=xdy$, evaluate on ${\mathrm{\partial }}_x,{\mathrm{\partial }}_y$ to obtain the displayed two derivatives. For example $u=y$ gives ${\mathrm{\nabla }}_{{\mathrm{\partial }}_y}(ye)=(1+xy)e$, which equals $2e$ at $(1,1)$. At $x=0$ this coefficient form vanishes but the derivative of $y$ still contributes $e$. With $a=0$ the connection is $d$; with $u=0$ it is zero. These formulas apply to an empty or zero-dimensional base using the unique empty or zero one-form, and require no choices.

Since $e^f>0$, $e^{\mathrm{\prime }}$ is a frame everywhere. Scalar coefficients commute, and $d(e^f)=e^{f}df$, so [F1] gives ${\omega }^{\mathrm{\prime }}=\omega +df$. This changes coordinates of the same derivative, not the intrinsic connection.

On the trivial line over $\mathbb{R}$ with $\omega =0$ and $f=x$, the new form is $dx$. In particular ${\mathrm{\nabla }}_{{\mathrm{\partial }}_x}{e}^{\mathrm{\prime }}=e^{\mathrm{\prime }}$ whereas ${\mathrm{\nabla }}_{{\mathrm{\partial }}_x}e=0$. The old constant section has new coefficient $e^{-x}$, and its new covariant derivative is $d(e^{-x})+dx{e}^{-x}=0$, confirming agreement on an actual section. Constant $f$ gives $df=0$; the frame never vanishes even when $f=0$.

The identification sends $(q,(f(q),v))$ to $(q,v)$, with smooth inverse $(q,v)\mapsto (q,(f(q),v))$. It takes the pullback frame to the constant frame. By [F1] and [F2] the new matrix is $f^{\ast }0=0$, hence ${\mathrm{\nabla }}_X^f({\sum }_a{u}^a{e}_a)={\sum }_{a}X(u^a)e_a$ for arbitrary smooth functions on $N$.

In particular, even if $f:\mathbb{R}\to M$ is constant, the pullback section $s(t)=t{e}_1$ has ${\mathrm{\nabla }}_{{\mathrm{\partial }}_t}^{f}s=e_1$ when $r\ge 1$. It need not be a section pulled back from $M$. Constant coefficients, in contrast, have zero derivative. Rank zero gives the unique zero operator and an empty source gives the empty bundle; no immersion, injectivity or nonzero differential is required.

Put $C(t)={\int }_a^{t}c(u)du$ and $v(t)=e^{-C(t)}{v}_0$. The ordinary fundamental theorem of calculus on each continuity piece and chain rule give $v^{\mathrm{\prime }}=-cv$. The function $C$ is continuous across the finite subdivision, so $v$ matches at every corner. Also $C(a)=0$, whence $v(a)=v_0$. By [F1] and [F2] this is the parallel solution.

Evaluating at $b$ proves the formula. If $c=2$ and $[a,b]=[0,1]$, the multiplier is $e^{-2}$. Zero initial value remains zero, $c=0$ gives the identity, and $a=b$ gives the empty integral and identity. The exponential multiplier is always positive and nonzero. For the reversed curve $h(u)=a+b-u$, its coefficient is $-c(h(u))$; substitution changes the integral's sign, giving the reciprocal multiplier. These finite scalar integrations require no selection of solution branches.

Every ${\mathrm{\partial }}_k{g}_{ij}$ is zero, so [F1] gives ${\mathrm{\Gamma }}^k{}_{ij}=0$ for every index. Expanding $Y={\sum }_j{Y}^j{\mathrm{\partial }}_j$ in the connection Leibniz law gives the claimed derivative. For example in two dimensions, $X=x{\mathrm{\partial }}_x+y{\mathrm{\partial }}_y$ and $Y=x^2{\mathrm{\partial }}_x+xy{\mathrm{\partial }}_y$ give ${\mathrm{\nabla }}_{X}Y=2x^2{\mathrm{\partial }}_x+2xy{\mathrm{\partial }}_y$.

By [F2] the parallel equation is $v^{\mathrm{\prime }}=0$. Each component is constant on each smooth segment, and continuity identifies its constants across the finitely many corners. Thus transport sends ${\sum }_j{v}^j{\mathrm{\partial }}_j{\mathrm{\mid }}_p$ to ${\sum }_j{v}^j{\mathrm{\partial }}_j{\mathrm{\mid }}_q$, including constant curves and singleton intervals. Zero components stay zero. For $n=0$ this is the unique map of zero tangent spaces, and $n=1$ is the ordinary scalar derivative.

Differentiating the coordinate map gives vectors $(\cos \theta ,\sin \theta )$ and $(-r\sin \theta ,r\cos \theta )$, whose inner products are $1,0,r^2$. Thus $G=\mathrm{diag}(1,r^2)$, $G^{-1}=\mathrm{diag}(1,r^{-2})$, and the only nonzero metric derivative is ${\mathrm{\partial }}_r{g}_{\theta \theta }=2r$. Formula [F1] gives ${\mathrm{\Gamma }}^r{}_{\theta \theta }=-(2r)\mathrm{/}2=-r$ and ${\mathrm{\Gamma }}^{\theta }{}_{r\theta }={\mathrm{\Gamma }}^{\theta }{}_{\theta r}=(2r)\mathrm{/}(2r^2)=1\mathrm{/}r$.

The remaining entries are ${\mathrm{\Gamma }}^r{}_{rr}={\mathrm{\Gamma }}^r{}_{r\theta }={\mathrm{\Gamma }}^r{}_{\theta r}={\mathrm{\Gamma }}^{\theta }{}_{rr}={\mathrm{\Gamma }}^{\theta }{}_{\theta \theta }=0$. For the first and fourth, every metric derivative is zero. In the middle two the only potentially nonzero term ${\mathrm{\partial }}_r{g}_{\theta \theta }$ is multiplied by $g^{r\theta }=0$; in the last, the term $-{\mathrm{\partial }}_r{g}_{\theta \theta }$ is multiplied by $g^{\theta r}=0$. At $r=1$ the three displayed nonzero entries are $-1,1,1$. None of these formulas applies at $r=0$: there the angular coordinate vector vanishes and the coordinate map is not a chart. There is therefore no singularity of the Euclidean metric asserted at the origin.

The metric and inverse matrices are $g_{ij}=e^{2u}{\delta }_{ij}$ and $g^{ij}=e^{-2u}{\delta }^{ij}$. Since ${\mathrm{\partial }}_i{g}_{j\mathrm{\ell }}=2e^{2u}{u}_i{\delta }_{j\mathrm{\ell }}$, substitution in [F1] cancels the factors $2$, $e^{2u}$ and $e^{-2u}$ and gives ${\sum }_{\mathrm{\ell }}{\delta }^{k\mathrm{\ell }}(u_i{\delta }_{j\mathrm{\ell }}+u_j{\delta }_{i\mathrm{\ell }}-u_{\mathrm{\ell }}{\delta }_{ij})$, exactly the asserted formula.

For $u=x^2$, one has $u_x=2x,u_y=0$. The formula gives the four listed entries and ${\mathrm{\Gamma }}^x{}_{xy}={\mathrm{\Gamma }}^x{}_{yx}={\mathrm{\Gamma }}^y{}_{xx}={\mathrm{\Gamma }}^y{}_{yy}=0$. At $x=0$ all entries vanish, whereas at $x=1$ the four listed entries are $2,-2,2,2$. Constant $u$ gives zero coefficients everywhere. The conformal factor is strictly positive for every real $u$, so this calculation never inverts a degenerate metric.

For completeness the sphere's local geometry is supplied here. On each of its six open coordinate hemispheres, projection onto the other two coordinates has inverse obtained by inserting the chosen-sign function $\sqrt{1-u^2-v^2}$ on the open unit disk. These smooth inverse graphs cover the sphere and their overlapping coordinates are restrictions of smooth projections and graph maps. Their differentials identify tangent vectors with the plane $p^{\perp }$: differentiation of $⟨p,p⟩=1$ gives inclusion, and the graph differential is injective from a two-dimensional domain into the two-dimensional plane. The Euclidean product restricted to that plane is positive definite, and in each graph its coefficients are dot products of the two smooth differential columns. It therefore defines the stated smooth round metric by [F1].

Differentiating $⟨Y,p⟩=0$ gives $⟨D_{X}Y,p⟩=-⟨Y,X⟩$, so $D_{X}Y+⟨X,Y⟩p$ is tangent by step 1.1. This formula is smooth, real-linear, function-linear in $X$ and obeys ${\mathrm{\nabla }}_X(fY)=X(f)Y+f{\mathrm{\nabla }}_{X}Y$ by the component product rule; hence it is an affine connection. The normal correction is orthogonal to every tangent $Z$, and differentiation of the Euclidean product gives $X⟨Y,Z⟩=⟨{\mathrm{\nabla }}_{X}Y,Z⟩+⟨Y,{\mathrm{\nabla }}_{X}Z⟩$. Finally, if $p^a$ are the three restricted ambient coordinate functions, then $Y^a=Y(p^a)$ and $(D_{X}Y-D_{Y}X{)}^a=X(Y(p^a))-Y(X(p^a))=[X,Y](p^a)$ by [F4]. The symmetric normal terms cancel, proving torsion zero. Thus [F3] identifies this connection with Levi–Civita.

For an arbitrary tangent field $V$ along a curve, the formula in step 2.1 gives $D_t^{\mathrm{\nabla }}V=V^{\mathrm{\prime }}+⟨\dot{\gamma },V⟩\gamma$: in any local tangent frame expand $V={\sum }_a{v}^a{e}_a(\gamma )$ and apply [F5] and the ordinary product rule to each ambient component. This derives the formula for arbitrary along-curve fields, without requiring an ambient extension of $V$. On the equator, $T^{\mathrm{\prime }}=-\gamma$ and $⟨\dot{\gamma },T⟩=1$, so $D_t^{\mathrm{\nabla }}T=0$. Also $N^{\mathrm{\prime }}=0$ and $⟨\dot{\gamma },N⟩=0$, so $D_t^{\mathrm{\nabla }}N=0$.

The vectors $T,N$ form an orthonormal tangent basis at each time. For initial vector $aT(0)+bN(0)$, the field $aT(t)+bN(t)$ is parallel, and uniqueness in [F5] makes it the transported field. At $t=2\pi$ both basis vectors equal their initial values, proving identity transport, including the zero vector. The same formula gives identity on a singleton interval and handles any number of whole equatorial turns. All data are explicit and no global tangent frame on the sphere is assumed.

On $\mathbb{R}$ with $g=d{x}^2$, prescribe ${\mathrm{\Gamma }}^1{}_{11}=1$, equivalently the matrix $dx$ in tangent frame ${\mathrm{\partial }}_x$. This gives a smooth affine connection by [F1], with ${\mathrm{\nabla }}_{f{\mathrm{\partial }}_x}(u{\mathrm{\partial }}_x)=f(u^{\mathrm{\prime }}+u){\mathrm{\partial }}_x$. Its sole lower-index pair is symmetric, so [F2] gives torsion zero.

Set $X=Y=Z={\mathrm{\partial }}_x$. The left side in [F3] is ${\mathrm{\partial }}_{x}1=0$, while its right side is $1+1=2$, so compatibility with $d{x}^2$ fails. This does not assert failure for every metric: with $\tilde{g}=e^{2x}d{x}^2$, the one-dimensional compatibility equation is ${\mathrm{\partial }}_x{e}^{2x}=2{\mathrm{\Gamma }}^1{}_{11}{e}^{2x}$, which holds. For arbitrary local multiples the additional derivatives of their coefficients match by the scalar product rule, so this last test indeed gives compatibility with $\tilde{g}$.

For two distinct standard basis vectors $a,b$, the quarter circle $\gamma (t)=\cos ta+\sin tb$, $0\le t\le \pi \mathrm{/}2$, has unit tangent $T(t)=-\sin ta+\cos tb$. Formula [F1] gives $D_t^{\mathrm{\nabla }}T=-\gamma +\gamma =0$. A constant vector perpendicular to $a,b$ has zero derivative and zero inner product with $\dot{\gamma }$, so it too is parallel. This determines transport on a tangent basis by [F3].

Traverse the three arcs $e_1\to e_2$, $e_2\to e_3$, $e_3\to e_1$ in that order. The initial tangent vector $e_2$ is $T(0)$ on the first arc and therefore becomes $-e_1$ at $e_2$. On the second arc $-e_1$ is a constant perpendicular vector, so remains $-e_1$ at $e_3$. On the third arc $T(0)=e_1$ and $T(\pi \mathrm{/}2)=-e_3$; hence the transported input $-e_1$ becomes $e_3$ at $e_1$. Composition in [F2] thus sends $e_2$ to $e_3$ around this closed loop.

The constant loop at $e_1$ sends $e_2$ to $e_2$ by [F2], which differs from $e_3$. Both are unit tangent vectors at $e_1$ and the loops have identical endpoints, proving the failure. The nonzero input detects the difference; zero is fixed by both. The three arcs join continuously and each is smooth up to its endpoints, so they are admissible even at the corners. No curvature or area formula is used.

Insert [F2] into [F1] to obtain the general Cartesian formulas. For the given $f$, the first derivatives are ${\mathrm{\partial }}_{x}f=2xy$ and ${\mathrm{\partial }}_{y}f=x^2$. Differentiating again gives ${\mathrm{\partial }}_x^{2}f=2y$, ${\mathrm{\partial }}_x{\mathrm{\partial }}_{y}f=2x$, ${\mathrm{\partial }}_y{\mathrm{\partial }}_{x}f=2x$, ${\mathrm{\partial }}_y^{2}f=0$, exactly the displayed symmetric matrix.

The coordinate derivatives of the vector components contributing to divergence are ${\mathrm{\partial }}_x(x^2)=2x$ and ${\mathrm{\partial }}_y(xy)=x$, whose sum is $3x$. At $(0,0)$ both the Hessian and divergence vanish; at $(1,1)$ they are respectively the matrix with rows $(2,2)$ and $(2,0)$, and the scalar $3$. Constant $f$ gives zero Hessian and zero $X$ gives zero divergence. In dimension zero the general formulas use empty sums, while dimension one gives $f^{\mathrm{\prime }\mathrm{\prime }}$ and $(X^1{)}^{\mathrm{\prime }}$.