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An isometric immersion is an immersion $f:M\to N$ between Riemannian manifolds whose derivative preserves the inner products of tangent vectors; equivalently, $f^{\ast }{g}_N=g_M$. Its derivative is injective and the target may have larger dimension. A local isometry additionally requires a local diffeomorphism, so its tangent maps are linear isomorphisms. For example, a sphere with its induced Riemannian metric is isometrically immersed in Euclidean space, although its positive intrinsic sectional curvature differs from that of the ambient space.

A local isometry between Riemannian manifolds is a local diffeomorphism $f$ with $f^{\ast }{g}_N=g_M$. Equivalently, its differentials are isometric linear isomorphisms between tangent spaces. It preserves the Levi-Civita connection, intrinsic sectional curvature, and affinely parametrized geodesics. A map whose differentials are merely isometric injections is an isometric immersion; the local diffeomorphism requirement distinguishes these notions.

The displayed estimate has two independent defects. Even for $a(v)=v$, its right side cannot use $f(t)$. At $t=1$, choose nonnegative smooth functions $\psi ,\chi$ with compact support and prescribe $f(1,x,v)=\psi (x)\chi (v)$. This is a legitimate transported solution with $f_{\mathrm{in}}(x,v)=\psi (x+v)\chi (v)$. The proposed inequality would requireVelocity dilation makes $\Vert \chi {\Vert }_1/\Vert \chi {\Vert }_{\infty }$ arbitrarily large, while the other factors stay fixed. Thus no universal $C_a$ works with the same-time mixed Lebesgue norm.

There is a second issue after replacing $f(t)$ by $f_{\mathrm{in}}$: the Jacobian determinant bounds give a local diffeomorphism, not necessarily a one-to-one map. Suppose additionally that $a$ is injective. With $y=x-ta(v)$, the change of variables formula and $|\det Da|\ge {\alpha }_1$ giveSurjectivity is not needed because the domain of integration can be enlarged. With injectivity, the corrected estimate has $C_a=1/{\alpha }_1$. The upper bound ${\alpha }_2$ is unnecessary.

More generally, dispersion with a nonlinear velocity map uses the area formula to sum over inverse branches. Define the weighted inverse multiplicityIf ${\mathrm{esssup}}_w{m}_a(w)<\infty$, the same argument gives the corrected estimate with $C_a=\Vert m_a{\Vert }_{\infty }$. At most $N$ inverse branches give $C_a\le N/{\alpha }_1$. Thus the missing global assumption concerns multiplicity, not merely local volume distortion.

For a noninjective map with constant Jacobian determinant giving a counterexample in two dimensions, write $v=(s,r)$ and setThe polar coordinates calculation gives $\det Da=e^{2s}{e}^{-2s}=1$, yet $a(0,2\pi k)=(1,0)$ for every integer $k$. To turn this into a failure of the estimate, take a small open disk $W$ about $(1,0)$ that avoids the origin. For each of $N$ inverse branches over $W$, choose a smooth cutoff ${\chi }_k(v)=\chi (a(v))$ supported on that branch, where $\chi$ has support strictly inside $W$ and is extended by zero. These velocity supports are disjoint. SetThen $\Vert f_{\mathrm{in}}{\Vert }_{L_x^1{L}_v^{\infty }}=\Vert \psi {\Vert }_1\Vert \chi {\Vert }_{\infty }$, independent of $N$, whereas at a fixed $t>0$ and suitable $x$,The last integral is positive and independent of $N$. No finite $C_a$ exists for this fixed $a$, although ${\alpha }_1={\alpha }_2=1$. In one dimension, by contrast, a nonvanishing derivative has a constant sign, so the lower derivative bound makes $a$ a global diffeomorphism.

The degree and its local signs. Orient $S^n$ and let $[S^n]$ generate its top reduced homology ${\tilde{H}}_n(S^n;\mathbb{Z})\cong \mathbb{Z}$. The mapping degree is the integer characterized byFor $n\ge 1$ this is the usual top homology definition using the fundamental class; using reduced homology also covers $S^0$.

Suppose $f$ is smooth and $p$ is a regular value. Each $x\in f^{-1}(p)$ has invertible tangent map $D{f}_x$, so the inverse function theorem makes $f^{-1}(p)$ discrete. It is also closed in the compact sphere, hence finite. Choose disjoint small neighborhoods of these points on which $f$ is a local diffeomorphism. The Excision theorem identifies the source local homology with the direct sum of one copy of $\mathbb{Z}$ for each inverse image. The induced local map is multiplication by $+1$ or $-1$ according as $D{f}_x$ preserves or reverses orientation. The map from the global fundamental class to these local orientation classes therefore proves the degree as a sum of local degrees formula:Here the determinant is computed in positively oriented tangent bases. Thus the mapping degree counts inverse images with signs, rather than just their cardinality.

The quotient map. Put $m=n+1$ and write $f=F{|}_{S^{m-1}}$. Give $D^m$ its standard orientation and its boundary the induced orientation. The connecting homomorphismis an isomorphism: the disk has zero positive reduced homology. It sends the relative fundamental class $[D^m,S^{m-1}]$ to $[S^{m-1}]$. If the map on relative homology induced by $F$ multiplies this class by $d$, naturality givesThus $d=\mathrm{deg}f$. Collapsing the boundary gives a sphere $D^m/S^{m-1}\cong S^m$, and the quotient map identifies its top reduced homology with the top relative homology of the disk pair. Give this quotient sphere the orientation determined by that identification. The relation $qF=\tilde{F}q$ now showsThis is the quotient-sphere degree identity.

The graph intersection. Orient $D^m\times D^m$ by the product orientation, orient $N=D^m\times \{0\}$ by its first factor, and orient the graph of a function $\Gamma$ by $x\mapsto (x,F(x))$. An intersection is precisely a zero of $F$, and no such zero lies on the boundary because $F(S^{m-1})\subset S^{m-1}$. At an intersection, transverse intersection means that $D{F}_x$ is surjective, hence invertible. The zeros are consequently isolated and finite.

For the smooth intersection number use the ordered tangent spaces $T_{x}N$ first and $T_{(x,F(x))}\Gamma$ second. Relative to the product basis their concatenated basis has matrixIts determinant is $\det D{F}_x$, so the intersection sign is ${ϵ}_x=\mathrm{sgn}\det D{F}_x$. By the same local Excision theorem argument, now in relative homology at the interior point $0$, the sum of these signs equals the multiplier of $F_{\ast }$ on $H_m(D^m,S^{m-1};\mathbb{Z})$. Therefore the graph intersection formula for mapping degree isIf the tangent spaces are ordered $\Gamma$ first and $N$ second, every sign changes by $(-1{)}^m$; the order above specifies the appropriate convention.

The one-dimensional model $F(x)=2x^3-x$ on $[-1,1]$ illustrates the graph intersection formula for mapping degree: its three zeros have signs $+1,-1,+1$, and its endpoint map has mapping degree $1$ on reduced homology.

First work at the level of differential forms. The fiber description implies $\pi \circ f=\pi$, so every pullback ${\pi }^{\ast }\beta$ is $f$-invariant. Since $\pi$ is a surjective local diffeomorphism, its pullback on forms is injective: at any $y\in N$, choose a lift $x$ and use the inverse of $d{\pi }_x$ to evaluate the form on arbitrary tangent vectors at $y$.

Conversely let $\alpha$ be an $f$-invariant form on $M$. For a local inverse $s:V\to M$ of $\pi$, define $\beta {|}_V=s^{\ast }\alpha$. If two inverses select the same lift they agree locally; if they select different lifts, the fiber condition identifies the second with $f$ composed with the first. Locally these alternatives are stable, using disjoint inverse neighborhoods at distinct lifts. The equality $f^{\ast }\alpha =\alpha$ therefore makes the local definitions agree. They glue to a unique smooth form $\beta$ on $N$ with ${\pi }^{\ast }\beta =\alpha$. Thusis an isomorphism of complexes, with the exterior derivative on both sides.

It remains to compare the cohomology of the invariant complex with invariant de Rham cohomology classes. Put $P=(I+f^{\ast })/2$ on forms. If $f^{\ast }[\alpha ]=[\alpha ]$ and $d\alpha =0$, then $P\alpha$ is invariant, closed and represents the same class, because its class is the average of two equal classes. If an invariant form is exact, $\alpha =d\eta$, then $\alpha =d(P\eta )$ with an invariant primitive. This proves both surjectivity and injectivity on cohomology. For $p=0$ there are no negative-degree primitives, and the same conclusion follows directly from closed functions.

Combining the two steps givesThis is the de Rham cohomology of a finite quotient argument specialized to an involution. It does not require orientation. The hypotheses do not explicitly exclude $f=\mathrm{id}$; then all fibers are singletons, $\pi$ is a bijective local diffeomorphism and the conclusion is the ordinary diffeomorphism invariance. If $f$ has one fixed point, local injectivity of $\pi$ makes its fixed set open as well as closed; connectedness of $M$ forces this identity case. Otherwise the fibers have two points and $\pi$ is the usual double covering.

For $v\in T_{p}S$ sufficiently small, let ${\gamma }_v$ be the geodesic with ${\gamma }_v(0)=p$, ${\dot{\gamma }}_v(0)=v$, and define ${\exp }_p(v)={\gamma }_v(1)$. The domain consists more generally of initial velocities whose geodesic exists through time one. Uniqueness and affine rescaling give ${\exp }_p(tv)={\gamma }_v(t)$ near zero, soThus the differential is the identity under the natural identification of tangent spaces, and the inverse function theorem makes the exponential map a local diffeomorphism near zero.

A punctured plane has geodesics reaching its missing point in finite time, so its exponential map need not be global; adding the point remedies this example. But an extension cannot always remedy the failure. On the punctured circular cone $S=\{(r\cos \theta ,r\sin \theta ,r):r>0\}$, radial generators are geodesics and reach the missing point in finite intrinsic length. Any globally defined extension would have to include their limiting point $(0,0,0)$. The cone's tangent planes have different limits as $\theta$ varies; no smooth embedded surface through that point can contain this punctured circular cone. Hence no such extension exists for this $S$.

For an oriented surface and a closed topological disc $D$ with smooth positively oriented boundary, the Gauss-Bonnet theorem with boundary saysHere $k_g$ is signed geodesic curvature with the disc on the left. A smooth boundary has no corner terms; piecewise smooth boundaries require exterior-angle terms.

For the flat disc in the problem this gives ${\int }_{\partial D}{k}_{g}ds=2\pi$. Conditions (i) and (ii) mean that the proposed replacement disc has the same boundary and agrees with the original smooth surface from the outside. Smoothness therefore gives the same tangent planes, metric derivatives and boundary geodesic curvature on the replacement, with consistent orientation. Applying Gauss-Bonnet to that disc forcesThus these gluing conditions fix the total curvature of any smooth replacement disc, regardless of its detailed interior shape.

The nonnegative continuous Gaussian curvature on the compact replacement disc, together with its zero integral established by Gauss-Bonnet, forces it to vanish at every point of that disc. Outside it the replacement surface agrees with the original flat surface, so its curvature also vanishes there; smoothness covers the common boundary. Consequently no point of the replacement surface can have positive Gaussian curvature. Therefore no surface can satisfy all three conditions. A disc with some positive curvature could only have compensating negative curvature or alter the smooth boundary gluing, both excluded here.

An abstract smooth surface is a Hausdorff second-countable topological space with an atlas of charts to open subsets of ${\mathbb{R}}^2$ whose transition maps are smooth. It is orientable when it has such an atlas for which every transition map has positive Jacobian determinant. A map $f:S_1\to S_2$ is a smooth map when every coordinate representationis smooth wherever it is defined.

The map $a:C\to C$ is a smooth involution with no fixed point: the equationwould force $x=y=0$, which is impossible on $x^2+y^2=1$. For each $p\in C$, choose a sufficiently small coordinate neighbourhood $U_p$ such thatThen the quotient projection restricts to a homeomorphismTransporting a smooth chart ${\varphi }_p:U_p\to {\mathbb{R}}^2$ across this homeomorphism gives the quotient chartOn an overlap, a lift lies either in another chosen neighbourhood or in its image under $a$. The corresponding transition map is therefore a transition map on $C$, possibly composed with the diffeomorphism $a$, and is smooth. Since this is a free action of the finite group $\{1,a\}$, the quotient is Hausdorff and second countable. This constructs the smooth quotient by a free finite group action, and in these charts $\pi$ is locally the identity. Hence $\pi$ is a local diffeomorphism, in particular smooth.

To test orientability, parametrize the cylinder byThe involution acts in these coordinates aswhose derivative has determinant $-1$. Thus $a$ is an orientation-reversing diffeomorphism of the cylinder.

If the quotient surface $S$ were orientable, its orientation would pull back through the local diffeomorphism $\pi$ to an orientation of $C$. The identity $\pi \circ a=\pi$ would then force $a$ to preserve that pulled-back orientation, contradicting the negative determinant above. Therefore

For a smooth local diffeomorphism $a:{\mathbb{R}}^d\to {\mathbb{R}}^d$, each inverse branch contributes an inverse Jacobian determinant in the area formula. Their sum is the weighted inverse multiplicity. A lower determinant bound controls each branch separately, not the number of branches. Bounded weighted inverse multiplicity is sufficient for dispersion with a nonlinear velocity map.