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An immersion of a smooth $d$-manifold in ${\mathbb{R}}^{d+1}$ supplies a rank-one normal bundle $\nu$ with $TM\oplus \nu \cong {\epsilon }^{d+1}$. The Whitney product formula for Stiefel–Whitney classes therefore requires $w(\nu )=w(TM{)}^{-1}$. A real line bundle has no positive-degree classes beyond $w_1$, so every higher-degree part of the inverse total class must vanish. For ${\mathbb{RP}}^2\times {\mathbb{RP}}^2$, the stabilized tangent bundle of real projective space gives $w(TM{)}^{-1}=(1+a)(1+b)$ in ${\mathbb{F}}_2[a,b]/(a^3,b^3)$. Its nonzero degree-two part $ab$ excludes an immersion in ${\mathbb{R}}^5$ even though the normal line need not be orientable.

An injective immersion from a compact smooth manifold into a Hausdorff smooth manifold is a smooth embedding. Indeed, it is a continuous bijection onto its image, and compactness makes its inverse continuous.

Imagine folding a sheet along its horizontal axis by the smooth map $F:{\mathbb{R}}^2\to {\mathbb{R}}^2$, $F(x,y)=(x,y^2)$. The inputs $(a,r)$ and $(a,-r)$ have the same output. Along $y=0$, horizontal motion still changes the output, but vertical motion produces no first-order change: $F(a,\epsilon )-F(a,0)=(0,{\epsilon }^2)$. These inputs are critical points of a smooth map even though their differential is not the zero linear map. The missing direction in the target tangent space is what matters.

Precisely, let $f:X^m\to Y^n$ be a smooth map between smooth manifolds without boundary. An input $p\in X$ is a critical point of a smooth map when its differentialis not surjective, equivalently when its rank of a linear map is less than $n$. An input where it is surjective is a regular point; a map with no critical points of a smooth map is a submersion. The threshold is the target dimension, not $\min (m,n)$. In manifold charts, compute the matrix rank of the $n\times m$ Jacobian matrix; changes of charts preserve that matrix rank. ^[ref]

For the fold,Thus its set of critical points of a smooth map is the entire horizontal axis. Away from that axis, its differential can produce every infinitesimal target displacement. At the axis, it can produce only horizontal ones. This diagnoses where the two inverse branches in the picture merge.

For a scalar smooth function $f:X\to \mathbb{R}$, the target dimension is one. A linear map to $\mathbb{R}$ fails to be surjective exactly when it is zero, so here a critical point of a smooth map means $d{f}_p=0$. In Euclidean space this is precisely the familiar critical point condition $\nabla f(p)=0$: every first partial derivative vanishes. On a Riemannian manifold, it is equivalently ${\nabla }_{g}f(p)=0$ for the chosen Riemannian metric; the Riemannian gradient depends on that metric, but whether it vanishes does not. See Loring W. Tu, An Introduction to Manifolds, Proposition 8.23 for the scalar coordinate criterion. The scalar smooth function $f(t)=t^3$ has a critical point at zero without a local maximum or local minimum.

A source of lower dimension gives a different warning. For the smooth map $G:\mathbb{R}\to {\mathbb{R}}^2$, $G(t)=(t,t^2)$, the Jacobian matrix is the column $(1,2t{)}^T$, of matrix rank one at every input. Every input is a critical point of a smooth map, since the target dimension is two, even though this parametric curve has nonzero velocity everywhere and is an immersion. More generally, if $m<n$, no differential $T_{p}X\to T_{f(p)}Y$ can be surjective, so every input is critical. The terminology measures the ability to fill target directions, rather than whether the parametrized curve looks smooth.

The map descends to an injective immersion from $\mathbb{R}\times (\mathbb{R}/\mathbb{Z})$ into the three-dimensional torus. Injectivity follows because an integer difference in $s$ with an integer difference in $\alpha s$ must be zero. Its image is dense by an irrational rotation of the circle, and is tangent to the commuting independent fields ${\partial }_x+\alpha {\partial }_z$ and ${\partial }_y$. It is a leaf of a regular foliation, but cannot be an embedded submanifold: in an embedding chart the coordinate plane would be locally closed yet contain a dense ambient subset.

A finite-dimensional commutative real algebra with a Euclidean multiplicative norm has dimension at most two. Automatic real bilinearity from a multiplicative norm supplies bilinearity if it was not part of the initial description. On its unit sphere, squaring has derivative $v\mapsto 2xv$, injective because there are no zero divisors. The only equal squares are antipodal pairs, so the map induces an injective immersion ${\mathbb{RP}}^{n-1}\to S^{n-1}$. Compactness and the inverse function theorem make it a diffeomorphism when $n\ge 2$. For $n\ge 3$, the fundamental groups $\mathbb{Z}/2$ and zero contradict this. The real and complex number algebras attain dimensions one and two.

The isothermal coordinates $(u,v)$ give the immersion $(u-u^3/3+u{v}^2,v-v^3/3+u^{2}v,u^2-v^2)$. Its coordinate functions are harmonic and its metric is $(1+u^2+v^2{)}^2(d{u}^2+d{v}^2)$, so it is a complete immersed minimal surface with self-intersections.

Choose finitely many coordinate charts on a compact manifold and smooth functions supported in them, positive on a cover. Extend the displayed chart products by zero. A positive first coordinate and its matching chart coordinates separate points, while their derivatives detect every nonzero tangent vector. Thus the map is an injective immersion; compactness makes it a smooth embedding. Extra zero coordinates make its codimension positive if required.

An immersion $F:M\to N$ into a Riemannian manifold $(N,h)$ inherits the Riemannian metric $g=F^{\ast }h$: measure two tangent vectors after mapping them into the surrounding tangent space. This is a pullback of a Riemannian metric, withThe injective derivative of an immersion makes this inner product positive-definite. For a parametrized surface in Euclidean space, its coefficients are simply the scalar products of the coordinate tangent vectors: the first fundamental form.

For the circular cylinder $F(\theta ,z)=(R\cos \theta ,R\sin \theta ,z)$, the tangent vectors $F_{\theta }$ and $F_z$ are orthogonal, with squared norms $R^2$ and $1$. ConsequentlyThe unrolled coordinate $x=R\theta$ gives $d{s}^2=d{x}^2+d{z}^2$. This calculation explains intrinsic flatness of a circular cylinder and makes Riemannian distance on a circular cylinder a planar problem with repeated endpoint lifts.

Preserving a Riemannian metric locally preserves the arc length of every mapped curve, but it need not preserve distances measured by unrestricted paths in the surrounding space. The great-circle distance between two sphere points, for example, exceeds their straight chord length unless they coincide. ^[ref].

For irrational $\alpha$, the mapis an injective immersion with dense image. It is not a smooth embedding, since points with arbitrarily large parameter return arbitrarily close to its initial point in the subspace topology.

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 Lagrangian immersion is an immersion $\iota :L^n\to (M^{2n},\omega )$ satisfying ${\iota }^{\ast }\omega =0$.

A Lie subgroup is a subgroup carrying a Lie group structure such that its inclusion is an injective immersion and a group homomorphism. Its Lie algebra includes in that of the ambient Lie group.

Put $p=\gamma (0)$, $v=\dot{\gamma }(0)$, $a=J(0)$ and $b=D_{t}J(0)$. Choose the short curve $c(s)={\exp }_p(sa)$ and let $P_s:T_{p}M\to T_{c(s)}M$ be parallel transport along $c$. The tangent vectorssatisfy $V(0)=v$ and ${\nabla }_{s}V(0)=b$. Define a geodesic variation byFor each fixed $s$ this is an affinely parametrized geodesic with initial point $c(s)$ and initial velocity $V(s)$. Although the Riemannian manifold need not be complete, smooth dependence for the geodesic equation on a neighborhood of the compact reference segment provides a single $\epsilon >0$ on which $F$ exists for all $|s|<\epsilon$ and $0\le t\le 1$. Thus completeness is unnecessary.

Set $X(t)={\partial }_{s}F(0,t)$. The torsion-free Levi-Civita connection gives ${\nabla }_s{\partial }_{t}F={\nabla }_t{\partial }_{s}F$. Differentiating ${\nabla }_t{\partial }_{t}F=0$ with respect to $s$, and commuting the two covariant derivatives, yieldsAlso $X(0)=c^{\prime }(0)=a$ and $D_{t}X(0)={\nabla }_{s}V(0)=b$. Uniqueness for the Jacobi field equation now gives $X(t)=J(t)$ on the whole segment. This is realization of Jacobi fields by geodesic variations.

The printed formula switches the order of its two variables. The consistent notation here is $F(0,t)=\gamma (t)$ and $J(t)={\partial }_{s}F(0,t)$. A parametrized surface here means a smooth variation map; it cannot be required to be an immersion, since the allowed Jacobi field $J=0$ would already violate that condition along the reference curve.

The image is a ring torus: a circle of radius $b$ is rotated about the vertical axis at distance $a$ from that axis. Put $r=\sqrt{x^2+y^2}$. Since $a>b$, every image point has $r>0$, and the image is preciselyConversely, on this level set the pairs $((r-a)/b,z/b)$ and $(x/r,y/r)$ are unit vectors, so choosing their angles gives a preimage.

The specified angular rectangle removes two closed circles from the torus. More explicitly its image isHere $C_u$ is the outer equatorial circle and $C_v$ is a meridian. The given openness property makes $U$ open in $T$. On $U$, the angles of $(r-a)+iz$ and $x+iy$ are uniquely chosen in $(0,2\pi )$; their continuous argument branches give the inverse $(x,y,z)\mapsto (u,v)$. Thus the restriction is a homeomorphism onto $U$, and the inverse is locally smooth.

For the immersion condition, the two derivatives have inner productsThey are linearly independent everywhere, since $a+b\cos u\ge a-b>0$. Hence this is a smooth embedded surface parametrization.

The angular cuts do not represent singularities of the embedded torus of revolution. To give a direct proof covering all points, define $F=(\sqrt{x^2+y^2}-a{)}^2+z^2-b^2$ on the open region $r>0$. On $T$,The implicit function theorem therefore describes $T$ near every point as a smooth graph of two coordinates. The whole torus is a smooth embedded surface.

An immersion $M^4\to {\mathbb{R}}^5$ would inject $TM$ into the trivial rank-five bundle. Taking its orthogonal complement gives a normal real line bundle $\nu$ with $TM\oplus \nu \cong {\epsilon }^5$. The Whitney product formula for Stiefel–Whitney classes requiresThe real line bundle need not be orientable, but it has no Stiefel–Whitney classes above degree one.

Let $a,b$ be the pulled-back degree-one generators from the two real projective planes. Their mod-two cohomology ring, by the Künneth theorem, is ${\mathbb{F}}_2[a,b]/(a^3,b^3)$. To compute the tangent classes, at a line $L$ the projective tangent space is $\mathrm{Hom}(L,L^{\perp })$. Adding $\mathrm{Hom}(L,L)\cong \mathbb{R}$ gives $\mathrm{Hom}(L,{\mathbb{R}}^3)\cong 3L$, with the real line identified with its dual using a metric. Thus the stabilized tangent bundle of real projective space gives $T{\mathbb{RP}}^2\oplus {\epsilon }^1\cong 3\gamma$ andThe tangent bundle of the product is the direct sum of the pulled-back tangent bundles. Since $(1+a+a^2)(1+a)=1$ in this truncated ring, and similarly for $b$,The degree-two class $ab$ is nonzero by the product-ring basis. It cannot be a class of the rank-one bundle $\nu$. Equivalently $w_1(TM{)}^2+w_2(TM)=ab\ne 0$, violating the codimension-one Stiefel–Whitney immersion obstruction. Hence ${\mathbb{RP}}^2\times {\mathbb{RP}}^2$ admits no immersion in ${\mathbb{R}}^5$.

Let $\zeta =u+iv$. Direct differentiation givesThese are holomorphic polynomials and their squared sum isMoreover direct calculation of the metric gives $E=G=(1+u^2+v^2{)}^2>0$ and $F=0$, so the map is genuinely an immersion everywhere. The preceding characterizations prove that it is an isothermal parametrization of a minimal surface, the Enneper surface.

The null condition follows by expansion:It vanishes exactly when $E=G$ and $F=0$; the immersion hypothesis ensures their common value is positive. Thus it characterizes isothermal coordinates for an immersion.

With complex coordinate $\zeta =u+iv$, mixed derivatives commute andand similarly for $y,z$. By the Cauchy-Riemann equations, each ${\phi }_j$ is holomorphic exactly when the corresponding coordinate is harmonic. Combining with the preceding part gives an isothermal immersion is minimal exactly when all three ${\phi }_j$ are holomorphic. Without immersion the null condition could also describe a degenerate map, so the nondegeneracy hypothesis matters.

An isothermal parametrization is an immersion with first fundamental form $Ed{u}^2+2Fdudv+Gd{v}^2={\lambda }^2(d{u}^2+d{v}^2)$, where $\lambda >0$. Thus $E=G={\lambda }^2$ and $F=0$.

Write $W={\varphi }_{uu}+{\varphi }_{vv}$. Differentiating the metric identities givesFor example ${\varphi }_{vv}\cdot {\varphi }_u=({\varphi }_v\cdot {\varphi }_u{)}_v-{\varphi }_v\cdot {\varphi }_{uv}=-G_u/2$; the other identities are analogous. Hence $W$ is normal to the surface. For a unit normal $N$, its normal component is $e+g$. The supplied mean curvature formula reduces to $H=(e+g)/(2{\lambda }^2)$. ThereforeA surface is a minimal surface when its mean curvature vector vanishes. Since ${\lambda }^2>0$, the preceding identity proves minimality is equivalent to $\Delta \varphi =0$ in isothermal coordinates.

Use the metric on $M$ induced by its given Euclidean embedding. For two smooth maps define the C1 metric on a smooth mapping space bywhere the differential has domain $T_{x}M$ with the induced inner product. Compactness makes both suprema finite. The norm properties give symmetry and the triangle inequality, and the first term makes distance zero imply equality of maps. Thus convergence in $d_1$ is uniform convergence of both maps and differentials.

Because $i$ is an immersion, $D{i}_x$ is injective at every point. The continuous function $|D{i}_{x}v|$ on the compact unit tangent bundle has a positive minimum $a$. ThereforeFor all sufficiently large $n$, ${\sup }_x\Vert D{f}_{n,x}-D{i}_x\Vert <a/2$, and thenConsequently every sufficiently late $f_n$ is an immersion. If $M$ has dimension zero it is a finite set, the immersion assertion is automatic, and eventual injectivity follows directly from separation of its finitely many image points; we henceforth treat positive dimension.

For injectivity, choose finitely many coordinate maps ${\varphi }_b$ on convex balls, with smaller coordinate neighborhoods covering $M$ and closures lying inside the larger balls. Write $I_b=i\circ {\varphi }_b$ and $F_{n,b}=f_n\circ {\varphi }_b$. On the finitely many relevant compact coordinate closures, the differentials of the $I_b$ have a common positive lower bound $s$, and convergence in $d_1$ givesfor all late $n$. The assumed uniform second-derivative bound gives a common bound $L$ for $D^2{F}_{n,b}$ on those larger coordinate neighborhoods. If the bound is expressed using an intrinsic Hessian, the coordinate chain rule also involves first derivatives and derivatives of the fixed charts; their bounds are uniform by C1 convergence and compactness. Thus $L$ can be larger than the printed constant $C$, but is independent of $n$.

For $u,v$ whose segment lies in one such coordinate ball, Taylor formula with integral remainder givesChoose $\delta >0$ with $L\delta \le s/2$. Then, whenever $0<|u-v|<\delta$,The Lebesgue number lemma applied to the finite smaller-chart cover, together with uniform continuity of its coordinate inverses on the relevant compact sets, gives $\rho >0$ such that every pair $x,y\in M$ with $0<|x-y|<\rho$ lies in one such chart with coordinate distance below $\delta$. The displayed estimate therefore excludes all sufficiently close pairs of equal images, uniformly in late $n$.

The remaining pairs form the compact setIf it is nonempty, injectivity of the smooth embedding $i$ gives a positive minimum $b$ of $|i(x)-i(y)|$ on it. Uniform convergence makes ${\sup }_x|f_n(x)-i(x)|<b/3$ for all sufficiently large $n$, and hence on $K_{\rho }$,If $K_{\rho }$ is empty there is nothing left to check. Combining the near-pair and distant-pair bounds makes $f_n$ injective for every sufficiently late $n$. It is already an immersion, and a continuous injection from compact $M$ into a Hausdorff space is a homeomorphism onto its image. By compact injective immersion is an embedding,This proves the requested stability of compact smooth embeddings using the stated second-derivative assumption. That assumption supplies a convenient Taylor constant; the stronger C1 openness result can replace it by the uniform continuity of the limiting differential on finitely many small coordinate balls.

An immersion $\varphi :U\to {\mathbb{R}}^3$ has differential of rank $2$, equivalently linearly independent ${\varphi }_u,{\varphi }_v$. It has isothermal coordinates when its first fundamental form has $E=G>0$, $F=0$. A minimal surface has zero mean curvature.

For an isothermal immersion, differentiation of $E=G$ and $F=0$ givesand similarly its scalar product with ${\varphi }_v$ is zero. Its normal component is $e+g$. The given mean curvature formula reduces to $H=(e+g)/(2E)$, soThus an isothermal immersion is minimal exactly when each coordinate is a harmonic function.

An explicit example is the Enneper surface, parametrized on ${\mathbb{R}}^2$ byEvery coordinate has zero Laplacian. Direct differentiation gives ${\varphi }_u\cdot {\varphi }_v=0$ and $|{\varphi }_u{|}^2=|{\varphi }_v{|}^2=(1+u^2+v^2{)}^2>0$. Thus it is an isothermal immersion and is minimal everywhere.

It is not an open subset of any of the three listed surfaces. The parameter points $(\sqrt{3},0)$ and $(-\sqrt{3},0)$ both map to $(0,0,3)$, but their tangent planes differ: ${\varphi }_u=(-2,0,\pm 2\sqrt{3})$ while ${\varphi }_v=(0,4,0)$. A plane, catenoid, or helicoid is an embedded smooth surface with a unique tangent plane at each point. If our whole image lay in any one of them, both rank-$2$ tangent planes at this point would have to equal its tangent plane, a contradiction.