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Choose a real vector subspace $\mathfrak{m}$ complementary to $\mathfrak{h}$, so $\mathfrak{g}=\mathfrak{m}\oplus \mathfrak{h}$. The smooth maphas differential $(U,V)\mapsto U+V$ at $(0,0)$, an isomorphism. The inverse function theorem makes $F$ a diffeomorphism from a sufficiently small product neighborhood onto an open neighborhood of the identity in $G$.

There is a neighborhood of zero in $\mathfrak{m}$ in which $\exp U\in H$ implies $U=0$. Otherwise choose nonzero $U_j\in \mathfrak{m}$ tending to zero with $\exp U_j\in H$. A normalized subsequence tends to a unit vector $U\in \mathfrak{m}$. The limit directions of a closed subgroup from part (a) give $U\in \mathfrak{h}$, contradicting $\mathfrak{m}\cap \mathfrak{h}=0$. Shrink the product neighborhood so that this exclusion holds throughout its first factor.

If $F(U,V)\in H$, then $\exp V\in H$ by the definition of $\mathfrak{h}$, and hence $\exp U=F(U,V)\exp (-V)\in H$. The exclusion forces $U=0$. Conversely, every $F(0,V)=\exp V$ belongs to $H$. Thus the local product coordinates for a closed subgroup identify $H$ near the identity exactly with the smooth coordinate slice $U=0$.

Left translation by each element of $H$ gives the same description around that element, so $H$ is an embedded submanifold of $G$ in its subspace topology. The multiplication and inversion maps of $G$ restrict to smooth maps on this embedded submanifold, making $H$ a Lie group and an embedded Lie subgroup. Its tangent space at the identity is $\mathfrak{h}$, because $d{\exp }_0$ is the identity on this coordinate slice. It remains closed by hypothesis. This proves the closed-subgroup theorem, with no prior smoothness assumption on $H$.

The local flow of $X$ is the smooth map $\Phi :D\to M$, where $D\subset \mathbb{R}\times M$ is the maximal open domain of initial-point/time pairs, whose curve $t\mapsto \Phi (t,q)={\varphi }_t(q)$ is the maximal integral curve of a vector field through $q$, with ${\varphi }_0(q)=q$. Smooth dependence in the ordinary differential equation theorem gives joint smoothness and openness of $D$. Each time slice is a diffeomorphism between its domain and image, with inverse ${\varphi }_{-t}$.

Here is the group law, including its domain qualification. Fix $s,q$ where the relevant trajectories exist. Both curvessolve the same autonomous ordinary differential equation and have value ${\varphi }_s(q)$ at $t=0$. The uniqueness from the first part therefore giveson their common interval. These are local identities; neither the vector field nor its local flow has been assumed complete.

For the commuting local flows ${\varphi }_t$ and ${\psi }_s$, differentiate ${\varphi }_t({\psi }_s(q))={\psi }_s({\varphi }_t(q))$ with respect to $s$ at zero. The chain rule givesThus the pushforward of a vector field $Y$ by the $X$ flow equals $Y$. Interchanging the two fields gives the analogous identity for $X$ under ${\psi }_s$.

Now let ${\mathcal{D}}_q=\mathrm{span}(X_q,Y_q)$. Pointwise linear independence makes this a rank-two smooth distribution. The identity just proved implies $[X,Y]=0$: in coordinates, differentiating $D_q{\varphi }_t{Y}_q=Y_{{\varphi }_t(q)}$ at $t=0$ gives $DX(q)Y_q=DY(q)X_q$, precisely the vanishing Lie bracket of vector fields. The bracket of two local sections is also in $\mathcal{D}$, sinceand the analogous terms for $[fX,gX]$ and $[fY,gY]$ remain in the same span. Therefore $\mathcal{D}$ is an involutive distribution.

The Frobenius theorem says that a smooth constant-rank involutive distribution has local coordinates $(u^1,\dots ,u^n)$ in which it is spanned by ${\partial }_{u^1},{\partial }_{u^2}$, and has unique maximal connected integral manifolds through its points. These are the leaves of a regular foliation. Let $S$ be the maximal leaf through $p$. The global confinement to an immersed leaf gives the answer: $S$ is a two-dimensional immersed integral manifold and every smooth $\mathcal{D}$-tangent curve starting at $p$ stays in $S$.

For the curve assertion, in each Frobenius chart the transverse coordinate functions $u^3,\dots ,u^n$ have zero derivative along such a curve. Hence its segment in that chart stays in one plaque. On any compact subinterval of the curve's parameter interval, a finite chain of overlapping such segments connects the initial plaque to the last; all belong to the same maximal leaf. Exhausting the parameter interval proves the assertion for every time.

The two commuting local flows also give the leaf parametrization directly near $p$:Its partial derivatives are ${\partial }_{s}F=X_F$ and ${\partial }_{t}F=Y_F$, by the pushforward identity. Its differential has rank two. The constant rank theorem makes a sufficiently small image an embedded local plaque with tangent plane $\mathcal{D}$. This is a concrete local construction of the integral manifold.

There is a global distinction in the word “submanifold”. If it is required to mean an embedded submanifold, the assertion for all times is not valid in general. On the three-dimensional torus ${\mathbb{R}}^3/{\mathbb{Z}}^3$, takeThe fields are independent and commute, and their dense immersed cylinder in a three-dimensional torus through zero isIt is an injectively immersed cylinder $\mathbb{R}\times (\mathbb{R}/\mathbb{Z})$ and is dense in the torus. Indeed, fixing $x\mathrm{mod}1$ and varying $s$ by integers makes $\alpha s\mathrm{mod}1$ dense by an irrational rotation of the circle; $t$ is unrestricted. Every point of this leaf is reached by a tangent curve $r\mapsto (rs,rt,\alpha rs)\mathrm{mod}{\mathbb{Z}}^3$. An embedded surface containing it would, in a submanifold chart near one of its points, be a closed coordinate plane containing a dense subset of the ambient open chart, an impossibility. Thus the global conclusion uses an immersed leaf; the embedded conclusion is local, for curve segments remaining in a suitable chart.

The fibers must retain their base labels. In particular, a literal untagged union of fibers in $E$ is not sufficient when $f$ is not injective. Use the pullback vector bundleThis makes precise the indexed-union notation in the PDF. For a vector bundle trivialization ${\Phi }_U(e)=(\pi (e),v)$, defineIt is a fiberwise linear isomorphism, with inverse $(p,v)\mapsto (p,{\Phi }_U^{-1}(f(p),v))$. The transition functions are $g_{VU}\circ f$, hence smooth. They satisfy the same cocycle identities as those of $E$ and give a smooth total space of manifold dimension $\dim N+r$.

One may also see that this is an embedded submanifold of $N\times E$: in a bundle chart the constraint is the graph of the smooth map $f$ in the base coordinates, leaving the $r$ fiber coordinates free. Thus the chart construction agrees with its natural subspace smooth structure. No immersion or surjectivity hypothesis on $f$ is required. Therefore

For $p\in \mathrm{Fix}(\sigma )$, differentiation gives $(d{\sigma }_p{)}^2=I$ and $(d{\sigma }_p{)}^{\ast }{\omega }_p=-{\omega }_p$. Thus $d{\sigma }_p$ is an anti-symplectic involution of the symplectic vector space $T_{p}M$. By part (a), its $+1$ eigenspace is a Lagrangian subspace.

In the local manifold chart supplied in the question, $\sigma$ is linear, so its fixed-point set is the intersection of the chart with this $+1$ eigenspace. Therefore the fixed-point set is an embedded submanifold andEvery such tangent space has half the ambient dimension and the symplectic form vanishes on it. Hence every nonempty fixed-point set is a Lagrangian submanifold. The fixed-point set itself need not be connected.

Use ${\mathbb{C}}^2$ with the standard symplectic form ${\omega }_0=d{x}_1\wedge d{y}_1+d{x}_2\wedge d{y}_2$ and the compatible almost complex structure given by multiplication by $i$. Consider the embedded submanifoldIn coordinates $(x_1,y_1,x_2,y_2)$, its tangent space is $W=\{(a,b,b,0):a,b\in \mathbb{R}\}$. Applying $J$ gives $J(a,b,b,0)=(-b,a,0,b)$. If this is again in $W$, its fourth coordinate forces $b=0$, and its second and third coordinates then force $a=0$. Thus $W\cap JW=0$ and ${\dim }_{\mathbb{R}}L=2$: $L$ is a totally real submanifold.

However, the pullback of a differential form under the parametrization isThus this totally real submanifold is not a Lagrangian submanifold.

On $N={\mu }^{-1}(-1/4)$, the diagonal fundamental vector field $X$ is nonzero, since the fixed-point set lies at moment map values $0$ and $-1/2$. The nondegenerate bilinear form ${\omega }_p$ and $d\mu ={\iota }_X\omega$ imply $d\mu \ne 0$, so $-1/4$ is a regular value. By the regular level set theorem, $N$ is a three-dimensional embedded submanifold of the four-dimensional Complex projective plane.

At $p\in N$, set $W=T_{p}N=\ker d{\mu }_p$. Then $\omega (X,v)=0$ for every $v\in W$, so $X\in W^{\omega }$, the symplectic orthogonal complement. This complement is one-dimensional and $d\mu (X)=\omega (X,X)=0$, givingThus the level set is a coisotropic submanifold, neither an isotropic submanifold nor a Lagrangian submanifold. Indeed an isotropic subspace of a symplectic vector space in dimension four has dimension at most two, whereas $\dim N=3$. Its characteristic line field is generated by the diagonal circle action.

An immersed submanifold of $M$ is a manifold $Y$ equipped with an injective immersion $\psi :Y\to M$. It is an embedded submanifold when $\psi$ is also a homeomorphism onto its image with the subspace topology, equivalently when it is a smooth embedding. For irrational $\alpha$, the irrational winding of the torusis an injective immersion $\mathbb{R}\to T^2$ with dense image, and therefore is not an embedding. If $Y$ is compact, however, an injective immersion is a continuous bijection from a compact space to its image in the Hausdorff manifold $M$; its inverse is continuous. Thus the compact injective immersion is an embedding theorem makes $\psi (Y)$ embedded.

Now let $X\subset M$ be embedded, with $\dim M=m$, $\dim X=m-k$, and $p\in X$. Apply the constant rank theorem to its inclusion. After choosing coordinates and reordering them, there is a neighborhood $U$ of $p$ with coordinates $(x^1,\dots ,x^m)$ for whichThis is the slice chart for an embedded submanifold.

It is false that every embedded submanifold is the inverse image of a regular value of a map to a Euclidean space. If a codimension-$k$ submanifold is $f^{-1}(y)$ for a regular value of $f:M\to {\mathbb{R}}^k$, the differentials of the component functions give a global frame of its conormal bundle, so its normal bundle is trivial. The core circle of the Möbius band is embedded but has the nontrivial Möbius normal line bundle. This is the normal-bundle obstruction to being a regular level set.

Finally, an inductive spinning construction gives the requested torus. Place an embedding $F:N^m\hookrightarrow {\mathbb{R}}^{m+1}$ in the half-space $F_{m+1}>0$ and defineThe positive radius makes this map injective, and its differential is injective in both the $N$ and circle directions; compactness then makes it an embedding. Starting with $S^1\hookrightarrow {\mathbb{R}}^2$ and iterating proves the embedding of the n-dimensional torus in codimension one:

The curve is symmetric in both coordinate axes and under interchange of $x$ and $z$. Its shape, including its eight inflection points, is:

Writing $r^2=x^2+y^2$, the surface of revolution has equationLetIts gradient isIf this vanished, then $r^2\in \{0,1\}$ and $z^2\in \{0,1\}$, but at such a point $(r^2-1{)}^2+(z^2-1{)}^2\le 2$, not five. Hence zero is a regular value, and the regular level set theorem shows that $S=F^{-1}(0)$ is a smooth embedded surface.

For a surface of revolution, the two principal curvatures are the curvature of the meridian and the normal curvature of a parallel. Along $r=1$,so the normal is vertical in the meridian plane and the parallel principal curvature vanishes. Therefore the Gaussian curvature, the product of the principal curvatures, is zero there.

These are not the only zero-curvature points. The meridian itself has inflection points, at which its principal curvature vanishes. Setting $u=r^2$ and $v=z^2$, their positive squared coordinates satisfyBesides interchanging $u$ and $v$, the solution is approximatelyEach corresponding meridian inflection point sweeps out another circle on $S$, so $r=1$ does not exhaust the zero set of the Gaussian curvature.

For an embedded submanifold $X\subseteq {\mathbb{R}}^N$ and $p\in X$, the tangent space isEquivalently, in a local parametrization $\phi :U\subseteq {\mathbb{R}}^d\to X$ with $\phi (u)=p$, one has $T_{p}X=\mathrm{im}D{\phi }_u$.

Define the smooth mapAt $(M,v)$ its derivative in the direction $(A,w)$ isThe restriction to variations $(0,w)$ is the surjective linear map $M:{\mathbb{R}}^n\to {\mathbb{R}}^m$, since $M$ has full row rank. Thus $0$ is a regular value of $\Phi$. The regular level set theorem now shows thatis an embedded submanifold of $X_{m,n}\times {\mathbb{R}}^n$, of dimension $mn+n-m$.

A submersion is a smooth map between manifolds $F:X^n\to Y^m$ for whichis surjective at every $p\in X$.

The local submersion theorem says that around every $p\in X$ there are coordinates $(x^1,\dots ,x^n)$ centered at $p$ and $(y^1,\dots ,y^m)$ centered at $F(p)$ in whichTo prove it, surjectivity lets us choose $m$ source coordinates such that $d{F}^1,\dots ,d{F}^m$ are independent. Complete them by source coordinates $x^{m+1},\dots ,x^n$ and defineThe derivative of $G$ is invertible at $p$, so the inverse function theorem makes $G$ a local coordinate system; in these coordinates $F$ is the displayed projection.

For $q\in Y$, the fiber is locally given byThese are slice coordinates, so $F^{-1}(q)$ is an embedded submanifold of dimension $n-m$.

Properness is essential. The projectionis a submersion but is not proper. Its fiber over $x\ne 0$ is diffeomorphic to $\mathbb{R}$, whereas its fiber over zero is $\mathbb{R}\setminus \{0\}$ and has two connected components.

Merely requiring every fiber to be a submanifold is also insufficient. The surjective maphas every fiber finite and therefore a zero-dimensional embedded submanifold. Some regular values have three preimages and others have one, so the fibers are not all diffeomorphic. The map fails to be a submersion at $x=\pm 1/\sqrt{3}$.

An immersed submanifold of $M$ is a manifold $N$ with an injective immersion $\iota :N\to M$. It is an embedded submanifold when $\iota$ is also a homeomorphism onto its image with the subspace topology, equivalently when it is a smooth embedding.

The restriction of $J$ to the $J$-invariant bundle $TY$ is an almost complex structure. Its Nijenhuis tensor is the restriction of the ambient Nijenhuis tensor because vector fields tangent to an embedded submanifold have tangent Lie bracket. The ambient tensor vanishes since $X$ is a complex manifold, so the Newlander-Nirenberg theorem makes the induced structure on $Y$ integrable. The inclusion has complex-linear differential and is therefore holomorphic; hence $Y$ is a complex submanifold.

For a complex submanifold, the holomorphic normal bundle isIf $Y$ is a smooth hypersurface, taking top exterior powers in the holomorphic conormal sequencegivesThe normal bundle of a hypersurface is $N_{Y/X}\cong {\mathcal{O}}_X(Y){|}_Y$, so the adjunction formula is

On $X={\mathbb{CP}}^n\times {\mathbb{CP}}^m$, a bihomogeneous polynomial of bidegree $(d_1,d_2)$ is a section of the holomorphic line bundle $\mathcal{O}(d_1,d_2)$. Its zero locus is smooth precisely when the section is transverse to the zero section, equivalently when $F$ and all of its homogeneous first partial derivatives have no common projective zero. Sincethe adjunction formula yieldsThus a smooth $F$ with $d_1=n+1$ and $d_2=m+1$ defines a complex submanifold with trivial canonical bundle.

Let $\overline{\nabla }$ be the Euclidean connection and split the ambient tangent bundle along the embedded submanifold as $T{\mathbb{R}}^{n+m}{|}_M=TM\oplus NM$. The second fundamental form is the normal-bundle-valued bilinear formIt is symmetric because $\overline{\nabla }$ is torsion-free and the Lie bracket of tangent vector fields remains tangent:This is the symmetry of the second fundamental form.

Ambient matrix multiplicationis bilinear and therefore smooth. Its restriction to the embedded submanifold $G\times G$ is smooth, and part a shows that its image lies in $G$. By the defining smooth structure on an embedded submanifold, this restriction is a smooth map $G\times G\to G$.

For a Riemannian metric on an ambient manifold and an embedded submanifold, let $\pi :TN{|}_M\to TM$ be orthogonal projection. Project the restriction of a connection to an embedded submanifold to obtain $D_{X}Y=\pi {\nabla }_{X}Y$. Since $\pi Y=Y$, the Leibniz rule and linearity over smooth functions hold, making $D$ an affine connection. If the ambient connection is the Levi-Civita connection, pairing with tangent vectors shows that $D$ is a metric connection; projecting the zero-torsion identity shows it is a torsion-free connection. By uniqueness, $D$ is the induced Levi-Civita connection.

The coordinate hyperplane is an embedded submanifold, as standard projective charts flatten its defining homogeneous coordinate. Its complement is the affine chart ${\mathbb{R}}^n$ and is connected. The lemma regular zero hypersurfaces have disconnected complements therefore excludes its being the zero set of a smooth real function with zero a regular value.

For an embedded submanifold $i:M\hookrightarrow N$, the ambient affine connection induces a pullback connection on $i^{\ast }TN=TN{|}_M$. Extend a local section off $M$ and differentiate in a direction tangent to $M$. If two extensions agree on $M$, their difference has zero coefficients on $M$, whose tangential derivatives are zero. Thus the restricted derivative is independent of the extension. Its value may have a normal component, so obtaining a connection on $TM$ requires a further projection.