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A critical point of a smooth map is an input; a critical value is an output reached from at least one such input. For the fold $F(x,y)=(x,y^2)$, all inputs $(a,0)$ are critical points of a smooth map, and all outputs $(a,0)$ are critical values. The identical coordinate pairs live in different copies of ${\mathbb{R}}^2$. To classify an output $(a,b)$, inspect its entire fiber of a function:For $b>0$ both inputs have matrix rank two, so the output is a regular value. For $b=0$ its input has matrix rank one, so the output is a critical value. For $b<0$ the fiber of a function is the empty set, and the output is a regular value despite being outside the image of a function.

In general, for a smooth map $f:X\to Y$, write $C_f$ for its set of critical points of a smooth map. Its set of critical values is $f(C_f)\subseteq Y$. Equivalently, $q\in Y$ is a critical value if some $p\in f^{-1}(q)$ has a differential that is not surjective. A regular value requires a surjective differential at every input in that fiber of a function. The universal condition holds vacuously for an empty fiber of a function: every output outside $f(X)$ is a regular value. These quantifiers are explicit in Loring W. Tu, An Introduction to Manifolds, Definition 8.22 and its two following remarks.

The distinction is useful when solving equations. For a smooth map $f:X^m\to Y^n$ between smooth manifolds without boundary and a regular value $q$ with nonempty fiber of a function, the regular level set theorem makes $f^{-1}(q)$ an embedded submanifold of dimension $m-n$, with tangent spaceThe implicit function theorem explains the freedom count: locally, $n$ independent equations fix $n$ coordinates and leave $m-n$ free. The rank-nullity theorem gives the same count for the displayed kernel of a linear map. See Eckhard Meinrenken, Differentiable Manifolds, Theorems 1.11 and 1.15 for the local projection description and its consequence for level sets.

For example, $H:{\mathbb{R}}^2\to \mathbb{R}$, $H(x,y)=x^2+y^2$, has gradient $(2x,2y)$ and just one critical point, the origin, with critical value zero. For $c>0$, its fiber of a function is the circle of radius $\sqrt{c}$, a one-dimensional embedded submanifold; negative $c$ are regular values with empty fibers. At $c=0$, the fiber of a function is a single point instead of a circle. A critical value does not force its fiber of a function to be nonsmooth: that singleton is itself a zero-dimensional embedded submanifold, but the regular dimension conclusion $2-1=1$ does not apply.

Nor does a single regular input certify a regular value. The scalar smooth function $h(t)=t^2(t-1)$ has $h^{-1}(0)=\{0,1\}$ and derivative $h^{\prime }(t)=3t^2-2t$. The input zero is a critical point, while the input one is regular because $h^{\prime }(1)=1$. Their shared output zero is nevertheless a critical value. Conversely, different critical points of a smooth map can share one critical value: $f(t)=(t^2-1{)}^2$ sends both critical inputs $-1$ and $1$ to zero. Counting inputs and counting critical outputs are different tasks.

Finally, for $G(t)=(t,t^2):\mathbb{R}\to {\mathbb{R}}^2$, every output on the parabola is a critical value, and every output off it is a regular value with empty fiber of a function. There is no nonempty regular fiber of a function here, so the negative expression $1-2$ never becomes a claimed manifold dimension. Both diagrams show exact mathematical examples, not measured data.

A level set of a real-valued function on three-dimensional space. Where the function is continuously differentiable and its gradient is nonzero, the regular level set theorem makes this a smooth surface with normal vector parallel to the gradient. The equation alone can also have singular points, or fail to describe a two-dimensional surface at all.

If zero is a regular value of the joint smooth aggregate excess demand $Z(p,t)$, almost every parameter gives a regular economy. The regular level set theorem makes the zero set a manifold. Its projection to parameter space fails to have full rank exactly where the derivative with respect to prices is singular; Sard theorem makes these exceptional parameter values a null set.

A subset $X\subset M^m$ is a $d$-dimensional embedded submanifold if, with the subspace topology, every point has a smooth ambient chart in which $X$ is the intersection of the coordinate domain with ${\mathbb{R}}^d\times \{0\}$. These slice charts give its smooth structure, and its inclusion is an embedding.

The inverse function theorem says that a smooth map $F:U\subset {\mathbb{R}}^m\to {\mathbb{R}}^m$ with invertible derivative at $a$ restricts to a diffeomorphism between open neighborhoods of $a$ and $F(a)$. A point $q\in N$ is a regular value of a smooth map $f:M^m\to N^n$ when $d{f}_x:T_{x}M\to T_{q}N$ is surjective at every $x\in f^{-1}(q)$.

Choose local coordinates centered at $x$ and $q$. At a fiber point, rank $n$ means that, after relabeling coordinates, the first $n$ columns of the coordinate derivative form an invertible matrix. DefineIts derivative is block upper triangular with that invertible block and an identity block, hence invertible. The inverse function theorem makes $F$ a new ambient coordinate chart in which $f^{-1}(q)$ is exactly $v_1=\cdots =v_n=0$. Therefore the regular level set theorem giveswhen the fiber is nonempty. An empty fiber is regular vacuously; if $m<n$, it is the only possible regular fiber. The construction does not assign a negative dimension to a nonempty manifold.

For the projective hyperplane, the mapis injective with inverse $[x_0:x_1:0]\mapsto [x_0:x_1]$ on its image. At every image point, either $x_0$ or $x_1$ is nonzero; the corresponding standard projective chart uses $x_2/x_i$ as one coordinate and cuts the image out by setting it to zero. The map and inverse are smooth in these charts, and the slice condition provesNevertheless, ${\mathbb{RP}}^2\setminus X$ is the single affine chart $x_2\ne 0$, diffeomorphic to ${\mathbb{R}}^2$, so it is connected. If a smooth real function had exactly this zero set with zero regular, the inverse function theorem near any point of $X$ would give both positive and negative values off $X$. On the connected complement it never vanishes, so continuity requires its sign to be constant. This contradiction is the obstruction that regular zero hypersurfaces have disconnected complements. ThusIt illustrates why local defining equations for an embedded hypersurface need not glue to a global real defining function, as in real projective hyperplane is not a global regular zero set.

Normalize the second price to one and write $p>0$ for the first price and $t>0$ for the varying endowment. Let $Z(p,t)$ be first-good aggregate excess demand. Under Walras law, the second excess-demand coordinate equals $-pZ(p,t)$, so one market-clearing equation suffices. A regular exchange economy at fixed $t$ has $Z_p(p,t)\ne 0$ at every equilibrium.

A sufficient joint condition is that zero be a regular value of the smooth map $Z:(0,\infty {)}^2\to \mathbb{R}$:To prove the almost-everywhere conclusion, the regular level set theorem makes $M=Z^{-1}(0)$ a smooth one-dimensional manifold. Project $M$ to the parameter axis by $\pi (p,t)=t$. Its tangent space consists of vectors $(h,k)$ with $Z_{p}h+Z_{t}k=0$. If $Z_p\ne 0$, every $k$ is possible, so the derivative of $\pi$ is surjective. If $Z_p=0$, the joint condition gives $Z_t\ne 0$, forcing $k=0$; the point is critical for $\pi$.

Thus the parameters with a singular equilibrium are exactly the critical values of this projection. Sard theorem makes that set a null set. Therefore almost every parameter gives a regular economy. No claim of existence at every parameter is needed for this sufficient regularity criterion.

A $d$-dimensional embedded submanifold $F\subseteq M^m$ has the subspace topology and, near each of its points, a manifold chart identifying it with a coordinate slice ${\mathbb{R}}^d\times \{0\}$. These slice charts for an embedded submanifold give its smooth structure, and its inclusion is a smooth embedding.

A regular value $q\in N^n$ of a smooth map $f:M^m\to N^n$ is one for which $d{f}_p:T_{p}M\to T_{q}N$ is surjective at every $p\in f^{-1}(q)$. If the preimage is nonempty this forces $m\ge n$.

The inverse function theorem used here says that if $G:U\subseteq {\mathbb{R}}^m\to {\mathbb{R}}^m$ is smooth and $D{G}_p$ is an invertible matrix, then its restriction to a suitable neighbourhood of $p$ is a diffeomorphism onto an open neighbourhood of $G(p)$, with smooth inverse.

To prove the regular level set theorem, fix $p\in f^{-1}(q)$ and choose coordinates centred at $p,q$. The coordinate representative $\tilde{f}$ has a rank-$n$ Jacobian matrix. After permuting domain coordinates, its first $n$ columns form an invertible matrix. DefineIts Jacobian matrix is block triangular with invertible diagonal blocks, so the inverse function theorem makes $G$ a local coordinate change. In these coordinates, $f^{-1}(q)$ is precisely the slice with first $n$ coordinates zero. Thus it is an embedded submanifold of dimension $m-n$, withAn empty preimage is harmless; the local assertion is then vacuous.

Three equivalent formulations of orientability of a smooth manifold are: an oriented atlas whose transition Jacobian determinants are positive; a smoothly varying orientation of a vector space on each tangent space; and a nowhere-vanishing smooth top-degree differential form. An oriented atlas gives compatible oriented coordinate frames. A partition of unity glues their positive local top forms, and a nonzero top form declares a frame positive precisely when it evaluates positively on that frame. These constructions prove the equivalence.

For the orientability of a regular fibre, write $F=f^{-1}(q)$. The derivative gives a short exact sequence of vector bundlesThis is the reason the normal bundle of a regular fibre is trivial. Choose any fixed positive basis $b_1,\dots ,b_n$ of $T_{q}N$ and an ambient orientation on $M$. Declare a frame $v_1,\dots ,v_{m-n}$ of $T_{p}F$ positive whenis positive in $T_{p}M$, where $d{f}_p({\tilde{b}}_i)=b_i$. Changing a lift adds a tangent vector and leaves the determinant sign unchanged. Local smooth lifts exist because $df$ is surjective, so this rule varies smoothly and agrees on overlaps. It gives a global orientation on $F$. Only an orientation of the single vector space $T_{q}N$ was chosen; $N$ itself need not be orientable.

Regard ${\mathrm{Mat}}_n(\mathbb{C})$ as a real vector space of dimension $2n^2$, and let ${\mathrm{Herm}}_n(\mathbb{C})$ be the real vector space of Hermitian matrices, of dimension $n^2$. The smooth Gram maphas $F^{-1}(I)=U(n)$ and differential $D{F}_A(K)=A^{\ast }K+K^{\ast }A$. At a unitary $A$, any Hermitian matrix $H$ is obtained by taking $K=AH/2$. Indeed $A^{\ast }K=H/2$ and $K^{\ast }A=H/2$. Hence $I$ is a regular value, and the regular level set theorem proves the unitary group as a regular level set result:

Similarly let $G(A)=A^{T}A$ from ${\mathrm{Mat}}_n(\mathbb{R})$ to the symmetric real matrices, whose dimension is $n(n+1)/2$. At an orthogonal $A$ its differential is $D{G}_A(K)=A^{T}K+K^{T}A$, again onto by choosing $K=AH/2$ for a symmetric $H$. Therefore the orthogonal group as a regular level set hasMatrix multiplication is smooth on both groups, and inversion is conjugate transpose or transpose, respectively, so these manifold structures also make them Lie groups. Both groups are closed and bounded in their finite-dimensional ambient matrix spaces, hence compact.

Realification identifies a unitary complex matrix with an orthogonal transformation of ${\mathbb{R}}^{2n}$. This is a smooth injective group homomorphism, with closed image, so $U(n)$ is the stated Lie subgroup of $O(2n)$. The assumed maximal-rank quotient map has fibers diffeomorphic to $U(n)$. Consequently the compact homogeneous space, also known as the space of orthogonal complex structures, has dimensionThe original PDF's sphere is $S^{n^2}$, with a square on the exponent; its dimension exceeds $d$ by $n$.

For completeness, maps to a sphere above the dimension of a compact smooth manifold are null-homotopic. Let $X$ be a compact smooth manifold of dimension $d<q$, with $q\ge 1$, and let $f:X\to S^q\subset {\mathbb{R}}^{q+1}$ be continuous. Choose a finite open cover on which $f$ is uniformly close to its value at a chosen point of that open set, and take a smooth partition of unity subordinate to the cover. The weighted sum of those chosen sphere values is a smooth map $g_0:X\to {\mathbb{R}}^{q+1}$ uniformly within $ϵ<1$ of $f$. It never vanishes. Normalize it to $g=g_0/\Vert g_0\Vert$. Normalizing $(1-s)f+s{g}_0$ gives a homotopy from $f$ to $g$, since that vector remains within $ϵ$ of the unit vector $f$.

Every value of $g$ is critical because its differential has rank at most $d<q$. The Sard theorem makes its image have measure zero in $S^q$, so it misses some point $p$. The punctured sphere $S^q\setminus \{p\}$ is homeomorphic to ${\mathbb{R}}^q$ and contractible. Thus $g$, and hence $f$, is null-homotopic. This reasoning does not require $X$ to be connected, and all constant maps are homotopic because $S^q$ is path connected.

Apply it with $X=O(2n)/U(n)$ and $q=n^2$, for positive integer $n$. We obtainThe notation means homotopy classes of maps; no homotopy equivalence between the two spaces is asserted.

For an invertible $A$, factor $\det (A+tB)=\det A\det (I+t{A}^{-1}B)$. The determinant's permutation expansion gives $\det (I+tC)=1+t\mathrm{tr}C+O(t^2)$. Differentiating at zero yieldsAt a determinant-one matrix choose $B=A/n$; the derivative is one. Thus one is a regular value of the smooth determinant map on $GL(n)$. The regular level set theorem makes $SL(n)$ a submanifold of codimension of a submanifold one, withThis includes $n=1$, when $SL(1)=\{1\}$ has dimension zero.

The regular level set theorem says that a regular level of a smooth map from an $m$-manifold to an $r$-manifold is an $(m-r)$-manifold. The corresponding sublevel-set theorem says that $\{F\le c\}$ is a manifold with boundary $F^{-1}(c)$ when $c$ is a regular value of a smooth real-valued function, with the interior $F<c$.

Apply this to $F(x)=|x{|}^2$ on ${\mathbb{R}}^{n+1}$. On $F^{-1}(1)$ its gradient $2x$ is nonzero. Thus $S^n$ is an $n$-manifold and $B^{n+1}$ is an $(n+1)$-manifold with boundary $\boxed{\partial B^{n+1}=S^n}$. Locally the implicit function theorem supplies $F-1$ as one coordinate; the ball occupies its nonpositive side, which is a half-space after reversing that coordinate. This also covers $n=0$: the sphere consists of two points and the ball is the interval with those endpoints.

Put $m=\dim M$ and $n=\dim N$. The hypothesis says that $q$ is a regular value of $F$. Since its preimage is nonempty and the differential of a smooth map is surjective there, $m\ge n$.

We use the following finite-dimensional inverse function theorem: if a smooth map $H:O\to {\mathbb{R}}^m$, with $O\subset {\mathbb{R}}^m$ open, has invertible derivative at $a$, there are open neighborhoods of $a$ and $H(a)$ on which $H$ is a diffeomorphism, with smooth inverse.

Fix $p\in F^{-1}(q)$. Choose coordinate charts centered at $p$ and $q$, and shrink the source neighborhood so that its image under $F$ lies in the target chart. Its coordinate expression is $f=(f^1,\dots ,f^n)$ with $f(0)=0$ and derivative of rank $n$. Choose $n$ domain coordinates giving an invertible $n\times n$ minor, and relabel them as $x^1,\dots ,x^n$. DefineIts derivative at zero has block formso it is invertible. The inverse function theorem makes $u=H(x)$ a new coordinate chart near $p$. In this chart $F=q$ is exactly $u^1=\cdots =u^n=0$, because the target chart sends $q$ to zero. These are slice charts for an embedded submanifold; reordering coordinates if desired puts the free coordinates first. Giving the level set the subspace topology, these restricted charts are smoothly compatible, as in part (a). This proves the regular level set theorem rather than merely invoking it.

The level set is an embedded submanifold of dimension $\boxed{m-n}$. The same coordinates show $T_{p}Z=\ker d{F}_p$: tangent vectors to the slice have zero first $n$ components, which are exactly the components of $d{F}_p$. The empty-dimensional case $m=n$ gives a discrete zero-dimensional embedded submanifold.

Subtract the sphere equation from the last equality to obtain $x_2^2=1/2$. Thus the equality locus is the disjoint union of two sphereseach a smooth two-dimensional manifold. The two equality gradients are independent there: $x_2\ne 0$ and at least one of $x_1,x_3,x_4$ is nonzero.

On this locus the inequality becomes $1-2x_4^2\le a$, orSince $0<a<1$, this cuts out two closed caps on each sphere. At the cap boundary, $x_1^2+x_3^2=a/2>0$, so the restricted differential of $x_4$ is nonzero. The regular level set theorem for a sublevel set therefore gives a manifold with boundary. Explicitly it consists of four disks, with four boundary circles, and

A point $x$ is critical for a smooth map $f:X\to Y$ when $D{f}_x:T_{x}X\to T_{f(x)}Y$ is not surjective. A critical value is the image of a critical point; a regular value has no critical point in its fiber, including the case of an empty fiber.

To prove the regular level set theorem, take $x_0\in f^{-1}(y)$ and write $n=\dim X$, $m=\dim Y$. In local coordinates, surjectivity supplies an invertible $m\times m$ derivative minor; relabel source coordinates so it uses the first $m$ columns. The maphas an invertible derivative. The inverse function theorem makes it a local diffeomorphism. In these coordinates the fiber is obtained by fixing its first $m$ coordinates, so it is locally an open subset of ${\mathbb{R}}^{n-m}$. These smooth charts prove that a nonempty regular fiber is a smooth embedded manifold of dimension $n-m$, with tangent space $\ker D{f}_x$.

For the special orthogonal group, use $F:M_n(\mathbb{R})\to {\mathrm{Sym}}_n(\mathbb{R})$ defined by $F(A)=A{A}^T$. Its derivative is $D{F}_A(H)=H{A}^T+A{H}^T$. At an orthogonal $A$, any symmetric matrix $B$ is obtained by taking $H=BA/2$, so $I$ is a regular value. Therefore $O(n)=F^{-1}(I)$ has dimension $n^2-n(n+1)/2=n(n-1)/2$. The determinant takes only the values $\pm 1$ there, so $SO(n)$ is an open and closed part of $O(n)$ and has the same manifold dimension:It is closed in matrix space and bounded because every orthogonal matrix column has norm one. The Heine-Borel theorem proves compactness. Finally the tangent space is the derivative kernel,Thus its translated tangent space is the vector space of skew-symmetric matrices.

Regard complex matrices as a real vector space of dimension $2n^2$ and Hermitian matrices as a real vector space of dimension $n^2$. Set $F(A)=A{A}^{\ast }$. Its derivative is $D{F}_A(H)=H{A}^{\ast }+A{H}^{\ast }$. At a unitary $A$, any Hermitian $S$ is reached by $H=SA/2$. Hence $I$ is a regular value, and the regular level set theorem makes $\mathrm{U}(n)=F^{-1}(I)$ a smooth real manifold of dimension $\boxed{n^2}$.

The tangent space is the kernel of this derivative. Equivalently,At $I$, therefore, $H$ is skew-Hermitian. The explicit geodesic with initial point $I$ and velocity $H$ is $\gamma (t)=e^{tH}$. The matrix exponential is unitary because $e^{t{H}^{\ast }}{e}^{tH}=I$, and $\dot{\gamma }(0)=H$. Its acceleration is $\gamma H^2$. Since $H^2$ is Hermitian and every tangent vector at $\gamma$ is $\gamma K$ with $K$ skew-Hermitian, the induced real inner product satisfiesIndeed the trace of a Hermitian matrix times a skew-Hermitian matrix is purely imaginary. Its acceleration is therefore always normal to the manifold, which is the geodesic equation for an induced Euclidean metric. Diagonalizing $H=W\mathrm{diag}(i{\lambda }_j)W^{\ast }$ gives the equivalent explicit construction $W\mathrm{diag}(e^{it{\lambda }_j})W^{\ast }$.

The real vector space of symmetric matrices has dimension $n(n+1)/2$. On its invertible part the derivative of the determinant isAt a determinant-one symmetric matrix, this derivative is nonzero: the symmetric direction $C=B$ gives value $n$. The regular level set theorem therefore yieldsThis applies to every signature component, not only positive-definite matrices.

The special linear congruence action on symmetric matrices preserves symmetry because $(AB{A}^T{)}^T=AB{A}^T$, and preserves determinant because $\det (AB{A}^T)=(\det A{)}^2\det B=1$. The identity acts trivially andThese verify a smooth left Lie group action of the special linear group. By Sylvester's law of inertia, it also preserves the numbers of positive and negative eigenvalues.

For $n=2$, parametrize the symmetric matrices byThus $M$ is exactly the two-sheeted hyperboloidThe two sheets consist respectively of positive-definite and negative-definite matrices. Each is preserved by the special linear congruence action on symmetric matrices. They are individually transitive: on the positive sheet $B=A{A}^T$ with $A=B^{1/2}$ and $\det A=1$; on the negative sheet use $B=-A{A}^T$. The stabilizer of either $I$ or $-I$ is $SO(2)$.

Differentiating the left Lie group action along $\exp (sX)$ gives the fundamental vector fieldFor the three matrices in the question, their coordinate components areA sign convention matters here. With the usual Lie bracket of vector fields $[U,V]f=U(Vf)-V(Uf)$, the fundamental fields of this left action obey $[u_X,u_Y]=-u_{[X,Y]}$. Indeed, for linear coordinate fields $u_X(z)=R_{X}z$, the bracket has coefficient $(R_Y{R}_X-R_X{R}_Y)z$. To obtain a Lie algebra representation rather than an anti-representation, use the infinitesimal left-action sign conventionAn explicit answer is thereforeThese are tangent to the two-sheeted hyperboloid: applying each to $Q=t^2-x^2-y^2$ gives zero. For example,They can be written on either sheet using $t=\pm \sqrt{1+x^2+y^2}$:Here the omitted ${\partial }_t$ component is determined by tangency, and derivatives of $t(x,y)$ must be included when computing brackets in these coordinates.

Direct matrix multiplication givesDirect differentiation of the displayed vector fields gives exactlyFor instance, in ambient $(t,x,y)$ coordinates, $[v_1,v_3]=(-2x,-2t,0)=v_2$. These are the defining relations of the sl2 Lie algebra. The three fields are linearly independent over constant real coefficients: if $a{v}_1+b{v}_2+c{v}_3$ vanishes on a sheet, its ${\partial }_x$ coefficient $(-a+c)y-2bt$ forces $b=0,c=a$, and its ${\partial }_y$ coefficient then equals $-2at$, forcing $a=c=0$. Thus the representation is faithful. Their pointwise span need only have dimension two, consistent with the dimension of each sheet.

An embedded smooth $d$-dimensional manifold $X\subset {\mathbb{R}}^N$ is locally carried by a smooth ambient coordinate change to ${\mathbb{R}}^d\times \{0\}$. A map between such manifolds is a smooth map between manifolds if its coordinate representatives are smooth. Its differential $d{f}_x:T_{x}X\to T_{f(x)}Y$ sends the tangent vector represented by a curve $\gamma$ to $(f\circ \gamma {)}^{\prime }(0)$; equivalently it is the derivative in charts.

A regular value $y$ is one for which $d{f}_x$ is surjective at every $x\in f^{-1}(y)$, including vacuous regularity when the fibre is empty. For example zero is not a regular value of $f:\mathbb{R}\to \mathbb{R}$, $f(x)=x^2$, because $d{f}_0=0$.

For an invertible matrix $A$, the derivative of the determinant isAt a determinant-one matrix it is nonzero, since taking $H=A/n$ gives value one. The regular level set theorem therefore makes $S{L}_n(\mathbb{R})={\det }^{-1}(1)$ an embedded manifold of dimension $n^2-1$, also for $n=1$ when it is a single point. Matrix multiplication has polynomial ambient coordinates and restricts to the group, so it is smooth. At the identity,Finally suppose the singular-matrix set were an embedded smooth manifold near zero for $n\ge 2$. Every curve $t\mapsto t{E}_{ij}$ lies in it, since its rank is at most one. Thus every matrix unit belongs to its tangent space at zero. These span ${\mathbb{R}}^{n^2}$, forcing its local dimension to be $n^2$. An embedded manifold of full ambient dimension is locally open, but arbitrarily small scalar multiples of the identity are invertible. This contradiction proves the singular-matrix set is not an embedded smooth manifold. The argument addresses the subset-manifold convention specified at the start.

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.

Define the smooth functionAt $x\in F^{-1}(1)$ its derivative isThis linear map is surjective because $x\ne 0$; for example, $D_{x}F(x)=2$. Hence one is a regular value, and the regular level set theorem makesan embedded codimension-one submanifold of ${\mathbb{R}}^{n+1}$. Its tangent space is the kernel of the derivative:This is the unit sphere as a regular level set.

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.

LetAt $x\in F^{-1}(-1)$,This differential is nonzero because $x\ne 0$, so $-1$ is a regular value. The regular level set theorem makes $F^{-1}(-1)$ a three-dimensional embedded submanifold; intersecting it with the open set $\{x_0>0\}$ preserves that property. Therefore $X$ is the three-dimensional hyperboloid model, and

Consider the smooth mapIts derivative isAt $A\in O(1,3)$ write $H=AY$. ThenThis derivative is onto the ten-dimensional space of symmetric matrices: for a symmetric $S$, taking $Y=\frac{1}{2}MS$ gives $Y^{T}M+MY=S$. Hence $M$ is a regular value. Since $M_4(\mathbb{R})$ has dimension 16, the regular level set theorem gives

The tangent vectors are exactly those for which the derivative vanishes:Writing $Y$ in one-plus-three block form shows explicitly thatThe three entries of $v$ and three independent entries of the skew-symmetric matrix $\Omega$ again display dimension six. This is the Lie algebra of the Lorentz group.