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For a hypersurface $N$ of a symplectic manifold, the restriction of the symplectic form has a one-dimensional kernel, its characteristic line field. On a regular value level $N=H^{-1}(c)$, this line is generated by the Hamiltonian vector field of $H$.

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.

For $k\ge 1$, the power map has $k$ distinct preimages of every nonzero finite regular value, all with positive local degree of a map between oriented manifolds. Integrating the pullback of a differential form of $\Omega =id\zeta \wedge d\bar{\zeta }/(1+|\zeta {|}^2{)}^2$ gives the same answer, since $\int \Omega =2\pi$ and $\int f^{\ast }\Omega =2\pi k$. For $k=0$ use the constant extension $1$ at infinity; its degree of a map between oriented manifolds is zero.

For a Hamiltonian group action, let $c$ be a regular value of the moment map fixed by the coadjoint action, and assume the action on ${\mu }^{-1}(c)$ is free and proper. The quotient is a symplectic manifold of dimension $\dim M-2\dim G$. Its unique reduced symplectic form is characterized by ${\pi }^{\ast }{\omega }_c={\iota }^{\ast }\omega$, where $\iota$ includes the level set and $\pi$ is the quotient projection. The kernel of the restricted two-form consists precisely of the orbit directions. More general noncentral levels use their stabilizer group; the stated central-level version suffices for circle group reduction.

If an embedded codimension-$k$ submanifold $X\subset M$ is $f^{-1}(y)$ for a regular value $y$ of a smooth map $f:M\to {\mathbb{R}}^k$, then $df$ identifies its normal bundle with the trivial bundle $X\times {\mathbb{R}}^k$. A submanifold with nontrivial normal bundle therefore cannot be such a regular level set. The core circle of the Möbius band is a codimension-one example.

For a regular value $q$ of a smooth map $f:M\to N$, let $F=f^{-1}(q)$. The differential of a smooth map identifies the quotient vector bundle $TM{|}_F/TF$ with the fixed target tangent space $T_{q}N$ over every point of $F$. After choosing a Riemannian metric, that quotient identifies with the normal bundle. Hence the normal bundle is trivial even when the target manifold has no global orientation.

Choose an orientation of a vector space on $T_{q}N$. The quotient identification for a regular value gives a smoothly oriented normal bundle along $F=f^{-1}(q)$. An oriented normal frame followed by a frame of $TF$ defines the latter's orientation through the ambient orientation. Changing normal lifts adds only tangent components, so the determinant sign is unchanged. No orientation on all of $N$ is required.

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.

For compact connected oriented $n$-manifolds without boundary, the degree is characterized byfor every smooth top-degree form $\eta$. Equivalently, at any regular value $y$,The preimage is finite by compactness and the inverse function theorem. Taking a form supported in a small neighbourhood of a regular value and applying change of variables proves the signed-count formula. The top de Rham cohomology of a compact connected oriented manifold is one-dimensional, with integration giving its coordinate; pullback and Stokes theorem make the integral characterization independent of the chosen normalized form. Thus this signed count is independent of the regular value. In particular a nonzero degree forces surjectivity, and the number of points in any regular fibre is at least the absolute degree. Degree is invariant under homotopy, as follows by applying Stokes theorem to the pullback on $M\times [0,1]$.

For the polynomial assertion take $k\ge 1$. The map $P:\mathbb{C}\to \mathbb{C}$ extends smoothly to a map of the Riemann sphere fixing infinity: in reciprocal coordinates its target coordinate isThe homotopy $P_t(z)=z^k+t(a_1{z}^{k-1}+\cdots +a_k)$ extends smoothly in the same way. The map $z\mapsto z^k$ has $k$ distinct inverse images of the regular value one, each orientation-preserving, since the real Jacobian of multiplication by a nonzero complex derivative is its squared absolute value. Its degree is $k$, so $\mathrm{deg}P=k\ne 0$. It is onto the sphere and must attain zero at a finite point. This proves every nonconstant complex polynomial has a root, without assuming the result to count roots of a general polynomial.

For a proper smooth $F:{\mathbb{R}}^n\to {\mathbb{R}}^n$, choose a compactly supported $n$-form $\eta$ with $\int \eta =1$ and defineProperness makes its pullback compactly supported. If ${\eta }^{\prime }$ is another choice, the given result gives $\eta -{\eta }^{\prime }=d\beta$ with compact support. The difference of the defining integrals is $\int d(F^{\ast }\beta )=0$, proving independence. Sard's theorem supplies a regular value. A proper map has finitely many preimages of it, and a sufficiently small neighbourhood has no additional inverse sheets outside the chosen inverse-function neighbourhoods: otherwise a sequence of extra preimages would have a convergent subsequence by properness and limit in that fibre. A normalized form there therefore gives the same finite signed-count formula, proving that the degree is an integer.

Finally $x\mapsto x^k$ is proper for $k\ge 1$. For odd $k$, a positive regular value has one inverse image with positive derivative. For positive even $k$, it has two inverse images with opposite derivative signs. ConsequentlyFor $k=0$ it is the constant map one, whose inverse image of the compact singleton $\{1\}$ is all of $\mathbb{R}$. It is not proper, so its degree is not defined by the proper-map construction requested here.

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.

The topological degree assigns an integer to a continuous map $f:M\to N$ between closed, connected, oriented manifolds of the same dimension $d$. Their top homology groups are infinite cyclic, generated by their fundamental classes. The induced map definesFor a smooth map and a regular value $y$, its equivalent local description iswhere the sign records preservation or reversal of orientation. The preimage is a finite set because the manifolds are compact and $y$ is regular. This is a signed count; opposite-orientation preimages cancel. If $\omega$ is a normalized target volume form with ${\int }_N\omega =1$, the pullback of a differential form gives the equivalent integral $\mathrm{deg}f={\int }_M{f}^{\ast }\omega$.

The homotopy invariance of mapping degree follows because homotopic maps induce the same homology map. In the smooth description, apply Stokes theorem to a homotopy $F:M\times [0,1]\to N$:since a top-dimensional form on $N$ is closed. Thus degree cannot change during a smooth deformation preserving the relevant boundary conditions. Degrees multiply under composition, and a nonzero-degree map is surjective: a missed target point would give a regular value with an empty preimage and hence degree zero.

For maps $S^d\to S^d$, the homotopy group ${\pi }_d(S^d)=\mathbb{Z}$ is completely detected by degree. This supplies integer topological sectors for many classical field-theory solitons. It is important to distinguish two ways spheres arise. In a nonlinear sigma model with the field approaching one fixed target value at infinity, compactify all of ${\mathbb{R}}^d$ to $S^d$ and classify the resulting map into the target. For a defect whose vacuum field varies with direction, retain the sphere $S_{\infty }^{d-1}$ surrounding the core and map it into the vacuum manifold. This is the vacuum-boundary degree as a defect charge. An asymptotic limit is a boundary hypothesis; finite energy alone in every possible model need not imply one.

For an Abelian Higgs vortex, the phase at infinity defines $S^1\to U(1)\cong S^1$. Its degree is the winding numberNonzero $n$ forces a zero of the Higgs field inside: a nowhere-vanishing field would extend its normalized phase through the disk and make the boundary loop contractible. This is also the flux classification of the Bogomolny vortex equations.

For a 't Hooft-Polyakov monopole, the normalized adjoint Higgs field $\hat{\Phi }$ gives $S_{\infty }^2\to SU(2)/U(1)\cong S^2$. With outward orientation its degree isThe integrand is the normalized target area form, so the hedgehog ansatz for a monopole has degree one. Nonzero degree obstructs extending the normalized Higgs direction through the ball; the full Higgs field must leave the vacuum manifold, normally vanishing in its core. The asymptotic unbroken Abelian magnetic flux measures the same integer with a convention-dependent sign.

For the Skyrme model, a field $U:{\mathbb{R}}^3\to SU(2)$ with $U\to 1$ at infinity gives $S^3\to S^3$, because SU(2) as the three-sphere identifies the target. A choice of orientations gives its topological baryon number in the Skyrme model as the winding integral $\pm (24{\pi }^2{)}^{-1}\int \mathrm{tr}(U^{-1}dU{)}^3$; specifying the orientation fixes the sign. Similarly, the transition map at infinity of a four-dimensional Yang-Mills instanton has degree in ${\pi }_3(SU(2))$, measured by its Chern charge.

A scalar-field kink illustrates the discrete-vacuum case: its endpoint data lie in ${\pi }_0(\{\pm 1\})$, rather than being a map between connected positive-dimensional spheres. For the double-well field its topological charge is $Q=[\varphi (+\infty )-\varphi (-\infty )]/2$. A kink has $Q=1$, an antikink $Q=-1$; they cannot individually deform to vacuum while their endpoints are fixed.

Degree classifies topology, rather than every solution. Smooth motion or local perturbations with fixed boundary data preserve it, and a nonzero degree obstructs decay to the vacuum within that class. It does not guarantee existence of a smooth energy minimizer, a fixed size, uniqueness, or dynamical stability. The degree and energetic stability of a field configuration are distinct: a field can shrink towards a singular limit at fixed degree, as Derrick's theorem diagnoses in many two-derivative theories. Moreover degree need not classify all homotopy classes for an arbitrary target; other homotopy groups or invariants may be needed. Within one degree there can be continuous moduli spaces and several stationary solutions.

Fix the sign convention ${\iota }_{{\xi }_M}\omega =-d{\mu }^{\xi }$, where ${\xi }_M$ is the fundamental vector field and ${\mu }^{\xi }=⟨\mu ,\xi ⟩$. A Hamiltonian group action is a symplectic Lie group action together with an equivariant moment map $\mu :M\to {\mathfrak{g}}^{\ast }$ satisfying this identity for every $\xi \in \mathfrak{g}$. Equivariance means $\mu (gx)={\mathrm{Ad}}_{g^{-1}}^{\ast }\mu (x)$. Merely requiring each fundamental field to have some Hamiltonian is the weaker condition of a weakly Hamiltonian action.

For a circle group action, identify its Lie algebra with $\mathbb{R}$, and let $Y$ generate the action. Put $Z={\mu }^{-1}(t)$ and let $i:Z\hookrightarrow M$. Since $t$ is a regular value,The action is free, so $Y_x\ne 0$. The symplectic orthogonal complement of this hyperplane is exactly $\mathbb{R}{Y}_x$, which is contained in $T_{x}Z$. ThusThe free action of the compact circle group is proper, and the quotient $Q=Z/S^1$ is a smooth manifold with projection $\pi :Z\to Q$ a principal bundle. The form $i^{\ast }\omega$ is invariant and kills the orbit directions, hence is basic and descends to a unique two-form ${\omega }_Q$ satisfying ${\pi }^{\ast }{\omega }_Q=i^{\ast }\omega$. It is closed because pullback by a submersion detects differential forms. If a quotient tangent vector annihilates ${\omega }_Q$, a lift lies in the displayed kernel and hence projects to zero. Therefore ${\omega }_Q$ is nondegenerate. This proves symplectic reduction directly.

If $L\subset Q$ is a Lagrangian submanifold, its inverse image $\tilde{L}={\pi }^{-1}(L)$ is a smooth circle bundle over $L$. The restricted symplectic form is ${\pi }^{\ast }({\omega }_Q{|}_L)=0$. If $\dim M=2N$, then $\dim Q=2N-2$, $\dim L=N-1$, and $\dim \tilde{L}=N$, proving that $\tilde{L}$ is Lagrangian. This is the Lagrangian lift through circle reduction.

For scalar multiplication on ${\mathbb{C}}^{n+1}$, the generator is $Y={\sum }_j(-y_j{\partial }_{x_j}+x_j{\partial }_{y_j})$. Direct contraction givesChoose $t>0$. Its level is the sphere of radius $R=\sqrt{2t}$ and its quotient is ${\mathbb{CP}}^n$. Complex conjugation on the sphere reverses the ambient symplectic form and descends to an anti-symplectic involution of the quotient. Its fixed set is ${\mathbb{RP}}^n$: a projective point fixed by conjugation has a real representative after a phase change. For two tangent vectors to this fixed set, anti-symplecticity makes the reduced form equal to its negative, so it vanishes. The real dimension is $n$, half the quotient dimension, proving the Lagrangian fixed locus of an anti-symplectic involution property.

The lifted Lagrangian submanifold can be written explicitly asThe only redundancy in this parametrization is the displayed free involution: two nonzero real representatives on the same complex ray differ by a real sign. The quotient embeds smoothly in the sphere, and projection to $[x]\in {\mathbb{RP}}^n$ exhibits its circle bundle structure. Its Lagrangian property follows from the reduction argument, including the case $n=0$.

Let $\mu :P\to \mathbb{R}$ be the moment map for the circle group, assume $c$ is a regular value, and assume the circle action on $N={\mu }^{-1}(c)$ is free. The action is proper because the circle is compact. Then $N$ has dimension $2n-1$ and $P_c=N/SO(2)$ is a smooth manifold of dimension $2n-2$.

If ${\xi }_P$ generates the action, $\omega ({\xi }_P,v)=d\mu (v)=0$ for every $v\in TN$. In fact $(TN{)}^{\omega }$ is exactly the line spanned by ${\xi }_P$, since $TN=\ker d\mu$ is a hyperplane and $\omega$ is nondegenerate. Therefore the kernel of the restricted two-form is precisely the orbit direction. The restriction is invariant and horizontal, so it descends to a unique nondegenerate closed two-form ${\omega }_c$ withThis proves the Marsden-Weinstein theorem in the circle case. If the Hamiltonian $H$ is invariant, then $\{\mu ,H\}=0$: its Hamiltonian flow stays on $N$, projects to the quotient and is generated there by the descended function $H_c$. This is a reduced Hamiltonian system in $2n-2$ local variables. At a critical level or a nonfree orbit the quotient can be singular, so the smooth dimension assertion needs the stated hypotheses.

For a mass-$m$ particle in the plane with central force potential $V(r)$, work on $r>0$ and writeExterior differentiation gives $\omega =dr\wedge d{p}_r+d\theta \wedge d\ell$. Rotation translates $\theta$ and has moment map $\ell$. Fix $\ell ={\ell }_0$ and quotient out $\theta$ to obtain the planar rotational symplectic reductionThe reduced equations are $\dot{r}=p_r/m$ and ${\dot{p}}_r={\ell }_0^2/(m{r}^3)-V^{\prime }(r)$. The centrifugal term arises from the conserved angular momentum; it is not an extra force imposed by hand. Reconstruct the angle from $\dot{\theta }={\ell }_0/(m{r}^2)$. For ${\ell }_0\ne 0$ the origin cannot lie on the level; for ${\ell }_0=0$ the same local reduction holds away from the origin, while the full phase-space origin is a singular fixed orbit.

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.

When both consumers have the same quasilinear utility on $(0,\infty )\times \mathbb{R}$, the same strictly concave one-variable maximization gives $x_1=x_2=1/p$, independently of their wealth. If their first-good endowments are $a,b$, thenThus the joint regular-value condition always holds. With positive total first-good endowment, the equilibrium isBoth first-good markets and second-good markets clear: second-good demands sum to $p^{\ast }(a+b)+t+c-2=t+c$. If a nonstandard endowment choice has $a+b\le 0$, there is no positive-price zero, so the regular value condition is vacuous. In the usual positive-endowment case every parameter has the displayed unique regular exchange economy.

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 the space of real $n\times n$ matrices as ${\mathbb{R}}^{n^2}$, and use the smooth determinant map. For invertible $A$, factor $A+tH=A(I+t{A}^{-1}H)$. Expanding the determinant to first order givessince the linear coefficient of $\det (I+tK)$ is the sum of the diagonal entries of $K$. At every $A$ with determinant one, taking $H=A$ makes this derivative $n\ne 0$. Thus one is a regular value, and part (ii) makes $\mathrm{SL}(n,\mathbb{R})$ a submanifold of dimension $n^2-1$.

Its tangent space is the kernel of this differential. At the identity,Indeed tangent derivatives of curves with constant determinant lie in that kernel, and the local regular-value coordinates show every kernel vector is tangent to the level set.

A point $x\in X$ is a critical point of a smooth map if $d{f}_x:T_{x}X\to T_{f(x)}Y$ is not surjective. A critical value is the image of at least one critical point. A regular value is a point $y\in Y$ for which every point of $f^{-1}(y)$ has surjective differential. A point outside the image is therefore a regular value vacuously. These definitions use rank relative to $\dim Y$, not merely whether a derivative vanishes in chosen coordinates.

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.

The Marsden-Weinstein theorem needs a Hamiltonian action, not merely an action by symplectomorphisms. Assume an equivariant moment map $\mu :P\to {\mathfrak{g}}^{\ast }$ exists, choose a coadjoint-fixed regular value $a$ (zero is the usual choice), and assume the action on $C={\mu }^{-1}(a)$ is free and proper. Regularity makes $C$ a submanifold of codimension $g=\dim G$. Equivariance and invariance of $a$ make $C$ invariant under $G$, and freeness and properness give a smooth quotient $P^{\prime }=C/G$. Thus

Let $\iota :C\hookrightarrow P$ and $\pi :C\to P^{\prime }$. For $v\in T_{p}C$ and $\xi \in \mathfrak{g}$,Moreover the conormal of $C$ is spanned by the independent $d{\mu }_{\xi }$, so nondegeneracy of $\omega$ gives $(T_{p}C{)}^{\omega }=\{{\xi }_P(p)\}$. The coadjoint-fixed value makes these orbit directions tangent to $C$. Hence the kernel of ${\iota }^{\ast }\omega$ is exactly the orbit tangent space. The restricted form is invariant and horizontal, so descends uniquely to a two-form ${\omega }^{\prime }$ withIt is closed because $\omega$ is closed. If a quotient tangent vector annihilates ${\omega }^{\prime }$, any lift lies in the kernel just identified and is vertical; the quotient vector is therefore zero. This proves nondegeneracy and completes symplectic reduction.

If the action is not free, singular strata or orbifold phenomena may occur. At a general noncentral regular value one quotients by its coadjoint stabilizer $G_a$, not by all of $G$; the regular dimension is $2n-g-\dim G_a$. These are genuine hypotheses behind the stated dimension formula.

Both proposed examples can be given explicitly. For symplectic reduction of an isotropic oscillator, take two modes withIts Hamiltonian flow is $z_j\mapsto e^{-it}{z}_j$, a circle action with moment map $H$. For $E>0$, $H^{-1}(E)$ is the sphere $S^3$ of radius $\sqrt{2E}$, and the action is free. The Hopf fibration identifies its quotient with ${\mathbb{CP}}^1\cong S^2$. On the chart $z_1\ne 0$, use $w=z_2/z_1$ and choose a section with $z_1=\sqrt{2E/(1+|w{|}^2)}$ real positive. Pulling back $\omega$ to this section givesWriting $w=x+iy$ gives $2Edx\wedge dy/(1+x^2+y^2{)}^2$, whose integral is $2\pi E$. This is the Fubini-Study form scaled by $E$ in the convention of area $2\pi$ at unit scale. The reduced dimension is $4-2=2$. At $E=0$, the circle fixes the origin and the smooth regular-level hypotheses fail.

For a unit-mass free particle in the plane, use the rotation circle action with angular-momentum moment map $\ell =x{p}_y-y{p}_x$. Choose ${\ell }_0\ne 0$ so the entire level avoids $r=0$ and the action is free. Polar canonical coordinates haveFixing $\ell ={\ell }_0$ and quotienting the rotation angle leavesThis planar rotational symplectic reduction replaces a planar free trajectory by radial motion with the centrifugal potential. A nonzero moment value is allowed because the circle is abelian and every value is coadjoint-fixed.