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Let ${\mathcal{C}}_{\mathbb{C}}^{\infty }$ be the additive sheaf of complex-valued smooth functions, and ${\mathcal{C}}_{\mathbb{C}}^{\infty ,\ast }$ its multiplicative sheaf of nowhere-zero functions. The smooth exponential sequence isIts kernel is the integer-valued locally constant functions, namely the integer constant sheaf. It is surjective on stalks because a nowhere-zero smooth function has a smooth logarithm on a sufficiently small neighbourhood.

The additive smooth-function sheaf is a fine sheaf, using a partition of unity, and the smooth manifold is a paracompact space. Its positive-degree sheaf cohomology vanishes. The long exact sequence in sheaf cohomology consequently gives an isomorphism

To connect this with bundles, trivialize a smooth complex line bundle over an open cover. Its transition functions $g_{ij}$ form a multiplicative Čech cocycle, satisfying $g_{ij}{g}_{jk}=g_{ik}$. Changing trivializations changes it by a Čech coboundary. Conversely, any such cocycle glues trivial line bundles, and cohomologous cocycles give isomorphic bundles. Thus the isomorphism classes are $H^1(X,{\mathcal{C}}_{\mathbb{C}}^{\infty ,\ast })$.

The connecting isomorphism is the First Chern class. Locally choose logarithms $g_{ij}=e^{2\pi i{f}_{ij}}$; on triple intersectionsrepresents that class. Combining the gluing classification with $\delta$ gives the requested bijection:

Moser theorem says that if $M$ is a compact smooth manifold without boundary and ${\omega }_t$, $0\le t\le 1$, is a smooth family of symplectic forms with constant de Rham cohomology class, then there is a smooth isotopy ${\psi }_t$, with ${\psi }_0=\mathrm{id}$, such thatThus the endpoints are symplectomorphic. The hypothesis is a whole path of symplectic forms, not merely equality of the endpoint de Rham cohomology classes.

Take the usual convention that the smooth isotopy starts at ${\rho }_0=\mathrm{id}$. The exterior derivative commutes with a pullback of a differential form, soConsequentlyThe right-hand condition says precisely that ${\rho }_t$ is a symplectic isotopy.

Here an exact symplectic manifold is necessarily noncompact in positive dimension: if $\dim M=2n$ and $\omega =-d\lambda$, then ${\omega }^n=-d(\lambda \wedge {\omega }^{n-1})$, contradicting the positive symplectic volume and the Generalized Stokes theorem on a compact smooth manifold without boundary. Thus the paper's introductory word “closed” must here be read as “without boundary”, as its parenthesis suggests, rather than as including compactness.

The level equation is $|z_1{|}^2+|z_2{|}^2=|z_0{|}^2$. In particular $z_0\ne 0$, so the global affine coordinates on this level are $w_j=z_j/z_0$. They identify it withThe diagonal circle action becomes $(w_1,w_2)\mapsto (t{w}_1,t{w}_2)$ and is free: at least one coordinate is nonzero, so fixing a point forces $t=1$. Since the circle is compact, the Free proper Lie-group action theorem makes the quotient a smooth two-dimensional smooth manifold.

The map $(w_1,w_2)\mapsto [w_1,w_2]$ is the Hopf fibration. It is onto the complex projective line, and two unit representatives determine the same complex line precisely when they differ by a scalar of modulus one, hence by a circle orbit. Local sections, for example $[1,u]\mapsto (1,u)/\sqrt{1+|u{|}^2}$, show that the induced identification is a diffeomorphism. ThereforeThis identifies the symplectic reduction as a smooth manifold; no computation of its reduced symplectic form is needed.

A Lie group is a group and smooth manifold whose multiplication and inversion are smooth. A Lie algebra is a vector space with a bilinear alternating bracket satisfying the Jacobi identity. For a Matrix Lie group $G\subset GL(m,\mathbb{C})$, a logarithmic chart of a matrix Lie group near $g$ sends $h$ to $\log (g^{-1}h)\in \mathfrak{g}=T_{I}G$; the matrix exponential is its local inverse, and the Baker--Campbell--Hausdorff formula makes the local group operations smooth.

For $B_1,B_2\in \mathfrak{g}$, consider the group commutatorIts logarithm takes values in the vector space $\mathfrak{g}$, and expansion at $(0,0)$ givesThus $\mathfrak{g}$ is closed under the commutator. Bilinearity, alternation, and the Jacobi identity follow from matrix multiplication, so this proves that the Lie algebra of a matrix Lie group has

A principal bundle $\pi :P\to M$ with structure group $G$ is a smooth fiber bundle with a free right $G$-action, each fiber a single orbit, and equivariant local trivializations ${\pi }^{-1}(U)\cong U\times G$.

For the right action of the unitary group on $GL(n,\mathbb{C})$, defineThe matrix $g{g}^{\ast }$ is a positive-definite matrix and a Hermitian matrix, and $q(gu)=q(g)$ for $u\in U(n)$. The polar decomposition of an invertible complex matrix gives the unique factorizationConsequentlyis a diffeomorphism. If $W\subset H(n)$ is a neighborhood of zero and $N=q^{-1}(W)$, then $N$ is an open neighborhood of $I$, it is a union of complete orbits, andMoreover, two matrices have the same value of $q$ exactly when they differ by right multiplication by a unitary matrix. Thus $q$ induces the smooth identificationunder which $\pi$ is projection $H(n)\times U(n)\to H(n)$. This is the general linear group modulo the unitary group, and $\pi$ is in fact a globally trivial principal $U(n)$-bundle.

A Lie group is a group that is also a smooth manifold, with differentiable multiplication and inversion. A Lie algebra is a vector space with a bilinear alternating Lie bracket satisfying the Jacobi identity.

For a Matrix Lie group $G$, put $\mathfrak{g}=T_{I}G$. If $X,Y\in \mathfrak{g}$, the matrix commutator $[X,Y]=XY-YX$ again lies in $\mathfrak{g}$: the group commutator $e^{tX}{e}^{sY}{e}^{-tX}{e}^{-sY}$ lies in $G$, and the coefficient of $ts$ in its matrix logarithm is $[X,Y]$. Bilinearity and antisymmetry are immediate, while associativity of matrix multiplication gives the Jacobi identity. Thus $T_{I}G$, with the commutator bracket, is the Lie algebra of a matrix Lie group $\mathcal{L}(G)$.

The special linear group $SL(2,\mathbb{R})$ is the inverse image of the regular value $1$ under the smooth determinant map, and multiplication and inversion are smooth. Differentiating $\det (I+tA)=1+t\mathrm{tr}A+O(t^2)$ shows thatThe Cayley-Hamilton theorem applied to a trace-zero two-by-two matrix gives

The exponential map of a matrix Lie group is the matrix exponentialSince $\det (e^A)=e^{\mathrm{tr}A}=1$, its image lies in $SL(2,\mathbb{R})$. Put $d=\det A$. The identity $A^2=-dI$ sums the series explicitly. If $d>0$, with $r=\sqrt{d}$,If $d=0$, the trace is $2$, while if $d<0$, with $r=\sqrt{-d}$,HenceBut $\mathrm{diag}(-2,-1/2)\in SL(2,\mathbb{R})$ has trace $-5/2$. It is therefore outside the image, so the exponential map is not surjective.

Let $N=(0,\dots ,0,1)$ and $S=(0,\dots ,0,-1)$ be the poles of $S^n$. The open sets $U_N=S^n\setminus \{N\}$ and $U_S=S^n\setminus \{S\}$ cover the sphere. Stereographic projection gives coordinate chartsfrom these sets to ${\mathbb{R}}^n$. Their inverses areOn the overlap, the transition map iswhich is a smooth diffeomorphism of ${\mathbb{R}}^n\setminus \{0\}$. These two compatible charts make $S^n$ a smooth manifold of dimension $n$.

Now let $G$ be an $m$-dimensional Lie group and choose a basis $e_1,\dots ,e_m$ of its tangent space $T_{e}G$ at the identity. Definewhere $L_g$ is Left translation on a Lie group. Smoothness of multiplication makes each $E_i$ a smooth left-invariant vector field, and invertibility of $(d{L}_g{)}_e$ makes $E_1(g),\dots ,E_m(g)$ a basis of $T_{g}G$ at every point. Thus the $E_i$ form a global frame and every Lie group is a parallelizable manifold.

The columns of a matrix in the special unitary group $SU(2)$ are orthonormal and its determinant is one. Consequently every element has the unique formThe pair $(z_1,z_2)\in {\mathbb{C}}^2\cong {\mathbb{R}}^4$ therefore identifies $SU(2)$ diffeomorphically with $S^3$. The Lie-group construction then proves that $S^3$ is parallelizable; this is the SU(2) as the three-sphere identification.

Another example is $S^1$, which is the Lie group $U(1)$. Explicitly, at $(x,y)\in S^1$ the vectoris smooth, tangent, and nowhere zero, so it is a global one-vector frame. Thus $S^1$ is another parallelizable sphere.

An $m$ by $n$ matrix has matrix rank $m$ exactly when at least one of its $m$ by $m$ minors has nonzero determinant. For each multi-index $I$, the setis open in the vector space of all real $m$ by $n$ matrices, because the determinant is continuous. Their union is $X_{m,n}$, so $X_{m,n}$ is itself open in ${\mathbb{R}}^{mn}$. It therefore inherits the standard smooth manifold structure and has dimension $mn$. This is the full-row-rank matrix manifold.

An almost complex structure on a smooth manifold $M$ is a smooth bundle endomorphism $J:TM\to TM$ satisfying $J^2=-1$. On the complexified tangent bundle, $J$ has the eigenbundle decompositionwith eigenvalues $i$ and $-i$. A differential form of type (p, q) is a section of ${\bigwedge }^p(T^{1,0}M{)}^{\ast }\otimes {\bigwedge }^q(T^{0,1}M{)}^{\ast }$. The operators $\partial$ and $\bar{\partial }$ are the type-$(1,0)$ and type-$(0,1)$ components of the exterior derivative.

We prove the three stated conditions are equivalent. If $X,Y$ have type $(1,0)$ and $\alpha$ has type $(0,1)$, the formula for the exterior derivative givesThus closure of $T^{1,0}M$ under the Lie bracket is equivalent to the vanishing of the $(2,0)$ component of $d\alpha$ for every $(0,1)$-form. Complex conjugation of differential-form type gives the corresponding vanishing of the $(0,2)$ component on $(1,0)$-forms. Since every complex one-form is a sum of these two types, this is preciselyThis proves (i)$⟺$(ii).

Under (ii), the $(2,0)$ component of $d^{2}f=0$ is ${\partial }^{2}f$, proving (iii). Conversely, for $X,Y$ of type $(1,0)$,If this vanishes for every smooth complex-valued $f$, then $[X,Y{]}^{0,1}=0$, so (iii) implies (i). These conditions define an integrable almost complex structure.

For a complex manifold, a holomorphic chart $z^j=x^j+i{y}^j$ defines $J_M({\partial }_{x^j})={\partial }_{y^j}$ and $J_M({\partial }_{y^j})=-{\partial }_{x^j}$. The derivative of a holomorphic coordinate change is complex linear, hence commutes with multiplication by $i$; the definitions therefore glue and are independent of coordinates. Locally $T^{1,0}M$ is spanned by the commuting fields $\partial /\partial z^j$, so it is closed under brackets and $J_M$ is integrable. This is the almost complex structure induced by a complex atlas.

We next prove the local Dolbeault-Poincaré lemma. Write a $\bar{\partial }$-closed $(0,q)$-form on a slightly larger polydisc aswhere neither $\alpha$ nor $\beta$ contains $d{\bar{z}}_n$. Apply the supplied one-variable Cauchy-Green operator coefficientwise to $\beta$, obtaining $\gamma$ with $\partial \gamma /\partial {\bar{z}}_n=\beta$. Then $\phi -\bar{\partial }\gamma$ contains no $d{\bar{z}}_n$, and $\bar{\partial }\phi =0$ says that its coefficients are holomorphic in $z_n$ and $\bar{\partial }$-closed in the first $n-1$ variables. Induction on $n$, with the one-variable formula as the base case, makes this remainder $\bar{\partial }$-exact. Hence every $\bar{\partial }$-closed $(0,q)$-form with $q>0$ is $\bar{\partial }$-exact on each bounded polydisc.

Finally, translation by $i$ leaves $d\bar{z}$ unchanged, so this form descends to $\mathbb{C}/(z\sim z+i)$. It is in fact exact on this noncompact complex cylinder, because the invariant function $z+\bar{z}=2\mathrm{Re}z$ descends and satisfies

A smooth manifold of dimension $k$ is a Hausdorff, second-countable space with an atlas of smoothly compatible charts to open subsets of ${\mathbb{R}}^k$. A value $y\in Y$ is a regular value of a smooth map $f:X\to Y$ when every $x\in f^{-1}(y)$ has surjective derivative $d{f}_x$.

The inverse function theorem says that a smooth map between equal-dimensional manifolds whose derivative is invertible at a point is a diffeomorphism between neighbourhoods of that point and its image. If $y$ is a regular value of a map from an $m$-manifold to an $n$-manifold, choose coordinates corresponding to an invertible $n$ by $n$ minor of $df$. Applying the inverse theorem to $f$ together with the remaining $m-n$ coordinates makes $f$ the coordinate projection onto the first $n$ coordinates. Its level set is therefore locally ${\mathbb{R}}^{m-n}$. This proves the preimage theorem.

The critical-point set is closed because failure of full rank is the simultaneous vanishing of all maximal minors of $df$. If $X$ is compact, this set is compact, so its image under $f$ is compact and hence closed in the manifold $Y$. Its complement, the set of regular values, is open.

For the printed equations putA rank calculation shows that $(a,0)$ is a regular value when $a>0$: if
$\nabla F_2=\lambda \nabla F_1$, thenAt a point of the level set these relations force $a\le 0$. Hence for $a>0$, the preimage theorem makes $X_a$ a two-dimensional manifold.

As printed, however, the assertion for every $a\ne 0$ is false. If $a<0$, the pointsbelong to $X_a$. Near either point the first equation solves smoothly for $w$, while the second equation isIts positive-$z$ and negative-$z$ sheets meet only at the origin, so deleting the meeting point disconnects every sufficiently small neighbourhood. A punctured neighbourhood in a two-manifold is connected. Thus $X_a$ is not a manifold there.

For $a=0$, Cauchy--Schwarz givesThe first equation requires $w^2=x+y-z^2\ge 0$, so equality must hold. ThereforeThis is the image of a smooth embedding with nonzero derivative, and hence
$\boxed{X_0\text{ is a one-dimensional manifold}}$.

The antipodal action fixes the origin. Away from the origin it acts freely, so the quotient is locally Euclidean there. A small punctured neighborhood of the image of the origin, however, has linkthe real projective plane, whereas the link of a point in a three-dimensional smooth manifold is $S^2$. Equivalently, the neighborhood is a cone on ${\mathbb{RP}}^2$ and cannot be a three-ball. Thus

A Lie group is a smooth manifold $G$ equipped with smooth maps $m:G\times G\to G$, $m(g,h)=gh$, and $\iota :G\to G$, $\iota (g)=g^{-1}$, for which multiplication is associative, there is an identity $e$, and every $g$ has the inverse $g^{-1}$. In equations,Thus the group axioms and the smooth structure are compatible.

As a smooth manifold, $G\cong {\mathbb{R}}^{\ast }\times \mathbb{R}$. It has two connected components, distinguished by the sign of $a$: each of $a>0$ and $a<0$ is connected, and a continuous path in $G$ cannot cross $a=0$.

For a connected Riemannian manifold $(M,g)$, its Riemannian distance iswhere the infimum is over the piecewise smooth curves from $p$ to $q$. Connectedness of a smooth manifold implies path connectedness, so this set of curves is nonempty.

The Gauss lemma says that the differential of ${\exp }_p$ preserves the radial inner product: for $v,w\in T_{p}M$,Consequently radial geodesics from $p$ are orthogonal to the images of tangent vectors to spheres centred at the origin in $T_{p}M$. In a sufficiently small normal neighbourhood of $p$, this impliesevery competing curve has length at least the total variation of its radial coordinate, and the radial geodesic has that length.

The axioms $d_g(p,q)\ge 0$, symmetry, and the triangle inequality follow directly from length and concatenation. Certainly $d_g(p,p)=0$. If $q\ne p$, choose a normal ball $B_g(p,r)$ that does not contain $q$. Every curve from $p$ to $q$ first meets its boundary, and its initial part has length at least $r$ by the Gauss lemma. Hence $d_g(p,q)\ge r>0$. Thus $d_g(p,q)=0$ if and only if $p=q$.

For a compact oriented smooth manifold $M$, the Euler class of its tangent bundle satisfiesEquivalently, the sum of the indices of the isolated zeros of a vector field equals the Euler characteristic.

Products of manifold charts form a smooth atlas on the topological product of two smooth manifolds. The transition maps are products of the two smooth transition maps, so the dimension is the sum of the dimensions. The smooth structure is independent of the chosen compatible atlases.

A pseudo-Riemannian manifold is a smooth manifold with a smooth metric tensor that is a nondegenerate bilinear form on each tangent space. Positivity is not required. A Riemannian manifold is the positive-definite case, while a Lorentzian manifold has exactly one temporal direction in its metric signature.

On a compact oriented smooth manifold $N$ of dimension four without boundary, set $Q([\alpha ],[\beta ])={\int }_N\alpha \wedge \beta$ for closed real two-forms. The Stokes theorem makes the expression independent of representatives, and the even degrees make it symmetric. It is the real de Rham cohomology version of the intersection form. The Hodge decomposition theorem and Hodge star operator express it as $Q(h,k)=(h,\ast k{)}_{L^2}$ on harmonic representatives.

On a smooth manifold, ${\mathcal{A}}^k(U)$ is the space of smooth complex-valued degree-$k$ differential forms on $U$. Restriction of forms supplies the sheaf maps. Smooth partitions of unity make this a fine sheaf.