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The exterior derivative is an $\mathbb{R}$-linear map $d:{\Omega }^p(M)\to {\Omega }^{p+1}(M)$, agrees with the differential $df$ of a smooth function, satisfies the graded Leibniz rule $d(\alpha \wedge \beta )=d\alpha \wedge \beta +(-1{)}^p\alpha \wedge d\beta$ for $\alpha \in {\Omega }^p(M)$, and has $d^2=0$. These properties determine it locally and hence globally. To see locality directly, if a differential form $\alpha$ vanishes near $x$, choose a smooth bump function $\chi$ equal to one near $x$ and supported where $\alpha =0$. The identity $d(\chi \alpha )=d\chi \wedge \alpha +\chi d\alpha$ gives $(d\alpha {)}_x=0$. Thus global forms can be computed using local extensions in a manifold chart.

In coordinates write $\alpha ={\sum }_I{a}_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$. Since $d(d{x}^i)=d^2{x}^i=0$, the graded Leibniz rule forcesThis is the uniqueness of the exterior derivative from its axioms. The coordinate formula also establishes existence: it has the stated properties, and the chain rule shows that coordinate changes give the same operator.

The de Rham cohomology is the real quotient vector spaceThus a cohomology class records a closed differential form modulo an exact differential form. In degree zero there are no exact forms; closed functions are locally constant. The Poincare lemma says that every closed form of positive degree on a star-shaped open subset of ${\mathbb{R}}^n$ is exact, and hence that positive-degree closed forms are locally exact on a smooth manifold.

For the first de Rham cohomology of the two-sphere, let $U=S^2\setminus \{N\}$ and $V=S^2\setminus \{S\}$. Stereographic projection identifies each with ${\mathbb{R}}^2$, so a closed one-form $\alpha$ has primitives $f_U,f_V$. On the connected overlap $U\cap V$, the derivative of $f_U-f_V$ is zero, so this difference is a constant. Subtracting that constant from $f_U$ makes the primitives agree. They glue to a global smooth primitive. Therefore $H_{\mathrm{dR}}^1(S^2)=0$.

Let $q:S^2\to {\mathbb{RP}}^2$ be the double covering map and $a$ the antipodal map. The pullback of a differential form identifies forms downstairs with $a$-invariant forms upstairs. Averaging $(\eta +a^{\ast }\eta )/2$ commutes with $d$. If an invariant form is exact upstairs, averaging its primitive proves it exact downstairs. Conversely an invariant cohomology class has an invariant representative by the same averaging. This proves the de Rham cohomology of a finite quotient identification $H_{\mathrm{dR}}^p({\mathbb{RP}}^2)\cong H_{\mathrm{dR}}^p(S^2{)}^{a^{\ast }}$. In degree zero the sphere is connected and $a^{\ast }$ fixes constants; degree one is zero; in degree two the granted action is multiplication by $-1$, whose invariant real subspace is zero. Forms of degree greater than two vanish. HenceThis is real de Rham cohomology; it does not detect the integral two-torsion of the real projective plane.

A Lie group is a finite-dimensional smooth manifold with a group structure for which multiplication and inversion are smooth. The group defined here is the compact symplectic group $Sp(n)=U(2n)\cap Sp(2n,\mathbb{C})$, rather than the full complex symplectic group. Also, the displayed matrix expression requires $B$ to be $2n\times 2n$; the printed $n\times n$ size is incompatible with $J$.

Since $J^2=-I$, we have $J^{-1}=-J$. The matrix exponential commutes with conjugation, as follows term by term from its absolutely convergent power series. ConsequentlyTo construct logarithm charts for the compact symplectic group, consider the real vector spaceDifferentiating the defining identities at $I$ gives precisely these conditions. Conversely if $X\in \mathfrak{k}$, $e^X$ is unitary andso $e^X\in Sp(n)$. These are the infinitesimal conditions of the compact symplectic Lie algebra.

Near $I$, the convergent matrix logarithm series $\log (I+C)={\sum }_{m\ge 1}(-1{)}^{m+1}{C}^m/m$ is smooth and inverse to the matrix exponential near zero. These local inverses respect transpose, conjugate transpose, and conjugation; also $\log (A^{-1})=-\log A$ when both matrices are sufficiently close to $I$. Shrink their neighborhoods accordingly. If $A\in Sp(n)$ there, unitarity givesThe symplectic identity is equivalent to $A^t=J^{-1}{A}^{-1}J$, so, putting $X=\log A$, it gives $X^t=-J^{-1}XJ=JXJ$, equivalently $XJ+J{X}^t=0$. Thus this local logarithm restricts to a bijection between a neighborhood of $I$ in $Sp(n)$ and an open neighborhood of zero in $\mathfrak{k}$. Its inverse is the restricted matrix exponential. These restrictions are manifold charts; left multiplication translates them to every $A_0\in Sp(n)$ via $A\mapsto \log (A_0^{-1}A)$. The chart overlaps are smooth compositions of multiplication, exponential and logarithm. The subspace topology is Hausdorff and second countable, inherited from the finite-dimensional matrix space, so these charts give a smooth manifold.

Closure under products and inverses follows from $AJ{A}^t=J$ and unitarity. Matrix multiplication is polynomial in real and imaginary entries, and inversion on the unitary group is $A\mapsto A^{\ast }$, a real linear operation. Their restrictions are smooth in the charts just constructed. Hence $Sp(n)$ is a Lie group without needing a closed-subgroup theorem.

Write $X$ in $n\times n$ blocks. The two infinitesimal conditions giveThe skew-Hermitian matrix $P$ has $n^2$ real parameters; the complex symmetric matrix $Q$ has $n(n+1)/2$ complex parameters, hence $n(n+1)$ real parameters. ThereforeFor $n=1$, any two-by-two matrix satisfies $AJ{A}^t=(\det A)J$, so the group is $SU(2)$. Explicitly,The rows are orthonormal and the determinant is one; conversely unitarity and determinant one force this form. This is the SU(2) as the three-sphere parametrization. The map and its inverse, extraction of the first row, are smooth. Thus $Sp(1)$ is diffeomorphic to $S^3$.

A fiber metric on a real vector bundle $E\to M$ is a smoothly varying positive-definite inner product on each fiber. Choose a trivializing open cover $\{U_i\}$ and a smooth partition of unity $\{{\rho }_i\}$ subordinate to it. The partition theorem gives nonnegative functions summing to one, a locally finite family of subsets of supports, and $\mathrm{supp}{\rho }_i\subset U_i$; the usual Hausdorff second-countable smooth manifold hypotheses ensure this theorem applies. Transfer the Euclidean inner product to each local vector bundle trivialization, obtaining $h_i$, and setEach weighted term extends smoothly by zero outside $U_i$, and local finiteness makes the sum smooth in every vector bundle trivialization. At each $x$ some weight is positive, so $h_x(v,v)>0$ for every nonzero $v$. This proves existence of a fiber metric, with no orientability or triviality assumption.

A vector bundle morphism covering the identity is a smooth map $F:E^{\prime }\to E^{\prime \prime }$ that preserves base points and is a linear map on each fiber. Its induced map on the module of smooth sections is $s\mapsto F\circ s$, and is $C^{\infty }(M)$-linear. We prove the converse by constructing bundle morphisms from maps of smooth sections.

First the given map $\alpha$ is local. If a global section $s$ vanishes on a neighborhood of $x$, take a smooth bump function $\chi$ supported there with $\chi =1$ near $x$. Then $\chi s=0$, so $\chi \alpha (s)=\alpha (\chi s)=0$, and hence $\alpha (s)(x)=0$. Thus sections agreeing near $x$ have images agreeing at $x$.

Choose a local frame $e_1,\dots ,e_r$ on $U$, and a bump function equal to one on a smaller neighborhood $V$ of $x$ and supported in $U$. Multiplying the frame by that bump and extending by zero gives global smooth sections $t_j$ whose restrictions to $V$ are the frame. If $s(x)=0$, write $s=\sum a_j{e}_j$ on $V$. A second bump extends each $a_j$ to a global smooth function ${\tilde{a}}_j$ agreeing near $x$. By locality and $C^{\infty }(M)$-linearity,Every fiber vector $v\in E_x^{\prime }$ is the value of a global smooth section, by the same bumped-frame construction. Define $F_x(v)=\alpha (s)(x)$ for any such section. The just-proved vanishing statement makes this well-defined. The maps $F_x$ are linear maps, and locally their matrix columns are the smooth sections $\alpha (t_j)$ in a frame of $E^{\prime \prime }$. Thus $F$ is smooth, is a vector bundle morphism, and satisfies $\alpha (s)=F\circ s$. Fiberwise evaluation also proves uniqueness.

Finally apply the fiber metric construction to the tangent bundle. A Riemannian metric $g$ defines the musical isomorphismPositive definiteness makes it a fiberwise bijection; its inverse is smooth because inverse metric matrices vary smoothly. Hence $TM$ and $T^{\ast }M$ are isomorphic as real smooth vector bundles on every such manifold. The isomorphism depends on the chosen metric and is not canonical.

Let $Q$ be the configuration smooth manifold, and let $L(t,q,v)$ be a smooth Lagrangian on its tangent bundle. For a path with fixed endpoints, its action isThe principle of stationary action requires the first variation to vanish for every fixed-endpoint variation. In a coordinate chart take $q_s=q+s\eta$, with $\eta (a)=\eta (b)=0$. Differentiation under the integral and integration by parts giveRepeated coordinate indices are summed. The endpoint term vanishes. Since compactly supported variations can be chosen independently in each coordinate, the fundamental lemma of the calculus of variations yields the Euler-Lagrange equationsConversely these equations make the displayed first variation zero for every fixed-endpoint variation. Variations localized in coordinate charts establish the same assertion for paths on $Q$. Thus the Euler-Lagrange equations are exactly the stationary-path condition, not necessarily a condition for an action minimum.

A symplectic manifold is a smooth manifold $M$ equipped with a differential two-form $\omega$ such thatThe first condition says it is a closed differential form; the second says it is pointwise nondegenerate. A nondegenerate skew-symmetric matrix has even size, so the dimension is $2n$. Equivalently, ${\omega }^n$ is a nowhere-zero top-degree differential form. It supplies an orientation and the volume form ${\omega }^n/n!$. These are the defining conditions on the symplectic form.

A real Lie group is a group $G$ equipped with a finite-dimensional real smooth manifold structure, conventionally Hausdorff and second countable, for which multiplication $G\times G\to G$ and inversion $G\to G$ are smooth.

Its tangent space at the identity $e$ is defined by smooth curves through $e$: two curves represent the same tangent vector when their derivatives in a local chart agree at zero. For $X\in T_{e}G$, left translation $L_g(h)=gh$ determines the left-invariant vector fieldThe commutator of these derivations on smooth functions is another left-invariant vector field, so define the Lie bracketThis construction supplies the Lie algebra structure. In a Matrix Lie group, $X^L(g)=gX$ and differentiating the two fields gives $[X,Y]=XY-YX$.

For the special linear group, differentiate the determinant along a curve $g(t)$ through $I$. Expansion of the determinant, or its differential $D{\det }_A(H)=\det (A)\mathrm{tr}(A^{-1}H)$, givesThe determinant differential is surjective at $I$, so the level set $\det =1$ has tangent space equal to its differential's kernel. Equivalently, every tangent matrix has zero trace, and every zero-trace matrix supplies a curve $e^{tX}$ of determinant $e^{t\mathrm{tr}X}=1$. Hencewith the matrix commutator bracket. The same calculation over $\mathbb{C}$ gives the complex special linear Lie algebra ${\mathfrak{sl}}_n(\mathbb{C})$; as a real group, the complex group has that space viewed as a real Lie algebra.

The matrix exponential is the everywhere-convergent serieswhose inverse matrix is $\exp (-X)$. The matrix logarithm is locally defined near $I$ byThese maps are inverse on suitable neighbourhoods of zero and $I$, giving a logarithmic chart of a matrix Lie group. A logarithm is not a globally single-valued inverse of the exponential.

Every invertible complex matrix nevertheless has at least one matrix logarithm. Put it in Jordan normal form. For a block $J=\lambda I+N$, $\lambda \ne 0$ and $N^s=0$, choose any complex scalar logarithm $\ell$ of $\lambda$ and setThe finite logarithm and exponential identities in the nilpotent variable give $e^L=\lambda (I+N/\lambda )=J$. Combine the blocks and conjugate back. Thus the exponential map is surjective on ${\mathrm{GL}}_n(\mathbb{C})$, by existence of a logarithm for every invertible complex matrix.

A connected counterexample is ${\mathrm{SL}}_2(\mathbb{R})$. It is connected: polar decomposition of an invertible real matrix writes each element as $KP$, with $K\in \mathrm{SO}(2)$ and $P$ positive definite symmetric of determinant one; $\mathrm{SO}(2)$ is connected and $P$ is connected to $I$ through $P^t$.

But $g=\mathrm{diag}(-2,-1/2)$ belongs to this group and is not $e^X$ for any real $X$. Such an $X$ would commute with $e^X=g$. Since $g$ has two distinct real eigenvalues, direct commutation makes $X$ a real diagonal matrix. Its exponential has positive diagonal entries, a contradiction. ThereforeThis is an exponential-surjectivity obstruction from distinct negative eigenvalues; connectedness does not eliminate the obstruction.

A tangent vector at $p$ is a derivation at a point: a real-linear map $v:C_p^{\infty }(M)\to \mathbb{R}$ on germs of smooth functions satisfyingAddition and scalar multiplication preserve this identity, defining the tangent space $T_{p}M$ as a vector space. A derivation annihilates constants. In a manifold chart $x=(x^1,\dots ,x^n)$ centered near $p$, the local factorization $f-f(p)={\sum }_i(x^i-x^i(p))f_i$, with $f_i(p)={\partial }_{i}f(p)$, gives $v(f)={\sum }_{i}v(x^i){\partial }_{i}f(p)$. The factorization follows by integrating the first derivatives along a straight segment in a sufficiently small coordinate ball. ThereforeConversely these coordinate derivatives satisfy the derivation identity, so they really give a basis rather than merely a spanning family.

The cotangent space is the dual vector space $T_p^{\ast }M=(T_{p}M{)}^{\ast }$. Its basis $d{x}^i{|}_p$ is characterized by $d{x}^i({\partial }_j)={\delta }_j^i$. If $\alpha =a_{i}d{x}^i=b_{j}d{y}^j$, the chain rule gives the cotangent coordinate transitionThe cotangent bundle is the disjoint union $T^{\ast }M={\coprod }_{p\in M}{T}_p^{\ast }M$, with projection $\pi (p,\alpha )=p$. For each base manifold chart $(U,x)$, defineIts inverse sends $(z,a)$ to $(x^{-1}(z),a_{i}d{x}^i{|}_{x^{-1}(z)})$. On overlaps the transition is $(x,a)\mapsto (y(x),b)$, with the displayed fibre-linear transformation. Its coefficients and inverse are smooth, so these maps give a smooth atlas in dimension $2n$. Give the total space the topology obtained by transporting the product topology through these charts. Different base points are separated by disjoint base neighborhoods; distinct points over the same base point are separated within one bundle chart. A countable base smooth atlas and countable product bases give second countability. Thus the total space is a Hausdorff, second-countable smooth manifold. The same charts are vector bundle trivializations, since $\pi$ is product projection and their fibre changes are invertible linear maps.

The intrinsic canonical one-form on a cotangent bundle isIn the bundle chart it is $\lambda ={\sum }_i{a}_{i}d{x}^i$. This definition is coordinate independent, hence its exterior derivative is the globally defined smooth differential two-formThis sign follows the fibre-first order in this problem; the equally common position-first symplectic form is $-d\lambda$. The volume formnever vanishes. A smooth manifold is orientable exactly when it admits a nowhere-zero top-degree form; its positive ordered bases determine a consistent orientation. Consequently $T^{\ast }M$ is an orientable smooth manifold, even when $M$ is not. This is cotangent bundle orientation. In dimension zero the same conclusion uses the nowhere-zero zero-form $1$.

Use real differential forms and the usual convention that a smooth manifold has no boundary unless one is specified. On a noncompact manifold the formal-adjoint identity below uses compactly supported forms. On a compact manifold it has no boundary contribution; if a boundary were allowed, extra boundary conditions would be necessary.

The Riemannian metric gives the dual inner product on $T_x^{\ast }M$ and hence the inner product on exterior powers of a cotangent space:Extend this bilinearly. Equivalently, the increasing wedge products of an orthonormal coframe are an orthonormal basis of ${\Lambda }^p{T}_x^{\ast }M$. The Hodge star operator is the unique map $\ast :{\Lambda }^p{T}_x^{\ast }M\to {\Lambda }^{n-p}{T}_x^{\ast }M$ satisfying $\alpha \wedge \ast \beta =⟨\alpha ,\beta ⟩{\mathrm{vol}}_g$. If $e^1,\dots ,e^n$ is a positively oriented orthonormal coframe, $\ast e^I=\epsilon (I,I^c)e^{I^c}$. Moving the $p$ factors of $I$ across the $n-p$ factors of its complement changes the sign by $(-1{)}^{p(n-p)}$, soThe basis formula also proves that $\ast$ is an isometry.

For $\beta \in {\Omega }^{p-1}(M)$ and $\alpha \in {\Omega }^p(M)$ with compactly supported product, the Stokes theorem and the Leibniz rule for the exterior derivative giveThus $(d\beta ,\alpha {)}_{L^2}=(-1{)}^p{\int }_M\beta \wedge d\ast \alpha$. The formal adjoint operator $\delta$ satisfies $\ast \delta \alpha =(-1{)}^{p}d\ast \alpha$. Applying the inverse star and its square on degree $p-1$ gives the codifferentialThe exponent is congruent to $p+(p-1)(n-p+1)$ modulo two, so this convention is consistent in every degree. Define the Hodge Laplacian by $\Delta =d\delta +\delta d$ and a harmonic differential form by $\Delta \alpha =0$. On a compact manifold,Both nonnegative terms vanish for a harmonic form, giving $d\alpha =\delta \alpha =0$. Conversely these two equations imply harmonicity.

Now let $n=4$. The printed eigenvalue relation omits the subscript on its right-hand side; the intended equation is $\ast {\alpha }_{\pm }=\pm {\alpha }_{\pm }$. On two-forms ${\ast }^2=1$, soThese are respectively self-dual two-forms and anti-self-dual two-forms, and sum to $\alpha$. Their uniqueness follows because the $+1$ and $-1$ eigenspaces have zero intersection. Star is an orthogonal involution here, hence is self-adjoint and its opposite eigenspaces are orthogonal.

For closed two-forms define $Q([\alpha ],[\beta ])={\int }_N\alpha \wedge \beta$. Adding $d\xi$ to $\alpha$ changes the integral by ${\int }_{N}d(\xi \wedge \beta )=0$; adding an exact form to $\beta$ has the same effect. This is the real de Rham intersection form in dimension four. The wedge product of differential forms in degrees two and two is symmetric because $(-1{)}^{2\cdot 2}=1$; hence $Q$ is a symmetric bilinear form on $H_{\mathrm{dR}}^2(N;\mathbb{R})$.

In four dimensions $\delta =-\ast d\ast$ in every degree. If $h$ is a harmonic two-form, $\delta h=0$ gives $d\ast h=0$, and $dh=0$ gives $\delta \ast h=-\ast dh=0$. Thus $\ast h$ is harmonic, and ${\mathcal{H}}^2={\mathcal{H}}^{+}\oplus {\mathcal{H}}^{-}$ is the harmonic star-eigenspace decomposition. Use the permitted unique harmonic representative for each de Rham cohomology class. For harmonic $h,k$,The mixed terms vanish by orthogonality. The first restriction is positive definite and the second negative definite. In particular a nonzero harmonic $h$ has $Q(h,\ast h)=\Vert h{\Vert }_{L^2}^2>0$, proving nondegeneracy. Hence the signature of the intersection form from harmonic duality isThe finite dimensions and the harmonic description also follow from the Hodge decomposition theorem.

For completeness, the Hodge decomposition theorem on a compact oriented Riemannian manifold without boundary states the finite-dimensional harmonic space and the $L^2$-orthogonal direct sumThe three summand components are unique, although their potentials need not be. Given an exact three-form $\beta =d\omega$ on $N$, decompose its smooth two-form potential as $\omega =h+d\xi +\delta \psi$, with $\psi \in {\Omega }^3$. Since $dh=0$ and $d^2=0$, $\beta =d\delta \psi$. Put $a=\delta \psi$. In dimension four, $a=-\ast d\ast \psi$, and ${\ast }^2=1$ on two-forms gives $\ast a=-d\ast \psi$, so $d\ast a=0$. ConsequentlyThis constructs a self-dual primitive of an exact three-form. The sum contains no factor $1/2$: it is twice the self-dual projection of the coexact potential. Using the unscaled projection alone would give $d\sigma =\beta /2$. This proves the final unheaded request.

A holomorphic atlas on a real smooth manifold of dimension $2n$ consists of compatible smooth charts ${\phi }_a:U_a\to V_a\subseteq {\mathbb{C}}^n$ whose transition maps ${\phi }_b{\phi }_a^{-1}$ are biholomorphic on their domains. An almost complex structure is a smooth real vector bundle endomorphism $J:TX\to TX$ with $J^2=-\mathrm{id}$. In each chart defineThe differential of a holomorphic map is a complex-linear map, so the transition differentials commute with multiplication by $i$. Consequently the local definitions agree on overlaps. They are smooth and square to $-\mathrm{id}$, yielding the natural almost complex structure induced by a complex atlas. This construction uses the atlas, rather than an arbitrary choice of coordinates on the underlying real smooth manifold.

A real Lie group is a finite-dimensional real smooth manifold with a group structure whose multiplication and inversion are smooth. Its tangent space at the identity element $e$ is the vector space $\mathfrak{g}=T_{e}G$ of velocities of smooth curves through $e$. For the general linear group ${\mathrm{GL}}_n(\mathbb{R})$ it identifies with $M_n(\mathbb{R})$. The matrix exponential and local matrix logarithm areThe exponential is globally defined, and each matrix $\exp X$ is invertible with inverse $\exp (-X)$. The logarithm here is a local map near $I$; it is not a globally defined inverse on every real invertible matrix. The differential of $\exp$ at zero is the identity, and the two power series are inverse locally.

For a general $G$, the assumed Exponential map of a Lie group with $d{\exp }_0=I$ has a smooth local inverse by the inverse function theorem, giving a local exponential chart. Define the Lie bracket from local group commutators byFor small $s,t$ the group commutator lies in that chart. More explicitly, the map $F(u,v)=\log (\exp u\exp v\exp (-u)\exp (-v))$ is smooth and has $F(u,0)=F(0,v)=0$. Its mixed second differential at $(0,0)$ is a bilinear map of $X,Y$, proving bilinearity. Swapping the two group elements inverts the commutator, and $\log (g^{-1})=-\log g$ near $e$. Thus $F(v,u)=-F(u,v)$ and $[Y,X]=-[X,Y]$. In the matrix group, expansion to the mixed term $st$ gives the familiar commutator $XY-YX$.

For $g\in G$, let $c_g(h)=gh{g}^{-1}$ and define the Adjoint representation of a Lie group by ${\mathrm{Ad}}_g=d(c_g{)}_e\in \mathrm{GL}(\mathfrak{g})$. It is smooth and satisfies ${\mathrm{Ad}}_{gh}={\mathrm{Ad}}_g{\mathrm{Ad}}_h$. Its derived representation is $\mathrm{ad}=d(\mathrm{Ad}{)}_e:\mathfrak{g}\to \mathrm{End}(\mathfrak{g})$. Naturality of the Exponential map of a Lie group under conjugation givesAt $t=0$, the derivative of the logarithm of the commutator is ${\mathrm{Ad}}_{\exp (sX)}Y-Y$: the derivative of multiplication at $(e,e)$ adds tangent vectors, and $d{\log }_e=I$. Differentiate in $s$ to concludeTo prove the Jacobi identity without assuming it in the bracket construction, first establish that the differential of a Lie group homomorphism preserves Lie brackets. For a homomorphism $f:G\to H$, the allowed exponential identity gives ${\log }_H(f({\exp }_{G}u))=d{f}_e(u)$ locally. Apply this identity to the group commutator and take the mixed derivative to obtain $d{f}_e([X,Y])=[d{f}_{e}X,d{f}_{e}Y]$. In particular $f=\mathrm{Ad}:G\to \mathrm{GL}(\mathfrak{g})$ givesApplying both sides to $Z$ yields $[[X,Y],Z]=[X,[Y,Z]]-[Y,[X,Z]]$. Rearranging with antisymmetry provesNeither injectivity of $\mathrm{ad}$ nor the existence of a global logarithm was used.

Finally let $H$ be a normal Lie subgroup of $G$, with its corresponding Lie algebra $\mathfrak{h}\subseteq \mathfrak{g}$. For each $g\in G$, conjugation restricts to $H$, so its differential preserves $\mathfrak{h}$: ${\mathrm{Ad}}_g\mathfrak{h}=\mathfrak{h}$. Differentiate along $g=\exp (tX)$ to get $[X,Y]=\mathrm{ad}(X)Y\in \mathfrak{h}$ for every $X\in \mathfrak{g}$ and $Y\in \mathfrak{h}$. Thus the Lie algebra of a normal Lie subgroup is an ideal of a Lie algebra. The assertion concerns a subgroup carrying the corresponding Lie-subgroup structure, for example any closed subgroup; no arbitrary abstract subgroup is being assigned a tangent space.

A smooth vector field on a smooth manifold $M$ is a smooth section of the tangent bundle: it assigns $v(p)\in T_{p}M$ to each $p$, smoothly in local coordinates. In a coordinate chart it has the form $v={\sum }_i{a}_i\partial /\partial x_i$, with smooth coefficients $a_i$. Equivalently, it acts on smooth functions as a derivation, $v(fh)=v(f)h+fv(h)$.

For a Lie group $G$, write $L_a(h)=ah$ for Left translation on a Lie group. A left-invariant vector field satisfiesThus it is determined by its value at the identity. Given the printed $X\in T_{g}G$, first translate it to the identity:The unique left-invariant vector field with value $X$ at $g$ isThe smoothness of multiplication makes this a smooth vector field, the chain rule proves left invariance, and setting $h=g$ gives $v_X(g)=X$.

We prove the completeness of left-invariant vector fields: this left-invariant vector field is a complete vector field. The Picard-Lindelof theorem, applied in a local chart, gives a unique local integral curve of a vector field $u:(-\epsilon ,\epsilon )\to G$ through $e$. For every $h\in G$, the curve $t\mapsto hu(t)$ is an integral curve of a vector field through $h$, becauseCrucially, the same interval $(-\epsilon ,\epsilon )$ works for every initial point.

Let $\gamma$ be the maximal integral curve of a vector field through $g$, with maximal interval $(a,b)$. If $b<\infty$, choose $t_0\in (a,b)$ with $b-t_0<\epsilon /2$. The curveexists on $(t_0-\epsilon ,t_0+\epsilon )$ and agrees with $\gamma$ on the overlap by uniqueness of the local ordinary differential equation. It extends $\gamma$ past $b$, a contradiction. The same argument at a finite $a$ excludes that possibility. Thus $(a,b)=\mathbb{R}$, and uniqueness on overlapping intervals gives uniqueness on all of $\mathbb{R}$.

After establishing completeness, uniqueness also gives the one-parameter subgroup law for the global curve through $e$:Both sides, as curves in $t$, are integral curves of a vector field through $u(s)$ at $t=0$. Defining the Exponential map of a Lie group by $u(t)=\exp (t\xi )$, the answer is

The identity component $G^{\circ }$ is an open normal subgroup, and every open identity neighbourhood generates it. Let $G^{\circ }$ be the connected component of $e$. The product of connected spaces is connected, so the image of $G^{\circ }\times G^{\circ }$ under multiplication is connected and contains $e$. It is therefore contained in $G^{\circ }$. Inversion has the same property. Thus $G^{\circ }$ is a subgroup.

A smooth manifold is locally connected. In particular, a coordinate neighbourhood of $e$ can be chosen homeomorphic to an open ball, so there is a connected open neighbourhood $O$ of $e$ contained in $G^{\circ }$. Its translates $hO$, $h\in G^{\circ }$, are open and lie in $G^{\circ }$, and cover $G^{\circ }$. Therefore the identity component of a Lie group is open in $G$. Conjugation by any $a\in G$ is a homeomorphism fixing $e$, so it maps $G^{\circ }$ into itself; conjugation by $a^{-1}$ gives the reverse inclusion. Hence $G^{\circ }$ is a normal subgroup.

If $U$ is an open neighbourhood of $e$ in $G^{\circ }$, let $H=⟨U⟩$, allowing inverses in the meaning of generated subgroup. For each $h\in H$, $hU$ is open and lies in $H$, so $H={\bigcup }_{h\in H}hU$ is open in $G^{\circ }$. Every other left coset of $H$ is also open. Thus $H$ is both open and closed in the connected space $G^{\circ }$. It is nonempty, so $H=G^{\circ }$.

A quadratic form on $V={\mathbb{R}}^n$ is a function $Q(v)=B(v,v)$ for a symmetric bilinear form $B$. Equivalently, $Q(tv)=t^{2}Q(v)$ and the polarizationis a bilinear map. In coordinates there is a unique real symmetric matrix $A$ with $Q(v)=v^{T}Av$. No assumption of nondegeneracy or positive definiteness is needed.

An element $g\in {\mathrm{GL}}_n(\mathbb{R})$ stabilizes $Q$ when $Q(gv)=Q(v)$ for every $v$. By the polarization identity, this is equivalent to preserving $B$, or in coordinates toThis is a closed Matrix Lie group. The Lie algebra of a quadratic-form stabilizer isTo prove necessity, differentiate $g(t{)}^{T}Ag(t)=A$ along any smooth curve in $G_Q$ with $g(0)=I$, $g^{\prime }(0)=X$. The derivative at zero is $X^{T}A+AX$.

To prove sufficiency, suppose $X^{T}A+AX=0$ and consider the matrix exponential $g(t)=e^{tX}$. ThenIts value at zero is $A$, so $e^{tX}\in G_Q$ for every real $t$. This curve has derivative $X$ at zero, proving the claimed tangent space description even for a degenerate quadratic form.

Equivalently, the condition is $B(Xu,v)+B(u,Xv)=0$ for all $u,v$. For a nondegenerate quadratic form it is the corresponding Special orthogonal Lie algebra. To see the degenerate case explicitly, choose a basis withWriting $X=\left(\begin{array}{cc}P& R\\ S& T\end{array}\right)$, the condition becomeswhile $S,T$ are arbitrary. Thus the nullspace of the quadratic form is invariant, but arbitrary infinitesimal maps into it are allowed. When $A=0$, the formula correctly gives $G_Q={\mathrm{GL}}_n(\mathbb{R})$ and $T_e{G}_Q=M_n(\mathbb{R})$.

In a manifold chart, write a smooth differential form as $\eta ={\sum }_I{a}_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$, with $I$ increasing. Define its exterior derivative byTo check that this definition is global, change coordinates to $y$. The chain rule expresses $d{x}^i$ as ${\sum }_j({\partial }_j{x}^i)d{y}^j$. Differentiating it gives zero because the second partial derivatives of $x^i$ are symmetric and $d{y}^k\wedge d{y}^j$ is antisymmetric. The coordinate definition satisfies the graded Leibniz rule, so differentiating the transformed expression for $\eta$ leaves exactly ${\sum }_{I}d{a}_I\wedge d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$. This agrees with the original formula and proves independence of the manifold chart. The same cancellation of mixed derivatives gives $d^2=0$.

The de Rham cohomology is thereforethe vector space of closed differential forms modulo exact differential forms.

Here is a Green-theorem proof of exactness on a disc. Identify the disc with a round disc centred at zero, and let $\eta =Pdx+Qdy$ be a closed differential form. Set $f(x)={\int }_{[0,x]}\eta$. For $x$ and $x+h$ in the disc, Green theorem on the oriented triangle with vertices $0,x,x+h$ gives zero boundary integral, since ${\partial }_{x}Q-{\partial }_{y}P=0$. HenceTaking $h=\epsilon v$ and dividing by $\epsilon$ yields $d{f}_x(v)={\eta }_x(v)$. The radial integral defining $f$ is smooth, so $\eta =df$ andDegenerate triangles follow by continuity. The same proof works on any star-shaped plane domain, including the whole plane.

Integration requires an orientation: take $M$ to be an oriented compact $n$-dimensional smooth manifold without boundary. The Generalized Stokes theorem says ${\int }_{M}d\beta =0$ for every smooth $(n-1)$-form $\beta$. Every $n$-form is closed by dimension, and changing it by an exact differential form leaves its integral unchanged. Thusis a well-defined linear map $H_{\mathrm{dR}}^n(M;\mathbb{R})\to \mathbb{R}$. Compactness alone does not supply the orientation omitted from the printed integration statement.

For an $n$-dimensional Lie group, choose a nonzero $\lambda \in {\Lambda }^n{T}_e^{\ast }G$ and define the invariant volume form on a Lie group byEach $d{L}_{g^{-1}}$ is an isomorphism, so $\omega$ is nowhere zero and defines an orientation. The composition identity $L_{(hg{)}^{-1}}\circ L_h=L_{g^{-1}}$ gives $L_h^{\ast }\omega =\omega$. Moreover, every left-invariant differential form of degree $n$ is determined by its value at $e$, so the space of such forms is one-dimensional.

Under the supplied invariant-representative assumption, all top-degree de Rham cohomology classes are multiples of $[\omega ]$. For compact $G$, orient it so $\omega$ is positive; then ${\int }_G\omega >0$, whereas the Generalized Stokes theorem makes every exact top form have integral zero. Therefore $[\omega ]\ne 0$ andunder that assumption. For actual compact Lie groups, this argument applies when $G$ is connected. The supplied assumption fails for disconnected groups: on $S^1\times \mathbb{Z}/2$, the two components have independent degree-one classes, while globally left-invariant top forms form only a one-dimensional space. In general the top-degree de Rham cohomology of a compact Lie group is ${\mathbb{R}}^c$, where $c$ is its number of connected components.