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The multiplicative norm first rules out zero divisors: if $xy=0$ and $x\ne 0$, then $0=|x||y|$ forces $y=0$. If the dimension is zero the claimed bound already holds, so suppose it is positive.

It is worth checking that the printed description of a ring with an underlying real vector space really permits differentiation of multiplication. This follows from the norm identity even if compatibility with real scalars is not separately stated. For fixed $x$, the map $L_x(y)=xy$ is additive by distributivity, andIt is therefore continuous. Additivity gives rational linearity. Approximating a real number by rationals and using continuity proves $L_x(ty)=t{L}_x(y)$ for every real $t$. The same argument holds in the other variable. This is automatic real bilinearity from a multiplicative norm, so multiplication is a bilinear map between finite-dimensional vector spaces, and hence is smooth.

Let $S=\{x\in A:|x|=1\}$ be the unit sphere of the given positive-definite bilinear form. It is a copy of $S^{n-1}$. The squaring mapis well defined because $|x^2|=|x{|}^2=1$, and is smooth. Commutativity gives its derivativeFor $x\in S$ and $v\ne 0$, the norm identity gives $|D{Q}_x(v)|=2|v|$, so the derivative is injective, including on the tangent space $T_{x}S=x^{\perp }$. It takes tangent vectors into $T_{x^2}S$ since $Q$ maps the sphere to itself. Alternatively, polarization of $|xy{|}^2=|x{|}^2|y{|}^2$ gives $⟨xy,xz⟩=|x{|}^2⟨y,z⟩$, so $⟨x^2,2xv⟩=2⟨x,v⟩=0$ directly.

The only identifications under squaring are antipodal:The factorization uses commutativity, and the last implication uses the absence of zero divisors. Conversely $Q(-x)=Q(x)$. It follows that $Q$ induces an injective smooth mapThe antipodal quotient map is a local diffeomorphism, so the injectivity of $DQ$ implies injectivity of $Dq$. Since Real projective space is compact, the stated compact injective immersion is an embedding theorem makes $q$ a closed embedding.

For $n\ge 2$ its source and target have the same dimension $n-1$. The inverse function theorem makes its image open as well as closed in the connected sphere $S^{n-1}$. It is nonempty, so the image is the whole sphere; consequently $q$ is a diffeomorphism. But for $n\ge 3$ this is impossible:Indeed, the antipodal two-sheeted covering map $S^{n-1}\to {\mathbb{RP}}^{n-1}$ has simply connected total space in this range, and is its universal cover, with deck group $\mathbb{Z}/2$. A sphere of dimension at least two is simply connected. These fundamental groups cannot be those of homeomorphic spaces.

We have obtained the dimension bound for a commutative Euclidean normed algebra:The real numbers and complex numbers with their usual products and Euclidean norms attain dimensions one and two. In dimension two, the induced map ${\mathbb{RP}}^1\to S^1$ is entirely compatible with the argument: the circle squaring map identifies exactly the two antipodal preimages. We neither assume a multiplicative identity nor use a classification of real division algebras; the squaring argument proves the bound from the supplied hypotheses.

For differentiable maps $F$ at $x$ and $G$ at $F(x)$, the chain rule assertsWrite $F(x+h)=F(x)+Ah+r(h)$ with $\Vert r(h)\Vert =o(\Vert h\Vert )$, and $G(F(x)+u)=G(F(x))+Bu+s(u)$ with $\Vert s(u)\Vert =o(\Vert u\Vert )$. Since $u=Ah+r(h)=O(\Vert h\Vert )$, substitution givesBoth remaining terms are $o(\Vert h\Vert )$, proving differentiability and the stated chain rule. The estimate for $s$ also holds when its argument is zero, by setting $s(0)=0$.

For the circle constraint take $H(u)=u_1^2+u_2^2$. The composite $H\circ F$ is constant, so the chain rule givesSince $\Vert F(x)\Vert =1$, this is a nonzero row vector annihilating $D{F}_x$. Thus $D{F}_x$ cannot be invertible: $\det D{F}_x=0$ at every $x$. Geometrically, the derivative of a map into a level set takes values in the tangent space to the circle.

A Riemannian metric is a smooth field of positive-definite symmetric bilinear forms $g_x$ on the tangent spaces, equivalently a smooth section of ${\mathrm{Sym}}^2{T}^{\ast }M$ with that positivity. Its musical isomorphism isNondegeneracy makes every fiber map an isomorphism. In coordinates the map has matrix $g_{ij}$, and its inverse ${♯}_g$ has matrix $g^{ij}$. Smoothness of the inverse follows from the inverse-matrix formula and the nonvanishing determinant. Both maps cover the identity on $M$, giving a smooth vector bundle isomorphism.

For local flattening of a Riemannian metric, choose a manifold chart neighborhood $V$ around $p$ with closure contained in $U$, and let $h={\varphi }^{\ast }\delta$ on $V$. Choose a smooth bump function $0\le \chi \le 1$, with compact support in $V$ and equal to one on a smaller neighborhood $W$ of $p$. SetThe formulas glue smoothly because the modification is supported strictly inside $V$. A convex combination of positive-definite forms is positive definite. On $W$ the metric is exactly $h$, so $\varphi$ gives an isometry to a Euclidean open subset. Outside $U$ the original metric remains unchanged.

Geodesic completeness means that every geodesic with arbitrary initial point and tangent vector extends for every real affine parameter. On each connected component, the Hopf-Rinow theorem equates this with completeness of the Riemannian distance.

The Euclidean-end claim needs a compact-core interpretation that is absent from its literal hypotheses. Indeed, $M=\{x\in {\mathbb{R}}^n:|x|>1\}$ with $n\ge 2$, $X=\varnothing$ and $g=\delta$ satisfy those hypotheses. The geodesic $\gamma (t)=(2-t)e_1$ reaches the missing unit sphere at $t=1$ and cannot continue within $M$. Thus the printed assumptions alone do not prove completeness.

Here is the intended compact-core completeness for a Euclidean end. Write $\Phi :M\setminus X\to \{|x|>1\}$ for the end coordinates and assume additionally thatis compact for every sufficiently large $R$. Choose such an $R$ beyond the metric transition and large enough to include the initial point of a geodesic. Its speed $s$ is constant. On every segment outside $K_R$, the Euclidean radius changes at a rate at most $s$, so over a finite parameter interval it cannot exceed $R+sT$. The full segment therefore stays in the compact set $K_{R+sT}$. Its velocities also stay in a compact subset of $TM$, since their metric norm is fixed and the base set is compact. The smooth geodesic ordinary differential equation consequently extends past any supposed finite endpoint. The reversed-time argument is identical. This proves completeness under the compact-core condition, and explains exactly what the exterior-of-a-ball counterexample lacks.

For upward stability of Riemannian completeness, lengths of all curves satisfy $L_{\tilde{g}}\ge L_g$, hence $d_{\tilde{g}}\ge d_g$. A $d_{\tilde{g}}$-Cauchy sequence is therefore $d_g$-Cauchy and has a limit $p$ because $g$ is complete. Smooth positive-definite metrics induce the manifold topology. More explicitly, on a small coordinate ball around $p$, $\tilde{g}$ has a bounded largest matrix eigenvalue, so the coordinate straight segment gives $d_{\tilde{g}}(p,q)\le C|\varphi (q)-\varphi (p)|$. Thus convergence to $p$ is also in $d_{\tilde{g}}$. That distance is complete, and Hopf-Rinow yieldsFor disconnected $M$, apply this argument on each component.

The divisor determines the bundle through its local equations. Choose an open cover on which $V$ has a reduced holomorphic defining equation $f_i$, using $f_i=1$ on sets disjoint from $V$. Because $V$ is a complex submanifold of codimension of a submanifold one, its ideal is locally generated by a coordinate; consequently $f_i/f_j$ is a nowhere-zero holomorphic function on each overlap. Inside the sheaf of meromorphic functions on a complex manifold, consider the locally free sheaf of rank oneThese local descriptions agree on overlaps. With $e_i=1/f_i$, its transition functions of a vector bundle are specified byThe ratios satisfy the Čech cocycle condition, and the associated holomorphic line bundle associated to a divisor is $[V]$. The section $1=f_i{e}_i$ is holomorphic and vanishes to order one along $V$. Replacing $f_i$ by $u_i{f}_i$, with $u_i$ a holomorphic unit, replaces the frame by $u_i^{-1}{e}_i$ without changing the subsheaf of meromorphic functions. Refining the cover changes no local sections either. This proves choice independence up to isomorphism rather than merely producing a bundle for one cover.

For a complex manifold of dimension $n$, the canonical bundle isIt is the holomorphic line bundle of holomorphic top-degree differential forms. To compute it on Complex projective space, let $S=\mathcal{O}(-1)$ be the complex tautological line bundle, so that $S^{\ast }=[H]=\mathcal{O}(1)$ is the hyperplane line bundle. At a line $\ell \subset {\mathbb{C}}^{n+1}$, the tangent space is $\mathrm{Hom}(\ell ,{\mathbb{C}}^{n+1}/\ell )$. Tensoring the tautological quotient sequence by $S^{\ast }$ gives the Euler sequence on complex projective spaceTaking the top exterior power gives $\det T^{1,0}{\mathbb{P}}^n=\mathcal{O}(n+1)$; taking its dual bundle yieldsHere a negative tensor power means the corresponding positive tensor power of the dual bundle.

The proposed sections over $U$ form a sheaf: compatible maps into $L$ glue uniquely, and their projections remain $\varphi$. Addition and multiplication by holomorphic functions are defined in the fibres of $L$, making this a sheaf of ${\mathcal{O}}_X$-modules. If $e_i$ is a holomorphic local frame of $L$ on $W_i\subset Y$, every section over $U\subset {\varphi }^{-1}(W_i)$ has the unique expressionThus the sheaf is a locally free sheaf of rank one. Its transition functions of a vector bundle are those of $L$ composed with $\varphi$, so the corresponding pullback vector bundle is precisely ${\varphi }^{\ast }L$. This proves invertibility locally and compatibility globally.

For the local blowup of a complex manifold at a point, work on the projective chart $Y_i\ne 0$ and writeAll the incidence equations reduce to $x_j=u{t}_j$. Therefore $(u,(t_j{)}_{j\ne i})$ gives a holomorphic coordinate chart isomorphic to ${\mathbb{C}}^n$. On an overlap with $Y_{\ell }\ne 0$,which is a biholomorphism where $t_{\ell }\ne 0$. The incidence subset is closed in the product of two Hausdorff spaces and is covered by these charts; hence it is a complex manifold of dimension $n$. In each chart the exceptional fibre is $u=0$, so it is a smooth complex submanifold, with its induced projective coordinates givingAway from the origin the inverse of the projection is $x\mapsto (x,[x])$, which is holomorphic; the projection is therefore a biholomorphism outside $E$.

To find the canonical bundle, pull back the nowhere-zero top form on ${\mathbb{C}}^n$. In the chart above, its Jacobian determinant givesThe sign depends only on the ordering of the coordinates. This is a section of $K_X$ with divisor on a complex manifold exactly $(n-1)E$: it has that vanishing order in every exceptional chart and no zeros elsewhere. A nonzero meromorphic section of a holomorphic line bundle with divisor $D$ identifies its holomorphic line bundle with $[D]$, by sending the local generator $1/f_i$ of $\mathcal{O}(D)$ to the nowhere-zero frame obtained by dividing the section by $f_i$. ConsequentlyThis also includes $n=1$, when the local map is an isomorphism and the exponent is zero.

For a general complex manifold, choose a coordinate neighbourhood $U$ of $P$, with $P$ sent to $0$, and replace $U$ by the inverse image of this coordinate neighbourhood in the local blowup of a complex manifold at a point. Glue this space to $M\setminus \{P\}$ along $U\setminus \{P\}$ using the preceding biholomorphism. The resulting charts give a complex manifold $\tilde{M}$ and a holomorphic map $\varphi$; the exceptional fibre is ${\mathbb{P}}^{n-1}$ and the map is a biholomorphism off that fibre. For completeness, the local projection is proper because its inverse image over a compact subset is closed in that subset times compact Complex projective space. This ensures the gluing is Hausdorff: separate points with distinct images downstairs, and points over $P$ inside the local blowup. The atlas is second countable as well.

The top exterior power of the differential defines a global holomorphic bundle map ${\varphi }^{\ast }{K}_M\to K_{\tilde{M}}$. Equivalently, it is a section of $K_{\tilde{M}}\otimes ({\varphi }^{\ast }{K}_M{)}^{-1}$. The same local Jacobian determinant has vanishing order $n-1$ on $E$, and the differential is invertible elsewhere. Applying the holomorphic line bundle associated to a divisor construction gives the canonical bundle formula for a point blowup

Odd-dimensional spheres cannot carry a symplectic form, by even-dimensionality. For $S^{2m}$ with $m\ge 2$, the second de Rham cohomology group vanishes. Any hypothetical symplectic form would therefore be $\omega =d\alpha$. The symplectic cohomology obstruction givesby the Generalized Stokes theorem. But the symplectic orientation makes ${\omega }^m$ a positive volume form, whose integral on this nonempty compact manifold is positive. This is a contradiction.

The oriented area form on $S^2$ is nondegenerate and automatically closed, since a two-dimensional manifold has no nonzero three-forms. Thus among positive-dimensional spheres, exactly $S^2$ admits a symplectic structure:If zero-dimensional symplectic manifolds are admitted, $S^0$ also qualifies: its zero two-form is nondegenerate on the zero tangent spaces. This is a convention-dependent additional case, not another positive-dimensional example.

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 $1$.

A vector field is a smooth section $X:M\to TM$ of the tangent bundle. It differentiates smooth functions by $Xf(p)=X_p(f)$. Define the Lie bracket of vector fields intrinsically byExpanding this expression on a product $fh$ shows that the two mixed terms cancel, giving $[X,Y](fh)=f[X,Y]h+h[X,Y]f$. It is therefore a derivation, hence a vector field. In local coordinates,Its coefficients are smooth. The intrinsic definition depends on no chart, which proves coordinate independence of this formula and defines the bracket on the whole manifold.

For a Lie group $G$, let $L_g(h)=gh$ be Left translation on a Lie group. A left-invariant vector field satisfies $(d{L}_g{)}_h{X}_h=X_{gh}$ for every $g,h$. Evaluation $X\mapsto X_e$ is linear and injective, since $X_g=(d{L}_g{)}_e{X}_e$. Conversely any $B\in T_{e}G$ gives a smooth left-invariant vector field $X_B(g)=(d{L}_g{)}_{e}B$. Smoothness follows from smooth multiplication and its differential. ThusFor any diffeomorphism $F$, the intrinsic bracket identity on functions gives $F_{\ast }[X,Y]=[F_{\ast }X,F_{\ast }Y]$. Applying this naturality of the Lie bracket to $F=L_g$ proves that the bracket of two left-invariant vector fields remains left invariant. Evaluation at $e$ therefore defines the Lie algebra bracket on $\mathfrak{g}=T_{e}G$.

For the special orthogonal group, differentiation of $Q(t{)}^{T}Q(t)=I$ at $Q(0)=I$ gives $B^T+B=0$, so its tangent space is contained in the skew-symmetric matrices. Conversely, for any such $B$, the matrix exponential $Q(t)=e^{tB}$ satisfies $Q(t{)}^{T}Q(t)=I$ and lies in the determinant-one component, because $Q(0)=I$ and its determinant varies continuously in $\{\pm 1\}$. Its initial derivative is $B$, provingFor matrices, $X_B(Q)=QB$. Extend these fields to the open matrix group $\mathrm{GL}(n,\mathbb{R})$, where the differential of $Q\mapsto QB$ in direction $H$ is $HB$. HenceRestriction to the embedded special orthogonal group preserves this bracket because the fields are tangent. Evaluating at $I$ gives $[B_1,B_2]=B_1{B}_2-B_2{B}_1$, the required bracket on the Special orthogonal Lie algebra. The commutator is again skew symmetric. This calculation is differentiating left-invariant matrix fields; reversing the order of the two differentials would give the wrong sign.

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.

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 $W\mathrm{diag}(e^{it{\lambda }_j})W^{\ast }$.

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 $T_e{G}_Q=M_n(\mathbb{R})$.

Over $\mathbb{R}$ or $\mathbb{C}$, a Lie algebra is a vector space with a bilinear map $[,]$ which is alternating and obeys the Jacobi identity:Bilinearity and alternation imply $[X,Y]=-[Y,X]$.

For a Matrix Lie group, identify the tangent space $\mathfrak{g}=T_{I}G$ with derivatives ${\gamma }^{\prime }(0)$ of smooth curves through $I$. Product curves show that the sum of two such derivatives is again tangent, and reparametrization supplies scalar multiples. In particular this is a real vector space, even when the matrices have complex entries. For $X,Y\in \mathfrak{g}$, choose a curve $\gamma (s)$ with ${\gamma }^{\prime }(0)=X$. Conjugating a curve with derivative $Y$ shows thatDifferentiate this curve in the finite-dimensional vector space $\mathfrak{g}$. The result isThe matrix commutator is bilinear and alternating, and expanding the six terms proves its Jacobi identity. Thus this construction gives the Lie algebra of a matrix Lie group, with the appropriate bracket, using actual group curves rather than an assumed commutator closure.

For the unitary group, differentiating $\gamma (t{)}^{†}\gamma (t)=I$ gives $X^{†}+X=0$. Conversely, if $X^{†}=-X$, then $e^{tX}$ is unitary and is a curve with derivative $X$. ConsequentlyThis is the unitary Lie algebra. The diagonal entries are purely imaginary, contributing $N$ real parameters, and each upper off-diagonal entry contributes two real parameters. A matrix-unit basis of the unitary Lie algebra isThese $N+2\left(\genfrac{}{}{0ex}{}{N}{2}\right)=N^2$ anti-Hermitian matrix units and their combinations are linearly independent over $\mathbb{R}$ and span every allowed entry.

The symplectic stabilizer is a subgroup: the identity preserves $J$, and if $MJ{M}^T=J$ and $NJ{N}^T=J$, thenMultiplying $MJ{M}^T=J$ on the left by $M^{-1}$ and on the right by $M^{-T}$ gives $M^{-1}J{M}^{-T}=J$. Products and inverses remain unitary. This is the compact symplectic group, often denoted $USp(2n)$ or $Sp(n)$.

Differentiating the stabilizer equation gives $XJ+J{X}^T=0$. For $X=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$, this saysTogether with $X^{†}=-X$, these are equivalentlyThese conditions are also sufficient: $e^{tX}$ is unitary, andHence it stays in the subgroup. They characterize its compact symplectic Lie algebra without adding any trace condition. In fact the trace automatically vanishes. The free anti-Hermitian block $A$ has $n^2$ real parameters, and the complex symmetric matrix $B$ has $n(n+1)$ real parameters. Thus