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The relation $[x,y]=-[y,x]$ for a bilinear map proposed as a Lie bracket. An Alternating Lie bracket always satisfies it. Conversely, setting $y=x$ gives $2[x,x]=0$, so antisymmetry implies alternation over a field of characteristic different from two.

In characteristic two, the sign disappears and this relation just says $[x,y]=[y,x]$. It does not force $[x,x]=0$, even when the Jacobi identity holds. For example, on a two-dimensional vector space with basis $u,v$, set $[u,u]=v$ and every other basis bracket to zero, then extend bilinearly. Every nested bracket vanishes, so the Jacobi identity holds, and the product is symmetric. Nevertheless it is not a Lie algebra, because $[u,u]\ne 0$.

Over a field of characteristic zero, an associative Lie product is a bilinear map that is simultaneously an associative algebra product and a Lie bracket. Such a product has every triple product zero. To prove this, write multiplication by juxtaposition. By associativity and antisymmetry of a Lie bracket,since moving $z$ left through $y$ and then through $x$ introduces two minus signs. Applying antisymmetry of a Lie bracket instead to the two elements $xy$ and $z$ gives $(xy)z=-z(xy)$. Thus $xyz=-xyz$, and $A^3=0$ in characteristic zero.

Conversely, an alternating multilinear map defining a product whose triple products all vanish is associative, because both associations give zero, and satisfies the Jacobi identity, because each nested product gives zero. A nonzero example has basis $p,q,z$, products $pq=z=-qp$, and all other products of basis vectors zero. It is the Heisenberg Lie algebra with its Lie bracket used as the associative algebra product. Its image $A^2=kz$ annihilates $A$ on both sides, and it has no identity element.

Suppose a ring has an underlying real normed vector space and its norm is multiplicative, without initially assuming real bilinearity of multiplication. For fixed $x$, distributivity makes $y\mapsto xy$ additive, while $\Vert xy-xz\Vert =\Vert x\Vert \Vert y-z\Vert$ makes it continuous. Additivity gives rational scalar compatibility, and continuity extends it to every real scalar. Apply the same argument in the other variable to obtain a bilinear map. The norm identity also excludes nonzero zero divisors.

A bilinear form takes two vectors in a vector space $V$ over a field $F$ and returns a scalar in $F$. Its defining property is separate linearity: holding either input fixed gives a linear functional of the other. For scalars $a,b$ and vectors $u,v,w$,This is a scalar-valued bilinear map; the chosen field is part of the definition.^[ref]

Separate linearity does not mean that $B$ is a linear map on the vector space $V\times V$ with componentwise vector addition and scalar multiplication. Over the real numbers, $B(2v,2w)=4B(v,w)$, whereas a linear map on the pair would give $2B(v,w)$. The simplest example is $B(x,y)=xy$ on the one-dimensional vector space $\mathbb{R}$: fixing $y=3$ gives $x\mapsto 3x$, but doubling both $x$ and $y$ quadruples their product.

A bilinear form measures how two inputs interact. The dot product measures alignment; a symmetric bilinear form can describe a quadratic form; an alternating bilinear form can measure signed area. A nondegenerate bilinear form turns each vector into a unique linear functional and so connects geometry with the dual space. The same short definition therefore supports lengths, bilinear orthogonal complements, and symplectic vector spaces, depending on the additional properties imposed.

Over the complex numbers, a bilinear form is linear in both inputs without conjugating either scalar. A Hermitian form instead is a sesquilinear form and conjugates when its inputs are exchanged. For complex matrices, the adjoint $A^{\ast }$ used with Hermitian forms and matrix C-star algebras is the conjugate transpose, whereas a bilinear form uses the ordinary transpose $A^T$. For example, $B(z,w)=zw$ has $B(i,i)=-1$, while the Hermitian form $H(z,w)=z\overline{w}$ has $H(i,i)=1$.^[ref]

The commutator of two elements of an associative algebra over a field isIt measures how much the order of multiplication matters: $[A,B]=0$ exactly when $AB=BA$. It is a bilinear map, it is alternating since $AA-AA=0$, and associativity makes it satisfy the Jacobi identity. Hence every associative algebra becomes a Lie algebra on the same vector space when equipped with this product. This holds in every characteristic of a field.

For square matrices, this is the matrix commutator, and it is defined even for singular matrices. The group commutator of invertible matrices is instead $AB{A}^{-1}{B}^{-1}$: it equals the identity element when the matrices commute, whereas $AB-BA$ equals the zero matrix. The Lie bracket from local group commutators explains how differentiation connects the two constructions.

For nonzero finite-dimensional complex vector spaces $U,V$, suppose a bilinear map never vanishes on a pair of nonzero vectors. Equivalently, each map obtained by fixing one nonzero argument is injective. If its associated map on $U\otimes V$ has image dimension $r$, projectivization gives a map to ${\mathbb{CP}}^{r-1}$ whose positive degree-two class pulls back to $x+y$. The nonzero top power $(x+y{)}^{\dim U+\dim V-2}$ in the product cohomology ring of complex projective space proves the bound. Multiplication of polynomials of bounded degree attains equality. Nonzero hypotheses matter: with a zero factor, slice conditions can be vacuous.

A Lie algebra over a field $k$ is a vector space $\mathfrak{g}$ with a bilinear map $[,]:\mathfrak{g}\times \mathfrak{g}\to \mathfrak{g}$, called the Lie bracket, that is alternating and satisfies the Jacobi identity. Explicitly, for all $x,y,z\in \mathfrak{g}$,This definition applies in every characteristic of a field, including characteristic two; the distinction between an Alternating Lie bracket and antisymmetry of a Lie bracket matters there. See Pavel Etingof, Lie Groups and Lie Algebras, Definition 8.4 and Remark 8.5.

The motivating question is how to describe very small symmetries. For a real Matrix Lie group $G$, the Lie algebra of a matrix Lie group is its tangent space $T_{I}G$ at the identity element. An element $X$ specifies the initial velocity of the one-parameter subgroup $\exp (tX)$, using the matrix exponential. The Lie bracket records the first nonzero effect of changing the order of two small transformations. Even when first-order motions cancel, a second-order motion can remain: the small rotation commutator below leaves a rotation about a third axis.

This turns local symmetry questions into computations with linear maps. A basis and its structure constants of a Lie algebra determine every Lie bracket by bilinearity. For example, three matrices suffice for the sl2 matrix commutator example, while the familiar cross product describes the SO(3) Lie algebra. These descriptions help calculate rotations in three dimensions and build Lie algebra representations; the same bracket structure underlies angular momentum commutation relations. The striking feature is that a nonlinear family of motions has a local description by a vector space and one product. Global information still needs care: the Lie algebra alone does not distinguish the circle group from the additive real line.

Historically, the connection with differential equations is visible in Sophus Lie's Theorie der Transformationsgruppen, volume I (1888), prepared with Friedrich Engel. Chapter 6 interprets first-order differential operators as infinitesimal transformations, linking their action on functions to the transformations those functions admit. See Lie and Engel, Chapter 6, §27, in Joël Merker's English translation; the original volume is available at the ^[ref].

For trivial coefficients in a field $k$, a Lie algebra two-coboundary is a bilinear map $\beta (x,y)=t([x,y])$ for a linear functional $t$ in the dual space ${\mathfrak{g}}^{\ast }$. Some conventions put a minus sign in this formula; the resulting vector subspace $B^2(\mathfrak{g},k)$ is the same. Each such map is a Lie algebra two-cocycle: it is an alternating bilinear form, and its cyclic expression is $t([[x,y],z]+[[y,z],x]+[[z,x],y])=0$ by the Jacobi identity. Replacing a linear section of a central extension of a Lie algebra by $s^{\prime }(x)=s(x)+t(x)z$ changes its Lie algebra two-cocycle by $-\beta$. Thus $H^2(\mathfrak{g},k)=Z^2(\mathfrak{g},k)/B^2(\mathfrak{g},k)$ in Lie algebra cohomology records exactly the obstruction to a section that is a Lie algebra homomorphism.

The Lie bracket is the product in a Lie algebra: a bilinear map with an Alternating Lie bracket and the Jacobi identity. Bilinearity means, for $a,b\in k$,Unlike matrix multiplication, a Lie bracket need not satisfy associativity. In the sl2 matrix commutator example, $[H,[E,F]]=0$ but $[[H,E],F]=2H$. The replacement compatibility law is the Jacobi identity, which makes bracketing with a fixed element a derivation of a Lie algebra.

For Matrix Lie groups, the Lie bracket is the matrix commutator $[X,Y]=XY-YX$ on their tangent space. For the cross-product Lie algebra, it is $[x,y]=x\times y$. Both products measure a failure of infinitesimal motions to commute. In a real Lie group, a zero Lie bracket means that the two corresponding one-parameter subgroups commute for all real parameters.

In a local exponential chart, the mixed second differential of the group commutator defines a bilinear map on the tangent space at the identity. Inversion after swapping the two group elements proves antisymmetry. Differentiating conjugation gives $[X,Y]=\mathrm{ad}(X)Y$; applying naturality of the Lie bracket to the Adjoint representation of a Lie group then proves the Jacobi identity. For a matrix group this recovers $XY-YX$ by expanding through the mixed term.

Since the general linear Lie algebra ${\mathfrak{gl}}_1(k)$ has zero Lie bracket, identify it with $kz$. A central extension of a Lie algebra is a short exact sequenceof Lie algebra homomorphisms with $\iota (kz)$ contained in the center of a Lie algebra $Z(\tilde{\mathfrak{g}})$. Thus $\pi$ is surjective, its kernel is precisely the indicated line, and that line commutes with all of $\tilde{\mathfrak{g}}$.

Choose a linear map $s:\mathfrak{g}\to \tilde{\mathfrak{g}}$ with $\pi s=\mathrm{id}$. Such a section exists by lifting a basis. Identify $\tilde{\mathfrak{g}}$ with the direct sum of vector spaces $\mathfrak{g}\oplus kz$ by $(x,tz)\mapsto s(x)+t\iota (z)$. The difference $[s(x),s(y)]-s([x,y])$ belongs to the kernel of $\pi$, so there is a unique bilinear map $\alpha :\mathfrak{g}\times \mathfrak{g}\to k$ such thatIt is an alternating bilinear form: $\alpha (x,x)=0$, equivalently $\alpha (x,y)=-\alpha (y,x)$ in characteristic zero. Taking the central component of the Jacobi identity gives the Lie algebra two-cocycle conditionConversely, any alternating bilinear form satisfying this identity defines a Lie algebra byIndeed its Lie bracket is an alternating multilinear map and a bilinear map, and the two components of its Jacobi identity are respectively the Jacobi identity in $\mathfrak{g}$ and the displayed Lie algebra two-cocycle condition. The line $kz$ is central, and projection to $\mathfrak{g}$ gives the required central extension of a Lie algebra.

For later use, replacing $s$ by $s^{\prime }(x)=s(x)+t(x)\iota (z)$, where $t\in {\mathfrak{g}}^{\ast }$ is in the dual space, changes the Lie algebra two-cocycle toThus changes of section add a Lie algebra two-coboundary, up to the conventional sign. A section that is a Lie algebra homomorphism has ${\alpha }^{\prime }=0$ and gives a direct sum of Lie algebras.

Let $A$ be a three-dimensional vector space with basis $p,q,z$. Define a bilinear map $\bullet :A\times A\to A$ bywith all other products of basis vectors zero. Equivalently,Every product belongs to $kz$, and every product with an element of $kz$ is zero. Hence both $(u\bullet v)\bullet w$ and $u\bullet (v\bullet w)$ vanish for every $u,v,w\in A$. This proves associativity, so $A$ is an associative algebra. The product is an alternating multilinear map, and each nested product in the Jacobi identity is zero, so the same product is a Lie bracket. It is nonzero because $p\bullet q=z\ne 0$.

There is no multiplicative identity element: any candidate $1_A$ would satisfy $1_A\bullet z=0$, whereas an identity element would require $1_A\bullet z=z$. ThusThis is an associative Lie product; its underlying Lie algebra is the Heisenberg Lie algebra.

Continuity on the compact product of unit spheres bounds the bilinear map: there is $C$ such that $\Vert f(h,k){\Vert }_3\le C\Vert h{\Vert }_1\Vert k{\Vert }_2$. This follows by scaling each nonzero vector to unit norm; if one factor space is zero, the map and the relevant assertion are trivial.

Bilinearity gives the exact expansionThe proposed derivative $L(h,k)=f(h,b)+f(a,k)$ is linear and bounded, with $\Vert L(h,k){\Vert }_3\le C(\Vert b{\Vert }_2\Vert h{\Vert }_1+\Vert a{\Vert }_1\Vert k{\Vert }_2)$. Set $\rho =(\Vert h{\Vert }_1^2+\Vert k{\Vert }_2^2{)}^{1/2}$. Its remainder obeysThus the derivative of a continuous bilinear map is the Fréchet derivative

If a linear map satisfies $\Vert Tx\Vert \le M\Vert x\Vert$, then $\Vert Tx-T{x}^{\prime }\Vert \le M\Vert x-x^{\prime }\Vert$, so it is continuous. Conversely continuity at zero gives $\delta >0$ such that $\Vert x\Vert <\delta$ implies $\Vert Tx\Vert <1$. Apply this to $\delta x/(2\Vert x\Vert )$ for $x\ne 0$ to obtain $\Vert Tx\Vert \le 2\Vert x\Vert /\delta$. Hence a linear map is continuous exactly when it is bounded.

For the separately continuous bilinear map, put $T_x(y)=F(x,y)$ for $\Vert x\Vert \le 1$. Every $T_x:Y\to Z$ is a bounded linear map by the first result. For each fixed $y$, continuity in the other variable gives $\Vert F(x,y)\Vert \le C_y\Vert x\Vert$, so ${\sup }_{\Vert x\Vert \le 1}\Vert T_{x}y\Vert <\infty$. The Uniform boundedness principle therefore gives ${\sup }_{\Vert x\Vert \le 1}\Vert T_x\Vert =M<\infty$, and bilinearity then yields

To spell out the uniform-boundedness step, define the closed subsets $E_m=\{y:{\sup }_{\Vert x\Vert \le 1}\Vert T_{x}y\Vert \le m\}$ of the Banach space $Y$. They cover $Y$. The Baire category theorem gives an $E_m$ containing a ball $B(y_0,r)$. For $\Vert h\Vert <r$, both $y_0+h$ and $y_0$ lie in $E_m$, so ${\sup }_x\Vert T_{x}h\Vert \le 2m$. Scaling $h=(r/2)y$ for $\Vert y\Vert \le 1$ bounds all the operator norms by $4m/r$. This proves the claimed uniform estimate without presuming joint continuity.

Let $p=(x_0,y_0)$, $a=f_x(p)$ and $b=f_y(p)$. For sufficiently small $h,k$, decompose the increment into an $x$-segment at height $y_0+k$ and a $y$-segment at $x_0$. The one-variable mean value theorem applies on each segment because existence of its partial derivative gives continuity along the segment. Thus, when both increments are nonzero,for some $0<\theta ,\eta <1$. The same formula with the absent segment omitted covers zero increments. Continuity of both partial derivatives at $p$ makes the remainder after subtracting $ah+bk$ at most $\epsilon (|h|+|k|)$ for small increments, and hence at most $\sqrt{2}\epsilon \sqrt{h^2+k^2}$. Therefore the Fréchet derivative is $(h,k)\mapsto ah+bk$. Continuity throughout the disc is not required; continuity at the point suffices.

For the determinant, expand by multilinearity of the determinant in the columns of $I+H$. Taking no perturbed column gives one; taking exactly one gives $\mathrm{tr}H$; all other terms have degree at least two in the entries of $H$. Since matrix-entry functionals are bounded in any finite-dimensional norm, their sum is $O(\Vert H{\Vert }^2)$. HenceFor invertible $A$, multiplicativity of the determinant gives $\det (A+H)=\det A\det (I+A^{-1}H)$. The bounded linear map $H\mapsto A^{-1}H$ transfers the preceding remainder estimate, giving the derivative of the determinantTo justify an open invertible neighborhood, let $\Vert K\Vert <1$. Completeness and submultiplicativity make the Neumann series $S={\sum }_{j=0}^{\infty }(-K{)}^j$ converge in matrix norm. Multiplying finite partial sums by $I+K$ gives $I-(-K{)}^{N+1}$, which tends to $I$. Continuity of multiplication proves $(I+K)S=S(I+K)=I$. Thus $I+K$ is invertible, and the series gives $(I+H{)}^{-1}=I-H+O(\Vert H{\Vert }^2)$ near zero.

Apply the first-derivative formula at $I+H$ to a test matrix $K$:The last estimate is uniform for $\Vert K\Vert \le 1$, so it proves differentiability of the derivative in the operator norm, rather than merely a directional second derivative. The second derivative of the determinant is the symmetric bilinear mapIts symmetry follows from $\mathrm{tr}(HK)=\mathrm{tr}(KH)$.

The Uniform boundedness principle states that a pointwise bounded family $\mathcal{T}$ of bounded linear maps from a Banach space $X$ to a normed space $Y$ is uniformly bounded in operator norm. Define closed sets $E_n=\{x:{\sup }_{T\in \mathcal{T}}\Vert Tx\Vert \le n\}$. Pointwise boundedness makes their union $X$. The Baire category theorem gives a ball $B(x_0,r)\subset E_n$ for some $n$. For $\Vert v\Vert <r$, both $x_0+v$ and $x_0$ lie in $E_n$, so $\Vert Tv\Vert \le 2n$ for every $T$. Scaling yields ${\sup }_T\Vert T\Vert \le 2n/r$, proving the principle.

For the separately continuous bilinear map, consider $T_x(y)=F(x,y)$ with $\Vert x\Vert \le 1$. Each $T_x$ is bounded and linear. For fixed $y$, continuity of the other linear map gives ${\sup }_{\Vert x\Vert \le 1}\Vert F(x,y)\Vert <\infty$. Apply the Uniform boundedness principle to this family to getBilinearity then gives $F(x,y)-F(x_0,y_0)=F(x-x_0,y)+F(x_0,y-y_0)$; the bound proves joint continuity.

For the trilinear map, fix $w$. The just-proved bilinear result bounds $G(\cdot ,\cdot ,w)$. Thus the bounded linear maps $T_{x,y}:w\mapsto G(x,y,w)$, for $\Vert x\Vert ,\Vert y\Vert \le 1$, are pointwise bounded. Apply the Uniform boundedness principle once more to obtain $\Vert G(x,y,w)\Vert \le C\Vert x\Vert \Vert y\Vert \Vert w\Vert$. Expansion of the difference in three slots proves joint continuity. Separate continuity suffices for these multilinear maps.

Without linearity it does not suffice. On ${\mathbb{R}}^2$ set $H(x,y)=xy/(x^2+y^2)$ away from the origin and $H(0,0)=0$. Each coordinate section is continuous, but $H(t,t)=1/2$ for $t\ne 0$, so $H$ is not continuous at the origin.

The bundle is on $A$, so the printed, undefined target $\mathrm{Pic}(X)$ must be read as the Picard group $\mathrm{Pic}(A)$. Write this Picard group additively and set $q(h)=[h^{\ast }L]$. The zero morphism pulls back $L$ to a constant one-dimensional vector space tensored with ${\mathcal{O}}_A$, so $q(0)=0$. The Theorem of the Cube, pulled back along $(f,g,h):A\to B^3$, givesThis is the bilinear cross-effect of a line bundle on an abelian variety: set $d(f,g)=q(f+g)-q(f)-q(g)$. Substitution into this identity givesThe definition gives $d(f,g)=d(g,f)$, so additivity holds in both arguments. Additivity includes negative multiples and the zero morphism. Hence $D_L$ is a symmetric bilinear map with values in $\mathrm{Pic}(A)$. The equalities refer to isomorphism classes of line bundles; without chosen rigidifications they are not asserted to be specified canonical isomorphisms of bundles.

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.

Construct the tensor product of sheaves by first forming the presheaf $U\mapsto \mathcal{F}(U){\otimes }_{{\mathcal{O}}_X(U)}\mathcal{G}(U)$, with restriction maps induced by those of the two sheaves of modules, and then applying sheafification. The local module actions are compatible with restrictions and therefore give the sheaf an ${\mathcal{O}}_X$-module structure. Its stalks areIndeed, a finite collection of germs of sheaf sections can be represented on a common neighbourhood, and every finite tensor relation holds on a sufficiently small neighbourhood. Equivalently, this construction represents bilinear maps of sheaves of modules that are balanced over the structure sheaf. Sections of the resulting sheaf need not themselves be tensors of global sections: that is why the sheafification step matters.

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.