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For vector subspaces $U,W$ of a common vector space, this is a surjective linear map. Its kernel of a linear map is $\{(x,-x):x\in U\cap W\}$, which has a linear isomorphism to $U\cap W$ by $x\mapsto (x,-x)$. The rank-nullity theorem and the direct sum basis give the dimension formula for a sum of subspaces. The paired domain keeps the two contributions separate even when their underlying vectors overlap.

For a vector subspace $W$ of $X$, its codimension is the dimension of the quotient vector space $X/W$. This definition also applies when $X$ is infinite-dimensional. If $X=W\oplus E$ is a direct sum, the map $e\mapsto e+W$ is a linear isomorphism from $E$ to $X/W$, so ${\mathrm{codim}}_{X}W=\dim E$. In particular, the kernel of a nonzero linear functional has codimension one: normalize $Te=1$ and write $x=(x-(Tx)e)+(Tx)e$.

The algebraic double dual of a vector space is the dual space of its algebraic dual. An element assigns a scalar linearly to each linear functional on the original space. The bidual evaluation map sends $v$ to the rule $\ell \mapsto \ell (v)$. It is a canonical linear isomorphism in finite dimension, unlike an identification of $V$ with $V^{\ast }$, which generally requires an additional choice. For spaces with topology, using the continuous dual space instead defines the topological bidual; that is a distinct construction in infinite dimension.

An $m\times n$ matrix has full row rank when its matrix rank is $m$, equivalently when its rows are linearly independent vectors. This requires $m\le n$. Its column space is all of ${\mathbb{F}}^m$, so the represented linear map is surjective and every linear system $Ax=b$ is solvable. If $n>m$, its nullity is $n-m>0$, so solutions are not unique.^[ref]

For example, the matrix transpose of the matrix in full column rank isIt has full row rank two. Every output $(p,q{)}^T$ has solutions $(p-t,q-t,t{)}^T$, with $t\in \mathbb{R}$. A square matrix of size $n$ has full row rank and full column rank together exactly when its matrix rank is $n$; then it is an invertible matrix. A nonsquare matrix cannot have both properties, and has no two-sided inverse: such an inverse would make its linear map a linear isomorphism between vector spaces of different dimensions.

An isometric immersion is an immersion $f:M\to N$ between Riemannian manifolds whose derivative preserves the inner products of tangent vectors; equivalently, $f^{\ast }{g}_N=g_M$. Its derivative is injective and the target may have larger dimension. A local isometry additionally requires a local diffeomorphism, so its tangent maps are linear isomorphisms. For example, a sphere with its induced Riemannian metric is isometrically immersed in Euclidean space, although its positive intrinsic sectional curvature differs from that of the ambient space.

A local isometry between Riemannian manifolds is a local diffeomorphism $f$ with $f^{\ast }{g}_N=g_M$. Equivalently, its differentials are isometric linear isomorphisms between tangent spaces. It preserves the Levi-Civita connection, intrinsic sectional curvature, and affinely parametrized geodesics. A map whose differentials are merely isometric injections is an isometric immersion; the local diffeomorphism requirement distinguishes these notions.

The annihilator of a vector subspace $W\le V^{\ast }$ isIf $\dim W=k$, extend a basis of $W$ to one of $V^{\ast }$. In the resulting dual basis of $V^{\ast \ast }$, the last $n-k$ coordinate functionals span $W^{\circ }$. Hence $\boxed{\dim W+\dim W^{\circ }=n}$.

The bidual evaluation map $j:V\to V^{\ast \ast }$ is $j(x)(\varphi )=\varphi (x)$. It is a linear isomorphism in finite dimension. Its inverse image of the annihilator isThis identification uses evaluation only; no choice of inner product is involved.

Define the bidual evaluation map by evaluation:For fixed $v$, this is a linear functional of $\ell$, so it belongs to the double dual. It is also linear as a function of $v$. If $v\ne 0$, some coordinate of $v$ in a basis is nonzero, and the corresponding dual basis functional detects it. Hence $J(v)\ne 0$, proving injectivity.

The spaces $V,V^{\prime },V^{\prime \prime }$ all have the same finite dimension. Therefore an injective linear map $J:V\to V^{\prime \prime }$ is surjective, proving the required linear isomorphism.

This linear isomorphism is natural because evaluation requires no basis. More explicitly, for any linear map $A:V\to W$, its dual map satisfiesso $A^{\prime \prime }{J}_V=J_{W}A$.

Write $V^{\ast }=V^{\prime }$ for the dual space. Define $\theta (v)(f)=f(v)$. For fixed $v$, this is a linear functional on $V^{\ast }$; evaluation is also linear in $v$, so $\theta :V\to V^{\ast \ast }$ is a linear map. If $v\ne 0$, extend $v$ to a basis and take its first coordinate linear functional $f$; then $f(v)=1$. Thus $\theta$ is injective. A dual basis has the same number of elements as the original basis, so $\dim V=\dim V^{\ast }=\dim V^{\ast \ast }$. The rank-nullity theorem now makes $\theta$ an linear isomorphism.

The annihilator of a vector subspace of a linear subspace $U\le V$ isExtend a basis $v_1,\dots ,v_k$ of $U$ to a basis $v_1,\dots ,v_n$ of $V$, with dual basis $f_1,\dots ,f_n$. A linear functional belongs to $U^{\circ }$ exactly when its first $k$ coordinates in this dual basis vanish. Hence $U^{\circ }$ has basis $f_{k+1},\dots ,f_n$, provingThe dual map is ${\alpha }^{\ast }(h)=h\circ \alpha$; here this is the algebraic dual map, denoted ${\alpha }^{\prime }$ in the source. If $v\in \ker \alpha$, then $({\alpha }^{\prime }h)(v)=h(\alpha v)=0$. Consequently $\mathrm{im}{\alpha }^{\prime }\subseteq (\ker \alpha {)}^{\circ }$. Moreover,because coordinate linear functionals separate vectors. This proves $(\mathrm{im}{\alpha }^{\prime }{)}^{\circ }=\theta (\ker \alpha )$, including the requested second inclusion. Taking dimensions givesThe first inclusion has equal dimensions on both sides, soFor a matrix $A=(a_{ij})$ representing $\alpha$, the matrix of ${\alpha }^{\prime }$ in the corresponding dual bases is $A^{\mathsf{T}}$: its $ji$ entry is $h_i(\alpha v_j)=a_{ij}$. Thus the column rank of $A^{\mathsf{T}}$ is the row rank of $A$, while the column rank of $A$ is $\dim \mathrm{im}\alpha$. Applying the established equality to these two linear maps proves row rank equals column rank over every field.

The adjoint on an inner product space ${\alpha }^{\ast }:V\to V$ is characterized byHere is a finite-dimensional construction that supplies every required basis fact. If $v_1,\dots ,v_n$ is any basis, its coordinate linear functionals form a basis of $V^{\ast }$: a linear functional is uniquely specified by its values on these $n$ vectors. Hence $\dim V^{\ast }=n$. The linear mapis injective, because $J(w)=0$ implies $⟨w,w⟩=0$, and positive definiteness implies $w=0$. Equal dimensions make $J$ an linear isomorphism. If ${\alpha }^{\prime }$ is the dual map, thenThis establishes existence and linearity. If two possible vectors $u,w$ represent the same linear functional through the inner product, then $J(u-w)=0$, so $u=w$. Applied to every argument, this proves uniqueness of the adjoint on an inner product space.

For linear maps $S,T$, two applications of the defining inner product identity giveUniqueness therefore gives $(ST{)}^{\ast }=T^{\ast }{S}^{\ast }$. Likewise $(S+T{)}^{\ast }=S^{\ast }+T^{\ast }$, $(cT{)}^{\ast }=c{T}^{\ast }$ for real $c$, and $I^{\ast }=I$. An induction gives $({\alpha }^n{)}^{\ast }=({\alpha }^{\ast }{)}^n$, and applying additivity to a polynomial with real coefficients givesSymmetry of the real inner product also gives $({\alpha }^{\ast }{)}^{\ast }=\alpha$. If $m$ is the minimal polynomial of $\alpha$, the displayed identity gives $m({\alpha }^{\ast })=0$, so the minimal polynomial of ${\alpha }^{\ast }$ divides $m$. Applying the same argument to ${\alpha }^{\ast }$ gives reverse divisibility. Both polynomials are monic, hence their minimal polynomials are equal.

The distinct eigenvalues are exactly the roots of the minimal polynomial in the chosen field. An eigenvector $v$ with eigenvalue $\lambda$ satisfies $m(\alpha )v=m(\lambda )v$, so $m(\lambda )=0$. Conversely, if $m(X)=(X-\lambda )q(X)$ and $\alpha -\lambda I$ were a linear isomorphism, then $m(\alpha )=0$ would imply $q(\alpha )=0$, contradicting minimal degree. Thus $\alpha -\lambda I$ is not injective, giving an eigenvector. It follows that $\alpha$ and ${\alpha }^{\ast }$ have the same distinct eigenvalues. This includes the complex eigenvalues after complexification of a real vector space; over the real field it states equality of their real eigenvalues.

The annihilator of a vector subspace isChoose a basis $e_1,\dots ,e_r$ of $U$ and extend it to a basis $e_1,\dots ,e_d$ of $V$. In the dual basis, precisely $f_{r+1},\dots ,f_d$ span $U^{\circ }$. ThereforeThe canonical restriction linear transformationhas kernel of a linear map $U^{\circ }$. It is surjective: extend any linear functional on $U$ by assigning arbitrary values, for instance zero, to the additional basis vectors. The First isomorphism theorem for vector spaces gives the explicit linear isomorphismIt is natural because the displayed map itself makes no basis choice; a basis was used only to prove surjectivity. In particular, $U^{\prime }$ is canonically a quotient vector space of $V^{\prime }$, rather than canonically a subspace of $V^{\prime }$.

For the given nondegenerate bilinear form, define the linear mapBy part (a), $\Phi$ is a linear isomorphism. The paper's bilinear orthogonal complement uses $U$ in the first argument; with this convention,Here $U^0$ is the annihilator of a vector subspace $U$. Since $\Phi$ is a linear isomorphism, its restriction gives a linear isomorphism $U^{\perp }\to U^0$.

For completeness, choose a basis $u_1,\dots ,u_k$ of $U$ and extend it to a basis $u_1,\dots ,u_n$ of $V$. In its dual basis $u_1^{\ast },\dots ,u_n^{\ast }$, a linear functional vanishes on $U$ exactly when its first $k$ coefficients are zero. ThereforeConsequentlyThe argument includes $U=0$ and $U=V$, and requires no symmetry assumption on the bilinear form.

A smooth vector field on a smooth manifold $M$ is a smooth section of a vector bundle $X:M\to TM$ of the tangent bundle projection $\pi$, so $\pi \circ X={\mathrm{id}}_M$. In a coordinate chart it has the form $X={\sum }_i{X}^i{\partial }_i$ with smooth component functions $X^i$.

Suppose a smooth vector bundle isomorphism $\Phi :M\times {\mathbb{R}}^n\to TM$ covering the identity is given. For the standard basis $e_1,\dots ,e_n$ of ${\mathbb{R}}^n$, put $X_i(p)=\Phi (p,e_i)$. These are smooth vector fields, and the linear isomorphism on each fibre makes $X_1(p),\dots ,X_n(p)$ a basis of $T_{p}M$.

Conversely, given such a global frame of a vector bundle, defineIt is smooth, covers the identity and is a linear isomorphism on each fibre. Its inverse is smooth as well: in any coordinate chart, the component columns of the $X_i$ form an invertible smooth matrix $C(p)$, and the fibre coordinates of the inverse are $C(p{)}^{-1}v$. matrix inversion is smooth on the invertible matrices. Thus the tangent bundle is trivial exactly when a global frame exists, the defining property of a parallelizable manifold.

For a Lie group $G$, write $L_h(g)=hg$. A left-invariant vector field is a field satisfyingSuch a field is determined by $v=X(e)\in T_{e}G$, since $X(g)=(d{L}_g{)}_{e}v$. Conversely, this formula defines a left-invariant field, by the chain rule and $L_h{L}_g=L_{hg}$.

It also proves smoothness even if smoothness was not initially assumed. Choose a smooth curve $c(t)$ with $c(0)=e$, $c^{\prime }(0)=v$. ThenMultiplication is a smooth map on a Lie group, so differentiating its coordinate functions in the second variable yields a smooth function of $g$. This gives smooth components for $X$ in every local chart.

Choose a basis $v_1,\dots ,v_m$ of $T_{e}G$, where $m=\dim G$, and left translate it. Each $L_g$ is a diffeomorphism, with inverse $L_{g^{-1}}$, so $(d{L}_g{)}_e$ is a linear isomorphism and the translated fields are a basis at every $g$. The explicit parallelization of a Lie group by left translations isIts inverse sends $w\in T_{g}G$ to $(g,(d{L}_{g^{-1}}{)}_{g}w)$. Identifying $T_{e}G$ with ${\mathbb{R}}^m$ gives the required product bundle. The differential facts used are linearity, the chain rule, $d(\mathrm{id})=\mathrm{id}$ and smooth dependence of the differential of a smooth map on its base point.

A Riemannian isometry $F:(M,g)\to (N,h)$ is a diffeomorphism satisfying $F^{\ast }h=g$. A local isometry is a local diffeomorphism with the same pullback identity; equivalently $d{F}_p$ is an isometric linear isomorphism for every $p$. An isometric immersion into a higher-dimensional manifold does not satisfy this local-diffeomorphism requirement.

Let ${\varphi }_t$ be the local flow of the generating vector field $K$. The identity ${\varphi }_t^{\ast }g=g$ differentiates to ${\mathcal{L}}_{K}g=0$, where $\mathcal{L}$ is the Lie derivative of a tensor field. For arbitrary vector fields $X,Y$, metric compatibility and the torsion-free connection giveThis proves the Killing equation. Conversely, a smooth Killing field has a local flow, andSince ${\varphi }_0$ is the identity, ${\varphi }_t^{\ast }g=g$ wherever the flow is defined. Its inverse is ${\varphi }_{-t}$ locally, so the local flow consists of local isometries. Neither direction assumes that $K$ is complete.

Put $H(X,Y)={\nabla }_X({\nabla }_{Y}K)-{\nabla }_{{\nabla }_{X}Y}K$ and $B(X,Y,Z)=⟨H(X,Y),Z⟩$. These expressions are tensorial in $X,Y,Z$. Take the covariant derivative in direction $X$ of the Killing equation, subtracting the terms from differentiating $Y$ and $Z$. It follows thatWith the curvature convention $\mathcal{R}$ of Solution 1, the torsion-free connection also givesDefine $C(X,Y,Z)=⟨\mathcal{R}(X,K)Y,Z⟩$. It is skew in $Y,Z$ by skew-adjointness of the Riemann curvature tensor. The first Bianchi identity givesso $C$ has the same antisymmetrization in $X,Y$ as $B$. Consequently $D=B-C$ is symmetric in its first two slots and skew in its last two. These symmetries force it to vanish:We have proved the second covariant derivative of a Killing vector identity, including its sign:The printed identity uses the opposite curvature-sign convention in Killing derivative identities. If $R=-\mathcal{R}$, namely $R(X,Y)Z={\nabla }_Y{\nabla }_{X}Z-{\nabla }_X{\nabla }_{Y}Z+{\nabla }_{[X,Y]}Z$, the same result reads ${\nabla }_{X,Y}^{2}K=R(K,X)Y$, exactly as printed. The convention must change with the sign; under $\mathcal{R}$, writing $\mathcal{R}(K,X)Y$ would be incorrect.

Finally let $H_0=K-\tilde{K}$ have zero value and zero first covariant derivative at $p$. Along any geodesic issuing from $p$, the identity just proved saysThus $H_0$ is a Jacobi field with zero initial value and derivative. Uniqueness for this linear ordinary differential equation implies $H_0=0$ on each such geodesic, hence on a convex normal neighborhood of $p$. Consider the set of points where both $H_0$ and $\nabla H_0$ vanish. It is closed by continuity and open by this same local argument; on the resulting neighborhood $H_0$ vanishes identically, so its derivative does too. It is nonempty, and $M$ is connected, therefore it is all of $M$. This proves $K=\tilde{K}$ everywhere and the general statement that a Killing field is determined by its value and first covariant derivative.

For $v=(v_1,v_2,v_3)$ define the skew-symmetric matrixThen $\hat{v}x=v\times x$. This is a linear isomorphism ${\mathbb{R}}^3\to \mathfrak{so}(3)$, with inverse $\varphi (\xi )=({\xi }_{32},{\xi }_{13},{\xi }_{21})$. Since rotations preserve the cross product,Thus $\varphi$ intertwines the Adjoint representation with the standard representation. The vector triple-product identity gives, for every $x$,Consequently $[\hat{v},\hat{w}]=\widehat{v\times w}$ and the chosen normalization satisfiesNo undetermined scale remains in this explicit matrix convention.

Identify $SU(2)$ with the unit quaternions usingThe relations $jz=\overline{z}j$ and $j^2=-1$ verify multiplication, and every matrix in $SU(2)$ has this form. If $v$ is imaginary, then $\overline{v}=-v$. Quaternionic conjugation reverses product order and $q^{-1}=\overline{q}$, soThus conjugation preserves $\mathrm{Im}\mathbb{H}\cong {\mathbb{R}}^3$. Multiplicativity of the quaternion norm shows that this action preserves the inner product. The unit quaternions form the connected three-sphere, so its determinant, equal to one at the identity, is always one. We obtain a homomorphism $C:SU(2)\to SO(3)$.

Its group kernel consists of unit quaternions commuting with every imaginary quaternion. Commuting with both $i$ and $j$ forces a quaternion to be real, so the group kernel is exactly $\{\pm 1\}$. To prove surjectivity explicitly, let $u$ be a unit imaginary quaternion and put $q=\cos (\theta /2)+u\sin (\theta /2)$. For imaginary $u,v$, multiplication satisfies $uv=-u\cdot v+u\times v$. Expanding $qv{q}^{-1}$ therefore givesThis is the Rodrigues rotation formula about axis $u$. Every element of $SO(3)$ has such an axis-angle description: an odd-dimensional orthogonal matrix of determinant one has an eigenvector of eigenvalue one, and its action on the perpendicular plane is a plane rotation. Hence $C$ is onto, andThis is the quaternionic form of the Adjoint double cover from SU(2) to SO(3). Its differential sends an imaginary quaternion $u$ to $2\hat{u}$, because $[u,v]=2u\times v$, consistent with the bracket normalization used above.

A smooth rank-$r$ vector bundle over a smooth manifold $M$ consists of a smooth total space $E$, a smooth projection $\pi :E\to M$, and an $r$-dimensional real vector space structure on each fibre $E_p={\pi }^{-1}(p)$. Each point has a neighborhood $U$ with a vector bundle trivializationthat is a diffeomorphism, commutes with projection to $U$, and is a linear isomorphism on every fibre. The overlap maps have the form $(p,v)\mapsto (p,h(p)v)$ with $h:U\cap V\to G{L}_r(\mathbb{R})$ smooth. For a complex vector bundle, replace $\mathbb{R}$ by $\mathbb{C}$ and require complex-linear fibre maps.

A connection on a vector bundle is an $\mathbb{R}$-linear mapwhere $\Gamma (E)$ is the module of smooth sections, ${\Omega }^1(M;E)$ consists of vector-bundle-valued differential forms of degree one, and $f$ is a smooth function. Equivalently, evaluating on a vector field $X$ gives operators ${\nabla }_X$ with ${\nabla }_{fX}s=f{\nabla }_{X}s$ and ${\nabla }_X(fs)=X(f)s+f{\nabla }_{X}s$. A connection on a vector bundle is local: its value at a point depends on the germ of the section of a vector bundle, so the same definition applies to local sections.

A quadratic form on a finite-dimensional real vector space is a function $q(v)=B(v,v)$ for a symmetric bilinear form $B$. Conversely $B(u,v)=\frac{1}{2}(q(u+v)-q(u)-q(v))$ recovers that bilinear form. Two quadratic forms are congruent if a linear isomorphism $T$ satisfies $q_2(v)=q_1(Tv)$; in coordinates this is matrix congruence, $A_2=P^T{A}_{1}P$, with $P$ invertible.

Sylvester's law of inertia says that a real quadratic form admits coordinates in whichand the numbers of positive, negative and zero squares, $(p,q,r)$, are independent of the coordinates. Here $p+q+r=\dim V$; $p$ and $q$ are the maximal dimensions of subspaces on which the quadratic form is positive definite and negative definite, while $r$ is the dimension of its radical of a bilinear form. These numbers are the inertia of a bilinear form. The rank of a quadratic form is $p+q$, and the signature of a quadratic form, in the signed-difference convention, is $p-q$.

Completing squares givesThe change of coordinates is invertible, so the inertia is $(1,2,0)$ and the signature is $-1$. If signature is recorded as an ordered pair instead, that pair is $(1,2)$.

True. Use the lifted basis $u_1,\dots ,u_r$ above and put $W=\mathrm{span}\{u_1,\dots ,u_r\}$. If $\sum c_i{u}_i=0$, applying $\alpha$ gives $\sum c_i{v}_i=0$, so all $c_i=0$. Thus the $u_i$ form a basis of $W$, and $\alpha {|}_W$ takes this basis to the basis of $V$. It is a linear isomorphism.

In fact the right inverse gives the stronger splittingFor every $u\in U$, write $u=(u-\beta \alpha u)+\beta \alpha u$; the first term is in the kernel and the second in $W$. Their intersection is zero because $\alpha {|}_W$ is injective. This is a direct sum decomposition.

A smooth real vector bundle of vector bundle rank $r$ consists of a smooth manifold $E$, a smooth projection $\pi :E\to M$, real vector space structures on its fibers, and local vector bundle trivializationsthat commute with projection to $U$ and are linear isomorphisms on each fiber. On overlaps the change of trivialization is $(p,v)\mapsto (p,g_{VU}(p)v)$, with $g_{VU}:U\cap V\to G{L}_r(\mathbb{R})$ smooth into the general linear group. The usual smooth manifold convention is Hausdorff and second countable.

For an $n$-dimensional $M$, take the disjoint union of its tangent spaces. A manifold chart $x:U\to {\mathbb{R}}^n$ suppliesIf $y$ is another chart, the resulting change of coordinates on the total space isThis is smooth, its inverse is the same construction with $x,y$ exchanged, and the fiber map is invertible and linear. Thus these charts define a smooth structure with the required bundle trivializations.

For completeness, the topology so constructed is Hausdorff: different base points can be separated downstairs, while vectors over one point can be separated inside one product chart. A countable base atlas for $M$ and countable bases for its products with ${\mathbb{R}}^n$ give second countability. Hence the total space is genuinely a smooth manifold, not just a collection of fibers. We obtain

The fibers must retain their base labels. In particular, a literal untagged union of fibers in $E$ is not sufficient when $f$ is not injective. Use the pullback vector bundleThis makes precise the indexed-union notation in the PDF. For a vector bundle trivialization ${\Phi }_U(e)=(\pi (e),v)$, defineIt is a fiberwise linear isomorphism, with inverse $(p,v)\mapsto (p,{\Phi }_U^{-1}(f(p),v))$. The transition functions are $g_{VU}\circ f$, hence smooth. They satisfy the same cocycle identities as those of $E$ and give a smooth total space of manifold dimension $\dim N+r$.

One may also see that this is an embedded submanifold of $N\times E$: in a bundle chart the constraint is the graph of the smooth map $f$ in the base coordinates, leaving the $r$ fiber coordinates free. Thus the chart construction agrees with its natural subspace smooth structure. No immersion or surjectivity hypothesis on $f$ is required. Therefore