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Let $E\to M$ have rank $r$. A vector bundle covariant derivative is a real-linear operatorPairing its one-form slot with a vector field $X$ gives ${\nabla }_{X}s$, with ${\nabla }_{fX}s=f{\nabla }_{X}s$ and ${\nabla }_X(fs)=X(f)s+f{\nabla }_{X}s$.

In the horizontal description, a connection on a vector bundle $A$ is a smooth splitting $T_{a}E=V_{a}E\oplus H_{a}E$, where $V_{a}E=\ker (d\pi {)}_a$, and the horizontal choice is compatible with fibre addition and scalar multiplication. This linearity is essential; an arbitrary nonlinear horizontal splitting is not a vector-bundle connection. Identify a vertical vector at $a\in E_x$ with the element of $E_x$ obtained by differentiating the curve $a+tv$. Let $K_A:T_{a}E\to E_x$ be vertical projection followed by this identification. The induced vector bundle covariant derivative isIn a local frame of a vector bundle $e=(e_1,\dots ,e_r)$, a linear horizontal splitting has the formfor a matrix-valued connection one-form $\Gamma$. Thus for $s=ev$, ${\nabla }^{A}s=e(dv+\Gamma v)$, which directly verifies the Leibniz rule.

Conversely, the Leibniz rule makes $\nabla$ local: if a section vanishes near a point, multiply it by a smooth cutoff function supported in that neighborhood and equal to one near the point to see that its derivative vanishes there too. Local sections can therefore be differentiated by any extension agreeing near the point. Given $\nabla$, define its local connection one-form by $\nabla e_b={\sum }_a{e}_a{\Gamma }_b^a$. The Leibniz rule gives $\nabla (ev)=e(dv+\Gamma v)$. Under $e^{\prime }=eg$, $v^{\prime }=g^{-1}v$, one hasThe tangent fibre coordinate transforms by $w^{\prime }=g^{-1}w-g^{-1}(dg)(\xi )g^{-1}v$, so $w=-\Gamma (\xi )v$ transforms precisely to $w^{\prime }=-{\Gamma }^{\prime }(\xi )v^{\prime }$. Consequently these local horizontal subspaces of a vector bundle connection glue to a smooth linear splitting inducing the given $\nabla$. Its local formula also shows uniqueness. This proves the horizontal connection associated to a covariant derivative construction.

A horizontal lift $\tilde{\gamma }$ of $\gamma$ is a smooth curve in $E$ with $\pi \circ \tilde{\gamma }=\gamma$ and $\dot{\tilde{\gamma }}(t)\in H_{\tilde{\gamma }(t)}E$. In a local frame write $\tilde{\gamma }(t)=e(\gamma (t))v(t)$. Horizontality is exactly the linear ordinary differential equationWe use existence and uniqueness for an ODE with smooth coefficients, and the continuation theorem that a solution extends across a finite interval when it remains bounded inside the coordinate domain. For a linear system $\dot{v}=-C(t)v$ with continuous coefficients bounded by $L$ on a compact time interval, the Gronwall inequality bounds $|v(t)|$ by its initial norm times $e^{L|t-t_0|}$, preventing finite-time fibre escape. Cover a compact subinterval of $I$ by finitely many bundle charts and partition it so each piece lies in one chart. Solve successively, changing fibre frames at the partition points. Coordinate invariance of the horizontal equation and uniqueness make the pieces agree. Exhausting $I$ by compact intervals containing $0$ produces a unique horizontal lift on all of $I$ for every prescribed $a_0$. This is global continuation of linear parallel transport.

The lift equation is linear in its initial value. Hence the endpoint map $P_{\gamma }:E_{\gamma (0)}\to E_{\gamma (1)}$ is linear. Solving the same equation backwards, or transporting along $t\mapsto \gamma (1-t)$, gives its inverse, soThis is parallel transport. For two parallel lifts $s_1(t),s_2(t)$, the assumed metric connection identity extends along the embedded curve and becomesHere $D_t$ is the covariant derivative along a curve. The permitted extension of sections and the tangent vector along the embedded curve justifies applying the bundle identity; equivalently it follows in each local frame. Thereforeso parallel transport is an isometry. This is the principle that parallel transport preserves a fibre metric.

Extend $J$ complex-linearly to the complexification of a real vector bundle $T_{\mathbb{C}}X=TX{\otimes }_{\mathbb{R}}\mathbb{C}$. Since $J^2=-\mathrm{id}$, its $+i$ and $-i$ eigenbundles have smooth projectionsEach has complex rank $n$: complex conjugation interchanges them, and together they have rank $2n$. These are the type decomposition of the complexified tangent bundle. For an arbitrary almost complex structure, they are smooth complex vector bundles; a holomorphic vector bundle structure requires integrability.

When $J$ comes from a holomorphic atlas, write a holomorphic coordinate as $z_j=x_j+i{y}_j$. Its Wirtinger derivatives areThe induced almost complex structure has $J{\partial }_{x_j}={\partial }_{y_j}$ and $J{\partial }_{y_j}=-{\partial }_{x_j}$. Hence the displayed vector fields are respectively $+i$ and $-i$ eigenvectors. They are linearly independent, and each collection has $n$ elements. Thus they give local frames for $T^{1,0}X$ and $T^{0,1}X$ respectively, with $T^{1,0}X$ the holomorphic tangent bundle. The repeated $T^{0,1}X$ in the first sentence of the printed item must be read as the two complementary eigenbundles.

With the normalization used here, $N=2N_J$, where $N_J$ is the standard Nijenhuis tensor. First check that it is a tensor: the Lie bracket identity $[fU,V]=f[U,V]-V(f)U$ givesSkew symmetry gives the same $C^{\infty }$-linearity in the second input. In coordinates from a holomorphic atlas, the real coordinate vector fields ${\partial }_{x_j},{\partial }_{y_j}$ commute, and $J$ sends each to another such field with a constant sign. Every Lie bracket in $N_J$ therefore vanishes on these coordinate fields. Tensoriality then gives $\boxed{N=0}$ on all smooth vector fields. The factor two in this paper does not change this vanishing conclusion.

Use the complex-bilinear extension of the Nijenhuis tensor expression to complex vector fields. If $U,V\in T^{1,0}X$, then $JU=iU$ and $JV=iV$, soFor $U,V\in T^{0,1}X$, the same substitution with $-i$ gives $N(U,V)=-8[U,V{]}^{1,0}$. For inputs of opposite types, the two leading Lie brackets cancel, as do the two terms involving $J$, so $N(U,V)=0$. Applying the type decomposition of the complexified tangent bundle to both inputs consequently givesThus $N=0$ precisely when the Lie bracket of any two sections of either eigenbundle remains in that same eigenbundle. In other words, $N=0$ if and only if both tangent-type distributions are involutive distributions. This proves the equivalence directly and does not assume that an arbitrary almost complex structure already has holomorphic coordinates.

Choose local coordinates $y^1,\dots ,y^{D-1}$ on the hypersurface $\Sigma$, and let ${\Phi }_s$ be the local flow of the vector field $X$. The map $F(s,y)={\Phi }_s(p(y))$ follows the integral curves of a vector field starting at $p(y)\in \Sigma$. At $s=0$, its differential maps the $s$ direction to $X$ and the remaining directions to a basis of $T_p\Sigma$. Transversality makes these vectors linearly independent, so the inverse function theorem makes $F$ a local manifold chart.

Set $x^0=s$ and transport the $y^i$ unchanged along each integral curve of a vector field. Then the adapted coordinates satisfyThe flow-box theorem gives this construction locally. The parameter is the flow parameter, and need not be arc length; a global manifold chart would require additional hypotheses.

Contract the Christoffel symbol formula for the Levi-Civita connection. The first and third derivative terms cancel after relabelling their summed indices:The Jacobi determinant derivative formula gives ${\partial }_{\mu }g=g{g}^{\rho \sigma }{\partial }_{\mu }{g}_{\rho \sigma }$. In a Lorentzian chart $g<0$, and hence ${\partial }_{\mu }\sqrt{-g}=\frac{1}{2}\sqrt{-g}{g}^{\rho \sigma }{\partial }_{\mu }{g}_{\rho \sigma }$. Thus the trace is the logarithmic derivative of the volume density:In particular, the covariant divergence formula for a vector field is ${\nabla }_{\mu }{A}^{\mu }=(-g{)}^{-1/2}{\partial }_{\mu }(\sqrt{-g}{A}^{\mu })$.

Rescale the ambient coordinates by $q_i=a_i{r}_i$. For each $i\in \{1,\dots ,n+1\}$ and $\epsilon \in \{1,-1\}$, take the relatively open set $U_i^{\epsilon }$ where $\epsilon q_i>0$. Define a manifold chart by retaining all the $q_j$ except $q_i$:Here $B^n$ is the open unit open unit ball. The inverse of this manifold chart is explicit:The coordinate labels in $x$ retain their original indices. All the $a_i$ are nonzero, so this inverse exists; the strictly positive radicand makes it smooth on the whole open open unit ball. Projection and this inverse are continuous, so each manifold chart is a homeomorphism onto an open subset of ${\mathbb{R}}^n$.

These $2(n+1)$ manifold charts cover the ellipsoid, because at every point at least one $q_i$ is nonzero. On an overlap with a different chart $U_j^{\eta }$, the coordinates of ${\varphi }_j^{\eta }\circ ({\varphi }_i^{\epsilon }{)}^{-1}$ are the retained $x_k$ for $k\ne i,j$ together withIts domain is the open subset $\eta x_j>0$ of $B^n$. The displayed functions, and the reverse transition functions obtained by exchanging $i,j$, are smooth. Transitions between identical charts are the identity; opposite-sign charts with the same omitted coordinate do not overlap. The ambient subspace topology is Hausdorff and second countable, as it is inherited from Euclidean space. Thus this coordinate-projection atlas of an ellipsoid is a smooth atlas, proving the ellipsoid is a smooth manifold of dimension $n$. The rescaling $r\mapsto q$ also gives a diffeomorphism with the unit sphere $S^n$.

For the three-sphere, use its identification with SU(2) rather than separate local constructions. With the Pauli matrices ${\tau }_i$, put $T_i=-i{\tau }_i$. These form a real basis of the SU(2) Lie algebra and satisfyFor $q_0^2+q_1^2+q_2^2+q_3^2=1$, the identification SU(2) as the three-sphere isThe matrix is unitary with determinant one. Conversely, every SU(2) matrix has this form, and both the correspondence and its inverse are smooth.

The left translations $L_g(h)=gh$ are diffeomorphisms. Define three left-invariant vector fields byMatrix multiplication makes these vector fields smooth globally. Since $(d{L}_g{)}_e$ is an isomorphism from the SU(2) Lie algebra to $T_g\mathrm{SU}(2)$, the three vectors are linearly independent at every point, and each is nowhere zero. This is the parallelization of a Lie group by left translations.

Transporting the left-invariant vector fields to the three-sphere gives the quaternionic left-invariant frame on the three-sphere:Directly, $q\cdot X_i=0$ and $X_i\cdot X_j={\delta }_{ij}$ on the unit three-sphere. Thus the vectors are tangent and form a global orthonormal frame of a vector bundle. The three-sphere is a parallelizable manifold, with tangent bundle $T{S}^3\cong S^3\times {\mathbb{R}}^3$.

The characteristic polynomials of a linear multistep method areThe consistency of a numerical method conditions hold for every real $a$:Both roots of a polynomial of $\rho$, namely $1$ and $-1$, are simple and have modulus one. The root condition for a multistep method therefore gives zero-stability for every $a$. The Dahlquist equivalence theorem now gives convergence for every fixed $a\in \mathbb{R}$. As usual, this means convergence of a numerical method on each fixed finite interval for a sufficiently regular ordinary differential equation with a Lipschitz continuous vector field and starting values tending to the exact starting values. When $a\ne 0$, the implicit time-stepping method update is locally uniquely solvable for sufficiently small $h$, for example by a contraction mapping if $|ha|L<1$. Zero-stability does not assert that a large fixed step is suitable for a stiff differential equation.

Consider an initial value problem $y^{\prime }=f(t,y)$ on $[0,T]$, with a Lipschitz continuous vector field in the solution variable and enough smoothness for the stated error expansions. Convergence of a numerical method meanswith consistent starting values. An order of a numerical method $p$ quantifies this approximation: with sufficiently accurate starting data and the required stability, the global error is $O(h^p)$. Stability of a numerical method controls the amplification and accumulation of perturbations; it is the link between a small local defect and small global error.

For a one-step formula $Y_{n+1}={\Phi }_h(t_n,Y_n)$, define the unscaled local truncation errorThe method has local order $p$ if $\Vert d_{n+1}\Vert \le C{h}^{p+1}$ uniformly along sufficiently smooth solutions. If the defect is divided by $h$, its order is instead $O(h^p)$; the convention must be specified. Assume also the perturbation estimateThe global error $e_n=Y_n-y(t_n)$ then satisfiesIterating, or applying the discrete Gronwall inequality, givesThere are $O(1/h)$ steps on a fixed interval, so stable propagation turns $O(h^{p+1})$ local defects into $O(h^p)$ accumulated error. Starting error $O(h^p)$ is needed to retain order $p$. This explains why matching an exact Taylor expansion is insufficient unless errors are controlled during repeated steps.

For a fixed-coefficient linear multistep method, writeThe consistency of a numerical method conditions are $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)$. The higher order conditions for a linear multistep method arewith right side zero for $q=0$; exact order $p$ means first failure at $q=p+1$. They come from inserting the exact solution and comparing its Taylor expansion.

Errors at $h=0$ follow the homogeneous recurrence determined by $\rho$. Zero-stability is equivalent to the root condition for a multistep method: all roots of a polynomial have modulus at most one, and every unit-modulus root is simple. Roots outside the disk give exponential growth in the number of steps; repeated unit roots give polynomial growth in that number. Repeated interior roots are harmless because their polynomial factors are dominated by decay. Under the usual Lipschitz continuity and initialization assumptions, the Dahlquist equivalence theorem givesFor a zero-stable order-$p$ linear multistep method, local defects and starting errors of the stated orders yield global error $O(h^p)$ by a recurrence bound and the discrete Gronwall inequality. Every required starting value matters.

A concrete failure of “consistency implies convergence” isHere $\rho =(\zeta -1)(\zeta -2)$ and $\sigma =-1$, so $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)=-1$: the method is consistent, of order one. But its root $2$ violates zero-stability. For $y^{\prime }=0$ with exact solution zero and starting values $Y_0=0$, $Y_1=h^2$, the computed solution is $Y_n=h^2(2^n-1)$. These starting errors vanish, yet at $n\approx T/h$ the error diverges. Local accuracy cannot compensate for an unstable recurrence.

A different meaning of stability concerns a fixed step on decaying equations. The Dahlquist test equation $y^{\prime }=\lambda y$ introduces $z=h\lambda$. A one-step method gives $Y_{n+1}=R(z)Y_n$ and is absolutely stable when $|R(z)|\le 1$. For a linear multistep method, the corresponding amplification polynomial of a multistep method is $\rho (\zeta )-z\sigma (\zeta )$, and every root must satisfy the bounded root condition, not just the root approximating $e^z$. Strict asymptotic decay requires the roots to have modulus less than one for $\mathrm{Re}z<0$. This distinction matters at reducible methods with an undamped parasitic mode, as in question 1(c).

The Forward Euler method has order one and $R(z)=1+z$, so its linear stability domain is $|1+z|\le 1$. For $y^{\prime }=-\kappa y$, with $\kappa >0$, it requires $0\le h\kappa \le 2$. It converges as $h\to 0$, yet an excessively large step can produce growing oscillations on an exactly decaying problem. The Backward Euler method has the same order, but $R(z)=(1-z{)}^{-1}$ makes it A-stable and L-stable. For a stiff differential equation, fast decay can impose a severe explicit time-step restriction even when accuracy of the slow dynamics would permit a much larger step.

The trapezoidal rule has order two and is A-stable, but is not L-stable, since its stability function tends to $-1$ along the negative real axis. The Second Dahlquist barrier limits an irreducible A-stable linear multistep method to order at most two. This does not limit implicit Runge-Kutta methods: question 4's Lobatto IIIA method is fourth-order and A-stable. Conversely, question 1's fourth-order member is convergent but not A-stable. These examples show that order of a numerical method, zero-stability and A-stability answer different questions.

For nonlinear dissipative problems, B-stability controls distances between numerical solutions; algebraic stability of a Runge-Kutta method is a sufficient coefficient criterion. Question 4 shows that A-stability alone does not imply that coefficient criterion. Variable time steps require additional analysis of step ratios and error propagation, and stiff problems may require uniform-in-stiffness error estimates beyond the fixed-problem convergence result. The useful chain is local accuracy plus the appropriate perturbation bound, giving global accuracy; absolute and nonlinear stability then determine which practical time steps preserve the intended dynamics.

The flow of the vector field is obtained by solvingDefine $g^{\epsilon }(x,u)$ to be this solution at parameter $\epsilon$. Uniqueness implies $g^0=\mathrm{id}$ and $g^{\epsilon +\delta }=g^{\epsilon }\circ g^{\delta }$ wherever both sides are defined: shifting the solution parameter gives the same initial-value problem. In particular $g^{-\epsilon }$ is the local inverse. This constructs the one-parameter group generated by the vector field, generally as a local group if the flow is not complete.

Symmetry of the derivative matrix means ${\partial }_i{F}_j={\partial }_j{F}_i$. Contracting this symmetric pair with the antisymmetric Levi-Civita symbol givesEvery circle bounds its planar disk, and the vector field is smooth on all of ${\mathbb{R}}^3$, including that disk. Stokes theorem therefore gives for either orientationThe global domain is essential to this argument; the same local symmetry condition on a domain with a hole need not make every closed-curve line integral vanish.

Away from the origin, the given vector field is the gradient of the globally single-valued radius:The sphere has centre $(5,3,2)$, at distance $\sqrt{38}$ from the origin, and radius one. Its intersection circle is therefore wholly away from the origin, since every point on the sphere has distance at least $\sqrt{38}-1>0$. Along the closed curve the chain rule gives $\mathbf{F}\cdot d\mathbf{x}=dr$. Consequently, independently of orientation,The intersection is nondegenerate: the centre's distance from the plane is $4/\sqrt{35}<1$. No parametrization of the resulting circle is needed because $r$ is a potential of a conservative vector field throughout the punctured space.

Set $s=x^2+y^2>0$. The first two components of the curl vanish since the horizontal components are independent of $z$ and the last component depends only on $z$. For the third,so this is a curl-free vector field everywhere in its domain:Choose the orientation in which the projection onto the $xy$-plane runs counterclockwise, and parametrizeOn this curve the horizontal contribution is $(-\sin t,\cos t)\cdot (-\sin t,\cos t)dt=dt$, while the vertical contribution is $zdz=d(z^2/2)$. Its integral over a closed curve is zero. Thereforewith $-2\pi$ for the opposite orientation. This does not contradict Stokes theorem: the planar disk spanning this curve meets the excluded $z$-axis at $(0,0,200)$, where the vector field is undefined. The horizontal one-form is the angular differential and records one winding about that axis; thus this curl-free vector field has no single-valued global potential of a conservative vector field on its domain.

A conservative vector field on ${\mathbb{R}}^n$ is a vector field $\mathbf{V}=\nabla F$ for a globally defined smooth potential of a conservative vector field $F$. The chain rule givesso its line integral is path independent.

Green theorem states that, for a bounded planar region $D$ with piecewise smooth boundary and continuously differentiable $P,Q$ on a neighbourhood of its closure,The boundary is positively oriented: the region lies on the left while it is traversed, so outer components run counterclockwise and hole boundaries clockwise.

For the proposed potential of a conservative vector field, the fundamental theorem of calculus and differentiation under the integral sign giveSince $Q_x=P_y$, the last integral equals $P(x,y)-P(x,0)$, including negative $y$ with the usual signed-integral convention. Hence the proposed potential is global andThe fact that the vector field is defined on all of ${\mathbb{R}}^2$ ensures both straight integration segments stay inside its domain; a hole in the domain could obstruct a global potential.

Integrating the oscillatory term together with the constant components gives a potential of a conservative vector fieldDirect partial derivatives recover both components of $\mathbf{V}$, proving it is a conservative vector field. Its line integral is therefore path independent. At an integer endpoint $(m,n)$, the sine term vanishes, so for any of the specified pathsAny other potential has the same gradient, so differs from $F$ by a constant on the connected plane. In particular every potential changes by $1$ under $x\mapsto x+1$ and by $2$ under $y\mapsto y+1$. Equivalently the integral from $(0,0)$ to $(1,0)$ is $1$, whereas a unit-periodic potential would make that difference zero. There is no potential periodic with period one in both coordinate directions. This is an example of how periods obstruct a periodic potential: a periodic vector field need not have a periodic potential.

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