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

Associate to $f$ the symmetric bilinear form $B(u,v)=\frac{1}{2}(f(u+v)-f(u)-f(v))$. If $f$ is identically zero, every basis is diagonal. Otherwise choose $v$ with $f(v)\ne 0$. Every $w$ has the decompositionwhose second term lies in $v^{\perp }$. Thus $V=\mathbb{R}v\oplus v^{\perp }$, with no cross term in $f$. Induction on dimension gives a diagonal basis. Rescaling its nonzero entries reduces the form towith $k$ additional zero coordinates. Define the rank of a quadratic form $r=p+q$ and its signature $s=p-q$. Rank is intrinsic, since it is the rank of $v\mapsto B(v,\cdot )$, equivalently $n-\dim \mathrm{rad}B$. Also $p$ is the greatest dimension of a subspace on which the restriction of $f$ is a positive-definite quadratic form: any such subspace intersects the negative-plus-zero coordinate space only at zero, so its dimension is at most $p$, while the positive coordinate space attains $p$. The corresponding negative statement identifies $q$. This proves the basis independence of $r$ and $s$, the Sylvester's law of inertia conclusion.

For a totally isotropic subspace $W$ on which $f$ vanishes, polarization identity also makes $B$ vanish on $W\times W$. Quotient by the radical of a bilinear form. In the resulting nondegenerate form, projection of an isotropic subspace to either the positive or negative coordinate space is injective, so its dimension is at most $\min (p,q)$. Consequently $\dim W\le k+\min (p,q)$. The radical of a bilinear form together with the vectors $e_i^{+}+e_i^{-}$, $1\le i\le \min (p,q)$, attains this bound. The maximum dimension of a totally isotropic subspace is thereforeThe criterion for membership in a diagonal basis follows similarly: a nonzero vector can belong to a diagonal basis if it has nonzero quadratic value, by the same orthogonal splitting, or if it lies in the radical of a bilinear form, by choosing a radical basis containing it. An isotropic vector in a diagonal basis must be orthogonal to every basis vector, and hence must lie in that radical. For the specified form, its radical is the $z$-axis, giving exactlyThe zero vector never belongs to a basis.

Galois cohomology and the finite division quotient. Let $K$ be a number field, $G_K=\mathrm{Gal}(\overline{K}/K)$ with its profinite topology, and $M$ a discrete continuous $G_K$-module. Its Galois cohomology is continuous group cohomology. In degrees zero and one,Thus a crossed homomorphism measures the failure of a choice to be Galois-invariant, and changing the choice changes the cocycle by a coboundary. A short exact sequence of such modules gives a long exact sequence in group cohomology. The connecting map sends an invariant element to the cocycle obtained by lifting it and comparing its Galois translates.

For an elliptic curve $E/K$ and $n\ge 2$, multiplication by $n$ is surjective on $E(\overline{K})$ and has kernel $E[n]$. The Kummer exact sequence of an elliptic curve therefore givesExplicitly, choose $nQ=P$ and set $\kappa (P)(\sigma )=\sigma Q-Q$. Replacing $Q$ changes this by a coboundary; replacing $P$ by $P+nR$, $R\in E(K)$, does not change it. If its class is zero, some torsion translate of $Q$ is Galois-invariant, so $P\in nE(K)$. This proves injectivity of the Kummer map of an elliptic curve.

The whole group $H^1(K,E[n])$ need not be finite. The crucial restriction is ramification. Let $S$ contain the finite primes of bad reduction of an elliptic curve and those dividing $n$. At $v\notin S$, good reduction and prime-to-residue-characteristic multiplication imply that every division point of a local point is in an unramified extension, by the formal-group lifting argument. Thus every global Kummer class restricts trivially to inertia there.

Choose a finite Galois extension $L/K$ containing $E[n]$ and the $n$th roots of unity, and enlarge $S$ to include its ramified primes. Over $L$, the torsion module is constant and can be identified with ${\mu }_n^2$. By Kummer theory,For classes unramified outside the primes $S_L$ above $S$, valuations outside $S_L$ must be divisible by $n$. Their restriction images therefore lie in $L(S_L,n{)}^2$, the square of an S-unramified power class group. Its finiteness can be seen without assuming the Mordell-Weil theorem: the finiteness of S-unramified Kummer classes follows fromThe S-unit group is finitely generated by the Dirichlet unit theorem and its valuations at $S_L$. The localized ideal class group is a quotient of the ordinary ideal class group of $L$, which is finite by the finiteness of the ideal class group. The Minkowski bound for ideal classes supplies an integral ideal of bounded norm in each class, and only finitely many integral ideals have bounded norm.

The kernel of restriction from $H^1(K,E[n])$ to $H^1(L,E[n])$ is also finite: a cocycle trivial after restriction can, after subtracting a coboundary, be made to factor through the finite group $\mathrm{Gal}(L/K)$. There are only finitely many maps from that group to the finite module $E[n]$. This is the finite-extension kernel of a Kummer map mechanism. Consequently the possible Kummer classes form a finite set, provingThis is the Weak Mordell-Weil theorem. Imposing membership in the local Kummer image at every place refines the finite unramified collection to the n-Selmer group. Its relation with the obstruction to a globally rational point iswhere the last group is the $n$-torsion of the Tate–Shafarevich group. Thus local solubility gives a finite computable upper bound but can leave a genuine global obstruction.

Heights and finite generation. For the other essay, normalize the absolute values on a field $K$ so that the product formula holds. The Absolute logarithmic Weil height of $[x_0:x_1]\in {\mathbb{P}}^1(K)$ isAt the infinite places the local degrees are one or two and the usual real or complex modulus is used. The product formula makes this independent of scaling; it is also unchanged by extending $K$. Over $\mathbb{Q}$, using coprime integral coordinates gives $h([u:v])=\log \max (|u|,|v|)$, the naive height on the projective line. The Northcott theorem says that points of bounded height and bounded field degree form a finite set; in particular this holds over the fixed field $K$. One way to see the finiteness is the height-Mahler measure formula: for an algebraic number of degree $d$, the coefficients of its primitive minimal integer polynomial are bounded by binomial factors times $\exp (dh)$. Bounds on $d$ and $h$ therefore leave only finitely many integer polynomials and hence only finitely many algebraic numbers.

Set $h_E(P)=h(x(P))$ for $P\ne O$, and $h_E(O)=0$. Since $x:E\to {\mathbb{P}}^1$ has degree two, bounded $h_E$ gives only finitely many points of $E(K)$. The height growth under a morphism of the projective line states that a degree-$d$ morphism gives $h(f(t))=dh(t)+O(1)$. The upper bound follows from its homogeneous coordinate polynomials; the lower bound follows from their having no common zero, using a resultant identity at each place. Applied to the degree-four duplication map, it gives a uniform constant $C$ withConsequently $4^{-j}{h}_E(2^{j}P)$ is Cauchy: the absolute difference of consecutive terms is at most $C{4}^{-(j+1)}$. Define the canonical height of an elliptic curve byIt is nonnegative and satisfies $\hat{h}(2P)=4\hat{h}(P)$. The factor one-half is the conventional normalization for the divisor $(O)$, since the first-coordinate height belongs to $2(O)$.

The addition-divisor identity also gives the bounded-error height identityuniformly in $P,Q$. This is the height form of the identity for the sum and difference pullbacks of the line bundle associated to $2(O)$; it includes points where the affine addition formulas have a zero denominator. Apply it to $2^{j}P,2^{j}Q$, divide by $2\cdot 4^j$, and take the limit. Thus $\hat{h}$ satisfies the parallelogram law, and its polarization is a bilinear form, the height pairing. Nonnegativity implies the Cauchy-Schwarz inequality for that pairing, by evaluating its nonnegative quadratic polynomial on $mP+nQ$ for all integers $m,n$ and approximating real ratios by rational numbers. In particularThese properties do not presuppose finite generation. Bounded canonical height gives a finite set by the bounded difference and the Northcott theorem. Torsion points have canonical height zero. Conversely, if $\hat{h}(P)=0$, all $2^{j}P$ have bounded naive height, so two coincide; their difference makes $P$ torsion. Thus the height detects precisely the free part, once finite generation is proved.

Use the Weak Mordell-Weil theorem with $n=2$ and choose finitely many coset representatives $R_i$. Write $P=2Q+R_i$, and put $H={\max }_i\hat{h}(R_i)$. The height inequality yieldsIteration enters the finite set $\{P:\hat{h}(P)\le H+1\}$, since after $j$ steps the height is at most $H+2^{-j}\max (\hat{h}(P)-H,0)$. Unwinding $P=2Q+R_i$ expresses every point using that finite set and the $R_i$. This proves the canonical-height proof of Mordell-Weil finite generation and henceThe Fundamental theorem of finitely generated abelian groups supplies this decomposition. The Mordell-Weil theorem combines a finite quotient from arithmetic Galois cohomology with a contracting height descent; either ingredient alone would not establish finite generation.

Use the real Lie-algebra convention of the question, with a gauge covariant derivative $D_{\mu }={\partial }_{\mu }+A_{\mu }$. The gauge field strength is its curvature:Write $a_{\mu }={\delta }_X{A}_{\mu }/ϵ=-{\partial }_{\mu }X+[X,A_{\mu }]$. To first order in $ϵ$,Substitute $a_{\mu }$. The mixed second derivatives of $X$ cancel. The terms containing first derivatives of $X$ cancel in pairs. The remaining terms areThe Jacobi identity combines the last two into $[X,[A_{\mu },A_{\nu }]]$. ConsequentlyThe transformation is homogeneous even though the connection transformation contains an inhomogeneous derivative term.

For a finite-dimensional Lie algebra, define its Adjoint representation by ${\mathrm{ad}}_X(Y)=[X,Y]$ and its Killing form byIt is a symmetric bilinear form by cyclicity of the trace. The Jacobi identity gives ${\mathrm{ad}}_{[Z,X]}=[{\mathrm{ad}}_Z,{\mathrm{ad}}_X]$. Set $A={\mathrm{ad}}_Z$, $B={\mathrm{ad}}_X$, $C={\mathrm{ad}}_Y$. ThenThis proves the invariant bilinear form on a Lie algebra property, without assuming simplicity or nondegeneracy.

For definiteness use the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ and take a real compact semisimple Lie algebra as the gauge algebra. Its positive internal metric is $B=-\kappa$. A Killing-form Yang-Mills Lagrangian with the coupling absorbed into the connection isThe spacetime metric is unchanged by internal gauge transformations. HenceThus gauge invariance follows directly from invariance of the Killing form. Other overall conventions are possible, but the energy sign must be checked rather than inferred from a prefactor in isolation.

For physical kinetic terms the internal form must be real, nondegenerate and positive definite after choosing the overall sign. If ${\mathcal{E}}_i=F_{0i}$ and ${\mathcal{B}}_i=\frac{1}{2}{ϵ}_{ijk}{F}_{jk}$, the above convention has Lagrangian $[B({\mathcal{E}}_i,{\mathcal{E}}_i)-B({\mathcal{B}}_i,{\mathcal{B}}_i)]/(2g_{\mathrm{YM}}^2)$ and physical energy densityThe canonical Hamiltonian density also contains the nondynamical multiplier and a spatial divergence. With ${\Pi }_i={\mathcal{E}}_i/g_{\mathrm{YM}}^2$ and $D_i{\Pi }_i={\partial }_i{\Pi }_i+[A_i,{\Pi }_i]$, integration by parts using the invariant bilinear form on a Lie algebra givesThe Gauss law constraint in gauge theory sets $D_i{\Pi }_i=0$. If the boundary flux vanishes, or the appropriate boundary contribution is included, the integrated physical Hamiltonian is the positive energy displayed above.

An indefinite internal form would give gauge-field polarizations with opposite kinetic signs. A degenerate form would fail to supply a kinetic term for some directions. The compactness criterion from the Killing form says that negative-definiteness of the Killing form of a real finite-dimensional algebra is equivalent to compact semisimplicity; thus the pure Killing-form construction selects compact semisimple real forms. One cannot use a complex-bilinear Killing form on arbitrary complex field components as if it were a positive Hermitian metric.

This does not prohibit Abelian gauge theories. A compact Abelian factor has zero Killing form, so it needs a separately chosen positive invariant bilinear form on a Lie algebra, rather than the Killing form. More generally an algebra with a positive invariant metric on a Lie algebra is compact reductive, namely a direct sum of a compact semisimple algebra and an Abelian center. A noncompact group can also share the same compact Lie algebra through global covering choices in an Abelian factor; positivity is a statement about the algebra and internal metric, not by itself a classification of global gauge-group topology. Quantum matter anomalies and global restrictions would require additional input; no matter content is specified here.

For the finite matrix transformation, let $g(x)\in SU(N)$ and regard $g$ as a multiplication operator. The identity ${\partial }_{\mu }{g}^{-1}=-g^{-1}({\partial }_{\mu }g)g^{-1}$ givesTaking commutators of these differential operators cancels the adjacent multiplication operators $g^{-1}g$, yieldingThis argument keeps the derivatives acting on test fields and avoids treating $D_{\mu }$ as just a matrix.

There is a normalization issue in the printed last paragraph. The standard Killing form defined through the adjoint trace on $\mathfrak{su}(N)$ isnot simply $\mathrm{tr}(XY)$. For example, with $N=2$ and $X=Y=-i{\sigma }_3/2$, the defining trace is $-1/2$, whereas the adjoint trace is $-2$. The printed trace formula can be used as a rescaled invariant form, with the constant absorbed into the gauge coupling; it has exactly the invariance needed here. For either normalization,by cyclicity. This proves finite Yang-Mills gauge transformation invariance, with the healthy sign chosen for anti-Hermitian gauge fields. The normalization discrepancy is not a failure of gauge invariance.

Finally let $g(x)=e^{ϵX(x)}=I+ϵX(x)+O({ϵ}^2)$. Since $X\in \mathfrak{su}(N)$ is traceless and skew-Hermitian, this exponential lies in $SU(N)$. Expanding the finite formula givesThus the stated infinitesimal transformation is the derivative of the finite transformation at the identity. It describes transformations in the identity component; arbitrary global or large gauge transformations need not be generated by one globally defined infinitesimal parameter.

Use the real Hilbert space $V=H_0^1(0,1)$, the zero-boundary Sobolev space, with norm $\Vert v{\Vert }_V=\Vert v^{\prime }{\Vert }_{L^2}$. The Poincare inequality makes this equivalent to its usual Sobolev space norm. Define the symmetric bilinear form and continuous linear functionalThe Cauchy-Schwarz inequality and the sharp interval Poincare inequality implyThus $a$ is a bounded bilinear form and a coercive bilinear form, and $\ell$ is bounded because $f$ is square-integrable. The Lax-Milgram theorem supplies a unique weak solution:This is obtained from the differential equation by integration by parts; the test functions have zero boundary trace. Conversely, the weak solution satisfies $y^{\prime \prime }=xy-f$ in the sense of distributions. Since $xy-f\in L^2$, it belongs to $H^2(0,1)\cap H_0^1(0,1)$ and solves the equation almost everywhere.

Equivalently, the Ritz method minimizesover $V$. Indeed $J(y+w)-J(y)=a(w,w)/2$, since the mixed term is $a(y,w)-\ell (w)=0$. This proves existence and uniqueness of the minimizer directly from the weak solution, as well as its equivalence to the variational problem.

A quadratic form on a finite-dimensional real vector space $V$ is a function $q(v)=B(v,v)$ for some symmetric bilinear form $B$. A positive-definite quadratic form satisfies $q(v)>0$ for every $v\ne 0$.

Completing squares givesSet $u=x+y$, $v=y+z$, $w=z$. The corresponding ordered basis in the original coordinates isand the diagonal form is $u^2+v^2+2w^2$. All diagonal coefficients are positive, so the form is positive definite.

The positive uniform ellipticity condition from part (i) is needed here. It is not explicitly imposed in the printed preamble. Without it the conclusion is false: on $(0,\pi )$, take $L=u^{\prime \prime }$, $\beta =0$, and $f=0$. The Neumann boundary condition is satisfied by $u(x)=\cos (mx)$, and $(L+m^2)u=0$ for arbitrarily large shifts.

There is also an actual error in the PDF's supplied trace inequality: a nonzero constant has nonzero trace and zero gradient. Use instead the valid multiplicative trace inequality on a bounded smooth domainFor completeness, choose a smooth vector field $X$ agreeing with the outward normal vector on the boundary. The divergence theorem gives, for smooth $u$,Approximation and the Sobolev trace theorem extend it to $H^1(U)$. Thus the intended existence result remains true after correcting the hint.

Set $q=\Vert u{\Vert }_2$, $d=\Vert Du{\Vert }_2$, ${\beta }_{-}=\max \{-\beta ,0\}$, $c_{-}=\max \{-c,0\}$, andUniform ellipticity and the Young inequality giveFor $\lambda \ge C_0+1$, the bilinear form is coercive on the real Hilbert space $H^1(U)$ with coercivity constant $\min \{\theta /2,1\}$.

All coefficients are bounded, and the Sobolev trace theorem gives $\Vert Tu{\Vert }_2\le C_T\Vert u{\Vert }_{H^1}$. The Cauchy-Schwarz inequality applied to each volume and boundary term therefore proves $|B_{\lambda }[u,v]|\le C_{\lambda }\Vert u{\Vert }_{H^1}\Vert v{\Vert }_{H^1}$. Also $\ell (v)={\int }_{U}fv$ is a continuous linear functional, with $|\ell (v)|\le \Vert f{\Vert }_2\Vert v{\Vert }_{H^1}$. Every hypothesis of the Lax-Milgram theorem is now verified, and it gives

On the zero-boundary Sobolev space $H_0^1(U)$, the associated bilinear form isThe homogeneous Dirichlet boundary condition removes the boundary term, so $(Lu,v{)}_{L^2}=B[u,v]$ for sufficiently regular $u$ and $v\in H_0^1$.

Formal self-adjointness means that $L$ equals its formal adjoint, or equivalently $(L\varphi ,\chi )=(\varphi ,L\chi )$ for compactly supported smooth functions. HereWith the symmetric real principal coefficients in the question, $L=L^{\ast }$ is equivalent to $b=0$. It makes $B$ a symmetric bilinear form.

For the assertions in part (b), positivity must mean strict Dirichlet positivity:This is the positive-definite operator convention. The weaker positive semidefinite operator convention $B[u,u]\ge 0$ does not suffice: $L=-\Delta -{\lambda }_1$ has zero energy on a first Dirichlet eigenfunction. Under strict positivity, the compactness argument in part (b)(i) gives a uniform lower bound as well.

Use the decomposition into Lagrangian subspaces $V=V_{+}\oplus V_{-}$ from part (a). The bilinear form $\Omega$ pairs $V_{+}$ and $V_{-}$ nondegenerately: a vector in either summand annihilating the other already annihilates its own summand and hence all of $V$, so is zero.

Choose a basis $e_1,\dots ,e_n$ of $V_{+}$. The linear mapis an linear isomorphism. Let $f_j$ correspond to the dual basis of the $e_i$. Then $\Omega (e_i,f_j)={\delta }_{ij}$, and the pairings within each summand vanish. Thus

A bilinear form $B$ on $V$ is nondegenerate whenIn finite dimensions this is equivalent to nondegeneracy in the first argument.

Define the linear mapNondegeneracy makes ${\beta }_1$ injective. Since $V$ and its dual space have the same finite dimension, ${\beta }_1$ is an isomorphism. Similarly define ${\beta }_2(w)=B_2(-,w)$ and setThen $\alpha$ is linear and

If $w\in \ker \alpha$, this identity gives $B_2(v,w)=0$ for every $v$. Conversely, if $B_2(v,w)=0$ for every $v$, then $B_1(v,\alpha w)=0$ for every $v$, so nondegeneracy gives $\alpha w=0$. HenceThis is the representation of a bilinear form relative to a nondegenerate bilinear form.

Define on $H$ the bilinear formand the linear functionalThe Hardy inequality on an interval, Cauchy-Schwarz inequality, and the one-sided Poincare inequality show that $a$ is a bounded bilinear form and that

For $v\in H$, $v(x)={\int }_0^x{v}^{\prime }(t)dt$, so Cauchy--Schwarz and Fubini's theorem giveHenceThus $a$ is a coercive bilinear form. The Lax-Milgram theorem supplies a unique $u\in H$ satisfying $a(u,v)=\ell (v)$ for every $v\in H$. This is precisely the unique weak solution described by the weak boundary value problem with an inverse-square potential.

For a finite-dimensional Lie algebra representation $\rho$ on $V$, the Trace form of a Lie algebra representation isWrite again $A=\rho (a)$, $B=\rho (b)$, and $C=\rho (c)=[A,B]$. The operator $C$ commutes with both $A$ and $B$. Direct use of the cyclic property of the trace givesIn the last line, cyclicity and $AC=CA$ turn $\mathrm{tr}(ABC)$ into $\mathrm{tr}(BAC)$. Thus the nonzero vector $c\in \mathfrak{g}$ is orthogonal to the basis $a,b,c$, and hence to all of $\mathfrak{g}$. The bilinear form is therefore degenerate.

Let $V$ be the real vector space with basis $(e_i{)}_{i\in I}$. The Coxeter Gram matrix defines the symmetric bilinear formIts diagonal entries are $2$. The Geometric representation of a Coxeter group is generated by the reflectionsEach has square one, and on $\mathrm{span}\{e_i,e_j\}$ the product $\sigma (x_i)\sigma (x_j)$ has order $m_{ij}$. The reflections therefore satisfy the Coxeter relations and define a group representation $\sigma :W\to GL(V)$.

All edges in the Type E(p,q,k) Coxeter graph are unlabelled, so the factor $a^2$ in part c is one. Separate the arm of length $p$ from the central vertex. The remaining two arms form a type $A_{q+k+1}$ chain, while deleting the central vertex leaves the disjoint type $A_q$ and type $A_k$ chains. Using $\det G(A_r)=r+1$ in the formula from part c yieldsThe associated bilinear form is degenerate exactly whenThe positive-integer solutions of $1/a+1/b+1/c=1$, up to permutation, are $(3,3,3)$, $(2,4,4)$, and $(2,3,6)$. Consequently the degenerate arm-length triples are the permutations ofFor every other allowed $(p,q,k)$ the determinant is nonzero, so the form is nondegenerate.