Past exam of the mathematics course of the University of Cambridge / 2019 / iii

An Ideal of a Lie algebra is a vector subspace $I\subseteq \mathfrak{g}$ such that $[\mathfrak{g},I]\subseteq I$. The derived series of a Lie algebra is defined by ${\mathfrak{g}}^{(0)}=\mathfrak{g}$ and ${\mathfrak{g}}^{(r+1)}=[{\mathfrak{g}}^{(r)},{\mathfrak{g}}^{(r)}]$.

Suppose that $I$ is an ideal and $x\in \mathfrak{g}$, $u,v\in I$. The Jacobi identity givesBoth terms on the right lie in $[I,I]$, because $[x,u],[x,v]\in I$. Thus $[I,I]$ is again an ideal. Starting from the ideal $\mathfrak{g}$ and applying this observation inductively proves that every term ${\mathfrak{g}}^{(r)}$ of the derived series is an ideal of $\mathfrak{g}$.

A Simple Lie algebra is a nonabelian Lie algebra whose only Ideals of a Lie algebra are $0$ and the whole algebra.

Use the standard basis $e,f,h$ of the sl2 Lie algebra, withLet $I\ne 0$ be an ideal and choose $0\ne x=ae+bf+ch\in I$. Since $I$ is invariant under the Adjoint representation, it is invariant under the linear operator $\mathrm{ad}h$. The three basis vectors are eigenvectors of $\mathrm{ad}h$ with distinct eigenvalues $2,-2,0$. Applying the corresponding polynomial spectral projections to $x$ shows that $I$ contains at least one nonzero multiple of $e$, $f$, or $h$.

If $e\in I$, then $[f,e]=-h\in I$ and $[f,h]=2f\in I$; the cases $f\in I$ and $h\in I$ are identical after taking brackets with the other basis vectors. Hence $e,f,h\in I$, so $I={\mathfrak{sl}}_2(\mathbb{C})$. Therefore ${\mathfrak{sl}}_2(\mathbb{C})$ is simple.

The Killing form of a finite-dimensional Lie algebra $\mathfrak{g}$ is the symmetric bilinear formThe cyclicity of the trace makes it invariant:Consequently its radical of a bilinear form $R=\{x:\kappa (x,\mathfrak{g})=0\}$ is an ideal, since $x\in R$ implies $\kappa ([y,x],z)=\kappa (x,[z,y])=0$ for all $y,z\in \mathfrak{g}$.

If $\mathfrak{g}$ is simple, then $R$ is either $0$ or $\mathfrak{g}$. In the second case the Killing form vanishes identically, so the stated solvability criterion makes $\mathfrak{g}$ a Solvable Lie algebra. A nonabelian simple Lie algebra cannot be solvable: its first derived algebra is a nonzero ideal and hence equals $\mathfrak{g}$, after which the derived series never reaches zero. Thus $R=0$, and the Killing form of a simple Lie algebra is nondegenerate.

Write an element of the diagonal Cartan subalgebra asand define the linear functionals ${\epsilon }_i(t)=t_i$. The root-space decomposition of the Symplectic Lie algebra ${\mathfrak{sp}}_6(\mathbb{C})$ then has the C3 root system

A convenient root basis isThe corresponding Cartan matrix isThus the Dynkin diagram is the three-node $C_3$ chain, with a double edge between ${\alpha }_2$ and ${\alpha }_3$ and its arrow pointing toward the shorter root ${\alpha }_2$.

The Weyl reflection formula $s_i(\beta )=\beta -⟨\beta ,{\alpha }_i^{\vee }⟩{\alpha }_i$, together with the Cartan matrix, givesThese are the images of every simple root under each simple reflection.

With the Euclidean inner product used above, the simple coroots areIndeed, if a positive root is $\beta ={\sum }_i{n}_i{\alpha }_i$, thenwhose coefficients are nonnegative. The negative roots give the negatives of these combinations, so the displayed coroots form a fundamental system of a root system for the dual root system.

Duality reverses root lengths. The dual of $C_3$ is therefore the B3 root system: its Dynkin diagram is again a three-node chain with a double final edge, but its arrow points toward the now-short root ${\alpha }_3^{\vee }$.

For $t\in \mathfrak{t}$, the root-space decomposition gives its centralizerEvery root space is one-dimensional, so this centralizer has the minimum possible dimension $\ell =\dim \mathfrak{t}$ exactly when no summand on the right occurs. By the Regular element criterion in a Cartan subalgebra,

Choose a positive system of a root system in which $\alpha$ is a simple root; this is possible after applying an element of the Weyl group. Let $\theta$ be the highest root. Since the rank is greater than one, $\theta \ne \alpha$, and the maximality of $\theta$ implies that $\theta +\alpha$ is not a root.

For $0\ne x\in {\mathfrak{g}}_{\alpha }$, the $(\ell -1)$-dimensional space $\ker \alpha \subset \mathfrak{t}$ centralizes $x$. The line $\mathbb{C}x$ also centralizes $x$, and $[x,{\mathfrak{g}}_{\theta }]\subseteq {\mathfrak{g}}_{\alpha +\theta }=0$. These independent spaces giveHence a nonzero simple-root vector is not regular when $\ell >1$.

Put $h_0=\psi (h)$ and $e_0=\psi (e)={\sum }_{i=1}^{\ell }{e}_i$, where $0\ne e_i\in {\mathfrak{g}}_{{\alpha }_i}$. Since $[h_0,e_0]=2e_0$ and the root spaces are a direct sum, ${\alpha }_i(h_0)=2$ for every simple root. Every root has simple-root coefficients of one sign, so no root vanishes on $h_0$. The Regular element criterion in a Cartan subalgebra therefore shows that $h_0$ is regular.

Restrict the Adjoint representation of $\mathfrak{g}$ along $\psi$. By Complete reducibility of semisimple Lie algebra representations, it is a direct sum of finite-dimensional sl2 Lie algebra modules. The $h_0$-eigenvalues on a root space are twice the heights of the roots, so they are all even. Each irreducible summand consequently has even highest weight, contains exactly one zero-weight vector, and has a one-dimensional kernel for the raising operator $e_0$ by the Classification of finite-dimensional sl2 representations.

Because $h_0$ is regular, its zero-weight space in $\mathfrak{g}$ is precisely $\mathfrak{t}$ and has dimension $\ell$. There are therefore exactly $\ell$ irreducible summands, whenceThus $e_0$ is regular; equivalently it is a principal nilpotent element in the given Principal sl2 subalgebra.

For a dominant integral highest weight $\lambda$, the Weyl dimension formula iswhere ${\Phi }^{+}$ is the chosen positive system of a root system, ${\alpha }^{\vee }$ is a coroot, and $\rho$ is the Weyl vector.

For the B2 root system with ${\alpha }_1$ short, the positive roots areSubstituting $\lambda =a{\omega }_1+b{\omega }_2$ into the Weyl dimension formula for B2 gives

Let $V=V({\omega }_2)$ be the five-dimensional defining representation of the so5 Lie algebra. The tensor square splits into its symmetric square and exterior square:The invariant symmetric form spans a trivial subrepresentation of $S^{2}V$, while its traceless complement is the irreducible $V(2{\omega }_2)$ of dimension $14$. The identification ${\Lambda }^{2}V\cong {\mathfrak{so}}_5$ makes the exterior square the ten-dimensional Adjoint representation, whose highest weight is the highest root $2{\omega }_1$. Therefore the Tensor-square decomposition of the defining so5 representation is

The Poincare-Birkhoff-Witt theorem shows that the weights of the Verma module $M({\omega }_2)$ arewith multiplicities given by the corresponding Kostant partition function.

The Dynkin labels of ${\omega }_2$ are $(0,1)$. Hence the two simple-root singular vectors are $f_1{v}_{{\omega }_2}$, of weight ${\omega }_2-{\alpha }_1$, and $f_2^2{v}_{{\omega }_2}$, of weight ${\omega }_2-2{\alpha }_2$. They generate the Maximal proper submodule of a dominant integral Verma module. Its set of weights is consequently

Let $C(\Delta )$ be the fundamental chamber of a root system determined by $\Delta$. Every positive root is a nonnegative linear combination of the simple roots, so every point in the interior of $C(\Delta )$ pairs strictly positively with every positive root. In particular, the interior meets none of the reflecting hyperplanes belonging to the subsystem ${\Phi }_0$ of roots of maximal length.

The connected set $C(\Delta )$ therefore lies in one chamber of the long-root subsystem. Let ${\Delta }_0$ be the unique fundamental system of a root system defining that chamber. Taking closures gives $C(\Delta )\subseteq C({\Delta }_0)$. Uniqueness follows because the interiors of two distinct chambers are disjoint. Thus there is a unique such ${\Delta }_0$.

Write the simple roots of the G2 root system as $\alpha$ short and $\beta$ long. The Long-root A2 subsystem of G2 has simple rootsIf ${\omega }_1,{\omega }_2$ are its fundamental weights, then the $A_2$ root-weight relations giveFor the $A_2$-dominant weight $\lambda =a{\omega }_1+b{\omega }_2$, therefore,The dominant weight inequalities for $G_2$ are thus equivalent tobecause $A_2$-dominance already gives $a,b\ge 0$.

The weights of the seven-dimensional irreducible $G_2$ representation are zero and the six short roots. In the $A_2$ weight lattice, the three positive short roots areTogether with their negatives, they split into the weight sets of the two dual three-dimensional Fundamental representations of sl3; the zero weight supplies a trivial representation. By the Restriction of the seven-dimensional G2 representation to long-root A2,

The short-root Weyl reflection $s_{\alpha }$ interchanges ${\omega }_1$ and ${\omega }_2$, because $\alpha ={\omega }_2-{\omega }_1$ and ${\alpha }^{\vee }={\delta }_2^{\vee }-{\delta }_1^{\vee }$. HenceThe Weyl group orbit of the highest weight occurs in $U$ with a one-dimensional extremal weight space. Let $v$ be a nonzero vector of weight $s_{\alpha }\lambda$. The same reflection exchanges the long simple roots: $s_{\alpha }{\delta }_1={\delta }_2$ and $s_{\alpha }{\delta }_2={\delta }_1$. If an $A_2$ raising operator did not annihilate $v$, then $s_{\alpha }\lambda +{\delta }_i$ would be a weight of $U$; applying $s_{\alpha }$ would make $\lambda +{\delta }_j$ a weight, contradicting the fact that $\lambda$ is the highest weight.

Thus $v$ is an $A_2$ highest-weight vector of weight $b{\omega }_1+a{\omega }_2$. The Complete reducibility of semisimple Lie algebra representations then supplies the corresponding irreducible summand, so

Write $u_Q=|Q{|}^{-1}{\int }_{Q}u$. Since $u-u_Q$ has arithmetic mean zero,and the pairwise-difference identity givesFirst take $u$ smooth. Join $x$ to $y$ by changing one coordinate at a time and apply the Cauchy-Schwarz inequality:For a one-dimensional slice $v$, the fundamental theorem of calculus yieldsIntegrating the $i$th summand over $x,y\in Q$ therefore gives at most $L^{n+2}\Vert D_{i}u{\Vert }_{L^2(Q)}^2$. HenceThe density of smooth functions in a Sobolev space extends the estimate to every $u\in H^1({\mathbb{R}}^n)$. Thus

With the standard Sobolev space inner product, $u_i\rightharpoonup u$ weakly in $H^1(U)$ meansfor every $v\in H^1(U)$. Equivalently, every bounded linear functional on $H^1(U)$ takes convergent values on the sequence.

The Sobolev space $H^1(U)$ is a separable Hilbert space. A bounded sequence therefore has a weakly convergent subsequence by the weak subsequence of a bounded Hilbert-space sequence; write $u_{i_j}\rightharpoonup u$ in $H^1(U)$. Since $U$ is bounded with smooth boundary, the Rellich-Kondrachov compactness theorem says that $H^1(U)\hookrightarrow L^2(U)$ is compact. Passing to a further subsequence gives

Suppose the claimed Poincare-Wirtinger inequality were false. There would be $u_k\in H^1(U)$ such that, after settingwe have $(v_k{)}_U=0$, $\Vert v_k{\Vert }_2=1$, and $\Vert D{v}_k{\Vert }_2\to 0$. The sequence is bounded in $H^1(U)$, so part 1(b)(ii) supplies a subsequence converging strongly in $L^2(U)$ and weakly in $H^1(U)$ to some $v$.

The weak gradient of $v$ is zero. Because $U$ is connected, $v$ is a constant function; its mean is zero, so $v=0$. Strong convergence would then give $\Vert v_k{\Vert }_2\to 0$, contradicting $\Vert v_k{\Vert }_2=1$. Therefore

If the Poincare inequality with a boundary trace failed, after normalization there would be $u_k\in H^1(U)$ withAs in part 1(c)(i), a subsequence converges strongly in $L^2(U)$ and weakly in $H^1(U)$ to a constant function $u$. The Sobolev trace theorem is a bounded linear map, so the traces converge weakly while their norms tend to zero; hence the trace of $u$ is zero. A constant with zero trace is zero, contradicting $\Vert u{\Vert }_2=1$. Consequently

The Lax-Milgram theorem states that if $H$ is a real Hilbert space, $B:H\times H\to \mathbb{R}$ is a bounded bilinear form, and there is an $\alpha >0$ such thatthen for every bounded linear functional $F\in H^{\prime }$ there is a unique $u\in H$ satisfyingMoreover, $\boxed{\Vert u{\Vert }_H\le {\alpha }^{-1}\Vert F{\Vert }_{H^{\prime }}}$. Symmetry of $B$ is not required.

The Dirichlet boundary condition is built into the first component's space, while the Neumann boundary condition is natural. Thus a weak solution is a pairsuch that for every $(v,z)\in H_0^1(U)\times H^1(U)$,If $u,w$ are $C^2$ up to the boundary, taking compactly supported test functions and applying the fundamental lemma of the calculus of variations gives both differential equations pointwise in $U$. Membership of $H_0^1(U)$ gives $u=0$ on $\partial U$. Applying integration by parts to the second identity and using its differential equation leavesfor every smooth boundary trace $z$. Hence $\partial w/\partial \nu =0$ on $\partial U$, so the equations and both boundary conditions hold classically.

On the product Hilbert space $H=H_0^1(U)\times H^1(U)$ defineandThe Cauchy-Schwarz inequality makes $B$ and $F$ bounded. On the diagonal,The Poincare inequality controls $\Vert u{\Vert }_2$ by $\Vert Du{\Vert }_2$, andThe displayed diagonal value therefore controls the full product $H^1$ norm, so $B$ is coercive. The Lax-Milgram theorem now gives exactly one pair $(u,w)\in H$ satisfying the weak identities. Hence a unique weak solution exists for every $f,g\in L^2(U)$.

The function $f$ is real analytic at $y$ if there is a neighborhood of $y$ on which its multivariable Taylor seriesconverges to $f(x)$. Here $\alpha$ is a multi-index; equivalently, $f$ agrees locally with a convergent real power series centred at $y$.

Locally write the real analytic hypersurface as $\Sigma =\{F=0\}$ with $dF\ne 0$. The conormal bundle is spanned by $dF$. The surface is a characteristic hypersurface at $p$ precisely when the principal symbol vanishes on that conormal:Only the matrix $(a_{ij}+a_{ji})/2$, which is a symmetric matrix, contributes to this expression.

One prescribes analytic Cauchy data: the value of $u$ and one first derivative transverse to $\Sigma$, for examplewhere $\varphi$ and $\psi$ are real analytic on $\Sigma$. Being a non-characteristic hypersurface allows the equation to solve for the second derivative in the transverse direction. The Cauchy-Kovalevskaya theorem then gives one and only one local real analytic solution near $p$.

Expanding the equation givesso its principal symbol isIf a characteristic curve is locally a graph $y=y(x)$, its conormal is proportional to $(-y^{\prime },1)$. The characteristic equation is thereforeOn each region separated by $y=\pm 1$, separation of variables givesThus all the characteristic curves areThe inner curves approach the horizontal characteristics $y=\pm 1$ as $x\to \pm \infty$; the outer hyperbolic-cotangent branches have vertical asymptotes and also approach $y=\pm 1$. This describes the requested sketch.

The initial line $x=0$ has conormal $(1,0)$, and $p(y,1,0)=1$, so it is a non-characteristic hypersurface at every point. Since the coefficients and prescribed data are real analytic, the Cauchy-Kovalevskaya theorem gives a unique real analytic solution in a neighborhood of each point of $\{x=0\}$, hence in a neighborhood of that line.

Put $a(y)=(1-y^2{)}^2$. Multiply $u_{xx}={\partial }_y(a{u}_y)$ by $2u_x$ and integrate over $-1<y<1$. An integration by parts givesbecause $a(1)=a(-1)=0$. Therefore the energy estimate is in fact the conservation lawIf $u(0,y)=u_x(0,y)=0$, then differentiating the first identity in $y$ also gives $u_y(0,y)=0$, so the conserved nonnegative energy is zero. Hence $u_x=0$ and $(1-y^2)u_y=0$ throughout the open strip. There $|y|<1$, so both derivatives vanish; connectedness and the initial value now give

Assume $0<\epsilon <1$ and setThe travel-time coordinate for a one-dimensional variable-speed wave equationsends the initial interval $I_{\epsilon }=(-a,a)$ to $(-S_{\epsilon },S_{\epsilon })$. By the characteristic curves for speed one minus y squared, the two characteristic coordinates are $s-x$ and $s+x$. The finite propagation speed and uniqueness theorem for hyperbolic partial differential equations therefore give the maximal characteristic diamondEquivalently,In the $(x,s)$ plane this is a diamond with vertices $(0,\pm S_{\epsilon })$ and $(\pm S_{\epsilon },0)$; transforming back bends its four sides into the characteristic curves found in part 3(c). Beyond any one of those sides, a point's backward characteristics meet $x=0$ outside $I_{\epsilon }$, where no Cauchy data were prescribed, so uniqueness cannot be extended farther.