Combined interior and boundary elliptic regularity for the Dirichlet Laplacian on a smooth bounded domain states that, for every integer $k\ge 0$,when $\Delta u=f$ and $u$ has zero boundary trace. More generally one first has an additional $\Vert u{\Vert }_{L^2}$ term, which uniqueness and the Poincare inequality remove here.

For the shifted equation, write $\Delta u=f+u$. The weak estimate gives $u\in H^1$. Applying the displayed estimate first with an $L^2$ right side gives $u\in H^2$. Repeating,until $u\in H^{k+2}$. The lower-order term is controlled at each stage, yieldingThis is boundary elliptic regularity for the shifted Dirichlet Laplacian.

After integration by parts, the weak formulation isThe left side is the inner productwhich is positive definite and induces the usual $H^1$ norm. The right side is bounded in this norm. The Riesz representation theorem, equivalently the Lax-Milgram theorem, gives a unique weak solution. This is the weak Dirichlet problem for the massive Laplacian with mass one and the signs multiplied by $-1$.

The functionalis bounded on $H_0^1(\Omega )$ by the Cauchy-Schwarz inequality and the Poincare inequality:The Riesz representation theorem therefore supplies a unique $u\in H_0^1(\Omega )$ satisfyingfor every test function. This is exactly the weak identity from part (a). Uniqueness also follows by testing the homogeneous difference with itself.

The zero-boundary Sobolev space isOn it defineIf this quadratic form vanishes, then $\nabla u=0$. The Poincare inequality givesso $u=0$. It is therefore an inner product, and its norm is equivalent to the usual $H^1$ norm.

A function $u\in H_0^1(\Omega )$ is a weak solution whenThis follows from integration by parts and incorporates the homogeneous Dirichlet boundary condition through membership in $H_0^1$.

For compatible $C^2$ boundary functions $a$ and $b$, use exactly the same Volterra series. The factorial estimate holds in the $C^2$ norm after differentiating the integral formula, so the series converges to a $C^2$ function on the full square. It satisfiesand the two boundary values. This directly proves existence. One can equivalently approximate $a,b$ in $C^2$ by compatible analytic functions; the same estimates make their analytic solutions Cauchy in $C^2$.

The same Volterra series converges uniformly on the entire compact characteristic square $0\le \xi ,\eta \le 1$, becauseThe boundary functions are analytic on neighbourhoods of the compact axis segments, so finitely many complex neighbourhoods give uniform Cauchy estimates for their derivatives. Applying $T^m$ adds the two factorial denominators above, and the corresponding derivative series converges on a neighbourhood of every point of the closed square. Thus the local analytic solutions continue across the whole square and agree on overlaps by uniqueness.

Equivalently, the integral equation bounds $u$ and every differentiated equation on each smaller rectangle; no norm can blow up at a first missing corner. The local analytic existence theorem therefore extends the solution through that corner. This is Global continuation for the analytic Goursat problem.

Introduce the null coordinatesThen $\square =-4{\partial }_{\xi }{\partial }_{\eta }$, so the equation becomesWrite the compatible boundary values asTwice integrating the equation gives the equivalent Volterra integral equation

Let $T$ denote the double-integral operator including the factor $-1/4$, and put $g=a+b-c$. Successive approximation gives the Neumann seriesOn a rectangle $|\xi |\le R$, $|\eta |\le R$,The series and its differentiated series converge locally uniformly. Since $a$ and $b$ are analytic, the sum is analytic and solves the equation and data near the origin.

If two solutions have the same data, their difference $w=Tw$. Iterating and using the same factorial estimate gives $\Vert w{\Vert }_{\infty }=0$ on every sufficiently small rectangle. This proves uniqueness. The argument is the Analytic Goursat problem for a Klein--Gordon equation.

Consider an analytic quasilinear partial differential equationLet the analytic initial hypersurface be $\Gamma =\{\varphi =0\}$ and prescribe $u=u_0$ and one transverse derivative ${\partial }_{N}u=u_1$ on $\Gamma$. The tangential derivatives of $u_0$ together with $u_1$ determine the full first jet $Du$ on $\Gamma$. The hypersurface is non-characteristic at $x_0$ with respect to these data whenThis is precisely the principal symbol of a partial differential equation evaluated on the conormal $d\varphi (x_0)$.

The Cauchy-Kovalevskaya theorem then gives a unique real-analytic solution near $x_0$. In coordinates flattening $\Gamma$ to $x_n=0$, non-characteristicity lets the equation solve analytically for ${\partial }_n^{2}u$, after which the analytic equation and the two initial jets determine every higher Taylor coefficient.

Choose a short simple root ${\alpha }_1$ and a long simple root ${\alpha }_2$. The G2 root system has positive rootsThe two hexagons formed by the short and long roots give the usual twelve-root diagram. The fundamental weights areThe second is the highest root, so the irreducible module $L({\omega }_2)$ is the Adjoint representation of a Lie algebra.

The seven weights of $L({\omega }_1)$ are zero and the six short roots, each with multiplicity one. Its crystal, with arrows denoting the lowering operators ${\tilde{f}}_i$, is the chain

Applying the Weyl dimension formula to $\lambda =n_1{\omega }_1+n_2{\omega }_2$ gives

For a dominant integral weight $\lambda$, the Weyl character formula isHere $W$ is the Weyl group, $\ell (w)$ its Coxeter length, $\rho$ the Weyl vector, and $\mathrm{ch}$ the formal character of a weight module. Taking the limit at the identity gives the Weyl dimension formula

For every root, the Weyl reflection isFor $\alpha ={\epsilon }_i-{\epsilon }_j$, it swaps the $i$th and $j$th coordinates. For $\alpha ={\epsilon }_i+{\epsilon }_j$, it sendsand fixes all other coordinates. The Weyl group of $D_n$ is therefore the group of signed permutations with an even number of sign changes,

Let $V={\mathbb{C}}^4$. The Special linear Lie algebra $\mathfrak{sl}(V)$ acts on ${\Lambda }^{2}V$. The wedge productis a nondegenerate symmetric bilinear form, and the action preserves it because $\mathfrak{sl}(V)$ acts trivially on ${\Lambda }^{4}V$. This gives an injective homomorphismBoth Lie algebras have dimension $15$, so the map is an isomorphism. This realizes the Isomorphism between so6 and sl4.

Write an element of the diagonal torus asand let ${\epsilon }_i$ extract $t_i$. The root-space decomposition iswith one-dimensional root spaces. Thus the $D_n$ root system is

The upper-triangular choice givesA compatible simple system isThe highest root and Weyl vector areThe fundamental weights are

Using the paper's letters, the root lattice and weight lattice are respectivelyTheir quotient is

The Dynkin diagram is the $D_n$ diagram: a chain ${\alpha }_1-\cdots -{\alpha }_{n-2}$ whose last node is joined to both ${\alpha }_{n-1}$ and ${\alpha }_n$. The Extended Dynkin diagram adds ${\alpha }_0=-\theta$ joined to ${\alpha }_2$. For $D_4$, the central node ${\alpha }_2$ consequently has the four leaves ${\alpha }_0,{\alpha }_1,{\alpha }_3,{\alpha }_4$.