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Write $\rho ={\sum }_n{r}_n{\sigma }_n$ in the Generalized Bloch representation. Linearity of the Lindbladian gives ${\dot{r}}_m={\sum }_n\mathrm{Tr}({\sigma }_m\mathcal{L}({\sigma }_n))r_n$. For the Hamiltonian term, cyclicity yieldsFor one Lindblad operator $V_d$, the jump term becomes $\mathrm{Tr}({\sigma }_m{V}_d{\sigma }_n{V}_d^{†})=\mathrm{Tr}(V_d^{†}{\sigma }_m{V}_d{\sigma }_n)$, and the two remaining terms giveThusEvery coefficient is real because $\mathcal{L}({\sigma }_n)$ is Hermitian. In addition, $L_{mn}=-L_{nm}$, although a dissipative $D^{(d)}$ need not be symmetric.

The Hamiltonian commutator has zero trace, and each dissipator obeys $\mathrm{Tr}(V\rho V^{†})=\mathrm{Tr}(V^{†}V\rho )$, so $\mathrm{Tr}\mathcal{L}(\rho )=0$. Consequently $\boxed{{\dot{r}}_{N^2}=0,r_{N^2}=1/\sqrt{N}}$ for normalized density operators. Orthogonality to $I/\sqrt{N}$ makes every other ${\sigma }_k$ traceless. Define $s=(r_1,\dots ,r_{N^2-1}{)}^T$. With $M=L+{\sum }_d{D}^{(d)}$, trace preservation gives the block formHere the indices of $A$ range from $1$ to $N^2-1$. Since $\mathcal{L}(I)={\sum }_d(V_d{V}_d^{†}-V_d^{†}{V}_d)$, the offset can also be written $c_m=N^{-1}\mathrm{Tr}({\sigma }_m{\sum }_d[V_d,V_d^{†}])$. This is the Affine Bloch equation; its offset is zero for unital dynamics, which preserve the identity.

Assume the usual real Hamiltonian controls with freely chosen finite duration, rather than a fixed-time or amplitude-restricted task. Define the Dynamical Lie algebrathe real linear span closed under all nested commutators. Replacing each generator by $-i{H}_m$ gives the same algebra. The associated connected reachable group acts on propagators, on each unitary orbit of a density operator, and on pure states. For $N\ge 2$, the necessary and sufficient conditions are as follows.

For exact unitary operator controllability, $\boxed{\mathfrak{g}=\mathfrak{u}(N)}$. For operator control up to global phase, the possibilities are $\mathfrak{su}(N)$ or $\mathfrak{u}(N)$. If every Hamiltonian is traceless, $\det U(T)=\exp [-i{\int }_0^T\mathrm{Tr}H(t)dt]=1$, so the central phase direction is absent and exact access is to $SU(N)$. The full special unitary Lie algebra has dimension $N^2-1$ and the Lie algebra of the unitary group has dimension $N^2$.

For density operator controllability on every spectral orbit,Scalar phases do not affect conjugation. If only a particular density spectrum is being considered, a smaller group can suffice: for distinct eigenvalue multiplicities $n_j$ and centralizer ${\mathfrak{c}}_{\rho }=\{X\in \mathfrak{u}(N):[X,\rho ]=0\}$, the orbit criterion is $\dim \mathfrak{g}-\dim (\mathfrak{g}\cap {\mathfrak{c}}_{\rho })=N^2-{\sum }_j{n}_j^2$. A maximally mixed state has a one-point orbit and needs no controls. This restricted-orbit condition is weaker than controllability for every density spectrum.

For pure-state controllability, up to a fixed unitary change of basis,where the last two alternatives occur only for even $N$. Here $\mathfrak{sp}(n)$ is the standard compact symplectic Lie algebra acting on ${\mathbb{C}}^{2n}$, not the real noncompact symplectic algebra. These are precisely the unitary matrix algebras whose connected groups act transitively on normalized complex vectors, and also give the ray version of controllability. For $N=2$, $\mathfrak{sp}(1)=\mathfrak{su}(2)$. For even $N\ge 4$, the symplectic cases show why pure-state control can be weaker than density control: $\dim \mathfrak{sp}(n)=n(2n+1)$ is smaller than the nondegenerate density-orbit dimension $N^2-N$, and its scalar center acts trivially on that orbit. For $N=1$, the physical density and pure-state spaces are single points, while exact operator controllability still requires the phase algebra $\mathfrak{u}(1)$.

Assume $c_1{c}_2\ne 0$, as required for a connected chain and the supplied divisions. The single-excitation subspace has dimension three, so its Dynamical Lie algebra ${\mathfrak{g}}_1={\mathrm{Lie}}_{\mathbb{R}}\{iA,iB\}$ is contained in $\mathfrak{u}(3)$. The intended conclusion from nine independent skew-Hermitian directions is $\boxed{{\mathfrak{g}}_1=\mathfrak{u}(3)}$, giving unitary operator controllability, density operator controllability and pure-state controllability on this sector. Here is a direct generation argument that does not assume an independent identity control or rely on the normalization of the printed intermediate matrices.

Here $c_n$ denotes the hopping entry of the supplied reduced $A$. With the stated Pauli matrices, restricting $c_n(X_n{X}_{n+1}+Y_n{Y}_{n+1})$ instead gives hopping $2c_n$; absorbing that common factor into the reduced coupling does not change the Dynamical Lie algebra.

Write $R_{mn}=E_{mn}-E_{nm}$ and $Y_{mn}=i(E_{mn}+E_{nm})$ using matrix units. The physical generators giveNext isolate the second transition and generate its other direction:Finally $R_{13}=[R_{12},R_{23}]$ and $Y_{13}=[R_{12},Y_{23}]$. The three $R_{mn}$, three $Y_{mn}$ and two diagonal differences are eight independent traceless skew-Hermitian matrices, hence span $\mathfrak{su}(3)$. Because $\mathrm{Tr}(iB)=i\ne 0$, subtracting the traceless part of $iB$ also supplies $i{I}_3$, completing $\mathfrak{u}(3)$. This is the trace-direction step in identity augmentation in quantum controllability and the three-site instance of Endpoint control of an XX spin chain. If a coupling vanishes, the chain disconnects and the full-sector conclusion fails.

There is a redundant direction in the printed final pair of commutators: with the printed intermediate definitions, both are proportional to $E_{13}-E_{31}$. Thus that pair cannot itself establish the stated independence. The mixed commutator $[R_{12},Y_{23}]=i(E_{13}+E_{31})$ above provides the missing direction and establishes the required conclusion independently.

Let ${\mathfrak{g}}_{\mathrm{full}}$ be the algebra on the full eight-dimensional space. Restriction to the invariant single-excitation sector is a surjective Lie algebra homomorphism $R:{\mathfrak{g}}_{\mathrm{full}}\to \mathfrak{u}(3)$. If $\dim {\mathfrak{g}}_{\mathrm{full}}=9$, the rank-nullity theorem gives $\ker R=0$, soThis is an abstract algebra isomorphism; on the full Hilbert space it is a reducible eight-dimensional representation preserving sectors of dimensions $1,3,3,1$. The sector actions are linked by the same nine generators, rather than independently controllable blocks. In particular, this is not $\mathfrak{u}(8)$ or $\mathfrak{su}(8)$, so it does not imply full-system controllability. The full spin Hamiltonians are traceless, so their algebra is embedded in $\mathfrak{su}(8)$ even though its abstract type is $\mathfrak{u}(3)$; its central direction is not an unrestricted overall phase on the full space.

Both $A$ and $C$ are real symmetric and couple only neighboring sites whose diagonal entries in $J$ have opposite signs. For an entry $x_{mn}$ of $x=iA$ or $iC$, symmetry gives $(x^{T}J+Jx{)}_{mn}=(J_{nn}+J_{mm})x_{mn}$. Every nonzero entry joins signs $+1$ and $-1$, while the diagonal entries vanish. Hence $\boxed{x^{T}J+Jx=0\text{ for }x=iA,iC}$.

The condition is preserved by real linear combinations and commutators, so the entire Dynamical Lie algebra satisfies it. For the physical generator $x(t)=-i(A+f(t)C)$, differentiation gives $\frac{d}{dt}(U^{T}JU)=U^T(x^{T}J+Jx)U=0$, and therefore $\boxed{U(t{)}^{T}JU(t)=J}$. This is an orthogonal dynamical symmetry in quantum control: a preserved symmetric bilinear form. It need not give eigenspaces of a commuting operator; indeed $J$ does not commute with $A$ or $C$.

Set $D=\mathrm{diag}(1,i,1)$. Then $D^{T}JD=I$, so the transformed generators $x^{\prime }=D^{†}xD$ satisfy $x^{\prime T}+x^{\prime }=0$ as well as $x^{\prime †}+x^{\prime }=0$. They are consequently real antisymmetric, lying in $\mathfrak{so}(3)$. The original generators already supply the independent directions $p=iC=i(E_{12}+E_{21})$ and $q=i(A-C)=i(E_{23}+E_{32})$, and their commutator is $[p,q]=E_{31}-E_{13}$. These three independent directions exhaust the three-dimensional algebra. ThusHere $E_{mn}$ denotes a matrix unit. The reachable connected group is a unitary conjugate of $SO(3)$, a proper subgroup of $SU(3)$, so there is no unitary operator controllability.

Pure-state control also fails. For every reachable state ${\psi }^{\prime }=U\psi$, the quantity ${\psi }^{\prime T}J{\psi }^{\prime }={\psi }^{T}J\psi$ is invariant, and its absolute value is unchanged even by an overall state phase. The normalized states $e_1$ and $(e_1+e_2)/\sqrt{2}$ have respectively $|{\psi }^{T}J\psi |=1$ and $0$, so no allowed control connects their rays. Therefore $\boxed{\text{neither pure-state nor density operator controllability holds}}$ for this reduced three-level system, despite the absence of a simple excitation-sector obstruction within the sector.

A Killing spinor is a nonzero spinor parameter for which the fermionic supersymmetry variations vanish on a bosonic background. In minimal four-dimensional supergravity without a cosmological term this meansIt is an overdetermined equation and expresses an unbroken supersymmetry of the background. In the cosmological deformation the corresponding equation is ${\mathcal{D}}_{\mu }ϵ=0$; in theories with additional bosonic fields their contributions also enter the fermion variations.

For the undeformed theory the connection preserves both the spinor adjoint and gamma matrices. Consequently the Dirac currentobeys ${\nabla }_{\nu }{K}_{\mu }=0$. In particular ${\nabla }_{(\nu }{K}_{\mu )}=0$, so $K$ is a Killing vector, and in fact is parallel. For the real cosmological deformation, ${\nabla }_{\mu }ϵ=-(m/2){\gamma }_{\mu }ϵ$ and ${\nabla }_{\mu }\overline{ϵ}=(m/2)\overline{ϵ}{\gamma }_{\mu }$ instead give ${\nabla }_{\mu }{K}_{\nu }=(m/2)\overline{ϵ}[{\gamma }_{\mu },{\gamma }_{\nu }]ϵ$, whose symmetric part again vanishes.

The causal property uses an ordinary commuting spinor. Take an orthonormal frame with $({\gamma }^0{)}^{†}=-{\gamma }^0$, $({\gamma }^i{)}^{†}={\gamma }^i$, and $\overline{ϵ}={ϵ}^{†}{\gamma }^0$. Then $K^0=-{ϵ}^{†}ϵ$. Reverse its time orientation and write $V=-K$, soThe matrices ${\alpha }^i$ are Hermitian and obey $\{{\alpha }^i,{\alpha }^j\}=2{\delta }^{ij}$. For every spatial unit vector $u$, $(u_i{\alpha }^i{)}^2=1$, whence $|u_i{V}^i|\le V^0$. Choose $u$ along the spatial part of $V$ to obtainA nonzero Killing spinor cannot vanish at an isolated point: its defining connection equation transports its value along every curve. Thus on a connected background this is a nonzero, nowhere-spacelike causal Dirac spinor current. The estimate does not need the Majorana condition, but it does need commuting components; Grassmann bilinears have no such ordinary positivity order.

For the positive energy theorem, take a complete asymptotically flat spacelike hypersurface with a spin structure $\Sigma$, with suitable decay, Einstein constraint equations, and matter satisfying the dominant energy condition. Assume there are no inner boundaries, or that any inner-boundary contribution has the required nonnegative sign. The auxiliary commuting spinor approaches an arbitrary constant ${ϵ}_{\infty }$ at spatial infinity. Define the real Nester two-formAntisymmetry removes second derivatives except through the Riemann curvature tensor. Differentiation and the three-gamma spinor curvature identity used above giveThe first term comes from differentiating the two spinor factors. For the second term, replace ${\nabla }_{\nu }{\nabla }_{\rho }$ by half their commutator; its spin-curvature contraction is $G^{\mu \nu }{\gamma }_{\nu }/2$ from each conjugate term. This is the reason the Einstein tensor, and hence the matter energy density, enters the boundary-charge identity.

In an orthonormal frame adapted to the future normal of $\Sigma$, put ${\chi }_i={\nabla }_iϵ$. Since ${\gamma }^0{\gamma }^{0ij}=-{\gamma }^{ij}$,Here ${\nabla }_i$ is the spacetime spin connection pulled back to $\Sigma$, including its extrinsic curvature. Choose the auxiliary spinor to solve the elliptic Witten spinor equationThe standard asymptotic-flatness and completeness hypotheses permit this boundary-value problem; the positivity identity also excludes a decaying homogeneous zero mode. It is this hypersurface equation that removes the negative square. A spacetime Killing spinor is not assumed for a general initial-data set.

Stokes' theorem and $G_{\mu \nu }={\kappa }^2{T}_{\mu \nu }$ now giveThe dominant energy condition makes the second integrand nonnegative because $n$ and $V$ are future causal. Evaluating the two-form with the asymptotic spin connection identifies its surface integral with the ADM energy and momentum:For example its energy term uses $B^{0i}=\frac{1}{2}({\partial }_j{h}_{ij}-{\partial }_i{h}_{jj})|{ϵ}_{\infty }{|}^2$ to first order in the asymptotic spatial metric tensor perturbation. With ${\kappa }^2=8\pi G$ this is exactly the ADM normalization. Since $P_i{\alpha }^i$ has eigenvalues $\pm |\mathbf{P}|$, positivity for every boundary spinor provesKilling spinors describe saturation: their spatial derivative term vanishes, and the contracted matter term must also vanish for the corresponding charge to be zero. They characterize preserved supersymmetry rather than being prerequisites for positivity. Under the usual rigidity hypotheses, zero total four-momentum makes the nonnegative integral vanish for a full basis of asymptotic spinors; their parallel extensions force a vanishing Riemann curvature tensor and give the Minkowski vacuum. A single Killing spinor with null current alone does not justify asserting that every component of the total momentum vanishes.

Put $\delta =\Delta t$ and $L=A+B$, keeping the spatial grid fixed for the temporal error assertion. The matrix exponential expansions giveHence the Lie-Trotter splitting commutator error isThe energy estimate from part 1 makes $e^{tA}$, $e^{tB}$ and $e^{tL}$ contractions, so both $S_{\delta }$ and $E_{\delta }$ have norm at most one. Telescope their $n$th powers:For $n\delta \le T$, this givesThus the Lie-Trotter splitting has global temporal error $O(\Delta t)$. The constant here may depend on the fixed spatial operators; a mesh-uniform joint-limit error estimate needs additional regularity and commutator bounds. If $A$ and $B$ commute, this particular exponential splitting is exact, but variable diffusion coefficients generally remove that commutativity.

The Clifford algebra relation impliesIndeed move $C(v)$ past $C(b)$ and then past $C(a)$ using the two anticommutator identities; the three-generator products cancel.

Since all $C(w)$ are self-adjoint,Each scalar vanishes because $T$ is skew-adjoint. This proves $\boxed{\pi (T{)}^{\ast }=-\pi (T)}$.

For the requested commutator, the first identity givesThe first sum is $C(Tv)$ by the orthonormal basis expansion of $v$. For the second, the skew-adjoint property gives ${\sum }_i(T{e}_i,v)e_i=-Tv$. By real linearity of $C$, the difference is thereforeThis calculation explains the factor $1/4$: it makes infinitesimal rotations of $V$ act correctly on its Clifford generators.

To distinguish the two normalizations, denote the physicists' Hermite polynomial by $h_n(x)=(-1{)}^n{e}^{x^2}(d/dx{)}^n{e}^{-x^2}$. Define the normalized Hermite functions byIntroduce the differential operators $a=(x+d/dx)/\sqrt{2}$, $a^{\ast }=(x-d/dx)/\sqrt{2}$ on the Schwartz space. Direct multiplication gives $[a,a^{\ast }]=1$ andThe Gaussian $H_0={\pi }^{-1/4}{e}^{-x^2/2}$ satisfies $a{H}_0=0$, and repeated differentiation gives $H_n=(a^{\ast }{)}^n{H}_0/\sqrt{n!}$. The commutator identity implies $a{H}_n=\sqrt{n}{H}_{n-1}$ and $a^{\ast }{H}_n=\sqrt{n+1}{H}_{n+1}$. ThereforeFor normalization, integration by parts makes $a^{\ast }$ the adjoint operator of $a$ on these rapidly decreasing functions, so $\Vert (a^{\ast }{)}^n{H}_0{\Vert }^2=n!\Vert H_0{\Vert }^2=n!$. Distinct eigenvalues of the symmetric quantum harmonic oscillator give orthogonality.

The generating identity follows directly from the Rodrigues definition of the Hermite polynomials:The Taylor expansion of the entire Gaussian justifies the middle equality for every complex $w$. Substitute $w=z/\sqrt{2}$ and the Bargmann-Fock space basis $e_n(z)=z^n/\sqrt{n!}$ to obtainThe convergence is locally uniform in $z$ by the same entire generating series.

For $0<t<1$, $A_x(\sqrt{t}z)$ belongs to the Bargmann-Fock space: after writing $z=u+iv$, its squared modulus times $e^{-|z{|}^2}$ has strictly negative quadratic coefficients $-(1+t)u^2-(1-t)v^2$. Taking the Bargmann-Fock space inner product of the two generating functions and using the orthonormal basis proved in part (b) givesThe coefficient series converges absolutely by the Cauchy-Schwarz inequality. Substitution factors the right side intoUse the Gaussian integral $\int e^{-a{u}^2+bu}du=\sqrt{\pi /a}{e}^{b^2/(4a)}$ for $a>0$ and complex $b$; the complex case follows from the real case by analyticity. The resulting exponent isThus the Mehler formula for Hermite functions isAs a normalization check, its $t\downarrow 0$ limit is $H_0(x)H_0(y)$. Setting $t=e^{-2s}$ and multiplying by $e^{-s}$ yields the harmonic oscillator transition kernel with eigenvalues $2n+1$.

The canonical commutation relations are $[\hat{x},\hat{p}]=i\hslash I$ (in several dimensions $[x_i,p_j]=i\hslash {\delta }_{ij}I$, with the position-position and momentum-momentum commutators zero). For Hermitian $x,p$, the commutator is anti-Hermitian, just as $i\hslash I$ is; there is no conflict with their self-adjointness. The annihilation operator and its adjoint obeyThus the quantum harmonic oscillator Hamiltonian is $\boxed{H=\hslash (N+1/2)}$. The commutators $[N,a]=-a$, $[N,a^{†}]=a^{†}$ and the normalized ground state constructNormalization follows inductively from $a{a}^{†}=N+1$. Positivity of $N$ and repeated lowering exclude a noninteger eigenvalue: lowering it until it becomes negative would produce a nonzero vector of negative $N$ eigenvalue. Every eigenstate therefore lowers to the unique ground state, so these normalized ladder states exhaust the oscillator eigenstates.

For two independent oscillators, the tensor products $|n_1,n_2⟩$ have $E=\hslash (n_1+n_2+1)$. The level $n=n_1+n_2$ has degeneracy $n+1$. In its degenerate subspace the perturbation $\lambda x_1{x}_2$ reduces tosince terms that raise or lower both numbers leave the subspace. For $n=0$ its matrix is zero. For $n=1$, in $|1,0⟩,|0,1⟩$, it is $(\lambda \hslash /2)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. For $n=2$, in $|2,0⟩,|1,1⟩,|0,2⟩$, it is $(\lambda \hslash /2)\left(\begin{array}{ccc}0& \sqrt{2}& 0\\ \sqrt{2}& 0& \sqrt{2}\\ 0& \sqrt{2}& 0\end{array}\right)$. Their eigenvalues giveThe $n=1$ states are the symmetric and antisymmetric combinations. At $n=2$, the shifted states are $(|2,0⟩\pm \sqrt{2}|1,1⟩+|0,2⟩)/2$, and the zero-shift state is $(|2,0⟩-|0,2⟩)/\sqrt{2}$.

For the Adjoint representation of a Lie algebra, the operator associated with $x\in L$ isThe Jacobi identity gives $[{\mathrm{ad}}_x,{\mathrm{ad}}_y]={\mathrm{ad}}_{[x,y]}$, so this really is a Lie algebra representation. Define the lower central series of a Lie algebra by $L^1=L$ and $L^{r+1}=[L,L^r]$. A nilpotent Lie algebra is one for which $L^{c+1}=0$ for some integer $c$. If this holds, ${\mathrm{ad}}_x$ carries $L^r$ into $L^{r+1}$, and consequently $({\mathrm{ad}}_x{)}^{c}L=0$. This proves one implication without any condition on the field.

For the converse, we prove the linear form of the Engel theorem: if $A\subseteq \mathrm{End}(V)$ is a finite-dimensional Lie algebra consisting entirely of nilpotent linear maps, with $V\ne 0$ finite-dimensional, then there is a nonzero vector annihilated by all of $A$. This argument works over any field.

Induct on $\dim A$. The assertion is immediate when $A=0$. First observe that if $b^m=0$, then its commutator action on $\mathrm{End}(V)$ is nilpotent, becauseLet $B$ be a proper Lie subalgebra of $A$. The commutator action of $B$ on $A/B$ therefore consists of nilpotent linear maps. Its image has dimension at most $\dim B<\dim A$, so induction supplies a nonzero coset $a+B$ annihilated by $B$. Thus $[B,a]\subseteq B$, and the normalizer $N_A(B)=\{a\in A:[a,B]\subseteq B\}$ strictly contains $B$. This is the Engel normalizer lemma.

Choose $B$ maximal among proper Lie subalgebras. Its normalizer must be $A$, so $B$ is a Lie algebra ideal. Moreover $\dim (A/B)=1$: otherwise the inverse image of a one-dimensional subalgebra of $A/B$ would lie strictly between $B$ and $A$. By induction,Since $B$ is an ideal, $W$ is $A$-invariant: $b(av)=a(bv)+[b,a]v=0$. Take $x\in A\setminus B$. Its restriction to $W$ is nilpotent and hence has a nonzero kernel. A nonzero vector in that kernel is annihilated by $B$ and $x$, and therefore by all of $A$. This completes the induction.

Apply this result successively to $V$, then its quotient by a common annihilated line, and so on. We obtain a complete flag $0=V_0\subset V_1\subset \cdots \subset V_n=V$ with $A{V}_i\subseteq V_{i-1}$. Equivalently every element of $A$ is strictly upper triangular in one common basis; any product of $n$ such operators is zero.

Now take $V=L$ and $A=\mathrm{ad}(L)$. The hypothesis supplies exactly the nilpotence needed above. Every iterated bracket of length $n+1$, where $n=\dim L$, is an application of $n$ adjoint operators and vanishes. These brackets span $L^{n+1}$, giving $L^{n+1}=0$. The zero algebra is already nilpotent. Thus $L$ is nilpotent if and only if every ${\mathrm{ad}}_x$ is nilpotent. The common flag is essential: separate nilpotence of unrelated operators would not imply that their mixed products vanish.

Use the derived series of a Lie algebraA solvable Lie algebra has $L^{(r)}=0$ for some $r$. For a two-dimensional algebra with basis $u,v$, alternating bilinearity shows that $[L,L]$ is contained in the line spanned by $[u,v]$. A one-dimensional Lie algebra is abelian, so $[[L,L],[L,L]]=0$. Every two-dimensional complex Lie algebra is solvable, with derived length at most two. This dimension argument actually works over any field with the alternating definition of a Lie bracket.

There is a genuine qualification in the last request: the assertion about all Irreducible Lie algebra representations is true for finite-dimensional representations. Without that restriction, the infinite-dimensional simple module for the two-dimensional affine Lie algebra is a counterexample. Take $L=\mathbb{C}x\oplus \mathbb{C}y$, with $[x,y]=y$, and let it act on $V=\mathbb{C}[t]$ byThen $[x,y]f=tf(t-1)-(t-1)f(t-1)=yf$, so this is a Lie algebra representation. Any nonzero submodule is stable under multiplication by $t$, hence is an ideal $p(t)\mathbb{C}[t]$. Stability under $y$ implies $p(t)\mid p(t-1)$. The two polynomials have the same degree; their leading coefficients also agree, so $p(t-1)=p(t)$. In characteristic zero this forces $p$ to be constant. Thus $V$ is infinite-dimensional and irreducible, while $L$ is solvable. The literal unrestricted assertion is false.

Here is a proof of the intended finite-dimensional assertion, including the essential Lie theorem argument. We prove that a nonzero finite-dimensional representation $\rho :L\to \mathrm{End}(V)$ of a complex solvable algebra has a common eigenvector, by induction on $\dim L$. If $L=0$, any nonzero vector works. Otherwise $[L,L]\ne L$; choose a hyperplane $K$ containing $[L,L]$. It is a solvable ideal, and $L=K\oplus \mathbb{C}x$. Induction gives $v\ne 0$ and $\lambda \in K^{\ast }$ with $\rho (y)v=\lambda (y)v$ for all $y\in K$. Write $X=\rho (x)$ and suppress $\rho$ on elements of $K$.

Let $v_j=X^{j}v$ and let $m$ be the first index for which $v_m$ lies in the span of $v_0,\dots ,v_{m-1}$. The space $W=\mathrm{span}(v_0,\dots ,v_{m-1})$ is $X$-invariant. Repeatedly using $yX=Xy+[y,x]$ and $[y,x]\in K$ proves inductively thatThus $W$ is $K$-invariant, and every $y\in K$ has constant diagonal $\lambda (y)$ in this basis. The trace of a commutator is zero, soSince we work over $\mathbb{C}$, $\lambda ([x,y])=0$ for all $y\in K$. Therefore the nonzero simultaneous eigenspaceis $X$-invariant: $yXw=\lambda (y)Xw+\lambda ([y,x])w=\lambda (y)Xw$. A complex linear operator on a nonzero finite-dimensional space has an eigenvector. An eigenvector of $X{|}_{V_{\lambda }}$ is consequently a common eigenvector for $L$.

Its span is a one-dimensional submodule. If $V$ is irreducible, that span must be all of $V$. Hence every finite-dimensional irreducible representation of a finite-dimensional complex solvable Lie algebra has dimension one. Conversely a one-dimensional representation is given by a linear functional on $L/[L,L]$, because every commutator acts by zero. Passing repeatedly to quotients also gives the simultaneous triangularization of a Lie algebra representation. Neither the eigenvector step nor the trace argument extends to the infinite-dimensional counterexample above.

Here root systems are finite, reduced and crystallographic, as for complex semisimple Lie algebras. For a base of a root system $\Delta =\{{\alpha }_1,\dots ,{\alpha }_r\}$, the Dynkin diagram has one vertex for each simple root. Two distinct vertices have $a_{ij}{a}_{ji}$ bonds, with the arrow on a multiple bond pointing toward the shorter root. The classification of finite crystallographic Dynkin diagrams consists of the following connected diagrams; the integers describing arms count vertices away from the unique branching vertex.

This is classification up to diagram isomorphism and overall rescaling of the root system. It includes the conventional identifications $D_3\cong A_3$ and $B_1\cong A_1$ when those low-rank notations are used. The weighting here encodes root lengths and bonds; it is not the separate theory of vertex labels attached to nilpotent orbits. Dropping the crystallographic or reduced hypothesis changes the classification.

Precisely, a base of a root system is a vector-space basis $\Delta \subset \Phi$ such that each root has a unique expansion $\beta ={\sum }_i{n}_i{\alpha }_i$, with integer coefficients all nonnegative or all nonpositive. This defines the positive and negative roots. We use the column-coroot convention for the Cartan matrix:Some conventions transpose this matrix; specifying which index labels the coroot resolves that difference. The Weyl group is the subgroup of orthogonal transformations generated by the Weyl reflections

For the C3 root system, take an orthonormal basis $e_1,e_2,e_3$ of ${\mathbb{R}}^3$ andThere are six long roots of squared length $4$ and twelve short roots of squared length $2$. A base is ${\alpha }_1=e_1-e_2$, ${\alpha }_2=e_2-e_3$, ${\alpha }_3=2e_3$. The Cartan matrix convention for C3 above givesThus the double bond connects ${\alpha }_2$ to ${\alpha }_3$, with its arrow directed toward ${\alpha }_2$.

An explicit complex Lie algebra realizing this root system is the symplectic Lie algebraIt is closed under commutators, since the two identities $X^{T}J=-JX$, $Y^{T}J=-JY$ imply $[X,Y{]}^{T}J=-J[X,Y]$. Equivalently,so its dimension is $9+6+6=21$. A Cartan subalgebra isWith $E_{ab}$ denoting a matrix unit, the root spaces have generators $E_{ij}-E_{3+j,3+i}$ for $e_i-e_j$ ($i\ne j$), $E_{i,3+j}+E_{j,3+i}$ for $e_i+e_j$ ($i<j$), and $E_{i,3+i}$ for $2e_i$; the analogous lower-left generators give the negative roots. These are all eighteen root spaces, each one-dimensional, and together with $H$ they exhaust the algebra.