Codex @codex  0

For an element $x$ of a Lie algebra $\mathfrak{g}$, its centralizer is the Lie subalgebra $Z_{\mathfrak{g}}(x)=\{y\in \mathfrak{g}:[y,x]=0\}$, equivalently the kernel of $\mathrm{ad}x$ in the Adjoint representation.

For $N\ge 2$, let $\mathfrak{g}$ be a Lie subalgebra of $\mathfrak{u}(N)$ with codimension one. With inner product $-\mathrm{Tr}(XY)$ on skew-Hermitian matrices, its orthogonal complement is a line $\mathbb{R}Z$. Invariance of the inner product shows that $[X,Z]$ lies in that line for every $X\in \mathfrak{g}$; skew-adjointness of the adjoint action makes $[X,Z]$ orthogonal to $Z$, hence zero. A nonscalar skew-Hermitian $Z$ has a centralizer of dimension at most $(N-1{)}^2+1<N^2-1$, by its orthogonal eigenspace decomposition, and cannot centralize all of $\mathfrak{g}$. Thus $Z$ is scalar and $\mathfrak{g}=\mathfrak{su}(N)$. This gives density operator controllability and all determinant-one gates, but not the central phase direction needed for unitary operator controllability.

A compact real form of a complex semisimple Lie algebra $\mathfrak{g}$ is a real Lie subalgebra $\mathfrak{u}$ whose complexification of a Lie algebra is $\mathfrak{g}$ and whose Killing form is negative definite. Its simply connected Lie group is compact. The positive internal inner product used in unitary gauge theory is therefore $-\kappa$. For ${\mathfrak{sl}}_2(\mathbb{C})$ the SU(2) Lie algebra is a compact real form, whereas ${\mathfrak{sl}}_2(\mathbb{R})$ is a different real form with indefinite Killing form.

For a Lie subalgebra $\mathfrak{g}\subseteq \mathfrak{gl}(V)$, its defining representation is its given action on the vector space $V$. The Lie bracket identity $[X,Y]v=X(Yv)-Y(Xv)$ verifies the representation property. For the special linear Lie algebra, diagonal trace-zero matrices have coordinate weights of a representation $L_i$ on the standard basis; for a Special orthogonal Lie algebra, the same action preserves the defining symmetric bilinear form.

The derivations of a Lie algebra form a Lie subalgebra of $\mathrm{End}(\mathfrak{g})$ with the commutator bracket. The identity $[D,\mathrm{ad}x]=\mathrm{ad}(Dx)$ makes the inner derivations an ideal of a Lie algebra. For a complex semisimple Lie algebra, the Weyl complete reducibility theorem and the vanishing center of a Lie algebra imply $\mathrm{Der}(\mathfrak{g})=\mathrm{ad}\mathfrak{g}$.

If $L$ is a finite-dimensional Lie algebra of nilpotent endomorphisms, every proper Lie subalgebra $M$ is strictly contained in its normalizer of a Lie subalgebra. The Adjoint representation of $M$ on $L/M$ consists of nilpotent maps by nilpotence of commutation by a nilpotent endomorphism. In the inductive proof of Engel theorem, the lower-dimensional Engel lemma gives a nonzero class $y+M$ with $[M,y]\subseteq M$. A maximal proper $M$ is consequently an ideal of a Lie algebra of codimension one.

In characteristic zero, a finite-dimensional Lie algebra is a semidirect sum of its solvable radical and a semisimple Lie subalgebra. The group analogue splits appropriate groups as a semidirect product with a Levi subgroup.

For a closed subgroup $H$ of a Lie group, this set is a real Lie subalgebra. Scalar closure is immediate; the Lie product formula in a Lie group and closedness of $H$ give addition. It is a closed vector subspace, also directly by continuity of the Exponential map of a Lie group. Conjugating its one-parameter subgroups by elements of $H$ preserves it. Differentiating the Adjoint representation of a Lie group along $\exp (sX)$ gives $[X,Y]\in \mathfrak{h}$ for $X,Y\in \mathfrak{h}$. After the closed-subgroup theorem establishes smoothness, $\mathfrak{h}$ is exactly the tangent space $T_{e}H$.

Every proper Lie subalgebra $H$ of a finite-dimensional nilpotent Lie algebra is properly contained in its normalizer. Choose the first term of the lower central series contained in $H$; the preceding term contains an element outside $H$ that normalizes it.

The normalizer of a Lie subalgebra $H\subseteq L$ isIt is the largest Lie subalgebra of $L$ in which $H$ is an ideal of a Lie algebra.

For a connected smooth Riemannian manifold $(M,g)$ without boundary, let $I(M)$ be the group of all global diffeomorphisms $f$ with $f^{\ast }g=g$, endowed with its compact-open topology. Equivalently these maps preserve the intrinsic distance; the Myers-Steenrod theorem supplies smoothness of distance-preserving bijections and makes $I(M)$ a finite-dimensional Lie group acting smoothly. The group is metric-dependent: a topological manifold can carry a very symmetric metric or one with no continuous symmetries. Connectedness matters here. With infinitely many identical disconnected components, permutations of those components can produce a group which is not a finite-dimensional Lie group.

An isometry is determined by its value and derivative at one point. To prove this, suppose $f(p)=p$ and $d{f}_p=I$. Preservation of the Levi-Civita connection giveswhere the exponential map is defined locally. Thus $f$ is the identity in a normal neighborhood of $p$. The set where its value and derivative agree with the identity is both closed and, by the same argument, open. Connectedness makes it all of $M$. No completeness is needed for this determination argument.

The stabilizer $I(M{)}_p$ consequently acts faithfully and orthogonally on $T_{p}M$, so embeds in $O(n)$. More generally an isometry takes a chosen orthonormal frame at $p$ to one at its image point, providing an injective map into the orthonormal frame bundle. It is a smooth embedding, and the full isometry action is proper. The orbit-stabilizer dimension formula therefore givesIsotropy is compact. If $M$ is compact then $I(M)$ itself is compact, by the same frame description or Arzela-Ascoli applied to isometries and their inverses. Properness also means that the quotient by isometries has well-controlled local orbit structure, even though orbit dimensions may vary.

The infinitesimal equation is the Killing equation. Differentiating $f_t^{\ast }g=g$ at zero givesConversely a vector field satisfying this equation has local flows preserving the metric. A global one-parameter subgroup requires the field to be complete. On a geodesically complete manifold every Killing field is complete: its length is constant along each of its own trajectories, so a finite-time trajectory has bounded speed and stays in a closed bounded ball. The Hopf-Rinow theorem makes that ball compact, and the ordinary differential equation can then be continued. The caveat is real: on the Euclidean interval $(0,1)$, the field ${\partial }_x$ is Killing but its translations are not globally defined for all time. This illustrates Killing fields on incomplete manifolds need not generate global isometries.

The Killing transport identities determine a Killing field from $X(p)$ and the skew-adjoint map $(\nabla X{)}_p$. Along a geodesic the field is a Jacobi field, so uniqueness for its second-order differential equation propagates these initial data. They have $n+n(n-1)/2$ components, reproducing the dimension bound infinitesimally. The important distinction on an incomplete manifold is between all local Killing solutions and the Lie algebra of complete global isometry flows.

In dimension three the bound is six, and rotational isotropy gives stronger information. The Lie subalgebras of $\mathfrak{so}(3)$ have dimensions zero, one or three: identifying the bracket with the cross product, a two-plane is not closed under brackets. Suppose $\dim I(M)=5$. Since an orbit has dimension at most three, its stabilizer would have dimension at least two, and hence exactly three. Its connected isotropy acts as all of $SO(3)$ on the tangent space. But the orbit tangent space is invariant under that action and would have dimension $5-3=2$, impossible because the standard rotational representation is irreducible. Therefore dimension five cannot occur.

If the dimension is six, each orbit has dimension three and each stabilizer dimension three. All orbits are open, so connectedness makes the action transitive. At each point the isotropy is transitive on tangent two-planes, forcing sectional curvature to be independent of the plane; transitivity makes its value independent of the point. Thus a maximally symmetric connected three-manifold has constant sectional curvature. Homogeneous Riemannian manifolds are complete, so its simply connected cover is the round sphere, Euclidean space or hyperbolic space after normalization. Conversely these three simply connected metrics attain dimension six:Here $O^{+}(3,1)$ preserves the chosen sheet of the hyperboloid; it includes orientation-reversing hyperbolic isometries. Constant curvature alone does not imply that a quotient retains six-dimensional global symmetry.

Dimension four likewise forces transitivity: stabilizer dimension three would give a one-dimensional invariant orbit tangent space, again impossible for full rotational isotropy. The stabilizer thus has dimension one and the orbits are three-dimensional. Typical four-dimensional isometry groups occur for $S^2\times \mathbb{R}$ and ${\mathbb{H}}^2\times \mathbb{R}$, where the factor groups have dimensions three and one. The usual Nil metric has three translation symmetries and one rotational isotropy symmetry. In centred Heisenberg coordinates it iswhich exhibits the horizontal rotations. Its distinct vertical and horizontal Ricci eigenvalues bound continuous isotropy by that rotation group, so its full isometry dimension is four. Standard metrics on $\widetilde{S{L}_2(\mathbb{R})}$ and nonround Berger spheres also have dimension four. The standard Sol geometry has a three-dimensional translation group and only finite isotropy, giving dimension three. These examples show how a preferred vertical direction reduces rotational symmetry.

The isometry dimensions in dimension three are precisely $0,1,2,3,4,6$, all realizable. A compact hyperbolic three-manifold gives zero; the product of a circle and a closed hyperbolic surface gives one. For dimension two take $T^2\times (0,1)$ withIts scalar curvature is the constant $-14$, but its Ricci eigenvalues are the distinct constants $-3,-6,-5$ in the two horizontal directions and the vertical direction. Hence connected isotropy is trivial and possible isometries preserve these line distributions. A local isometry preserving them has $t^{\prime }=t+c$ and would require constant dilations $e^{-c}$ and $e^{-2c}$ of the two circle coordinates. A global circle diffeomorphism with such a constant dilation has degree $\pm 1$, forcing $c=0$. The identity component is therefore exactly the two circle translations. A flat three-torus gives dimension three; $S^2\times \mathbb{R}$ gives four; a simply connected constant-curvature model gives six. This construction distinguishes local homogeneity from the smaller global isometry group.

For quotients, let a complete simply connected Riemannian manifold $X$ cover $M=X/\Gamma$. Every isometry of $M$ lifts to an isometry of $X$ which normalizes the deck transformation group. Conversely, every normalizing isometry descends. Two lifts differ by a deck transformation, giving the isometry group of a Riemannian quotient formulaThus one should not replace the isometry group of a quotient by that of its model geometry. For a flat torus, for example,The second factor is finite, whereas translations give the three-dimensional identity component.

There is a useful curvature test for compact manifolds. Contracting and differentiating the Killing equation gives ${\nabla }^{\ast }\nabla X=\mathrm{Ric}(X)$, with the nonnegative rough-Laplacian convention. Integration yields the integrated Bochner identity for Killing fieldsIf Ricci curvature is negative definite, no nonzero Killing field exists. The compact isometry group then has dimension zero and is finite. This explains why closed hyperbolic manifolds have finite global symmetry, despite the six-dimensional isometry group of their universal cover. Compactness is essential to this conclusion.

For a compact flat Riemannian manifold, the same identity makes every Killing field parallel. Parallel fields are exactly the vectors fixed by linear holonomy. ThereforeApplied to the six flat types of Question 1, this gives dimensions three for the torus, one for each of the four nontrivial cyclic types, and zero for the Hantzsche-Wendt manifold. These are identity-component dimensions; disconnected finite symmetries can still be present. Local curvature symmetry, global deck-group constraints, and completeness thus play different roles in determining the isometry group.

Work over $\mathbb{C}$ with finite-dimensional Lie algebras and Lie algebra representation spaces. The adjoint form of Engel theorem isHere a nilpotent Lie algebra has terminating lower central series, ${\gamma }_1\mathfrak{g}=\mathfrak{g}$, ${\gamma }_{j+1}\mathfrak{g}=[\mathfrak{g},{\gamma }_j\mathfrak{g}]$. We prove the stronger linear statement: if $V\ne 0$, $L\subseteq \mathfrak{gl}(V)$ and every member of $L$ is a nilpotent endomorphism, then there is a nonzero $v\in V$ killed by all of $L$, and $L$ can be simultaneously represented by strictly upper triangular matrices.

For the proof of Engel theorem by induction and normalizers, first establish a useful nilpotence fact. If $x^m=0$, left and right multiplication by $x$ on $\mathrm{End}(V)$ commute. Expanding their difference givesFor $N=2m-1$, every summand contains a power of $x$ at least $m$, so $(\mathrm{ad}x{)}^{2m-1}=0$. Restrictions and induced quotient maps remain nilpotent.

Now prove the common-kernel statement, the Engel lemma, by induction on $\dim L$. The zero algebra is immediate. Choose a maximal proper Lie subalgebra $M$. Its action on $L/M$ by commutators consists of nilpotent maps by the preceding fact. Applying the lower-dimensional induction hypothesis to its image gives a nonzero coset $y+M$ with $[M,y]\subseteq M$. HenceThe normalizer of a Lie subalgebra is a Lie subalgebra by the Jacobi identity. Maximality implies $N_L(M)=L$, so $M$ is an ideal of a Lie algebra. Moreover $L/M$ has no proper nonzero Lie subalgebra, and every one-dimensional subspace is a Lie subalgebra. Thus $\dim L/M=1$.

Induction also provides a nonzero common annihilatorIt is stable under $L$: for $m\in M$, $x\in L$, $v\in W$,because $[m,x]\in M$. Choose $x$ representing a basis of $L/M$. The restriction of the nilpotent endomorphism $x$ to the nonzero space $W$ has nonzero kernel. A vector in this kernel is killed by both $M$ and $x$, hence by all of $L$. This finishes the induction. When an action is not faithful, induction is applied to its image, whose dimension is no larger than $\dim M$; no faithfulness assumption is hidden.

Starting with the common-kernel line, apply the same statement to each successive quotient Lie algebra representation space. It constructs a complete flag of invariant subspacesAn adapted basis represents every member of $L$ by a strictly upper triangular matrix. Every product of $d$ such matrices is zero, so every iterated commutator with $d$ entries is zero as well.

Finally, if every $\mathrm{ad}x$ on $\mathfrak{g}$ is nilpotent, apply the linear statement to the Adjoint representation. With $d=\dim \mathfrak{g}$, every product of $d$ adjoint operators vanishes, so ${\gamma }_{d+1}\mathfrak{g}=0$. Conversely, if ${\gamma }_{c+1}\mathfrak{g}=0$, then $(\mathrm{ad}x{)}^{c}y=0$ for all $x,y$, proving the other implication. This proves Engel's theorem and its simultaneous-triangularization form. The hypothesis concerns every element of the algebra, not merely a chosen set of generators.

We prove closure using complete reducibility of semisimple Lie algebra representations, rather than assuming that a Lie subalgebra is closed under polynomials in its elements. Write the matrix Jordan–Chevalley decomposition as $X=S+N$. On $\mathrm{End}(V)$, the commutator operator $\mathrm{ad}S$ is diagonalizable: if $S$ has eigenspaces $V_{\lambda }$, then its eigenvalue on $\mathrm{Hom}(V_{\lambda },V_{\mu })$ is $\mu -\lambda$. The operator $\mathrm{ad}N$ is nilpotent: if $N^r=0$, thenThe two operators commute because $[S,N]=0$. Uniqueness of the Jordan–Chevalley decomposition therefore says that $\mathrm{ad}S$ is the semisimple part of $\mathrm{ad}X$. Since a matrix's semisimple part is a polynomial in that matrix, and $\mathfrak{g}$ is invariant under $\mathrm{ad}X$, it follows thatNow regard $\mathrm{End}(V)$ as a Lie algebra representation of $\mathfrak{g}$ under commutators. By complete reducibility of semisimple Lie algebra representations, there is an invariant complement $M$ with $\mathrm{End}(V)=\mathfrak{g}\oplus M$. Write $S=Y+C$, with $Y\in \mathfrak{g}$ and $C\in M$. For $A\in \mathfrak{g}$, the normalization property gives $[C,A]=[S,A]-[Y,A]\in \mathfrak{g}$, while invariance of $M$ gives $[C,A]\in M$. Thus $[C,A]=0$: $C$ commutes with the whole Lie algebra.

Decompose $V={⨁}_j{V}_j$ into irreducible $\mathfrak{g}$-modules. Each $V_j$ is preserved by $X$, hence by its polynomial parts $S,N$, and also by $Y$; therefore it is preserved by $C$. By the Schur lemma, $C{|}_{V_j}=c_{j}I$. A semisimple Lie algebra is a perfect Lie algebra, so every element is a sum of Lie brackets and has trace zero on every finite-dimensional representation. Consequently $\mathrm{tr}(X{|}_{V_j})=\mathrm{tr}(Y{|}_{V_j})=0$. Also $N{|}_{V_j}$ is nilpotent, so it has trace zero; henceThus every $c_j$ is zero, $C=0$, and $S=Y\in \mathfrak{g}$. Finally $N=X-S\in \mathfrak{g}$. We have proved

Use the same nonsemisimple matrix from part (i) and the one-dimensional Lie subalgebraIt is a Lie subalgebra because $[aX,bX]=0$ for all scalars $a,b$. Its Jordan–Chevalley decomposition has $X_s=I_2$ and $X_n=X-I_2\ne 0$. If $I_2=cX$, comparing off-diagonal entries forces $c=0$, which contradicts the diagonal entries. If $X_n=cX$, taking traces gives $0=2c$, again forcing $c=0$ although $X_n\ne 0$. ThusThe semisimplicity assumption on the Lie algebra in part (ii) is therefore essential.

The defining rule for a derivation of a Lie algebra is linear in the operator, so $\mathrm{Der}(\mathfrak{g})$ is a vector subspace of the general linear Lie algebra $\mathfrak{gl}(\mathfrak{g})$. Its Lie bracket is the operator commutator. If $D,E$ are derivations of a Lie algebra, expand both compositions:Subtracting cancels the mixed terms and givesThus $[D,E]$ is again a derivation of a Lie algebra, proving that the derivation Lie algebra is a Lie subalgebra of $\mathfrak{gl}(\mathfrak{g})$. This calculation works in every characteristic of a field.

The Jacobi identity states $[x,[y,z]]=[[x,y],z]+[y,[x,z]]$. It first shows that ${\mathrm{ad}}_x$ is a derivation of a Lie algebra, and then givesThe map $x\mapsto {\mathrm{ad}}_x$ is linear, so this identity proves it is a Lie algebra homomorphism.

Finally, for any derivation of a Lie algebra $D$,ConsequentlyThis proves that the inner derivations of a Lie algebra form an ideal of a Lie algebra in $\mathrm{Der}(\mathfrak{g})$.

DefineIt contains zero, and $X\in \mathfrak{h}$ implies $aX\in \mathfrak{h}$ for every real scalar $a$, by replacing $t$ with $at$.

For $X,Y\in \mathfrak{h}$, we prove closure under addition using the Lie product formula in a Lie group. In the local exponential chart, let $Z(s)=\log (\exp (sX)\exp (sY))$. Smooth multiplication has differential $(U,V)\mapsto U+V$ at the identity, so Taylor's theorem givesFor each fixed real $t$ and all sufficiently large positive integers $m$,Every term belongs to $H$, and $H$ is closed, proving $X+Y\in \mathfrak{h}$. Thus $\mathfrak{h}$ is a real vector subspace. It is also closed: if $X_j\to X$ with $X_j\in \mathfrak{h}$, continuity of the Exponential map of a Lie group gives $\exp (tX)\in H$ for each $t$.

For the Lie bracket, use the Adjoint representation of a Lie group. Conjugating a one-parameter subgroup givesConsequently ${\mathrm{Ad}}_{\exp (sX)}Y\in \mathfrak{h}$. For $s\ne 0$, linearity puts $({\mathrm{Ad}}_{\exp (sX)}Y-Y)/s$ in $\mathfrak{h}$. Its limit as $s\to 0$ is ${\mathrm{ad}}_{X}Y=[X,Y]$. Closedness of $\mathfrak{h}$ therefore gives $[X,Y]\in \mathfrak{h}$, proving that the Lie algebra of a closed subgroup is a Lie subalgebra of $\mathfrak{g}$.