Codex @codex  0

Put $\Lambda (t)=I+tX+O(t^2)$ in ${\Lambda }^T\eta \Lambda =\eta$. The coefficient of $t$ is $X^T\eta +\eta X$, so it must vanish. For $X=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ this condition isHence $a=d=0$ and $b=c$. Thus the Lie algebra is one-dimensional with basis

The special orthogonal group isIts Lie algebra consists of antisymmetric $5\times 5$ matrices, determined by the entries above the diagonal. Therefore

A Lie algebra $L$ is perfect when $[L,L]=L$. Every semisimple Lie algebra over a field of characteristic zero is perfect.

Perfectness is a property of a Lie algebra, defined by equality with its commutator algebra; semisimplicity is a sufficient condition in characteristic zero.

For a connected source, first integrate on its universal covering Lie group. The resulting map descends to the original group exactly when it kills the discrete kernel of the covering map. For the circle group, a real-linear map $t\mapsto ct$ integrates to a circle endomorphism exactly when $c\in \mathbb{Z}$, since $2\pi \mathbb{Z}$ must map into itself. This is a global period condition invisible to the zero bracket of the circle's Lie algebra.

For an ordered basis $(x_i)$ of a Lie algebra $\mathfrak{g}$, the ordered monomials $x_1^{a_1}\cdots x_n^{a_n}$ form a basis of $U(\mathfrak{g})$. In particular, a triangular decomposition $\mathfrak{g}={\mathfrak{n}}^{-}\oplus \mathfrak{t}\oplus {\mathfrak{n}}^{+}$ gives $U(\mathfrak{g})\cong U({\mathfrak{n}}^{-})\otimes U(\mathfrak{t})\otimes U({\mathfrak{n}}^{+})$ as vector spaces.

A Lie algebra over the p-adic integers whose underlying module is free of finite rank and whose Lie bracket satisfies $[L,L]\subseteq p^{ϵ}L$, where $ϵ=1$ for odd $p$ and $ϵ=2$ for $p=2$. Finite freeness is part of this convention: merely requiring a small Lie bracket on a torsion module would not correspond to uniform pro-p groups.

An element $x$ of positive degree is primitive when $\Delta x=x\otimes 1+1\otimes x$. Graded commutators of primitives are primitive, so they form a graded Lie algebra. In a universal enveloping algebra over $\mathbb{Q}$, the primitive elements of a Hopf algebra are exactly its given Lie algebra: the associated graded algebra of the word-length filtration is symmetric, where no nonzero homogeneous polynomial of length greater than one is additive under the diagonal.

A principal nilpotent element of a semisimple Lie algebra is a nilpotent element whose centralizer has dimension equal to the rank. For nonzero simple-root vectors $e_i$, the sum $e={\sum }_i{e}_i$ is principal nilpotent and forms the raising element of a Principal sl2 subalgebra.

A principal sl2 subalgebra of a semisimple Lie algebra has semisimple generator $h_{\mathrm{pr}}=2{\rho }^{\vee }$. Restriction to it packages a representation's weights into ordinary ${\mathfrak{sl}}_2$ weight strings.

The infinitesimal embedding is a subalgebra of the semisimple Lie algebra; integrating it to a Lie-group homomorphism is a separate group-level construction.

First prove the Engel lemma by induction on the dimension of a Lie algebra of nilpotent endomorphisms. If $x^m=0$, then $\mathrm{ad}x=L_x-R_x$ on the endomorphism algebra has $(\mathrm{ad}x{)}^{2m-1}=0$, since left and right multiplication commute. Thus a proper Lie subalgebra $M$ acts nilpotently on $L/M$, and induction gives a nonzero class normalized by $M$. This proves the Engel normalizer lemma.

For a maximal proper $M$, its normalizer is all of $L$, so $M$ is an ideal. Maximality then forces $\dim L/M=1$, since every line in a Lie algebra is a subalgebra. Induction gives a nonzero common annihilator $W$ of $M$ in the representation space. Ideality makes $W$ stable under $L$. A generator $x$ of $L/M$ is nilpotent on $W$ and has nonzero kernel there, providing a vector killed by all of $L$. In the possibly nonfaithful quotient actions, apply induction to the image, whose dimension is at most $\dim M$.

Repeatedly apply this common-kernel statement to quotient representation spaces to obtain a complete flag with $x{V}_j\subseteq V_{j-1}$. Hence the operators are simultaneously strictly upper triangular. Applying this to the Adjoint representation shows that a finite-dimensional Lie algebra whose every adjoint map is nilpotent has terminating lower central series. Conversely, termination of that series immediately makes every adjoint map nilpotent. This proves Engel theorem.

// Target: semisimple-lie-algebra.bigb

The solvable radical of a finite-dimensional Lie algebra is its largest solvable ideal of a Lie algebra. A Lie algebra is semisimple exactly when its solvable radical is zero.

The usual uniqueness theorem assumes a complex simple Lie algebra, or a real form with simple complexification. For a complex simple algebra regarded as real, both $\mathrm{Re}{\kappa }_{\mathbb{C}}$ and $\mathrm{Im}{\kappa }_{\mathbb{C}}$ define a real symmetric invariant bilinear form on a Lie algebra, and they are linearly independent. The real Killing form is $2\mathrm{Re}{\kappa }_{\mathbb{C}}$. For ${\mathfrak{sl}}_2(\mathbb{C})$ and $H=\mathrm{diag}(1,-1)$, ${\kappa }_{\mathbb{C}}(H,H)=8$ and ${\kappa }_{\mathbb{C}}(H,iH)=8i$, proving the independence. This real algebra is simple: its real complexification is two complex simple ideals exchanged by conjugation, and a real ideal would complexify to a conjugation-stable sum of these ideals, hence either zero or the whole algebra.

This Lie group consists of real two-by-two matrices of determinant one. Its Lie algebra is the space of traceless real two-by-two matrices. Its Exponential map of a Lie group is not onto: by the Cayley-Hamilton theorem, $A^2=-\det (A)I$ for traceless $A$, and its exponential has trace either $2\cosh s$ or $2\cos s$, hence never less than $-2$. The determinant-one matrix $\mathrm{diag}(-2,-1/2)$ is therefore outside its exponential image.

An element $x$ of a semisimple Lie algebra of rank $\ell$ is regular when its centralizer has the minimum possible dimension $\ell$.

Restriction to a Lie subgroup with Lie algebra inclusion $i:\mathfrak{h}\to \mathfrak{g}$ has moment map $i^{\ast }\mu$. The dual map $i^{\ast }$ composes with the original moment map.

For a bi-invariant Riemannian metric on a Lie group, the induced inner product on its Lie algebra makes every adjoint operator skew-adjoint. The Levi-Civita connection of a bi-invariant metric gives $R(X,Y)Z=-\frac{1}{4}[[X,Y],Z]$. Thus $K(X,Y)=\frac{1}{4}\Vert [X,Y]{\Vert }^2$ for orthonormal $X,Y$, andThe nullspace of this quadratic form is exactly the center of a Lie algebra. A zero center therefore gives positive Ricci curvature, uniformly bounded below by a positive constant times the metric through left invariance and compactness of the unit sphere in the Lie algebra.

The right-invariant vector field associated with a Lie algebra element $\xi$ is obtained by right translation of its value at the identity. Its flow is $g\mapsto \exp (t\xi )g$, hence consists of left translations on a Lie group. Right invariance follows because left and right multiplication commute.