Semisimple Lie Algebras, Cohomology, and Levi Theory — Examples

In the ordered basis $(e,f,h)$, the relations $[h,e]=2e$, $[h,f]=-2f$, and $[e,f]=h$ give ${\mathrm{ad}}_e=\left(\begin{array}{ccc}0& 0& -2\\ 0& 0& 0\\ 0& 1& 0\end{array}\right),{\mathrm{ad}}_f=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 2\\ -1& 0& 0\end{array}\right),{\mathrm{ad}}_h=\left(\begin{array}{ccc}2& 0& 0\\ 0& -2& 0\\ 0& 0& 0\end{array}\right)\mathrm{.}$

Squaring the last matrix gives trace $8$. Multiplying the first two matrices in either order gives trace $4$. Their squares have trace zero, and multiplying either by ${\mathrm{ad}}_h$ in either order also has trace zero. By [L1], these are precisely the claimed pairings.

The Gram matrix has determinant $\det \left(\begin{array}{ccc}0& 4& 0\\ 4& 0& 0\\ 0& 0& 8\end{array}\right)=-128.$ Characteristic zero makes this scalar nonzero, so $K$ is nondegenerate. In particular [L2] recovers semisimplicity. Every basis pairing was calculated, including the zero pairings, and no choice principle is used.

On $\mathrm{End}(k^n)$, ${\mathrm{ad}}_X=L_X-R_X$. The trace identities $\mathrm{tr}(L_X{L}_Y)=n\mathrm{tr}(XY),\mathrm{tr}(R_X{R}_Y)=n\mathrm{tr}(XY),\mathrm{tr}(L_X{R}_Y)=\mathrm{tr}(X)\mathrm{tr}(Y)$follow by applying the maps to the matrix units $E_{ij}$. Hence${\mathrm{tr}}_{{\mathfrak{g}\mathfrak{l}}_n}({\mathrm{ad}}_X{\mathrm{ad}}_Y)=2n\mathrm{tr}(XY)-2\mathrm{tr}(X)\mathrm{tr}(Y)\mathrm{.}$ For traceless $X,Y$, the central line in ${\mathfrak{g}\mathfrak{l}}_n=kI\oplus {\mathfrak{s}\mathfrak{l}}_n$ contributes zero to the adjoint trace, proving the first formula.

Use the standard split Cartan and root matrices, all defined over $k$. For a Cartan element write its coordinates as $(t_1,\dots ,t_r)$. The roots of ${\mathfrak{s}\mathfrak{o}}_{2r}$ are $\pm {\epsilon }_i\pm {\epsilon }_j$; those of ${\mathfrak{s}\mathfrak{o}}_{2r+1}$ add $\pm {\epsilon }_i$. Summing $\alpha (H)\alpha (H^{\mathrm{\prime }})$ over the roots gives respectively $4(r-1){\sum }_i{t}_i{s}_i$ and $(4r-2){\sum }_i{t}_i{s}_i$. Since the defining matrix trace is $2{\sum }_i{t}_i{s}_i$, both results equal $(n-2)\mathrm{tr}(H{H}^{\mathrm{\prime }})$.

The roots of ${\mathfrak{s}\mathfrak{p}}_{2r}$ are $\pm {\epsilon }_i\pm {\epsilon }_j$ and $\pm 2{\epsilon }_i$. Their sum gives $4(r+1){\sum }_i{t}_i{s}_i$; dividing by the same defining trace $2{\sum }_i{t}_i{s}_i$ yields $2(r+1)\mathrm{tr}(H{H}^{\mathrm{\prime }})$. Invariance pairs a root space only with its opposite root space, and direct multiplication of the standard root matrices gives the same nonzero scalar there. Thus the identities on the Cartan and opposite-root pairs determine the displayed orthogonal and symplectic forms on the whole algebra.

The trace pairing is nondegenerate on each displayed matrix algebra: diagonal Cartan coordinates pair coordinatewise, while every root matrix pairs nontrivially with its opposite. The scalar multipliers $2n$, $n-2$, and $2(n+1)$ are nonzero in characteristic zero, so all three Killing forms are nondegenerate.

The excluded orthogonal ranks explain the endpoints: ${\mathfrak{s}\mathfrak{o}}_1=0$, ${\mathfrak{s}\mathfrak{o}}_2$ is one-dimensional abelian and has zero Killing form, and ${\mathfrak{s}\mathfrak{o}}_4\cong {\mathfrak{s}\mathfrak{l}}_2\oplus {\mathfrak{s}\mathfrak{l}}_2$ is semisimple but not simple (its formula is still $2\mathrm{tr}(XY)$). Thus none is silently included among the simple orthogonal cases. All calculations are finite and choice-free.

Since $n$ is invertible in $k$, every matrix has the unique decomposition $X=\frac{\mathrm{tr}X}{n}I+(X-\frac{\mathrm{tr}X}{n}I),$ whose second term is traceless. Thus ${\mathfrak{g}\mathfrak{l}}_n=kI\oplus {\mathfrak{s}\mathfrak{l}}_n$. The first summand is the center and the second is semisimple for $n\ge 2$; for $n=1$ it is zero, which is semisimple by convention. Hence [L1] makes ${\mathfrak{g}\mathfrak{l}}_n$ reductive for every $n\ge 1$.

Since $I$ is central, ${\mathrm{ad}}_I=0$. Therefore [L2] gives $K(I,X)=0$ for every $X$. The nonzero vector $I$ lies in the radical of the form, so it is degenerate; at $n=1$ it is identically zero. The calculation is finite and uses no choice.

The standard matrix-unit commutator argument shows that ${\mathfrak{s}\mathfrak{l}}_2$ and ${\mathfrak{s}\mathfrak{l}}_3$ are nonabelian simple in characteristic zero. Hence ${\mathfrak{s}\mathfrak{l}}_2\oplus 0$ and $0\oplus {\mathfrak{s}\mathfrak{l}}_3$ are simple ideals whose direct sum is $\mathfrak{g}$.

Applying [L1], the complete ideal list is $0,{\mathfrak{s}\mathfrak{l}}_2\oplus 0,0\oplus {\mathfrak{s}\mathfrak{l}}_3,\mathfrak{g}\mathrm{.}$ This includes the empty and full subcollections.

For $x\in {\mathfrak{s}\mathfrak{l}}_2$ and $y\in {\mathfrak{s}\mathfrak{l}}_3$, the two adjoint maps ${\mathrm{ad}}_{(x,0)}$ and ${\mathrm{ad}}_{(0,y)}$ act on opposite blocks, so their product is zero. Restriction to a block is its own adjoint trace. Thus $K_{\mathfrak{g}}=K_{{\mathfrak{s}\mathfrak{l}}_2}\perp K_{{\mathfrak{s}\mathfrak{l}}_3}$. Everything is finite and choice-free.

A linear map $\lambda :\mathfrak{g}\to k$ is a derivation precisely when $\lambda ([x,y])=x\cdot \lambda (y)-y\cdot \lambda (x)=0$. Thus $Z^1(\mathfrak{g},k)$ is exactly the space of linear forms vanishing on $[\mathfrak{g},\mathfrak{g}]$.

Every inner derivation into the trivial module has the form $x\mapsto x\cdot a=0$, so $B^1(\mathfrak{g},k)=0$ and $H^1=Z^1$.

Put $I=[\mathfrak{g},\mathfrak{g}]$. If $\lambda$ is in the space from step 1.1, define $\bar{\lambda }(x+I)=\lambda (x)$. This is well-defined because $x+I=y+I$ implies $x-y\in I$ and hence $\lambda (x-y)=0$; it is plainly linear and satisfies $\lambda =\bar{\lambda }\circ q$. Conversely every linear form on $\mathfrak{g}\mathrm{/}I$ pulls back along the linear map $q$ from [L2] to a form vanishing on $I$. These constructions are linear and inverse, proving the displayed natural isomorphism together with steps 1.1–1.2. If the abelianization is zero, both sides are zero; no basis or choice is used.

Since both the bracket of $\mathfrak{a}$ and its action on $kz$ are zero, every term in the Chevalley–Eilenberg differential of a $2$-cochain vanishes. Equivalently, the possible target $C^3(\mathfrak{a},kz)$ is already zero because ${\bigwedge }^3\mathfrak{a}=0$. Hence $d\omega =0$.

The same triviality makes the differential $C^1\to C^2$ zero, so the nonzero form $\omega$ is not a coboundary. Indeed $C^2(\mathfrak{a},kz)$ is one-dimensional, generated by $x^{\ast }\wedge y^{\ast }\otimes z$.

On $kz\oplus \mathfrak{a}$ define $[(a,u),(b,v)]=(\omega (u,v),0)$; this is the usual cocycle-extension formula here because the action and the bracket of $\mathfrak{a}$ are both zero. It is bilinear and alternating, and Jacobi holds because every bracket lies in $kz\oplus 0$, which brackets to zero. Its kernel $kz\oplus 0$ is abelian, its quotient is $\mathfrak{a}$, and the induced action is trivial, so it is the abelian extension represented by $\omega$. Its only nonzero basis bracket is $[(0,x),(0,y)]=(z,0)$. Renaming the three basis vectors gives exactly the displayed Heisenberg relations. This checks the zero and one-dimensional boundary spaces explicitly and uses no choice.

Put $V={\mathbb{R}}^3$. In the semidirect product, $[(v,A),(w,B)]=(Aw-Bv,[A,B])$. Hence $V\oplus 0$ is an abelian ideal, and the quotient by it is $\mathfrak{s}\mathfrak{o}(3)$.

Under the vector-space identification $u\mapsto A_u$, $A_u(v)=u\times v$, one has $[A_u,A_v]=A_{u\times v}$. If an ideal of $\mathfrak{s}\mathfrak{o}(3)$ contains a nonzero $A_u$, then the vectors $u\times v$ as $v$ varies span $u^{\perp }$, and a further bracket supplies the $u$-direction. Thus the ideal is all of $\mathfrak{s}\mathfrak{o}(3)$. The algebra is nonabelian and $[\mathfrak{s}\mathfrak{o}(3),\mathfrak{s}\mathfrak{o}(3)]=\mathfrak{s}\mathfrak{o}(3)$, so it is simple and not solvable.

Let $I$ be a solvable ideal of $\mathfrak{e}(3)$. Its image in the quotient is a solvable ideal of $\mathfrak{s}\mathfrak{o}(3)$, hence is zero by step 1.2. Thus $I\subseteq V$. Conversely $V$ is itself a solvable ideal by step 1.1, so it is the radical.

The subalgebra $0\oplus \mathfrak{s}\mathfrak{o}(3)$ is semisimple, meets $V$ in zero, and together with $V$ spans the whole algebra. It is therefore a Levi factor by [L1]. The zero intersections and full quotient are explicit, and no choice is used.

Since the action is nonzero, choose $v\in V$ and $x\in {\mathfrak{s}\mathfrak{l}}_2$ with $xv\mathrm{\ne }0$. By [L1], $[(0,v),(y,w)]=(0,-yv)$ lies in $0\oplus V$, and $[0\oplus V,0\oplus V]=0$. Therefore $({\mathrm{ad}}_{(0,v)}{)}^2=0$ and $\exp ({\mathrm{ad}}_{(0,v)})=1+{\mathrm{ad}}_{(0,v)}$.

In particular, $\exp ({\mathrm{ad}}_{(0,v)})(x,0)=(x,-xv)\mathrm{.}$ Its second component is nonzero for the chosen pair, so the image subalgebra is not $\mathfrak{s}$. An automorphism carries a Levi factor to a Levi factor, so both are Levi subalgebras.

This explicit automorphism is a one-factor instance of the finite products in [L2], with $(0,v)$ in the abelian nilpotent ideal $0\oplus V$. Thus the example witnesses both literal nonuniqueness and Malcev conjugacy. Selecting one pair from “the action is nonzero” uses no choice family.

For a unit quaternion $a$, the map $u\mapsto au{a}^{-1}$ preserves the norm and orientation on $\mathrm{Im}\mathbb{H}$, so it gives $q(a)\in \mathrm{SO}(3)$. It is a homomorphism. If it fixes every imaginary quaternion, then $a$ commutes with $i,j,k$, hence is real; unit length gives $a=\pm 1$. Thus $\ker q=\{\pm 1\}$.

Differentiating at $1$ sends an imaginary quaternion $u$ to $A_u(v)=uv-vu=2u\times v$. This map is injective and both real vector spaces have dimension three, so it is an isomorphism. Its bracket compatibility follows by differentiating the homomorphism. The image of $q$ is therefore an open subgroup of connected $\mathrm{SO}(3)$, hence all of it; $q$ is a two-sheeted covering.

Since $\mathrm{SU}(2)\cong S^3$ is simply connected, [L2] identifies $q$ as the universal cover. Its deck group is its kernel, so ${\pi }_1(\mathrm{SO}(3))\cong \{\pm 1\}\cong \mathbb{Z}\mathrm{/}2$. A Lie-group isomorphism is a diffeomorphism and would preserve the fundamental group; hence the groups are not isomorphic even though step 2.1 gives isomorphic Lie algebras. This realizes [L1].

The quaternionic calculation itself is finite. The declared ${\mathsf{A}\mathsf{C}}_{\omega }$ is propagated from the library's covering-group suppliers [L1]–[L2]; it is not silently strengthened and no additional choice is used here.

If $\mathfrak{n}$ has class $c$, every Lie monomial of bracket length greater than $c$ vanishes. Thus the BCH expression supplied globally by [L3] is a finite polynomial. Its universal identities give $\mathrm{BCH}(x,0)=x=\mathrm{BCH}(0,x)$ and $\mathrm{BCH}(x,-x)=0$.

Let $N$ be the connected simply connected integration supplied by [L2]. By [L3], $\exp :\mathfrak{n}\to N$ is a diffeomorphism and the transported global product $x\ast y=\log (\exp x\exp y)$ is the truncated BCH polynomial; this agrees with the local formula in [L1]. Associativity, identity $0$, and inverse $-x$ follow from the laws of $N$.

The underlying manifold is $\mathfrak{n}\cong {\mathbb{R}}^{\dim \mathfrak{n}}$, hence is connected and simply connected, including dimension zero. The antisymmetric part of the quadratic BCH term is $[x,y]$, so differentiating the commutator recovers the original bracket.

In the Heisenberg algebra, $(ae+bf+cz)\ast (a^{\mathrm{\prime }}e+b^{\mathrm{\prime }}f+c^{\mathrm{\prime }}z)=(a+a^{\mathrm{\prime }})e+(b+b^{\mathrm{\prime }})f+(c+c^{\mathrm{\prime }}+\frac{1}{2}(a{b}^{\mathrm{\prime }}-b{a}^{\mathrm{\prime }}))z,$ because brackets of length three vanish. This is a concrete nonabelian instance. The declared ${\mathsf{A}\mathsf{C}}_{\omega }$ is exactly that inherited through [L2]–[L3]; the finite calculation adds no choice.

For $u=ax+by$, $[u,x]=-by,[u,y]=ay\mathrm{.}$ If $u$ is central, both brackets vanish, so $a=b=0$. Hence $Z(\mathfrak{a})=0$.

On the other hand, ${\mathfrak{a}}^{(1)}=[\mathfrak{a},\mathfrak{a}]=ky$ and ${\mathfrak{a}}^{(2)}=[ky,ky]=0$. Thus $\mathfrak{a}$ is solvable by [L1]. As a solvable ideal of itself, its radical is all of $\mathfrak{a}$, which is nonzero; by [L2] it is not semisimple.

Step 1.1 satisfies the proposed centerless hypothesis and step 1.2 fails the semisimplicity conclusion. The witness is nonzero, two-dimensional, and completely explicit; no choice is used.

Both groups are one-dimensional and abelian, so their tangent brackets at the identity are zero. Sending the tangent vector $1\in T_0\mathbb{R}$ to $i\in T_1{S}^1$ is therefore an isomorphism of their real Lie algebras. Concretely, the local homomorphism is $t\mapsto e^{it}$.

The circle is compact. The open cover $\{(-m,m):m\in \mathbb{N},m\ge 1\}$ of $\mathbb{R}$ has no finite subcover, so $\mathbb{R}$ is not compact. A Lie-group isomorphism is a homeomorphism and preserves compactness. Therefore the groups are not isomorphic.

In the language of [L1], both arise from the simply connected group $\mathbb{R}$: the line uses the zero kernel, while the circle uses the nonzero discrete central kernel $2\pi \mathbb{Z}$. Thus the same one-dimensional Lie algebra does not determine the connected group. The declared ${\mathsf{A}\mathsf{C}}_{\omega }$ is propagated from [L1]; the explicit witness and compactness argument introduce no additional choice.