Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 102 (Finite Dimensional Lie and Associative Algebras)

The space $\mathrm{End}(V)$ becomes a Lie algebra under the commutatorA Lie subalgebra $L$ is abelian when $[L,L]=0$; nilpotent when its Lower central series of a Lie algebra ${\gamma }_{r+1}=[L,{\gamma }_r]$ reaches zero; and soluble when its derived series of a Lie algebra $L^{(r+1)}=[L^{(r)},L^{(r)}]$ reaches zero.

If $H⊊L$ and $L$ is nilpotent, choose the least $r$ with ${\gamma }_r(L)\subseteq H$. Then ${\gamma }_{r-1}(L)⊈H$, whileThus the normalizer of a Lie subalgebra strictly contains $H$. If $H$ is maximal proper, its normalizer must be all of $L$, so $H$ is an ideal. Solubility is insufficient: in the two-dimensional affine Lie algebra $L=⟨x,y⟩$ with $[x,y]=y$, the maximal subalgebra $\mathbb{C}x$ is not an ideal.

A derivation of a Lie algebra is a linear map $D$ satisfyingIt is inner when $D=\mathrm{ad}z$ for some $z\in L$.

Every nonzero finite-dimensional nilpotent Lie algebra has an outer derivation of a nilpotent Lie algebra. Choose a codimension-one maximal subalgebra $K$; it is an ideal by the result above, and write $L=K\oplus \mathbb{C}z$. The centralizer $C_L(K)$ is nonzero because it contains $Z(L)$. Let $n$ be largest such thatand choose $z_0\in C_L(K)\setminus {\gamma }_{n+1}(L)$. DefineBecause $K$ is an ideal and $z_0$ centralizes $K$, the derivation identity holds on $K\times K$ and on $z\times K$, hence everywhere. If $D=\mathrm{ad}x$, then $D(K)=0$ would put $x$ in $C_L(K)\subseteq {\gamma }_n(L)$, socontrary to the choice of $z_0$. Thus $D$ is outer.

The analogous assertion fails for soluble algebras. In the affine example, a derivation hasand equals $\mathrm{ad}(dx-by)$. Thus every derivation is inner although $L$ is nonzero and soluble.

A Lie algebra is semisimple when its soluble radical is zero. Its Killing form isThe radical of this invariant symmetric form is an ideal. The solvability result behind the Cartan criterion for semisimplicity applied to that ideal shows that it is soluble; semisimplicity therefore makes it zero. Hence $B_L$ is nondegenerate.

An abelian subalgebra $H$ is a Cartan subalgebra when its elements are semisimple and it is maximal toral, equivalently when $C_L(H)=H$. An arbitrary abelian subalgebra need not lie in one: in ${\mathfrak{sl}}_2$, the line spanned by the nilpotent matrix $e$ is abelian, whereas every element of a Cartan subalgebra is semisimple.

Let $H=C_L(h_0)$ for a regular $h_0\in H$, as allowed. Generalized eigenspaces of $\mathrm{ad}{h}_0$ giveIf $x\in H$ is orthogonal to $H$, invariance givesThus $x$ is orthogonal to all of $L$, and nondegeneracy gives $x=0$. Therefore $B_L{|}_H$ is nondegenerate.

The commuting semisimple maps $\{\mathrm{ad}h:h\in H\}$ can be simultaneously diagonalized. ConsequentlyHere $L_0=C_L(H)=H$, the nonzero weights $\Phi \subseteq H^{\ast }$ are the roots, and the Jacobi identity gives $[L_{\alpha },L_{\beta }]\subseteq L_{\alpha +\beta }$.

A finite-dimensional Lie algebra representation is a homomorphism $\rho :L\to \mathfrak{gl}(V)$. It is irreducible when $V$ has no invariant subspaces other than $0$ and $V$.

The algebra ${\mathfrak{sl}}_2$ has basis $e,f,h$ withFor every $n\ge 0$, its $(n+1)$-dimensional irreducible module $V(n)$ has basis $v_0,\dots ,v_n$ and actionwith out-of-range vectors zero. Any nonzero invariant subspace contains a weight vector; repeated application of $e$ reaches $v_0$, and repeated application of $f$ then generates the whole module, proving irreducibility. The adjoint module of ${\mathfrak{sl}}_2$ is $V(2)$, so every ideal is an invariant subspace and ${\mathfrak{sl}}_2$ is simple.

For a root $\alpha$, nondegeneracy of the Killing pairing between $L_{\alpha }$ and $L_{-\alpha }$ allows choices $e_{\alpha },f_{\alpha }$ withAfter rescaling, $(e_{\alpha },f_{\alpha },h_{\alpha })$ obey the ${\mathfrak{sl}}_2$ relations, giving a copy of ${\mathfrak{sl}}_2$ in $U_{\alpha }$.

Restrict the adjoint representation of $L$ to this copy. Finite-dimensional ${\mathfrak{sl}}_2$ theory shows that the $\alpha$-string through zero has one nontrivial summand $V(2)$, with weight spaces $L_{-\alpha }$, $\mathbb{C}{h}_{\alpha }$, and $L_{\alpha }$. Any additional vector in $L_{\alpha }$ would generate another weight-two summand and another independent zero-weight coroot, contradicting nondegeneracy of the root--coroot pairing on $H$. Hence

For $\beta \notin \mathbb{Z}\alpha$, the spaceis stable under $e_{\alpha },f_{\alpha },h_{\alpha }$. Adjacent raising and lowering maps are nonzero until the endpoints, and every root space is one-dimensional, so $W$ is a simple ${\mathfrak{sl}}_2$-module of highest weight $p+q$. Its lowest and highest $h_{\alpha }$-weights giveand therefore $\beta (h_{\alpha })=q-p$. Since $[L_{\alpha },L_{-\alpha }]=\mathbb{C}{h}_{\alpha }$, every $x$ in this bracket line satisfies

The Jacobson radical $J=J(R)$ is the intersection of all maximal right ideals, equivalently the largest ideal annihilating every simple right module. The Artin–Wedderburn theorem givesbecause $k$ is algebraically closed.

The descending chain $J\supseteq J^2\supseteq \cdots$ stabilizes since $R$ is finite-dimensional. If $J^m=J^{m+1}$, Nakayama lemma applied to the finite right module $J^m$ gives $J^m=0$. Thus $J$ is nilpotent.

For $f\in {\mathrm{End}}_R(P_i)$, the Fitting lemma givesfor large $n$. Indecomposability makes one summand zero, so $f$ is either invertible or nilpotent. In the latter case $1-f$ is invertible. This is the criterion that ${\mathrm{End}}_R(P_i)$ is a local ring.

Let $M={⨁}_i{P}_i$. Reduction modulo $MJ$ definesSince $M/MJ={⨁}_i{S}_i$ and the $S_i$ are pairwise nonisomorphic simples,by Schur lemma. Arbitrary scalars on the direct summands lift to scalar identity maps on the $P_i$, so $\theta$ is surjective.

If $f\in \ker \theta$, then $f(M)\subseteq MJ$. A product of $r$ such maps sends $M$ into $M{J}^r$, so $\ker \theta$ is nilpotent. A nilpotent ideal lies in the Jacobson radical, while the semisimplicity of the quotient $k^n$ gives the reverse inclusion. HenceThis is exactly the definition of a basic algebra.

A simply-laced positive-definite Coxeter graph has no multiple edges and has positive-definite symmetric Cartan matrix, with diagonal entries $2$ and entry $-1$ precisely across an edge. Its connected components are exactly the ADE classification

The graph underlying $1\to 2$ is $A_2$. For simple roots ${\alpha }_1,{\alpha }_2$ with $({\alpha }_i,{\alpha }_i)=2$ and $({\alpha }_1,{\alpha }_2)=-1$, its roots areThere is one base for each Weyl chamber, hence six bases. The Weyl group is generated by the two root reflections withso $W\simeq S_3$. Relative to a chosen base, the two Coxeter elements are $s_1{s}_2$ and $s_2{s}_1$; both have order three.

A representation of the quiver is one linear map $f:V_1\to V_2$. Choosing bases that put $f$ in rank normal form decomposes it into copies ofThese are indecomposable and have dimension vectors $(1,0)$, $(0,1)$, and $(1,1)$, the three positive roots of $A_2$. More generally, Gabriel theorem says that for any orientation of a simply-laced positive-definite Dynkin graph, indecomposable quiver representations are in bijection with its positive roots. The finite ADE root system therefore gives finitely many indecomposables.