Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 102 (Finite Dimensional Lie and Associative Algebras)

Write $D=\mathrm{ad}y$. The generalized eigenspace decomposition of the linear map $D$ isfor any sufficiently large $N$. Because $D$ is a derivation, the generalized-eigenspace bracket lemma givesConsequently $L_{0,y}$ is a Lie subalgebra.

The set $I(K)=\{x\in L:[x,K]\subseteq K\}$ is the normalizer of a Lie subalgebra $K$. Certainly $L_{0,y}\subseteq I(L_{0,y})$. Conversely, if $x\in I(L_{0,y})$, then $y\in L_{0,y}$ givesOn the direct sum of the nonzero generalized eigenspaces, $D$ is invertible. Hence the nonzero-eigenvalue component of $x$ vanishes, and

Now let $K$ be a Lie subalgebra containing $L_{0,y}$. Since $y\in K$, the subspace $K$ is $D$-invariant. The generalized zero eigenspace of the induced map on $L/K$ is the image of $L_{0,y}$, hence is zero. If $x\in I(K)$, then $Dx=[y,x]\in K$, so $x+K$ lies in that zero eigenspace. Thus $x\in K$ and

A Nilpotent Lie algebra is one whose lower central serieseventually reaches zero. Suppose $L$ is nilpotent and $K⊊L$. Choose the least $r\ge 2$ for which ${\gamma }_r(L)\subseteq K$. Then ${\gamma }_{r-1}(L)⊈K$, and anysatisfies $[x,K]\subseteq [L,{\gamma }_{r-1}(L)]={\gamma }_r(L)\subseteq K$. Therefore $x\in I(K)\setminus K$, proving the normalizer condition for a nilpotent Lie algebra

It remains to prove the converse needed here. The Engel lemma states that if a finite-dimensional Lie algebra of linear maps consists of nilpotent maps, then the maps have a common nonzero vector in their kernels. To prove it, induct on the dimension of the algebra. For a maximal proper subalgebra $H$, induction applied to the action of $H$ on $L/H$ produces $x\notin H$ with $[H,x]\subseteq H$. Thus $H$ is an ideal of codimension one. Induction also gives a nonzero common kernelThe ideal property makes $W$ invariant under $L$; a nilpotent representative of a basis of $L/H$ has a nonzero kernel on $W$, yielding a vector killed by all of $L$.

Apply the lemma to the Adjoint representation. It produces a nonzero element of the Center of a Lie algebra. Induction on $\dim L$, followed by passage to the quotient by this center, proves Engel theorem: if every $\mathrm{ad}y$ is nilpotent, then $L$ is nilpotent. The hypothesis $L_{0,y}=L$ says exactly that every $\mathrm{ad}y$ is nilpotent, so

A finite root system in a real Euclidean vector space $E$ is a finite spanning set $\Phi \subset E\setminus \{0\}$ such that, for every $\alpha \in \Phi$, the root reflectionpreserves $\Phi$, and the Cartan integer $2(\beta ,\alpha )/(\alpha ,\alpha )$ is an integer for all $\alpha ,\beta \in \Phi$. It is reduced when the only scalar multiples of $\alpha$ in $\Phi$ are $\alpha$ and $-\alpha$. Its Weyl group is the subgroup of the orthogonal group generated by the reflections $s_{\alpha }$. A base of a root system $\Delta$ is a basis of $E$ such that every root is an integer combination of elements of $\Delta$ whose nonzero coefficients all have the same sign.

The coroot of $\alpha$ isLet $C$ be the Weyl chamber determined by $\Delta$:The roots $\alpha$ and ${\alpha }^{\vee }$ are positive scalar multiples, so their reflecting hyperplanes and their positive half-spaces are identical. The same chamber $C$ therefore defines positivity in the coroot system ${\Phi }^{\vee }$. Its walls correspond exactly to the rays ${\mathbb{R}}_{>0}{\alpha }^{\vee }$ for $\alpha \in \Delta$. Hence its simple roots arewhich is therefore a base of ${\Phi }^{\vee }$.

For an arbitrary finite-dimensional complex Lie algebra $L$, a Cartan subalgebra is a nilpotent Lie subalgebra $H$ equal to its own normalizer. When $L$ is semisimple, this is equivalently a maximal abelian subalgebra consisting of elements that act semisimply in the Adjoint representation.

Choose such an $H$. Simultaneous diagonalization gives the root-space decompositionThe nonzero weights $\alpha \in H^{\ast }$ are the roots. The restriction of the Killing form $B$ to $H$ is a nondegenerate bilinear form, so each $\alpha$ corresponds to a unique $t_{\alpha }\in H$ with $\alpha (h)=B(t_{\alpha },h)$. On the real span of these $t_{\alpha }$, the restriction of $B$ supplies a positive-definite inner product after choosing the standard real form. The sl2 subalgebra associated with a root givesand shows that these reflections preserve the finite set $\Phi$. Thus the roots form a finite reduced crystallographic root system, whose Weyl group is generated by these reflections.

A finite-dimensional Lie algebra $L$ is semisimple when its solvable radical is zero, equivalently when it has no nonzero solvable ideals.

We prove the Weyl complete reducibility theorem. Induct on the dimension of a finite-dimensional $L$-module $V$. It is enough first to split a submodule $W$ of codimension one. The one-dimensional quotient is trivial because a semisimple Lie algebra is perfect. By induction, $W$ is a direct sum of irreducible modules. A Casimir element formed using the Killing form commutes with the $L$-action, acts as zero on every trivial summand, and acts by a nonzero scalar on every nontrivial irreducible summand. Its image is therefore the sum $W_1$ of the nontrivial summands, while its kernel contains the trivial summands $W_0$ and maps onto $V/W$. ThusInside $\ker \Omega$, choose a lift $v$ of a basis of $V/W$. For $x\in L$, $xv\in W_0$, and $L$ acts trivially on $W_0$. Hence $[x,y]v=0$ for all $x,y\in L$. Since $L=[L,L]$, actually $xv=0$ for every $x$, so $\mathbb{C}v$ is the required invariant complement.

This codimension-one case implies the general case. For an arbitrary submodule $W\subseteq V$, letwith the natural Hom representation. The maps vanishing on $W$ form an $L$-submodule $X_0$ of codimension one. Splitting $X_0$ supplies an $L$-equivariant $f$ with $f{|}_W=I_W$. Thenso every invariant subspace has an invariant complement and every finite-dimensional representation is completely reducible.

For the requested example, embed ${\mathfrak{sl}}_2$ as the upper-left $2\times 2$ block in the Special linear Lie algebra ${\mathfrak{sl}}_3$. Under the restricted Adjoint representation,Here $V_n$ is the irreducible ${\mathfrak{sl}}_2$-module of highest weight $n$. The first summand is the three-dimensional adjoint module, the second is trivial, and the last two are the two-dimensional defining module and its dual, which are isomorphic. This explicit direct sum demonstrates complete reducibility.

Because $M$ is a finite-dimensional semisimple module, it has an isotypic decompositionwhere the $S_i$ are pairwise nonisomorphic simple right $R$-modules. By Schur lemma,is a division ring, while ${\mathrm{Hom}}_R(S_i,S_j)=0$ for $i\ne j$. Consequently every endomorphism preserves the isotypic summands and is a matrix of entries from $D_i$ on each one. ThereforeIf the ground field is algebraically closed and the $S_i$ are finite-dimensional over it, Schur lemma gives $D_i=k$.

A vertex $i$ of a quiver is a sink when no arrow starts at $i$. A representation of a quiver assigns a vector space $V_j$ to every vertex and a linear map $V_a:V_j\to V_{\ell }$ to every arrow $a:j\to \ell$. A morphism $f:V\to W$ is a family of linear maps $f_j:V_j\to W_j$ such that $f_{\ell }{V}_a=W_a{f}_j$ for every arrow. The quiver has finite representation type when it has only finitely many isomorphism classes of indecomposable finite-dimensional representations.

At a sink $i$, the Bernstein–Gelfand–Ponomarev reflection functor replacesand reverses the arrows ending at $i$; the new arrow maps are the kernel inclusion followed by the coordinate projections. Every representation is a direct sum of copies of the simple representation $S_i$ and a representation for which the displayed incoming map is surjective. On the latter representations, reflection at the resulting source, using the corresponding cokernel, is inverse up to natural isomorphism. Thus reflection gives a bijection between indecomposable representations other than $S_i$ on the two sides. Adding the one omitted simple representation on each side proves that reversing all arrows into a sink preserves finite representation type.

For the four-arrow star $Q_1$, take $V_1=\cdots =V_4=k$, $V_5=k^2$, and let the four arrows have imagesAn endomorphism must preserve the first two lines, so its map on $V_5$ is diagonal. Preserving the third forces its diagonal entries to agree, and then all vertex maps are multiplication by that same scalar. The endomorphism ring is therefore $k$, so this representation is a brick module and hence indecomposable. Any isomorphism between parameters preserves the first three labelled lines; the induced projective linear transformation is therefore the identity, and the fourth line gives $\lambda =\mu$. Since $k$ is infinite, this is an infinite family of pairwise nonisomorphic indecomposables. Hence $Q_1$ is not of finite representation type. Repeatedly applying the reflection result to sinks or, dually, to sources shows that every orientation obtained by reversing some of its arrows also has infinite representation type.