用户: Fyx1123581347/扶磊 Lie 代数/Killing Form

定义 0.1. For any Lie algebra $\mathfrak{g}$, the Killing form is defined to be $$B:\mathfrak{g}\times \mathfrak{g}\to \mathbb{C},B(X,Y)=\mathrm{Tr}(a{d}_{X}a{d}_Y:\mathfrak{g}\to \mathfrak{g})$$

Clearly $B(X,Y)=B(Y,X)$. We shall write $B(X,Y)$ as $(X,Y)$.

命题 0.2. For all $Z\in \mathfrak{g}$, $$(a{d}_{Z}X,Y)+(X,a{d}_{Z}Y)=0.$$

Notice they may be interpreted as $Z(X,Y)=Zc=0$.

推论 0.3. If $\mathfrak{h}$ is an ideal of $\mathfrak{g}$, then $${\mathfrak{h}}^{\perp }=\{x\in \mathfrak{g}\mid \beta (X,Y)=0,\forall Y\in \mathfrak{h}\}$$is also an ideal of $\mathfrak{g}$.

定理 0.4 (Cartan Criterion). Let $\mathfrak{g}\in \mathfrak{gl}(V)$ be a sub lie algebra. Suppose for all $X,Y\in \mathfrak{g}$, we have $\mathrm{Tr}(XY:V\to V)=0$. Then $\mathfrak{g}$ is sovable.

命题 0.5.

•	$\mathfrak{g}$ is solvable if and only if $B(\mathfrak{g},[\mathfrak{g},\mathfrak{g}])=0$.
•	$\mathfrak{g}$ is semisimple if and only if $B$ is nondegenerate as a bilinear form.

证明. Suppose $\mathfrak{g}$ is solvable. Consider the adjoint representation $ad:\mathfrak{g}\to \mathfrak{gl}(\mathfrak{g})$. By Lie theorem, there exists a basis of $\mathfrak{g}$ such that $ad(X)\in \mathfrak{gl}(\mathfrak{g})$ are upper triangular matrices for all $X\in \mathfrak{g}$. So $$ad([X,Y])=a{d}_{X}a{d}_Y-a{d}_{Y}a{d}_X$$is strict upper triangular. Thus $$B(Z,[X,Y])=Tr(a{d}_{Z}a{d}_{[X,Y]})=0.$$Conversely, it suffices to show $[\mathfrak{g},\mathfrak{g}]$ is sovable. Consider the restriction of $[\mathfrak{g},\mathfrak{g}]$ under the adjoint representation, whose kernel is $Z(g)$. That is, we have $$0\to Z(g)\cap [\mathfrak{g},\mathfrak{g}]\to [\mathfrak{g},\mathfrak{g}]\to ad([\mathfrak{g},\mathfrak{g}])\to 0$$So we are going to show $ad([\mathfrak{g},\mathfrak{g}])$ is solvable. This follows from Cartan criterion. (In fact, it suffices that $B([\mathfrak{g},\mathfrak{g}],[\mathfrak{g},\mathfrak{g}])=0$)

Suppose $\mathfrak{g}$ is semisimple. If $X\in \mathfrak{g}$ is such that $B(X,Y)=0$ for all $Y$. But ${\mathfrak{g}}^{\perp }$ is a solvable ideal: $$0\to Z(\mathfrak{g})\cap {\mathfrak{g}}^{\perp }\to {\mathfrak{g}}^{\perp }\to ad({\mathfrak{g}}^{\perp })\to 0$$This follows again from Cartan criterion.

Suppose $B$ is nondegenerate. Let $\mathfrak{a}$ be an abelian ideal of $\mathfrak{g}$. Let us prove $\mathfrak{a}=0$. Since $B$ is nondegenerate, it suffices to show $\mathfrak{a}\subseteq {\mathfrak{g}}^{\perp }$. That is, for $X\in \mathfrak{a},Y\in \mathfrak{g}$, $$Tr(a{d}_{X}a{d}_Y;\mathfrak{g})=0.$$Notice that $a{d}_{X}a{d}_Y{\mid }_{\mathfrak{a}}=0$ and $a{d}_{X}a{d}_Y(\mathfrak{g})\subset \mathfrak{a}$. Thus the trace map is zero on $\mathfrak{a}$ and $\mathfrak{g}/\mathfrak{a}$.

$\square$

推论 0.6. A semisimple lie algebra is a direct product of simple lie algebras.

证明. Simple lie algebra is clearly semisimple. The direct product of simple lie algebras is also semisimple. Now suppose $\mathfrak{g}$ is semisimple. Let $\mathfrak{h}$ be an ideal of $\mathfrak{g}$, then ${\mathfrak{h}}^{\perp }$ is also an ideal. As $B$ is nondegenerate, we have $$\mathfrak{g}=\mathfrak{h}\oplus {\mathfrak{h}}^{\perp }.$$(Once we can prove $B$ is nondegenrate on $\mathfrak{h}$).

Claim: $\mathfrak{h}\cap {\mathfrak{h}}^{\perp }$ is solvable. This is because $ad:\mathfrak{h}\cap {\mathfrak{h}}^{\perp }\to \mathfrak{gl}(\mathfrak{g})$ has solvable image.

$\square$

推论 0.7. If $\mathfrak{g}$ is semisimple, then $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$.

1Cartan Criterion

To prove $\mathfrak{g}$ is solvable, it suffices to show $[\mathfrak{g},\mathfrak{g}]$ is sovable. By Engel theorem, it suffices to show elements in $[\mathfrak{g},\mathfrak{g}]$ are nilpotent. Let $X\in [\mathfrak{g},\mathfrak{g}]$ and ${\lambda }_1,\dots ,{\lambda }_n$ be eigenvalues of $X$ (counted with multiplicities). Take basis of $V$ so that the matrix of $X$ is in Jordan form. Let $\bar{D}\in \mathfrak{gl}(V)$ be the linear transformation such that with respect to the basis have diagonal form as $X$ except has eigenvalues conjugate to those of $X$. We are going to show $Tr(\bar{D}X)=0$. Denote $$X=\sum _i[Y_i,Z_i]$$with $Y_i,Z_i\in \mathfrak{g}$. Then$$Tr(\bar{D}X)=\sum _{i}Tr(\bar{D}[Y_i,Z_i])=-\sum _{i}Tr([Y_i,\bar{D}]Z_i).$$

Recall for $X,Y,Z\in End(V)$,$$Tr([X,Y]Z)+Tr(X[Y,Z])=0.$$It suffices to show $a{d}_{\bar{D}}$ is a polynomial of $a{d}_X$. The Jordan decomposition of $X$ is $$X=D+(X-D).$$The Jordan decomposition of $a{d}_X$ is$$a{d}_X=a{d}_D+a{d}_{X-D}$$(identities in $\mathfrak{gl}(End(V))$). Then $a{d}_D$ is a polynomial of $a{d}_X$. Let us prove $a{d}_{\bar{D}}$ is a polynomial of $a{d}_D$.

$$a{d}_D(E_{ij})=({\lambda }_i-{\lambda }_j)E_{ij}$$

Take $g\in \mathbb{C}[t]$ so that $g({\lambda }_i-{\lambda }_j)={\bar{\lambda }}_i-{\bar{\lambda }}_j$. We win.

2Representations of semisimple lie algebra

Casimir operator

命题 2.1. Let $V$ be a finite dimensional vector space, $\mathfrak{g}\subset \mathfrak{gl}(V)$ subalgebra. Suppose $\mathfrak{g}$ is semisimple. Then the pairing $$B_V:\mathfrak{g}\times \mathfrak{g}\to \mathbb{C},(X,Y)\mapsto Tr(XY;V)$$is nondegenerate.

证明. We need to show $${\mathfrak{g}}^{\perp }=\{X\mid {\beta }_V(X,Y)=0,\forall Y\in \mathfrak{g}\}=0.$$To see this, it suffices to show ${\mathfrak{g}}^{\perp }$ is a solvable ideal.

$\square$

Let $\mathfrak{g}\subset \mathfrak{gl}(V)$ be a semisimple lie algebra. Let $v_1,\dots ,v_n$ be a basis of $\mathfrak{g}$ as a vector space, and $v_1^{\prime },\dots ,v_n^{\prime }$ be the dual basis of $\mathfrak{g}$ with respect to the nondegenrate bilinear map $B_V$. Define the Casmir operator$$C_V:V\to V$$by$$C_V=\sum _{i=1}^n{v}_i{v}_i^{\prime }.$$

命题 2.2.

1.	$C_V$ is independent of the choice of the basis.
2.	$C_V:V\to V$ is a homomorphism of $\mathfrak{g}$-modules.
3.	$Tr(C_V)=\dim V$.

证明. (1) Let $u_i$ be another basis and $u_i^{\prime }$ the corresponding dual basis. $$\sum _{i=1}^n{v}_i{v}_i^{\prime }=\sum _{i=1}^n{u}_i{u}_i^{\prime }.$$Suppose $v_i={\sum }_k{a}_{ik}{u}_k$, $v_j^{\prime }={\sum }_l{b}_{jl}{u}_l^{\prime }$. We have $${\delta }_{ij}={\beta }_V(v_i,v_j^{\prime })=\sum _{k,l}{a}_{ik}{b}_{jl}{\delta }_{k,l}=\sum _k{a}_{ik}{b}_{jk}.$$Let $A=(a_{ik})$, $B=(b_{jl})$. Then $A{B}^t=I$. $$\begin{array}{rl}\sum _{i=1}^n{v}_i{v}_i^{\prime }& =\sum _{k,l}\sum _i{a}_{ik}{b}_{il}{u}_k{u}_l^{\prime }\\ & =\sum _{k,l}{\delta }_{k,l}{u}_k{u}_l^{\prime }\\ & =\sum _k{u}_k{u}_k^{\prime }.\end{array}$$(2) $$\begin{array}{rl}{C}_{V}X-X{C}_V& =\sum _i{v}_i{v}_i^{\prime }X-\sum _{j}X{v}_j{v}_j^{\prime }\\ & =\sum _i{v}_i[v_i^{\prime },X]+\sum _j[v_j,X]v_j^{\prime }.\end{array}$$Suppose $[v_i^{\prime },X]={\sum }_k{a}_{ik}{v}_k^{\prime }$, $[v_j,X]={\sum }_j{b}_{jl}{v}_l$. By $$B([v_i^{\prime },X],v_j)+B(v_i^{\prime },[v_j,X])=0.$$

$\square$

3Semisimplicity

命题 3.1. Let $\mathfrak{g}$ be a semisimple lie algebra, let $\rho :\mathfrak{g}\to \mathfrak{gl}(V)$ be a representation, and $W$ be a subrepresentation. Then there exists a subrepresentation $W^{\prime }$ such that $V\simeq W\oplus W^{\prime }$.

证明. The image $\rho (\mathfrak{g})$ is also semisimple. Replace $\mathfrak{g}$ by $\rho (\mathfrak{g})$, we may assume $\mathfrak{g}\subset \mathfrak{gl}(V)$.

Notice that any $1$-dimensional representation of a semisimple lie algebra is trivial ($0$ representation). In fact, the kernel contains $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$.

Claim: If $W$ is a subrepresentation of $V$ such that $\mathrm{codim}(W,V)=1$, then there exists a subrepresentation $W^{\prime }$ of $V$ such that $V=W\oplus W^{\prime }$.

Suppose $W$ is irreducible. Let $C_V:V\to V$ be the Casmir operator. Recall that $C_V$ is a homomorphism of representation. Since $W$ is a subrepresentation of $V$, we must have $C_V(W)\subseteq W$, as $C_V=\sum v_i{v}_i^{\prime }$. So $C_V:W\to W$ is a homomorphism of the representation $W$. By Schur’s lemma, $C_V$ must be a scalar $\lambda$. It also induces a homomorphism $C_V:V/W\to V/W$. As $V/W$ is a trivial representation, $C_V{\mid }_{V/W}$ is zero. We have $Tr(C_V)=\dim \mathfrak{g}\ne 0$, so $\lambda \ne 0$. That is, $\ker (C_V)\cap W=0$. This follows from

注 3.2. Invariant subspace of $C_V$ is a subrepresentation of $V$. This shows the importance of Casmir operator.

In general, we prove the claim by induction on $\dim V$. For codimension $1$ subspace $W$ of $V$. If $W$ is not irreducible, it contains an irreducible subrepresentation $Z$ (e.g. with the minimal dimension). Apply the induction hypothesis to $V/Z$, we can find a subrep $Y$ of $V$ such that $Y\supset Z$ and $$V/Z\simeq W/Z\oplus Y/Z.$$We have $\dim Y/Z=1$. Thus there exists a subrepresentation $W^{\prime }\subset Y$ such that $$Y=Z\oplus W^{\prime }.$$It now follows that $$V=W\oplus W^{\prime }.$$

注 3.3. $$X⟨v,\varphi ⟩=⟨Xv,\varphi ⟩+⟨v,X\varphi ⟩.$$We have $${\mathrm{Hom}}_{\mathfrak{g}}(k,V)=\{v\in V\mid Xv=0,\forall X\in V\}.$$It follows that $${\mathrm{Ext}}^i(V,W)={\mathrm{Ext}}^i(k,\underline{\mathrm{Hom}}(V,W)).$$(Notice that $\underline{\mathrm{Hom}}$ and $\otimes$ have no derived functor) It suffices to show ${\mathrm{Ext}}^1(k,V)=0$ for any $V$. This is exactly what we have done for the codimension $1$ case.

$\square$

命题 3.4. Let $\mathfrak{g}\subset \mathfrak{gl}(V)$ be a semisimple lie algebra. For any $X\in \mathfrak{gl}(V)$, let $X=X_s+X_n$ be the Jordan decomposition. For $X\in \mathfrak{g}$, we have $X_s,X_n\in \mathfrak{g}$.

证明. Say lie subalgebras with the desired property “good”. Since $\mathfrak{g}$ is semisimple, $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]\subseteq \mathrm{sl}(V)$. In fact, for any representation $\rho \to \mathfrak{gl}(W)$ of a semisimple lie algebra $\mathfrak{g}$, we have $\rho (\mathfrak{g})\subseteq \mathfrak{sl}(W)$.

For any subrepresentation $W\subseteq V$ of $\mathfrak{g}$, define $$S_W=\{Y\in \mathfrak{gl}(V)\mid Y(W)\subseteq W,Tr(Y{\mid }_W)=0\}.$$Then $S_W$ is a good subalgebra of $\mathfrak{gl}(V)$. As $Y_s,Y_n$ are polynomials of $Y$, $W$ is invariant under them. In fact, they are Jordan decompositions of $Y{\mid }_W$.

$\square$