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# 数学代考|李群李代数代考Lie groups and Lie algebras代写|MATH251 Representations

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## 数学代考|李群李代数代考Lie groups and Lie algebras代写|Representations

An endomorphism $D$ of $\mathcal{A}$ is called a derivation of $\mathcal{A}$ if $D(X \cdot Y)=(D X) \cdot Y+$ $X \cdot(D Y), \forall X, Y \in \mathcal{A}$. Note that if $D_1, D_2$ are derivations of $\mathcal{A}$ then $D_1 \circ D_2-D_2 \circ D_1$ is also a derivation.

A representation of a Lie algebra $\mathcal{G}$ in a vector space $V$ is a linear map $\varphi: \mathcal{G} \rightarrow$ End $V$ such that $\varphi([X, Y])=[\varphi(X), \varphi(Y)] \equiv \varphi(X) \circ \varphi(Y)-\varphi(Y) \circ \varphi(X)$.

A representation $\varphi$ is called a reducible representation or simple if $V$ contains a nontrivial subspace $V^{\prime}, 0 \neq V^{\prime} \neq V$, which is invariant under the action of $\varphi$, i.e., $\varphi(Y) v \in V^{\prime}$ for all $Y \in \mathcal{G}, v \in V^{\prime}$; otherwise $\varphi$ is called an irreducible representation.
A representation is called a completely reducible or semisimple if $V$ can be represented as the direct sum of invariant subspaces.

A representation is called an indecomposable representation if it is reducible but not completely reducible.

An important representation of $\mathcal{G}$ is the so-called adjoint representation of $\mathcal{G}$ acting in $\mathcal{G}$ itself:
$$\operatorname{ad} X: \mathcal{G} \rightarrow \mathcal{G} ; \operatorname{ad} X: Y \mapsto[X, Y], X, Y \in \mathcal{G} .$$
It is easy to see that $\varphi: X \mapsto a d X$ is a representation, i.e., that $a d[X, Y]=(\operatorname{ad} X) \circ$ $(a d Y)-(a d Y) \circ(a d X)$.

It is also clear that $a d X$ is a derivation of $\mathcal{G}$, i.e., that $a d X([Y, Z])=[\operatorname{ad} X(Y), Z]+$ $[Y, a d X(Z)]$; the last equality is actually the Jacobi identity.
The derivations $a d X$ are called inner derivations of $\mathcal{G}$.
For further use, we shall note the equality ad $D Y=[D, a d Y]$ valid for every derivation $D$ and $\forall Y \in \mathcal{G}$.

## 数学代考|李群李代数代考Lie groups and Lie algebras代写|Solvable Lie Algebras

Let $\mathcal{G}$ be a Lie algebra. Define inductively $\mathcal{G}^{(1)}=[\mathcal{G}, \mathcal{G}], \mathcal{G}^{(k+1)}=\left[\mathcal{G}^{(k)}, \mathcal{G}^{(k)}\right]$, for $k=1,2, \ldots$

It is clear that $\mathcal{G}^{(k+1)} \subset \mathcal{G}^{(k)}$ and that all $\mathcal{G}^{(k)}$ are ideals of $\mathcal{G}$. (For the latter first note that $\mathcal{G}^{(1)}$ is an ideal as the commutator of two ideals; then by induction.)
If $\mathcal{G}^{(k)}=0$ for some $k$ then $\mathcal{G}$ is called solvable. (Then all $\mathcal{G}^{(i)}$ are solvable.)
In particular, if $\mathcal{G}^{(1)}=0$ then $\mathcal{G}$ is an Abelian algebra.
Note that a solvable Lie algebra contains an Abelian ideal.

It can also be shown that a finite-dimensional Lie algebra is solvable iff its adjoint representation is triangular (i.e., there exists a basis in which all operators $a d X$ are given by upper (or equivalently lower) triangular matrices with zeros below (above) the main diagonal).

## 数学代考|李群李代数代考Lie groups and Lie algebras代写|Representations

$$\operatorname{ad} X: \mathcal{G} \rightarrow \mathcal{G} ; \operatorname{ad} X: Y \mapsto[X, Y], X, Y \in \mathcal{G} .$$

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