# 数学代写|李群和李代数代写lie group and lie algebra代考|Math222

## 数学代写|李群和李代数代写lie group and lie algebra代考|THE LEVI DECOMPOSITION OF A LIE ALGEBRA

The study of Lie algebras in general can be reduced to the study of two special classes of Lie algebras: solvable $\mathrm{Lie}$ algebras and semisimple $\mathrm{Lie}$ algebras. These types of Lie algebras may be defined in terms of ideals as follows: With any Lie algebra $L$, there is associated a derived series of ideals defined recursively by
$$L^{\prime}=[L, L]$$
and
$$L^{(k+1)}=\left[L^{(k)}, L^{(k)}\right]$$
for any positive integer $k$. It is readily proved by induction on $k$ that $L^{(k)}$ is an ideal of $L$ and that the inclusion $L^{(k+1)} \subset L^{(k)}$ holds. We say that $L$ is a solvable Lie algebra if the derived series of ideals
$$L \supset L^{\prime} \supset L^{\prime \prime} \supset \cdots$$
is eventually zero. Since a Lie algebra $L$ is Abelian if and only if $[L, L]=0$, it follows that every Abelian Lie algebra is solvable. The simplest solvable Lie algebra which is not Abelian is the two-dimensional affine Lie algebra. If we choose a basis $e_1, e_2$ for this Lie algebra such that
$$\left[e_1, e_2\right]=e_1,$$
then $L^{\prime}$ consists of the multiples of $e_1$, and $L^{\prime \prime}=0$, proving that $L$ is solvable. In any Lie algebra, the solvable ideals form a sublattice of the lattice of all its ideals, because the sum and intersection of solvable ideals are again solvable ideals. In particular, the sum of all the solvable ideals in a Lie algebra is its unique maximal solvable ideal, called the radical of the Lie algebra. We may define a semisimple Lie algebra as a Lie algebra which has no Abelian ideals, other than 0 . It is easy to see that a semisimple Lie algebra cannot have any solvable ideals either, and hence a Lie algebra is semisimple if and only if its radical is zero.

## 数学代写|李群和李代数代写lie group and lie algebra代考|SEMISIMPLE LIE ALGEBRAS

To study the structure of semisimple Lie algebras, one uses the fact that they can be decomposed in terms of simple Lie algebras. A simple Lie algebra is a non-Abelian Lie algebra which has no proper ideals at all. Every simple Lie algebra is, as one would expect, semisimple, for the only nonzero ideal is the whole algebra; if $L^{\prime}=L$, this is not solvable, while if $L^{\prime}=0$, then $L$ is Abelian and consequently not simple. An example of a simple Lie algebra is the real Lie algebra $s o(3, \mathbb{R})$, the familiar three-dimensional vector space $\mathbb{R}^3$ equipped with the usual vector cross product. To see this, we must argue that there are no proper ideals in the Lie algebra $s o(3, \mathbb{R})$. Such an ideal would be a proper subspace $S$ of $\mathbb{R}^3$, a line or a plane passing through the origin such that the vector cross product $\mathbf{x} \times \mathbf{y}$ of any vector $\mathbf{x}$ in $S$ by a vector $\mathbf{y}$ in $\mathbb{R}^3$ must lie in $S$. Since the cross product of two vectors is perpendicular to both and is nonzero unless the two vectors are collinear or at least one is zero, no line or plane can have this property.

The fact that the rotation group $S O(3, \mathbb{R})$ is simple (has no proper normal subgroups) almost follows from the simplicity of its Lie algebra. Since the closure of a normal subgroup is again normal and since the Lie algebra of a closed normal subgroup of $S O(3, \mathbb{R})$ is an ideal of its Lie algebra, we only have to rule out discrete and dense normal subgroups. A normal subgroup of $S O(3, \mathbb{R})$ is dense if its closure is all of $S O(3, \mathbb{R})$. If $g$ is an element of any normal subgroup $S$ of $S O(3, \mathrm{R})$, then all elements $\mathrm{hgh}^{-1}$ which are conjugate to $g$ must also be in $S$. But this is equivalent to saying that every rotation about any axis by an angle equal to that of the rotation $g$ about its axis must be in $S$. If $g$ is not the identity, multiplying it by the inverse of a rotation $h g h^{-1}$ about an axis which has been shifted a small amount, we obtain an element $g h g^{-1} h^{-1}$ in $S$ which is not the identity, but as close to it as we wish. In this way one can further argue that any proper normal subgroup contains a whole neighborhood of the identity element. Since any neighborhood of the identity generates the whole rotation group, it follows that $S O(3, \mathbb{R})$ cannot have any proper normal subgroups at all.

Any semisimple Lie algebra $S$ can be written as a direct sum of simple ideals,
$$S=S_1 \oplus \cdots \oplus S_k .$$

## 数学代写|李群和李代数代写lie group and lie algebra代考|THE LEVI DECOMPOSITION OF A LIE ALGEBRA

$$L^{\prime}=[L, L]$$

$$L^{(k+1)}=\left[L^{(k)}, L^{(k)}\right]$$

$$L \supset L^{\prime} \supset L^{\prime \prime} \supset \cdots$$

$$\left[e_1, e_2\right]=e_1,$$

## 数学代写|李群和李代数代写lie group and lie algebra代考|SEMISIMPLE LIE ALGEBRAS

$$S=S_1 \oplus \cdots \oplus S_k .$$

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