Basic concepts of Lie algebras

References:

  1. Karin Erdmann & Mark J. Wildon, Introduction to Lie Algebras, Springer Undergraduate Mathmatics Series;
  2. Roger W. Carter, Lie Algebras and Root Systems, in the book Lectures on Lie groups and Lie algebras, by Roger William Carter, Graeme Segal, Ian Grant Macdonald;
  3. Jean-Pierre Serre, Complex Semisimple Lie Algebras, translated from the French by G. A. Jones;
  4. James E. Humphreys, Introduction to Lie Algebras and Representation Theory, Graduate Texts in Mathematics 9 (GTM 9), Springer-Verlag;
  5. William Fulton & Joe Harris, Representation theory, a first course, GTM 129, Springer-Verlag;
  6. J. Dixmier, Algèbres de Lie, Les Cours de Sorbonne, rédaction de A. Pereira Gomes.

1 and 3 are highly recommended.

\S 1. Basic concepts of Algebras.

Let F be a field and A a vector space over F. We say that A is an Falgebra, if A is simultaneously a ring, and the ring structure and the vector space structure on A is compatible, in the sense that: for any x, y \in A and \lambda \in F^{\times}, we have

\displaystyle \lambda (xy)=(\lambda x)y=x(\lambda y).

The dimension of A over F as a vector space, is called the dimension of the algebra A

The algebra A is called

  • associative, iff (short for if and only if) for all x, y, z \in A, we have (xy)z=x(yz);
  • commutative, iff for all x, y \in A, one has xy=yx;
  • a Lie algebra, iff the multiplication xy, usually written as [xy], fulfilling that
  • (i) [xx]=0,\; \forall x \in A; and
  • (ii) for any x, y, z \in A, we have the Jacobi identity:

\displaystyle [x[yz]]+[y[zx]]+[z[xy]]=0.

Exercise. Check that the Lie bracket [,] on a Lie algebra A satisfies that for all x, y \in A, we have [xy]=-[yx] (anti-symmetry). And [xy]=-[yx] \Leftrightarrow [xx]=0, provided that the characteristic of F is not 2.

\S 2. Examples of Lie algebras.

1. Any F-vector space V has a Lie bracket defined by [uw]=0 for all u, w \in V. This is the abelian Lie algebra structure on V. In particular, the field F may be regarded as a one dimensional abelian Lie algebra.

2. Let A be an associative F-algebra. Define the bracket product [,]: A \times A \to A by [xy]=xy-yx for all x, y \in A. Then (A, [,]) becomes a Lie algebra.

By this construction, we noted that \mathrm{M}(n, F), the collection of all the n \times n matrix over F, with the bracket product defined by [PQ]=PQ-PQ\; \forall P, Q \in \mathrm{M}(n, F), is a Lie algebra. This Lie algebra is usually denoted by \mathfrak{gl}(n, F). We will illustrate this notation later.

Furthermore, given a vector space V over F of dimension n, we can identify End(V), the collection of all the endomorphism of V, to \mathrm{M}(n, F) (by choosing a certain F-basis of V). Hence End(V) is also a Lie algebra. We denote it by \mathfrak{gl}(V).

3. Let A be an F-algebra. The derivated algebra of A, consisting of all the F-linear mapping D: A \to A, such that D(ab)=aD(b)+D(a)b, \forall a, b \in A, is a Lie algebra. We denote this algebra of derivations of A by DerA.

4. Poisson algebra. This example originates in Physics. Let A be a commutative ring, and the usual multiplication of A is simply written by xy\; \forall x,y \in A. Define a bilinear mapping \displaystyle \{, \}: A \times A \to A,\; (x, y) \mapsto {xy}. This bilinear mapping is called a Poisson bracket, if it fulfills the following three conditions:

(i) {xy}=-{yx};

(ii) {x, yz}={xy}z+y{xz};

(iii) {x{yz}}={{xy}z}+{y{xz}}

for all the x, y, z \in A.

Exercise. Verify that every Poisson algebra is a Lie algebra!

We give a concrete example of Poisson algebra. Denote the set of all the smooth functions (a function is smooth if it is infinitely differentiable) over \mathbb{R}^{2n} by C^{\infty}(\mathbb{R}^{2n}). For f, g \in C^{\infty}(\mathbb{R}^{2n}), defined the Poisson bracket of  as follows:

\displaystyle \{f, g\}=\sum_{i=1}^n\,\big(\frac{\partial f}{q_i}\frac{\partial g}{p_i} - \frac{\partial f}{p_i}\frac{\partial g}{q_i}\big),

where the coordinates are denoted by (q_1, \ldots, q_n, p_1, \ldots , p_n) \in \mathbb{R}^{2n}.

One can check that such defined {f, g} satisfies all the requirements of a Poisson algebra, thus it is also a Lie algebra.

\S 3. Subalgebras, ideals and centre of a Lie algebra.

Given a Lie algebra L, we can define a Lie subalgebra of L to be a subspace L_1, such that [ab] \in L_1, \forall a, b \in L_1, or formally, [L_1, L_1] \subset L_1.

Examples of Lie subalgebras. Let \mathfrak{sl}(n, F) be the subspace of \mathfrak{gl}(n, F) consisting of all the matrices of trace 0. Let \mathfrak{b}(n, F) (resp. \mathfrak{n}(n, F)) be the upper triangular (resp. strictly upper triangular) matrices in \mathfrak{gl}(n, F). Then \mathfrak{sl}(n, F), \mathfrak{b}(n, F) and \mathfrak{n}(n, F) are all Lie subalgebra of \mathfrak{gl}(n, F).

We can also define an ideal of a Lie algebra L to be a subspace I of L such that [LI] \subset I or [IL] \subset I. The fact that [xy]=-[yx] for a Lie algebra L release us to distinguish the left and right ideal.

We point out that \mathfrak{sl}(n, F) is an ideal of \mathfrak{gl}(n, F), and \mathfrak{n}(n, F) is an ideal of \mathfrak{b}(n, F).

Note. An ideal is always an subalgebra. However, a subalgeba need not to be an ideal. Remember in mind that \mathfrak{n}(n, F) is a subalgebra of \mathfrak{gl}(n, F). But if n>2, then it is not an ideal. Verification is left as an exercise.

Immediately we note that L  itself and \{0\} are always ideals of L. They are called the trivial ideals. If the only ideals of L are the trivial ones, we say that L is simple.

Another important example of  an ideal is the centre. For a Lie algebra L, the centre Z(L) of L is defined by Z(L) = \{x \in L : [xy]=0\; \forall y \in L\}. Thus L is abelian iff L=Z(L).

Exercise. Let I, J be two ideals of a Lie algebra L. Show that

\displaystyle I+J:=\{a+b : a \in I, \; b \in J\}

and

\displaystyle [IJ]:=\{\sum_i\,[a_ib_i] : a_i \in I, \; b_i \in J\}

are both ideals of L.

\S 4. Homomorphism and quotient algebras.

Let L_1 and L_2 be two Lie algebras over F. An F-linear mapping f: L_1 \to L_2 is called a homomorphism of Lie algebras if f([xy])=[f(x)f(y)], \; \forall x, y \in L_1. A homomorphism is called an isomorphism if in addition, it is bijective. It is easy to check that the kernel of a homomorphism f, Ker f, is an ideal of L_1, and the image of f, Imf is a subalgebra of L_2.

Example of isomorphism. Recall that we have \mathfrak{gl}(V) \simeq \mathfrak{gl}(n, F), if V is an F-vector space of dimension n.

An utmost important example of homomorphism is the famous adjoint homomorphism of a Lie algebra L. It is defined by

\displaystyle \mathrm{ad}: L \to \mathfrak{gl}(L), \mathrm{ad}\, (y) = [xy],\; \forall y \in L.

The kernel of ad L is Z(L), the centre of L.

For an ideal I of a Lie algebra L, the quotient vector space L/I has a Lie algebra structure; for any x, y \in L, define the Lie bracket in L/I by [x+I, y+I]=[xy]+I. To check this Lie bracket is well-defined we use the bilinearlity of the Lie bracket in L and the definition of an ideal. We left it as an exercise to check that L/I is a Lie algebra with the bracket defined as above.

Remark. 关于同态(同构)的基本定理,以及在同态之下,理想的对应关系,略!

靠,打了这么多英文,视力都降低啊… 我决定以后尽量打中文.

Problems.

以下,假设L为域F上的n维Lie algebra.

1. 设x_1, \ldots, x_nL的一组基,因此[x_ix_j]=\sum_{k=1}^n\, a_{ij}^k\,x_k. 这些a_{ij}^k, i, j, k\in \{1, \ldots, n\}称为结构常数(structure constant). 证明:

(i) a_{ii}^k=0=a_{ij}^k+a_{ji}^k;

(ii) (Cartan-Maurer euqation) \sum_k \, (a_{ij}^ka_{kl}^m+a{jl}^ka_{ki}^m+a_{li}^ka_{kj}^m)=0.

(iii)每个 a_{ij}^k都满足(1,2)-张量的变换规则.

2. 设A是一个代数, D, EA的导出代数(简称导数), 即D, E \in \mathrm{Der}(A). 证明:

(i) [D, E]=DE-EDA的导数. 这里乘法是运算的复合.

(ii) 举例说明, DE可以不是一个导数.

3. 对任意x\in L, 伴随同态adxL的一个导数. 从而adL \subset \mathrm{Der}(L).

4. 设KL的子代数(甚至仅仅是子空间即可). KL中的正规化子(normaliser) N_L(K)定义为N_L(K)=\{x \in L : [xK] \subset K\}KL中的中心化子(centraliser) C_L(K)定义为N_L(K)=\{x \in L : [xK]=0\}. 证明: N_L(K) 与 C_L(K)都是子代数.

5. Show that \mathfrak{sl}(2, F) is simple provided that char F \neq 2.

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