Derivations, vector fields, Lie groups and their Lie algebras

October 28, 2012

In this post, we will add a bunch of examples to those we saw in last post. We will do a brief excursion into the world of differential geometry, where we will see how the idea of a Lie algebra is related to Lie groups and geometry. If you’ve never seen the definition of a smooth manifold, I refer you to the Wikipedia page.

First, remember that given a $\mathbb{K}-$ algebra $A,$ a derivation of $A$ is a linear operator $D \in \text{End}_{\mathbb{K}}A$ satisfying Leibniz rule: for every $a,b \in A,$

$D(ab) = D(a)b + aD(b).$

For example, the usual directional derivatives in $\mathbb{R}^n$ from calculus are obviously derivations of the algebra of smooth real valued functions $C^{\infty}(\mathbb{R}^n).$ We also saw in my first post on Lie algebras that for $\mathfrak{g}$ a Lie algebra and $x \in \mathfrak{g}$ , $\text{ad}(x)$ is a derivation of $\mathfrak{g}.$ We will note $\text{Der}(A)$ the set of all derivations of an algebra $A.$

It is easy to see that if $A$ is an algebra over the field $\mathbb{K},$ $\text{Der}(A)$ is a $\mathbb{K}-$ vector subspace of $\text{End}_{\mathbb{K}}A.$ In fact, when we equip $\text{End}_{\mathbb{K}}A$ with the bracket operation $[X,Y] = XY -YX$ (making it a Lie algebra), the subspace $\text{Der}(A)$ is a Lie subalgebra of $\text{End}_{\mathbb{K}}A.$ Indeed, for $D,E \in \text{Der}(A)$ and $a,b \in A,$ we have

$[D,E](a,b) = (DE)(ab) - (ED)(ab)$

$= D(E(a)b + aE(b)) - E(D(a)b + aD(b))$

$= D(E(a))b + E(a)D(b) + D(a)E(b) + aD(E(b))$
$- E(D(a))b - D(a)E(b) - E(a)D(b) - aE(D(b))$

$= (DE)(a)b - (ED)(a)b + a(DE)(b) - a(ED)(b)$

$= [D,E](a)b + a[D,E](b).$

Let’s now digress a little bit to describe what is a vector field on a smooth manifold.

Consider a smooth surface $S$ sitting in $\mathbb{R}^3,$ a sphere for example. Then for any parametrized curve $\gamma : [-1,1] \to S$ such that $\gamma(0) = p \in S,$ the derivative $\gamma'(0)$ is a vector sitting in the plane tangent to $S$ at $p.$ Intuitively, this plane is actually the space of all vectors that are the derivative of some curve $\gamma : [-1,1] \to S$ with $\gamma(0) = p,$ i.e. $T_pS = \{ \gamma'(0) : \gamma : [-1,1] \to S, \gamma(0) = p \}.$ With this point of view, the tangent space to a point $p \in \mathbb{R}^n$ can be viewed as the set of “arrows” emanating from $p.$

Alternatively, we can adopt kind of a dual point of view. Define a derivation at p to be a a linear map $f : C^{\infty}_p(\mathbb{R}^n) \to \mathbb{R}$ satisfying the Leibniz rule, where $C^{\infty}_p(\mathbb{R}^n)$ is the vector space of germs of smooth real valued functions at $p$ (i.e. the set of equivalence classes of functions under the relation $f \sim g$ iff $f = g$ on some open neighborhood of $p).$ It can be shown that the derivations at $p$ are exactly the directional derivatives at $p,$ given by

$v_p(f) = \dfrac{d}{dt} (f \circ \gamma)(0)$

for $\gamma : [-1,1] \to \mathbb{R}^n$ some smooth curve with $\gamma'(0) = v$ and $\gamma(0) = p.$ Thinking of $v$ again as an “arrow” emanating from $p,$ directing the derivation, we can think of the tangent space of $\mathbb{R}^n$ at $p$ as the actual set of derivations at $p.$ This is the easiest way to define the tangent space of an arbitrary smooth manifold.

For $M$ a smooth manifold and $p \in M,$ define the tangent space at $p,$ noted $T_pM,$ to be the space of derivations at $p.$ Equivalently, we can define it as the space of vectors

$\gamma'(0) = \dfrac{d}{dt}(\varphi \circ \gamma)(0)$

where $\varphi$ is a coordinate map defined on a chart around p and $\gamma : [-1,1] \to M$ a curve such that $\gamma(0) = p.$

For some manifold $M,$ the (disjoint) union of all tangent spaces is called the tangent bundle,

$TM = \displaystyle\bigcup_{p \in M} T_pM.$

There is a natural projection map $\pi : TM \to M$ defined by $\pi (p,v) = p$ for $p$ a point of $M$ and $v$ a vector in $T_pM.$ The space $TM$ can be given a canonical smooth structure making it itselft a differentiable manifold. A smooth vector field on a differentiable manifold $M$ is then a smooth section of $TM,$ i.e. a smooth map $\phi : M \to TM$ such that $\pi \circ \phi = \text{Id}_M.$

The important result that connects this all with what we’ve done so far is that the space of smooth vector fields on $M$ is exactly the space of derivations of the real algebra of smooth real-valued functions $\text{Der}(C^{\infty}(M)).$ Therefore, what we have said at the beginning of this post applies :

the set of smooth vector fields on a differentiable manifold form a Lie algebra.

We can now describe Lie groups and their Lie algebras. A Lie group is a group that also has the structure of a smooth manifold in which the operations of multiplication and taking inverses are diffeomorphisms. So on the one hand we can use what we know from group theory on them, but on the other the notions of vector fields and all the machinery from differential geometry can be used to study them.

We can show that left multiplication in a Lie group is a diffeomorphism. We thus have the notion of a left invariant vector field on a Lie group : noting left multiplication by $g \in G$ by $l_g,$ a vector field $X$ is left invariant if $X(f \circ l_g) = X(f) \circ l_g$ for all $g \in G$ and $f \in C^{\infty}(M).$ It can be shown that left invariant vector fields are smooth and closed under the bracket operation and so form a Lie subalgebra of the space of smooth vector fields. This is the Lie algebra $\mathfrak{g}$ of the Lie group $G.$

A left invariant vector field is determined locally. For example, if you know how it acts on functions at the identity $1 \in G,$ then you know that for any $g \in G$ and $f \in C^{\infty}_g(G),$ it acts on $f$ at $g$ by

$X(f)(g) = (X(f) \circ l_g)(1) = (X(f \circ l_g)(1).$

This observation in fact gives us a vector space isomorphism between the space of left invariant vector fields on $G$ and the tangent space at the identity : for a vector field $X$ take $X_1$ its component in $T_1G$ and from a tangent vector at the identity create a left invariant field like above. We can thus transport the Lie algebra structure of left invariant vector fields to the tangent space at the identity and this gives us a more concrete way to view the Lie algebra of a Lie group.

Indeed, consider all smooth curves $\gamma : [-1,1] \to G$ with $\gamma(0) = 1_G.$ Then the Lie algebra of $G,$ being identified with $T_1G,$ can be described as $\mathfrak{g} = \{ \gamma'(0) \}$ with the Lie algebra structure induced by the isomorphism above. When $G$ is one of the classical Lie groups of matrices, we recover in this way the classical linear Lie algebras described in my last post. Here are some examples:

The set of invertible matrices $GL(n,\mathbb{R})$ can be given a differentiable structure in a natural way : we just embedd it in $\mathbb{R}^{n^2}$ and use the induced differentiable structure. Then $GL(n, \mathbb{R}$ is an open subset, being the preimage of $\mathbb{R}^{n^2} - \{0\}.$ Moreover, multiplication and inversion are given by polynomial functions so they are differentiable operations and $GL(n, \mathbb{R})$ forms a Lie group. (Note that this discussion is also valid for $\mathbb{C}$ instead of $\mathbb{R}).$ A closed subgroup of some $GL(n, \mathbb{R})$ is called a closed linear group. Examples of closed linear groups are

$SO(n) = \{x \in GL(n, \mathbb{R}) : xx^t = 1, \det x = 1 \},$

$U(n) = \{ x \in GL(n, \mathbb{C}) : xx^* = 1 \},$

$SL(n, \mathbb{C}) = \{ x \in GL(n, \mathbb{C}) : \det x = 1 \}.$

These are all groups of continuous symmetries. For example, $SO(3)$ is the group of rotations in 3 dimensional Euclidean space. Thus a curve in $SO(3)$ represents a rotational motion and the set $\{ \gamma'(0) \}$ for all curves $\gamma : [-1,1] \to SO(3)$ with $\gamma(0) = 1_{SO(3)},$ ie. the Lie algebra of $SO(3),$ contains all the information about infinitesimal rotational motion. For example, this defines a smooth curve in $SO(3)$ :taking the derivative at 0 gives

which is indeed a skew-symmetric matrix (i.e. a matrix whose transposed is its own additional inverse). Using other curves, we can show that the Lie algebra $\mathfrak{so}(3, \mathbb{R})$ of skew-symmetric matrices is contained in the tangent space at the identity of $SO(3).$ In fact, since for a curve in $SO(3)$ we have $\gamma(t) \gamma(t)^t = \text{Id},$ we find that

$0 = \gamma'(t)\gamma(t)^t + \gamma(t) \gamma'(t)^t$

$= \gamma'(0) \text{Id}^t + \text{Id} \gamma'(0)^t$

$= \gamma'(0) + \gamma'(0)^t.$

So the Lie algbera of $SO(3)$ is the Lie algebra of skew-symmetric real $3 \times 3$ matrices. The bracket operation induced by the left invariant vector fields on $SO(3)$ is just the usual bracket for linear Lie algebras.

Similar considerations shows that the Lie algebra of $U(n)$ is $\mathfrak{u}(n)$ the space of matrices such that $X + X^* = 0,$ the Lie algebra of $SL(n, \mathbb{C})$ is $\mathfrak{sl}(n, \mathbb{C})$ the space of matrices with trace $0,$ etc…

This very intriguing stuff. On the one hand, the study of Lie algebras is very algebraic and on the other, Lie groups are very geometric. There is a nice dance between these two notions and I would like to investigate the geometric side of things a little bit more but the task I have given myself is to understand the classification of (semi-)simple Lie algebras. In the next posts, I will thus resume the algebraic investigation but hopefully I will come back to this when I’m done.

Reference:

Knapp – Lie Groups Beyond an Introduction
Humphreys – Introduction to Lie Algebras and Representation Theory
L. W. Tu – An Introduction to Manifolds
Lee – Introduction to Smooth Manifolds.

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Derivations, vector fields, Lie groups and their Lie algebras

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