Basic concepts
📜
The statement of the theorem
To every Lie group we can associate a Lie algebra whose underlying vector space is the tangent space of the Lie group at the identity element and which completely captures the local structure of the group. Informally we can think of elements of the Lie algebra as elements of the group that are "infinitesimally close" to the identity, and the Lie bracket of the Lie algebra is related to the commutator of two such infinitesimal elements. Before giving the abstract definition we give a few examples:
- The Lie algebra of the vector space R^{n} is just R^{n} with the Lie bracket given by
[A, B] = 0.
(In general the Lie bracket of a connected Lie group is always 0 if and only if the Lie group is abelian.)
- The Lie algebra of the general linear group GL(n, C) of invertible matrices is the vector space M(n, C) of square matrices with the Lie bracket given by
[A, B] = AB − BA.
- If G is a closed subgroup of GL(n, C) then the Lie algebra of G can be thought of informally as the matrices m of M(n, C) such that 1 + εm is in G, where ε is an infinitesimal positive number with ε^{2} = 0 (of course, no such real number ε exists). For example, the orthogonal group O(n, R) consists of matrices A with AA^{T} = 1, so the Lie algebra consists of the matrices m with (1 + εm)(1 + εm)^{T} = 1, which is equivalent to m + m^{T} = 0 because ε^{2} = 0.
- The preceding description can be made more rigorous as follows. The Lie algebra of a closed subgroup G of GL(n, C), may be computed as
where exp(tX) is defined using the matrix exponential. It can then be shown that the Lie algebra of G is a real vector space that is closed under the bracket operation, .
The concrete definition given above for matrix groups is easy to work with, but has some minor problems: to use it we first need to represent a Lie group as a group of matrices, but not all Lie groups can be represented in this way, and it is not even obvious that the Lie algebra is independent of the representation we use. To get around these problems we give
the general definition of the Lie algebra of a Lie group (in 4 steps):
- Vector fields on any smooth manifold M can be thought of as derivationsX of the ring of smooth functions on the manifold, and therefore form a Lie algebra under the Lie bracket [X, Y] = XY − YX, because the Lie bracket of any two derivations is a derivation.
- If G is any group acting smoothly on the manifold M, then it acts on the vector fields, and the vector space of vector fields fixed by the group is closed under the Lie bracket and therefore also forms a Lie algebra.
- We apply this construction to the case when the manifold M is the underlying space of a Lie group G, with G acting on G = M by left translations L_{g}(h) = gh. This shows that the space of left invariant vector fields (vector fields satisfying L_{g}_{*}X_{h} = X_{gh} for every h in G, where L_{g}_{*} denotes the differential of L_{g}) on a Lie group is a Lie algebra under the Lie bracket of vector fields.
- Any tangent vector at the identity of a Lie group can be extended to a left invariant vector field by left translating the tangent vector to other points of the manifold. Specifically, the left invariant extension of an element v of the tangent space at the identity is the vector field defined by v^_{g} = L_{g}_{*}v. This identifies the tangent spaceT_{e}G at the identity with the space of left invariant vector fields, and therefore makes the tangent space at the identity into a Lie algebra, called the Lie algebra of G, usually denoted by a Fraktur Thus the Lie bracket on is given explicitly by [v, w] = [v^, w^]_{e}.
This Lie algebra is finite-dimensional and it has the same dimension as the manifold G. The Lie algebra of G determines G up to "local isomorphism", where two Lie groups are called locally isomorphic if they look the same near the identity element.
Problems about Lie groups are often solved by first solving the corresponding problem for the Lie algebras, and the result for groups then usually follows easily.
For example, simple Lie groups are usually classified by first classifying the corresponding Lie algebras.
We could also define a Lie algebra structure on T_{e} using right invariant vector fields instead of left invariant vector fields. This leads to the same Lie algebra, because the inverse map on G can be used to identify left invariant vector fields with right invariant vector fields, and acts as −1 on the tangent space T_{e}.
The Lie algebra structure on T_{e} can also be described as follows:
the commutator operation
(x, y) → xyx^{−1}y^{−1}
on G × G sends (e, e) to e, so its derivative yields a bilinear operation on T_{e}G. This bilinear operation is actually the zero map, but the second derivative, under the proper identification of tangent spaces, yields an operation that satisfies the axioms of a Lie bracket, and it is equal to twice the one defined through left-invariant vector fields.
If G and H are Lie groups, then a Lie group homomorphism f : G → H is a smooth group homomorphism. In the case of complex Lie groups, such a homomorphism is required to be a holomorphic map. However, these requirements are a bit stringent; every continuous homomorphism between real Lie groups turns out to be (real) analytic.
The composition of two Lie homomorphisms is again a homomorphism, and the class of all Lie groups, together with these morphisms, forms a category. Moreover, every Lie group homomorphism induces a homomorphism between the corresponding Lie algebras. Let be a Lie group homomorphism and let be its derivative at the identity. If we identify the Lie algebras of G and H with their tangent spaces at the identity elements, then is a map between the corresponding Lie algebras:
which turns out to be a Lie algebra homomorphism (meaning that it is a linear map which preserves the Lie bracket). In the language of category theory, we then have a covariant functor from the category of Lie groups to the category of Lie algebras which sends a Lie group to its Lie algebra and a Lie group homomorphism to its derivative at the identity.
Two Lie groups are called isomorphic if there exists a bijective homomorphism between them whose inverse is also a Lie group homomorphism. Equivalently, it is a diffeomorphism which is also a group homomorphism. Observe that, by the above, a continuous homomorphism from a Lie group to a Lie group is an isomorphism of Lie groups if and only if it is bijective.
Isomorphic Lie groups necessarily have isomorphic Lie algebras; it is then reasonable to ask how isomorphism classes of Lie groups relate to isomorphism classes of Lie algebras.
The first result in this direction is Lie's third theorem, which states that every finite-dimensional, real Lie algebra is the Lie algebra of some (linear) Lie group. One way to prove Lie's third theorem is to use Ado's theorem, which says every finite-dimensional real Lie algebra is isomorphic to a matrix Lie algebra. Meanwhile, for every finite-dimensional matrix Lie algebra, there is a linear group (matrix Lie group) with this algebra as its Lie algebra.
On the other hand, Lie groups with isomorphic Lie algebras need not be isomorphic. Furthermore, this result remains true even if we assume the groups are connected. To put it differently, the global structure of a Lie group is not determined by its Lie algebra; for example, if Z is any discrete subgroup of the center of G then G and G/Z have the same Lie algebra (see the table of Lie groups for examples). An example of importance in physics are the groups SU(2) and SO(3). These two groups have isomorphic Lie algebras, but the groups themselves are not isomorphic, because SU(2) is simply connected but SO(3) is not.
On the other hand, if we require that the Lie group be simply connected, then the global structure is determined by its Lie algebra: two simply connected Lie groups with isomorphic Lie algebras are isomorphic. (See the next subsection for more information about simply connected Lie groups.) In light of Lie's third theorem, we may therefore say that there is a one-to-one correspondence between isomorphism classes of finite-dimensional real Lie algebras and isomorphism classes of simply connected Lie groups.
A Lie group is said to be simply connected if every loop in can be shrunk continuously to a point in . This notion is important because of the following result that has simple connectedness as a hypothesis:
Theorem: Suppose
and
are Lie groups with Lie algebras
and
and that
is a Lie algebra homomorphism. If
is simply connected, then there is a unique Lie group homomorphism
such that , where
is the differential of
at the identity.
Lie's third theorem says that every finite-dimensional real Lie algebra is the Lie algebra of a Lie group. It follows from Lie's third theorem and the preceding result that every finite-dimensional real Lie algebra is the Lie algebra of a unique simply connected Lie group.
An example of a simply connected group is the special unitary group SU(2), which as a manifold is the 3-sphere. The rotation group SO(3), on the other hand, is not simply connected. (See Topology of SO(3).) The failure of SO(3) to be simply connected is intimately connected to the distinction between integer spin and half-integer spin in quantum mechanics. Other examples of simply connected Lie groups include the special unitary group SU(n), the spin group (double cover of rotation group) Spin(n) for , and the compact symplectic group Sp(n).
Methods for determining whether a Lie group is simply connected or not are discussed in the article on fundamental groups of Lie groups.
The exponential map from the Lie algebra of the general linear group to is defined by the matrix exponential, given by the usual power series:
for matrices . If is a closed subgroup of , then the exponential map takes the Lie algebra of into ; thus, we have an exponential map for all matrix groups. Every element of that is sufficiently close to the identity is the exponential of a matrix in the Lie algebra.
The definition above is easy to use, but it is not defined for Lie groups that are not matrix groups, and it is not clear that the exponential map of a Lie group does not depend on its representation as a matrix group. We can solve both problems using a more abstract definition of the exponential map that works for all Lie groups, as follows.
For each vector in the Lie algebra of (i.e., the tangent space to at the identity), one proves that there is a unique one-parameter subgroup such that . Saying that is a one-parameter subgroup means simply that is a smooth map into and that
c(s+t)=c(s)c(t)\$$
for all $s$ and $t$ . The operation on the right hand side is the group multiplication in $G$ . The formal similarity of this formula with the one valid for the exponential function justifies the definition
\exp(X)=c(1).
This is called the exponential map, and it maps the Lie algebra ${\mathfrak {g}}$ into the Lie group $G$ . It provides a diffeomorphism between a neighborhood of 0 in ${\mathfrak {g}}$ and a neighborhood of $e$ in $G$ . This exponential map is a generalization of the exponential function for real numbers (because $\mathbb {R}$ is the Lie algebra of the Lie group of positive real numbers with multiplication), for complex numbers (because $\mathbb {C}$ is the Lie algebra of the Lie group of non-zero complex numbers with multiplication) and for matrices (because $M(n,\mathbb {R} )$ with the regular commutator is the Lie algebra of the Lie group $\mathrm {GL} (n,\mathbb {R} )$ of all invertible matrices).
Because the exponential map is surjective on some neighbourhood $N$ of $e$ , it is common to call elements of the Lie algebra infinitesimal generators of the group $G$ . The subgroup of $G$ generated by $N$ is the identity component of $G$ .
The exponential map and the Lie algebra determine the local group structure of every connected Lie group, because of the Baker–Campbell–Hausdorff formula: there exists a neighborhood $U$ of the zero element of ${\mathfrak {g}}$ , such that for $X,Y\in U$ we have
\exp(X)\,\exp(Y)=\exp \left(X+Y+{\tfrac {1}{2}}[X,Y]+{\tfrac {1}{12}}[\,[X,Y],Y]-{\tfrac {1}{12}}[\,[X,Y],X]-\cdots \right),
where the omitted terms are known and involve Lie brackets of four or more elements. In case $X$ and $Y$ commute, this formula reduces to the familiar exponential law $\exp(X)\exp(Y)=\exp(X+Y)$ .
The exponential map relates Lie group homomorphisms. That is, if $\phi :G\to H$ is a Lie group homomorphism and $\phi _{*}:{\mathfrak {g}}\to {\mathfrak {h}}$ the induced map on the corresponding Lie algebras, then for all $x\in {\mathfrak {g}}$ we have
\phi (\exp(x))=\exp(\phi _{*}(x)).$$
In other words, the following diagram commutes,
(In short, exp is a natural transformation from the functor Lie to the identity functor on the category of Lie groups.)
The exponential map from the Lie algebra to the Lie group is not always onto, even if the group is connected (though it does map onto the Lie group for connected groups that are either compact or nilpotent). For example, the exponential map of SL(2, R) is not surjective. Also, the exponential map is neither surjective nor injective for infinite-dimensional (see below) Lie groups modelled on C^{∞}Fréchet space, even from arbitrary small neighborhood of 0 to corresponding neighborhood of 1.
A Lie subgroup $H$ of a Lie group $G$ is a Lie group that is a subset of $G$ and such that the inclusion map from $H$ to $G$ is an injectiveimmersion and group homomorphism. According to Cartan's theorem, a closed subgroup of $G$ admits a unique smooth structure which makes it an embedded Lie subgroup of $G$ —i.e. a Lie subgroup such that the inclusion map is a smooth embedding.
Examples of non-closed subgroups are plentiful; for example take $G$ to be a torus of dimension 2 or greater, and let $H$ be a one-parameter subgroup of irrational slope, i.e. one that winds around in G. Then there is a Lie group homomorphism $\varphi :\mathbb {R} \to G$ with $\mathrm {im} (\varphi )=H$ . The closure of $H$ will be a sub-torus in $G$ .
The exponential map gives a one-to-one correspondence between the connected Lie subgroups of a connected Lie group $G$ and the subalgebras of the Lie algebra of $G$ . Typically, the subgroup corresponding to a subalgebra is not a closed subgroup. There is no criterion solely based on the structure of $G$ which determines which subalgebras correspond to closed subgroups.
- ^Helgason 1978, Ch. II, § 2, Proposition 2.7
- ^Hall 2015
- ^Hall 2015 Theorem 3.20
- ^But see Hall 2015, Proposition 3.30 and Exercise 8 in Chapter 3
- ^Hall 2015 Corollary 3.50
- ^Hall 2015 Theorem 5.20
- ^Hall 2015 Example 3.27
- ^Hall 2015 Section 1.3.4
- ^Hall 2015 Corollary 5.7
- ^Hall 2015 Theorem 5.6
- ^Hall 2015 Section 13.2
- ^Hall 2015 Theorem 3.42
- ^"Introduction to Lie groups and algebras : Definitions, examples and problems"(PDF). State University of New York at Stony Brook. 2006. Archived from the original(PDF) on 28 September 2011. Retrieved 11 October 2014.
- ^Hall 2015 Theorem 5.20
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