Lie Groups and Their Lie Algebras
Lie groups describe symmetries that vary smoothly. Their group operation lets us translate tangent vectors from the identity to any other point, reducing many questions about vector fields to finite-dimensional linear algebra. For the definition, refer to Example 31. To motivate this study, note that a group was originally coined to describe symmetries. The collection of symmetries of any object is a group, and every group is the symmetries of some object. For example, the symmetry group of a square contains four rotations, two reflections in the midlines and two reflections in the diagonals, which is a finite group of eight elements (usually denoted by \(D_4\), with the convention that its order is eight). Another simple example is a circle in the plane, whose symmetry group consists of all rotations and reflections, and can be identified with \(O(2)\), the set of all two by two orthogonal matrices. The circle example leads to the compatibility between group operations and smooth structure. More precisely, a Lie group admits three structures
an algebraic structure as a group,
a geometric structure as a topological manifold, and
a smooth structure so that one can do analysis on them.
We construct the Lie algebra from left-invariant vector fields, identify its bracket for matrix groups, and explain how a group homomorphism acts on these infinitesimal data.
1. Basic Definitions and Operations
Lemma 1. If \(G\) is a smooth manifold with a group structure such that \(\eta : G \times G \to G : (g,h) \mapsto gh^{-1}\) is smooth, then \(G\) is a Lie group.
Proof. Let \(f: G \to G \times G : g \mapsto (e,g)\) be a map. \(f\) is smooth because of its smoothness in both components. Clearly, the inverse map \(i : G \to G : g \mapsto g^{-1}\) can be written as \(i = \eta \circ f\), and \(i\) is smooth as the composition of smooth maps. It follows that \(\operatorname{id} \times i : G \times G \to G \times G : (g,h) \mapsto (g,h^{-1})\) is smooth, as it is smooth in both components. Finally, set \(m = \eta \circ (\operatorname{id} \times i)\), which is clearly the multiplication map, and is smooth as the composition of smooth maps. ◻
Let \(G\) be a Lie group. On \(G\), we can define some operations. In particular, for all \(a,b \in G\),
denote left translation (left multiplication) and right translation (right multiplication), respectively. It is clear that \(L_a^{-1} = L_{a^{-1}}\) and \(R_b^{-1} = R_{b^{-1}}\), i.e. \(L_a\) and \(R_b\) are diffeomorphisms. Furthermore, these operations are commutative: \(L_a R_b = R_b L_a\). Finally, collecting the maps introduced in the proof of Lemma 1, we have
are the multiplication and inverse maps.
Lemma 2. For any \(a,b \in G\) and any \((X_a,Y_b) \in T_a G \times T_b G \cong T_{(a,b)} (G \times G)\), we have
Proof. The tangent space of a product splits as a direct sum, so linearity gives
On the first slice multiplication is \(R_b\), and on the second it is \(L_a\). This proves the formula, with both terms in \(T_{ab}G\). Equivalently, testing on a smooth function \(f\) gives \(X(f\circ R_b)+Y(f\circ L_a)\); derivatives act on these functions before they are evaluated at the base points. ◻
Lemma 3. If a smooth manifold is an abstract group and its multiplication is smooth, then its inverse map is smooth. Thus inversion need not be separately assumed in this equivalent definition of a Lie group.
Proof. For each fixed \(a\), translation \(L_a\) and its set-theoretic inverse \(L_{a^{-1}}\) are smooth by smoothness of multiplication. This does not yet assume smooth dependence of \(a^{-1}\) on \(a\). Let \(F(a,b)=(a,ab)\). It is bijective, and
Given \((U,V)\), solve uniquely \(X=U\) and \(Y=(dL_a)^{-1}(V-dR_bU)\); hence the differential is invertible. The inverse function theorem makes \(F\) a local diffeomorphism everywhere. Its local inverses agree with its unique global set-theoretic inverse, making that inverse smooth. Finally
is a composition of smooth maps. ◻
Definition 4. Let \(G\) be a Lie group. A subset \(H \subset G\) is a Lie subgroup if \(H\) is (immersed) submanifold of \(G\), and the group operation on \(H\) is smooth (with respect to its own differentiable structure).
Here \(H\) must be a subgroup as well as an injectively immersed manifold, and its manifold topology may be finer than the subspace topology. Multiplication is smooth in that topology; Lemma 3 then gives smooth inversion.
The Lie bracket of vector fields suggests an algebraic definition. It retains bilinearity, skew-symmetry, and the Jacobi identity, without requiring a manifold.
Definition 5. A Lie algebra is a vector space \(V\) with the bilinear bracket operation \([\cdot,\cdot] : V \times V \to V\) such that for all \(X,Y,Z \in V\),
(Skew-Symmetry) \([X,Y]=-[Y,X]\),
(Jacobi’s Identity) \([X,[Y,Z]] + [Y,[Z,X]] + [Z,[X,Y]] = 0\).
All Lie algebras here are over \(\mathbb R\), so skew-symmetry implies \([X,X]=0\). The bracket is bilinear over constants, not over arbitrary smooth functions when the vectors are vector fields.
Remark 6 (The Left-Invariant Property on Tangent Space at Identity). Let \(e \in G\) denote the identity element of the Lie group \(G\). Consider \(a \in G\), \(X_e \in T_e G\), and the left translation \(L_a\) acting on \(X_e\). By ongoing discussions, translations in \(G\) define canonical identifications between the tangent spaces to \(G\) at different point. In particular, \(\dif L_a : T_g G \to T_{ag} G\) for any \(g \in G\). This property allow us to consider invariant tensors, the most important case being that of vector fields.
We say a vector field \(X\) on \(G\) is left-invariant if \((\dif L_a)_b (X_b) = X_{ab}\) for all \(a,b \in G\). This definition is sensible because, by direct calculation and properties discussed, \((\dif L_a)_b (X_b) = (\dif L_a)_b (\dif L_b) (X_e) = (\dif L_{ab}) (X_e) = X_{ab}\). Clearly, this condition is saying \((\dif L_a) \circ X = X \circ L_a\) for every \(a \in G\); equivalently, \(X\) is \(L_a\)-related to itself, or \(L_{a^*} X = X\) (since \(L_a\) is a diffeomorphism), for all \(a \in G\). Collecting the discussion above, it follows that any \(X_e \in T_e G\) determines a left-invariant vector field \(X\) on \(G\). The converse also holds: any left-invariant vector field \(X\) admits a unique determination by its “value” at \(e \in G\), i.e. by \(X(a) = (\dif L_a) (X_e)\). Therefore, it makes sense to define the following object:
as a vector subspace of \(\Gamma(TG)\). Finally, by ongoing discussion, as a vector space, \(\mathfrak g \cong T_e G\) and \(\dim \mathfrak g = \dim G\), because the map \(\mathfrak g \ni X \mapsto X_e\) defines is a linear isomorphism between \(\mathfrak g\) and \(T_e G\).
We will not pursue the following point currently, but note that \(\Gamma(TG)\) admits a Lie bracket operation that turns it into an infinite-dimensional Lie algebra. For now, we show that \(\mathfrak g\) admits a bracket operation that turns it into a Lie algebra.
Proposition 7. If \(X, Y \in \mathfrak g\), then \([X,Y] \in \mathfrak g\).
Proof. Left invariance says \(X(f\circ L_a)=(Xf)\circ L_a\), and the analogous identity holds for \(Y\). Apply \(X\) to the identity for \(Y\), subtract the reversed identity, and obtain
At \(b\), the left side equals \(d(L_a)_b([X,Y]_b)f\), whereas the right side is \([X,Y]_{ab}f\). Equality for all smooth test functions proves left invariance. Bilinearity, antisymmetry and Jacobi are inherited from the bracket of vector fields. ◻
Note that there is a much more general statement: for any \(a \in G\),
This follows from general properties of the bracket along with the fact that \(L_a\) is a (local) diffeomorphism. By Proposition 7, we can state the following definition.
Definition 8. \(\mathfrak g\) in (1), together with the Lie bracket operation, is the Lie algebra of the Lie group \(G\).
Proposition 9 (Lie algebra is finite-dimensional). Let \(G\) be a Lie group, and \(\mathfrak g\) be the Lie algebra of \(G\). Then, the evaluation map \(\mathcal E: \mathfrak g \to T_e G : X \mapsto X_e\) is a vector space isomorphism. Therefore, \(\mathfrak g\) is finite-dimensional, with dimension equal to \(\dim G\).
Proof. Clearly, \(\mathcal E\) is linear over \(\mathbb R\), and this is an injective map. The latter follows because if \(\mathcal E(X) = 0\) for some \(X \in \mathfrak g\), then the left-invariance of \(X\) implies \(X_g = (\dif L_g)_e (X_e) = 0\), i.e. \(X = 0\). Now, it remains to show \(\mathcal E\) is surjective.
To do so, for every \(V \in T_e G\), define
Clearly, if there is a left-invariant vector field on G whose value at the identity is \(V\), it has to be given by this formula. Note that the smoothness property of \(\widetilde V\) is straightforward to show; we prove it separately, in Lemma 10. Finally, clearly, because \(L_h \circ L_g = L_{hg}\) for all \(h,g \in G\), we have
as desired. Therefore, \(\widetilde V \in \mathfrak g\). To conclude, note that \(L_e\) is the identity map of \(G\), it follows that \(\mathcal E(\widetilde V) = \widetilde V\big|_e = V\). This shows the surjectivity of \(\mathcal E\). ◻
The only remaining point is smoothness of the field obtained by translating a tangent vector. To complete the proof of Proposition 9, we provide the following result.
Lemma 10. Every left invariant vector field \(X\) on \(G\) is smooth.
Proof. Fix \(V=X_e\), and choose a smooth curve \(c(t)\) through \(e\) with velocity \(V\). In a coordinate neighborhood of an arbitrary \(a_0\), each coefficient of \(X\) is
For \(a\) near \(a_0\) and \(t\) sufficiently small the argument remains in the coordinate neighborhood. The map inside the derivative is jointly smooth in \((a,t)\), so its partial derivative at \(t=0\) is smooth in \(a\). Thus all coordinate coefficients are smooth. The test function must be defined near \(a\), not merely near \(e\). ◻
Example 11 (Basic Examples of Lie Algebras). The left-invariant vector fields on \(\mathbb R^n\) are precisely the constant vector fields, i.e. the linear combinations of coordinate vector fields (in the canonical coordinate system) with constant coefficients. The bracket of two constant vector fields on \(\mathbb R^n\) is zero. It follows that the Lie algebra of \(\mathbb R^n\) is \(\mathbb R^n\) itself with the null bracket. In general, a vector space equipped with the null bracket is called an abelian Lie algebra.
The Lie algebra of the direct product group \(G \times H\) is the direct sum of Lie algebras, \(\mathfrak g \oplus \mathfrak h\), where the bracket is taken component-wise.
Indeed \(T_{(e_G,e_H)}(G\times H)=T_{e_G}G\oplus T_{e_H}H\), and translating a pair of tangent vectors gives the two independent component fields. Their mixed derivatives vanish, so \([(X_1,Y_1),(X_2,Y_2)]=([X_1,X_2],[Y_1,Y_2])\).
Example 12 (Matrix Lie Group). Let \(G = \operatorname{GL}_n(\mathbb R)\). Recall this inherits its manifold structure as an open subset of the Euclidean space \(\mathcal M_n (\mathbb R)\), i.e. the space of \(n \times n\) matrices with real entries. In particular, the tangent space at the identity \(T_e G = \mathcal M_n (\mathbb R)\). Following convention, the Lie algebra associated with \(\operatorname{GL}_n (\mathbb R)\) is denoted \(\mathfrak{g}\).
Here, we first show a definition of a Lie algebra of \(\mathcal M_n (\mathbb R)\): following convention again, let \(\mathfrak{gl}\) denote the Lie algebra we obtain from associating \(\mathcal M_n (\mathbb R)\) with a bracket operation (to be defined below).
First, let \(A \in \mathfrak{gl} = \mathcal M_n (\mathbb R)\), and \(\widetilde A\) denotes the corresponding left-invariant vector field on \(G\). For any \(g \in G\), we have \(\widetilde A g = (\dif L_g)(A) = gA\) (matrix multiplication on the right hand side). In what follows, \(B \in \mathfrak{gl}\) and \(\widetilde B\) denotes the corresponding left-invariant vector field on \(G\).
We now determine the bracket operation. Using the ongoing notation, let the matrix entries \((X^{ij})\) denote the global coordinate system on the space of matrices. Then, \(\widetilde A = A_{ij} \pd{}{{X^{ij}}}\) is the corresponding tangent vector at \(T_e G\). Therefore, the corresponding left-invariant vector at \(X \in G\) is given by \(\widetilde A|_X = X^{ik} A_{kj} \pd{}{{X^{ij}}}\big|_{X}\). It follows that
where, again, the Einstein summation convention is used throughout. Evaluating this expression at the identity element (i.e. identity matrix), we have
Therefore, the Lie bracket operation on \(\mathfrak{gl}\) is the matrix commutator \([A,B] = AB - BA\) for \(A, B \in \mathcal M_n (\mathbb R)\). This is also known as the commutator bracket of matrices.
The matrix calculation identifies the two descriptions of the Lie algebra: left-invariant vector fields on the group and matrices with their commutator bracket.
Theorem 13. Evaluation at the identity identifies the Lie algebra of \(\operatorname{GL}_n(\mathbb R)\) with the vector space \(\mathcal M_n(\mathbb R)\) equipped with \([A,B]=AB-BA\). This matrix Lie algebra is denoted \(\mathfrak{gl}_n(\mathbb R)\). It is not the Lie algebra of the additive group of all matrices, whose bracket would be zero.
Proof. Proposition 9 identifies invariant fields with their value in \(T_I\operatorname{GL}_n=\mathcal M_n\). Under the open-subset tangent-space identification, \(\widetilde A(X)=XA\). Differentiating these matrix-valued fields gives
Evaluation at \(I\) therefore preserves the bracket. In particular the matrix commutator inherits Jacobi from vector fields (it can also be checked by expanding its twelve terms). ◻
Corollary 14. Every Lie group admits a left-invariant smooth global frame; therefore, every Lie group is parallelizable.
For the definition of parallelizable, see Appendix Vector Fields and Their Flows. Note that if \(G\) is a Lie group, a local or global frame consisting of left-invariant vector fields is called a left-invariant frame.
Proof. Choose a basis \(v_1,\ldots,v_n\) of \(T_eG\). The fields \(X_i(g)=d(L_g)_e v_i\) are smooth by Lemma 10. Because \(d(L_g)_e\) is a linear isomorphism, their values form a basis at every \(g\). They give a global frame, or explicitly the bundle trivialization \((g,t)\mapsto\sum_i t_iX_i(g)\). Its inverse is smooth by inversion of the frame matrix in each coordinate chart. ◻
2. Lie Algebras
A smooth group homomorphism can be differentiated at the identity. Compatibility with translations extends this linear map to the corresponding left-invariant fields, and compatibility with Lie brackets makes it a Lie algebra homomorphism.
Theorem 15 (Induced Lie algebra homomorphism). Let \(G\) and \(H\) be Lie groups, and \(\mathfrak g\) and \(\mathfrak h\) be their Lie algebras. Let \(F: G \to H\) be a Lie group homomorphism. For every \(X \in \mathfrak g\), there exists a unique vector field in \(\mathfrak h\) that is \(F\)-related to \(X\). Denote such vector field by \(F_* X\). The map \(F_* : \mathfrak g \to \mathfrak h\) is a Lie algebra homomorphism.
Proof. Distinguish the identities \(e_G,e_H\). Define
This is a smooth left-invariant field on all of \(H\). Differentiating \(F\circ L_g=L_{F(g)}\circ F\) at \(e_G\) gives
Conversely any invariant field related to \(X\) must have the displayed value at \(e_H=F(e_G)\), which determines it uniquely. Linearity follows from linearity of \(dF_{e_G}\).
To verify the bracket, if \(U=F_*X\) and \(V=F_*Y\), relatedness is the identity \(X(f\circ F)=(Uf)\circ F\) for every smooth local function \(f\) on \(H\). Taking successive derivatives and subtracting yields \([X,Y](f\circ F)=([U,V]f)\circ F\). Thus \([U,V]\) is related to \([X,Y]\), and it is invariant by Proposition 7. Uniqueness proves \(F_*[X,Y]=[F_*X,F_*Y]\). No surjectivity or injectivity of \(F\) was used. ◻
Remark 16. Note that in the proof of Theorem 15, \(F\) is not necessarily a diffeomorphism. Even in the absence of \(F\) being a diffeomorphism, \(F_* X\) is a well-defined smooth vector field on \(H\).
Lemma 17 (Induced homomorphisms).
The induced homomorphism of the identity map on \(G\), denoted \((\operatorname{id}_G)_* : \mathfrak g \to \mathfrak g\), is the identity of \(\mathfrak g\).
If \(F : G \to H\) and \(E : H \to K\) are Lie group homomorphisms, then
\[\begin{aligned} (E \circ F)_* = E_* \circ F_* : \mathfrak g \to \mathfrak k ~, \end{aligned}\]where \(\mathfrak g\) and \(\mathfrak k\) are Lie algebras of \(G\) and \(K\), respectively.
Isomorphic Lie groups have isomorphic Lie algebras.
Proof. At the identity these claims are \(d(\mathrm{id})=\mathrm{id}\) and the chain rule \(d(E\circ F)=dE\circ dF\). Evaluation is an isomorphism, so the same identities hold for invariant fields. Apply the composition identity to a group isomorphism and its smooth inverse to obtain inverse Lie algebra homomorphisms. ◻
Note that if \(H\) is a Lie subgroup of \(G\), we might hope that the Lie algebra of \(H\), denoted \(\mathfrak h\), is a Lie subalgebra of that of \(G\). However, elements of \(\mathfrak h\) are vector fields on \(H\), not on \(G\)! We can still view \(\mathfrak h\) is a subalgebra of \(\mathfrak g\) in some sense.
Theorem 18 (Lie algebra of a Lie subgroup). Let \(H \subset G\) be a Lie subgroup of a Lie group \(G\). Let \(\iota : H \xhookrightarrow{} G\) be the inclusion map. There exists a Lie subalgebra \(\mathfrak H \subset \mathfrak g\) that it is canonically isomorphic to \(\mathfrak h\), such that
Proof. Apply the preceding homomorphism theorem to the inclusion. Because the inclusion map \(\iota\) is a Lie group homomorphism, it follows that \(\iota_* (\mathfrak h)\) is a Lie subalgebra of \(\mathfrak g\). This subalgebra is precisely the set of left-invariant vector fields on \(G\), whose values at the identity are of the form \((\dif \iota)_e (v)\) for some \(v \in T_e H\). The two characterizations of \(\mathfrak H\) given in the statement of the theorem are equivalent, because the differential \((\dif \iota)_e : T_e H \to T_e G\) is the inclusion of \(T_e H\) as a subspace of \(T_e G\). By the injectivity of \((\dif \iota)_e\) on \(T_e H\), it follows that \(\iota_*\) is injective on \(\mathfrak h\). With surjectivity is obtained from definition, it follows that \(\iota_*\) is an isomorphism between \(\mathfrak h\) and \(\mathfrak H\). ◻
Remark 19. Theorem 18 tells us that whenever \(H\) is a Lie subgroup of \(G\), we often identify \(\mathfrak h\) as a subalgebra of \(\mathfrak g\). Note, again, elements of \(\mathfrak h\) are not themselves left-invariant vector fields on \(G\). But, every element of \(\mathfrak h\) corresponds to a unique element of \(\mathfrak g\), determined by its value at the identity. Furthermore, the injection of \(\mathfrak h\) into \(\mathfrak g\) determined by the association above respects Lie brackets.
Reading: Carmo (1992), Riemannian geometry.