Lie Subgroups, the Exponential Map, and Cartan’s Theorem
A subgroup of a Lie group need not be an embedded submanifold: its topology may interact poorly with the ambient topology. Cartan’s closed subgroup theorem gives a useful criterion. A subgroup that is closed as a subset automatically has an embedded Lie group structure.
We reach this theorem through the exponential map. One-parameter subgroups connect tangent vectors at the identity to curves in the group, allowing us to detect the tangent directions of a closed subgroup before we know that it is a manifold.
Definition 1. Let \(G\) and \(H\) be Lie groups. A Lie group homomorphism from \(G\) to \(H\) is a smooth map \(F : G \to H\) that is also a group homomorphism. Such a map \(F\) is called a Lie group isomorphism if it is also a diffeomorphism; in this case, we say \(G\) and \(H\) are isomorphic Lie groups.
Example 2 (Examples of Lie group homomorphisms). The determinant function \(\det : \operatorname{GL}_n (\mathbb R) \to \mathbb R \setminus \{0\}\) is smooth because \(\det A\) is a polynomial in the matrix entries of \(A \in \operatorname{GL}_n (\mathbb R)\). It is a Lie group homomorphism, because \(\det (AB) = (\det A) (\det B)\).
Here is an important example of a Lie group homomorphism. Let \(G\) be a Lie group, and let \(g \in G\). Define the conjugation map by \(g\) as \(C_g : G \to G : x \mapsto g x g^{-1}\). Via the notation introduced in the question, \(C_g (x) = (L_g \circ R_{g^{-1}}) (x)\). Note that \(C_g\) is smooth as the composition of smooth maps, and for all \(x,y \in G\), \(C_g (xy) = g (xy) g^{-1} = g(xg^{-1}g y)g^{-1} = (gxg^{-1})(gyg^{-1}) = C_g (x) C_g (y)\); therefore, \(C_g\) is a homomorphism. Furthermore, because \(C_{g^{-1}}\) exists, it follows that \(C_g\) is a Lie group isomorphism.
Definition 3. Let \(H \subset G\) be a subgroup of a Lie group \(G\). \(H\) is normal if \(C_g (H) = H\) for every \(g \in G\).
Theorem 4. Every Lie group homomorphism has constant rank.
Proof. Let \(F : G \to H\) be a Lie group homomorphism, and let \(e\) and \(\widetilde e\) denote the identity elements of \(G\) and \(H\), respectively. Take any \(g_0 \in G\). We claim \((\dif F)_e\) and \((\dif F)_{g_0}\) have the same rank. Since \(F\) is a homomorphism, it follows from direct calculations that
or \(F \circ L_{g_0} = L_{F(g_0)} \circ F\). By the chain rule,
Clearly, both \((\dif L_{g_0})_{e}\) and \((\dif L_{F(g_0)})_{\widetilde e}\) are isomorphisms, because the left multiplication map is a diffeomorphism. Composition of a map with an isomorphism does not change the rank of the original map. Therefore, \((\dif F)_e\) and \((\dif F)_{g_0}\) must have the same rank. ◻
Theorem 4 allows us to apply constant-rank arguments to Lie group homomorphisms. Here are two consequences.
Corollary 5. A Lie group homomorphism is a Lie group isomorphism if and only if it is bijective.
Proof. Suppose \(F\) is bijective. By constant rank and injectivity, its rank is \(\dim G\): otherwise a positive-dimensional coordinate fiber would contain distinct points with the same image. If this rank were less than \(\dim H\), cover \(G\) by countably many compact coordinate pieces on which the constant-rank form holds. Each image is compact with empty interior in \(H\), since locally it lies in a lower-dimensional slice. Surjectivity would express the locally compact Hausdorff manifold \(H\) as a countable union of closed nowhere-dense sets, contradicting the Baire theorem. Thus the differential is bijective. The inverse function theorem gives smooth local inverses, which agree with the global inverse. The converse follows from the definition. ◻
Proposition 6. Let \(G\) be a Lie group, and suppose \(H \subset G\) is a subgroup that is also an embedded submanifold. Then H is a Lie subgroup.
Proof. Restrict multiplication to \(H\times H\) and inversion to \(H\). These maps are smooth as maps into \(G\), and their values lie in \(H\). In a slice chart for an embedded submanifold, a smooth map into \(G\) with image in \(H\) has its last coordinates identically zero; its first coordinates prove smoothness as a map into \(H\). This establishes the Lie group operations in the submanifold structure. ◻
For the definition of Lie subgroup, see Definition 4. The simplest example of embedded Lie subgroups are the open subgroups. Their cosets are open as well, which forces an open subgroup to be closed.
Proposition 7. Let \(G\) be a Lie group, and \(H \subset G\) be an open subgroup. Then, \(H\) is an embedded Lie subgroup. Furthermore, \(H\) is closed, so it is a union of connected components of \(G\).
Proof. Embeddedness follows from standard results. In addition, every left coset \(gH = \{gh : h \in H\}\) is open in \(G\), as the image of \(H\) under the diffeomorphism \(L_g\). \(G \setminus H\) is open as the union of the cosets of \(H\) other than \(H\) itself, hence \(H\) must be closed in \(G\). The second desired conclusion then follows. ◻
Definition 8. For any subset \(E\) of a group \(G\), the subgroup \(\langle E\rangle\) generated by \(E\) is the intersection of all subgroups containing \(E\). Equivalently it consists of finite products of elements of \(E\cup E^{-1}\), including the empty product \(e\).
Proposition 9. Let \(G\) be a Lie group, and \(W \subset G\) is any neighbourhood of the identity. Then, the following are true:
\(W\) generates an open subgroup of \(G\).
If \(W\) is connected, it generates a connected open subgroup of \(G\).
If \(G\) is connected, then \(W\) generates \(G\).
Proof. Let \(S=W\cup W^{-1}\), so \(e\in S\) and \(\langle W\rangle=\bigcup_{n\geq1}S^n\). It contains an open identity neighborhood \(U\subset W\); translating \(U\) by each subgroup element shows that the subgroup is open. If \(W\) is connected, \(S\) is connected because its two connected pieces meet at \(e\). Each \(S^n\) is a continuous image of a connected product, and these sets are increasing, so their union is connected. Finally every open subgroup is also closed by Proposition 7; a nonempty clopen subset of connected \(G\) is all of \(G\). ◻
For any Lie group \(G\), the connected component of \(G\) containing the identity is called the identity component of \(G\). This turns out to make further rigid the open subgroups of a Lie group \(G\).
Lemma 10. Let \(G\) be a Lie group and let \(G_0\) be its identity component. Then, \(G_0\) is a normal subgroup of \(G\), and is the only connected open subgroup. Every connected component of \(G\) is diffeomorphic to \(G_0\).
Proof. Manifolds are locally path connected, so connected components are open. The image of connected \(G_0\times G_0\) under \((a,b)\mapsto ab^{-1}\) is connected and contains \(e\), hence lies in \(G_0\); thus \(G_0\) is a subgroup. Conjugation is a homeomorphism fixing \(e\), so it preserves the identity component, proving normality. A connected open subgroup lies in \(G_0\); being also closed, it must fill the connected space \(G_0\). Left translation maps the component of \(e\) to the component of any \(g\), giving the claimed diffeomorphisms. ◻
For more general Lie groups, the following result shows how to produce many more examples of embedded Lie subgroups.
Proposition 11.
Let \(F : G \to H\) be a Lie group homomorphism. The kernel of \(F\) is a properly embedded Lie subgroup of \(G\), whose codimension is equal to the rank of \(F\).
Let \(F : G \to H\) be an injective Lie group homomorphism. The image of \(F\) has a unique smooth manifold structure such that \(F(G)\) is a Lie subgroup of \(H\), and \(F:G \to F(G)\) is a Lie group isomorphism.
Proof. Write \(r=\operatorname{rank}F\), constant by Theorem 4. Constant-rank coordinates make the fiber \(F^{-1}(e_H)\) a slice of codimension \(r\). It is closed by continuity and the Hausdorff property of \(H\). Its inclusion is proper: intersecting the closed fiber with a compact subset of \(G\) remains compact. Proposition 6 supplies the group structure.
If \(F\) is injective, the constant-rank form cannot have a positive-dimensional fiber, so \(r=\dim G\). Transport the topology and all charts of \(G\) across the bijection \(F:G\to F(G)\). This makes it a diffeomorphism by construction, the inclusion into \(H\) is an injective immersion, and the transported operations are smooth because \(F\) is a homomorphism. Uniqueness refers to the structure making \(F\) a diffeomorphism; it does not assert that the image has the ambient subspace topology. ◻
We first examine subgroups already equipped with an immersed Lie group structure. In this setting, closedness and embeddedness are equivalent.
Theorem 12. Let \(G\) be a Lie group, and \(H \subset G\) is a Lie subgroup. Then \(H\) is closed in \(G\) if and only if \(H\) is embedded.
Remark 13. In general, smooth submanifolds can be closed without being embedded, or embedded without being closed (as is the open unit ball in \(\mathbb R^n\)). Theorem 12 uses the fact that translations tie the local topology at every point of a Lie subgroup to that at the identity. The closed subgroup theorem goes further: every subgroup of a Lie group that is topologically a closed subset (but not assumed to be a submanifold!) is automatically a properly embedded Lie subgroup.
Proof. Suppose first that \(H\) is embedded. A slice chart \(U\) about \(e\) makes \(H\cap U\) closed in \(U\). Choose identity neighborhoods \(W,V\) with \(\overline W\subset U\) and \(VV^{-1}\subset W\). For \(h_i\in H\) converging to \(g\), discard finitely many terms so that \(h_i g^{-1}\in V\). Then \(h_i h_j^{-1}\in W\). Fix \(j\) and take the limit: \(g h_j^{-1}\in\overline W\subset U\), and relative closedness gives \(g h_j^{-1}\in H\), hence \(g\in H\).
For the converse we record carefully the difference between the two topologies. The inclusion of the Lie subgroup is an immersion, so an intrinsic identity neighborhood \(V\subset H\) is embedded. Choose a transverse submanifold \(P\) through \(e\) with \(T_eG=T_eH\oplus T_eP\). Multiplication \(\psi(v,p)=vp\) has derivative \((X,Y)\mapsto X+Y\) at \((e,e)\), so, after shrinking, it is a diffeomorphism from \(V_0\times P_0\) to an ambient open neighborhood \(U_0\). Put \(K=P_0\cap H\). The subgroup law gives
For each \(h\in K\), the intrinsic open set \(V_0h\subset H\) intersects \(K\) only in \(h\), by injectivity of \(\psi\). Thus \(K\) is discrete in the intrinsic topology of \(H\), and is countable: choose a distinct basis element from a countable basis of \(H\) isolating each of its points.
Closedness of \(H\) now makes \(K\) closed in the manifold \(P_0\), and hence locally compact Hausdorff. If this nonempty countable space had no isolated point, its singleton subsets would be closed and nowhere dense; their countable union would contradict Baire's theorem. Thus some \(h_1\in K\) is isolated in the ambient transverse topology. Choose open \(P_1\subset P_0\) with \(P_1\cap H=\{h_1\}\). Then
is a single slice, with the given intrinsic smooth structure. Left translation by \(h h_1^{-1}\) carries this slice neighborhood to a slice neighborhood of any \(h\in H\). These charts show that the existing immersed subgroup is embedded. ◻
Remark 14. Without closedness, the countable transverse set \(K\) need not have an isolated point in its ambient topology. For the irrational line \(t\mapsto(e^{it},e^{i\alpha t})\) in the two-torus, \(\alpha\notin\mathbb Q\), the subgroup is dense and the transverse intersections are countable and dense. Intrinsic discreteness of \(K\) does not imply ambient discreteness; closedness was used precisely at the Baire step.
1. Exponential Map
Let \(G\) be a Lie group. Since left-invariant vector fields are naturally defined in terms of the group structure of \(G\), one might reasonably expect to find some relationship between the group law for the flow of a left-invariant vector field, and the group multiplication in \(G\). The definition of a one-parameter subgroup makes this relationship precise.
Definition 15. A one-parameter subgroup of \(G\) is a Lie group homomorphism \(\gamma : \mathbb R \to G\), with \(\mathbb R\) considered as a Lie group under addition.
A one-parameter subgroup here is the map \(\gamma:\mathbb R\to G\), not just its image, and it need not be onto \(G\). Its image may be a periodic circle, an injectively immersed line, or the identity alone. We identify these maps by their initial velocities.
Theorem 16. Let \(G\) be a Lie group. The one-parameter subgroups of \(G\) are the maximal integral curves of left-invariant vector fields starting at the identity element of \(G\).
Proof of Theorem 16. Differentiate the homomorphism identity to obtain one direction. For the converse, uniqueness of integral curves gives the homomorphism identity locally, and translation extends the curve to all time.
First, let \(\gamma : \mathbb R \to G\) be a one-parameter subgroup. By definition, \(\gamma\) is a group homomorphism, meaning \(\gamma(s+t) = \gamma(s)\gamma(t)\) for all \(s,t \in \mathbb R\), and \(\gamma(0) = e\). Let \(X \in \mathfrak g\) be the unique left-invariant vector field on \(G\) such that \(X_e = \dot \gamma(0)\). We want to show that \(\gamma\) is an integral curve of \(X\). Taking the derivative with respect to \(s\) at \(s=0\), we have
\[\begin{aligned} \dot \gamma(t) &= \left. \frac{\dif}{\dif s} \right|_{s=0} \gamma(t+s) \nonumber \\ &= \left. \frac{\dif}{\dif s} \right|_{s=0} \gamma(t)\gamma(s) \nonumber \\ &= \left. \frac{\dif}{\dif s} \right|_{s=0} L_{\gamma(t)} (\gamma(s)) \nonumber \\ &= (\dif L_{\gamma(t)})_e (\dot \gamma(0)) = (\dif L_{\gamma(t)})_e (X_e) = X_{\gamma(t)} ~. \end{aligned}\]This implies \(\gamma\) is an integral curve of \(X\) starting at \(e\). Since its domain is \(\mathbb R\), it is a maximal integral curve.
Conversely, let \(X \in \mathfrak g\) be a left-invariant vector field on \(G\), and let \(\gamma : I \to G\) be the maximal integral curve of \(X\) starting at \(\gamma(0) = e\). We want to show that \(I = \mathbb R\) and \(\gamma(s+t) = \gamma(s)\gamma(t)\). Let \(s \in I\) be arbitrary, and consider the two curves \(\alpha(t) = \gamma(s+t)\) and \(\beta(t) = L_{\gamma(s)} (\gamma(t)) = \gamma(s)\gamma(t)\). They both satisfy the initial condition \(\alpha(0) = \gamma(s)\) and \(\beta(0) = \gamma(s)e = \gamma(s)\). Computing their derivatives, we get
\[\begin{aligned} \dot \alpha(t) &= \dot \gamma(s+t) = X_{\gamma(s+t)} = X_{\alpha(t)} ~, \\ \dot \beta(t) &= \frac{\dif}{\dif t} L_{\gamma(s)}(\gamma(t)) = (\dif L_{\gamma(s)})_{\gamma(t)} (\dot \gamma(t)) = (\dif L_{\gamma(s)})_{\gamma(t)} (X_{\gamma(t)}) = X_{\beta(t)} ~, \end{aligned}\]where the last equality follows from the fact that \(X\) is left-invariant. Therefore, both \(\alpha\) and \(\beta\) are integral curves of \(X\). By the fundamental uniqueness of integral curves, they must be equal wherever they are both defined, implying \(\gamma(s+t) = \gamma(s)\gamma(t)\).
Finally choose \(\delta>0\) with \((-\delta,\delta)\subset I\). If the maximal right endpoint \(b\) were finite, choose \(s\in I\) with \(b-\delta/2<s<b\). The curve \(t\mapsto\gamma(s)\gamma(t-s)\), defined for \(|t-s|<\delta\), is an integral curve agreeing with \(\gamma(t)\) near \(s\); uniqueness makes them agree throughout their overlapping intervals. It extends the curve past \(b\), a contradiction. The same argument at the left endpoint proves \(I=\mathbb R\). Now the identity \(\gamma(s+t)=\gamma(s)\gamma(t)\) holds for all real \(s,t\). This uses smooth ODE uniqueness, not analytic continuation.
This shows the desired statement. ◻
Remark 17. Given \(X \in \mathfrak g\), the one-parameter subgroup determined by \(X\) in this way is called the one-parameter subgroup generated by \(X\). Because left-invariant vector fields are uniquely determined by their values at the identity, it follows that each one-parameter subgroup is uniquely determined by its initial velocity in \(T_e G\), and thus there are one-to-one correspondences between the collection of one-parameter subgroups of \(G\), \(\mathfrak g\), and \(T_e G\).
Theorem 16 leads to the exponential map: evaluate the one-parameter subgroup at time one. Rescaling its initial velocity recovers every other time on that subgroup.
Definition 18. Given a Lie group \(G\) with Lie algebra \(\mathfrak g\), the exponential map of \(G\) is a map \(\exp : \mathfrak g \to G\) such that
where \(\gamma\) is the (maximal) integral curve of \(X\) starting at the identity.
Proposition 19. Let \(G\) be a Lie group. For any \(X \in \mathfrak g\), \(\gamma(s) = \exp (sX)\) is the one-parameter subgroup of \(G\) generated by \(X\).
Proof. Write \(c_X\) for the complete integral curve with initial velocity \(X\). For each real \(s\), differentiation shows \(t\mapsto c_X(st)\) solves the equation for \(sX\), starts at \(e\), and hence equals \(c_{sX}(t)\) by uniqueness. At \(t=1\), \(\exp(sX)=c_X(s)\), which proves the claim, including \(s=0\) and negative \(s\). ◻
Proposition 20. Let \(G\) be a Lie group and \(\mathfrak g\) be its Lie algebra. The following statements are true.
The exponential map is a smooth map from \(\mathfrak g\) to \(G\).
(Group law) For any \(X \in \mathfrak g\) and \(s,t \in \mathbb R\), \(\exp ((s+t)X) = \exp(sX) \exp(tX)\).
For any \(X \in \mathfrak g\), \((\exp X)^{-1} = \exp(-X)\).
For any \(X \in \mathfrak g\) and \(n \in \mathbb N\), \((\exp X)^n = \exp(nX)\).
The differential \((\dif \exp)_0 : T_0 \mathfrak g \to T_e G\) is the identity map, once we identify both \(T_0 \mathfrak g\) and \(T_e G\) canonically with \(\mathfrak g\).
The exponential map restricts to a diffeomorphism from some neighbourhood of \(0 \in \mathfrak g\) to a neighbourhood of \(e \in G\).
Let \(H\) be another Lie group, \(\mathfrak h\) is its Lie algebra, and \(F: G \to H\) be a Lie group homomorphism, then the following diagram commutes:
A Lie-group homomorphism commutes with the exponential maps and its differential at the identity. The flow of a left-invariant vector field \(X\) is given by \(\kappa_t = R_{\exp(tX)}\), i.e. by right multiplication by \(\exp (tX)\).
Proof of Proposition 20. We proceed by proving the statements sequentially.
Define a vector field \(\Xi\) on the product manifold \(G \times \mathfrak g\) by \(\Xi_{(g,X)} = (X_g, 0)\). Since the map \((g,X) \mapsto X_g = (\dif L_g)_e (X)\) is smooth (it depends smoothly on \(g\) and linearly on \(X\)), \(\Xi\) is a smooth vector field. The integral curve of \(\Xi\) starting at \((e,X)\) is given by \(t \mapsto (\gamma_X(t), X)\), where \(\gamma_X\) is the maximal integral curve of \(X\) starting at \(e\). Because left-invariant vector fields are complete, this flow is defined for all \(t \in \mathbb R\). By the fundamental theorem on flows, the global flow map \(\Theta_t(g,X)\) depends smoothly on its initial conditions. In particular, evaluating at \(t=1\) and \(g=e\), the map \(X \mapsto \pi_G (\Theta_1(e,X)) = \exp(X)\) is smooth, implying that \(\exp : \mathfrak g \to G\) is a smooth map.
By Proposition 19, the curve \(\gamma(s) = \exp(sX)\) is the one-parameter subgroup generated by \(X\). Since it is a group homomorphism from \(\mathbb R\) to \(G\), we straightforwardly have \(\exp((s+t)X) = \gamma(s+t) = \gamma(s)\gamma(t) = \exp(sX)\exp(tX)\).
From (b), taking \(s=1\) and \(t=-1\) yields \(\exp(X)\exp(-X) = \exp(0X) = \exp(0)\). The integral curve of the zero vector field is constantly \(e\), so \(\exp(0) = e\). Therefore, \((\exp X)^{-1} = \exp(-X)\).
This follows directly from (b) by mathematical induction on \(n\).
Let \(X \in \mathfrak g\). Under the canonical identification of \(T_0 \mathfrak g\) with \(\mathfrak g\), we can evaluate the differential \((\dif \exp)_0 (X)\) by considering the smooth curve \(c(t) = tX\) in \(\mathfrak g\), which has \(c(0)=0\) and initial velocity \(X\). Then,
\[\begin{aligned} (\dif \exp)_0 (X) = \left. \frac{\dif}{\dif t} \right|_{t=0} \exp(tX) ~. \end{aligned}\]By definition, \(t \mapsto \exp(tX)\) is the integral curve of \(X\) starting at \(e\), so its derivative at \(t=0\) is exactly \(X_e\). Under the standard identification of \(T_e G\) with \(\mathfrak g\), this is precisely \(X\). Thus, \((\dif \exp)_0\) is the identity map.
Since \((\dif \exp)_0 : T_0 \mathfrak g \to T_e G\) is the identity map, it is a linear isomorphism. The inverse function theorem then implies that \(\exp\) restricts to a diffeomorphism from some open neighbourhood of \(0 \in \mathfrak g\) to an open neighbourhood of \(e \in G\).
For any \(X \in \mathfrak g\), the curve \(\alpha(t) = \exp_G(tX)\) is a one-parameter subgroup of \(G\). Since \(F : G \to H\) is a Lie group homomorphism, the composition \(F \circ \alpha : \mathbb R \to H\) is also a Lie group homomorphism, and thus a one-parameter subgroup of \(H\). Its initial velocity at \(t=0\) is
\[\begin{aligned} \left. \frac{\dif}{\dif t} \right|_{t=0} F(\exp_G(tX)) = (\dif F)_e (\dot \alpha(0)) = (\dif F)_e (X_e) = (F_* X)_e ~. \end{aligned}\]Therefore, \(F \circ \alpha\) is the one-parameter subgroup of \(H\) generated by \(F_* X\), meaning \(F(\exp_G(tX)) = \exp_H(t F_* X)\). Evaluating at \(t=1\) gives \(F(\exp_G(X)) = \exp_H(F_* X)\), which establishes that the diagram commutes.
Let \(X \in \mathfrak g\). We want to find the flow \(\kappa_t\) of \(X\). Fix \(g \in G\) and consider the curve \(\beta(t) = R_{\exp(tX)}(g) = g \exp(tX) = L_g(\exp(tX))\). At \(t=0\), we have \(\beta(0) = g e = g\). The tangent vector to this curve at time \(t\) is
\[\begin{aligned} \dot \beta(t) &= \frac{\dif}{\dif t} L_g(\exp(tX)) \nonumber \\ &= (\dif L_g)_{\exp(tX)} \left( \frac{\dif}{\dif t} \exp(tX) \right) \nonumber \\ &= (\dif L_g)_{\exp(tX)} (X_{\exp(tX)}) ~. \end{aligned}\]Because \(X\) is a left-invariant vector field, \((\dif L_g)_h (X_h) = X_{gh}\) for all \(h \in G\). Setting \(h = \exp(tX)\) yields \(\dot \beta(t) = X_{g \exp(tX)} = X_{\beta(t)}\). Therefore, \(\beta\) is an integral curve of \(X\) starting at \(g\). By the uniqueness of integral curves, it must be the flow of \(X\), so \(\kappa_t(g) = R_{\exp(tX)}(g)\).
This concludes the proof. ◻
Remark 21. The Lie-group exponential and the Riemannian exponential both arise from ODEs, but are different constructions. The former uses complete invariant vector fields and is defined on all of \(\mathfrak g\). The latter depends on a metric and its geodesics, and its domain can be smaller without completeness. They agree at the identity for a bi-invariant Riemannian metric, but no such metric has been assumed here.
Lemma 22. Let \(G\) be a Lie group, and let \(H \subset G\) be a Lie subgroup. With \(\mathfrak h\) considered as a subalgebra of \(\mathfrak g\) in the usual way, by Theorem 18, the exponential map of \(H\) is the restriction of the exponential map of \(G\) to \(\mathfrak h\). Furthermore, \(\mathfrak h = \{X \in \mathfrak g : \exp (tX) \in H \text{ for all } t \in \mathbb R\}\).
Proof. Apply the exponential naturality in Proposition 20 to the smooth inclusion \(H\hookrightarrow G\). It gives \(\exp_G(X)=\exp_H(X)\) for \(X\in\mathfrak h\), proving both the first assertion and one inclusion in the second.
For the reverse inclusion suppose the smooth ambient curve \(c(t)=\exp_G(tX)\) lies in \(H\). Use the product neighborhood \(\psi:V_0\times P_0\to U_0\) from the proof of Theorem 12; its construction and the countability of \(K=P_0\cap H\) did not use closedness. For \(t\) in a small interval about zero, the transverse component of \(\psi^{-1}(c(t))\) is a continuous map into countable \(K\). A connected subset of a countable Euclidean set has only one point: if it had two, a separating coordinate projection would contain a nontrivial, uncountable interval. The transverse component is therefore constant, equal to \(e\). Hence \(c\) lies in the single smooth plaque \(V_0\), and is smooth as an \(H\)-valued curve there. Differentiating at zero gives \(X\in T_eH\). This supplies the missing justification even when the subgroup is only immersed. ◻
2. Cartan’s Closed Subgroup Theorem
We showed in Theorem 12 that a Lie subgroup is embedded if and only if it is closed. We now have all the tools needed to prove a much stronger statement: if a subgroup of a Lie group is topologically a closed subset, then it is automatically an embedded Lie subgroup. We proceed in a few steps.
Proposition 23. Let \(G\) be a Lie group and let \(\mathfrak g\) be its Lie algebra. For any \(X,Y \in \mathfrak g\), there exists a smooth function \(Z : (-\varepsilon,\varepsilon) \to \mathfrak g\) for some \(\varepsilon > 0\) such that for all \(t \in (-\varepsilon,\varepsilon)\),
Proof. On a sufficiently small interval let \(\varphi(t)=\log(\exp(tX)\exp(tY))\), using the local inverse of \(\exp\) at zero. It is smooth, \(\varphi(0)=0\), and the product differential at \((e,e)\) together with \(d\exp_0=\mathrm{id}\) gives \(\varphi'(0)=X+Y\). Taylor's formula with integral remainder gives the explicit smooth function
This includes \(t=0\) without division by zero and proves the asserted exponential identity. ◻
Remark 24 (Baker–Campbell–Hausdorff). The BCH expansion applies to \(\log(\exp(tX)\exp(tY))\), not directly to the remainder \(Z\). With the normalization of Proposition 23 it yields
In particular \(Z(0)=\frac12[X,Y]\); the terms \(X+Y\) have already been removed. The expansion is local in \(t\). The first terms are recorded in Etingof, §7.6, Theorem 7.19 and Example 7.20, p. 46. The following limit and closed subgroup proof need only the smooth remainder proved above, not convergence of an infinite BCH series.
Corollary 25. Let \(G\) be a Lie group and let \(\mathfrak g\) be its Lie algebra. For any \(X,Y \in \mathfrak g\) and fixed \(t \in \mathbb R\),
For completeness, for a fixed \(t\), the arguments \(t/n\) eventually lie in a compact subinterval of the domain of \(Z\). Continuity makes \(Z(t/n)\) bounded, so \(t^2Z(t/n)/n\to0\). Continuity of \(\exp\) then justifies the limit in (1).
Proof. It follows from Proposition 23 that
Furthermore, applying the property of exponential map gives us
For a fixed \(t \in \mathbb R\), pass (1) to the limit and \(n\to \infty\) to conclude. ◻
Theorem 26 (Cartan’s Closed Subgroup Theorem). Let \(G\) be a Lie group, and let \(H \subset G\) be a subgroup that is also a closed subset of \(G\). Then, \(H\) is an embedded Lie subgroup.
Proof. Define
It contains zero and is closed under real scalar multiplication. If \(X,Y\in\mathfrak H\), every term \((\exp(tX/n)\exp(tY/n))^n\) belongs to the group \(H\). The product limit and closedness put \(\exp(t(X+Y))\) in \(H\), for every real \(t\). Thus \(\mathfrak H\) is a vector subspace.
Choose a linear complement \(\mathfrak j\) and consider \(\Phi:\mathfrak H\times\mathfrak j\to G\), \(\Phi(X,Y)=\exp(X)\exp(Y)\). Its differential at \((0,0)\) is \((X,Y)\mapsto X+Y\), an isomorphism. On small product neighborhoods \(A\times B\) it is a diffeomorphism onto an ambient open set. We claim that \(B\) can be shrunk so that
If not, choose nonzero \(Y_i\in\mathfrak j\) tending to zero with \(\exp(Y_i)\in H\). Set \(c_i=\|Y_i\|>0\), and pass to a subsequence with \(Y_i/c_i\to Y\), \(\|Y\|=1\). For each fixed real \(t\), choose the integer \(n_i=\lfloor t/c_i\rfloor\), allowing negative integers. Then
Since \(\exp(n_iY_i)=\exp(Y_i)^{n_i}\in H\), closedness implies \(\exp(tY)\in H\). This holds for every \(t\), so \(Y\in\mathfrak H\cap\mathfrak j=\{0\}\), contradicting \(\|Y\|=1\). The norm, not its square, is essential in this normalization.
Now for \((X,Y)\in A\times B\), the element \(\exp(X)\) lies in \(H\). Consequently \(\Phi(X,Y)\in H\) if and only if \(\exp(Y)\in H\), which by the claim is equivalent to \(Y=0\). The coordinates \(\Phi^{-1}\) therefore identify \(H\) near the identity with the slice \(A\times\{0\}\). Left translation gives the same slice description near every point of \(H\). Thus \(H\) is an embedded submanifold with its subspace topology, and Proposition 6 gives smooth group operations. It is properly embedded because it is closed. Its tangent space at the identity is exactly \(\mathfrak H\); closure under Lie brackets follows from the already proved Lie-subgroup algebra theorem. ◻
Reading: Carmo (1992), Riemannian geometry.