Lie Subgroups

Proof:

Since \(H\) already carries the structure of an embedded submanifold, hence of a smooth manifold, the only thing to verify is that its multiplication \(H \times H \to H\) and inversion \(H \to H\) are smooth. Multiplication on \(G\) is a smooth map \(m : G \times G \to G\), and its restriction to \(H \times H\) is smooth as a map into \(G\) — restriction of a smooth map to a submanifold of the domain is smooth, and this holds whether \(H\) is merely immersed or embedded. Because \(H\) is a subgroup, this restricted map takes values in \(H\). Since \(H\) is embedded, a smooth map into \(G\) whose image lies in \(H\) is smooth as a map into \(H\) by the restriction-of-codomain property for embedded submanifolds. Hence \(H \times H \to H\) is smooth. The identical argument applied to inversion shows \(H \to H\) is smooth, so \(H\) is a Lie group and therefore a Lie subgroup.

Proof:

As an open subset of \(G\), \(H\) is an open submanifold, and an open subset is an embedded submanifold of codimension zero; being a subgroup as well, it is an embedded Lie subgroup by the previous result. To see that \(H\) is closed, write \(G\) as the disjoint union of the left cosets of \(H\). Each coset \(gH\) is the image of the open set \(H\) under the left translation \(L_g\), which is a diffeomorphism, so every coset is open. The complement \(G \setminus H\) is the union of all cosets other than \(H\) itself, hence open, so \(H\) is closed. Being both open and closed, \(H\) is a union of connected components of \(G\).

Proof:

For subsets \(A, B \subseteq G\) write \(AB = \{ab : a \in A,\, b \in B\}\) and \(A^{-1} = \{a^{-1} : a \in A\}\). Let \(W_1 = W \cup W^{-1}\), and for each \(k > 1\) let \(W_k\) be the set of products of \(k\) or fewer elements of \(W_1\); the subgroup \(H\) generated by \(W\) is the union \(\bigcup_k W_k\), since this is exactly the set of finite products of elements of \(W\) and their inverses.

The set \(W^{-1}\) is the image of \(W\) under inversion, a diffeomorphism, so it is open; hence \(W_1\) is open. For \(k > 1\), \[ W_k = W_1 W_{k-1} = \bigcup_{g \in W_1} L_g(W_{k-1}), \] and each left translation \(L_g\) is a diffeomorphism, so by induction each \(W_k\) is open, and therefore \(H\) is open. This proves (1).

Suppose \(W\) is connected. Then \(W^{-1}\), a continuous image of \(W\), is connected, and \(W_1 = W \cup W^{-1}\) is a union of connected sets sharing the identity, hence connected. The set \(W_2 = m(W_1 \times W_1)\) is the image of a connected space under the continuous multiplication map, so it is connected, and by induction each \(W_k = m(W_1 \times W_{k-1})\) is connected. As a union of connected sets all containing the identity, \(H = \bigcup_k W_k\) is connected. This proves (2).

Finally, suppose \(G\) is connected. The subgroup \(H\) is open, hence closed by the preceding lemma, and it is nonempty because it contains the identity. A nonempty subset of a connected space that is both open and closed is the whole space, so \(H = G\), proving (3).

Proof Sketch:

That \(G_0\) is a subgroup follows from connectivity: the product map and inversion send connected sets containing the identity back into the component of the identity, so \(G_0 G_0 \subseteq G_0\) and \(G_0^{-1} \subseteq G_0\). It is open because in a manifold the components are open, and then the preceding generation result identifies it as the connected open subgroup generated by any connected identity neighborhood; uniqueness follows since any connected open subgroup contains the identity and is contained in its component. Normality holds because conjugation by any \(g\) is a diffeomorphism fixing the identity, hence carries \(G_0\) onto the identity component again. Each remaining component is a coset \(g G_0\), carried onto \(G_0\) by the diffeomorphism \(L_{g^{-1}}\).

Proof:

A Lie group homomorphism has constant rank, so its kernel \(F^{-1}(e)\) — the level set over the identity — is a properly embedded submanifold of codimension equal to the rank of \(F\), by the constant-rank level set theorem. Being a subgroup and an embedded submanifold, it is a Lie subgroup by the embedded-subgroup criterion established earlier.

Proof:

Since \(F\) has constant rank and is injective, the global rank theorem makes it a smooth immersion. The image of an injective immersion carries a unique smooth structure for which it is an immersed submanifold and the map onto it is a diffeomorphism; this is the uniqueness of the immersed smooth structure. With that structure \(F : G \to F(G)\) is a bijective smooth map with smooth inverse, hence a diffeomorphism and a group isomorphism, so a Lie group isomorphism. Because \(F(G)\) is a subgroup carrying a compatible Lie group structure as an immersed submanifold, it is a Lie subgroup. The analogous statement for the image of an arbitrary homomorphism — not necessarily injective — belongs to a later stage of the theory and is not needed here.

Examples:

(a) The positive-determinant subgroup of the general linear group is an open subgroup, hence an embedded Lie subgroup.

(b) The special linear group \(SL(n, \mathbb{R})\) is the kernel of the determinant homomorphism \(\det : GL(n, \mathbb{R}) \to \mathbb{R}^*\). The determinant is surjective, so it is a submersion by the global rank theorem, and its kernel is a properly embedded Lie subgroup of codimension one, of dimension \(n^2 - 1\). The complex case is identical: \(SL(n, \mathbb{C})\) is the kernel of \(\det : GL(n, \mathbb{C}) \to \mathbb{C}^*\), properly embedded of codimension two and dimension \(2n^2 - 2\).

(c) The complex general linear group embeds in a real one. Replacing each complex entry \(a + ib\) of an \(n \times n\) matrix by the \(2 \times 2\) real block \(\left(\begin{smallmatrix} a & -b \\ b & a \end{smallmatrix}\right)\) defines an injective Lie group homomorphism \(GL(n, \mathbb{C}) \to GL(2n, \mathbb{R})\). The image is the set of invertible real matrices whose \(2 \times 2\) blocks all have the form \(\left(\begin{smallmatrix} a & -b \\ b & a \end{smallmatrix}\right)\), a condition cut out by linear equations on the entries and hence closed in \(GL(2n, \mathbb{R})\); being a closed Lie subgroup, it is properly embedded by the closed subgroup criterion proved below. Thus \(GL(n, \mathbb{C})\) is realized as a Lie subgroup of \(GL(2n, \mathbb{R})\), the realization arising from the identification of \((x^1 + iy^1, \dots, x^n + iy^n)\) with \((x^1, y^1, \dots, x^n, y^n)\).

Example:

Let \(\alpha\) be irrational and let \(H \subseteq \mathbb{T}^2\) be the image of the injective immersion \(\gamma : \mathbb{R} \to \mathbb{T}^2\), \(\gamma(t) = \left( e^{2\pi i t}, e^{2\pi i \alpha t} \right)\), the dense line winding around the torus seen earlier. The map \(\gamma\) is an injective Lie group homomorphism from the additive group \(\mathbb{R}\), so by the image proposition \(H\) is a Lie subgroup of \(\mathbb{T}^2\) — immersed, but not embedded, since it is dense and a proper dense subset cannot be embedded. Here the immersed topology on \(H\), under which it is a copy of \(\mathbb{R}\), is genuinely finer than the subspace topology it inherits from the torus, which is exactly the gap the embedded-subgroup proof relied on being able to close. Whether a Lie subgroup is closed in the ambient group, and how closedness relates to embeddedness, is the question the next section settles.

Proof (first direction):

Assume \(H\) is embedded; we show it is closed. Let \(g\) be a point of the closure \(\overline{H}\), and choose a sequence \((h_i)\) in \(H\) converging to \(g\). Let \(U\) be the domain of a slice chart for \(H\) containing the identity, and let \(W\) be a smaller neighborhood of the identity with \(\overline{W} \subseteq U\). Because multiplication and inversion are continuous and send \((e, e)\) and \(e\) to \(e\), there is a neighborhood \(V\) of the identity small enough that \(g_1 g_2^{-1} \in W\) whenever \(g_1, g_2 \in V\).

Since \(h_i g^{-1} \to e\), all but finitely many terms lie in \(V\); discarding the rest, we may assume \(h_i g^{-1} \in V\) for every \(i\). Then for all \(i\) and \(j\), \[ h_i h_j^{-1} = \left( h_i g^{-1} \right)\left( h_j g^{-1} \right)^{-1} \in W. \] Fix \(j\) and let \(i \to \infty\): the left side converges to \(g h_j^{-1}\), which therefore lies in \(\overline{W} \subseteq U\). Now \(h_i h_j^{-1} \in H\), and \(H \cap U\) is a slice, hence closed in \(U\); so the limit \(g h_j^{-1}\) lies in \(H\). Thus \(g \in H h_j = H\), and \(H\) is closed.

Proof (second direction):

Assume \(H\) is closed, and write \(m = \dim H\), \(n = \dim G\). If \(m = n\), then \(H\) is an open subset, hence embedded, so assume \(m < n\). It suffices to find one point \(h_1 \in H\) and a neighborhood \(U_1\) of it in \(G\) such that \(H \cap U_1\) is an embedded submanifold of \(U_1\): for any other point \(h\), the right translation \(R_{h_1^{-1} h}\) is a diffeomorphism of \(G\) carrying \(H\) to \(H\) and \(h_1\) to \(h\), transporting the embedded slice to a neighborhood of \(h\).

Since \(H\) is an immersed submanifold, it is locally embedded: there is a neighborhood \(P\) of the identity in \(H\) and a slice chart \((U, \varphi)\) for \(P\) in \(G\), centered at the identity, with \(T_e G = T_e P \oplus T_e S\), where \(S\) is the slice complementary to \(P\) in the chart. Consider the map \(\psi : P \times S \to G\), \(\psi(v, s) = vs\). Its differential at \((e, e)\) restricts to the identity on each factor and so is an isomorphism onto \(T_e P \oplus T_e S = T_e G\); by the inverse function theorem \(\psi\) is a diffeomorphism from a product neighborhood \(P_0 \times S_0\) of \((e, e)\) onto a neighborhood \(U_0\) of the identity in \(G\), where \(P_0 \subseteq P\) and \(S_0 \subseteq S\).

Let \(K = S_0 \cap H\). Because \(H\) is a subgroup containing \(P_0\), one checks that \(\psi(P_0 \times K) = H \cap U_0\): a product \(vs\) lies in \(H\) precisely when \(s\) does, since \(v \in H\). It remains to see that \(K\) is discrete in \(H\), for then \(K\) has an isolated point \(h_1\) with a neighborhood \(S_1 \subseteq S_0\) meeting \(H\) only at \(h_1\), and \(U_1 = \psi(P_0 \times S_1)\) makes \(H \cap U_1 = \psi(P_0 \times \{h_1\})\) a slice — an embedded submanifold — completing the argument.

It remains to prove that \(K\) is discrete, and this is the step that uses closedness. We establish it in three stages: \(K\) is countable, \(K\) is closed in \(S_0\), and a countable closed subset of \(S_0\) must have an isolated point.

First, \(K\) is countable. The set \(H \cap U_0 = \psi(P_0 \times K)\) is an open subset of \(H\), and \(\psi\) carries the disjoint slices \(P_0 \times \{s\}\), one for each \(s \in K\), to disjoint open subsets of \(H \cap U_0\). A smooth manifold is second countable, and a second-countable space admits only countably many pairwise disjoint nonempty open sets; hence \(K\) is countable. Second, \(K\) is closed in \(S_0\): since \(H\) is closed in \(G\) and \(S_0 \subseteq G\), the intersection \(K = S_0 \cap H\) is closed in \(S_0\). Third, \(S_0\) is diffeomorphic to an open subset of a Euclidean space and so is locally compact and Hausdorff, and a nonempty countable closed subset of such a space contains an isolated point \(h_0\). Translating by \(h_0^{-1}\) within \(H\) carries \(h_0\) to the identity and the isolated point to an isolated point of \(K\) at \(e\); since every point of \(K\) can be moved to \(e\) this way, every point of \(K\) is isolated, so \(K\) is discrete in \(H\). It is precisely here that a dense subgroup would break the argument: if \(H\) were dense rather than closed, \(K\) would fail to be closed in \(S_0\), Baire's conclusion would not apply, and no slice could be extracted. The closed hypothesis is what rules the dense torus out.