6
$\begingroup$

This should be an elementary question in Lie group theory, although I'm pretty sure I'm mixing up concepts. Any help clarifying would be much appreciated.

Set up

Let $G$ be finite-dimensional real Lie group, and take it, for the sake of simplicity, to be connected. There is a three-fold correspondence between:

  1. tangent space at the identity element: $T_eG$,
  2. left-invariant vector fields: $\mathfrak{g}$,
  3. one-parameter subgroups of $G$: $\text{Hom}_{\text{LieGr}}(\mathbb{R},G)$,

which are interchangeably called the Lie algebra of $G$.

The exponential map of $G$ is defined as
$$ \exp_G:\mathfrak{g}\to G\,,\quad X\mapsto \exp_G(X):=\phi_X(1)\,, $$
where $\phi_X$ is the one-parameter subgroup generated by the (left-)invariant vector field corresponding to the tangent vector $X\in T_eG$. In particular, we have for all $t\in\mathbb{R}$ that $\phi_X(t)=\exp_G(tX)$ and therefore the exponential map sends the one-dimensional vector subspace through $X$ to the whole one-parameter subgroup generated by $X$.

(a) The map $\exp_G$ is just a local diffeomorphism. Nevertheless, its image $\exp_G(\mathfrak{g})$ is a (connected) neighbourhood of the identity. Hence, the subgroup $\langle \exp_G(\mathfrak{g})\rangle$ algebraically generated by this image is the whole Lie group $G$ back again (since $G$ is assumed connected) Prop A4.25.

(b) On the other hand, $\exp_G$ is but a local diffeomorphism and therefore shouldn't be able to retrieve global information about $G$. That is, Lie groups with isomorphic Lie algebras are not necessarily isomorphic, and the issue here is the usual simply-connectedness. However, it seems that in (a) we are perfectly able to distinguish the different Lie groups with a given Lie algebra. $$ $$

Question(s)

Where is the topological information coming from in the generating process in (a)? Or, where is it lost in (b)? Or, if the questions don't make sense, where is the confusion?
$$ $$

My thoughts is that in (a) we are really generating the Lie group $G$ by the image of one-parameter subgroups $\text{Hom}(\mathbb{R},G)$, which already contain enough topological information about $G$. In particular, by simple Yoneda if $\text{Hom}(\mathbb{R},G)\cong \text{Hom}(\mathbb{R},H)$ then $G\cong H$.[See comment.] However, the Hom is equally well isomorphic to the tangent space at the identity $T_eG$ and therefore it seems that we can always reproduce the whole group by this small bit of simply `vectorial' data.

In (b), when we worry about non-isomorphic Lie groups with isomorphic Lie algebras we are really completely forgetting about their underlying exponential maps, and just keeping the local diffeomorphism property. The Lie algebras are then purely algebraic objects with no a priori attached (global) Lie groups. The `corresponding' local Lie groups may be constructed via the BCH formula, which is defined entirely in terms of the Lie algebra structure.

Therefore, in (a) saying we reconstruct the whole Lie group from the exponential map may be seen as a tautology because the exponential map itself is defined globally with respect to a given Lie group. And, although Lie algebras may be isomorphic they are not the same, and the exponential map distinguishes just this.
$$ $$ Please, (re)tag more appropriately and/or edit the question if need be.

$\endgroup$
  • 1
    $\begingroup$ no time to answer but I think I see where you're getting turned around -- Yoneda's lemma tells you that if $\text{Hom}(\_, G) = \text{Hom}(\_, H)$ as functors, then $G = H$. Just knowing it for one specific $\_$ is not enough to conclude. There are many non-isomorphic Lie groups $G, H$ with $\text{Hom}(\mathbb{R}, G) = \text{Hom}(\mathbb{R}, H)$, for instance the real line and the circle. $\endgroup$ – hunter Feb 9 '15 at 14:49
  • 1
    $\begingroup$ @hunter Indeed! Thank you. That was definitely twisted. The usual `arguments' of trying to justify by all means what one is confusingly seeing (or wants to see). In any case, my confusion stands, because it seems that $\langle \exp_G(\mathfrak{g}) \rangle$ knows much more than $\text{Hom}(\mathbb{R}, G)$. The question is what it is exactly. $\endgroup$ – Carlos Feb 9 '15 at 15:17
3
$\begingroup$

Let's focus our attention on a straightforward special case: the quotient map $\mathbb{R} \to \mathbb{R} / \mathbb{Z}$, as a map of Lie groups, induces an isomorphism on Lie algebras, but it is not an isomorphism. So the Lie algebra cannot even tell whether a Lie group is compact or not (although it can get surprisingly close, as it turns out, modulo this example).

When you try to reconstruct a connected Lie group as the group generated by the image of the exponential map, the problem is that there are some relations between these generators which take place "far from the identity," and without more global information than just the Taylor expansion of the Lie group multiplication at the identity, you don't know which of these relations to impose or not. In the above example, one such relation is $\frac{1}{2} + \frac{1}{2} = 0$, which holds in $\mathbb{R} / \mathbb{Z}$ but not in $\mathbb{R}$.

(What's worse, you don't even know a priori that you can consistently impose these relations in such a way that you get a Lie group globally. That is, it's not at all obvious, although it is true, that every finite-dimensional Lie algebra is the Lie algebra of a Lie group.)

$\endgroup$
  • $\begingroup$ Right. I have no doubt the Lie algebra alone can't possibly distinguish the Lie groups of which it is the Lie algebra of, as the example $\mathbb{R}, S^1$ clearly shows. It is exactly this type of relations you mention that may (and do!) happen ``far away from the identity'' which are not accounted for by simply taking the group generated by the image of the exponential map that it is confusing me. Note that (a) in the question is a standard result, and says that $\langle \exp_G(\text{Lie}(G))\rangle =G$, when $G$ is connected, whether simply-connected or not. Maybe I'm misinterpreting it(?) $\endgroup$ – Carlos Feb 9 '15 at 18:36
  • 2
    $\begingroup$ @Carlos: that result says that the subgroup generated by exponentials of elements of the Lie algebra, inside $G$, is all of $G$. But to state this result you already need to know all of the relations holding among elements of $G$, including the ones "far away from the identity." That's just not information contained in the Lie algebra. $\endgroup$ – Qiaochu Yuan Feb 9 '15 at 18:49
  • $\begingroup$ So to say that $\langle \exp_G(\text{Lie}(G))\rangle =G$ is really like a tautology as I mentioned, since we need to know the Lie group $G$ beforehand. Perhaps, to make it "less tautological" is to interpret $\text{Lie}(G)$ not as the purely algebraic object we refer to as the Lie algebra, but as the Lie group homomorphisms $\text{Hom}(\mathbb{R},G)$ (which indirectly knows about $G$ globally). Then $\langle \exp_G(\text{Lie}(G))\rangle$ is the subgroup algebraically generated by all points lying on a one-parameter subgroup of $G$, inside $G$, which being open and closed, is all of $G$. $\endgroup$ – Carlos Feb 9 '15 at 19:33
  • $\begingroup$ If it makes sense what I just wrote, then the `topological/global information' I asked about is maybe contained in the isomorphism $\mathfrak{g}\cong \text{Hom}(\mathbb{R},G)$, which is completely forgotten in (b). $\endgroup$ – Carlos Feb 9 '15 at 19:37
  • $\begingroup$ @Carlos: yes, that's one way to say it. A Lie algebra, by itself, doesn't know how it sits as one-parameter subgroups of some Lie group (a priori). $\endgroup$ – Qiaochu Yuan Feb 9 '15 at 19:43
0
$\begingroup$

I guess I convinced myself of the remarks following and motivated by Qiaochu's answer, and I'll post it here as an answer in case someone gets similarly confused.

So indeed, the issue here was just a question of interpretation. The best, and probably the right, way of thinking about the exponential map of a Lie group $G$ is as a gadget meant to organize its one-parameter subgroups: $$ \exp_G:\text{Hom}(\mathbb{R},G)\to G\,,\quad \phi\mapsto \phi(1)\,. $$

The question in (a) is akin to the question about the surjectivity of $\exp_G$. This asks about how whether or not the one-parameter subgroups of $G$ cover $G$. In the same vein, in (a), when we say that the image of $\exp_G$ generates the Lie group $G$ (when connected) is telling exactly how the one-parameter subgroups of $G$ sit inside $G$ itself. The answer to the question "Where is the topological information coming from in the generating process in (a)?" is nowhere, because this information is all already there in $G$ which we haven't forgotten. What this does provide is more refined information about how the Lie group $G$ `behaves' in terms of maps of $\mathbb{R}$ into $G$. In particular, via the isomorphism $\mathfrak{g}\cong T_eG\cong \text{Hom}(\mathbb{R},G)$ (as given with a full knowledge of $G$), the result $\langle \exp_G(\mathfrak{g})\rangle=G$ comes as not so tautological because it tells we may write any group element $g\in G$ as a finite product of exponentials $g=\exp_G(X_1)\cdots \exp_G(X_k)$, where $X_1,\ldots, X_k\in T_eG$. This may be extremely useful in reducing a proof about $G$ to a proof about $\text{Hom}(\mathbb{R},G)$.

When we look at things such as in (b), and use $\exp_G$ to conclude that isomorphic Lie algebras may come from non-isomorphic Lie groups, we're simply using the local diffeomorphism property of $\exp_G$ (i.e. $(d \exp)_e=\text{id}_{\mathfrak{g}}$), but otherwise forgetting about $\exp_G$. This answers the question "Where is it lost in (b)?" At this point we simply have an abstract Lie algebra $\mathfrak{g}$ together with the BCH-formula (which equivalently encodes the local properties, i.e. around the identity, of a forgotten Lie group). The Lie algebra $\mathfrak{g}$ alone determines at most a local Lie group, but no more.

Note that there is a priori no notion of exponentiation of $\mathfrak{g}$ in the sense above. However, by Ado's theorem, any finite-dimensional Lie algebra $\mathfrak{g}$ is a subalgebra of $\mathfrak{gl}(n,\mathbb{R})$ for some $n$, and hence we may exponentiate $\mathfrak{g}$ as a matrix Lie algebra via the usual power series. For $U$ a neighbourhood of $0$ in $\mathfrak{g}\subseteq \mathfrak{gl}(n,\mathbb{R})$, we may form $\exp(U)\subseteq \text{GL}(n,\mathbb{R})$ (as a submanifold). The Baker-Campbell-Hausdorff formula in $\text{GL}(n,\mathbb{R})$ then gives $\exp(U)$ the structure of a local Lie group which can in fact be extended to a global Lie group. A useful reference on this latter part are Terry's notes on local groups.

$\endgroup$
  • $\begingroup$ Good note. I was just about to ask the same question, and also about symmetric subspaces. $\endgroup$ – Troy Woo May 21 '15 at 12:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.