Developing intuition for Lie groups and Lie algebras

Background: Until now, I've been able to motivate everything I've learned in mathematics, and develop some solid insights for everything. But I learned some Lie theory this summer, and while I have a good grasp of the elementary aspects and strong intuition for some or even most of what I've learned, there are some "holes" in my understanding of Lie algebras.

To give you an idea of what I'm looking for, I'll list some examples of things in Lie theory I DO understand and am able to motivate:

• The notion of a Lie group itself -- the idea comes from wanting to generalise what we know about discrete groups to more complicated contexts where the "manifold" structure of the group allows us to do so. Examples: compactness generalises finiteness, one-parameter groups generalise cyclic groups, etc.
• The exponential map -- For one-parameter groups to generalise cyclic groups, we need a "generalisation" of the group power to allow "real-index powers". The general way to define a real power is through the exponential map. Well, this real power stuff isn't always defined as it turns out (you need the exponential map to be surjective), but our motivation does explain why it "makes sense" that the exponential map is surjective in the connected abelian case (because then, the Lie algebra is basically a co-ordinate system on the Lie group -- I'm aware exponential co-ordinates are defined in more generality, but it's certainly more well-behaved here).
• The Lie algebra, i.e. "why is the logarithm/parameter space the tangent space?" We'd like to generalise the notion of a generator to a Lie group -- consider e.g. the circle group on the complex plane. An element near the identity generates a cyclic group, and as the element goes nearer to the identity -- as it becomes an infinitesimal generator, the cyclic group it approaches the entire group. Well, an element close to the identity is of the form $$1+\varepsilon t X$$, and generates a group element as $$(1+\varepsilon tX)^{1/\varepsilon}=e^{tX}$$. This is also intuition for the compound-interest limit, and for Euler's identity.
• The Lie bracket is the second-derivative of the commutator curve $$\gamma(t)=e^{tX}e^{tY}e^{-tX}e^{-tY}$$. Well, it's also the derivative of $$\gamma(\sqrt{t})$$, which proves closure under the Lie bracket.
• The real justification for the Lie bracket, however, comes from the fundamental fact that $$\mathrm{ad}:\mathfrak{g}\to\mathrm{Der}(\mathfrak{g}):=X\to[X,\cdot]$$ is the differential of the adjoint map $$\mathrm{Ad}:G\to\mathrm{Aut}(G):=g\mapsto\lambda x, gxg^{-1}$$, which is a group homomorphism. In particular, the preservation of the Lie Bracket by the differential of a group homomorphism is precisely the Jacobi identity: $$\mathrm{ad}([x,y])=[\mathrm{ad}(x),\mathrm{ad}(y)]$$. The basic point is that we are trying to reduce Lie group problems to Lie algebra ones as much as possible, and conjugation is an important idea that we'd like to see the map induced by on the Lie algebra -- we are seeing the result of the obvious fact that $$T\mathrm{Aut}(G)\subseteq\mathrm{Der}(TG)$$ (and also $$T\mathrm{Aut}(M)=\mathrm{Der}(M)$$ -- the fact that the automorphisms of an object form a group is equivalent to the derivations on an object forming a Lie algebra). Some more examples of the "study the Lie algebra approach":
• The uniqueness of the determinant as a map from $$G\to \mathbb{R}-\{0\}$$.
• An ideal is a subalgebra "induced" on the Lie algebra by a normal subgroup of the Lie group. This immediately provides the interpretation as "kernels of Lie algebra homomorphisms" as well as the condition $$[\mathfrak{g},\mathfrak{i}]\subseteq\mathfrak{i}$$.
• The idea behind the manifold-structure of a Lie group is that the flows are produced by left-multiplication by group elements, so those must be homeomorphisms. This motivation can be confirmed through various topological consequences, e.g.
• A neighbourhood of the identity generates the connected component. The idea behind the proof is this: if an entire open neighbourhood of the identity is contained in the subgroup, it means you can "flow in any direction" from the subgroup -- but to bring these flows to an arbitrary point of the manifold, you need left-multiplication to be a homeomorphism.
• The identity component is a (normal) subgroup. Because left-multiplication and inversion are continuous, they cannot tear the connected component apart (generalised "intermediate value theorem"), so it is closed under multiplication.
• Compact Lie groups -- How can a Lie group possibly "close in on itself"? Surely we keep "extending" an open neighbourhood $$W$$ of the identity by observing that $$xW$$ must be in the subgroup? The idea is that these translations of $$W$$ form an open cover of the group, if it has a finite subcover, then it makes sense for the group to close in on itself. By playing around with different open neighbourhoods $$W$$ and taking some suitable unions, one can see that this is equivalent to the condition that every open cover has a finite subcover, i.e. the group is compact.
• Characterisation of Abelian Lie groups -- "Compact Connected Abelian Lie Group is a torus" is a generalisation of "finite Abelian group is a product of cyclic groups" -- the idea is that the exponential map "wraps" the Lie algebra around into the Lie group -- this just gives the quotient of the Lie algebra by the kernel of the exponential map, which is topologically $$\mathbb{R}^n/\mathbb{Z}^n$$. The characterisation of a connected Abelian Lie group as a cylinder $$\mathbb{R}^{n+k}/\mathbb{Z}^k$$ follows similarly.

With that said, here are some stuff I DON'T (completely) understand, and would like to have a similar level of understanding for:

• Why is the structure of a Lie group characterised by its second-order structure? I know that this follows from the BCH formula, the local diffeomorphism nature of the exponential map and the fact that an open neighbourhood of 1 generates the group, but I have no intuition at all why the BCH formula "should" be true.
• What's the deal with simply-connected groups? I can certainly see why the Lie algebra cannot detect disconnectedness in a group -- I had expected that it could not detect compactness either (whether the group closes in on itself eventually), so the statement of Lie's third theorem would be "every Lie algebra has a corresponding unique connected, compact Lie group". Instead, the statement is "every Lie algebra has a corresponding simply connected Lie group".
• Non-surjectivity of the exponential map even in the connected case -- I'm not asking for counter-examples, I'm asking "what exactly goes wrong in groups like $$SL_\mathbb{R}(2)$$?", perhaps a hint about "what does the image of the exponential map look like?" (as an analogy, I would explain smooth functions failing to be analytic as "they are flatter than every polynomial at 0, and can be constructed as $$1/f(1/x)$$ where $$f$$ is any function that grows faster than every polynomial")
• Surjectivity when every element is contained in a maximal torus -- I read this here as a generalisation of "the exponential map is surjective in the connected compact case". Even if the generalisation isn't true, is there an intuitive way to understand why compactness makes the problem in the previous point go away.
• Characterisation of non-Abelian Lie groups -- Tell me if my understanding of simple and semisimple Lie algebras makes sense -- we want to classify non-Abelian Lie groups as products like we do Abelian Lie groups, and the only way to do so is as "semidirect products of simple Lie groups and Abelian Lie groups". A reductive Lie group is basically when this semidirect product is a direct product, and a semi-simple one is a reductive Lie group where there are no Abelian groups in the product. Is this right?
• Various abstract algebraic things -- I have no idea how to interpret things like nilpotent and solvable Lie algebras, radicals and so on in the context of Lie theory.
• At first when I heard of the Killing form, I presumed it would be some "natural" way to define a dot product on the Lie algebra -- but I honestly don't see how it is natural. Is it the only dot product that is invariant under Lie algebra automorphisms?

I've thought very hard about the theory, but I just can't seem to figure out how to fill these "holes". Am I missing some important central insight into Lie theory that are crucial to some of these questions?

• +1 for articulating what you know and don't know so well. But the “question” is very broad. Can you focus on one thing at a time? – Matthew Leingang Sep 24 '19 at 17:32
• The compactness can be detected. See negative definite Killing form. And solvable and nilpotent is not only restricted to Lie algebras. Groups, rings, algebras etc. have it, too, in particular also Lie groups. The Killing form is not a dot product. I suppose you will "miss several important central insights" without having worked with these topics. – Dietrich Burde Sep 24 '19 at 18:05
• Lie theory is a gigantic topic, there is tons and tons to say about it, and so it's not exactly surprising that some bits of it are still confusing at any particular stage. Try to have some patience! Isn't it wonderful that there's still so much more to learn? – Qiaochu Yuan Sep 24 '19 at 20:01
• Although this question surely falls in the category of too broad, I feel that if separated into multiple questions, they might go unanswered. Yes, MSE is platform for all the range of mathematical questions, but, somehow, something beyond common curriculum often doesn't attract as much as attention as it deserves. I would say that sheer amount of details that went into this question overweights the broadness, because if the OP included all of this context in any single question, it would be overkill. In the same spirit, I would count partial answers as good... and beneficial to community. – Ennar Sep 24 '19 at 22:36
• I really do encourage you, though, to ask your main questions separately, and then ask your follow-up questions in comments there. Not only will you be more likely to get coherent, organized answers (in ways that comments, mine included, are usually not), but it will make the site more useful for other people asking the same questions later. (In fact, I'd guess that at least some of the questions you're asking about already have answers on the site.) – Travis Willse Sep 25 '19 at 6:10