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Let $G$ be compact Riemann surface with the structure of a complex commutative Lie group, ie the multipliciation map $m:G \times G \to G$ is holomorphic (+certain usual diagrams satisfy axiomatic group law diagrams).

Problem: I have some troubles with filling gaps in the argument in proof of Thm. 2.3 presented in these notes that the exponential map $\text{exp}: \text{Lie}(X) \to X $ is surjective. (the notes refer to D. Mumford's Abelian varieties for a full proof but I can't find this source available).

The presented sketch there goes as follows: It is claimed that the exponential map $\text{exp}: \text{Lie}(G) \to G $ a local biholomorphism at $0$ (sending $0 \in \text{Lie}(G)$ to $e_G \in G$) and commutativity of $G$ implies that $\text{exp}$ is a morphism of complex Lie groups.

Next, it is claimed that $\text{exp}$ is surjective because it's image contains a neighborhood of $e_G = \text{exp}(0)$ and it's kernel discrete as it's injective in a neighborhood of $0$ (that's should be the consequence of local biholomorphism at $0$).

The two following sub-questions can be distilled out to complete all the gaps in the argumentation chain above:

(Q1) Why is $\text{exp}$ local biholomorphic at $0$ (especially locally homeomorphic at $0$)?
Rmk.: Here appears already following subtle issue: Note that in subsequent arguments - namely that image contains a neighborhood of $e_G = \text{exp}(0)$ and that it's kernel is discrete - are only exploited topological features of a local biholomorphism, namely that it's also a local homeomorphism.
On the other hand here seemingly we neccessarily need to take this detour around the complex analytic structure (for example the same statement for real compact connected group would fail) since without complex structure on $G$ $\text{exp}$ may be not a local homeomorphism even if we assume connectedness & compactness (see This discussion)
So the (Metaquestion(s)) popping up immediately at this point: How crucial is the complex structure on $G$ involved to conclude that the exponential map $\text{exp}$ is a local homeomorphism? The sketch shows that it sufficient as local biholom clearly imply local homeo, but it looks a bit over the top. Do we really need to assume complex structure to assure that the exp map is a local homeo at $e_G$? So this leads naturally to: Are there reasonable (...not too trivial) conditions on a connected compact (but not neccessarily complex) Lie group assuring that the $\text{exp}: \text{Lie}(G) \to G $ is still a local homeomorphism at $e_G$?
(#EDIT: As @user8268 pointed out in the comments below what I wrote above is plainly wrong. Exp map is allways locally homeo, but in general only in $e_G$.)

(Q2) When $G$ is commutative (as in our case), why is $\text{exp}$ is a morphism of complex Lie groups? (ie why it respects the group structure)?

Now having answered Q1 and Q2, the rest should come perfectly toghether: That the image contains a neighborhood of $e_G = \text{exp}(0)$ is a consequence of local biholomorphy (Q1), (...more precisely from local homeomorphy, which in turn follows from local biholomorphy; see Rmk. above) and that the kernel is discrete follows again from Q1 (locally biholo) and Q2 (preserve group structure)

user267839
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  • for Q1: it works for any complex Lie group and follows from the (complex) inverse function theorem and from the fact that the tangent map of $\exp$ at $0\in\mathrm{Lie}(G)$ is the identity – user8268 May 10 '24 at 17:03
  • @user8268: Re Q1: Wouldn't then the same argument work in real Lie group setting using inverse function theorem for differential geometry? On the other hand there seems to be some troubles with local homeo (...which would in turn follow from local diffeo) of the exp map; see https://math.stackexchange.com/questions/301504/on-surjectivity-of-exponential-map-for-lie-groups – user267839 May 10 '24 at 17:12
  • Or, where does my naive reasoning break down that this argument with complex/holomorphic inverse function theorem, can be imitated with (usual) differential inverse function theorem to show that exp is local diffeom for real Lie groups? – user267839 May 10 '24 at 17:15
  • Indeed it works also for real Lie groups; the question you link is about surjectivity and doesn't contradict this. – user8268 May 10 '24 at 17:18
  • @user8268: But isn't a local homeo always open? (But this would contract the answer in the linked thread) Note that it suffice to check openness of a map on the base sets of a topology which may be choosen small enough where the map is actually a homeo; or where is here the error in my reasoning? – user267839 May 10 '24 at 17:24
  • in Q1 the claim is that it is a local biholomorphism at $0$, not everywhere (though it is true that if $G$ is commutative then it's true everywhere) – user8268 May 10 '24 at 17:30
  • @user8268: oh right, sorry, So in turn it not matters which inverse function theorem one applies and what kind of Lie groups one considers, at $e$ specifically exp is always a local homeo, right? (What finally implies that the image of exp contains always certain open nbh of $e_G$...) – user267839 May 10 '24 at 17:37
  • @user8268: But knowing this now, do you see how one concludes there between the lines that exp is even surjective? (It is indeed even in real Lie algebra world true, see eg here: https://math.stackexchange.com/q/3859991) but there used machinery of Riemannian geometry (...concretely, having an open set in the image + connectedness & compactness one can hit "everything" by spreading geodesics), but I doubt that in the quoted notes one invoked exactly that kind of argument. Do you maybe see how to see surjectivity of exp using only complex analalytic tools once we know it's a biholom at $e$? – user267839 May 10 '24 at 18:08
  • Since your $G$ is abelian, the Q2 property holds, so being a local biholomorphism (or any-ism) around $0$ implies that it is true everywhere, and as you observed, it will be open and thus onto (for connected $G$). As for why Q2 holds - $G$ and $Lie(G)$ are both (complex) Lie groups with the same Lie algebra, $Lie(G)$ is 1-connected, so we have a canonical homomorphism $Lie(G)\to G$; you just need to verify that this morphism is actually exp. – user8268 May 10 '24 at 19:34
  • @user8268: on the argument that exp surjective: so using that exp is group homom (Q2), we conclude with (Q1) that exp is open, so the image is open in $G$. But in order to show that exp is onto using connectedness of $G$ we also need that the image is also closed, right? (so clopen + connected -> whole space) But an local homeom ( especially an open map) is in general not closed. So how do you conclude there that exp onto there? – user267839 May 11 '24 at 12:34
  • yes, what I should have said is that if you have 2 Lie groups with the same Lie algebra, one 1-connected and the other connected, then the 1st one is (in a canonical way) the universal covering of the 2nd one. It is a "standard" thing about Lie groups, if you haven't met it, the best thing is to look into some "standard" text (though I don't have a recommendation). – user8268 May 11 '24 at 17:32
  • @user8268: for sake of completeness here another "elementar" argument that exp is into using U1 and U2: image of exp contains an open subgroup $H$ of $G$, but then this subgroup must be also closed as it's the complment of the union of it's coset classes. Then we can use connectedness. – user267839 May 11 '24 at 19:18
  • *onto :) (...into is a complicated issue) – user267839 May 12 '24 at 11:04

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