There is a three-way correspondence between:

  • Real (connected and simply connected) Lie groups of dimension $n$;
  • $\mathbb R$-Lie algebras of dimension $n$;
  • Formal group laws in $n$ variables over the reals.

To get from the first to the second, one only really needs the fact that manifolds have tangent spaces which are vector bundles, and that the tangent space functor is strong (so that one can talk about left-invariant vector fields).

To get from the second to the third, one uses the Baker--Campbell--Hausdorff formula, which in the realm of formal power series is pure abstract algebra.

However, the only way I know of getting from the first to the third directly involves choosing analytic coordinates around the identity and Taylor expanding the Lie group multiplication.

My question is: can one give a direct geometric construction of a formal group law out of a Lie group without choosing coordinates? The sort of thing I had in mind might, for example, involve using jet bundles.

  • 4
    $\begingroup$ It would seem that passage from the second to the third involves a choice of basis for your Lie algebra. Since the exponential map is a local diffeomorphism, this choice of basis is essentially a choice of local coordinates at the identity of the group (and hence elsewhere). So, it looks like a choice of coordinates arises in other places. $\endgroup$ Oct 22, 2013 at 1:58
  • 1
    $\begingroup$ Also, I do not think passage from the first to the second is a correspondence. In general, there are many non-isomorphic Lie groups with a given (isomorphism class of) Lie algebra. If you require your Lie groups to be connected and simply-connected, then taking Lie algebras gives a correspondence. $\endgroup$ Oct 22, 2013 at 2:48
  • 4
    $\begingroup$ Maybe the answer is to look at formal groups (as opposed to formal group laws)? $\endgroup$
    – Drew Heard
    Oct 22, 2013 at 2:59
  • $\begingroup$ Thanks, Peter. You're right that (1) and (2) are not in correspondence in quite the form I stated. Have edited. As to your other comment, that is illuminating; really what is happening is that the equivalence between the categories of Lie algebras and formal group laws is only canonically defined in the direction from the latter to the former; the converse requires the axiom of choice (a choice of basis for each Lie algebra). $\endgroup$ Oct 22, 2013 at 3:04
  • $\begingroup$ Hi Richard, congrats on your first MO contribution! :-) $\endgroup$ Oct 22, 2013 at 9:01

2 Answers 2


Of course if one wants to avoid coordinates then one should replace formal group laws by the coordinate-invariant things that they are the coordinate-dependent versions of!

There are two dual ways to do this that I know of.

Commutative: First, observe that an $n$-dimensional formal group law over $k$ is precisely a comultiplication on $k[[x_1, ..., x_n]]$ (with respect to a suitably completed tensor product) with counit the map that kills all terms of positive degree and antipode the map that multiplies each monomial by $(-1)^{\text{deg}}$. This gives $k[[x_1, ..., x_n]]$ the structure of a commutative Hopf algebra (in a suitable category of power series algebras equipped with a suitably completed tensor product) and now we can ask how to construct this Hopf algebra from $G$. This can be done by considering the completion of the algebra of germs of smooth functions at the identity of $G$ with respect to the ideal generated by the germs vanishing at the identity. The comultiplication is induced from the multiplication $G \times G \to G$ via pullback of germs.

Cocommutative: First, recall that if $\mathfrak{g}$ is a Lie algebra then the universal enveloping algebra $U(\mathfrak{g})$ is naturally a cocommutative Hopf algebra. In fact over a field of characteristic zero the functor $\mathfrak{g} \mapsto U(\mathfrak{g})$, when regarded as taking values in Hopf algebras, is full; by PBW $\mathfrak{g}$ is precisely the Lie algebra of primitive elements of $U(\mathfrak{g})$. Hence we should be able to identify Lie algebras with a certain full subcategory of Hopf algebras.

Theorem (Cartier): Let $k$ be a field of characteristic zero. A Hopf algebra over $k$ is isomorphic to the universal enveloping algebra $U(\mathfrak{g})$ of a Lie algebra over $k$, namely its Lie algebra of primitive elements, if and only if it is cocommutative and conilpotent.

The definition of conilpotence is mildly involved but see, for example, Cartier's A primer of Hopf algebras.

Now I claim that $U(\mathfrak{g})$ should be seen as a version of the formal group law associated to $\mathfrak{g}$. The reason is that as a Hopf algebra it is dual in a suitable sense to the Hopf algebra above. So Cartier's theorem should be seen as a coordinate-independent version of the correspondence between Lie algebras and formal group laws which in particular does not require Baker-Campbell-Hausdorff.

It remains to figure out how to get $U(\mathfrak{g})$ directly from $G$, and the answer is the following:

Theorem (Schwartz): $U(\mathfrak{g})$ can naturally be identified with distributions on $G$ supported at the identity.

(The dual pairing between the two Hopf algebras above is then the natural pairing between distributions supported at the identity and germs at the identity.)

  • $\begingroup$ For working with completions of local rings it is "safer" to restrict to the analytic case, not the $C^{\infty}$ case, simply because local rings in the C$^{\infty}$ case are not noetherian and hence completions have to be treated with rather more care than in the noetherian case (e.g., the map from the local ring to its completion isn't even injective). But what is the intrinsic significance of the choice of $V$ in terms of your discussion with symmetric algebras? $\endgroup$
    – Marguax
    Oct 22, 2013 at 7:33
  • $\begingroup$ @Marguax: once you have the counit you have the augmentation ideal $I$ and then $V$ is canonically isomorphic to $I/I^2$. Or am I mistaken? $\endgroup$ Oct 22, 2013 at 7:42
  • $\begingroup$ It is totally non-canonical to lift $V$ back into the ring. To do that lifting is almost like choosing coordinates (not quite, but rather close). The completed local ring is canonically filtered (by powers of the maximal ideal), not canonically graded. $\endgroup$
    – Marguax
    Oct 22, 2013 at 9:18
  • $\begingroup$ Thanks, this is just the sort of thing I was looking for. I had the feeling that one should use something else than an actual formal group law. I was aware of the two dual Hopf algebras arising from O(G) and U(G) but not how to construct them directly from G. I had a few questions which I've appended to Marguax's comments above. $\endgroup$ Oct 22, 2013 at 9:21
  • $\begingroup$ @Marguax: yes, you are right. I'll edit. $\endgroup$ Oct 22, 2013 at 17:26

Sure: if $G$ is the given analytic group then consider the functor $F$ on finite local $\mathbf{R}$-algebras with residue field $\mathbf{R}$ given by defining $F(A)$ to be the set of maps of real-analytic spaces ${\rm{Sp}}(A) \rightarrow G$ "based" at the identity. This is a group-valued functor, and it is visibly pro-represented by the completion $O_{G,e}^{\wedge}$ of the noetherian analytic local ring $O_{G,e}$ of $G$ at the identity $e$. The group law on the functor thereby defines a group-object structure on the formal scheme Spf($O_{G,e}^{\wedge}$), or more specifically defines a formal Hopf algebra structure on $O_{G,e}^{\wedge}$ (i.e., a map $m^{\ast}:O_{G,e}^{\wedge} \rightarrow O_{G,e}^{\wedge} \widehat{\otimes} O_{G,e}^{\wedge}$ satisfying the usual conditions).

Voila, that's it. The same procedure works for analytic groups over non-archimedean fields in char. $p > 0$ (such groups might not be smooth, so "local coordinates" might not even exist, yet the method still works), as well as for group schemes of finite type over any field, and so on. No need for Lie algebras or a BCH formula. The notion of "formal group" makes good sense in terms of complete local noetherian rings even without any notion of formal smoothness (so no "local coordinates").

  • $\begingroup$ This is essentially an expanded version of an item mentioned in Qiaochu Yuan's answer. $\endgroup$
    – Marguax
    Oct 22, 2013 at 7:35
  • $\begingroup$ Nice!${}{}{}{}{}$ $\endgroup$ Oct 22, 2013 at 9:05
  • $\begingroup$ Thanks @Marguax. Can I check I follow? $O_{G,e}$ is the stalk at $e$ of the sheaf of analytic functions on $G$; to say that $F$ is pro-represented by the completion of $O_{G,e}$ w.r.t. its maximal ideal $I$ is to say that $F$ is a filtered colimit of representables, indexed by the cofiltered diagram of Weil algebras $n \mapsto O_{G,e}/(I^n)$. Thus $F$ can be seen as a cogroup in $\mathrm{Pro}(\mathrm{Weil})$. Taking cofiltered limits now sends coproduct to the (completed) tensor product of topological algebras, so finally we get a commutative Hopf algebra. Is that right? $\endgroup$ Oct 22, 2013 at 10:00
  • $\begingroup$ @Richard: I never heard of Weil algebras or cogroups before, but it sounds like we're thinking in the same way with different language. $\endgroup$
    – Marguax
    Oct 22, 2013 at 21:34

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