Groups pop up as automorphism groups in any category.

Rings pop up as endomorphism rings in any additive category.

Is there a similar way to attach a Lie algebra to an object in a category of a certain sort? Maybe even such that the attachment of a Lie algebra to a Lie group becomes a special case?

And yes, I have searched through the answers to Why study Lie algebras?.

  • $\begingroup$ Additive is a stronger condition than you need; you only need $\text{Ab}$-enriched. $\endgroup$ Jul 25 '14 at 10:01
  • 3
    $\begingroup$ @QiaochuYuan That's true, but in an additive category, the Ab-enriched structure is determined by the categorical property. $\endgroup$ Jul 25 '14 at 12:02
  • 3
    $\begingroup$ @Matthias, In my experience, Lie algebras pop up usefully when your category has some sort of linear structure. (Additive is a good place to start; better if you have something k-linear, or something stable.) In this setting, (coLie)Lie is Koszul dual to (commutative)cocommutative, and Koszul duality for augmented objects gives the interpretation of Lie objects as infinitesimal objects studied "at a point." I don't think the Koszul dual to cocommutative algebras in, say, spaces or sets (with whatever monoidal structure) has such a rich interpretation, though I'd be happy to hear about one! $\endgroup$ Jul 25 '14 at 12:11
  • 9
    $\begingroup$ The derivations of an algebra form a Lie algebra. $\endgroup$ Jul 25 '14 at 15:09
  • $\begingroup$ @Hiro: I'm mixed up about the "co-". Where does Quillen's equivalence between differential graded Lie algebras and differential graded (co)commutative co-algebras (over $\mathbb Q$) fit into what you are saying? $\endgroup$ Jul 25 '14 at 16:33

The category of Lie algebras is equivalent to a certain category of cocommutative Hopf algebras, with the equivalence given by sending a Lie algebra $\mathfrak{g}$ to its universal enveloping algebra $U(\mathfrak{g})$. These cocommutative Hopf algebras can in turn be thought of as group objects in a certain category of cocommutative coalgebras, and hence can potentially pop up as automorphism objects in any category enriched over cocommutative coalgebras.

You might object that you don't know any interesting examples of such categories, but in fact you do: the category of commutative algebras admits such an enrichment (see the nLab), and this is one abstract way to see why Lie algebras can act on commutative algebras (by derivations).

Speaking more philosophically, you should expect to be able to extract Lie algebras from any situation where you can cook up a sensible notion of infinitesimal automorphism or more generally an infinitesimal element of some group. Enriching over cocommutative coalgebras gives you one fairly general way to do this; if $X$ is an object in your category and $\text{End}(X)$ is the cocommutative bialgebra of endomorphisms of $X$, then the primitive elements of $\text{End}(X)$ (the ones satisfying $\Delta X = 1 \otimes X + X \otimes 1$, where $1 = \text{id}_X \in \text{End}(X)$) should be regarded as the infinitesimal endomorphisms of $X$, and indeed these naturally form a Lie algebra under the commutator bracket.

  • 4
    $\begingroup$ “The category of Lie algebras is equivalent to a certain category of cocommutative Hopf algebras” — do you know a good reference that works this out or states it more precisely? Or at least some specific term which readers can google for more details? $\endgroup$ Jul 25 '14 at 12:29
  • 4
    $\begingroup$ In the setting of (differential) graded modules, this comes from the Milnor-Moore theorem, but you have to add a conilpotence condition on your Hopf algebras to get the equivalence of categories. For instance, you can take a look at the chapter Hopf algebras, Theorem 7.2.19, which states a version slightly more general than the original result. $\endgroup$ Jul 25 '14 at 16:38
  • 2
    $\begingroup$ @Peter: as Sinan says, the extra condition is called conilpotence. See Theorems 3.6.1 and 3.8.1 in Cartier's A primer of Hopf algebras (people.math.osu.edu/kerler.2/VIGRE/InvResPres-Sp07/…). The underlying field needs to have characteristic $0$. $\endgroup$ Jul 25 '14 at 18:27
  • 5
    $\begingroup$ @PeterLeFanuLumsdaine I found it helpful to consider that the universal enveloping algebra $U: \text{LieAlg} \to \text{HopfAlg}$ has a right adjoint given by taking primitive elements $P: \text{HopfAlg} \to \text{LieAlg}$, and that in characteristic $0$ the unit $1 \to P U$ of the adjunction is an isomorphism. This implies that $U$ is fully faithful. Milnor-Moore identifies those Hopf algebras $H$ where the counit is an isomorphism. $\endgroup$
    – Todd Trimble
    Jul 31 '14 at 2:50

Lie algebras are equivalently groups internal to "infinitesimal geometry".

For instance when formalized in a topos for synthetic differential geometry then a Lie algebra of a Lie group is just the first-order infinitesimal neighbourhood of the unit element (e.g. Kock 09, section 6).

More generally in geometric homotopy theory, Lie algebras, being 0-truncated L-∞ algebras are equivalently "infinitesimal ∞-group geometric ∞-stacks" also called formal moduli problems (Lurie). See the link there for pointers to the proof of the equivalence, also (Pridham 07).


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.