# Lie groups vs Lie monoids

Does there exist a well developed theory of a class of objects which might rightfully be called Lie monoids? By this I mean with axioms similar to those of Lie grousp but with the axiomatic existence of inverses dropped. If so what is the analogous structure to lie algebras if one exists?

• Intuitively speaking, any element of a monoid infinitesimally close to the identity should be invertible (e.g. the group of units of any Banach algebra is open), so the obvious analogue of a Lie algebra here should still be a Lie algebra. – Qiaochu Yuan Dec 28 '13 at 3:05
• So that prompts the question: is there some analogue of the exponential map in this context? Also: does this mean that the tangent spaces at other monoid elements are "different" to the algebra? I guess that this is technically manifested as the tangent bundle not necessarily being trivial, as it is in a lie group? – Benjamin Dec 28 '13 at 4:08
• @QiaochuYuan, Actually there are many Lie algebras. To every idempotent e of a semigroup S there is the monoid eSe with identity e. In the Lie or algebraic settings eSe will again be Lie or algebraic. It has its own group of units and that group has a Lie algebra. My guess is there may be a lie algebroid lurking about. – Benjamin Steinberg Dec 28 '13 at 16:52
• One example of Lie monoid is that of invertible real matrices with positive entries. It's open in $GL_n$ so should have the same Lie algebra, and the exponential map cannot be defined everywhere. – YCor Jun 5 '14 at 21:18

There is a well developed theory of algebraic monoids, due principally to Putcha and Renner. I think Lie semigroups is less well developed but there is work by Hoffmann, Lawson and the thesis of Langlands was on this subject.

• Thanks for that. Are there any key words or good resources I should know? – Benjamin Dec 28 '13 at 0:14
• Look for wedges. This is one of the things the Lie semigroup guys look at. I know the algebraic monoid stuff better. – Benjamin Steinberg Dec 28 '13 at 0:18
• @Benjamin: What Putcha and Renner do is mostly in the spirit of linear algebraic groups rather than Lie groups, so I'm unsure how a close analogy with Lie groups would fit here. Maybe it's partly a question of what characteristic the underlying field has. – Jim Humphreys Dec 28 '13 at 18:06
• I'm actually only interested in the case of matrices as I have a specific application in mind. Work analogous to linear algebraic groups would be ideal. – Benjamin Dec 28 '13 at 19:57

See the following:

• MR1317811 Hilgert, Joachim; Neeb, Karl-Hermann Lie semigroups and their applications. Lecture Notes in Mathematics, 1552. Springer-Verlag, Berlin, 1993. xii+315 pp. ISBN: 3-540-56954-5 (Reviewer: Gestur Ólafsson)

• MR1179336 Neeb, Karl-Hermann On the foundations of Lie semigroups. J. Reine Angew. Math. 431 (1992), 165–189. (Reviewer: Jimmie D. Lawson)

• MR1235759 Mittenhuber, Dirk; Neeb, Karl-Hermann Remarks on our paper: "On the exponential function of an invariant Lie semigroup'' [Sem. Sophus Lie 2 (1992), no. 1, 21–30; MR1188629 (93j:22007)]. Sem. Sophus Lie 3 (1993), no. 1, 119–120.

Moreover, completions of infinite dimensional Lie groups (like diffeomorphism groups) with respect to right invariant Riemannian metrics tend to be semigroups, for an easy example see 4.8 of here.

Anders Kock mentions Lie monoids and some of their properties in his book on synthetic differential geometry. Basically, in SDG, a Lie monoid is a microlinear monoid object. It is easy to show that the tangent space at the identity of a Lie monoid is an $R$-Lie algebra, defined in the same way as for a Lie group, and isomorphic to the left- or right-invariant vector fields on the monoid (depending on your bracket convention). We also still have a Lie functor, taking Lie monoids to their Lie algebras, and Lie monoid homomorphisms to Lie algebra homomorphisms.

For example, for any microlinear space $M$ the mapping space $M^M$ is a Lie monoid, with the composition of maps as the multiplication. Then the Lie algebra of $M^M$ is the space $\frak{X}$$(M) of vector fields on M. Since each infinitesimal transformation is invertible, it follows that this is also the Lie algebra of \text{Diff}(M). One thing to note is that since we do not have an inversion map on a Lie monoid, then I do not see how would have a canonical isomorphism of Lie algebras between left- and right-invariant vector fields on the monoid. Usually the isomorphism would be given by pushing forward a left- or right-invariant vector field by the inversion map. In "Monoidal functors, species and Hopf algebras", by M. Aguiar and S. Mahajan (CRM Monograph Series, 29), they defined a Lie monoid (definition 1.25) in the following sense: Let (\mathsf{C}, \bullet, \beta) be a \mathbb{K}-linear symmetric monoidal category (possibly without unit). A Lie monoid in (\mathsf{C}, \bullet, \beta) is a pair (L, \gamma), where$$\gamma: L \bullet L \to L$$satisfies$$\gamma+ \gamma \circ \beta_{L,L}=0 \qquad \text{and} \qquad \gamma \circ (\gamma \bullet \text{id})\circ (\text{id}+\xi + \xi^2)=0,$$where \xi denotes the composite$$L \bullet L \bullet L\xrightarrow{\ \text{id}\,\bullet\,\beta_{L,L} \ }L\bullet L \bullet L \xrightarrow{\ \beta_{L,L}\,\bullet \,\text{id}\ }L \bullet L \bullet L.$$Here "\mathbb{K}-linear" means that each set \text{Hom}(A, B) of morphism in \mathsf{C} is a \mathbb{K}-module and composition of arrows$$\text{Hom}(A,B)\times \text{Hom}(B,C)\to \text{Hom}(A,C)$$is \mathbb{K}-bilinear. The application \beta is a natural isomorphism where for each A, B \in \text{Obj}(\mathsf{C}), the isomorphism$$\beta_{A,B}: A \bullet B \to B \bullet A$$satisfies$\beta_{A,B}\circ \beta_{B,A}=\text{id}$and two axioms of compatibility with$\bullet\$ (see definition 1.2).

• This isn't an answer to the question. It describes a notion of Lie algebra (as an algebra over the Lie operad; in particular, this definition is by no means due to Aguiar and Mahajan), not a notion of an analogue of Lie groups but where invertibility is dropped. – Qiaochu Yuan Jan 2 '16 at 7:17

The Kronecker products of real 2 by 2 matrices are a set of 4 by 4 matrices. Although not closed under addition of matrices, they are closed under the operation of 4 by 4 matrix multiplication. The 4 by 4 identity matrix is in the set of these Kronecker products. The matrix multiplication is continuous, but inverses don't always exist, so the set forms a topological monoid rather than a topological group. I don't know whether it would be a Lie monoid.