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A Lie monoid is a monoid, together with a structure of a smooth manifold (possibly with a boundary), such that the monoid multiplication is smooth.

If all left (or right) translations in a Lie monoid $M$ have full rank at the identity $e$, then the monoid is necessarily parallelizable, since any basis of $T_eM$ may then be extended to a left-invariant (or right-invariant) global frame for $TM$. In particular, this holds for any Lie group. Furthermore, any Lie monoid $M$ that admits an injective immersive homomorphism $\phi\colon M\rightarrow G$ into a Lie group $G$, is parallelizable. That is, any Lie submonoid of a Lie group is parallelizable.

However, left and right translations in a Lie monoid may not have full rank at identity, indicating that in general, a Lie monoid may not be parallelizable. The question hence reads: Do non-parallelizable Lie monoids exist?

A useful insight to this question stems from the theory of H-spaces. Since H-spaces are generalizations of topological monoids, and the only spheres that admit a H-space structure are parallelizable, spheres don't admit such a Lie monoid structure. Due to cohomology, projective spaces are also disqualified. Furthermore, there exist H-spaces which are not parallelizable (see e.g. Browder, Spanier: H-spaces and duality, Pac. J. Math. 12, 411-414 (1962). ZBL0112.14502.), but the given example is quite involved.

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  • $\begingroup$ I think any non-parallelizable sphere is an example of non-parallelizable semigroup, where monoid operation is just vector space addition compactified by infinity which becomes an absorbing element. You can freely adjoin a disjoint identity point to get a monoid. I'm not able to produce a connected (or even equidimensional) example, unfortunately. $\endgroup$
    – Denis T
    Commented Nov 27, 2022 at 19:23
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    $\begingroup$ The monoid operation (denote it $m\colon S^n\times S^n\rightarrow S^n$) you are talking about is not even continuous: take any sequence of vectors $v_n\rightarrow \infty$, and then $(v_n,-v_n) \rightarrow (\infty, \infty)$, but $m(v_n,-v_n)=0$ whereas $m(\infty, \infty)=\infty$. So $m$ does not map limits to limits. Also, $0$ is the identity element, so you didn't have to adjoin it. $\endgroup$
    – Žan Grad
    Commented Nov 27, 2022 at 20:52
  • $\begingroup$ Take a unital associative finite-dimensional algebra over $\mathbb{C}$, and consider it as a Lie monoid under multiplication. Translation by zero does not have full rank. $\endgroup$
    – Konrad Waldorf
    Commented Nov 28, 2022 at 7:38
  • $\begingroup$ Of course, but this is an example of a parallelizable manifold, a vector space. My question is whether we can find a non-parallelizable Lie monoid. $\endgroup$
    – Žan Grad
    Commented Nov 28, 2022 at 12:56
  • $\begingroup$ @ŽanGrad: Right, I misunderstood this. $\endgroup$
    – Konrad Waldorf
    Commented Nov 30, 2022 at 18:28

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