It's clearly more strongly a monoid. Is this a historical thing or something abstract I'm missing?

  • $\begingroup$ Whenever time evolution is involved, there is an underlying semigroup structure. If the system itself does not depend on time, then there is a solution operator $S(t)$ that takes the state $x$ of the system and evolves it forward in time to obtain $S(t)x$. If you then evolve that new state through time $t'$, the result is the same as evolving the original state through $t+t'$ seconds. In other words $S(t')S(t)x = S(t+t')x$. Then a $C^0$ condition is a stability condition at $t=0$ of the form $\lim_{t\downarrow 0} S(t)x = x$. $\endgroup$ – COVID-20 Apr 19 '17 at 21:42
  • 2
    $\begingroup$ Haha, you're right. I guess it's a historical thing. Note that the "semi-group theory in PDEs" deals only with one semi-group $\mathbb{R}^+$ (and its homomorphic images). $\endgroup$ – Michał Miśkiewicz Apr 19 '17 at 22:47
  • 1
    $\begingroup$ I don't understand the question really, if it is a monoid it is also a semigroup. We may say "the ring of integers" although it more strongly a Euclidean domain. $\endgroup$ – Jonas Meyer Apr 20 '17 at 3:43
  • 1
    $\begingroup$ @JonasMeyer I agree it's a somewhat silly question. But usually when I can make stronger statements with fewer syllables I do. $\endgroup$ – Funktorality Apr 24 '17 at 0:09

Using the term "semigroup" does not commit us to considering identity as a part of it. Sure, we give it notation $T_0$ or $S(0)$ or such when it's convenient notationally, but next minute we may want to exclude it from consideration. For example, one says "semigroup of compact operators on [a Banach space]" while knowing perfectly well that the identity isn't compact; it's just being ignored in this statement. Saying "monoid of compact operators", which explicitly includes identity, would be quite strange.

When I consider the Poisson semigroup on the circle (that's when you extend the function harmonically to the disk, and then restrict to a smaller concentric circle), it somehow doesn't occur to me that I should include identity as a part of semigroup, let alone to emphasize its membership by saying "monoid". That's because the Poisson semigroup is about convolution with the Poisson kernel, and the identity operator isn't. While the behavior as $t\to 0^+$ is important, including identity as a member of the semigroup does not help me the slightest bit in studying that behavior. (This goes for the heat semigroup, too.)

Summary: It makes sense to use terms that emphasize the structure that matters. The membership status of identity in operator semigroups does not matter from the analytic point of view. It typically does not share the properties of other operators in the semigroup.

| cite | improve this answer | |

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.