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Let $a,b:\mathbb{R}\rightarrow M$ be two smooth curves in a manifold $M$.
We define the product of functions $\cdot$ in the usual way so that $C^\infty(\mathbb{R}\rightarrow M)$ is a monoid.
The derivative of a curve $c:\mathbb{R}\rightarrow M$ is $$ \frac{d}{dt}\Big|_t c(t) := c_{*,t}(\frac{d}{dt}\Big|_t) $$

I would guess that

$$ \frac{d}{dt}\Big|_t (a\cdot b)(t) = \frac{d}{dt}\Big|_t a(t)\cdot b(t) + a(t) \cdot \frac{d}{dt}\Big|_t b(t) $$

I'm trying to prove this.

We have (with $m(a,b):=a\cdot b$)

$$ \frac{d}{dt}\Big|_t (a\cdot b)(t) = (a\cdot b)_{*,t}(\frac{d}{dt}\Big|_t)= \left[m_{*,(a(t),b(t))}\left( a_{*,t}, b_{*,t}\right) \right](\frac{d}{dt}\Big|_t)=m_{*,(a(t),b(t))}\left( \frac{d}{dt}\Big|_t a(t), \frac{d}{dt}\Big|_t b(t)\right)$$

First, I was thinking on using the derivative of the multiplication map like in Differential of the multiplication and inverse maps on a Lie group, but of course this is nonsensical, since $m$ is not a (smooth) group multiplication, but simply a monoid multiplication. What now, then?

soap
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  • What is the usual way to define multiplication for smooth functions in a manifold? – Jochen Feb 22 '18 at 12:28
  • @Jochen Oh, I would say $(a\cdot b)(t) = a(t)b(t)$, which, now that you say it, I see can only be defined if there is a multiplication in $M$ (which I'm implicitly assuming above that is also a Lie group). So $m(a,b)(t) = \mu(a(t),b(t))$ – soap Feb 22 '18 at 12:36
  • @Jochen But it seems to me that I still cannot use what I wanted to use (in the link), since what matters is if $m$ is a Lie group multiplication or not, and in this case it seems that it is not, even though it is closely related to one. – soap Feb 22 '18 at 12:57

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