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Let $M$ be a smooth manifold, and $\Gamma(TM)$ the space of all vector fields on $M$. Wikipedia says that an affine connection on $M$ is a map $\nabla:\Gamma(TM)^2\rightarrow \Gamma(TM)$ that satisfies: $$\nabla(X,Y)=\nabla_XY$$ $$f(\nabla_XY)=\nabla_{f(X)}Y$$ $$\nabla_Xf(Y)=\partial_Xf(Y)+f(\nabla_XY)$$ for all $f$ in the set of smooth functions on $M$... But the only way that these equations make sense is if the function $f$ satisfies $f:\Gamma(TM)\rightarrow \Gamma(TM)$, since the second equation can be rewritten as: $$f(\nabla(X,Y))=\nabla(f(X),Y)$$ and then, for this to be true, the manifold should be $M=\Gamma(TM)$, wich makes the definition completely circular. Maybe there is something I'm not getting but, by this logic, the definition isn't well-posed. If this definition is indeed incorrect, how is the affine connection properly defined when working with things like Riemann curvature tensors?

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Simón Flavio Ibañez
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    This is incorrectly parsed. You should have $f\nabla_XY$ and $fX$ — scalar multiplication pointwise. The third equation also should have no parentheses in either term. If Wikipedia wrote things this way, you should abandon it and read a good textbook. – Ted Shifrin Dec 31 '23 at 07:23
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    Try reading this: https://math.stackexchange.com/questions/354561/question-regarding-the-definition-of-affine-connection?rq=1 – Temoi Dec 31 '23 at 07:23
  • @TedShifrin, i'm sorry, my original idea was to put a link for the wikipedia article, but i forgot to do it. I've edited the post to include it. As for the parentheses... those are actually added by me... but i don't see why that matters since, clearly, $fX=f(X)$ unless the yuxtapostioning of elements is regarded as multiplication. – Simón Flavio Ibañez Dec 31 '23 at 08:04
  • @SimónFlavioIbañez and what does $fX$ mean if not the pointwise scalar multiplication? I guess that one has to first clarify that in order to understand that definition – agt Dec 31 '23 at 08:45
  • Actually in this case the parenthesis change the meaning of the expression. As @TedShifrin has already pointed out $fX$ means pointwise multiplication, while $f(X)$ is the derivative of $f$ along the vector field $X$. So in the definition it should be written without parenthesis. – Temoi Dec 31 '23 at 08:48
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    @MatteoIaccarino No, the derivative of $f$ along $X$ is $X(f)$. – Ted Shifrin Dec 31 '23 at 20:52
  • @TedShifrin yes sorry, you are right! – Temoi Jan 01 '24 at 11:01

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The definition is actually correct (without parenthesis, as already pointed out in the comments). Given $f$ a smooth function on $M$ and $X\in\Gamma(TM)$, you can define $fX\in\Gamma(TM)$ by pointwise multiplication, namely $(fX)(p):=f(p)X(p)$ for any $p\in M$. This is what the $fX$ (and $f\nabla_XY$) in the definition means.

Temoi
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