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I am reading The Rising Sea - Foundations of Algebraic Geometry written by Ravi Vakil, and I'm asking for help because there was a point that didn't touch to me. Maybe it's because I don't have that much knowledge about differential geometry and algebraic geometry: I stated to study algebraic geometry a couple of weeks ago.

For motivating example of sheaf/presheaf, writer suggest the sheaf of smooth function. Vakil let $\mathcal O(U)=\{f:U\to\mathbb R\mid f\mathrm{\,is \,smooth}\}$ be ring of smooth functions defined over open set $U\subset\mathbb R^n$. Then we can define equivalence relation $\sim$ over $O=\{(f,U)\mid p\in U, f\in\mathcal O(U)\}$ when $p\in \mathbb R^n$ fixed as $(f,U_f)\sim(g,U_g)$ if and only if there exists open $p\in W\subset U,V$ such that $f|_W=g|_W$. Then $\mathcal O_p=O/\sim$ is ring, and has unique maximal ideal $\mathfrak{m}_p=\{\overline{(f,U)}\mid f(p)=0\}$ so $\mathcal O_p$ is local.

In Rhetorical question for experts in exercise 2.1.B says:

What goes wrong if the sheaf of continuous functions is substituted for the sheaf of smooth functions? What goes wrong if you use the sheaf of $\mathcal C^1$ functions?

Above definition, I cannot see any error while replacing smoothness with continuity. Definition of $\mathcal O_p$, does not use differential an single second. But however, seeing the author raise these issues has made me think that there might be a problem.

What goes wrong? It would go wrong while handling cotangent space or else: because Vakil made a notice that "$\mathfrak m_p/\mathfrak m_p^2$ is $\mathcal O_p/\mathfrak m_p\cong\mathbb R$-vector space and it turns out to be naturally toe cotangent space to the differentiable or analytic manifold at $p$".

Any help would be graceful. Thank you.

KReiser
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    I guess that 2.1.B in Vakil's notes refers to his "Aside 2.1.2", which - as you have noted - is more than just the definition of a sheaf. It identifies the quotient $\mathfrak{m}/\mathfrak{m}^2$ with the cotangent space. And this is what you cannot always say. – ClemensB Apr 14 '25 at 07:47
  • Oh, understand it to some extent now. Thanks, @ClemensB. Can I bother you one more time? Why does $\mathfrak m/\mathfrak m^2$ become a cotangent space? Is there any reference related to that? I'm reading this article now, but it is slightly hard to understand. – Myungheon Lee Apr 14 '25 at 10:32
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    In this case $m=m^2$. Its a good exercise. I explained this in one of my old answers. https://math.stackexchange.com/questions/4878410/what-does-an-exotic-derivation-at-a-point-x-0-in-mathbbrn-look-like/4878635#4878635 – Moishe Kohan Apr 14 '25 at 20:32

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