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I am currently reading a book on supergeometry which also introduces sheaves and ringed spaces, and in particular it proposes a definition of a differential manifold as a locally ringed space $(M,\mathcal{O}_M)$ which is locally isomorphic to the locally ringed space $(\mathbb{R}^n,\mathcal{C}^\infty)$. One thing that I don't understand in this approach is how can we recover the concept of the value of a function at a point $x\in M$. One idea that I had as to how to do this was through the isomorphism.

Let's focus on a sufficiently small neighbourhood $\mathcal{V}$ of point $x$, which by definition is isomorphic to $\mathbb{R}^n$. We have a homeomorphism $\varphi:M\supset\mathcal{V}\rightarrow \mathbb{R}^n$ and a family of isomorphisms $\varphi_\mathcal{U}:\mathcal{C}^\infty(\mathcal{U})\rightarrow\mathcal{O}_M(\varphi^{-1}(\mathcal{U}))$ for all open $\mathcal{U}\subset\mathbb{R}^n$. We could use this to define the value of a section $s\in\mathcal{O}_M(\mathcal{U})$ as $s(x):=\varphi_\mathcal{U}^{-1}(s)(\varphi(x))$. This definition seems logical to me, however one thing which I cannot figure out is whether or not this definition is independent of our choice of isomorphism. The book discusses values of functions on supermanifolds without discussing this detail, so I would imagine this is a simple matter, but I have never worked with sheaves and ringed spaces before so this whole topic is a bit confusing to me.

Konrad Gębik
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Your definition works if by $\varphi$ you mean a chart, but you do have to check that it's independent of a choice of chart. An alternative definition which doesn't require making this choice is to use the stalks of the sheaf of functions $\mathcal{O}_M$. These are defined in terms of a suitable inverse limit and in particular are canonically defined.

At a point $x \in M$ the stalk $\mathcal{O}_{M, x}$ is the ring of germs of smooth functions at $x$. This is a local ring (that's what makes $M$ a locally ringed space) with unique maximal ideal $m_x$, the functions vanishing at $x$. The quotient $\mathcal{O}_{M, x}/m_x$ is called the residue field, which in this case is just $\mathbb{R}$, with the quotient map being given by evaluation at $x$.

Then the value of a section $f \in \mathcal{O}_M(U)$ (where $U$ is any open) at a point $x \in U$ is its image in $\mathcal{O}_{M, x}/m_x \cong \mathbb{R}$. This may seem extremely laborious but the point is that this all generalizes to arbitrary locally ringed spaces, including e.g. schemes; in this setting the local rings and residue fields can look quite different from each other.

Qiaochu Yuan
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  • Okay, I understand this definition but I am still not sure if this uniquely defines those values (as in the real number values). Do we know that there is only one induced isomorphism between $\mathcal{O}_{M,x}/m_x$ and $\mathbb{R}$? Otherwise we could pick another isomorphism which would give us a different real value at $x$ for some sections, right? – Konrad Gębik Jul 28 '24 at 11:41
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    @Konrad: funnily enough I originally had a few sentences about this but I dropped them. This can be addressed in two ways: 1) we can remember the $\mathbb{R}$-algebra structure on this whole sheaf. This gives us an $\mathbb{R}$-algebra structure on the residue field, and the $\mathbb{R}$-algebra structure on $\mathbb{R}$ is unique. 2) Even if we didn't do that, it doesn't matter because $\mathbb{R}$ has no nontrivial automorphisms already as an abstract ring (this is a nice exercise). So the isomorphism to $\mathbb{R}$ is unique! – Qiaochu Yuan Jul 28 '24 at 19:15
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    (The idea of the nice exercise is to argue that any automorphism fixes $\mathbb{Q}$ and also preserves squares, so it preserves non-negativity, so it preserves order, so it's continuous.) – Qiaochu Yuan Jul 28 '24 at 19:16
  • A related nice exercise is that the abstract ring structure on $C^{\infty}(M)$ remembers its $\mathbb{R}$-algebra structure, so this structure is in fact canonical (for example, every ring homomorphism between two such rings is automatically an $\mathbb{R}$-algebra homomorphism). The idea of the proof is similar to the above, you start with the observation that squares are canonically defined. This lets you define a canonical partial order on the ring, then use the canonical copy of $\mathbb{Q}$ to define $\mathbb{R}$ as the completion of $\mathbb{Q}$ wrt this order. – Qiaochu Yuan Jul 29 '24 at 18:05