topological revisionism

• hey wait don’t yell at me
• first cracks in the cultural wall
• topological headcanons for algebra enjoyers
  • general nonsense
  • ok but that’s not a manifold though
• spicy Lee fanfiction
  • ending the story arc
  • my favorite fanfiction
• coda: how deep does the rabbit hole go?

this past quarter i started my journey of learning ✨algebraic geometry✨! yippee

As of right now I’m planning on taking a year-long sequence following Hartshorne¹ primarily, supplemented by Liu’s Algebraic Geometry and Arithmetic Curves and Vakil’s The Rising Sea. This quarter we started from thinking about varieties cut out of \(k^n\) (for \(k\) characteristic zero, algebraically closed) and eventually made the Big Leap of Abstraction to schemes. I know some people like teaching algebraic geometry by jumping straight into thinking about sheaves and schemes, but I really liked working through the embedded varieties content in Chapter I of Hartshorne. Having worked with concrete examples before looking at Chapter II was super helpful for my visual intuition for schemes. Also, having nice results to replicate from Chapter I made me more invested in the lemmas we were proving when we got to more abstract stuff. Had I jumped straight into schemes first, I think I would have really struggled with both of these things, so I really appreciated our class’ emphasis on the classical viewpoint. Making the leap to schemes is really difficult because of all of the abstract machinery needed, and having seen reflections of some of the geometrical ideas with an embedded variety beforehand was nice.

I started learning algebraic geometry having done a good amount of differential topology but with a pretty weak commutative algebra background², so see below for my live reaction to learning that schemes are spaces that are “locally isomorphic to \(\spec\) of a ring”:

This mindset tends to do really well when dealing with manifolds! Working with charts and coordinates in \(\RR^n\) is nice since one can write down functions explicitly, or use matrices and linear algebra when talking about differentials and tangent spaces. So when you have one trick and every scheme looks like a manifold waiting to be hit by it, I really don’t think it’s necessarily a bad idea!

Unfortunately, I felt like I got burned by this because now instead of doing calculus and linear algebra, the right tool to use is now commutative algebra… which I’m currently bad at. And instead of being able to do all the analysis shenanigans I want in \(\RR^n\), now we’re dealing with prime ideals of a general ring and the Zariski topology, and working with \(\spec R\) directly is not as good of a time. I had to learn the hard way that thinking about the sheaf of sections and its stalks are more lucrative for schemes because of their nice descriptions in terms of localizations in the affine case. In his lead-up to talking about schemes in Chapter 3 of The Rising Sea, Vakil hints that this is a more useful way to think about a geometric space in general:

… The key insight behind this generalization from the notion of something like a manifold to a more versatile notion of a “geometric space” is the following: we can understand a geometric space (such as a manifold) well by understanding the functions on this space. More precisely, we will understand it through the sheaf of functions on the space.

To be honest, though, I found this a bit counterintuitive because while I’ve worked with \(C^\infty(M)\) on a manifold before as well as germs at a point \(C^\infty_{M, p}\) for a point \(p \in M\), these rings don’t seem relatively nice³. There are tons of zero-divisors (just think about functions making up of a partition of unity, for instance) and it doesn’t seem clear to me that this ring of functions has enough structure in it to tell you much directly about the about topological structure of the manifold.

But then I saw the following tweet from algebraic geometer Barbara Fantechi:

… I have told my students for decades that 1) one can work with embedded or abstract manifolds and varieties/schemes 2) they all can be described as space plus (equivalence chart of) atlas, or as locally ringed space, or by their functor of points. … Getting rid of such cultural barriers would make communication between algebraic and differential geometry MUCH easier!

Generally, it feels like the communication tends to go from differential geometry to algebraic geometry, which I think is really helpful in the learning process. A lot of algebraic geometry adjectives are described in analogy to certain concepts from manifolds, and having nice topological intuition when working with the Zariski topology is a good thing. But let’s try to go the other way now and see what it would take to impose the language of algebraic geometry onto differential geometry! This at first glance seems to be doable, since manifolds are a much more specialized kind of object in comparison to locally ringed spaces and schemes. But is there really a way to think about a manifold as a locally ringed space instead of as a space with an atlas?

hey wait don’t yell at me

Before we start, if you’re really looking to nitpick definitions in differential geometry or if you need a refresher on what manifold things mean, go ahead and expand this so you know what I’m talking about when I say various words:

the definitions that i’m choosing to use thanks

Let’s recall what a standard definition of a (smooth) manifold looks like:

Definition. A smooth manifold is a pair \((M, \mc A)\) where \(M\) is a second-countable Hausdorff topological space, and \(\mc A\) is a maximal atlas of charts. A chart is a map \(\vphi_i\) from \(U_i \subseteq M\) open to \(\RR^n\) that restricts to a homeomorphism on its image, and an atlas is a collection of charts that are compatible, so that the transition maps \(\vphi_j \circ \vphi_i^{-1} : \vphi_i(U_i \cap U_j) \to \vphi_j(U_i \cap U_j)\) is smooth as functions on Euclidean spaces.

I will very often drop “smooth” as an adjective for manifold. We also might as well define smooth maps between them too, just to fill out the morphisms in our category:

Definition. A smooth map between two manifolds \(M\) and \(N\) is a continuous map \(f : M \to N\) such that for every pair of charts \(\vphi : U \to \RR^m\) of \(M\) and \(\psi : V \to \RR^n\) of \(N\) with \(f(U) \subseteq V\), the map \(\psi \circ f \circ \vphi^{-1} : \vphi(U) \to \psi(V)\) is smooth.

Note that in both definitions, we have to deal with weird domain restriction business so the relevant charts overlap in the right places. We also see that “smoothness” of both of these definitions also essentially comes inherited from the smoothness of maps between Euclidean spaces, so we should expect any new definitions to do the same.

Some sources don’t require manifolds to be second-countable and Hausdorff, but I do because I want my partitions of unity. (Apologies to long line and double-origin enthusiasts.) Just as a reminder:

Definition. A partition of unity subordinate to an open cover \(\set{U_i}_{i \in I}\) of \(M\) is a collection of smooth real-valued functions \(\set{\rho_i}_{i \in I}\) such that (1) \(\rho_i \geq 0\), with \(\supp \rho_i \subseteq U_i\), (2) \(\set{\supp \rho_i}_{i \in I}\) is locally finite, i.e. any point is only contained in finitely many of these sets, and (3) \(\sum_{i \in I} \rho_i = 1_M\).

With the second-countable/Hausdorff requirement for a manifold, any open cover of \(M\) has a partition of unity suboordinate to it. In particular, in the construction of this partition of unity, one can make \(I\) countable and the supports \(\supp \rho_i\) all compact. This feels like it’s a little restrictive, but this is actually ok since manifolds are now paracompact and \(\sigma\)-compact (open covers have locally finite refinements/is union of countably many compact subsets). This will be useful later.

As a final remark, one can generalize everything to be a \(C^\alpha\)-manifold where the transition maps and morphisms are \(C^\alpha\) functions on Euclidean spaces instead of smooth ones, but today we generally don’t care about any values of \(\alpha\) other than \(\alpha = \infty\) and maybe \(\alpha = 0\). It would be interesting to think about how the below arguments change if you allow less smooth/more smooth (i.e. real analytic/\(C^\omega\)) functions instead.

…

are we all on the same page about what a manifold is? okay great let’s get to translating!

first cracks in the cultural wall

this translation idea is not an uncommon one, and more or less appears in Vakil’s The Rising Sea as Definition 4.3.9. We present here instead a slightly more stringent series of definitions and arguments coming from Torsten Wedhorn’s Manifolds, Sheaves, and Cohomology, Sections 4.1-4.2.

Definition. A smooth manifold is a pair \((M, \cOO_M)\) where \(M\) is a second-countable Hausdorff topological space and \(\cOO_M\) is a sheaf of local \(\RR\)-algebras, such that any point of \(M\) has an open neighborhood \(U\) with \((U, \cOO_M\vert_U)\) isomorphic to \((X, C^\infty_X)\) as a locally ringed space, where \(X \subseteq \RR^n\) is an open set and \(C^\infty_X\) is the sheaf of real-valued smooth functions on \(X\).

Vakil only requires that we have rings, but I think it’s not too hard to convince ourselves that we probably should elevate our ringed space to be one with a sheaf of \(\RR\)-algebras, since \(C^\infty(X)\) for \(X \subseteq \RR^n\) open is an \(\RR\)-algebra. As such, Wedhorn calls \(M\) a locally \(\RR\)-ringed space, since our sheaf is valued in the category of \(\RR\)-algebras instead of just rings, which I think is a sensible naming convention.

Note that this definition doesn’t actually tell us what the sheaf of \(\RR\)-algebras on \(M\) is, just that it’s locally isomorphic to the sheaf of real-valued smooth functions on \(\RR^n\). However, we’d like to say in general that our sheaf on \(M\) is just the sheaf of smooth real-valued functions on \(M\), as we might expect (Proposition 4.17). To help us, let’s also look at what smooth maps of manifolds should be now:

Definition. A morphism (smooth map) of manifolds \(f : M \to N\) is a morphism as locally ringed spaces.

To characterize our given sheaf \(\cOO_M\), let’s define the sheaf \(C^\infty_M\), where \(C^\infty_M(U)\) is the set of morphisms (smooth maps) \(h : U \to \RR\) for \(U \subseteq M\) open, and \(\RR\) having its sheaf of real-valued smooth functions \(C^\infty_\RR\). The underlying continuous maps \(U \to \RR\) have the structure of an \(\RR\)-algebra via pointwise addition/multiplication/scaling. The corresponding maps on sheaves \(h^\sharp (Y): C^\infty_\RR (Y) \to \cOO_M\vert_U(h^{-1}(Y))\) for any \(Y \subseteq \RR\) open have the structure of an \(\RR\)-algebra as well, since \(\cOO_M\vert_U\) is a \(\RR\)-algebra-valued sheaf on \(U\) and so one can naturally define for any morphisms \(h_1, h_2\) and a real \(c \in \RR\) the sheaf maps \((h_1 + h_2)^\sharp\), \((h_1 h_2)^\sharp\), and \((c h_1)^\sharp\) in terms of \(h_1^\sharp\) and \(h_2^\sharp\). Therefore, \(C^\infty_M\) is a sheaf of \(\RR\)-algebras on \(M\) as well.

Now we construct a morphism of sheaves \(\iota : C^\infty_M \to \cOO_M\) in the following way. Take a morphism \(h : U \to \RR\) in \(C^\infty_M(U)\) and consider the map on sheaves \(h^\sharp : C^\infty_\RR \to h_* (\cOO_M\vert_U)\). Then in \(C^\infty_\RR(\RR)\), we have the smooth function \(\id_\RR\), which under \(h^\sharp(\RR)\) gives an element \(h^\sharp(\RR)(\id_\RR) \in \cOO_M\vert_U(U) = \cOO_M(U)\). Then \(\iota(U)\) should be defined to send \(h \mapsto h^\sharp(\RR)(\id_\RR)\) for any \(U \subseteq M\) open. A direct computation shows

this is indeed a morphism :)

Take any \(V \subseteq U\) open. It suffices to verify that \(\iota(V) \circ \rho^U_{C^\infty_M, V} = \rho^U_{\cOO_M, V} \circ \iota(U)\), where the \(\rho^U_{\mc F, V}\)s are the appropriate restriction maps on each sheaf \(\mc F\) from \(U\) to \(V\). Taking an arbitrary \(h \in C^\infty_M(U)\), the left-hand side evaluates to \(\iota(V) (h \circ i) = (h \circ i)^\sharp(\RR)(\id_\RR)\), where \(i : V \to U\) is the natural inclusion. This decomposes further as \(i^\sharp(h^{-1}(\RR)) \circ h^\sharp(\RR)(\id_\RR)\), but this is now precisely the right-hand side, as the sheaf map \(i^\sharp\) is exactly the restriction map on the sheaf \(\cOO_M\), as desired.

…

Now, to show this is an isomorphism, we can check this locally. In particular, it suffices to check this on some \(U \subseteq M\) isomorphic to an open set \(X \subseteq \RR^n\). But then if we take some \(g : X \to \RR\) a morphism, then \(g^\sharp(\RR)(\id_\RR) = \id_\RR \circ g = g\), so this is obviously an isomorphism. This means that the sheaf we defined on \(M\) in general is always secretly just its sheaf of smooth functions!

Here, we used the fact that if we have \(X \subseteq \RR^n\) and \(Y \subseteq \RR^m\) open, for any morphism \(f : X \to Y\) viewed as locally ringed spaces with sheaves of smooth functions, the sheaf map \(f^\sharp\) must be given by precomposing by \(f\) (Example 4.5). Consider the following commutative diagram for any \(V \subseteq Y\) open:

We can see this with the following argument. Note that the map on the stalks \(f^\sharp_x : C^\infty_{Y, f(x)} \to C^\infty_{X, x}\) for any \(x \in X\) is constructed so that it commutes with the sheaf map \(f^\sharp\) via localization. This map then projects down to the identity map on the residue fields since it’s a nonzero map of \(\RR\)-algebras. Then, for any \(h : V \to \RR\) smooth in \(C^\infty_Y(V)\) for \(V\) open, \(f^\sharp(V)(h)(x) = h(f(x))\) along the outside of the commutative diagram, so any morphism is given by precomposition, as desired.

This allows us to immediately observe that a smooth map of manifolds \(f : M \to N\) is just given by a continuous map \(f : M \to N\) such that for any \(V \subseteq N\) open and section \(t \in \cOO_N(V)\), then \(t \circ f \in \cOO_M(f^{-1}(V))\) (Proposition 4.18). To see that \(f^\sharp\) is given by pre-composition by \(f\), note that it suffices to check this on any stalk, but then passing to the Euclidean case gives the desired result. Conversely, any continuous map satisfying this condition gives a map of sheaves, since we know that \(\cOO_M\) and \(\cOO_N\) are just the sheaves of smooth maps to \(\RR\) anyway. Note that the same result holds for a map of smooth manifolds defined on charts just by pulling back charts on \(N\) to \(M\), so this indicates that we’re onto something equivalent here!

In particular, we can now recover an atlas on \(M\) whose sheaf of smooth real-valued functions is the same as \(\mc O_M = C^\infty_M\)! We can just consider the isomorphisms \(\vphi_i : (U_i, \cOO_M\vert_{U_i}) \to (X_i, C^\infty_{X_i})\) for \(X_i \subseteq \RR^{n_i}\) which induce homeomorphisms \(\vphi_i : U_i \to \RR^n\). Checking these are compatible follows essentially immediately from the fact that \(\vphi_i^{-1}\) is a morphism, and our characterization of morphisms above shows that this gives an atlas. Using these charts to define a sheaf of smooth real-valued functions on \(M\) recovers the original sheaf, so these constructions are in fact equivalent. This shows that we can interpret the category of manifolds as a subcategory of the category of locally ringed spaces! Hooray!

I think we can be already be happy with this reframing of a manifold, but I also don’t think this was super impressive. We more or less already had our charts still sitting in our definition, and all we did was did a little trickery to spotlight the sheaf of smooth functions instead. Can we instead do this more algebraically? That is, can we instead take the algebraic structure of \(C^\infty(M)\) itself instead as the “smooth structure” on \(M\) and use it to tell us things about \(M\)?

topological headcanons for algebra enjoyers

Here things are going to get a little wacky. Rings of real-valued continuous functions \(X \to \RR\) for a general topological space \(X\) can be really strange, so it’s entirely possible that rings of smooth functions might be pretty weird as well. But, as it turns out, these rings of functions hold a lot of information about the space contained in their maximal ideals if \(X\) is “nice enough.” In particular, one can can recover the points of \(X\) and the topology on \(X\), and so this might be another way to get at a manifold without even needing the underlying topological space!

general nonsense

This is actually an exercise (Problem 1.26) in Atiyah-Macdonald⁴. Suppose \(X\) is a compact Hausdorff space, and let \(C(X)\) (what one sometimes might call \(C^0(X)\)) be its ring of real-valued continuous functions on \(X\), under pointwise addition/multiplication. (Before you yell at me again, it’s got a multiplicative unit, since the constant function \(1_X\) that evaluates to \(1\) for every point in \(X\) is continuous.)

For every point \(x \in X\), we have an ideal of functions \(\mfk m_x = \set{ f \in C(X) : f(x) = 0}\) vanishing at \(x\) (it’s not too hard to see this is an ideal). Note that the evaluation map at a point \(x\), \(\ev_x : C(X) \to \RR\) sending \(f \mapsto f(x)\) is a surjective ring homomorphism, and its kernel is exactly \(\mfk m_x\) by definition. So, by the First Isomorphism Theorem, \(C(X) / \mfk m_x \iso \RR\) so \(\mfk m_x\) is in fact maximal! That’s pretty neat and will become relevant in a second.

So far nothing too strange has happened yet, so let’s remember a little bit of algebraic geometry now. Recall that the maximal ideals of \(C(X)\) are points in \(\spec C(X)\), so we might want to consider the subspace of maximal ideals in \(\spec C(X)\). Let’s call this the maximal spectrum of \(C(X)\), \(\maxspec C(X)\) (and similarly for any ring \(A\)). Since this is a subset of \(\spec C(X)\), we can induce the Zariski topology on this subset, so \(\maxspec C(X)\) is a topological space now. And would you believe it, the mapping \(\mu\) from \(X \to \maxspec C(X)\) takind \(x \mapsto \mfk m_x\) turns out not only to be a continuous function between topological spaces, it’s a homeomorphism!

I think this is really wacky, because the Zariski topology is famously not even Hausdorff. But if this claim is true, this means that sitting inside \(\spec C(X)\), we have a compact Hausdorff space, which is a normal (\(T_4\)) space. Recall that normal spaces are really nice spaces that have tons of nice point-set topological properties. For instance, if a normal space \(X\) happens to also be second-countable, \(X\) now admits a metric by Urysohn’s Metrization Theorem. “Normal” is the adjective I therefore ascribe in my brain to “nice” topological spaces I see in analysis compared to the ones I see in a general point-set topology setting, and thinking about doing analysis to a Zariski thing feels wrong in a way that I can’t really articulate.

Anyways, we now we have three things to check to show this claim – we have to check that \(\mu\) is surjective, injective, and continuous with a continuous inverse.

We’ll do injectivity quickly first. For any two distinct points \(x, y \in X\), \(x\) and \(y\) are both closed points since \(X\) is Hausdorff, and so by Urysohn’s Lemma (applied to a compact Hausdorff and hence normal space) we know that there is a continuous function \(f : X \to [0, 1]\) with \(f(x) = 0\) and \(f(y) = 1\). This function \(f\) lies in \(\mfk m_x\) but not \(\mfk m_y\), so we see that \(\mfk m_x \neq \mfk m_y\) and hence \(\mu\) is injective.

Surjectivity is the trickiest part of the argument, but we’ll do it next. Letting \(\mfk m\) be any maximal ideal of \(C(X)\), we have to realize it as \(\mfk m_x\) for some \(x \in X\). To do this, let’s consider the set \(V = Z(\mfk m) = \set{x \in X : f(x) = 0, f \in \mfk m}\), i.e. the set of common zeroes of the functions in \(\mfk m\). The idea is now that \(V\) has to be non-empty, so we can pick out some \(x \in V\). But then \(\mfk m_x \subseteq \mfk m\), and since \(\mfk m_x\) was maximal, then \(\mfk m_x = \mfk m\), as desired. So we just have to make sure that we can always pick a point out of \(V\).

Suppose for the sake of contradiction that \(V\) were empty. If this is the case, then we will be able to find a unit \(g \in \mfk m\), which would be a contradiction because then \(\mfk m = C(X)\). First, for any \(f \in C(X)\), define the sets \(U_f = \set{x \in X : f(x) \neq 0}\). These are open since \(U_f = f^{-1}(\RR - \set 0)\), and they’re the preimage of an open set under a continuous map. We use these to form an open cover of \(X\). For any \(x \in X\), we can find some \(f_x \in \mfk m\) such that \(f_x(x) \neq 0\), so \(U_{f_x}\) is an open set containing \(x\). Clearly, \(X = \bigcup_{x \in X} U_{f_x}\), but then since \(X\) is compact, we can write \(X = \bigcup_{i=1}^n U_{f_{x_i}}\) for finitely many \(x_i \in X\). On this set, the function \(g = \sum_{i=1}^n f_{x_i}^2\) is non-negative and each \(x \in X\) is contained in some \(U_{f_{x_i}}\), on which \(f_{x_i}^2(x) > 0\). This is a unit since \(g > 0\) everywhere on \(X\), and moreover, since each \(f_{x_i} \in \mfk m\), \(g \in \mfk m\). This yields the desired contradiction, so \(V\) could not have been empty.

Finally, we have to show that \(\mu\) is continuous with continuous inverse. First, note that the \(U_f\)s form a basis for the topology on \(X\). We can do this using Urysohn’s Lemma again. For any \(U \subseteq X\) open and \(x \in U\), note that \(X - U\) is closed. Then, use Urysohn’s Lemma to construct a continuous function \(f_x\) such that \(f_x(x) = 1\) and \(f_x(X - U) = 0\). This makes \(U_{f_x} \subseteq U\) and contain \(x \in U\), so then \(U = \bigcup_{x \in U} U_{f_x}\) and therefore \(U\) is a union of these basic open sets.

Now, recall in the Zariski topology on \(\maxspec C(X)\), for any \(f \in C(X)\), the set \(D_f = \set{\mfk m \in \maxspec X : f \not \in \mfk m}\) is a basic open set by construction. Now, we check that \(\mu(U_f) = D_f\) for any \(f \in C(X)\). This follows since if \(x \in U_f\), \(f(x) \neq 0\). Then, \(\mfk m_x = \set{g \in C(X) : g(x) = 0}\), so \(f \not \in \mfk m_x\). This shows that \(\mu(x) = \mfk m_x \in D_f\). Conversely, given that \(\mu\) is surjective, write \(\mfk m = \mfk m_x \in D_f\) for some \(x \in X\). Since \(f \not \in \mfk m_x\), \(f(x) \neq 0\), so then \(x \in U_f\). This means that \(\mfk m = \mfk m_x \in \mu(U_f)\), which shows the sets are equal. Now, since \(\mu\) is a bijection, the preimage of any basic open set \(D_f\) under \(\mu\) is \(U_f\) open, so \(\mu\) is continuous. Moreover, \(\mu\) is clearly open-mapping since we just checked this on a basis, so \(\mu\) is a homeomorphism, as desired.

Shockingly, this means that \(C(X)\) completely determines the underlying space \(X\) if \(X\) is compact Hausdorff! Since \(X\) is homeomorphic to \(\maxspec C(X)\), if \(C(X) \iso C(Y)\) for some other compact Hausdorff space \(Y\), then necessarily \(X \iso Y\).

This is pretty cool! Note that we technically didn’t need the full power of Urysohn’s Lemma in this argument – we just needed that \(X\) was completely regular, i.e. has the property that points in \(X\) can be separated from closed sets by continuous real-valued functions. In particular, the comments of this stackexchange answer indicate that one can again reconstruct the topology on \(X\) if \(X\) is just Hausdorff and completely regular (i.e. Tychonoff). The thread is a little confusing on whether one can actually determine \(X\) uniquely up to isomorphism, but the Wikipedia page for Tychonoff spaces indicates that you probably want \(X\) to be something called realcompact also to make this true. In particular, in this case, you actually get the stronger result that if \(X\) and \(Y\) are spaces that are both Tychonoff and realcompact, then there is a bijection between \(\hom(X,Y)\) and \(\hom(C(Y), C(X))\). In particular, there is an anti-equivalence of categories between the category of these spaces and the corresponding category of rings of continuous functions on these spaces. It’s a little unclear to me whether or not this is stronger/weaker than being compact Hausdorff, but I’m no general topologist…

ok but that’s not a manifold though

fine. We can pursue the above further, but the main point of the exercise above is to indicate that something similar might also work in the case of a smooth manifold, but now we have smooth functions which are also probably nicer. Let’s specialize to the case of a smooth manifold \(M\) and run a similar argument to show that one can recover \(M\) from \(C^\infty(M)\) for any manifold \(M\). The following outline for the argument comes as an exercise from Characteristic Classes by Milnor and Stasheff (Problem 1C) and these stackexchange posts.

The argument proceeds very similarly. Note that again that \(C^\infty(M)\) is an \(\RR\)-algebra with a unit \(1_M\), the constant function on \(M\). Here, we will want to consider ring homomorphisms \(\vphi : C^\infty(M) \to \RR\), and we claim that \(\vphi\) extends to an \(\RR\)-algebra homomorphism. To do this, we need to establish a few properties about any such ring homomorphism \(\vphi\).

First, for an arbitrary function \(f \in C^\infty(M)\), if \(f > 0\), then \(\vphi(f) > 0\). \(\vphi(f) \neq 0\) since \(\frac 1f \in C^\infty(M)\) as \(f\) vanishes nowhere, so \(\vphi(f)\) is a unit in \(\RR\), i.e. not zero. Then, \(\sqrt f \in C^\infty (M)\), so \(\vphi(f) = \vphi(\sqrt f)^2 \geq 0\), and therefore \(\vphi(f) > 0\) since it is not zero. Next, from the fact that \(\vphi(1_M) = 1\), we can get \(\vphi(q 1_M) = q\) for any \(q \in \QQ\). Then, \(\vphi\) extends to a \(\QQ\)-algebra homomorphism, since

\[\vphi(q f) = \vphi(q 1_M \cdot f) = \vphi(q 1_M) \vphi(f) = q \vphi(f).\]

To boost this to an \(\RR\)-algebra homomorphism, we must show that \(\vphi\) preserves order on the copy of \(\RR \iso \RR 1_M \subseteq C^\infty(M)\). To see this, note that for any \(a > b \in \RR\), \((a-b) 1_M > 0\), so \(\vphi(a 1_M - b 1_M) > 0\) and therefore \(\vphi(a 1_M) > \vphi(b 1_M)\). If WLOG for some \(r \in \RR\) we had that \(\vphi(r 1_M) > r\), find some \(q \in \QQ\) with \(r < q < \vphi(r)\). Then \(\vphi(q1_M - r1_M) > 0\), but this is also equal to \(\vphi(q1_M) - \vphi(r1_M) = q - \vphi(r1_M) < 0\), contradiction. A similar argument shows that \(\vphi(r1_M)\) is not less than \(r\) either, so it must be equal to \(r\). Now the same argument as above extending \(\vphi\) to a \(\QQ\)-algebra homomorphism allows us to extend \(\vphi\) to an \(\RR\)-algebra homomorphism. As such, ring homomorphisms and \(\RR\)-algebra homomorphisms \(C^\infty(M) \to \RR\) are equivalent, which will be nice to allow us to conflate the two as needed.

Now, similar to the above, define an \(\RR\)-algebra homomorphism \(\ev_x : C^\infty(M) \to \RR\) for any \(x \in M\) that just evaluates a function at \(x\), i.e. \(f \mapsto f(x)\). As above, the set \(\mfk m_x = \set{f \in C^\infty(M) : f(x) = 0}\) is an ideal and is equal to \(\ker \ev_x\), and since \(\ev_x\) is surjective, \(C^\infty(M) / \mfk m_x \iso \RR\) and therefore \(\mfk m_x\) is maximal. Now, in this case, we show that every ring homomorphism \(\vphi : C^\infty(M) \to \RR\) is given by an evaluation map, using a method similar to the above.

Suppose that some homomorphism \(\vphi\) is not given by an evaluation map \(\ev_x\) for some \(x \in M\). Then, for every \(x \in M\), there is some function \(f_x \in C^\infty(M)\) with \(\vphi(f_x) \neq f_x(x)\). If we are given a compact set \(K \subseteq M\), we will show that we can construct some element \(F_K \in C^\infty(M)\) where \(\vphi(F_K) = 0\), but \(F_K \geq 0\) everywhere and is guaranteed to satisfy \(F_K > 0\) on \(K\). To do this, note that \(F_x = (f_x - \vphi(f_x) 1_M)^2\) for any \(x \in M\) gives us a start at this. Observe

\[\vphi(F_x) = \vphi((f_x - \vphi(f_x) 1_M)^2) = \vphi(f_x - \vphi(f_x) 1_M)^2 = (\vphi(f_x) - \vphi(f_x))^2 = 0\]

and clearly \(F_x \geq 0\). Moreover, at \(x\), \(F_x(x) = (f_x - \vphi(f_x) 1_M)^2(x) = (f_x(x) - \vphi(f_x))^2 > 0\). Then, take the open sets \(U_x = F_x^{-1}(0, \infty)\), each of which contains \(x\) by constuction. Use these open sets to form an open cover of \(K\) using by \(K = \bigcup_{x \in K}U_x\), but now since \(K\) is compact, \(K = \bigcup_{i=1}^n U_{x_i}\) for finitely many \(x_i \in K\). Then \(F_K = \sum_{i=1}^n F_{x_i}\) works. Note \(\vphi(F_K) = \sum_{i=1}^n \vphi(F_{x_i}) = 0\), and \(F_K\) is non-negative since it is a sum of non-negative functions. Finally, since for any \(x \in K\), \(x \in U_{x_i}\) for some \(i\), \(F_{x_i}(x) > 0\) for this \(i\), and so \(F_K(x) > 0\) since each \(F_{x_i}\) is at least non-negative.

But how do we get such a \(K\) to work with? Since \(M\) is no longer compact, we need to use a partition of unity of \(M\) to concoct a similar argument as in the compact case using finiteness. In particular, we will going to pick a partition of unity indexed by a countable set \(\set{\rho_i}_{i=1}^\infty\), with each \(\supp \rho_i\) compact.⁵ Then, let’s consider the function \(h = \sum_{i=1}^\infty i \rho_i\). This function is well-defined since \(\set{\supp \rho_i}_{i=1}^\infty\) is locally finite, and so in any neighborhood of a point \(x \in M\), \(h\) is locally a finite sum of smooth functions, so \(h\) is smooth and an element of \(C^\infty(M)\). Moreover, \(h\) will give us a compact subset of \(M\) by observing that \(h^{-1}([0, c])\) for any \(c \in \RR\) will be compact. In particular, \(h^{-1}([0, c]) \subseteq \bigcup_{i=1}^{\floor c} \supp \rho_i\), so it’ll be compact as it is a closed subset of a compact set. To see this, if a point \(x \not \in \bigcup_{i=1}^{\floor c} \supp \rho_i\), then

\[h(x) = \sum_{i=1}^\infty i \rho_i(x) = \sum_{i = \floor c + 1} i \rho_i(x) \geq (\floor c + 1) \sum_{i=1}^\infty \rho_i(x) = (\floor c + 1).\]

so \(x \not \in h^{-1}([0,c])\). Such an \(h\) constructed in this way will soon be critical in building our contradiction.

Our contradiction, as in the previous case, will be the construction of a badly-behaved smooth function that will again be a unit but vanishes under \(\vphi\). First, let \(h\) be as obtained above from the partition of unity and note that \(h > 1_M\), so it is non-negative. Let \(c = \vphi(h)\). Then, \(K = h^{-1}([0,c])\) is compact, and we can obtain an \(F_K\) non-negative everywhere with \(\vphi(F_K) = 0\) and \(F_K > 0\) on \(K\). In particular, since \(K\) is compact, \(F_K\) achieves some minimum value \(m > 0\) on \(K\). Then, \((\frac 1m + 1) F_K > 1_M\) on \(K\) by construction, is still non-negative, and still satisfies \(\vphi((\frac 1m + 1) F_K) = 0\). Now, consider the function \(g = h + c(\frac 1m + 1) F_K - c1_M\). First, note that \(g\) is strictly positive everywhere. For a point \(x \in K\), since \(h > 0\) and \(c(\frac 1m + 1) F_K - c 1_M > c1_M - c1_M = 0\) by construction, \(g(x) > 0\). If \(x \not \in K = h^{-1}([0, c])\), then \(h(x) > c\) since \(h(x) > 0\), so \(h(x) - c > 0\). Then, since \(c(\frac 1m +1)F_K > 0\), \(g(x) > 0\) again, so \(g\) is strictly positive everywhere. As we proved at the very start of this section, this means that \(g\) should be a unit. However,

\[\vphi(g) = \vphi(h) + \vphi\left(c \left(\frac 1m + 1\right) F_K\right) - \vphi(c 1_M) = c + 0 - c = 0,\]

contradiction. Hooray!

Wait, what did this prove again? We’ve now shown that every ring homomorphism \(C^\infty(M) \to \RR\) is given by \(\ev_x\) for some \(x \in M\). This means that \(\hat M = \hom_{Ring}(C^\infty(M), \RR)\) is in bijection with the points of \(M\) (again, using a smooth version of the Urysohn Lemma with bump functions, sending a point \(x \in M\) to \(\ev_x\) is injective). This gives us a lot of hope for now being able to reconstruct the manifold’s topological/smooth structure, since we’ve now been able to algebraically describe the manifold’s points.

While our strategy was pretty similar as to the case of \(C(X)\), it seems like our characterization of points is now no longer in terms of nice (in the last case, maximal) ideals, and is instead in terms of ring/\(\RR\)-algebra homomorphisms \(C^\infty(M) \to \RR\). However, we can canonically force ourselves to think in terms of ideals if we want to. Since all such \(\RR\)-algebra homomorphisms are surjective, the kernel of any such homomorphism \(\vphi : C^\infty(M) \to \RR\) is again maximal, since \(C^\infty(M) / \ker \vphi \iso \RR\). This at least gives a partial characterization of the points of \(M\) in terms of the ideals of \(C^\infty(M)\).

spicy Lee fanfiction

It feels like that if we really wanted to do algebraic geometry to a manifold, we should treat the points of a manifold described above as a particular subset of the maximal ideals of its ring of smooth functions. Unfortunately, there is a problem with this – maximal ideals don’t behave very functorially, since the preimage of a maximal ideal under a ring homomorphism need not be maximal. Regardless, the idea that points of a manifold correspond to \(\RR\)-algebra homomorphisms to \(\RR\), when dualized to spectra of rings, is kind of reminiscent of the idea of an \(\RR\)-point on a scheme! This will end up being the right idea to pursue if we want to think about a manifold purely in terms of algebra.

ending the story arc

But first, we should finish our algebraic reconstruction of a manifold \(M\) – we’ve so far been able to characterize the points in terms of \(C^\infty(M)\), but \(M\) also has a topology and a smooth structure (either of the sheaf-of-smooth-functions variety or of the atlas variety) that we still need to reconstruct. How do we get this from \(C^\infty(M)\)?

Since the smooth real-valued functions on \(M\) are the elements of \(C^\infty(M)\), and we are currently thinking of the points of \(\hat M = \hom(C^\infty(M), \RR)\) as the points of \(M\), we should again use \(C^\infty(M)\) to tell us topological information about \(M\). In particular, we need to think about a way to use an element \(f \in C^\infty(M)\) to define a map \(\hat M \to \RR\). Well, there’s kind of an obvious construction of such a map \(\hat f\) for any \(f \in C^\infty(M)\), just by taking \(\hat f (\ev_x) = f(x)\). We want to eventually think of \(\hat f\) as smooth, so surely we should at least make it continuous! Then, the natural way to define a topology on \(\hat M\) is to just pick the “coarsest one making the \(\hat f\)s continuous.”

Now, if you’re familiar with functional analysis, this might sound familiar. Thinking of \(C^\infty(M)\) and \(\hat M\) as \(\RR\)-vector spaces, think about the canonical bilinear pairing of \(\RR\)-vector spaces \(\ev : \hat M \times C^\infty(M) \to \RR\) by \(\ev(\ev_x, f) = f(x)\). Now, the topology we just defined on \(\hat M\) is just the weak topology!⁶ hooray functional analysis!

Does this topology on \(\hat M\) make it homeomorphic to \(M\) though? We’ve seen the map \(\mu : M \to \hat M\) sending \(x \mapsto \ev_x\) is a bijection, but we need to check that it is continuous and closed-mapping. We know the only open sets in the topology on \(\hat M\) are the exactly the ones needed to make any map \(\hat f\) continuous, so in particular the open sets are of the form \((\hat f)^{-1}(U)\) for some \(U \subseteq \RR\). We now want to check that \(\mu^{-1}((\hat f)^{-1}(U))\) is open in \(M\). Note that this preimage is really just \((\hat f \circ \mu)^{-1}(U)\), and wait, the map \(\hat f \circ \mu\) sends a point \(x \in M\) to \(f(x)\), so \(\hat f \circ \mu = f\). But \(f\) is continuous since it’s smooth, so \(\mu\) is continuous.

The only other thing we need to see is that \(\mu\) sends closed sets to closed sets. Suppose \(K\) is a proper closed subset of \(M\). Using the smooth version of Urysohn’s Lemma again, we can construct a smooth function \(f : M \to [0, 1]\) such that \(f\vert_K = 0\) (and \(f(x) = 1\) for some point \(x \not \in K\)). Now, \(\mu(K) = \set{ \ev_x : x \in K}\), and by the construction of this \(f\), \(f(K) = (\hat f \circ \mu)(K) = \set 0\), so \(\mu(K) = (\hat f)^{-1}(\set 0)\). Since \(\hat f\) is continuous, \(\mu(K)\) is closed, so \(\mu\) is now also closed-mapping. Since \(\mu\) is a bijection, \(\mu\) is also open-mapping, so its inverse is continuous. Therefore, \(\mu\) is a homeomorphism!

Finally, we need to reconstruct the smooth structure on \(\hat M\), which we now identify with \(M\). Since we literally have the smooth global sections \(C^\infty(M)\) as given, it seems like the right thing to do is to just define a sheaf of real-valued smooth functions on \(\hat M\) isomorphic to the sheaf of real-valued smooth functions on \(M\) under mapping by \(\mu\). We will do this by first constructing a suitable presheaf \(\mc S\) on \(\hat M = M\). (We’re only doing this identification in order to be able to compare sheaves on the same base space. However, throughout the following we are going to be super careful to make sure that anything we construct will be purely in terms of things we’ve constructed already, so some of our maps will look a little strange in order for them to actually make sense on the level of sets.)

First, we already have a candidate for the ring of global sections on \(\hat M\) – we can consider the \(\RR\)-algebra that we denote by \(\mc S(M) = \set{\hat f : f \in C^\infty(M)}\). In order to have any hope at making \(\mc S\) produce the right sheaf, we have to at least have \(C^\infty(M) \iso \mc S(M)\). Well, we at least have a natural surjective homomorphism \(\mu^\sharp : C^\infty(M) \to \mc S(M)\) sending \(f \mapsto \hat f\). This map is also injective – if \(f \in \ker \mu^\sharp\), then \(\hat f (\ev_x) = 0\) for any \(x \in M\), but then this implies that \(f(x) = 0\), so \(f = 0\). Therefore, \(\mu^\sharp\) is an isomorphism.

Now, we can define \(\mc S(U)\) for any \(U \subseteq M\) in the obvious way. Using the inclusion \(i_U : \mu(U) \to \hat M\) corresponding to the inclusion \(U \to M\), take \(\mc S(U) = \set{ \hat f \circ i_U : \hat f \in \mc S(\hat M)}\). In particular, any restriction map should correspond to precomposing by the appropriate inclusion of open sets, which automatically gives a presheaf. We can’t really demand that \(\mu^\sharp\) descends to a morphism of presheaves though, because locally on \(M\), we can have smooth functions that don’t extend globally to a smooth function everywhere. However, we can define a morphism \(\nu : \mc S \to C^\infty_M\) by taking \(\nu(U) : \mc S(U) \to C^\infty_M(U)\) to be given by precomposing by \(\mu\).

Now, we can use \(\mc S\) to recover \(C^\infty_M\). Again asking \(\mc S\) to be isomorphic to \(C^\infty_M\) isn’t going to work, again because of local-to-global issues. But, we claim that we have an isomorphism of sheaves \(\nu^+ : \mc S^+ \to C^\infty_M\), where \(\mc S^+\) is the sheafification of \(\mc S\). To construct this isomorphism, we use the morphism \(\nu\) defined above and show that for every \(x \in M\), \(\nu_x : \mc S_x \to C^\infty_{M,x}\) is an isomorphism. Then, we can finish by using the universal property of the sheafification to get our desired isomorphism \(\mc S^+ \to C^\infty_M\).

Recall that the stalks of a (pre)sheaf \(\mc F\) at a point \(x \in M\) are given by germs of sections at \(x\), i.e. \(\mc F_x = \set{(U, s) : x \in U \text{ open} \subseteq M, s \in \mc F(U)} / \sim\), where \(\sim\) is an equivalence relation generated by \((U_1, s_1) \sim (U_2, s_2)\) if \(s_1 \vert_V = s_2 \vert_V\) for some open \(V \subseteq U_1 \cap U_2\). Then \(\nu_x\) maps \([U, \hat f \circ i_U]\) to \([U, \nu(U)(\hat f \circ i_U)] = [U, \hat f \circ i_U \circ \mu]\), i.e. it acts by \(\nu\) on the section. To construct an inverse, we need a bump function \(\rho\) with \(x \in \supp \rho \subseteq U\), so that any smooth map \(f : U \to \RR\) can be extended to \(f \rho : M \to \RR\). Then we can construct the map \(\nu_x^{-1} : C^\infty_{M, x} \to \mc S^+\) via \([U, f] \mapsto [U, \widehat{f\rho} \circ i_U]\). To check that this is well-defined, suppose we had some other representative \([V, g] = [U, f]\), and we need to show that \([V, \widehat{g\rho} \circ i_V] = [U, \widehat{f\rho} \circ i_U]\). Note that since \(f\) and \(g\) agree on some open \(W \subseteq U \cap V\) containing \(x\), \(g\rho = f\rho\) on \(W\) as well. Then, \(\widehat{g\rho} = \widehat{f\rho}\) on \(\mu(W)\), so these germs are the same as these functions agree on \(\mu(W)\), so phew! \(\nu_x^{-1}\) is well-defined as a function.

Now, it’s clear by construction that \(\nu_x^{-1}\) is a homomorphism of \(\RR\)-algebras, so we only have to check that it’s an inverse to \(\nu_x\). Throughout, let \(V \subseteq U\) be the open set where \(\rho = 1\). To see this last step through, we first check that \(\nu_x^{-1} \circ \nu_x\) is the identity. If we pick some \([U, \hat f \circ i_U] \in \mc S(U)\) for some \(f \in C^\infty(M)\), then \(\nu_x^{-1}([U, \hat f \circ i_U \circ \mu])\) simplifies to \([U, \widehat{(\hat f \circ i_U \circ \mu) \rho} \circ i_U]\). We want this to be the same germ as \([U, \hat f \circ i_U]\), and in particular we claim these agree on \(\mu(V)\). For any \(y \in V\), \((\hat f \circ i_U)(\ev_y) = f(y)\). On the other hand, evaluating at \(\ev_y\), we unravel definitions to see that

\[(\widehat{(\hat f \circ i_U \circ \mu) \rho} \circ i_U)(\ev_y) = ((\hat f \circ i_U \circ \mu)\rho)(y) = (\hat f \circ i_U)(\ev_y) = f(y)\]

so these are both the same germ \([V, \hat f \circ i_V]\). Next, to check \(\nu_x \circ \nu_x^{-1}\) is the identity, for any \(f : U \to \RR\), we also claim that \(\widehat{f\rho} \circ i_U \circ \mu\) agrees with \(f\) on \(V\). This is an easier computation – for \(y \in V\),

\[(\widehat{f\rho} \circ i_U \circ \mu)(y) = (\widehat{f\rho}\circ i_U)(\ev_y) = (f\rho)(y) = f(y)\]

so indeed \(\nu_x\) is an isomorphism. By the argument above, this shows that the sheafification of \(\mc S\) recovers the smooth structure of \(M\)! In particular, if we have for manifolds \(M\) and \(N\) that \(C^\infty(M) \iso C^\infty(N)\), we have now shown that \(M \iso N\). The ring of smooth functions on a manifold therefore determines the entire manifold \(M\), its topological structure, and its smooth structure!

We can also talk about translating smooth maps into this setting. We already basically did this earlier though – a smooth map of manifolds \(\vphi : M \to N\) yields a map on their sheaves of smooth real-valued functions. In particular, we have a map of \(\RR\)-algebras \(\vphi^* : C^\infty(N) \to C^\infty(M)\) given by \(g \mapsto g \circ \vphi\). This also lifts to a map \(\vphi^{**} : \hat M \to \hat N\) given by \(\ev_x \mapsto \ev_x \circ \vphi^*\). In particular, \(\vphi^{**}\) commutes with the homeomorphism \(\mu_M : M \to \hat M\) in a nice way, so that \(\vphi^{**} \circ \mu_M = \mu_N \circ \vphi\). To see this, for any \(x \in M\), note that \((\mu_N \circ \vphi)(x) = \ev_{\vphi(x)}\), and \(\vphi^{**}(\ev_x) = \ev_x \circ \vphi^*\). On a function \(g \in C^\infty(N)\),

\[(\ev_x \circ \vphi^*)(g) = \ev_x \circ (g \circ \vphi) = g(\vphi(x)) = \ev_{\vphi(x)}(g)\]

so this equality holds.

Conversely, suppose we had a map of \(\RR\)-algebras \(\psi : C^\infty(N) \to C^\infty(M)\). As above, this yields a map \(\psi^* : \hat M \to \hat N\) given by \(\ev_x \mapsto \ev_x \circ \psi\). Taking global sections \(\mc S^+\) of these spaces, note that under the sheafification \(\mc S^+(M) \iso \mc S (M)\), which we now identify. Also, any \(\psi^*\) induces a map \(\psi^{**} : \mc S(N) \to \mc S(M)\) given by \(\hat g \mapsto \hat g \circ \psi^*\). Recall the isomorphism \(\mu^\sharp_M : C^\infty(M) \to \mc S(M)\) that precomposes by the homeomorphism \(\mu_M : M \to \hat M\). Then \(\psi^{**}\) also commutes with respect to this isomorphism, i.e. \(\psi^{**} \circ \mu^\sharp_N = \mu_M^\sharp \circ \psi\). To see this, pick a \(g \in C^\infty(N)\) again, and observe that \((\mu_M^\sharp \circ \psi)(g) = \psi(g) \circ \mu_M^{-1}\), and

\[(\psi^{**} \circ \mu^\sharp_N)(g) = \psi^{**}(g \circ \mu_N^{-1}) = \psi^{**}(\hat g) = \hat g \circ \psi^* = g \circ \mu_N^{-1} \circ \psi^*\]

To see this is equal to \(\psi(g) \circ \mu_M\), pick an \(x \in M\) and thereby an \(\ev_x \in \hat M\). Then we can evaluate this and get \((\psi(g) \circ \mu_M^{-1})(\ev_x) = \psi(g)(x)\). Similarly, \((g \circ \mu_N^{-1} \circ \psi^*)(\ev_x) = (g \circ \mu_N^{-1})(\ev_x \circ \psi)\). Now, \(\ev_x \circ \psi = \ev_y\) for some \(y \in N\), which allows us to rearrange a bit to get

\[(g \circ \mu_N^{-1})(\ev_y) = g(y) = \ev_y(g) = \ev_x \circ \psi \circ g = \psi(g)(x) .\]

This allows us to show that inducing a map on global sections from a smooth function and inducing a smooth map from a map on global sections are inverses, which tells us there is a bijection

\[\hom(M, N) \to \hom(C^\infty(N), C^\infty(M)).\]

This looks an awful lot like the the categorical anti-equivalence between affine schemes and rings – in this case, we have a categorical equivalence between manifolds and “\(\RR\)-algebras that look like \(C^\infty(M)\).” In particular, a manifold is some subset of \(\spec C^\infty(M)\), if we choose to think about the kernels of the maps in \(\hom(C^\infty(M), \RR)\) instead. Also, as we saw in the reconstruction of the topology on \(M\), every closed subset is a zero-locus of some smooth function by Urysohn’s Lemma, so in particular this topology is at least as fine as the Zariski topology on \(\spec C^\infty(M)\). This seems to indicate that we could really define a manifold instead via spectra of rings, as in schemes…

my favorite fanfiction

The above arguments were heavily inspired/directly come from Chapter 2 of \(C^\infty\)-Differentiable Spaces by Juan A. Navarro González and Juan B. Sancho de Salas, Chapters 3 and 7 of Smooth Manifolds and Observables by Jet Nestruev, and this stackexchange answer. At this point, I’d like to honor the approaches laid out in these books that basically argue what I have here in a much more straightforward fashion.

The approach of González and Sancho de Salas is to really think about a manifold as a kind of a spectrum in some sense. In particular, given a unital commutative \(\RR\)-algebra \(A\) giving a canonical homomorphism \(\RR \to A\), we say an ideal \(\mfk m \subseteq A\) is real if the composition with the quotient \(\RR \to A / \mfk m\) is an isomorphism. The collection of real ideals (which must all be maximal) is called the real spectrum of \(A\), \(\spec_\RR A\). This now solves our functoriality problem, because the preimage of a real ideal is indeed real under an \(\RR\)-algebra homomorphism. In algebraic geometry language, what we are really doing is restricting to the \(\RR\)-points of \(\spec C^\infty (M)\), which I think means the points whose residue field is \(\RR\)? I’m not quite familiar enough with algebraic geometry to be confident on this point.

For any \(f\) in \(A\), if we think about points of the real spectrum as maximal ideals, we can also see that \(f\) defines a real-valued function on \(A\) by passing to the quotient \(A / \mfk m \iso \RR\). As we did above, we topologize \(\spec_\RR A\) with what the book calls the Gelfand topology, i.e. the coarsest one that makes these maps on \(\spec_\RR A\) for any \(f\) continuous. If \(A\) is the algebra of global sections of a manifold, the book remarks that the Zariski topology on \(\spec_\RR A\) coincides with this, using the smooth version of Urysohn’s Lemma.

In general, the book goes on to talk about differentiable algebras, which are suitable quotients of \(C^\infty(\RR^n)\) by an ideal, and differentiable spaces, which are the analogues of rings and schemes in algebraic geometry. Differentiable spaces (as you might guess) are locally isomorphic to spaces called affine differentiable spaces, which are the real spectra of differentiable algebras. A lot of smooth manifold theory/ideas can be replicated in the realm of differentiable spaces, where a lot of the same arguments just kind of go through in very similar ways, but with a sprinkling of what look to me like functional analysis tools and some algebraic geometry ethos.

On the other hand, the approach of Nestruev⁷ is very similar, but is more focused on developing smooth manifold theory in a spectral way starting with the algebra smooth functions on a manifold. In particular, they restrict their attentions to a very specific class of \(\RR\)-algebras from which one can construct a manifold in the usual sense. As above, for an \(\RR\)-algebra \(A\), their manifolds are spaces of \(\RR\)-algebra homomorphisms \(A \to \RR\), and they build up the constraints on algebras needed to characterize rings of smooth functions and pull out a manifold. This is not at all an easy task and this, along with developing manifold theory from this point of view, is the bulk of a book.

Nestruev requires a “suitable” \(\RR\)-algebra to be geometrical, complete, and smooth. These names are a little mysterious to me, but they arise sort of naturally over the course of Chapters 3/4:

First, an algebra \(A\) is geometrical if \(\bigcap_{\vphi \in \hom(A, \RR)} \ker \vphi = 0\). To see why this is important, we need to think for a moment about \(\tilde A = \set{\tilde f : \hom(A, \RR) \to \RR : \tilde f(x) = x(f)}\), where \(x\) denotes an element of \(\hom(A, \RR)\). Here, \(x\) is meant to be the analogue of \(\ev_x\) above, and \(\tilde A\) is the analogue of the global sections \(\mc S(M)\) that we were thinking about. It was important in defining our presheaf \(\mc S\) and showing the eventual categorical equivalence that \(\mc S(M)\) was isomorphic to \(C^\infty(M)\). Here, Nestruev wants in general for \(A\) to be isomorphic to \(\tilde A\), but we can only enforce this if the above kernel condition is satisfied. The “geometrical” nature of this property feels like it comes from the idea that points of the dual \(\hom(A, \RR)\) should be like evaluating at points of the manifold, and if a function vanishes when evaluating algebraically at a point, it should be globally zero. This also feels like it maybe has to do with reducedness of a scheme, maybe. Anyways, after introducing the Gelfand topology on the dual space \(\hom(A, \RR)\), Nestruev is already able to show that it is Hausdorff and that this space recovers \(U \subseteq \RR^n\) open in the case of \(A = C^\infty(U)\).

The next thing we need to think about has to do with the sheaf of smooth functions on our space. Nestruev defines a notion of restriction of a geometrical algebra \(A\) to any subset \(K \subseteq \hom(A, \RR)\) as the set of functions \(f: K \to \RR\) such that for any point \(x \in K\), there is an open neighborhood \(U \subseteq K\) and an element \(\overline f \in A\) such that \(f\) and \(\overline f\) are equal on \(U\). Here, \(\overline f\) should be thought of as a function on \(\hom(A, \RR)\) because of the geometrical constraint, and as such this operation can be thought of as a topological restriction. This is a bit of a generalized version of how we defined the presheaf of smooth functions on our constructed space \(\hat M\). Recall that we couldn’t get a whole sheaf of smooth functions because not all smooth functions on an open set extend to a larger one, and therefore we were missing smooth functions in our presheaf.

This now brings us to our second constraint. A geometrical algebra \(A\) is complete if the restriction map \(\rho : A \to A\vert_{\hom(A, \RR)}\) is surjective. This feels like constraining \(A\) to now include all of necessary smooth functions on \(\hom(A, \RR)\) to make \(A\) into a sheaf with these restriction maps, which is something that we solved with sheafification. I’m not sure why this is called “complete” though. Maybe \(A\) is just complete in the “there are enough smooth functions” sense?

Finally, the last condition required encodes second-countability and the fact that the space is locally Euclidean in one go. A complete geometrical algebra \(A\) is smooth if there is a finite or countable open covering \(\set{U_k}\) of \(\hom(A, \RR)\) such that all of the algebras \(A\vert_{U_k}\) are isomorphic to the algebra of smooth functions \(C^\infty(\RR^n)\). This now recovers all of the properties of a smooth manifold we want. An interesting thing about this property of an algebra \(A\) is that it now requires \(A\) to be closed with respect to smooth composition. This means that for any finite collection of elements \(f_1, \dots, f_k \in A\) and \(g \in C^\infty(\RR^k)\), there is an \(f \in A\) such that \(f(a) = g(f_1(a), \dots, f_k(a))\) for all \(a \in \hom(A, \RR)\). This finaly property feels useful as a purely algebraic constraint that doesn’t really bring in topology at all.

These three conditions are then spun out into developing a theory of manifolds in terms of the dual spaces of these algebras, but interestingly using the idea of charts on a manifold rather than sheaves and locally ringed spaces. This feels better pedagogically as a possibly introductory book to the subject, but it does seem a little clunkier in some respects. Also, while spectra are mentioned, they stop just short of providing the correct fix as in González-Sancho de Salas to do things from an algebraic-geometrical perspective. However, they seem to be able to do a lot of cool cohomological stuff and other manifolds-related content later, which I haven’t looked at much and seems cool when done form this perspective.

A combination of the approaches from these two sources feels like it adequately bridges the gap between the chart-based intuition for manifolds that I had before and the modern sheaf/locally ringed space perspective that algebraic geometry embraces. I personally think that seeing this connection places the manifold in concert with the other geometric structures that I’ve now seen, but there are major drawbacks to taking this approach. For one, manifolds are too nice to be handled in this way – going from local-to-global is easy (and should be easy!) with a partition of unity, and as such the full power of a sheaf isn’t really needed to keep track of all the data coming from sections. Also, it feels like the niceness of doing things with linear algebra and calculus was all lost, and replacing it with algebra had us working with strange but very cool rings.

Also, we haven’t at all been able to translate structures defined on manifolds, which are arguably also important to the study of manifolds and differential geometry. In particular, we don’t have a good notion of vector bundles or differential forms on a manifold, and it’s not quite clear to me how exactly to algebraically encode those kinds of structures. But for now, being able to see manifolds on the same level as other geometric spaces feels good to have done once.

coda: how deep does the rabbit hole go?

While searching for sources to consult to answer this question, here are some extra odds and ends that I thought were cool/relevant but couldn’t fit in:

Recall that when we were thinking about manifolds as a locally ringed spaces of \(\RR\)-algebras, locally isomorphic to the smooth functions on \(\RR^n\), we enforced the constraint that our manifolds had to have their underlying topological spaces be Hausdorff and second-countable. This stackexchange post discusses whether it’s possible to drop Hausdorff and second-countable from the constraints on the topological space and encode them in the sheaf itself. It turns out you can, but it seems kind of more annoying than you’d like and translating topological properties into sheaf properties is probably more pain than it’s worth.

We also wondered what might happen if you used a different structure sheaf for \(\RR^n\) to induce a sheaf on your topological space. This stackexchange post asks about rings of continuous functions and if they behave as nicely as the ring of smooth functions. In short – nope! Unlike how one can recover the tangent space³ from the sheaf of rings of smooth functions on a manifold by thinking about \(\mfk m / \mfk m^2\), it happens that in the continuous case \(\mfk m = \mfk m^2\), so we don’t get any smoothness data. I wonder what happens for \(C^\alpha\) manifolds for \(0 < \alpha < \infty\) – do you get any other information? If you know, let me know!

Finally, the ultimate program to avoid talking about manifolds and topological spaces and just doing algebra is to think about these algebraic structures called \(C^\infty\)-rings! These are rings that are basically “closed with respect to smooth composition,” as mentioned in Nestruev’s book. This is the desired approach used to think about manifolds from the lens of synthetic differential geometry. Apparently Lawvere was thinking about this in the 60s, and here’s a link to the one non-nLab survey paper(?) I could find on the topic. This seems really cool! It feels like people who are up on HoTT and univalent mathematics would probably really like encoding a manifold in this way, and someday if agda-unimath gets to a point to where we could think about defining a manifold, this might be the way to do it.

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