A manifold whose functions are the smooth functions on the real line with rational period

Hi, I was reading in

<a href="http://books.google.com/books?id=N5mHmFiovgkC&lpg=PP1&pg=PP1#v=onepage&q=&f=false">Jet Nestruev: Smooth Manifolds and Observables, Springer, 2003</a>

about a month ago (after I stumbled over a question on MO) and there was an exercise that resisted solution for more than a week.

Well.... now I found out that I have just misread the exercise. However, this way I basically did several exercises at once. Here comes the problem and its solution:

The problem

(<emph>inspired</emph> by page 28, chapter 3, exercise 3.17.5 in Nestruev)

Find a smooth (real) manifold $M$ such that its algebra of smooth functions $C^\infty(M,\mathbb R)$ is isomorphic to the algebra of all smooth functions $f : \mathbb R \to \mathbb R$ that have some rational period $\tau$ (i.e. there exists $\tau \in \mathbb Q$ such that $f(x)=f(x+\tau)$ for all x). Note that we don't fix a period $\tau$ here. Let's call the algebra in question (smooth functions on the real line with some rational period) $A$.

You might want to stop reading here and think for a second (or minutes) about the solution or similar problems that have easier solutions. A more vague problem would be

Find a space $M$ such that the functions $M \to \mathbb R$ correspond to functions $\mathbb R \to \mathbb R$ that are periodic with some rational period.

The manifold version of the problem has no solution (there doesn't exist such a smooth manifold), as I will prove. I don't know if there is some precision of the vague version (e.g. some additional structure on the space) which provides a solution but I don't think so. If we fix a rational period (say 1), everything is nice and good in the world of smooth manifolds and smooth functions.

A simple sub-problem

If we would fix a rational number $\tau$, and look at all smooth functions $f : \mathbb R \to \mathbb R$ with period $\tau$, these f would factor through $[0,\tau]$ and provide functions $f : [0,\tau] \to \mathbb R$ such that $f(0)=f(\tau)$. Finally we see that they all factor through $\mathbb R / \tau \mathbb Z = S^1$ so there is an isomorphism $C^\infty(S^1,\mathbb R) \to \{\tau\text{-periodic smooth functions } f : \mathbb R \to \mathbb R\}$.
This even works for any non-rational $\tau$.
Let's call $A_\tau$ the algebra of smooth functions on the real line with period $\tau \in \mathbb R$.

From simple to hard

For a rational number $\tau \in \mathbb Q$, the algebra $A_\tau$ of smooth functions with period $\tau$ is clearly a sub-algebra of the algebra of all rational-periodic functions $A$. For two rational numbers $\tau,\mu$, we even have $A_\tau$ and $A_\mu$ isomorphic (via the pullback along $\cdot \frac \tau \mu : \mathbb R \to \mathbb R$).
On the geometric side, this should correspond to a homeomorphism $S^1 \to S^1$ and the induced morphism is indeed the identity. To see what I mean precisely by "induced morphism", look at the section "The Nestruev approach" below.

Furthermore, for each rational $\tau$ and each natural number $n$, $A_\tau$ is a sub-algebra of $A_{n\tau}$. Since every rational number $\tau$ can be written as reduced quotient $\tau = \frac a b$, every $\tau$-periodic function $f$ is also $b\cdot\tau$-periodic, hence $a$-periodic, and $a$ is the smallest natural number $n$ for which $f$ is $n$-periodic.

Looking at the geometric picture again, the inclusion $A_\tau \to A_{n\tau}$ induces a map $S^1 \to S^1$ (in the other direction). This map is the n-fold covering of $S^1$ by $S^1$ (proof left to the reader).

A maybe-not-solution

If we say a non-periodic function $f : \mathbb R \to \mathbb R$ has the rational period 0, the manifold we're looking at is just the real numbers. This notion of rational periodic function is surely not a useful one, since every function satisfies $f(x) = f(x+0)$, thus every function is 0-periodic. Note also that constant functions should have smallest natural period 1, not 0.

A misleading idea

Take the space $M := S^1 \cup \{\ast\}$, the disjoint union of a circle with a point. Every element of $A$ defines a function on $M$: it has a smallest natural period $n$, which we take as value on the point $\ast$, and we use the values on $[0,n]$ to define values on the $S^1$ (by factorizing through the quotient map $[0,n] \to S^1$ that identifies $0$ and $n$).

Now there are functions on $M$ that assign to $\ast$ some non-rational value, and this shows that this approach fails somehow. This cannot even be saved by taking for $M$ a locally ringed space, putting on $\ast$ the ring $\mathbb Q$ or $\mathbb Z$ because periods are always non-negative and $\mathbb N$ is not a ring. You might take a "locally monoided space"... but then you could as well just take as ringed space the pair $(\mathbb R, A)$. This can indeed be considered as one step in the "solution" of the problem by Nestruev below.

The Nestruev approach

Nestruev wants us to think of differential geometry in terms of observables and states. This approach looks at differential geometry in a kind-of-scheme-theoretic way, taking a manifold to be something glued from local function algebras.

An algebra $\mathcal F$ defines a topological space $|\mathcal F|$ with underlying set the dual space $Hom_{\mathbb R}(\mathcal F,\mathbb R)$ of all $\mathbb R$-algebra homomorphisms (functionals), and the weak topology for all maps $|\mathcal F| \to \mathbb R$ that come from evaluations at elements of $\mathcal F$.

Define an algebra $\tilde F := \{F : |\mathcal F| \to \mathbb R \mid \exists f \in \mathcal F\ \forall x \in |\mathcal F| : F(x) = x(f)\}.$
Now there is an obvious homomorphism $\phi : F \to \tilde F$, by sending each $f \in \mathcal F$ to the evaluation at $f$ and this $\phi$ is clearly surjective. It is not always injective. The necessary and sufficient condition to $\mathcal F$ is to be geometric, which is defined to be the property
$! \bigcap_{p \in |\mathcal F|} Ker(f) = 0,$
so a non-geometric algebra $\mathcal F$ has elements $f \in \mathcal F$ vanishing "on every point" $p \in |\mathcal F|$ yet not being the zero element of $\mathcal F$. Nestruev proves that the dual space of a geometric algebra is always Hausdorff, which is at least something into the direction of a manifold-like space.

Dual spaces

So let's have a look at the dual spaces of $A$ and $A_\tau$. This was the actual exercise I misread first

(page 28, chapter 3, exercise 3.17.4 and 3.17.5 in Nestruev)

Describe the dual spaces of $A_1$ and $A$.

The dual space of $A_1$ is simply $S^1$, as described above. A detailed proof is contained in Nestruev. The most difficult thing to prove is that $|A_1|$ doesn't contain more points than those coming from evaluations. To prove this, compactness of $[0,1]$ is used.

The dual space of $A$ consists of all $\mathbb R$-algebra homomorphisms $A \to \mathbb R$. At least we have all evaluations $ev_x$ at points $x \in \mathbb R$. In the case of $A_1$ these evaluations coincide for two $x,y \in \mathbb R$ if the difference $x-y$ was an integer. This can't happen here because for each two $x,y \in \mathbb R$ there is a function $f \in A$ with period greater than $max(x,y)$ such that $f(x) \neq f(y)$.

To see that there aren't more functionals $A \to \mathbb R$ than those coming from evaluations, suppose having such a functional $\Lambda$. Then for each $a \in A$, there is some $f_a \in A$ with $f_a(a) \neq \Lambda(f_a)$. Note that we could actually choose the $f_a$ to lie in $A_1$ and the whole proof works for $A_1$, too. The complements of the pre-images $U_a := \{f_a^{-1}(\Lambda(f_a))\}^c$ are non-empty open sets which form a cover of $[0,1]$. Choose a finite subcover $U_{a_1},...,U_{a_m}$ and define a function $g : \mathbb R \to \mathbb R$ by
$! g(x) := \sum_{k=1}^m (f_{a_k} - \Lambda(f_{a_k}))$
then we have $g \in A_1 \subset A$, $\Lambda(g)=0$ and furthermore: $g$ is nowhere vanishing on $[0,1]$.
From this we can derive that $\frac 1 g \in A_1 \subset A$, too. The requirement on $\Lambda$ to be a unital $\mathbb R$-algebra homomorphism now shows $! 1 = \Lambda(1) = \Lambda(g \cdot \frac 1 g) = \Lambda(g) \cdot \Lambda(\frac 1 g) = 0 \cdot \Lambda(\frac 1 g) = 0,$ which is a contradiction. Therefore, $\Lambda$ must have been the evaluation at a point $x \in \mathbb R$ and we conclude that $|A| = \mathbb R$. The topology given as the weak topology according to all evaluations coincides with the standard analytic topology on $\mathbb R$.

Geometric algebras

Nestruev also poses a follow-up problem (strangely, earlier in the text)

(page 24, chapter 3, exercise 3.9.5)

Decide whether $A_1$ and/or $A$ are geometric $\mathbb R$-algebras.

$A_1$ is clearly geometric, since as a set $|A_1|=\{\text{evaluations on } [0,1)\}$ and a 1-periodic function on $\mathbb R$ that vanishes everywhere on $[0,1)$ is always the constant zero function, so $\bigcap_{x \in |A_1|} Ker(x)$ contains only the zero function, thus is the zero ideal. So we know now that $\tilde{A_1}$ is isomorphic to $A_1$.

Similarly, if a rational-periodic function on $\mathbb R$ vanishes everywhere on $\mathbb R$, which is equivalent to being in the kernel of all elements of $|A_1|$, then it must be the constant zero function. So $A$ is geometric, too, and $\tilde{A}$ is isomorphic to $A$.

This is a good point to get confused. Take care of the definition of $\tilde{A}$ - these are exactly those maps $|A| \to \mathbb R$ that are \emph{given by evaluations}, nothing more. Of course, we would like to have some smooth-manifold-like object $M$ instead of $|A|$ and look at smooth functions on $M$ instead of this algebra $\tilde A$.

For geometric algebras $\mathcal F_1,\ \mathcal F_2$, an algebra homomorphism $\mathcal F_1 \to \mathcal F_2$ induces a dual morphism $|\mathcal F_2| \to |\mathcal F_1|$ (we have already looked at this in the case of the inclusion $A_\tau \to A_{n\tau}$, which has as dual map the n-fold covering of $S^1$, and the isomorphism $A_\tau \to A_\mu$, which has as dual map the identity of $S^1$). It is easy to prove that an isomorphism has a homeomorphism as dual map. The inclusion $\iota : A_\tau \to A$ has the dual map $|\iota| : \mathbb R \to S^1$ given by the quotient $\mathbb R \to \mathbb R / \tau\mathbb Z$, which is an $\infty$-fold covering map.

Complete algebras

The next step in the Nestruev definition of smooth manifolds is the notion of complete algebras, a technical notion required to be able to talk about restrictions which are required to define what a smooth algebra should be.
For any subset $X \subset |\mathcal F|$ of the dual space of a geometric $\mathbb R$-algebra $\mathcal F$, the elements $f \in \mathcal F = \tilde{\mathcal F}$ can be restricted to $X$, which yields the restriction homomorphism $\rho_X : \mathcal F \to \mathcal F_{\mid X}$. The space $F_{\mid X}$ has to be defined carefully, as the space of functions on $X$ that can locally be written as restrictions of functions in $\tilde{\mathcal F}$:
$! F_{\mid X} := \{ f : X \to \mathbb R \mid \exists U \subset X\ \exists \tilde f \in \tilde{\mathcal F} : \tilde{f}_{\mid U} = f_{\mid U}\}.$
An algebra $\mathcal F$ is called complete if the restriction to its dual space $\rho_{|\mathcal F|} : \mathcal F \to \mathcal F_{|\mathcal F|}$ is an isomorphism. This translates to: if a function locally stitched together from restrictions of functions from $\mathcal F$ is again an element of $\mathcal F$. The terminology complete can be understood because this map $\rho$ is always injective, but fails surjectivity on algebras which aren't complete (so these "miss" some functions, thus are incomplete).

Of course, Nestruev wants us to check if our algebras are complete:

(page 32, chapter 3, exercise 3.28.3)

Decide whether $A_1$ and/or $A$ are complete $\mathbb R$-algebras.

$A_1$ is clearly complete, since a function locally defined by the restriction of smooth 1-periodic functions can be seen as a smooth function $f$ on $[0,1]$ that satisfies $f(0)=f(1)$, so it's an element of $A_1$ again.

$A$ is incomplete. For example, define a function $f : \mathbb R \to \mathbb R$ by choosing for each interval $[n-1,n+1]$ with $n \in \mathbb Z$ a smooth periodic function $f_n \in A$ with period $n$ such that $f_n(x) = f_{n+1}(x)$ for $x \in [n,n+1]$. The functions $f_n$ can now be chosen such that $f$ is not periodic, for example by letting the maxima of the $f_n$ grow without boundary, letting $f$ be unbounded, too.

The algebras of smooth functions on manifolds are always complete (exercise!), so we have seen that the manifold we were looking for doesn't exist.

$C^\infty$-closed algebras

Returning to the general setting, we would like to have a homeomorphism $X \to |\mathcal{F}_{X}|$ for any $\mathbb R$-algebra $\mathcal F$ and subset $X \subset |\mathcal F|$. If $\mathcal F$ is geometric, there is always a continuous map $\mu_X : X \to |\mathcal{F}_{X}|,\ (\mu(x))(f) := f(x)$ which is a homeomorphism onto a subset of $|\mathcal{F}_{X}|$, but fails to be surjective in general.

A geometric $\mathbb R$-algebra $\mathcal F$ is $C^\infty$-closed, if for all finite families of functions $f_1,...,f_k \in \mathcal F$ and all smooth maps $g \in C^\infty(\mathbb R^k)$, the composite $x \mapsto g(f_1(x),...,f_k(x))$ is an element of $\mathcal F$ (considered as function on $|\mathcal F|$).

Nestruev proves that for each basis open set $X \subset |\mathcal F|$ of a $C^\infty$-closed geometric algebra $\mathcal F$, the map $\mu_X$ is surjective.

Smooth algebras

A complete, geometric $\mathbb R$-algebra $\mathcal F$ is called smooth if there exists a countable open cover $\{U_\alpha\}$ of $|\mathcal F|$ such that the algebras $\mathcal F_{\mid U_\alpha}$ are isomorphic to $C^\infty(\mathbb R^n)$ for some fixed natural number $n$, called the dimension of $\mathcal F$. Of course, the space $|\mathcal F| =: M$ has now a structure of a smooth manifold of dimension $n$ and $C^\infty(M) = \mathcal F$.

$A_1$ is a smooth algebra.

Repairing defects

Any $\mathbb R$-algebra $\mathcal F$ yields a geometric one, by quotienting out the ideal $\mathcal I := \bigcap_{x \in |\mathcal F|} Ker(x)$ (at the risk of getting just the trivial algebra). A geometric algebra is obviously not changed in this process.

Any geometric algebra $\mathcal F$ yields a complete one, by defining the completion to be the algebra $\mathcal F_{|\mathcal F|}$. A complete algebra remains untouched by this process. The completion of $A$ is just the algebra of smooth functions on the real line.

Any geometric algebra $\mathcal F$ yields a $C^\infty$-closed one, by adding all functions of the form $g(f_1,...,f_k)$ for $f_i \in A$ and $g \in C^\infty(\mathbb R^k)$. This closure is definable via abstract nonsense, too: it is the unique smooth envelope (which is defined by the universal property, that morphisms to $C^\infty$-closed algebras should factor uniquely through the smooth envelope). This way, the smooth envelope acts as mediator between non-closed algebras with smooth algebras.

Some comments

If anything here remains unclear, leave a comment. If something is wrong, please leave a comment. I also recommend reading Nestruev. It's a nice elementary textbook (translated from Russian) that could be interesting for anyone who does either differential geometry, algebraic geometry or theoretical physics and of course for those who like all of these topics and their intersection. The algebra $A_1$ is frequently used in other examples throughout the book, but $A$ never appears again.

Jet Nestruev is a collective pseudonym, like Nicolas Bourbaki but a little bit less influental (and they didn't write as much). The members of this group are A.Astashov, A.Bocharov, S.Duzhin, A.Sosinsky, A.Vinogradov and M.Vinogradov. If you like this kind of mathematics, take a look at this page about the works of A. Vinogradov.

The main theme, guessing from a rather philosophical paper, seems to be the notion of diffiety, a geometric object that plays for quantum field theories the same role that smooth manifolds play for classical mechanics. However, this is not just some abstract nonsense but actually something about partial differential equations. I guess I will have to learn about jet bundles some day...