A manifold whose functions are the smooth functions on the real line with rational period
Wednesday, March 31st, 2010 | Author: Konrad Voelkel
Hi, I was reading in
Jet Nestruev: Smooth Manifolds and Observables, Springer, 2003
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
(inspired by page 28, chapter 3, exercise 3.17.5 in Nestruev)
Find a smooth (real) manifold
such that its algebra of smooth functions
is isomorphic to the algebra of all smooth functions
that have some rational period
(i.e. there exists
such that
for all x). Note that we don't fix a period
here. Let's call the algebra in question (smooth functions on the real line with some rational period)
.
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 such that the
functions correspond to functions
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
A simple sub-problem
If we would fix a rational number , and look at all smooth functions
with period
, these f would factor through
and provide functions
such that
. Finally we see that they all factor through
so there is an isomorphism
.
This even works for any non-rational .
Let's call the algebra of smooth functions on the real line with period
.
From simple to hard
For a rational number , the algebra
of smooth functions with period
is clearly a sub-algebra of the algebra of all rational-periodic functions
. For two rational numbers
, we even have
and
isomorphic (via the pullback along
).
On the geometric side, this should correspond to a homeomorphism 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 and each natural number
,
is a sub-algebra of
. Since every rational number
can be written as reduced quotient
, every
-periodic function
is also
-periodic, hence
-periodic, and
is the smallest natural number
for which
is
-periodic.
Looking at the geometric picture again, the inclusion induces a map
(in the other direction). This map is the n-fold covering of
by
(proof left to the reader).
A maybe-not-solution
If we say a non-periodic function 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
, 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 , the disjoint union of a circle with a point. Every element of
defines a function on
: it has a smallest natural period
, which we take as value on the point
, and we use the values on
to define values on the
(by factorizing through the quotient map
that identifies
and
).
Now there are functions on that assign to
some non-rational value, and this shows that this approach fails somehow. This cannot even be saved by taking for
a locally ringed space, putting on
the ring
or
because periods are always non-negative and
is not a ring. You might take a "locally monoided space"... but then you could as well just take as ringed space the pair
. 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
An algebra defines a topological space
with underlying set the dual space
of all
-algebra homomorphisms (functionals), and the weak topology for all maps
that come from evaluations at elements of
.
Define an algebra
Now there is an obvious homomorphism , by sending each
to the evaluation at
and this
is clearly surjective. It is not always injective. The necessary and sufficient condition to
is to be
so a non-geometric algebra




Dual spaces
So let's have a look at the dual spaces of and
. 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
and
.
The dual space of is simply
, as described above. A detailed proof is contained in Nestruev. The most difficult thing to prove is that
doesn't contain more points than those coming from evaluations. To prove this, compactness of
is used.
The dual space of consists of all
-algebra homomorphisms
. At least we have all evaluations
at points
. In the case of
these evaluations coincide for two
if the difference
was an integer. This can't happen here because for each two
there is a function
with period greater than
such that
.
To see that there aren't more functionals than those coming from evaluations, suppose having such a functional
. Then for each
, there is some
with
. Note that we could actually choose the
to lie in
and the whole proof works for
, too. The complements of the pre-images
are non-empty open sets which form a cover of
. Choose a finite subcover
and define a function
by
then we have



![[0,1]](https://www.konradvoelkel.com/wp-content/plugins/latex/cache/tex_ccfcd347d0bf65dc77afe01a3306a96b.gif)
From this we can derive that







Geometric algebras
Nestruev also poses a follow-up problem (strangely, earlier in the text)
(page 24, chapter 3, exercise 3.9.5)
Decide whether
and/or
are geometric
-algebras.
is clearly geometric, since as a set
and a 1-periodic function on
that vanishes everywhere on
is always the constant zero function, so
contains only the zero function, thus is the zero ideal. So we know now that
is isomorphic to
.
Similarly, if a rational-periodic function on vanishes everywhere on
, which is equivalent to being in the kernel of all elements of
, then it must be the constant zero function. So
is geometric, too, and
is isomorphic to
.
This is a good point to get confused. Take care of the definition of - these are exactly those maps
that are \emph{given by evaluations}, nothing more. Of course, we would like to have some smooth-manifold-like object
instead of
and look at smooth functions on
instead of this algebra
.
For geometric algebras , an algebra homomorphism
induces a dual morphism
(we have already looked at this in the case of the inclusion
, which has as dual map the n-fold covering of
, and the isomorphism
, which has as dual map the identity of
). It is easy to prove that an isomorphism has a homeomorphism as dual map. The inclusion
has the dual map
given by the quotient
, which is an
-fold covering map.
Complete algebras
The next step in the Nestruev definition of smooth manifolds is the notion of
For any subset of the dual space of a geometric
-algebra
, the elements
can be restricted to
, which yields the restriction homomorphism
. The space
has to be defined carefully, as the space of functions on
that can
:
An algebra





Of course, Nestruev wants us to check if our algebras are complete:
(page 32, chapter 3, exercise 3.28.3)
Decide whether
and/or
are complete
-algebras.
is clearly complete, since a function locally defined by the restriction of smooth 1-periodic functions can be seen as a smooth function
on
that satisfies
, so it's an element of
again.
is incomplete. For example, define a function
by choosing for each interval
with
a smooth periodic function
with period
such that
for
. The functions
can now be chosen such that
is not periodic, for example by letting the maxima of the
grow without boundary, letting
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.
-closed algebras
Returning to the general setting, we would like to have a homeomorphism for any
-algebra
and subset
. If
is geometric, there is always a continuous map
which is a homeomorphism onto a subset of
, but fails to be surjective in general.
A geometric -algebra
is
-closed
and all smooth maps
, the composite
is an element of
(considered as function on
).
Nestruev proves that for each basis open set of a
-closed geometric algebra
, the map
is surjective.
Smooth algebras
A complete, geometric -algebra
is called
of
such that the algebras
are isomorphic to
for some fixed natural number
, called the
. Of course, the space
has now a structure of a smooth manifold of dimension
and
.
is a smooth algebra.
Repairing defects
Any -algebra
yields a geometric one, by quotienting out the ideal
(at the risk of getting just the trivial algebra). A geometric algebra is obviously not changed in this process.
Any geometric algebra yields a complete one, by defining the completion to be the algebra
. A complete algebra remains untouched by this process. The completion of
is just the algebra of smooth functions on the real line.
Any geometric algebra yields a
-closed one, by adding all functions of the form
for
and
. This closure is definable via abstract nonsense, too: it is the unique smooth envelope (which is defined by the universal property, that morphisms to
-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 is frequently used in other examples throughout the book, but
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
2021-05-23 (23. May 2021)
There is a mistake in "Dual spaces", where the function g is assumed to be nowhere vanishing. This seems to be wrong, as it is a sum of nowhere vanishing terms that might cancel each other out. Maybe this is fixable by some positivity assumptions, but I doubt it.