# Manifolds with polynomial transition maps

Title edited I thank მამუკა ჯიბლაძე and Corbennick for their suggestion on the title of this question. I changed the title based on the suggestion of Corbennick.

What is an example of a manifold $M$ which does not admit an atlas $\mathcal{A}$ with the following property?:

For every two charts $(\phi,U)$ and $(\psi,V)$ in $\mathcal{A}$, $\psi \circ \phi^{-1}$ is a polynomial map.(Its components are polynomial functions).

Motivation: The motivation for this question comes from the concept "Affine manifolds". I am indebted to Mike Cocos, for learning this concept and its related problem, that is Chern conjecture.

• Good question, bad title :D – მამუკა ჯიბლაძე Jul 17 '17 at 8:59
• @მამუკაჯიბლაძე May I ask you to suggest a new title? Thank you. – Ali Taghavi Jul 17 '17 at 10:59
• How about: Manifolds with polynomial transition maps? – user1688 Jul 17 '17 at 11:04
• @Corbennick Thank you!very good suggestion. – Ali Taghavi Jul 17 '17 at 12:01
• The Wikipedia page that you link to shows that there are lots of (necessarily non-simply connected but) very natural and important examples of compact manifolds with not only polynomial but even affine transition maps, like e. g. flat tori. I find it very interesting question whether there are any essentially non-affine examples, i. e. compact manifolds which do not admit atlases with affine transition maps but admit atlases with some higher degree polynomial transition maps. – მამუკა ჯიბლაძე Jul 18 '17 at 10:29

If I remember correctly, this is impossible for any (nonempty) simply-connected compact manifold of positive dimension. In particular, $S^2$ cannot have such an atlas. Off the top of my head, I don't remember where I saw this statement, but I'll try to find a reference.

Followup: I still don't remember the reference, but I think that the proof goes something like this:

By hypothesis, both $\tau=\psi\circ\phi^{-1}$ and $\sigma = \phi\circ\psi^{-1}$ are polynomial maps, and hence they are defined everywhere on $\mathbb{R}^n$. By hypothesis there is a nonempty open domain $D_1\subset\mathbb{R}^n$ such that $\sigma(\tau(x)) = x$ for all $x\in D_1$ and a nonempty open domain $D_2\subset\mathbb{R}^n$ such that $\tau(\sigma(y)) = y$ for all $y\in D_2$. Since $\sigma$ and $\tau$ are polynomial mappings, it follows that $\sigma(\tau(x)) = x$ for all $x\in\mathbb{R}^n$ and $\tau(\sigma(y)) = y$ for all $y\in\mathbb{R}^n$. Thus, $\sigma$ and $\tau$ are globally invertible. Moreover, by the chain rule, we have $\sigma'(\tau(x))\tau'(x) = I$ for all $x\in \mathbb{R}^n$, and, taking determinants, it follows that $\det(\tau'(x))$ and $\det(\sigma'(\tau(x)))$, which are polynomial in $x$ must actually be (nonzero) constants.

Consequently, it follows that $\Lambda^n(T^*M)$ carries a (unique) flat connection such that, if $\mathrm{d}x$ is the standard volume form on $\mathbb{R}^n$, then $\psi^*(\mathrm{d}x)$ is a (local) parallel section of $\Lambda^n(T^*M)$ for each $(\psi,U)$ in the atlas $\mathcal{A}$. By simple-connectivity, there is a global parallel volume form $\mu$ on $M$, and we can, composing the members of our atlas with linear transformations of $\mathbb{R}^n$, get a new atlas with polynomial transitions that satisfies $\psi^*(\mathrm{d}x) = \mu$ on $U$ for all $(\psi,U)$ in the new atlas.

There is also a way to analytically continue the map $\psi$ coming from a chart $(\psi,U)$ to the domain $U\cup V$ for any chart $(\phi,V)$ for which $U\cap V$ is non-empty: Since $\tau$ is defined on all of $\mathbb{R}^n$, and since $\psi = \tau\circ\phi$ on $U\cap V$, one can take $\psi(q) = \tau(\phi(q))$ for all $q\in V$, which extends $\psi$ to $V$.

Using this construction, any $(\psi,U)\in\mathcal{A}$ can be uniquely analytically continued along any smooth path in $M$, and one easily verifies that this extension depends only on the homotopy class of the path with fixed endpoints. Since $M$ is simply connected, this means that $\psi$ can be extended uniquely analytically to all of $M$, and hence $\psi:M\to\mathbb{R}^n$ is a smooth map that satisfies $\psi^*(\mathrm{d}x) = \mu$ on all of $M$. Obviously, this is impossible.

• Where are you using the volume form? – Tom Goodwillie Jul 17 '17 at 15:28
• @TomGoodwillie: I could have got away without it, but it was convenient to be able to write that $\psi^*(\mathrm{d}x)=\mu$ globally, which made the contradiction obvious, since, otherwise, it was not immediate (to me) that the analytic continuation of $\psi$ would always be a local diffeomorphism. – Robert Bryant Jul 17 '17 at 15:33
• What about if the transition functions are rational functions? This includes the sphere, for example. – Kevin Casto Jul 17 '17 at 15:36
• @AliTaghavi: The (flat) connection provides a global volume form with respect to which all the coordinate charts are unimodular. Then, yes, the fact that $\psi$ is analytic on $U$ and has a unique analytic continuation to all of $M$ coupled with the fact that $\mu$ is analytic on the whole of $M$ implies that the analytically continued $\psi$ satisfies $\psi^*(\mathrm{d}x)=\mu$ globally, since these analytic functions are equal on a nonempty open set. – Robert Bryant Jul 17 '17 at 17:42
• @AliTaghavi: The question is how do you do the rescaling? You need to produce a decision procedure to determine how to scale each member of the given atlas so as to get a new atlas with unimodular transitions. I am saying that you need to specify exactly how you are using the hypothesis of simple-connectivity to construct this decision procedure before I will believe that you know how to do it. The way I chose (although it could be avoided by an appeal to Cech cohomology) was to construct a flat connection and then appeal to simple-connectivity to provide the global flat section $\mu$. – Robert Bryant Jul 18 '17 at 13:34

Brief sketch of slight simplification of Bryant's answer:

Without loss of generality $\mathcal A$ is maximal with respect to the condition that all transition functions are polynomial. Now make a space $\tilde M$ by gluing together $U$ and $V$ whenever $(\phi, U)$ and $(\psi,V)$ are charts such that $\phi$ and $\psi$ agree in $U\cap V$. This is a smooth manifold (or rather each of its connected components is), because polynomials are so rigid. It is equipped with both a covering projection to $M$ and an immersion in $\mathbb R^n$. So if $M$ is simply connected then $M$ immerses in $\mathbb R^n$. If $M$ is also compact and $n>0$ then this is impossible.

• To explain why $\tilde M$ is a covering space of $M$: If a diffeomorphism between open sets of $\mathbb R^n$ is given by a polynomial formula and the same holds for the inverse diffeomorphism, then the same formulas give diffeomorphisms $\mathbb R^n\to \mathbb R^n$. This fails if we say "rational" or analytic" rather than "polynomial". – Tom Goodwillie Jul 17 '17 at 19:09
• Thank you very much for your answer. My apology if my question is elementary: May I ask you to elaborate the construction of $\tilde{M}$? – Ali Taghavi Jul 17 '17 at 19:46
• For each chart $(\phi,U)$ in the atlas, take a copy of the space $U$; call it $U_\phi$. Form the disjoint union of $U_\phi$ over all charts. Now make a quotient space as follows. Given charts $(U,\phi)$ and $(V,\psi)$, we identify a point $p\in U_\phi$ with a point $q\in V_\psi$ if and only if $p=q\in M$ and $\phi$ and $\psi$ coincide in some neighborhood of $p=q$. This quotient space is $\tilde M$. – Tom Goodwillie Jul 18 '17 at 2:05
• For each connected chart in $\mathcal A$ the canonical map $U_\phi\to\tilde M$ maps $U_\phi$ homeomorphically to an open subset. The space $\tilde M$ is Hausdorff. There is a map $\pi:\tilde M\to M$ given by sending each $U_\phi$ to $U$ by "the identity". To see that $\pi$ is a covering projection, argue as follows. Suppose that $(U,\phi)$ is a chart in $\mathcal A$. Then, for every polynomial map $h:\mathbb R^n\to \mathbb R^n$ with polynomial inverse, $(U,h\circ\phi)$ is again a chart in $\mathcal A$. I claim that $\pi^{-1}(U)$ is the disjoint union of the open sets $U_{h\circ\phi}$. – Tom Goodwillie Jul 18 '17 at 2:30
• You're welcome! Thanks for your interesting question. – Tom Goodwillie Jul 19 '17 at 12:11

Let me develop Robert Bryant's answer : if $M$ is a simply-connected manifold of dimension $n$ with a polynomial atlas, then there exists a local isomorphism $M \rightarrow \mathbb{R}^n$. Conversely, such a local isomorphism yields a polynomial atlas.

Indeed let $\mathcal{F} \subseteq \mathcal{C}^{\infty}_M$ be the subsheaf of polynomial functions (with respect to the given atlas). Then the pair $(M,\mathcal{F})$ is locally isomorphis to an open subset $U$ of $\mathbb{R}^n$ together with the constant sheaf associated to $\mathbb{R}[X_1,\dots,X_n]$. Thus $\mathcal{F}$ is a local system.

Since $M$ is simply-connected, the locally constant sheaf of $\mathbb{R}$-algebras $\mathcal{F}$ is constant. Thus there is an isomorphism $\mathbb{R}[X_1,\dots,X_n]_M \simeq \mathcal{F}$. The images of $X_1,\dots,X_n$ are smooth functions on $M$, which give a local isomorphism $M \rightarrow \mathbb{R}^n$ (compatible with the polynomial atlas).

Note that the map $M \rightarrow \mathbb{R}^n$ has open non-empty image, hence non-compact image if $n >0$. Thus such a $M$ is never compact for $n>0$.

• What exactly do you mean by a constant sheaf? The ones I know only admit sections corresponding to locally constant functions... – მამუკა ჯიბლაძე Jul 17 '17 at 14:40
• A constant sheaf is indeed the sheaf of locally constant functions with values in some given set. A locally constant sheaf (= local system) is a sheaf locally of this form. The latter are in $1$-$1$ correspondance with sets endowed with an action of the fundamental group $\pi$. The action is trivial (e.g. if $\pi = 1$) iff the sheaf is constant. – js21 Jul 18 '17 at 5:43
• But the sections we need must be all polynomial functions, they are rarely locally constant? – მამუკა ჯიბლაძე Jul 18 '17 at 5:44
• Ok. I am identifying polynomials functions on a given connected open subset of $\mathbb{R}^n$ with $\mathbb{R}[X_1,\dots,X_n]$. Is it what you wanted to point out ? – js21 Jul 18 '17 at 5:59
• Oh I see now, sorry. There is an isomorphism of sheaves depending on chosen polynomial atlas. In fact giving such isomorphism should be more or less equivalent to giving a polynomial atlas. Thanks! – მამუკა ჯიბლაძე Jul 18 '17 at 6:10