Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question concerns problem 1-B in the book of Milnor and Stasheff part a. They first define the set $F:=\{f:\mathbb{R}P^n\rightarrow\mathbb{R} \mid \text{$f\circ q$ is smooth}\}$ where $q:\mathbb{R}^{n+1}-{0}\rightarrow\mathbb{R}P^n$ that sends $x$ to $\mathbb{R}x$. The problem is to show that $F$ is a smoothness structure on $\mathbb{R}P^n$.

It is my understanding that to do this I must do the following: First I should show that the set of functions in $F$ separates points on $\mathbb{R}P^n$, then show that $i(\mathbb{R}P^n)\subset\mathbb{R}^{F}$ is a smooth manifold where $i_{f}(x)=f(x)$ for $f\in F$. Finally I should show that $F$ is the set of all smooth real valued functions on $\mathbb{R}P^n$.

1) What are some candidates for functions in $F$ that separate points? My first guess was maybe I should play with trignometric functions but then I think smoothness becomes an issue. Perhaps the $f_{ij}$ functions defined in part b work?

2) I dont know how to get my hands on $i(\mathbb{R}P^n)$ because $F$ could be infinite. What charts can you use? (I know what charts you should use if you use the definition of manifold given in a book like Lee's).

I hope things are clear, thanks!

share|improve this question
1  
Welcome Anette, a) shouldn't it be $f\circ g$? b) You understood what's to do correctly. c) As a hint for 1), how would you separate two points in $\mathbb{R}^{n+1}-0$ smoothly if they don't lie on a common line through the origin? Can you do this such that the map factors through $q$? If so, you are done. d) Ad 2), it should be possible to transport the standard charts to $i(\mathbb{R}P^n)$ (via $i$ of course) and show what's needed to show over there.Unfortunately I don't have the time to expand this to an answer right now. –  Ben A. Mar 28 '13 at 15:11
    
Thanks Ben! If I understand your hint correctly then I think the $f_{ij}({x})=x_ix_j/\Sigma{x_k^2}$ will do the trick since if the values of two points are equal under these maps for all $i$ and $j$ then the two points must be equal. I'm still thinking about 2) then... –  Anette Mar 28 '13 at 15:47
    
I really like your idea to pick the $f_{ij}$ to separate points! Now I thought about 2) too. In principle it should be possible to show, that the images of the $n+1$ standard open covering charts provide an atlas of $i(\mathbb{R}P^n)$ via the canonical homeomorphisms $\mathbb{R}^n\to i(\text{one of the std. charts})$. But I have to admit, the maximal rank condition to these maps are really struggling me. There may be a more intelligent choice of a covering, but then I don't see it. –  Ben A. Mar 31 '13 at 17:03

1 Answer 1

up vote 2 down vote accepted

You already got 1) in the comments, so we know that $i$ is injective. Now to get the idea: if we already knew that $\mathbb{R}P^n$ is a smooth manifold, then Milnor and Stasheff remark, that $i$ is an diffeomorphism onto its image. Therefore there must be a way to show that for an atlas of charts $U_j\subset\mathbb{R}P^n$ (in the abstract definition) of projective space, their images $i(U_j)\subset\mathbb{R}P^n$ provide an atlas of charts (in the embedded definition) too. Let's take the standard charts $$U_j = \{(x_0:x_1:\dots:x_{j-1}:1:x_{j+1}\dots:x_n)\in\mathbb{R}P^n\},\;j=0\dots n$$ with the obvious homeomorphism $\varphi_j\colon\mathbb{R}^n\to U_j$. We should show step by step, that

i) for all $j$, the map $i\circ\varphi_j\colon\mathbb{R}^n\to i(U_j)$ is a homeomorphism and considered as a map $\mathbb{R}^n\to i(U_j)$ it is smooth, and that

ii) the derivatives of this maps have maximal rank $n$ everywhere.

In general, injective closed maps are homeomorphisms onto their image. And as $\mathbb{R}P^n$ is compact and $\mathbb{R}^F$ is Hausdorff, $i$ indeed is closed, thus $i\colon\mathbb{R}P^n\to i(\mathbb{R}P^n)$ is an homeomorphism. Consequently, for all $j$, the composition $i\circ\varphi_j\colon\mathbb{R}^n\to i(U_j)$ is a homeomorphism as a composition of such. To see i) completely, just have a look at the definition of smothness of maps to $\mathbb{R}^F$.

Ad ii). For sake of simplicity, we shall only consider the case $j=0$ and denote $\psi:=i\circ\varphi_0$. (The other cases really are analogous.) Here the map is given by $\psi\colon (x_1,\dots,x_n)\mapsto (f(1,x_1,\dots,x_n))_{f\in F}$.

[Unfortunately I didn't find a more illustrative way to proof ii) in this example, so let me imitate how I would proof the equivalence of the definitions we come across here in general, though this is rather technical. I'll skip some details, let me know if you want some more.]

The claim is that for each $x\in\mathbb{R}^n$ the vectors $\partial\psi/\partial x_k$ are linearly independent at $x$. Therefore it suffices to find $n$ functions $f_1,\dots f_n\in F$ such that the derivative of the map $\mathbb{R}^n\to \mathbb{R}^{\{f_1,\dots,f_n\}}$, $(x_1,\dots,x_n)\mapsto (f_k(1,x_1,\dots x_n))_k$ is regular at $x$. Obeserve that this is the same as taking $D\psi$ at $x$ and then projecting to $\mathbb{R}^{\{f_1,\dots,f_n\}}\subset\mathbb{R}^F$ and therefore if this is surjective, $\psi$ has to be of full rank.

Taking a smooth bump function $\rho\colon\mathbb{R}^n\to\mathbb{R}$ at $x\in\mathbb{R}^n$, (i.e. such that for some compact neighborhoods $V\subset U\subset\mathbb{R}^n$ of $x$, $\rho|V=1$ and $\rho|_{\mathbb{R}^n-U}=0$ constantly, ) we find smooth functions $f_k\colon\mathbb{R}^n\to\mathbb{R}$ ($k=1, 2,\dots n$), via $f_k(x_1,\dots x_n) = x_k\rho(x)$, that behave like projections near $x$ and vanish sufficiently far away. Hence the induced maps $f_k\circ\varphi_0^{-1}\colon U_0\to\mathbb{R}$ extend smoothly (i.e. lying in $F$) to whole $\mathbb{R}P^n$ by zero; by abuse of notation let's denote the extensions by $f_k$ too.

Sufficiently close to $x$ we have $f_k(1,x_1,\dots x_n) = x_k$ by construction, thus $$\left(\frac{\partial f_k(1,x_1,\dots x_n)}{\partial x_j}(x)\right)_{j} = (0,\dots 0, 1, 0, \dots 0)$$ with the $1$ in the $k$-th component. For short, if you prefer, $\left(\frac{\partial f_k(1,x_1,\dots x_n)}{\partial x_j}(x)\right)_{j,k} = \delta_{j,k}$. This gives the desired and shows that $\psi$ is "immersive".

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.