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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!

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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 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 Mar 31 '13 at 17:03
up vote 3 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".

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Dear @Brian Ng, what particular part is unclear or needs more details? – Ben Jan 4 at 16:10

Certainly the induced embedding $i: \mathbb{P}^n \to \mathbb{R}^F$ separates points, e.g. homogenize any linear functional by the norm.

Let $U$ be the open subset of $\mathbb{R}^n$ given by specifying coordinates $(x_1, \dots, x_n)$. Then we can give coordinate patches to our embedded projective space by sending $(x_1, \dots, x_n)$ to $f(q(x_1, x_2, \dots, 1, \dots x_n))$ at the coordinate $f \in F$, where the $1$ occurs at the $i$th coordinate. These maps are smooth because $f \circ q$ is smooth, and they clearly cover $\mathbb{P}^n$.

Now, let $x_1, x_2, \dots, x_{n+1}$ be coordinates on $\mathbb{R}^{n+1}$, the domain of $q$, that we implicitly picked above. $n$ of the functions in $F$ are the coordinates$${{x_1}\over{|x|}},\,{{x_2}\over{|x|}},\,{{x_3}\over{|x|}}, \dots,\, {{x_{n+1}}\over{|x|}},$$and we can give a smooth inverse to the $i$th patch above by sending a point with values $a_1$, $\dots$, $a_{n+1}$ at those coordinates back to$${{a_1}\over{a_i}},\,{{a_2}\over{a_i}},\dots,\,{{a_{n+1}}\over{a_i}},$$where the term $a_i/a_i = 1$ is omitted, on the open subset away from $x_i = 0$.

Finally, for any function $f \in F$, the function is just projection otno the corresponding coordinate of our embedding, so certainly it is smooth. Conversely, if the function $f$ is smooth, $f \circ q$ is certainly somoth since $q$ is a smooth map, since its composition with the inverse coordinate patch functions from the last paragraph is clearly smooth.

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