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I am trying to solve an exercise about building an atlas for the projective space. For completeness, we work with the projective space as:

$\mathbb{RP}^n=\{L\subset\mathbb{R}^{n+1}|L\text{ 1-dimensional subspace of } \mathbb{R}^{n+1}\}$

We endorse $\mathbb{RP}^n$ with the topology induced by the distance:

$d(L,L')\colon = \text{inf}||x-x'||$, where $x\in L\cap S^n$, $x'\in L'\cap S^n$.

I.e., the distance is computed at the intersection between the lines in the projective space and the unit sphere $S^n$.

Now let's define our atlas on $\mathbb{RP}^n$. Let $L\in\mathbb{RP}^n$ and let $H\subset\mathbb{R}^{n+1}$ be a 1-codimensional linear subspace (hyperplane) such that $L\cap H = \{0\}$ (hence $\mathbb{R}^{n+1}=L\oplus H$). Let the open sets be:

$$U_{L,H}\colon =\{\Lambda\in\mathbb{RP}^n|\Lambda\cap H = \{0\}\}$$

And the coordinate maps:

\begin{align} \varphi_{L,H}\colon U_{L,H}&\longrightarrow \text{Hom}(L,H)\cong \mathbb{R}^{n}\\ \Lambda &\longmapsto \psi\colon L \rightarrow H \end{align} with $\psi$ defined by the condition that $\Lambda=\{(l,\psi(l))|l\in L\}$, i.e. $\Lambda$ being the graph of $\psi$.

I have to prove that

1) $U_{L,H}$ are open sets.

2) $(U_{L,H},\varphi_{L,H})$ is a chart.

3) $\{(U_{L,H},\varphi_{L,H})\}_{L,H}$ is an atlas on $\mathbb{RP}^n$.

So far, I think I have managed only 1).

How can I prove 2) and 3)? I am currently stuck on showing that $\varphi_{L,H}$ is a homeomorphism, i.e., continuous with continuous inverse.

Thanks a lot for your help.

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  • $\begingroup$ How is the topology on $\mathbb{RP}^n$ introduced? Via your metric $d$? $\endgroup$
    – Paul Frost
    Oct 7, 2019 at 13:47
  • $\begingroup$ Do you know that all norms on finite-dimensional real vector spaces are equivalent? $\endgroup$
    – Paul Frost
    Oct 7, 2019 at 14:06
  • $\begingroup$ @PaulFrost Yes, exactly. Maybe I was a bit misleading leaving it for the end. I can edit it if you think it's needed. $\endgroup$
    – Txordi
    Oct 7, 2019 at 16:43
  • $\begingroup$ @PaulFrost Yes, I know. But does it solve something? $\endgroup$
    – Txordi
    Oct 7, 2019 at 16:44
  • $\begingroup$ @PaulFrost I edited to introduce the distance at the beggining. $\endgroup$
    – Txordi
    Oct 7, 2019 at 18:05

1 Answer 1

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You define maps $\varphi_{L,H}\colon U_{L,H} \to \text{Hom}(L,H)$. These cannot be genuine coordinate maps because you need maps whose range is an open subset of $\mathbb R^n$. Thus we prolong $\varphi_{L,H}$ by a linear isomorphism $\text{Hom}(L,H) \to \mathbb R^n$. In fact, choose $\ell \in L \setminus \{ 0 \}$ and a linear isomorphism $f : H \to \mathbb R^n$. Define $A_\ell : \text{Hom}(L,H) \to H, A_\ell(\psi) = \psi(\ell)$. This gives us a plethora of maps $f \circ A_\ell \circ \varphi_{L,H} \colon U_{L,H} \to \mathbb R^n$ and we shall show these are all homeomorphisms. The vector spaces $\mathbb R^{n+1}, H \subset \mathbb R^{n+1}$ and $\mathbb R^n$ are endowed with the Euclidean norm $\lVert - \rVert$.

$f : H \to \mathbb R^n$ is a homeomorphism. This can be shown directly by simple computations or by invoking the fact that all norms on finite-dimensional real vector spaces are equivalent. It therefore suffices to show that $\varphi_{L,H,\ell} = A_\ell \circ \varphi_{L,H}$ is a homeomorphism.

Now let us return to the definition of $\varphi_{L,H}$. We have $\mathbb{R}^{n+1}=L\oplus H$, but this is an ambiguous notation: We may interpret $L\oplus H$ as the set of all pairs $(l,h)$ with $l \in K, h \in H$, but writing $\mathbb{R}^{n+1}=L\oplus H$ means that we have to understand it as in internal direct sum of subspaces (i.e. $L + H = \mathbb{R}^{n+1}, L \cap H = \{ 0 \}$). Thus let us identify the pair $(l,h)$ with $l + h$.

We have $\varphi_{L,H,\ell}(\Lambda) = h$ where $h \in H$ is the unique element such that $\ell + h \in \Lambda$. For $x \in \mathbb R^{n+1} \setminus \{ 0 \}$ let $L(x)$ be the one-dimensional subspace of $\mathbb R^{n+1}$ containing $x$. Clearly, for $t \ne 0$ we have $L(x) = L(tx)$. Define $g : H \to U_{L,H}, g(h) = L(\ell + h)$. By construction we see that $\varphi_{L,H,\ell} \circ g = id$ and $g \circ \varphi_{L,H,\ell} = id$. Thus $\varphi_{L,H,\ell}$ is a bijection whose inverse is $g$.

  1. If $(x_n)$ is sequence in $\mathbb R^{n+1} \setminus \{ 0 \}$ converging to $x \in \mathbb R^{n+1} \setminus \{ 0 \}$, then $L(x_n) \to L(x)$. In fact, we have $\lVert x_n \rVert \to \lVert x \rVert$, thus $\frac{x_n}{\lVert x_n \rVert} \to \frac{x}{\lVert x \rVert}$. We conclude $d(L(x_n),L(x)) = d(L(\frac{x_n}{\lVert x_n \rVert}),L(\frac{x}{\lVert x \rVert})) \le \lVert \frac{x_n}{\lVert x_n \rVert} - \frac{x}{\lVert x \rVert} \rVert \to 0$.

  2. If $(x_n)$ is sequence in $\mathbb R^{n+1} \setminus \{ 0 \}$ which has an accumulation point $x \in \mathbb R^{n+1} \setminus \{ 0 \}$, then $L(x)$ is an accumulation point of $(L(x_n))$. This follows by considering a convergent subsequence of $(x_n)$.

  3. $g$ is continuous. Let $(h_n)$ be a sequence in $H$ converging to $h \in H$. Hence $x_n = \ell + h_n \to \ell + h = x$ and 1. implies $g(h_n) = L(x_n) \to L(x) = g(h)$.

  4. $\varphi_{L,H.\ell}$ is continuous. Let $(\Lambda_n)$ be a sequence in $ U_{L,H}$ converging to $\Lambda \in U_{L,H}$. Set $h_n = \varphi_{L,H,\ell}(\Lambda_n), h = \varphi_{L,H.\ell}(\Lambda)$. Assume that $(h_n)$ is unbounded. W.l.o.g. we may assume that $\lVert h_n \rVert \to \infty$ (choose a subsequence if necessary). Define $x_n = \ell + h_n$. Then $x'_n = \frac{x_n }{\lVert x_n \rVert}$ forms a sequence in $S^n$ (which is a compact set) and therefore has an accumulation point $x' \in S^n$. We have $\lVert x_n \rVert \to \infty$ (note $\lVert x_n \rVert \ge \lVert h_n \rVert - \lVert \ell \rVert$), thus $\frac{\ell}{\lVert x_n \rVert} \to 0$. Therefore $x'$ is an accumulation point of $x''_n = x'_n - \frac{\ell}{\lVert x_n \rVert} = \frac{h_n}{\lVert x_n \rVert}$ which is a sequence in $H$. Because $H$ is closed, we conclude that $x' \in H$. Hence $L(x') \notin U_{L,H}$. By 2. $L(x'_n) = L(x_n) = \Lambda_n$ must have $L(x')$ as an accumulation point. Therefore $L(x') = \Lambda \in U_{L,H}$ which is a contradiction. We conclude that $(h_n)$ is bounded. Assume it is not convergent to $h$. Then it must have an accumulation point $h' \ne h$. This implies that $z_n = \ell + h_n$ has $\ell + h'$ as an accumulation point. Therefore $\Lambda_n = L(\ell + h_n)$ has $L(\ell + h')$ as an accumulation point. We conclude $L(\ell + h) = \Lambda = L(\ell + h')$. This means $\ell + h = t(\ell + h')$ for some $t \ne 0$. If $t = 1$, we get $h = h'$ which is impossible. Thus $t \ne 1$ and $(1-t)\ell = th' - h$ which is also impossible because $\mathbb{R}^{n+1}=L\oplus H$. This proves that $(h_n)$ is convergent to $h$.

Point 3) is now obvious: The $U_{L,H}$ form an open cover of $\mathbb{RP}^n$.

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  • $\begingroup$ Amazing. Thanks so much Paul, what a work! I was trying to prove everything by open sets. I am not used to this kind of proofs with sequences. They are really useful! I am still wondering if there is an easy way to do it only by open sets. Thanks a lot again, I was not expecting such a perfect and detailed answer! :) $\endgroup$
    – Txordi
    Oct 10, 2019 at 18:31

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