# Charts on projective space

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.

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

## 1 Answer

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 $$\mathbb{RP}^n$$.

• 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! :) Oct 10, 2019 at 18:31