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For a $\mathscr{C}^{1}$ curve $c:(-\epsilon,\epsilon) \to \mathbb{R}^{n+1} \setminus \{0\}$, let $\overline{c} = \pi \circ c$, where $\pi: \mathbb{R}^{n+1} \setminus \{0\} \to \mathbb{R}P^n$ is the natural projection.

Show that the 1-jets of two curves $\overline{c}_1, \overline{c}_2$ agree $\Leftrightarrow \: \exists \lambda \in \mathbb{R} \setminus \{0\}$, such that $c_2(0) = \lambda \cdot c_1(0)$ and $(\dot{c_2}(0)-\lambda \cdot \dot{c_1}(0) )\in \mathbb{R} \cdot c_2(0)$.

My ideas so far:
$"\Rightarrow"$ $c_1(0) \sim c_2(0)$ in $\mathbb{R}P^n$ $\Leftrightarrow c_1(0) = \lambda \cdot c_2(0)$ for some $\lambda \neq 0$.
And then for the first derivatives: $d \pi_{c_1(0)}\circ c'_1(0) = d\pi_{c_2(0)}\circ c'_2(0)$. But I don't see how this leads to the necessary condition.

Thanks in advance for any help !

Paul Joh
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  • How do you give tangent vectors at a point of $\Bbb RP^n$? – Ted Shifrin Mar 24 '22 at 01:25
  • @TedShifrin I have the chart $\phi_i:U_i \to \phi(U_i), : (x_1,...,x_n) \mapsto (\frac{x_1}{x_i}$,...,$\frac{x_{i-1}}{x_i},\frac{x_{i+1}}{x_i},...,\frac{x_{n+1}}{x_i})$. That should give me all vectors tangent to $U_i$, where $U_i = {x \in \mathbb{R}P^n | x_i \neq 0}$ – Paul Joh Mar 24 '22 at 08:55
  • Update: I just found a better interpretation: https://math.stackexchange.com/questions/941582/tangent-space-to-mathbbrpn?rq=1, that suggests that $\pi_{*}|{x}:T{x}\mathbb{S}^n \to T_p \mathbb{R} P^n$ is an isomorphism. – Paul Joh Mar 24 '22 at 09:16
  • Better yet (see a terse discussion in the complex case here, but it applies in the real situation, too), here's how you can think about $v\in T_{[p]}\Bbb RP^n$. Choose curve $\alpha$ with $\alpha(0)=[p]$ and $\alpha'(0)=v$. Let $\tilde\alpha$ be a lift to $\Bbb R^{n+1}-{0}$, and show that $\tilde\alpha'(0)\pmod p$ projects to $v$. The point is that this is all well-defined. – Ted Shifrin Mar 24 '22 at 16:02
  • Thanks for the answer. I thought about it pretty hard. Unfortunately in differential-geometry we did not talk about lifts. I have only heard about it from algebraic-topology. And I have never seen an Euler-exact sequence before. In that sense a lift would be $f:T\mathbb{R}P^n \to \mathbb{R}^{n+1} \setminus {0}$ where I can choose where to lift the basepoint along a ray from 0. And then I could choose the radial components everywhere along $\alpha(t)$. But I don't see how that leads to the projection of tangent vectors. – Paul Joh Mar 25 '22 at 21:30
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    The language of lifts is no big deal here. We just need a curve $\tilde\alpha$ in $\Bbb R^{n+1}-{0}$ so that $\pi(\tilde\alpha(t)) = \alpha(t)$. It is useful to think of tangent vector of $\tilde\alpha$ at $t=0$, say, at having a component parallel to $\tilde\alpha(0)$ and a component orthogonal. That orthogonal component maps to $\alpha'(0)\in T_{[p]}\Bbb RP^n$, independent of any choices. Thinking upstairs will make the geometry quite clear (rather than working in charts on projective space). If you still cannot sort this out, I will write more of an answer later. – Ted Shifrin Mar 25 '22 at 21:37
  • Ahh thank you @TedShifrin! I think I get the idea now. The projection of the two tangent vectors $d\pi_{[p]}(\dot{c}{1}(0)), d{[p]}\pi(\dot{c}2(0))$ are equal in $\mathbb{R}P^n$ if one is a scalar multiple of the other. But since we are losing the radial component when projecting with $\pi$ this means $\dot{c}_2(0) + \lambda{1}\cdot c_{2}(0) = \lambda_{2} \cdot \dot{c}{1}(0)$. For some $\lambda{1}, \lambda_{2} \in \mathbb{R}^{*}$. This is not very formal though so if you have a better answer that would be great. – Paul Joh Mar 26 '22 at 19:40

1 Answers1

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As you said, $\bar c_1(0)=\bar c_2(0)=\bar P$ if and only if $P=c_1(0)$ and $Q=c_2(0)$ are nonzero multiples of one another. Now we want to think about $d\pi_P$ and $d\pi_Q$. These both map $\Bbb R^{n+1}$ onto $T_{\bar P}\Bbb RP^n$, with kernel spanned by $P$.

If $Q=\lambda P$, then the curves $c_2$ and $\lambda c_1$ both pass through $Q$ at time $0$ and $d\pi_Q(\dot c_2(0)) = d\pi_Q(\lambda \dot c_1(0))$ if and only if $\dot c_2(0)=\lambda\dot c_1(0)+\mu Q$ for some scalar $\mu$. This is precisely the condition in the statement.

Ted Shifrin
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  • @PaulJoh: There's no reason to remove the origin from the domain of the differential mappings, and there is reason not to, since those maps are linear onto the tangent space. – Andrew D. Hwang Mar 26 '22 at 21:51
  • Yeah, please don’t edit my mathematics without confirming with me first. – Ted Shifrin Mar 26 '22 at 21:57
  • @AndrewD.Hwang You are right that makes sense. TedShifrin Sorry I won't. You are the expert. I thought you had to approve the edit before anyway. And thanks for all the help ! – Paul Joh Mar 27 '22 at 08:28