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I'm trying to prove the next:

To show that a regular curve $\alpha$ with $\alpha^{'}$ and $\alpha^{''}$ collinear is a pregeodesic, write $\alpha^{''}(s)=f(s)\alpha^{'}(s)$ and prove that

a) $\beta=\alpha\circ h$ is a geodesic if and only if $h''+ (f\circ h) (h')^2 = 0$.

b) If $\langle \alpha',\alpha'\rangle $ is never zero, then any constant speed reparametrization of $\alpha$ is a geodesic.

Suposse $\beta$ has unit speed, so $\langle \beta''(s), \beta'(s)\rangle = 0$ for all $s.$ From here $h'(s) ( h''(s) + f( h(s)) (h'(s))^2) \langle \alpha'(s),\alpha'(s)\rangle = 0,$ because $\beta^{''}$ is as a) and $\beta^{'}=(\alpha^{'}\circ h)h^{'}.$

c)$\langle \alpha',\alpha'\rangle$ is always zero or never zero.

To see this, $\langle \alpha'(s),\alpha'(s)\rangle' = 2f(s)\langle \alpha'(s),\alpha'(s)\rangle$, then $\langle \alpha'(s),\alpha'(s)\rangle = Ce^{2\int f(s)\,{\rm d}s}$ for some integration constant $C.$

d) If $\langle \alpha',\alpha'\rangle$ is always zero, then $\alpha$ is pre-geodesic.

I've proved a), b) and c). Such points follow by some computations with $\beta^{'}$ and $\beta^{''},$ the hypotesis that $\alpha$ is regular and the first at the proposition: $\alpha^{''}(s)=f(s)\alpha^{'}(s).$

My first doubt is: How is possible write $\alpha^{''}(s)=f(s)\alpha^{'}(s)?$ I don't get how to prove this. It is part of the hypotesis?

Also I'm stuck prove d).

Second: Why the proof of the behind ensures $\alpha$ is pregeodesic?

I think the next result ensures that, if the previous holds, then $\alpha$ is pregeodesic:

Let $\gamma:I\rightarrow M$ be a nonconstant geodesic. A reparametrization $\gamma\circ h:J\rightarrow M$ is a geodesic if and only if $h$ has the form $h(t)=at+b.$ If a curve has a reparametrization as a geodesic we call it pregeodesic.

Any kind of help is thanked in advanced.

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  • $\begingroup$ Can you elaborate second and third line ? I have a difficulty in grabing your intension. $\endgroup$ – HK Lee Apr 30 '18 at 2:57
  • $\begingroup$ @HKLee Ok. I add them in the post. $\endgroup$ – Squird37 Apr 30 '18 at 3:08
  • $\begingroup$ That's what you mean @HKLee? $\endgroup$ – Squird37 Apr 30 '18 at 3:21
  • $\begingroup$ If $c$ is a plane curve with $|c'|\neq 0$, and if $c'(s)=f(s)c''(s),\ f>0$, then $c$ is a image of geodesic ? $\endgroup$ – HK Lee Apr 30 '18 at 3:24
  • $\begingroup$ .I think that is. Utilizing $\alpha^{''}(s)=f(s)\alpha^{'}(s)$ we can prove a)-d), and then conclude $\alpha$ is pregeodesic. $\endgroup$ – Squird37 Apr 30 '18 at 3:43
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(1) If $c=\alpha \circ h(s)$ is a geodesic, then $ c''=0$ implies $$ \alpha '' (h')^2+\alpha' h'' =0 $$

That is, there is $f$ s.t. $\alpha '' =f\alpha'$

(2) If $\alpha '' =f\alpha '$, then $c:=\alpha \circ h$ has unit speed. Then $ c'=\alpha'h' $ and $c'\perp c''$. That is, $c''\perp \alpha'$ so that $$ c''=\alpha '' (h')^2+\alpha' h'' = \alpha '\{ f (h')^2+ h''\} =0 $$

Hence $c$ is a geodesic.

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  • $\begingroup$ Many thanks! My only doubt is how was used d) in your proof? I can't see it. $\endgroup$ – Squird37 Apr 30 '18 at 4:18
  • $\begingroup$ If $\alpha'(t)=0$ for all time $t$, then it is a point. What is a problem ? $\endgroup$ – HK Lee Apr 30 '18 at 4:23
  • $\begingroup$ You're right! My mistake. Thanks again. $\endgroup$ – Squird37 Apr 30 '18 at 4:27

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