Pedro Lauridsen Ribeiro
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 Nov 10 comment If $\alpha''$ and $\alpha'$ are collinear, then $\alpha$ is a pre-geodesic. $\alpha'(s)$ being null everywhere is not so great a restriction as it seems. Any $s$-dependent rescaling of $\alpha'$ will keep it null, and this is essentially all the freedom we have here. As for your second question, you are correct - the formula is not that pretty indeed, but it is consistent with the fact that affine reparametrizations keep the geodesic character (which amount to the freedom in choosing the constants in the primitives). Nov 10 comment If $\alpha''$ and $\alpha'$ are collinear, then $\alpha$ is a pre-geodesic. In my opinion, item (d) is written somewhat imprecisely. (by the way, it should be $f = (\log g')'$ in my previous comment - unfortunately, I can no longer edit it) Nov 10 comment If $\alpha''$ and $\alpha'$ are collinear, then $\alpha$ is a pre-geodesic. The hypothesis $\langle\alpha',\alpha'\rangle=0$ doesn't entail constant speed, as my counterexample above shows. To see the aforementioned equivalence, let $\beta:[a,b]\rightarrow M$ be a geodesic, and $g:[c,d]\rightarrow[a,b]$ a diffeomorphism (i.e. a reparametrization). Then $$(\beta\circ g)''(t)=\frac{g''(t)}{g'(t)}(\beta\circ g)'(t)=(\log g'(t))'(\beta\circ g)'(t)$$ for all $t$ by the chain and Leibniz rules, thus yielding $f(t)=(\log g)'(t)$. Reversing the above formula proves the converse statement. Nov 10 comment If $\alpha''$ and $\alpha'$ are collinear, then $\alpha$ is a pre-geodesic. No. This would imply that $\alpha$ is a geodesic, which is in general not true. Take for instance a null geodesic $\beta$ and a non-affine reparametrization $g$ (e.g. $g(t)=\tanh t$). Then $\alpha=\beta\circ g$ is a pre-geodesic which is not a geodesic, but still $\langle\alpha',\alpha'\rangle=0$ everywhere by the chain rule. Nov 10 answered If $\alpha''$ and $\alpha'$ are collinear, then $\alpha$ is a pre-geodesic. Jun 13 answered About timelike surfaces with non-diagonalizable shape operator. Mar 7 comment Translation of a polynomial Yep, you can sleep sound now ;-) Mar 7 comment Translation of a polynomial The second line is still wrong... :-P Mar 7 comment Translation of a polynomial In the second line of your last sequence of formulas, I believe that the exponent of $z_0$ is incorrect - it should be $k-j$ instead of $n-j$ since you are expanding $(z+z_0)^k$ with the inner sum (oops, Reveillark was faster than me, bummer) Dec 28 awarded Yearling Dec 15 awarded Caucus Nov 16 answered Gathering books on Lorentzian Geometry Feb 22 suggested rejected edit on Are $L_p$ spaces of functions with separable support separable? Jun 6 comment Open neighborhoods of a $G_\delta$ set @Cameron: nice indeed, but see my last comment on Asaf's answer... Jun 6 comment Open neighborhoods of a $G_\delta$ set It seems that disconnectedness of some of the sets involved plays a common role in all these counter-examples... My last move: what if $X$ is complete and connected, $A$ is connected and $A_n$ is connected for all $n\in\mathbb{N}$? Jun 6 awarded Commentator Jun 6 comment Open neighborhoods of a $G_\delta$ set Never mind, Asaf provided a suitable counter-example... Jun 6 comment Open neighborhoods of a $G_\delta$ set I've deleted a pair of messed-up comments before I could see your reply to one of them... You're right, $X$ here is totally disconnected. Jun 6 awarded Scholar Jun 6 awarded Supporter