Timeline for Uniqueness of tangent space given local injectivity of orthogonal projection onto it
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Feb 3, 2018 at 11:34 | comment | added | Dap | @Arrow: $\pi_{V'^\perp}(\lambda v)=o(\lambda)$ implies $\pi_{V'^\perp}(v)=o(1)$ as $\lambda\to 0,$ which means it's zero. (Sorry, the last version of this comment was nonsense.) | |
| Feb 3, 2018 at 10:31 | comment | added | Arrow | @Dap how to get $\pi_{V'^\perp}(v)=0$ from $g(\lambda v)=p+\lambda v+o(\lambda\|v\|)$? All I'm able to see is $\pi_{V'^\perp}(v+o(\|v\|))=0$. | |
| Jan 31, 2018 at 20:55 | comment | added | Qfwfq | @Dap: Yes, I see what you meant now. | |
| Jan 31, 2018 at 20:06 | comment | added | Dap | @Qfwfq: this argument doesn't just need an inverse from some $V_0,$ it needs $V_0$ to be a neighbourhood of $0$ so that you can approach $0$ from any direction. So the invertibility of $\pi_V|_D:D\to \pi_V(D)$ isn't enough by itself. | |
| Jan 31, 2018 at 17:05 | comment | added | Qfwfq | Just a remark: I think invariance of domain is not needed: since we know $X$ is locally Euclidean, there is a compact topological disk $D\subseteq X$ (of the same dimension as $X$) around $p$, and, up to shrinking $D$ if necessary, we can assume the restriction of the orthogonal projection to $V$ is injective on $D$. So we have a continuous bijection $\pi_{V}|_D:D\to V$ between a compact space and a Hausdorff space, hence it must be a homeomorphism onto its image $\pi_{V}(D)$. | |
| Jan 29, 2018 at 11:58 | vote | accept | Arrow | ||
| Jan 29, 2018 at 8:59 | comment | added | Dap | @Arrow: we have $\|v\|^2+\|g(v)-p-v\|^2=\|g(v)-p\|^2.$ If $s:=\frac{\|g(v)-p-v\|}{\|g(v)-p\|}\to 0$ (which follows from condition 2 + continuity of $g$) then $\frac{\|g(v)-p-v\|}{\|v\|}=\frac{s}{\sqrt{1-s^2}}\to 0$ since $s\mapsto \frac{s}{\sqrt{1-s^2}}$ is continuous at $s=0.$ | |
| Jan 28, 2018 at 22:05 | comment | added | Arrow | Understood. Forgive my ineptitude but I am not sure why $g(v)=p+v+o(\|v\|)$. We have $\| gv-p-v\|=\| \pi_{V^\perp} \circ (\theta_{-p}\circ g)(v) \|$ so it seems condition 2 only gives $gv-p-v\in o(\| \theta_{-p}\circ g(v) \|)$. What am I missing here? ($\theta_{-p}$ is subtraction by $p$.) | |
| Jan 28, 2018 at 19:23 | comment | added | Dap | @Arrow: 1) orthogonal projections restricted to subspace $X\subset\mathbb R^n$ aren't always open, so you need some extra property. 2) no, that doesn't make sense since $X_0$ isn't a vector space - try an example like the unit sphere. | |
| Jan 28, 2018 at 18:44 | comment | added | Arrow | Dear Dap, I have two questions regarding your elegant answer. 1) Do we really need invariance of domain? I think orthogonal projections are open so local injectivity gives local homeomorphy. 2) Won't the inverse $g:V_0\to X_0$ in fact be affine, as the set-theoretical inverse of an affine map? | |
| Jan 28, 2018 at 17:53 | history | edited | Dap | CC BY-SA 3.0 |
added 5 characters in body
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| Jan 28, 2018 at 17:32 | history | answered | Dap | CC BY-SA 3.0 |