Timeline for The (co)tangent sheaf of a topological space
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 18, 2019 at 0:21 | comment | added | Qfwfq | @IB: indeed, I left out the $\Omega_{X,\textrm{Kaehler}}$ from the discussion precisely because I had no idea if it had any relation with the usual ("correct") sheaf of de Rham differential forms in the differentiable case. Thank you for pointing out that in the diff. setting $(\Omega_{X,\textrm{Kaehler}})^\vee= T_X$ and $(\Omega_{X,\textrm{Kaehler}})^{\vee\vee}=\Omega_X$. (You mean the usual algebraic dual of sheaves of modules (no topology involved on each $\mathcal{O}_X$), right?) | |
| Apr 17, 2019 at 13:58 | comment | added | Ingo Blechschmidt | Just a comment, in the smooth situation, there is a third sheaf we could consider, namely the sheaf of Kähler differentials (as in algebraic geometry). This sheaf does not coincide with the sheaf of (correct) differential forms, but the dual of that sheaf is $T_X$ (with $\mathcal{O}_X$ the sheaf of smooth functions) and the dual of $T_X$ is the sheaf of (correct) differential forms. | |
| Apr 11, 2019 at 21:34 | vote | accept | Qfwfq | ||
| Apr 11, 2019 at 21:11 | answer | added | Tom Goodwillie | timeline score: 11 | |
| Apr 11, 2019 at 3:40 | comment | added | Theo Johnson-Freyd | @TomGoodwillie May I request that you post this as an "answer" so that the OP may accept it? | |
| Apr 10, 2019 at 23:49 | comment | added | Tom Goodwillie | $T_X$ is always $0$. If $D$ is a derivation and $f$ is a function, then for every point $x$ $Df$ vanishes at $x$; it suffices to prove this when $f(x)=0$, and in that case $f=gh$ where both $g(x)=0=h(x)$, so $Df=gDh+hDg$ vanishes at $x$. | |
| S Apr 10, 2019 at 22:38 | history | suggested | Ali Taghavi |
I add a tag.
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| Apr 10, 2019 at 22:30 | review | Suggested edits | |||
| S Apr 10, 2019 at 22:38 | |||||
| Apr 10, 2019 at 20:43 | comment | added | Qfwfq | Also, may it be the case that when $X$ is locally Euclidean $T_X$ is just a globally free sheaf of infinite rank? | |
| Apr 10, 2019 at 20:39 | comment | added | Qfwfq | Oh that's true, I didn't think about that | |
| Apr 10, 2019 at 20:31 | comment | added | André Henriques | $\mathcal I/\mathcal I^2$ is often zero. In particular, it is zero for $X=\mathbb R$. | |
| Apr 10, 2019 at 20:24 | history | asked | Qfwfq | CC BY-SA 4.0 |