Timeline for Dividing a n- cochain by a 1-cochain
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 28, 2015 at 14:28 | comment | added | Ali Taghavi | @NeilStrickland As I said to Alex Degtyarev, we need to a non vanishing type condition for our 1-cochain. Without this, even in derham cohomology, we can not divid by 1-forms. | |
| Jan 28, 2015 at 11:42 | comment | added | Thomas Rot | @AlexDegtyarev: note OP is taking coefficients in $\mathbb{R}$. | |
| Jan 28, 2015 at 10:37 | comment | added | Neil Strickland | The corresponding question in cohomology also has a negative answer. If $X$ is the plane with $n$ points removed, then $H^*(X)$ is freely generated over $\mathbb{Z}$ together with classes $\alpha_1,\dotsc,\alpha_n$ satisfying $\alpha_i\alpha_j=0$ for all $i$ and $j$. | |
| Jan 28, 2015 at 9:11 | comment | added | Ali Taghavi | @EricWofsey is it a good singular versionof non vanishing property? $\alpha(\gamma) \neq 0$ for a $\gamma$ ending at $x$? | |
| Jan 28, 2015 at 9:06 | comment | added | Ali Taghavi | @AlexDegtyarev according to your statment $\alpha \smile \alpha \neq 0$, what about if we consider the main question in cohomology classes? In cohomology $\alpha \smile \alpha=0$ | |
| Jan 28, 2015 at 9:06 | comment | added | Eric Wofsey | As Alex remarked, this question is rather less natural for singular cochains than it is for de Rham cochains; the condition that $\alpha\smile\beta=0$ is extremely strong for singular cochains. For instance, suppose $\alpha$ is "nowhere vanishing" in the fairly weak sense that for every $x\in X$, there is some path $\gamma$ ending at $x$ such that $\alpha(\gamma)\neq 0$. Then if $\alpha\smile\beta=0$, $\beta$ itself must be $0$! | |
| Jan 28, 2015 at 8:42 | comment | added | Ali Taghavi | such that the restriction of $\alpha$ to each $L_{t}$ is identically zero? | |
| Jan 28, 2015 at 8:40 | comment | added | Ali Taghavi | @AlexDegtyarev befor this post I was thinking to singular cohomological version of "Non vanishing one form" in term of cochain. Is it a good idea to define $\alpha$ vanish at a point $z\in X$ if for every curve passing $z$, the pull bach cochain is vanishing at a point of interval?(with an appropriate definition, for the intervalcase). In fact My main motivation was the following: Let $\alpha$ be a non vanishing 1-cochain on compact n dimensional space $X$(not necessarilly manifold) with $\alpha \smile \sigma \alpha=0$. can we divid $X$ to disjoint union of n-1 dim subspace $L_{t}$.... | |
| Jan 28, 2015 at 8:18 | comment | added | Alex Degtyarev | What if $\alpha=0$. In fact, it's completely unclear what the relation between the two conditions is as, in general, $\alpha\smile\alpha\ne0$ for a $1$-cochain (or even cocycle) $\alpha$. | |
| Jan 28, 2015 at 8:03 | history | asked | Ali Taghavi | CC BY-SA 3.0 |