Timeline for What is known about this type of generalisation of de Rham cohomology?
Current License: CC BY-SA 3.0
13 events
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| Mar 30, 2017 at 14:29 | comment | added | Josh Kirklin | You are of course correct. Your objection can be cured by the use of covariant derivatives, but it seems that before asking this question I hadn't sorted it out properly in my own head first. I will do so and then edit. | |
| Mar 30, 2017 at 12:44 | comment | added | Igor Khavkine | It seems to me that your definition for the $\eth$ operator is not coordinate invariant. In particular, $\partial_z \partial_z C \eth z \ne \partial_{z'} \partial_{z'} C \eth z'$. The difference will be proportional to $\partial^2 z'/\partial z^2$. I don't know whether this is by design, but it makes me suspicious of your statement that such formulas "can easily be extended to a higher dimensional space..." or even to a different coordinate system. | |
| Mar 30, 2017 at 9:55 | history | edited | Josh Kirklin | CC BY-SA 3.0 |
deleted 168 characters in body
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| Mar 30, 2017 at 9:46 | comment | added | Josh Kirklin | I've edited the question. I'm not sure why I put $\mathbb{C}^2$, I meant $\mathbb{C}$, sorry about that | |
| Mar 30, 2017 at 9:44 | history | edited | Josh Kirklin | CC BY-SA 3.0 |
Improved notation
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| Mar 30, 2017 at 7:20 | comment | added | Sebastian Goette | It would be nice if you edited your question to make notation clear. You are working on $\mathbb C^2$, but all your differential operators involve only one coordinate $z$. Are you avoiding to write indices, or do you mean to differentiate in only one direction? Since you say you want to consider symmetric tensors all the time, all we can do here is to guess. But then any answer you get might be an answer to a different question altogether ... | |
| Mar 29, 2017 at 23:02 | comment | added | Josh Kirklin | Yes that's right. Sorry if my notation is unclear. | |
| Mar 29, 2017 at 20:02 | comment | added | Qfwfq | So, by $\partial_z$ you actually mean the standard coordinate holomorphic vector field $\partial / \partial z$? | |
| Mar 29, 2017 at 15:18 | review | Close votes | |||
| Mar 29, 2017 at 22:56 | |||||
| Mar 29, 2017 at 15:12 | comment | added | Ben McKay | I have never seen this before. | |
| Mar 29, 2017 at 15:07 | comment | added | Josh Kirklin | I'm using $\partial_z$ to just mean a $1$-tensor, I suppose I could have written $\mathrm{d}z$ instead. My $1$-double-forms are symmetric $2$-tensors. | |
| Mar 29, 2017 at 14:57 | comment | added | Sebastian Goette | Do you mean $\partial z$ instead of $\partial_z$, or are your 1-forms differential operators? And since you always write $\partial_z$ and never $\partial_w$, is all the differentiation happening in one direction? But in case you mean $\partial z\otimes\partial w$ etc. and differentiation is $\partial\otimes\partial$ and $\bar\partial\otimes\bar\partial$, then your complex is not elliptic and therefore you cannot expect a Poincaré Lemma to hold. | |
| Mar 29, 2017 at 14:38 | history | asked | Josh Kirklin | CC BY-SA 3.0 |