Timeline for Why can we put these conditions on coordinates of worldsheet?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Dec 31, 2022 at 6:56 | vote | accept | particle-not good at english | ||
| Dec 24, 2022 at 21:59 | comment | added | particle-not good at english | Oops, I misunderstand something. As you said, this case is 2d, so, $d\theta$ is always can be expressed as $d\theta=\alpha\wedge\theta$. Then, tha last thing I should check is whether we can cover whole a world sheet by $\phi ^+ ,\phi ^-$. I'll check it. | |
| Dec 24, 2022 at 21:15 | comment | added | particle-not good at english | Sorry for late reply I was struggling with deriving the $θ=αdϕ$ from Frobenius theorem and I got stuck. I'd like you to give me advice. I think the theorem is theorem 3.2.1 in this joelshapiro.org/Pubvit/Downloads/Frobenius_RLyons.pdf page 23 and 24. By this theorem, I reached $\theta=A_1dx^1+A_2dx^2$ but I can't come up with idea how to construct $θ=αdϕ$ from the expression. Thanks for reading as always | |
| Dec 23, 2022 at 15:39 | comment | added | Bence Racskó | @particle This Frobenius theorem: en.wikipedia.org/wiki/… . Might update the answer later with more details. Also it is unlikely that the isothermal coordinates can cover the whole worldsheet unless the worldsheet has trivial topology. For closed strings this is certainly not the case. | |
| Dec 23, 2022 at 7:31 | comment | added | particle-not good at english | And I also want to know if I can use the coordinates for a whole world sheet. | |
| Dec 23, 2022 at 7:06 | comment | added | particle-not good at english | Thanks for reply! "$dθ∧θ=0$, which by the Frobenius theorem implies that there are functions α and ϕ such that $θ=αdϕ$." Sorry I dont understand which Frobenius theorem you used. Could you write the assumption and claim of the theorem? | |
| Dec 22, 2022 at 14:15 | comment | added | Bence Racskó | @particle Depending on whether you take the metric as a bilinear form or a quadratic form, the product $\theta\omega$ of two $1$-forms is either defined to be the symmetric tensor product $\theta\omega=(1/2)[\theta\otimes\omega+\omega\otimes\theta]$ or the "pointwise product" $(\theta\omega)(X)=\theta(X)\omega(X)$ valid for any vector field $X$. Both are commutative products and both reproduce the desired result. | |
| Dec 22, 2022 at 9:31 | comment | added | particle-not good at english | Thank you for answering! $ds^2=(\theta_1+\theta_0)(\theta_1-\theta_0)$ how do you get this? In other words, $θ_1θ_0＝θ_0θ_1$ is true? | |
| Dec 22, 2022 at 8:10 | history | edited | Bence Racskó | CC BY-SA 4.0 |
added 225 characters in body
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| Dec 22, 2022 at 8:04 | history | answered | Bence Racskó | CC BY-SA 4.0 |