Timeline for Integration of a function over 7-sphere
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 8, 2022 at 13:56 | comment | added | Deane Yang | No, $\hat{z} = rz$, where $r \in [0,\infty)$ and $z \in S^7$ is polar coordinates. Therefore, $\hat{z}^k = rz^k$ and $$ |\hat{z}|^2 = |\hat{z}^1|^2 + \cdots + |\hat{z}^4|^2 = r^2$$ and $|z^1|^2 + \cdots + |z^4|^12= 1$. But $|\hat{z}^1| \ne r |z^1|$. Then change of variables for polar coordinates says $$d\hat{z} = r^7\,dr\,dz, $$ where $d\hat{z}$ is the standard volume form on $\mathbb{C}^4$ and $dz$ is the standard volume form on $S^7$. | |
| Dec 7, 2022 at 23:41 | comment | added | Siddharth Prakash | @DeaneYang I was working on the same question and had a few questions about the formula. I understand that $\hat{z}_{1} \in C^4 $ but what is $z_1$? Is $|\hat{z}_{1}|=r|z_1|$ ? I want to know more about the change in variable formula you have used and learn how the Jacobian becomes $r^{7}$ ? Where can I find the derivation for the formula ? | |
| May 31, 2021 at 3:42 | comment | added | Hrushikesh Pawar | Oh, Yes, it works !!! I was not updating the power of $r$ in the RHS of the very first equation. Now, this helped me solve a side problem about the integration of type $$\int\limits_{\mathbb{S}^7} |z_1^nz_2^m|\,dz$$, where $z = (z_1,\,z_2,\,z_3,\,z_4)$. The next step would be to work it out for $|z_1z_4 - z_2z_3|^k$. Thank you for guiding me. | |
| May 29, 2021 at 15:33 | comment | added | Deane Yang | This should still work. You have to be careful with the powers of $r$. | |
| May 29, 2021 at 15:18 | comment | added | Hrushikesh Pawar | Problem. The method of integration (regarding the last two integrals) works only for this particular case. ACCORDING TO MATHEMATICA, if I take just $|z_1|$, instead of $|z_1z_4|$ or any other combination, the answer comes wrong. PS. Can I open a new question regarding this particular integral? | |
| May 29, 2021 at 13:51 | comment | added | Deane Yang | Yes. That is correct. | |
| May 29, 2021 at 8:40 | comment | added | Hrushikesh Pawar | Thank you for the Answer. This is by far the simplest to understand among the three. Just a quick question, we have set $\hat{z}\in \mathbb{C}^4$, but then while integrating, we are taking it over $\mathbb{R}^8$. First, is this allowed? If yes then, can the same be done for the last two integrals. Instead of an integral over $\mathbb{C}$, we integrate over $\mathbb{R}^2$, and then we get double integral over $S^1$ and $r$. | |
| May 28, 2021 at 20:28 | history | answered | Deane Yang | CC BY-SA 4.0 |