Timeline for Norm of Killing spinor
Current License: CC BY-SA 4.0
7 events
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| Feb 14, 2023 at 15:04 | comment | added | Partha | @SebastianGoette Thanks for the comment. I very much agree that the positive and negative spinors themselves need not be constant, my question is in your example are their norms constant? | |
| Feb 14, 2023 at 7:33 | comment | added | Sebastian Goette | No. $\Sigma^\pm$ are the $\pm i$ eigenspaces of Clifford multiplication with the radial vector of the cone. And the radial vector changes. Consider $S^2$. One spinor module can be identified with $\mathbb H$ where Clifford multiplication by the unit vectors in $\mathbb R^3$ acts as $I$, $J$, $K$ from the left and the complex unit $i$ acts on the right. The constant spinor $1$ lies in the $S^+$ at $e_1\in S^2$, and in $S^-$ at $e_2\in S^2$. | |
| Feb 13, 2023 at 19:01 | comment | added | Partha | @SebastianGoette Killing spinors on $S^{2n}$ comes from parallel spinor of the cone of $S^{2n}$, i.e., $\mathbb{R}^{2n+1}$. Parallel spinors on $\mathbb{R}^{2n+1}$ are just constants. So, doesn't it imply that the norm of the pullbacks also are constant for indivisual positive and negative parts? | |
| Feb 13, 2023 at 16:08 | comment | added | Sebastian Goette | A good example to test your intuition is a Killing spinor on $S^{2n}\subset\mathbb R^{2n+1}$. As far as I remember, the spinor bundle is trivial, with fibre the spinor module of $\mathbb R^{2n+1}$. The grading at $x\in S^{2n}$ is induced by Clifford multiplication with $x$, so as a graded bundle, the spinor bundle is no longer trivial. Killing spinors $\phi$ are constant, but $\phi_\pm$ is not. | |
| Feb 12, 2023 at 12:32 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Minor formatting and Math Jaxing
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| Feb 12, 2023 at 12:05 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title
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| Feb 12, 2023 at 11:54 | history | asked | Partha | CC BY-SA 4.0 |