Timeline for Weitzenböck Identities
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Sep 13, 2023 at 11:23 | comment | added | Liviu Nicolaescu | @EricArnéoVespiraKengne That is one motivation. The square of a Dirac is the sum of a positive operator + a zeroth order term that has a geometric meaning involving some curvature. If that 0th order term is also positive then the operator has no kernel. Many famous vanishing theorems in geometry have the form "if some form of curvature is positive then certain objects are trivial. E.g. Ricci positive implies b_1=0, Kodaira vanishing theorems are of this kind. | |
| Sep 12, 2023 at 17:16 | comment | added | Eric Arnéo Vespira Kengne | Thank you very much. So, the "Weitzenbock machinery" could be a means (perhaps the most efficient one) to achieve some kind of "geometric bootstrap"... | |
| Oct 25, 2013 at 11:23 | history | edited | Liviu Nicolaescu | CC BY-SA 3.0 |
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| Sep 18, 2012 at 9:47 | history | edited | Liviu Nicolaescu | CC BY-SA 3.0 |
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| Sep 17, 2012 at 8:28 | vote | accept | Michael Albanese | ||
| Sep 17, 2012 at 8:17 | history | edited | Liviu Nicolaescu | CC BY-SA 3.0 |
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| Sep 17, 2012 at 8:05 | comment | added | Liviu Nicolaescu | You're right The correct form is as in my book $\partial+\partial^*$. One needs the $1/{\sqrt{2}}$ factor only to get a Dirac operator. I will correct the post. | |
| Sep 17, 2012 at 0:50 | comment | added | Michael Albanese | Sorry if I am missing something, but on page 530 of your notes you define the Hodge-Dolbeault operator to be $\bar{\partial} + \bar{\partial}^*$ but above you define it to be $\mathscr{D} = \frac{1}{\sqrt{2}}(\partial + \bar{\partial})$. Furthermore, isn't $\mathscr{D}(\Omega^{0,even}(M)) \subseteq \Omega^{1,even}(M)\oplus\Omega^{0,odd}(M)$? | |
| Sep 16, 2012 at 23:28 | comment | added | Liviu Nicolaescu | Look at $\mathscr{D}=\frac{1}{\sqrt{2}}(\partial+\bar{\partial}): \Omega^{0,*}\to\Omega^{0,*}$.Then $\mathscr{D}^2$ is a Laplacian. The operator $\mathscr{D}$ is an example of a *graded Dirac, i.e., first order symmetric $\matscr{D}:C^\infty(E_0\oplus E_1)\to C^\infty(E_0\oplus E_1)$ such that $\mathscr{D}^2$ is Laplacian and $\mathscr{D}$ which maps sections of $E_0$ to sections of $E_1$ and vice-versa. It is uniquely determined by its restriction to $C^\infty(E_0)$. Check definition 11.1.1 of my notes and section 11.2.2 for the special case of Kahler manifolds. | |
| Sep 16, 2012 at 14:29 | comment | added | Michael Albanese | Also, in Chapter 11 of your notes, you say that a Dirac operator is one which squares to be a generalised Laplacian which (I'm pretty sure) does not agree with what you have written. Using this definition instead and setting $D_{1, 0} = \partial + \partial^*$ and $D_{0,1} = \bar{\partial} + \bar{\partial}^*$ we have $D_{1,0}^2 = \Delta_{\partial}$ and $D_{0,1}^2 = \Delta_{\bar{\partial}}$. Is that what you meant? These are just the complex anologues of your Example 11.1.5 on the Hodge-de Rham operator $d + d^*$. | |
| Sep 16, 2012 at 14:25 | comment | added | Michael Albanese | Thank you for your wonderful answer. I am not sure how the Hodge-Dolbeault operator (call it $D$ and ignore the constant factor of $\frac{1}{\sqrt{2}}$) fits in with the Laplacians $\Delta_{\partial}$ and $\Delta_{\bar{\partial}}$. You say that you can obtain a generalised Laplacian using $DD^*$ or $D^*D$, but neither of these give $\Delta_{\partial}$ or $\Delta_{\bar{\partial}}$. Instead, $DD^* + D^*D = \Delta_{\partial} + \Delta_{\bar{\partial}}$. | |
| Sep 14, 2012 at 18:25 | history | edited | Liviu Nicolaescu | CC BY-SA 3.0 |
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| Sep 14, 2012 at 15:07 | history | answered | Liviu Nicolaescu | CC BY-SA 3.0 |