Timeline for Atiyah's proof of the moduli space of SD irreducible YM connections
Current License: CC BY-SA 4.0
15 events
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| Jan 8, 2021 at 13:19 | comment | added | Quaere Verum | Sorry for the late comment - holiday break. I would just like to remark that the factor of two which I mentioned is actually correct - Atiyah's original publication appears to have had a typo or error in it. I will edit the OP to make this clear. | |
| Dec 16, 2020 at 20:02 | comment | added | Liviu Nicolaescu | @QuaereVerum Here is a subtility. On a compact oriented $spin^c$ manifold $M$ the bundle $E=\Lambda^* T^*M\otimes \mathbb{C}$ and be identified with $End(V)$, where $V=V_+\oplus V_-$. As such its admits two $\mathbb{Z}/2$ gradings. With one grading you get the Hodge-deRham operator with index Euler characteristic, and with another you get the signature operator. | |
| Dec 16, 2020 at 19:31 | comment | added | Quaere Verum | Ok! All clear then. I have deleted the remarks from my original post, as it seems clear the way it is now. | |
| Dec 16, 2020 at 19:28 | comment | added | Liviu Nicolaescu | I had an error. For a complex vector bundle $E$ $ch_2(E)=\frac{1}{2}(c_1^2-2c_2)$ When $E=Ad(P)\otimes \mathbb{C}$ we get $ch_2(E)=-c_2(E)=p_1(Ad(P)$ (the last equality is the definition of $p_1$) | |
| Dec 16, 2020 at 19:28 | history | edited | Liviu Nicolaescu | CC BY-SA 4.0 |
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| Dec 16, 2020 at 19:11 | comment | added | Quaere Verum | Although you've already helped me more than I could have hoped for, I made an edit to my original post explaining what I am talking about in more detail, because I think I did not make it clear (also the computation in my comment forgot about a factor of $1/2$). If you wouldn't mind, could you explain where the difference in our results comes from? Because your result seems to be off by a sign on the Pontrjagin class, whereas my result is off by a factor of $2$. | |
| Dec 16, 2020 at 17:55 | comment | added | Liviu Nicolaescu | You do not need to know what $ch(V_-)$ is. (The computation of $ch(V_-)$ is tricky. The clincher is that $D=d_-+d^*$ and you know what is the index $D$. | |
| Dec 16, 2020 at 17:00 | comment | added | Quaere Verum | Final remark incoming. According to what I computed, we get the $\dim G$ term, and $\text{rank}V_-(-2c_2(\text{Ad}(P))=-4c_2(\text{Ad}(P))=p_1(\text{Ad}(P))$, so I am not sure if the mistake is on my end, but I thought I would point out the discrepancy between what I got and what you've posted (namely the sign of the Pontrjagin class and possibly some minor details in the Chern character of the adjoint bundle). | |
| Dec 16, 2020 at 14:43 | comment | added | Liviu Nicolaescu | $Ad(P)$ is a real vector bundle $c_1(Ad(P)\otimes \mathbb{C})=0$. | |
| Dec 16, 2020 at 14:38 | comment | added | Quaere Verum | Trying to go through your explanation in detail - I have one small question. You immediately write $\text{ch}(\text{Ad}(P))=\dim G+\text{ch}_2(\text{Ad}(P))+\dots$. Is it valid to immediately omit the first Chern character? Since we're taking the product with another term that contains degree $2$ forms, could it not combine with terms there to add a term to the index density? The only candidate for that term would be $\text{ch}(\text{Ad}(P))\text{ch}_1(V_-)$. I guess that this term has to vanish somehow? | |
| Dec 16, 2020 at 14:10 | history | edited | Liviu Nicolaescu | CC BY-SA 4.0 |
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| Dec 16, 2020 at 0:59 | history | edited | Liviu Nicolaescu | CC BY-SA 4.0 |
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| Dec 15, 2020 at 23:27 | vote | accept | Quaere Verum | ||
| Dec 15, 2020 at 22:35 | history | edited | Liviu Nicolaescu | CC BY-SA 4.0 |
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| Dec 15, 2020 at 22:29 | history | answered | Liviu Nicolaescu | CC BY-SA 4.0 |