Timeline for Higgs fields whose determinant have only simple zeros
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 21, 2016 at 7:39 | vote | accept | Sebastian | ||
| Jul 21, 2016 at 0:33 | history | edited | Tony Pantev | CC BY-SA 3.0 |
Fixed an incorrect statement and suggested a different approach.
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| Jul 20, 2016 at 23:26 | comment | added | Tony Pantev | Ah, I see what you mean now. You are absolutely right! The statement I made about $\theta_{x}$ vanishing is incorrect. We can have a Higgs field $\theta$ with a multiple zero of the determinant at $x$ for which nevertheless the $\theta_{x}$ is a $2\times 2$ Jordan block. This happens when $\theta$ parametrizes a family of regular semisimple elements near $x$ specializing to a regular nilpotent element at $x$. I will correct the answer above to reflect this. | |
| Jul 20, 2016 at 8:39 | comment | added | Sebastian | Your statement ($\theta_x$ vanishes if and only if $det(\theta)$ has a zero of order at least 2) looked to me like a local statement, and my example is a counterexample for the local situation. Can I understand your first comment in the following way: If the spectral curve is irreducible the vanishing of the determinant to second order at x is equivalent to the vanishing of the Higgs field at x? | |
| Jul 18, 2016 at 12:22 | comment | added | Tony Pantev | I did not understand your last comment. If the the spectral cover is of the form $y^{2} = a^{2}$, then $\det(\theta) = - a^{2}$ so all zeroes are double. | |
| Jul 18, 2016 at 12:22 | comment | added | Tony Pantev | Since your $\theta$ is traceless the corresponding specral cover has equation $y^{2} - \det(theta) = 0$. So $\det(\theta)$ vanishes exactly at the branch points of the spectral cover. The determinant vanishes to first order at $x$ if the spectral cover is smooth and branched over $x$. This happens precisely when $\theta_{x}$ has a single eigenvector, i.e. when it has one $2\times 2$ Jordan block. The determinant vanishes to second order if and only if the spectral cover has a singularity at $x$, i.e $\theta_{x}$ has two eigenvectors , i.e. $\theta_{x} = 0$. | |
| Jul 18, 2016 at 7:19 | comment | added | Sebastian | Thank you very much for your answer. What I do not understand is why $det(\theta)$ vanishes to second order at $x$ only if $\theta_x=0.$ In fact, looking at the genus $2$ case and a very stable bundle of rank 2 with trivial determinant, it seems like there always exist Higgs fields whose determinant is the square of a holomorphic differential, and which does not vanish at the zeros of the zeros of its determinant. | |
| Jul 15, 2016 at 18:26 | history | edited | Tony Pantev | CC BY-SA 3.0 |
added 70 characters in body
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| Jul 15, 2016 at 13:04 | history | edited | Tony Pantev | CC BY-SA 3.0 |
edited body
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| Jul 15, 2016 at 12:58 | history | answered | Tony Pantev | CC BY-SA 3.0 |