Timeline for Hodge map and the Cohomology Ring of a Riemannian Manifold
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Jan 27, 2016 at 19:41 | comment | added | Sebastian Goette | To add some positive comment ... there is a formal adjoint to the de Rham operator on a Riemannian manifolds that you can write as $\delta=\pm *\circ d\circ *$. It satisfies $\delta^2=0$ as well, and it computes (the Poincaré dual of) $H^\bullet(M;\mathbb R)$, which is isomorphic to $H^\bullet(M;\mathbb R)$ if $M$ is orientable. You find some more on this here. The good news is that if $\omega$ is $d$-closed, then $*\omega$ is $\delta$-closed. This explains why you still get a map on cohomology. | |
| Jan 27, 2016 at 16:52 | comment | added | Andrea Pena | @geodude: Thanks a lot, that's quite interesting. | |
| Jan 27, 2016 at 16:51 | vote | accept | Andrea Pena | ||
| Jan 27, 2016 at 14:09 | comment | added | geodude | @AndreaPena [continuing] The other two Maxwell equations can be written as $d*F=J$, where $J$ is a 3-form whose components are the density of electric charge and current. If $[*F]$ were also a cocycle, electric charge would be zero (= physically not exist). Viceversa, if $F$ were not a cocycle just because neither $*F$ is, magnetic charge $dF$ would be non zero. I know this is about closeness and not about cohomology classes, but this is enough (see Johannes Huisman's answer). | |
| Jan 27, 2016 at 14:03 | comment | added | geodude | @AndreaPena You can write the Faraday tensor (electric and magnetic field together) as a 2-form $F$ on spacetime. Two of Maxwell's equations can be now written as $dF=0$, basically saying that magnetic charges and currents do not exist. On a contractible domain, this means that $f=dA$ for some $A$, which is a 1-form whose components are the scalar and vector potential. So, no magnetic charge $\Leftrightarrow$ $F$ is exact (or, on a generic domain, $F$ is closed). [continues] | |
| Jan 27, 2016 at 13:57 | answer | added | David E Speyer | timeline score: 2 | |
| Jan 27, 2016 at 12:25 | comment | added | Andrea Pena | @geodude Can you explain this in some more detail? | |
| Jan 27, 2016 at 12:24 | vote | accept | Andrea Pena | ||
| Jan 27, 2016 at 16:51 | |||||
| Jan 27, 2016 at 11:35 | comment | added | geodude | Physicist's answer: if $[\omega]=[*\omega]$, magnetic charges would exist. | |
| Jan 27, 2016 at 10:48 | answer | added | Johannes Huisman | timeline score: 8 | |
| Jan 26, 2016 at 21:04 | comment | added | Andrea Pena | If I restrict to harmonics then yes the map is defined, as explained at the start of the question. | |
| Jan 26, 2016 at 21:03 | comment | added | Andrea Pena | That is exactly the point of the question. I wanted to know if the map was well-defined. If, as you say, $\ast\omega$ need not be closed, then the map is of course not well-defined. Can you provide an example? | |
| Jan 26, 2016 at 20:45 | comment | added | Misha | Your question is impossible to answer. You did not define a map, since the choice of $\omega$ is not specified. Do you mean $\omega$ is closed? (Then $*\omega$ need not be closed.) Harmonic? Lastly, the question itself is better suited for MSE, not MO. | |
| Jan 26, 2016 at 20:15 | history | edited | Andrea Pena | CC BY-SA 3.0 |
added 12 characters in body
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| Jan 26, 2016 at 20:01 | history | asked | Andrea Pena | CC BY-SA 3.0 |