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Timeline for Are all vector-bosons gauge-bosons?

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May 27, 2020 at 23:33 comment added fewfew4 It's absolutely worth an answer though!
May 27, 2020 at 19:46 comment added Ramiro Hum-Sah I agree, tparker's answer and yours are both excellent and much more appropriate that the one of mine :) Examples from KK-like theories or string theories are exotic (exactly because the need of an infinite tower of massive states) and out of place. Thanks for your answer and comments.
May 27, 2020 at 18:38 comment added fewfew4 Thought I should comment that while this is possible, it necessarily requires an infinite tower of mass excitations. This might be unsavory for OP.
May 26, 2020 at 16:15 comment added Ramiro Hum-Sah I am sorry, I complicated the answer so much, that was not my intention :( I hope that the answer must still be useful.
May 26, 2020 at 16:14 comment added Davide Morgante @RamiroHum-Sah Yeah, i feel like string theory for now over my competences! Very cool though, thanks for the thorough explanation.
May 26, 2020 at 16:08 comment added Ramiro Hum-Sah Then, KR vectors are $U(1)$ gauge vectors with (local and unitary interactions) with extra higher derivative corrections (self-interactions and other coupling to other fields) suppressed by extra parameters that are not the gauge coupling. Then Maor is right, extra fields are required for consistency.
May 26, 2020 at 16:06 comment added Ramiro Hum-Sah Just to try to be clear: KR vector bosons are gauge bosons except for a detail. Higher polynomials in the YM 4-potential can not be added to the Yang-Mills lagrangian without violating the rule of minimal coupling and renormalizability, the same is true for the KR fields; except that in string theory there is other parameter (the string lenght) and KR fields can be "corrected" by higher polynomials in $B_{\mu \nu}$ without destroying unitarity and renormalizability.
May 26, 2020 at 15:50 comment added Ramiro Hum-Sah Physically: If you explore the behaviour of the effective Kalb-Ramond vector fields $B_{\mu i}$ at strong coupling, you should discover that they should start to behave as the components of a single multiplet is higher dimensions $B_{\mu \nu}$. Something not expected for gauge bosons.
May 26, 2020 at 15:49 comment added Ramiro Hum-Sah Dear Davide: No, it cannot be a gauge boson. Mathematically: $B_{\mu i}$ is not a connection on any principal bundle, is the component of a higher gauge field. Even if you restrict $B_{\mu \nu}$ to $B_{\mu i}$ you can not obtain nothing isomorphic to a gauge bundle because the deformation theories are not equal.
May 26, 2020 at 15:36 comment added myorbs In that answer it seems to be assumed that the vectors are gauge bosons, for some reason. Also I wonder if it is neccesary to include extra fields (graviton and scalar) to get the theory to be consistent? I may be asking a silly question since I am not very familiar with string theory
May 26, 2020 at 15:26 comment added Davide Morgante Interestingly enough, if i'm not mistaken, in the link you gave the Kalab-Ramond field is indicated as a gauge boson.
May 26, 2020 at 15:15 history answered Ramiro Hum-Sah CC BY-SA 4.0