Timeline for Product of two automorphic form is still automorphic form?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Aug 23, 2019 at 21:23 | comment | added | paul garrett | @robinz16, as in my other comment: a pointwise product really is a tensor product of the corresponding repns. The holomorphic discrete series behave anomalously... | |
| Aug 23, 2019 at 21:23 | comment | added | paul garrett | @Monty, yes, I assumed so: pointwise multiplication is tensor product of representations. It's anomalous that tensor products of holomorphic discrete series decompose discretely, and have a unique lowest among "lowest weight vectors", so in their discrete decomposition contain a copy of the predictable holomorphic discrete series. | |
| Aug 23, 2019 at 21:11 | comment | added | Monty | @paul, I mean multiplication of automorphic forms not tensor product of automorphic representations. | |
| Aug 23, 2019 at 19:33 | comment | added | Robin Zhang | Perhaps the question means “product” in the sense of multiplication? E.g. akin to how the product of two classical modular forms of weight $k$ and $\ell$ is a modular form of weight $k + \ell$. | |
| Aug 23, 2019 at 18:53 | comment | added | paul garrett | No. Very rarely would the tensor product of two irreducibles be irreducible. And, after all, already the Rankin-Selberg integral for two $GL2$ cuspforms against an Eisenstein series can be construed as computing the continuous-spectrum components of the tensor product, for example. | |
| Aug 23, 2019 at 18:17 | history | asked | Monty | CC BY-SA 4.0 |