I am wondering whether the automorphic forms in some automorphic representation is closed under product. I think it is true by definition of automorphic form. Am I right?
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3$\begingroup$ 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. $\endgroup$– paul garrettCommented Aug 23, 2019 at 18:53
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1$\begingroup$ 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 $k + \ell$. $\endgroup$– Robin ZhangCommented Aug 23, 2019 at 19:33
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1$\begingroup$ @paul, I mean multiplication of automorphic forms not tensor product of automorphic representations. $\endgroup$– MontyCommented Aug 23, 2019 at 21:11
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$\begingroup$ @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. $\endgroup$– paul garrettCommented Aug 23, 2019 at 21:23
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$\begingroup$ @robinz16, as in my other comment: a pointwise product really is a tensor product of the corresponding repns. The holomorphic discrete series behave anomalously... $\endgroup$– paul garrettCommented Aug 23, 2019 at 21:23
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