Timeline for Why can there be holomorphic modular forms of negative half integral weight?
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| Jun 12, 2019 at 6:53 | comment | added | Jesse Silliman | @Alice The formula (1.4) does not prove holomorphy, as it is not absolutely convergent, just like how classical $E_2$ is not holomorphic. Similarly, I don't know if the Fourier expansion converges absolutely for $k$ negative, so even if each individual term were holomorphic, I don't know if the sum is. I can't tell if the functional equations prove holomorphy, as they don't just switch positive and negative $k$, but also change $s$ and introduce a complex conjugate. | |
| Jun 12, 2019 at 5:20 | comment | added | GH from MO | I have no time to look into this paper, but let me point out that the usual Eisenstein series $E(z,s)$ is not holomorphic in $z$. Instead, it is an eigenfunction of the Laplace operator with eigenvalue $s(1-s)$. | |
| Jun 12, 2019 at 1:51 | comment | added | Alice | Yes, I mean the specialization at $s=0$. In the paper, Shimura computed the Fourier expansion of $E^*(z,0)$. For positive weights, we can also see that $E(z,0)$ is holomorphic from formula (1.4). For the negative weights, which are used to give the integral representation of the adjoint lifting, Shimura claims a functional equation relating positive and negative weight at p.93. One can also from the formulas of Lemma 3, p.288, Elementary theory of L-function and Eisenstein series to see that $E(z,0)$ should be holomorphic for positive and negative weight. | |
| Jun 12, 2019 at 1:17 | comment | added | Jesse Silliman | Shimura says that $E(z,s)$ is holomorphic in $s$, but that is not relevant to your question, as far as I can see. Do you mean to say that the specialization to $s = 0$, $E(z,0)$, is holomorphic? This is the only value of $s$ where the defining summation looks holomorphic in $z$. However, I could not see where in this paper Shimura claims that $E(z,0)$ is holomorphic. | |
| Jun 12, 2019 at 0:47 | history | asked | Alice | CC BY-SA 4.0 |