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In Shimura's paper "ON THE HOLOMORPHY OF CERTAIN DIRICHLET SERIES", he constructed a family of Eisenstein series $E(z,s)$ by summing factor of automorphy. $E(z,s)$ is of negative half-integral weight, and we can reconstruct it from the principal series of $\widetilde{SL_2}$. But $E(z,s)^2$ is a holomorphic modular form of negative integral weight, which must be $0$. I am very confused and would like to know what's the problem in my argument?

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    $\begingroup$ 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. $\endgroup$
    – Jesse Silliman
    Commented Jun 12, 2019 at 1:17
  • $\begingroup$ 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. $\endgroup$
    – Alice
    Commented Jun 12, 2019 at 1:51
  • $\begingroup$ 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 $s(1-s)$. $\endgroup$
    – GH from MO
    Commented Jun 12, 2019 at 5:20
  • $\begingroup$ @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. $\endgroup$
    – Jesse Silliman
    Commented Jun 12, 2019 at 6:53

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