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In p.312 of 'Rhoades, Robert C., Linear relations among Poincaré series via harmonic weak Maass forms. Ramanujan J. 29 (2012), no. 1-3, 311–320', the author defines the extended principal part at infinity as follows:

A weakly holomorhpic modular form is any meromorhpic modular form whose poles are supported at the cusps. The extended principal part at infinity of a weakly holomorphic modular form $f$ is the polynomial $P_{f,\infty} \in \mathbb C[q^{-1}]$ such that $f(z)-P_{f,\infty}(q^{-1}) = O(e^{-\epsilon y})$ as $y \to \infty$.

My questions are,

  1. what is the meaning of the extended principal part.
  2. What's the relation of this definition of extended principal part and the 'principal part of Laurent expansion'?
  3. Is the extended principal part of weakly holomorphic modular form determined uniquely?

In addition, the author defines the extended principal part at the cusps $x$ as follows:

If $x$ is a cusp, the extended principal part at $x$ is the finite sum of terms in the Fourier expansion around $x$ that do not have rapid decay toward $x$.

I didn't understand this sentence, but in my idea the natural definition of the extended principal part at $x$ is the following:

Let $\sigma \in \mathrm{SL}_2(\mathbb R)$ be an element satisfying $\sigma x = \infty$. Then $f|_k \sigma^{-1}$ has the Fourier expansion in $q$, where $q=e(z/h)$, $h$ is the cusp width at $x$. The extended principal part of $f$ at $x$ is a polynomial $P_{f,x}(q^{-1}) \in \mathbb C[q^{-1}]$ such that $f|_k \sigma^{-1} (z)-P_{f,x}(q^{-1}) = O(e^{-\epsilon y})$.

Is my understanding right?

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  • $\begingroup$ The extended principal part coincides with the principal part of $f$ as a Laurent series in $q$ (except maybe for the constant term, that you may or may not want to put in the definition). For arbitrary cusp your definition is correct. The extended principal part is then unique up to $z \to z+1$ which means $q \to \exp(2\pi i/h) q$. $\endgroup$ Commented Apr 7, 2020 at 8:43
  • $\begingroup$ You have written "Let $\sigma$ be the element" but it should be "an element". $\endgroup$ Commented Apr 7, 2020 at 16:35
  • $\begingroup$ Brunault // Could you let me know why? $\endgroup$
    – LWW
    Commented Apr 7, 2020 at 22:36
  • $\begingroup$ You can multiply $\sigma$ on the left by any matrix stabilizing $\infty$. Also, it should be $\sigma \in \mathrm{SL}_2(\mathbb{Z})$. $\endgroup$ Commented Apr 8, 2020 at 8:15
  • $\begingroup$ Brunault // I mean, the reason for the first comment you wrote. $\endgroup$
    – LWW
    Commented Apr 8, 2020 at 9:14

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