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Let $k\geq 2$ be an even integer and let $\Gamma=\Gamma_0(N)$. Let $f\in S_k(\Gamma)$. To $f$, one may associate an antiholomorphic cusp form of weight $k$ and level $\Gamma$ by defining $g(z):=f(-\bar{z})$. These satisfy the transformation property

$$g(\gamma z)=(c\bar{z}+d)^kg(z)\qquad \gamma\in\Gamma.$$

There is a related discussion here.

Denoting the space of such objects by $\overline{S_k(\Gamma)}$, it is clear that $S_k(\Gamma)\neq0$ iff $\overline{S_k(\Gamma)}\neq 0$.

I would like to know what happens in the situation $k\leq -2$. Clozel mentions this case in his article "Motifs et formes automorphes", on page 91, where he talks of the antiholomorphic case. Given an element of $\overline{S_k(\Gamma)}$, we can, by the above procedure, get a holomorphic cusp form. If $k<0$ then $S_k(\Gamma)=0$, so will this case ($k\leq -2$) ever occur?

EDIT: The definition $g(z):=\overline{f(z)}$ has the advantage of avoiding the restriction put above that the integer $k$ is even.

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  • $\begingroup$ Haven't you just proven that it can't? You clearly know that $S_k = 0$ for $k \le -2$ and that $S_k$ is non-zero if and only if $\overline{S_k}$ is; so what is left to say here? $\endgroup$
    – David Loeffler
    Commented Feb 19, 2016 at 1:54
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    $\begingroup$ @David: From the article of Clozel, I got the impression that he is considering a non vacuous case, hence, the question. I think I now understand what is going on. For a modular form $f$ of weight $k$, one can construct the function $\varphi_f$ on $G(\mathbb{A})$ which satisfies the property that $\varphi_f(xr(\theta))=e^{-ik\theta}\varphi_f(x)$ where $r(\theta)$ is the $2 \times 2$ matrix given by (Cos $\theta\,$ -Sin $\theta$; Sin $\theta\,$ Cos $\theta$). Now corresponding to $f$ one has $g$ as above, and the point seems to be that $\varphi_g(xr(\theta))=e^{ik\theta}\varphi_g(x)$. $\endgroup$
    – Rex
    Commented Feb 19, 2016 at 7:40

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I am not entirely sure what Clozel is trying to do here. He doesn't make it terribly precise what he means, and as you yourself have realised, if you read the article literally it seems that one case of his discussion is vacuous.

My best guess at what he intends is the following: if $f(z)$ is a holomorphic cusp form of weight $k$, then $\operatorname{Im}(z)^k f(-\overline{z})$ (or maybe $\operatorname{Im}(z)^{-k}$, I can't remember) transforms like a modular form of weight $-k$ (although it isn't holomorphic, of course); and it has fast decay at the cusps, is an eigenvector for the Laplacian, etc. So perhaps it is these forms he is trying to discuss on p91.

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  • $\begingroup$ Thanks for your answer. As you mention, including Im(z) has the problem that the new function will no longer be holomorphic or antiholomorphic. The word "weight" seems to be used in two different senses, one being the "weight" of a modular form, and the second being a "weight" vector for the representation of $K_\infty=SO(2)$. My guess as to what he wants to do is what I have mentioned in the comment to my question. I have now edited the definition (in the question) of how to go from holomorphic forms to antiholomorphic ones, though this does not seem to have much bearing upon the discussion. $\endgroup$
    – Rex
    Commented Feb 20, 2016 at 11:53
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It's reasonable to assume that he means $S_{-k}(\Gamma) = \overline{S_k(\Gamma)}$ for $k > 0$. So $k \ge 2$ are holomorphic and $k \le -2$ are antiholomorphic. If you treat them all as Maass forms with weight, as David pointed out, this is what you get.

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