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It is known (see for example the associated Wikipedia entry) that the Jacobi theta function $$\vartheta(z; \tau) = \sum_{n\in\mathbb{Z}} \exp(\pi in^2\tau + 2\pi inz)$$ arises from a certain representation of the Heisenberg group $H_3(\mathbb{R})$. Namely, we take the theta representation of $H_3(\mathbb{R})$, and look at the action of the discrete subgroup $H_3(\mathbb{Z})$. Then $\vartheta$ is invariant under this action.

By Stone–von Neumann, the theta representation is isomorphic to the Weyl representation of the Heisenberg group on $L^2(\mathbb{R})$. This seems to imply that there should be a function in $L^2(\mathbb{R})$ analogous to the Jacobi theta function. Using the Segal–Bargmann transform, one can try calculating the corresponding function explicitly, but the integrals involved seem to get very messy.

Question. Is there any short, explicit description of the function $f\in L^2(\mathbb{R})$ corresponding to $z\mapsto \vartheta(z;\tau)$?

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    $\begingroup$ @LSpice The Wikipedia page linked to by the OP says Weyl rather than Weil. The Weil representation is a representation of the metaplectic group, while the Weyl representation is a representation of the Heisenberg group. $\endgroup$ Commented Feb 27, 2022 at 19:26
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    $\begingroup$ @LSpice As the other commenter suggests, I did intend Weyl (I edited it back); the Weil representation is a closely related object, though, so no worries! $\endgroup$ Commented Feb 27, 2022 at 19:36
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    $\begingroup$ There can't be such a function in $L^2(\mathbb{R})$, since a translation operator (e.g. $T$ defined by $(Tf)(x) = f(x-1)$ has no eigenvectors in $L^2(\mathbb{R})$ . $\endgroup$
    – jjcale
    Commented Feb 27, 2022 at 20:41
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    $\begingroup$ Yes, the theta function has infinite norm, because $exp(-2\pi y^2 / Im \tau) \mid \theta(z)\mid^2$ is double periodic for $z \rightarrow z+1$ and $z \rightarrow z + \tau$ $\endgroup$
    – jjcale
    Commented Feb 28, 2022 at 21:12
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    $\begingroup$ @jjcale Oh I see now. Don't you want to post this as an answer? $\endgroup$ Commented Mar 1, 2022 at 19:28

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As noted in the comments, the element you're seeking is not in $L^2(\mathbb{R})$. Rather, it is a distribution (aka generalized function) that acts on Schwarz functions. Since you know it has to be invariant by $H_3(\mathbb{Z})$, and unique up to scale, it is immediate to verify that the answer is the Dirac comb: a $\delta$-function at each integer.

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