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A 2d rational conformal field theory (RCFT) gives rise to a modular tensor category (MTC) equipped with a Frobenius algebra object (see, for example, http://arxiv.org/abs/hep-th/0204148).

Is there an example of inequivalent RCFTs that give rise to the same underlying MTC and Frobenius algebra?

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The moonshine module VOA has trivial representation theory, i.e. the MTC is $Vec$ (and so the Frobenius algebra inside will be trivial). So this should give an example (i.e. take a trivial RCFT).

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    $\begingroup$ More simply, any even unimodular lattice gives a CFT whose MTC is trivial. $\endgroup$ – Noah Snyder May 22 '15 at 20:12
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This is just a small extension of Eric's answer. But the point is that for every MTC (arising from a VOA) there are infinitely many examples. So many, that it is even hopeless to classify them (the classification would include classifying all self-dual even lattices).

Every holomorphic VOA, including even self-dual lattices (E_8, Leech etc.), the moonshine etc, have by definition the trivial MTC $\mathrm{Vec}$ as representation category. But this implies that $V\otimes W$, with $W$ holomorphic has the same MTC tensor category as $V$.

For sure you can choose the same Frobenius algebras for $V$ and $V\otimes W$.

Sometimes one can distinguish them by their characters, but for example at rank 16 ($c=16$), there are two distinguished even self-dual lattices, namely $E_8\times E_8$ and $\mathrm{Spin}(32)/\mathbb{Z}_2$ (cf. André Henriques' answer here: https://mathoverflow.net/a/132858/10718), which even give rise to the same characters. So $V\otimes V_{E_8\times E_8}$ and $V\otimes V_{\mathrm{Spin}(32)/\mathbb{Z}_2}$ even have the same characters, but are different.

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