Vacuum Character in Compactified Boson Partition Function For a generic $c \ge 1$ 2D CFT, I (wrongly?) expect to be able to write its torus partition function as
$$Z(\tau, \bar\tau) = \chi_0(\tau) \bar \chi_0(\bar \tau) + \sum_{(h,\bar h) \ne (0,0)}n_{h,\bar h} \chi_h(\tau) \bar \chi_{\bar h}(\bar \tau)$$ with $$\chi_0(\tau) = \frac{1-q}{\eta(\tau)}q^{-\frac{c-1}{24}} \qquad \chi_h(\tau) = \frac{q^{h-\frac{c-1}{24}}}{\eta(\tau)}$$
The boson with compactification radius $R$ has torus partition function
$$Z(\tau,\bar \tau) = \frac{1}{|\eta(\tau)|^2}\sum_{n,m\in \mathbb Z}q^{(n/R+mR/2)^2/2}\bar q^{(n/R-mR/2)^2/2}=\sum_{n,m}\chi_{h_{n,m}}(\tau)\bar \chi_{\bar h_{n,m}}(\bar \tau)$$
where in the final equality the $(h,\bar h) = (0,0)$ characters take the form of non-degenerate characters with $h,\bar h=0$. This clearly fails my expectation, since the degenerate $(h,\bar h) = (0,0)$ character does not appear.
The answer to this similar question mentions that there are two representations of the Virasoro algebra with $h=0$: the degenerate one and the non-degenerate one. My question is then: why does the non-degenerate character appear here for the compactified boson? Is there something wrong with my assumption that a generic CFT can be decomposed in characters as I stated above?
 A: A generic CFT may or may not contain an identity representation, with character $\chi_0$. For example, Liouville theory does not. However, in the context of $c\geq 1$ CFT, one sometimes restricts to considering "compact" CFTs, which by definition contain the identity.
So, you will say, the compactified free boson at $c=1$ is not a compact CFT? Actually it is, but this is hard to tell from the partition function. The problem is that characters do not uniquely characterize representations, so you cannot read the representation content from the torus partition function. The character $\chi_{h_{0,0}}$ in your partition function is the character of a non-degenerate representation, but that representation does not appear in the model.
To know the representation content of the free boson you have to go back to its construction from the abelian affine Lie algebra. The vacuum state for that algebra obeys $L_{-1}|0\rangle = 2J_{-1}J_0|0\rangle=0$, where $L_{-1}$ is a Virasoro generator and $J_n$ are modes of the abelian affine Lie algebra. So definitely $|0\rangle$ is the vacuum, and you expect $\chi_0$ to contribute. However, the vacuum module also contains the non-vanishing level one state $J_{-1}|0\rangle$, so its character is not just $\chi_0$. It turns out that the affine vacuum module is a sum of infinitely many degenerate Virasoro modules. At the level of characters, the decomposition looks like
$$
1 = (1-q) + (q-q^4) + (q^4-q^9) + \cdots
$$
where each term in parentheses corresponds to one irreducible degenerate Virasoro module.
Then $\chi_{h_{0,0}}$ does appear, but it does not correspond to a non-degenerate Virasoro representation, and you can split off $\chi_0$ from it.
For a reference, see https://inspirehep.net/literature/244716 .
A: The confusion between degenerate and non-degenerate characters can be avoided if you think of the decomposition of the partition function in terms of $U(1)$-characters instead of Virasoro. The free boson theory has $U(1)$ conserved currents $J(z)=\partial X(z)$ and $\bar{J}(\bar{z})=\bar\partial X(\bar z)$. Therefore, the chiral algebra is $U(1)\times U(1)$. For this specific field theory, this description is more fundamental – as the stress tensor (whose modes generate the Virasoro algebra) arises from the Sugawara construction, $T(z)= :\!J(z)^2\!: $.
Now, the $U(1)$-characters of primaries, including the identity, is given by
$$ \chi_{h}  = \frac{q^{h-(c-1)/24}}{\eta(\tau)}~.$$
The $q^{1/24}/\eta(\tau)$ factor counts the $U(1)$ descendants of the primary, $J_{-1}^{k_1}J_{-2}^{k_2}\cdots|h\rangle$. And, there is no difference between the $U(1)$ vacuum character and $U(1)$ non-vacuum characters.
