# Difference between $c=1$ Dirac CFT and two copies of $c=\frac{1}{2}$ Majorana CFT?

The $c = \frac{1}{2}$ (non-chiral) Majorana CFT has six primary fields: the vacuum 1, the two Majorana fields $\eta, \bar \eta$ $\left( \Delta_\eta = \Delta_{\bar \eta} = \frac{1}{2}\right)$, the energy field $\varepsilon$ $\left( \Delta_\varepsilon =1\right)$, the twist field $\sigma$ $\left( \Delta_\sigma =\frac{1}{8}\right)$ and the disorder field $\mu$ $\left( \Delta_\mu =\frac{1}{8}\right)$. The latter two are non-local, whereas the rest are local. (This is relevant as it means that e.g. for anti-periodic spatial boundary conditions (APBC), the first four fields determine the finite-size energy spectrum.)

Now my question: to what extent can I think of the $c=1$ Dirac CFT as two copies of this $c = \frac{1}{2}$ CFT? I think that from a finite-size scaling point of view, this should be fine, hence I would expect both theories to have the same set of local primary fields (e.g., I guess the Dirac CFT has four primary fields with scaling dimension $\Delta =\frac{1}{2}$, is that true?). However, two copies of the $c=\frac{1}{2}$ CFT would, for example, have two twist fields $\sigma_{1,2}$, but I would be surprised if the Dirac CFT has two such fields?

First off, I suspect the $c=1$ Dirac CFT can indeed be seen as a stack of two $c=\frac{1}{2}$ Majorana CFTs. However, the subtlety lies elsewhere: the stack of two such fermionic CFTs is somewhat non-trivial. Indeed, fermions are never fully decoupled: they always anti-commute. For that reason, one can (as I was implicitly doing in my question above) consider a field $\boldsymbol{\sigma \times 1}$ (where we take a twist field of one of the $c=\frac{1}{2}$ CFTs and the vacuum of the other) but it is a most curious beast: it is neither bosonic or fermionic. For example, it anti-commutes with the fermionic field $\psi \times 1$ but commutes with the fermionic field $1\times \psi$. Now, I don't have enough affinity with these matters to see whether or not this means that we should exclude it all-together (honestly, I don't strictly see why), but at the very least it means it is not a conventional twist field. Instead, we can construct (parity-odd) twist fields $\boldsymbol{\sigma \times \mu}$ and $\boldsymbol{\mu \times \sigma}$. Hence we expect two parity-odd twist fields, each with dimension $\left( \frac{1}{8}, \frac{1}{8} \right)$ ---to be contrasted to the $\left( \frac{1}{16}, \frac{1}{16} \right)$ that we would have assigned to $\sigma \times 1$ and $1\times \sigma$. (Similarly we expect the two 'disorder'/parity-even twist fields $\mu \times \mu$ and $\sigma \times \sigma$.)
Now we can check whether this is compatible from the $c=1$ Dirac CFT point of view. As for example discussed in the beautiful lecture notes by McGreevy, one can consider the primary fields $$\mathcal V_{\alpha,\beta} (z, \bar z)= e^{\boldsymbol{\mathrm i} \left( \alpha \phi_L(z) + \beta \phi_R(\bar z) \right)}$$ for any (half-)integer $\alpha,\beta$ with the conditions $\alpha-\beta,\alpha+\beta \in \mathbb Z$. This has the conformal dimension $\left( \frac{\alpha^2}{2}, \frac{\beta^2}{2}\right)$.
Firstly, there are the four fermionic fields $\mathcal V_{\pm 1,0}, \mathcal V_{0,\pm 1}$, all with scaling dimension $\Delta = h + \bar h = \frac{1}{2}$. These operators are local (in the fermionic language) and correspond to the four fields I mentioned in the original post above (or more precisely, some linear combination thereof). So far, so good.
Then moving on to the twist fields. We have the two parity-odd $\mathcal V_{\frac{1}{2},\frac{1}{2}}$ and $V_{-\frac{1}{2},-\frac{1}{2}}$. This has the conformal dimension $\left( \frac{1}{8},\frac{1}{8} \right)$. This is compatible with the two twist fields I discussed in the first paragraph of this answer. Similarly, we have the parity-even $\mathcal V_{\frac{1}{2},-\frac{1}{2}}$ and $V_{-\frac{1}{2},\frac{1}{2}}$, consistent with the two 'disorder' fields derived at from the above 'stacked $c=\frac{1}{2}$' point of view.