Short answer: 1+1=1/2+1/2+1=1/2+1/2+1=1/2+1/2+1/2+1/2=2.
Long answer: the Weyl invariance is (almost) manifest. The proof is essentially a simple dimension counting. As the Weyl symmetry is about rescaling of the metric $h_{\alpha \beta}$, we have to make explicit the metric dependence in the various terms of the action. As the theory contains spinors, we have to use a 2d "tetrad" formalism with a "zweibein" $e^a_\alpha$ satisfying $h_{\alpha \beta} = e^a_\alpha e^b_\beta \eta_{ab}$, $e^\alpha_a e^b_\beta = \eta^b_a$ where $\alpha$, $\beta$... are curved 2d space-time coordinates indices and $a$, $b$... are 2d Minkowski coordinates indices. This formalism is implicit in the action written in the question but it is better to replace the curved metric dependant gamma matrices $\rho^\alpha$ by Minkowski, metric independant, gamma matrices $\rho^a$. So we rewrite the action as:
$S = k \int d\sigma \sqrt{-h} ( h^{\alpha \beta} \partial_\alpha X^\mu \partial_\beta X_\mu + 2i {\bar{\psi}}^\mu \rho^a e^\alpha_a \partial_\alpha \psi_\mu $
$- i {\bar \chi}_a \rho^b \rho^a \psi^\mu e^\beta_b \partial_\beta X_\mu - \frac{1}{4}{\bar \chi}_a \rho^b \rho^a \psi^\mu {\bar \chi}_b \psi_\mu )$
This action is invariant under the Weyl symmetry
$e^a_\alpha \mapsto \lambda e^a_\alpha$, $X \mapsto X$, $\psi \mapsto \lambda^{-1/2} \psi$, $ \chi \mapsto \lambda^{-1/2} \chi$
where $\lambda$ is an arbitrary function on the 2d spacetime.
Indeed:
As $e^a_\alpha \mapsto \lambda e^a_\alpha$, we have $h_{\alpha \beta} \mapsto \lambda^2 h_{\alpha \beta}$ and so $det(h) \mapsto \lambda^4 det(h)$ because we are in 2d and so $\sqrt{det(h)} \mapsto \lambda^2 \sqrt{det(h)}$. So we have to show that each term between the brackets in the expression giving $S$ is multiplied by $\lambda^{-2}$ under the Weyl symmetry.
First term: term in $h^{\alpha \beta} \partial X \partial X$. As $h_{\alpha \beta} \mapsto \lambda^2 h_{\alpha \beta}$, we have $h^{\alpha \beta} \mapsto \lambda^{-2} h^{\alpha \beta}$ hence the result because $X \mapsto X$.
Third term: term in $\chi \psi e^\beta_b$. We have $\chi \mapsto \lambda^{-1/2} \chi$, $\psi \mapsto \lambda^{-1/2} \psi$, $e^\beta_b \mapsto \lambda^{-1} e^\beta_b$ hence the result because $-1/2-1/2-1=-2$.
Fourth term: term in $\chi \psi \chi \psi$. We have $\chi \mapsto \lambda^{-1/2} \chi$, $\psi \mapsto \lambda^{-1/2} \psi$, hence the result because $-1/2-1/2-1/2-1/2=-2$.
Second term: term in $\overline{\psi} \rho^a e^\alpha_a \partial \psi$. It is the unique slightly subtle point because this term contains a derivative of a field charged under the Weyl symmetry. In fact, it is a general but not completely trivial fact that in any spacetime dimension, a fermion minimally coupled to gravity is (classically) Weyl invariant: the Dirac operator is conformally invariant. Let's see what happens explicitely in our 2d case. Under , $\psi \mapsto \lambda^{-1/2} \psi$, $e^\alpha_a \mapsto \lambda^{-1} e^\alpha_a$, the term $\overline{\psi} \rho^a e^\alpha_a \partial \psi$ gives
$\lambda^{-2}\overline{\psi} \rho^a e^\alpha_a \partial \psi + \lambda^{-3/2} \overline{\psi} \rho^a e^\alpha_a \psi \partial (\lambda^{-1/2})$.
We have to show that the undesired extra term vanishes:
$\overline{\psi} \rho^a \psi= 0$,
which is a special property of Majorana fermions in 2d (and which was already implicitely used in the expression of the action $S$ to put an apparently non covariant derivative instead of the full Dirac operator). There are many ways to prove this relation. One possibility: $\overline{\psi} \rho^a \psi = \psi^{T} \rho^0 \rho^a \psi$ because $\psi$ is real; if $a=0$, we obtain something proportional to $\psi^{T} \psi = \psi_{+}^2+\psi_{-}^2=0$ because the components $\psi_{\pm}$ of $\psi$ are anticommuting variables; if $a=1$, $\rho^0 \rho^1$ is the 2d "$\gamma_5$ matrix" and so we obtain something proportional to $\psi_{+}^2-\psi_{-}^2=0$ because the components $\psi_{\pm}$ of $\psi$ are anticommuting variables.
Q&A (4870)
