Here, $\chi = \alpha^{-m}$.

More generally, over all components, we write $$c_1^{\rm rep}({\cal O}_{\mathfrak{X}}(m)) \: = \: \left( \frac{m}{k} J, \cdots, 
\frac{m}{k} \alpha^{-m} J, \cdots \right).$$ Multiplication of components of ch$^{\rm rep}$ multiplies not only the cohomology classes, but also the coefficients. For example, $$\left( c_1^{\rm rep}({\cal O}(m))|_{\mathfrak{X} \times \{\alpha\} }
\right)^2 \: = \: 
\left( \frac{m}{k} J \right)^2 \alpha^{-2m}.$$ Now, for a line bundle $L$ on an ordinary space, $${\rm ch}_2(L) \: = \: (1/2) c_1^2(L),$$ but here, by contrast, $$\begin{aligned}
{\rm ch}_2^{\rm rep}({\cal O}(m))|_{\mathfrak{X} \times\{\alpha\}} & = & 
\frac{1}{2}
\left( \frac{m}{k} J \right)^2 \alpha^{-m}, \\
& = & \alpha^{+m} \frac{1}{2}
\left( c_1^{\rm rep}({\cal O}(m))|_{\mathfrak{X} \times \{\alpha\} }
\right)^2,\end{aligned}$$ so that the usual relation between Chern classes and Chern characters is modified on a stack. (In fact, if we were computing Chern classes of a bundle that split as several different eigenbundles, the relation would be much more complicated than just an additional complex phase.)

As a consistency check, let us compute the index of this line bundle, using Hirzebruch-Riemann-Roch. For any bundle ${\cal E} \rightarrow \mathfrak{X}$, the Hirzebruch-Riemann-Roch index theorem says $$\chi({\cal E}) \: = \: \int_{I_{\mathfrak{X}}} {\rm ch}^{\rm rep}({\cal E})
{\rm Td}(\mathfrak{X})$$ where $$\chi({\cal E}) \: = \: \sum_i (-)^i h^i(\mathfrak{X}, {\cal E}),$$ and $${\rm Td}(\mathfrak{X}) \: = \: \alpha_{\mathfrak{X}}^{-1} {\rm Td}( T 
I_{\mathfrak{X}} ),$$ where $$\alpha_{\mathfrak{X}} \: = \: {\rm ch}( d( \lambda_q) ), \: \: \:
\lambda_q \: = \: \sum_k (-)^k \wedge^k N_q^*,$$ for $N_q$ the normal bundle. (As $\lambda_q$ is not a pullback from $\mathfrak{X}$, but rather is defined intrinsically on $I_{\mathfrak{X}}$, ${\rm ch}^{\rm rep}(\lambda_q)$ is not well-defined, so instead the pertinent Chern character is defined via the diagonalization map $d$.)

In the present case, s