ince each component of the inertia stack $I_{\mathfrak{X}}$ is isomorphic to the original stack $\mathfrak{X}$, the normal bundle $N_q$ vanishes, and each component of ${\rm ch}(d(\lambda_q))$ is $1$. Furthermore, as $\mathfrak{X}$ is essentially a $k$-fold quotient of ${\mathbb P}^n$, $$\int_{\mathfrak{X}} \: = \: \frac{1}{k} \int_{{\mathbb P}^n}.$$ Plugging into the index formula, $$\begin{aligned}
\int_{I_{\mathfrak{X}}} {\rm ch}^{\rm rep}({\cal O}_X(m)) 
{\rm Td}(TI_{\mathfrak{X}}) 
& = & \sum_{\alpha} \int_{\mathfrak{X}} \alpha^{-m} 
{\rm ch}({\cal O}_{\mathfrak{X}}(m)) {\rm Td}(T\mathfrak{X}),
\\
& = & \sum_{\alpha} \alpha^{-m} \int_{\mathfrak{X}} 
\sum_i {\rm ch}_i({\cal O}_{\mathfrak{X}}(m)) 
{\rm Td}_{n-i}(T\mathfrak{X}).\end{aligned}$$ Now, since $\alpha$ is a $k$th root of unity, the sum $$\sum_{\alpha} \alpha^{-m}$$ will vanish unless $m$ is divisible by $k$. Thus, if $m$ is not divisible by $k$, we find that $\chi({\cal O}_{\mathfrak{X}}(m))$ vanishes. Next, suppose that $m=n k$ for some integer $n$. Then, $$\begin{aligned}
\int_{I_{\mathfrak{X}}} {\rm ch}^{\rm rep}({\cal O}_{\mathfrak{X}}(m)) 
{\rm Td}(I_{\mathfrak{X}}) 
& = & \sum_{\alpha} \int_{\mathfrak{X}} \alpha^{-m} 
{\rm ch}({\cal O}_{\mathfrak{X}}(m)) {\rm Td}(T\mathfrak{X}),
\\
& = & \sum_{\alpha} \int_{\mathfrak{X}} \pi^* 
{\rm ch}({\cal O}_{{\mathbb P}^n}(n))
{\rm Td}(T {\mathbb P}^n), \\
& = & \sum_{\alpha} \frac{1}{k} \int_{ {\mathbb P}^n }
 {\rm ch}({\cal O}_{{\mathbb P}^n}(n))
{\rm Td}(T {\mathbb P}^n), \\
& = &
 \int_{ {\mathbb P}^n }
 {\rm ch}({\cal O}_{{\mathbb P}^n}(n))
{\rm Td}(T {\mathbb P}^n), \\
& = & \chi\left({\mathbb P}^n, {\cal O}_{ {\mathbb P}^n }(n) \right).\end{aligned}$$

Now, let us compare to expectations. In the present case, if $m$ is not divisible by $k$, then all the sheaf cohomology groups of ${\cal O}_{\mathfrak{X}}(m)$ should vanish, so the Euler class $\chi({\cal O}_{\mathfrak{X}}(m))$ should vanish, exactly as we have computed. If $m$ is divisible by $k$, then $\chi({\cal O}_{\mathfrak{X}}(m)) = 
\chi({\cal 