Let ($v_1$,$v_2$,$v_3$) denote the vibrational state of $CO_2$.
Why is the transition $(0,0,0)\rightarrow (1,0,1)$ observed when the trasition $(0,0,0)\rightarrow(1,0,0)$ (asymmetric stretch) is not observed because it is IR inactive (not allowed)?
Is the following procedure correct for showing that the $(0,0,0)\rightarrow(1,0,1)$ transition is allowed?
The criteria for the transition being allowed is
$$\Gamma_{tot. sym.}\in\Gamma_{initial} \times \Gamma_{x,y,z}\times\Gamma_{final}$$
Where $\Gamma$ is the "irrep" of the group $D_{\infty h}$ and $\Gamma_{tot. sym.}=\Sigma_g^+$. The initial state is $(0,0,0)$ and the irrep is $\Gamma_{initial}=\Gamma_{(0,0,0)}=\Sigma_g^+$
The irreps of the three normal modes $v_1$, $v_2$ and $v_3$ are
$$\Gamma_1=\Sigma_g^+\text{, }\Gamma_2=\Pi_u\text{, }\Gamma_3=\Sigma_u^+$$
From a character table we get that
$$\Gamma_{x,y}=\Pi_u\text{ and }\Gamma_z=\Sigma_u^+$$
Now, this is the step I'm not sure about; we say that the irrep of the state $(1,0,1)$ is $$\Gamma_{(1,0,1)}=\Gamma_1\times\Gamma_3=\Sigma_g^+\times\Sigma_u^+=\Sigma_u^+$$
And finally, using a product table, we get
$$\Gamma_{(0,0,0)}\times\Gamma_{z}\times\Gamma_{(1,0,1)}=\Sigma_g^+\times\Sigma_u^+\times\Sigma_u^+=\Sigma_g^+\ni\Gamma_{tot.sym.}$$
And thus the transition is allowed and we observe it as a band in an IR spectrum.
