Why is the transition (0,0,0) -> (1,0,1) observed in a gas phase IR spectrum of CO2?

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.


[![enter image description here](https://i.stack.imgur.com/4r053.jpg)](https://i.stack.imgur.com/4r053.jpg) 


