Vanishing integral rule for two-electron integrals

I understand the vanishing integral rule for a one-electron integral to be:


$$\langle i | \hat{O} | j \rangle = 0 \hspace{1cm} \mathrm{if} \hspace{0.2cm} \Gamma\_O \neq \Gamma\_i \otimes \Gamma\_j$$


For electronic repulsion integrals, the Coulomb operator is totally symmetry, so the product must contain the totally symmetric irrep (TSIR). However, I am not sure what form the vanishing integral rule will have. Is it simply


$$\langle ij | kl \rangle = 0 \hspace{1cm} \mathrm{if} \hspace{0.2cm} \Gamma\_\mathrm{TSIR} \neq \Gamma\_i \otimes \Gamma\_j \otimes \Gamma\_k \otimes \Gamma\_l$$


or does it have a different form? Additionally, will a Schwarz-inequality matrix composed of:


$$Q\_{ij} = \sqrt{\langle ii | jj \rangle}$$


have an identical block-diagonal structure to the one-electron matrices constructed in this fashion?


