I want to show the red line. In fact, in another problem , with many help, I get the formal adjoint of divergence without restrict. Beforehand, I think I can deal the red line if I know how to get the formal adjoint of divergence. But turns out I was way too simple. I have no idea of proving red line.
What I try:
Assume $$ T=T_{ij} dx^i \wedge dx^j \in \Gamma(\wedge^2 T^* M),~~~G=dx^b $$ I try to show $$ \int_ M \langle \delta(T), G\rangle dV_g = \int_M \langle T, d(G) \rangle dV_g \tag{1} $$ First, I have $T_{ij}=-T_{ji}$ and $$ \delta(T)(\partial_k) = -g^{ij}(\nabla T)(\partial_i, \partial_j, \partial_k) \\ =(-g^{ij}\partial_i T_{jk} + g^{ij}\Gamma_{ij}^a T_{ak} + g^{ij}\Gamma_{ik}^a T_{ja}) dx^k $$ Therefore, there is $$ \langle \delta(T), G\rangle = -g^{ij}g^{kb}\partial_i T_{jk} + g^{ij}g^{kb}\Gamma_{ij}^a T_{ak} + g^{ij}g^{kb}\Gamma_{ik}^a T_{ja} $$ Besides, since $d(G)=d(dx^b)=0$, I have $$ \langle T, d(G)\rangle =0 $$ So, Proving (1) is equal to prove $$ \int_ M \langle \delta(T), G\rangle dV_g = 0 $$ In fact, there are (in fact, there's a lot of calculation left out, they are really fussy) $$ \int_ M \langle \delta(T), G\rangle dV_g = \int_M(-g^{ij}g^{kb}\partial_i T_{jk} + g^{ij}g^{kb}\Gamma_{ij}^a T_{ak} + g^{ij}g^{kb}\Gamma_{ik}^a T_{ja}) dV_g \tag{2} $$ and $$ \int_M (-g^{ij}g^{kb}\partial_i T_{jk} ) dV_g = \int_M (\partial_i g^{ij} g^{kb} T_{jk} + g^{ij} \partial_i g^{kb} T_{jk} )dV_g + g^{ij} g^{kb} T_{jk} \partial_{x_i}(dV_g) \\ = \int_M ( -g^{ak} g^{ij}\Gamma_{ia}^b T_{jk} -g^{ab} g^{ij}\Gamma_{ia}^k T_{jk} -g^{ai} g^{kb}\Gamma_{ia}^i T_{jk} ) dV_g \tag{3} $$ Combining (2) and (3), I have $$ \int_M\langle \delta(T), G\rangle = \int_M (-g^{ak}g^{ij}\Gamma_{ia}^b T_{jk}) dV_g =0 \tag{4} $$ the last equlity of (4) is because $T_{ij}=-T_{ji}$.
