Scattering cross sections from isospin invariance Considering isospin invariance, $[H,T_i]=0$,we can we show that the scattering cross sections for these $3$ pi mesons, $ \sigma_a(pp \longrightarrow \pi^+d)$, $ \sigma_b(np \longrightarrow \pi^0d)$ and $ \sigma_c(nn \longrightarrow \pi^-d)$, have the fractions: $ \sigma_a: \sigma_b:\sigma_c=1:\frac{1}{2}:1$
We consider that the $\pi$ mesons have isospin $1$  and $d$, representing deuterium, has isospin zero.
Why does $[H,T_i]=0$ imply that the transition amplitude can only depend on $T^2$?
 A: It's pretty much what you learned from rotations in space. The commutation relation you wrote simply implies that, further, the Hamiltonian is isorotation invariant,
$$
e^{i\theta \hat n \cdot \vec T} H e^{-i\theta \hat n \cdot \vec T}=H,
$$
for any angle θ and any direction n in isospace, that is, isospin is a symmetry of it.
The only invariant for this symmetry is its quadratic Casimir, $\vec T \cdot \vec T$, whose eigenvalues are dictated to be $T(T+1)$. The hamiltonian may thus be a function of this, alright, since it will commute with all three generators you wrote.
The transition amplitude for the reactions you wrote is driven by the strong hamiltonian in question, so, e.g.,
$$
\langle \pi^+ d|  e^{-iHt/\hbar} | pp \rangle,
$$
which you use in your QFT estimate.
While the details could be a mess, the various amps are related by symmetry, as you presumably appreciate.  The states are related to each other by suitable tensor products of $T_\pm$s, which we saw commute with H, and go on to connect to the bras. The cross sections go as the squares of the amplitudes and the phase space factors for the two reactions are very comparable.

Response to your request on C-Gs At the risk of doing your homework for you, so with a heavy heart... Adding the isosinglet d to the isotriplet π ensures the final state has T=1, and $T_3=M$, so $\pi^\pm \mapsto |1~~~\pm 1\rangle$, and  $\pi^0 \mapsto |1~~~ 0\rangle$. You dot these on a coproduct (isospin sum) of two nucleon isodoublets, so $\langle 1/2 ~1/2~~1/2~1/2|$, $\langle 1/2 ~-1/2~~1/2~1/2|$, $\langle 1/2 ~-1/2~~1/2~-1/2|$, respectively; so the Clebsches are
$ \langle 1/2 ~1/2~~1/2~1/2| 1 ~1\rangle=1$;  $ \langle 1/2 ~-1/2~~1/2~1/2| 1 ~0\rangle=1/\sqrt 2$;
$ \langle 1/2 ~-1/2~~1/2~-1/2| 1 ~-1\rangle=1$.
