If we have $(T_i)_{i=1}^N$, operators on a Hilbert space, that are also isometries and satisfy the following relation:
$$\sum_{i=1}^NT_iT_i^*=Id\quad (1)$$
How can you prove that they must also verify $T_i^*T_j=\delta_{i,j}Id\quad (2)$ ?
I think that the operators $T_i$ verify $T_i^*T_i=Id$ directly just because they are isometries. But how to prove that $T_i^*T_j=0$ for $i\neq j$?
Note first that we may assume that all $T_k$ are proper isometries; because if one of them, say $T_1$, is a unitary, we get $\sum_{k=2}^NT_kT_k^*=0$, which forces $T_k=0$ for all $k\geq2$.
Since $\sigma(T_k^*T_k)=\sigma(I)=\{1\}$, using that $\sigma(AB)\cup\{0\}=\sigma(BA)\cup\{0\}$ we deduce that $\sigma(T_kT_k^*)=\{0,1\}$ (the zero has to be there because otherwise $T_k$ is a unitary). It follows that $P_k=T_kT_k^*$ is a projection. The question is now whether these projections are pairwise orthogonal.
Fix indices $i$ and $j$ with $i\ne j$. The argument below only uses $$\sum_{k=1}^NP_k\leq I,$$which we write as $$\sum_{k\ne i}P_k\leq I-P_i.$$ Then $$ 0\leq P_iP_jP_i\leq P_i\left(\sum_{k\ne i} P_k\right)P_i\leq P_i(I-P_i)P_i=0. $$ So $P_iP_jP_i=0$. But then $$ 0=P_iP_jP_i=P_iP_j^2P_i=(P_jP_i)^*P_jP_i, $$ and this implies that $P_jP_i=0$.
This is $T_jT_j^*T_iT_i^*=0$. Multiplying by $T_j^*$ on the left and by $T_i $ on the right, we get $T_j^*T_i=0$.