Sequence of MMP with scaling cannot be isomorphism

Question

Let $(X,B)$ be a projective klt pair, H is an $\mathbb{R}$-divisor s.t. $K_X+B+H$ is nef. Suppose that we can run $(K_X+B)$-MMP with scaling $H$(that is, the flip exists) and denote $\alpha_i:X \dashrightarrow X_i$ to be the map to the i-step of the MMP. Then my question is how can I show that $\alpha_i\neq\alpha_j$ for any $i\neq j$ ?(That is, the induced birational map $\alpha_i\cdot\alpha_j^{-1}:X_j\dashrightarrow X_i$ is not an isomorphism.

Of course we can assume each step is the flip. Then from the construction of MMP with scaling we obtain a sequence of real numbers $1\geq t_0\geq t_1\geq t_2\geq...\geq 0$ . Can we prove the strict inequality : $t_i> t_{i+1}$?

Thanks for your comments!

Answer 1

I realized how to solve this problem now. As pointed above, we may assume each step of MMP is a flip. For $i<j$, We can choose an exceptional divisor $E$ over $X_i$ such that its center lies in the flipping locus, then by the negativity lemma we have that the discrepancy $a(X_i,B_i,E)<a(X_{i+1},B_{i+1},E)$. By the negativity of MMP we know that $a(X_{i+1},B_{i+1},E)\leq a(X_{j},B_{j},E)$. So we have $a(X_i,B_i,E)<a(X_{j},B_{j},E)$ and hence $X_i$ and $X_i$ cannot be isomorphism.

But it seems like that we do not really need MMP with scaling.