126 If $2-\sin ^{2}(x+2 y-1)=\frac{x^{2}+y^{2}-2(x+1)(y-1)}{x-y+1}$, then the minimum value of the product $x y$ is $\qquad$ .
$126 \frac{1}{9}$. Since it always holds that $1 \leqslant 2-\sin ^{2}(x+2 y-1) \leqslant 2$. And
$$
\left|\frac{x^{2}+y^{2}-2(x+y)(y-1)}{x-y+1}\right|=\left|x-y+1+\frac{1}{x-y+1}\right| \geqslant 2 .
$$

Therefore, the original equation can only hold when $x=y$ and $x+2 y-1=k \pi, k \in \mathbf{Z}$. From this, it is easy to see that the expression $x y=\left(\frac{k \pi+1}{3}\right)^{2}$ achieves its minimum value $\frac{1}{9}$ when $k=0$.