Let $T$ be a smooth projective variety and $f\_T : \mathbb {P}^N\_T \rightarrow \mathbb {P}^N\_T $ be a family of dominant rational maps. The dynamical degree of a dominant rational map $f $ is defined by $\lim\_{n \to \infty}(\mathrm {deg}(f^n))^{1/n}$.


Let $E $ be a set of $t \in T $ such that the dynamical degree of $f\_t$ is same as $f\_T $. Is $E $ is Zariski open? If yes, why?


It looks natural but I have no idea of proof since I have a very few knowledge of algebraic geometry maybe.



Is the exceptional set Zariski open?
It is not true that your exceptional set is always Zariski open. Here's an example. Consider the map
$$ f\_{a,b,c}(X,Y,Z) = [XY, XY+aZ^2, bYZ+cZ^2] : 
\mathbb P^2 \longrightarrow\mathbb P^2
$$
parametrized by $(a,b,c)\in\mathbb A^3$. Then generically $\delta(f\_{a,b,c})=2$, but it's a fun calculation to show that $\delta(f\_{a,b,c})\lt2$ (strict inequality) if and only if $c^2=-(\zeta+\zeta^{-1})ab$ for some root of unity $\zeta$. So your $E$ is the complement of a union of a countable union of Zariski closed subsets of $\mathbb A^3$, parameterized by roots of unity.


At the risk of self-promotion, Greg Call and I wrote a paper discussing specialization issues of the dynamical degree. 
Degeneration of Dynamical Degrees in Families of Maps
(<https://arxiv.org/abs/1609.02119>).


Generalizing a conjecture in that paper (for $\dim(T)=1$), one might guess:


**Conjecture** For all $\epsilon > 0$, the set
$$ \{ t\in T : \delta(f\_t)\le\delta(f\_T)-\epsilon \}$$
is a Zariski closed set.


