are minimal. We proved in a previous article that $\mu$ is then absolutely continuous with respect to Lebesgue measure [@Du]. We there followed the classical approach of Sinai-Pesin-Ledrappier, based on the construction of a suitable invariant partition which is dilated and realizes entropy (see [@P1], [@Le]). We propose in section \[appll\] a new proof, based on the CLT for the unbounded $\mu$-centered observable $J:= \log
{{\textsf {Jac} \,  }}f -  2 (\chi_1 + \ldots + \chi_k) \in \UU$. We obtain the following result, where ${\sigma}_J: = \lim_{n \to \infty} {1 \over \sqrt n}
\norm{S_n(J)}_2$.

[D:]{} If the Lyapunov exponents are minimal equal to $\log d^{1/2}$, then ${\sigma}_J = 0$, and $\mu$ is absolutely continuous with respect to Lebesgue measure.

A crucial fact for the proof is that for any holomorphic endomorphism of ${\mathbb {P}}^k$ and any $\mu$-generic point $x \in {\mathbb {P}}^k$, the minimal dilation rate of $f^n$ at $x$ (i.e. $\norm{(d_x
  f^n)^{-1}}^{-1}$) is bounded below by $d^{n/2}$ up to the multiplicative factor $1/n$. In other words, the usual $e^{-n{\epsilon}}$-correction, due to the non-uniform hyperbolicity of $({\mathbb {P}}^k,f,\mu)$, can be replaced here by $1/n$. This was proved by Berteloot-Dupont [@BeDu], using a pluripotential result of Briend-Duval [@BrDu1] and the fact 