Does $AD^{L(\mathbb{R})}$ directly implies projective determinacy? At least it certainely implies $PD$'s consistency.

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    $\begingroup$ Yes, of course it does. All projective sets are obviously in $L (\mathbb R) $ so they have winning strategies (in $L (\mathbb R) $ and therefore in $V $; note that any possible set of moves for either player is coded by a real). $\endgroup$ Oct 1 '17 at 12:11

Suppose $M$ is an inner model (of $\mathsf{ZF}$) with the same reals as $V$, and let $A\subseteq \mathbb R$ be a set of reals in $M$. Suppose further that $A$ is determined in $M$. Under these assumptions, $A$ is also determined in $V$. The point is that since winning strategies are coded by reals, and any possible run of the game for $A$ is coded by a real, then any run of the game in $V$ is already in $M$.

Similarly, since $M$ and $V$ share the same reals, they agree on what sets are first-order definable over the reals, that is, they agree on what sets of reals are projective. It follows that if projective determinacy holds in $M$ then it holds in $V$. In fact, much more is true in the sense that $M$ and $V$ agree on many more definitions over the reals than just first-order (for instance, $L(\mathbb R)^M=L(\mathbb R)$).

It follows immediately from these two observations (taking $M=L(\mathbb R)$) that if determinacy holds in $L(\mathbb R)$, then in $V$ all sets of reals in $L(\mathbb R)$ are determined. :-) That is, $\mathsf{AD}^{L(\mathbb R)}$ implies $L(\mathbb R)$-determinacy. In particular, projective determinacy holds in $V$.

Naturally, $V$ may have more sets of reals than $M$, some of which may not be determined. Also, in general, there may be notions of definability that do not coincide in $M$ and $V$ (for instance, by forcing we may easily change what sets of reals are in $\mathsf{HOD}$). It may be that some definition $\phi$ gives in $M$ a set $A$ that is determined in $M$. As long as we do not change the reals, that set $A$ will be determined in $V$, but running the same definition $\phi$ in $V$ may result in a different set $\hat A$ that may very well fail to be determined.


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