Show that $\bf AD_2 \Leftrightarrow \bf AD_{\omega} \nRightarrow\bf AD_{\mathbb{R}}$

How to show that $\bf AD_2 $ is equivalent to $\bf AD_{\omega}$, but not equivalent to $\bf AD_{\mathbb{R}}$?

$\bf AD$ is abbreviated for axiom of determinacy.$\bf AD_2 $, $\bf AD_{\omega}$, and $\bf AD_{\mathbb{R}}$ differ on the base of action space（which are $2$, $\omega$, $\mathbb{R}$ respectively） that is fixed at each stage.

My first unsuccessful attempt is to show the existence or non-existence of isomorphism among their corresponding games which result in another question that hasn't been solved yet.

Clearly, $\mathsf{AD}_\omega$ implies $\mathsf{AD}_2$: Given $A\subseteq2^\omega$, associate to it the game on integers where the first player to play a number not in $2$ loses, and if both play only $0$s and $1$s, then I wins iff the resulting real is in $A$. This game is $G_\omega(A^*)$ for some set $A^*$ (easily definable from $A$). Clearly, the games are equivalent, in the sense that from a strategy for either player in $G_\omega(A^*)$ one easily gets a strategy for the same player in $G_2(A)$.

(The same argument shows that $\mathsf{AD}_Y$ implies $\mathsf{AD}_X$ whenever $X,Y$ are nonempty, and $X$ injects into $Y$. In particular, $\mathsf{AD}_{\mathbb R}$ implies $\mathsf{AD}_\omega$.)

Conversely, one can associate to each $A\subseteq\omega^\omega$ an $A^*\subseteq2^\omega$ so that $G_\omega(A)$ and $G_2(A^*)$ are equivalent, in the sense that a strategy for either player in the latter game (easily) gives us a strategy for the same player in the former game.

For example, given $A$, we can consider the game where both players play $1$s and $0$s. Let $x\in 2^\omega$ be the real they produce. If I only plays finitely many $0$s, then we set $x\notin A^*$. If this is not the case, and II only plays finitely many $0$s, then $x\in A^*$. Finally, if both players played $0$ infinitely often, then:

We define a function $\pi$ by setting $\pi(x)=(n_0,n_1,\dots)\in\omega^\omega$. Say that the first $k$ moves of I are $1$s and the $k+1$-st is $0$. Then $n_0=k$. It is irrelevant what II has been moving so far. Say that, once I has played this first $0$, the next $j$ moves of II are $1$, and II's move after is $0$. Then $n_1=j$. It is irrelevant what I has been playing meanwhile. Say that, once II has played this $0$, the next $l$ moves of I are $1$s, and their $l+1$-st is $0$. Then $n_2=l$. Etc.

We then require $\pi^{-1}(A)\subseteq A^*$ and $\pi^{-1}(A^c)\subseteq (A^*)^c$. Again, it is easy to see that a strategy for either player in $G_2(A^*)$ gives us a strategy for that very player in $G_\omega(A)$.

One way to show that $\mathsf{AD}_{\mathbb R}$ is strictly stronger than $\mathsf{AD}_\omega$ is to consider the extent of uniformization:

Under $\mathsf{AD}_{\mathbb R}$, any $A\subseteq\mathbb R^2$ can be uniformized. Recall that this means that there is an $f:{\rm dom}(A)\to\mathbb R$ with $(x,f(x))\in A$ for all $x\in{\rm dom}(A)$, where ${\rm dom}(A)$ is the set of $x$ for which there is a $y$ with $(x,y)\in A$.

To see this, simply consider the (two-move!) game where I plays a real $x$ and II responds with a $y$, and II wins iff either $x\notin{\rm dom}(A)$, or else $(x,y)\in A$.

Clearly, I cannot have a winning strategy, so $\mathsf{AD}_{\mathbb R}$ gives us a function that uniformizes $A$.

On the other hand, if $\mathsf{AD}_\omega$ holds, then it holds in $L(\mathbb R)$. But a well-known argument of Solovay gives us that if $L(\mathbb R)$ is not a model of choice (as it is the case under $\mathsf{AD}_\omega$), then it satisfies that there is an $A\subseteq\mathbb R^2$ that cannot be uniformized: Work in $L(\mathbb R)$, and let $A=\{(x,y)\mid y$ is not ordinal definable from $x\}$.

If $A$ is uniformizable, let $f$ be a uniformization. Then $f$ is ordinal definable from some real $x_0$ (because in $L(\mathbb R)$ every set is ordinal definable from some real). But then $y_0=f(x_0)$ is ordinal definable from $x_0$, and $(x_0,y_0)\in A$, a contradiction.

There are other arguments: For example, one can also show that in fact $\mathsf{AD}_{\mathbb R}$ is strictly stronger consistency-wise than $\mathsf{AD}_\omega$; and Solovay showed that the former theory implies the existence of $\mathbb R^\sharp$, which of course is incompatible with $V=L(\mathbb R)$.