Cardinal arithmetic in $L(\mathbb{R})$

Question

I asked this on math.stackexchange but did not receive an answer, so I'm asking here.

Assume large cardinals. Can we have $\omega_2^{L(\mathbb{R})}=\omega_2$?

Note that $\omega_1=\omega_1^{L(\mathbb{R})}$ always: clearly $\omega_1\ge\omega_1^{L(\mathbb{R})}$, and since $L(\mathbb{R})$ contains every real, any countable ordinal in $V$ is countable in $L(\mathbb{R})$. Also, note that we can trivially have $\omega_2^{L(\mathbb{R})}<\omega_2$: just force with $Col(\omega_1,\omega_2)$. $L(\mathbb{R})$ doesn't change, but the old $\omega_2$ is collapsed.

As Asaf Karagila pointed out at my MSE question, in general $\omega_n^{L(\mathbb{R})}$ is singular in $L(\mathbb{R})$, and so can't be $\omega_n$. However, this doesn't work for $n=2$: $\omega_2^{L(\mathbb{R})}$ is measurable in $L(\mathbb{R})$.

More generally, I'm curious for what ordinals $\alpha$ we may have $\omega_\alpha=(\omega_\alpha)^{L(\mathbb{R})}$.

Answer 1

Should be possible, at least for $\omega_2$. Under large cardinals, $\omega_2^{L(\mathbb{R})}$ is the supremum of the lengths of the boldface $\Delta^1_2$-prewellorderings of the reals (i.e., $\delta^1_2$), which is the same whether it's computed in V or in $L(\mathbb{R})$. Woodin showed that, for example, if the nonstationary ideal on $\omega_1$ is $\omega_2$-saturated and $P(\omega_1)^\sharp$ exists, then $\delta^1_2$ is $\omega_2$.