*I asked this on [math.stackexchange](https://math.stackexchange.com/questions/2052956/cardinal-arithmetic-in-inner-models) 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 $\omega\_\alpha=(\omega\_\alpha)^{L(\mathbb{R})}$.



Cardinal arithmetic in $L(\mathbb{R})$
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 $\omega\_2$.


