TITLE: Is every ordinal potentially definable?
QUESTION [7 upvotes]: It is easy to see that, if $V\models\alpha>\omega_1^{CK}$, then $\alpha$ is not recursive in any forcing extension of $V$. The argument goes as follows:

The relation "$\Phi_e=r$" is $\Pi^0_2$.
The predicate "$r$ codes an ill-founded order" is $\Sigma^1_1$.
So if $V[G]\models $"$\Phi_e$ is a well-ordering," then so did $V$; and moreover, $V[G]$ and $V$ interpret $\Phi_e$ in the same way.

In the presence of generic absoluteness, we can extend this. For instance, it follows from Projective Determinacy that the set of ordinals which have projective copies in some set-generic extension of $V$ is precisely the set of ordinals which have projective copies already in $V$.
My question is about what happens when we don't have generic absoluteness.

Is it consistent with $ZFC$ that: "For every ordinal $\alpha$, there is a set-generic extension of the universe in which $\alpha$ has a projectively definable copy?"

If so, what is the least $n$ such that it is consistent with ZFC that every ordinal is "potentially $\Pi^1_n$"?
I am especially interested in what we know if $V=L$. In particular, I don't know that "there is a non-potentially-$\Pi^1_3$ ordinal" is even consistent with $ZFC+V=L$!
Conversely, I would be delighted (and amazed) if "Every ordinal is potentially $\Pi^1_3$" had nonzero consistency strength over $ZFC$, despite contradicting large cardinals.

REPLY [4 votes]: Great question!
The answer is yes. Start with $V=L$, and fix any ordinal $\alpha$.
Now go to the forcing extension $V[G]$ obtained by collapsing
$\aleph_\alpha^L$ to $\omega$, so that
$(2^\omega)^{V[G]}=\aleph_{\alpha+1}^L=\omega_1^{V[G]}$. In
$V[G]$, the ordinal $\alpha$ is definable in the structure
$H_{\omega_1}^{V[G]}$, since this structure sees all countable
objects, including all $L_\beta[G]$ for
$\beta<\omega_1^{V[G]}=\aleph_{\alpha+1}^L$, and therefore
$H_{\omega_1}^{V[G]}$ can identify the unique $\alpha$ for which
$\omega_1=\aleph_{\alpha+1}^L$. This definition is projective, since
we need to quantify only over countable objects.
I'll have to think a bit more to identify the exact complexity of
the definition.