Suppose $m, n\in\omega$ and $\kappa$ is a cardinal. Then $\kappa$ is *$\Pi^m\_n$-indescribable* if every $\Pi^m\_n$-sentence true about $\kappa$ is true about some $\lambda<\kappa$; formally, if for every $\Pi\_n$-sentence $\varphi$ in the language of set theory with a unary predicate and every $A\subseteq V\_\kappa$, there is some $\lambda<\kappa$ such that $$(V\_{\kappa+m}, \in, A)\models\varphi\implies (V\_{\lambda+m}, \in, A\cap V\_\lambda)\models\varphi.$$ For example, the $\Pi^1\_1$-indescribable cardinals are exactly the weakly compact cardinals. (Note that we can go beyond $\Pi^m\_n$-describability - [quite a ways](https://en.wikipedia.org/wiki/Shrewd_cardinal), even - but let's ignore that for now.)


I'm interested in a version of indescribability where we pay attention to what can happen in generic extensions of $V$ (I don't think this really qualifies as a "generic" version indescribability, but there's a vague connection). Specifically, for an ordinal $\alpha$ and a set $C\subseteq V\_\alpha$, write "$(V\_\alpha, \in, C)\models\_f\varphi$" if $((V\_\alpha)^{V[G]}, \in, C)\models\varphi$ for every $G$ which is set-generic over $V$ (note that the forcing for which $G$ is generic need *not* be an element of $V\_\alpha$). Now say that a cardinal $\kappa$ is *$\Pi^m\_n$-forcing indescribable* if for every $\Pi\_n$-sentence $\varphi$ in the language of set theory with a unary predicate and every $A\subseteq V\_\kappa$, there is some $\lambda<\kappa$ such that $$(V\_{\kappa+m}, \in, A)\models\_f\varphi\implies (V\_{\lambda+m}, \in, A\cap V\_\lambda)\models\_f\varphi.$$ *Note that $A$ is a fixed set in the ground model.*


My question is, roughly how "big" are the forcing-indescribable cardinals? For example, it is not clear to me what the relationship is between $\Pi^1\_1$-indescribable (= weakly compact) cardinals and $\Pi^1\_1$-forcing indescribable cardinals. I believe that *in $L$*, $\Gamma$-forcing indescribability implies $\Gamma$-indescribability (since we can replace $\varphi$ with "$\varphi$ holds in $L$"), but I don't see the converse, or how this holds in general $V$.



What is forcing indescribability?
Let me start things off by providing an upper bound. The bound is
very large, however, and I expect that it can be improved, perhaps
dramatically. But at least it shows the consistency of your large
cardinal relative to some other well-studied large cardinals.


**Theorem.** If $\kappa$ is $1$-$C^{(2)}$-extendible, then it is
forcing $\Pi^m\_n$-indescribable for every $m,n$.


**Definition.** A cardinal $\kappa$ is
[$1$-$C^{(2)}$-extendible](https://en.wikipedia.org/wiki/Extendible_cardinal),
if there is an elementary embedding $j:V\_{\kappa+1}\to
V\_{\theta+1}$, with critical point $\kappa$, such that the target
$j(\kappa)=\theta$ is $\Sigma\_2$-correct in $V$, meaning
$V\_\theta\prec\_{\Sigma\_2} V$.


This is a fairly strong large cardinal notion, far stronger than
the totally indescribable cardinals you mention in your question.
For example, every 1-extendible cardinal is superstrong and much
more. But these cardinals are weaker than Vopěnka's principle.


**Proof.** Assume that $\kappa$ is $1$-$C^{(2)}$-extendible. So
there is an elementary embedding $j:V\_{\kappa+1}\to V\_{\theta+1}$
with critical point $\kappa$ and $j(\kappa)=\theta$ is
$\Sigma\_2$-correct in $V$.


Suppose now that $A\subset V\_\kappa$ and $\langle
V\_{\kappa+m},\in,A\rangle\models\_f\varphi$, which means that
$\langle V[G]\_{\kappa+m},\in,A\rangle\models\varphi$ for every
forcing extension $V[G]$. This is a $\Sigma\_2$ property about $A$
and $\kappa$, since any violation of it would be revealed inside
some large enough $V\_\eta$, using forcing inside that $V\_\eta$.


Thus, by $\Sigma\_2$-correctness, we see that $V\_\theta$ agrees that
$\langle V\_{\kappa+m},\in,A\rangle\models\_f\varphi$. Since
$A=j(A)\cap\kappa$, we may pull this back by elementarity to
conclude that there is some $\lambda<\kappa$ with $\langle
V\_{\lambda+m},\in,A\cap V\_{\lambda+m}\rangle\models\_f\varphi$
inside $V\_\kappa$. But $\kappa$ itself must also be
$\Sigma\_2$-correct, and so actually $\langle
V\_{\lambda+m},\in,A\cap V\_{\lambda+m}\rangle\models\_f\varphi$ in
$V$, as desired. **QED**


I'll think some more about lower bounds and about pulling down the
strength of the hypothesis.


**Update.** I've realized that we can improve the upper bound as
follows. We don't really need the "$+1$", since that actually
provided a uniform version of the phenomenon, with the same
embedding working for every $A$.


Thomas Johnstone and I defined that a cardinal $\kappa$ is
*uplifting*, if it is inaccessible and $V\_\kappa\prec V\_\theta$ for
cofinally many inaccessible cardinals $\theta$. (J. D. Hamkins, T.
Johnstone, [Resurrection axioms and uplifting
cardinals](http://jdh.hamkins.org/resurrection-axioms-and-uplifting-cardinals/))
A boldface version is that $\kappa$ is *strongly uplifting*, if for
every $A\subset V\_\kappa$ there are cofinally many inaccessible
cardinals $\theta$ for which $\langle
V\_\kappa,\in,A\rangle\prec\langle V\_\theta,\in,A^\*\rangle$ for some
$A^\*\subset V\_\theta$. (J. D. Hamkins, T. Johnstone, [Strongly
uplifting cardinals and boldface
resurrection](http://jdh.hamkins.org/strongly-uplifting-cardinals-and-boldface-resurrection/))
These have diverse equivalent formulations, as I mention on the
linked blog post, connected with strengthenings of the strongly
unfoldable cardinals.


Let me now strengthen this a little more, for the present
application, with the following new large cardinal concept.


**Definition.** A cardinal $\kappa$ is *strongly
$C^{(n)}$-uplifting*, if for every $A\subset V\_\kappa$ there is a
$\Sigma\_n$-correct cardinal $\theta$ and $A^\*\subset V\_\theta$ with
$\langle V\_\kappa,\in,A\rangle\prec\langle
V\_\theta,\in,A^\*\rangle$.


This is what we really needed in the theorem above.


**Theorem.** If $\kappa$ is strongly $C^{(2)}$-uplifting, then it
is forcing $\Pi^m\_n$-indescribable for every $m$ and $n$.


**Proof.** Argue as in the first theorem above, but now we have
only $\langle V\_\kappa,\in,A\rangle\prec\langle
V\_\theta,\in,A^\*\rangle$, instead of $j$. If $\langle
V\_\kappa,\in,A\rangle+m\models\_f\varphi$, then this will be true
inside $V\_\theta$ since it is $\Sigma\_2$-correct, and so $V\_\theta$
thinks that this holds on an initial segment of $A^\*$, and so we
get $\lambda<\kappa$ with $\langle V\_\lambda,\in,A\cap
V\_\lambda\rangle\models\_f\varphi$ inside $V\_\kappa$, which is right
about this since $\kappa$ is itself $\Sigma\_2$-correct. **QED**


I think the strongly $C^{(2)}$-uplifting cardinals are
comparatively weak, and absolute to $L$, but I'll think more about
it.


