# Can one force there to be an elementary embedding $j:V_{\lambda}\rightarrow V_{\lambda}$ for some inaccessible $\lambda$?

Is it consistent that there exists an inaccessible cardinal $\lambda$ and a forcing extension $V[G]$ so that $$V[G]\models\text{There is some non-trivial elementary embedding j:V_{\lambda}^{V[G]}\rightarrow V_{\lambda}^{V[G]}}?$$

Any such forcing extension $V[G]$ will force the cofinality of $\lambda$ to be $\omega$.

Is it consistent that there exists an inaccessible cardinal $\lambda$ and a forcing extension $V[G]$ so that $$V[G]\models\text{There is some non-trivial elementary embedding j:V_{\lambda+1}^{V[G]}\rightarrow V_{\lambda+1}^{V[G]}}?$$

What if instead of having $\lambda$ be inaccessible, $\lambda$ satisfies some stronger large cardinal hypothesis? What is the consistency strength of these hypotheses?

• When you say $j:V_\lambda\to V_\lambda$, you mean $V_\lambda^V$ right? – Victoria Gitman Nov 7 '15 at 21:48
• Do you mean that the image of the embedding is $V_\lambda$ itself, right? – Asaf Karagila Nov 7 '15 at 21:58
• Victoria. I meant $V_{\lambda}^{V[G]}$ to make it look like an I3 and a I1 cardinal. Asaf. I simply put the quotes there to specify the statement that $V[G]$ models, but I now removed them so that they do not cause confusion. I edited the question accordingly. – Joseph Van Name Nov 7 '15 at 22:03
• You didn't answer my question; but I guess that the answer is "yes". – Asaf Karagila Nov 7 '15 at 22:04
• @EverettPiper, I am not sure that your argument that the situation Joe wants cannot occur in a small forcing extension is accurate. Laver showed that for a fixed $\kappa$, if I3 (say) holds in a $\kappa$-small extension for $\lambda$ and $\kappa$, then there is potentially another $\lambda'$, but same $\kappa$ such that I3 holds for them in $V$. In the case $\text{cf}(\lambda)=\omega$ in $V$, $\lambda'=\lambda$. – Victoria Gitman Nov 8 '15 at 19:37

Given Victoria's wonderful comment, I wonder if the following small forcing example is relevant to your question. The argument is from Laver's article "Certain very large cardinals are not created in small forcing extensions" (Annals of Pure and Applied Logic 149, 2007).

Call a a generic extension $V[G]$ $\kappa$-small if, in $V$, $\mathbb{P}$ has size less than $\kappa$. Laver denotes $I1$ by $E_\omega(\kappa, \lambda)$ where $\kappa$ is the critical point of the embedding $j$ and $\lambda$ is the sup of $j$'s critical sequence. $I3$ is simply $E_0(\kappa, \lambda)$ and there are intermediate hypotheses between these two extremes, denoted $E_n(\kappa, \lambda)$, where the subscript $n$ indicates that there is a non-trivial elementary embedding $j$ which is $\Sigma_{2n}^1$. Vincenzo Dimonte has some nice notes on these embeddings (and much more).

Theorem: If $V[G]$ is a $\kappa$-small extension of $V$, $n\leq\omega$, and $V[G]$ thinks there is a $\lambda$ satisfying $E_n(\kappa,\lambda)$, then $V$ thinks there is $\lambda$ satisfying $E_n(\kappa, \lambda)$.

edit: I should note that the following argument is not a proof of the theorem stated; the theorem just gives some context to the argument.

Laver notes that the embedding $k$ in $V[G]$ which witnesses $E_n(\kappa, \lambda)$ need not satisfy $k\restriction V_\lambda\in V$. There he gives an example where $\lambda$ does not have countable cofinality in $V$.

Consider the partial order which collapses $\omega_1$ and suppose $j$ witnesses $E_{n+2}(\kappa, \bar{\lambda})$. Then there are unboundedly many $\kappa'$ less than $\kappa$ with an embedding $j_{k'}$ satisfying $E_{n+1}(\kappa',\bar{\lambda})$.

Let, in $V$, $\lambda$ be the sup of the first $\omega_1$-many of the $\kappa'$s and note that this $\lambda$ has countable cofinality in the generic extension. So you can choose, in $V[G]$, an $\omega$-sequence $\kappa_0, \kappa_1, \dots$ converging to $\lambda$, along with their respective witness embeddings $\hat{j}_{\kappa_m}$ (the hat indicates that these are the lifts of the original embeddings $j_{\kappa_m}$ which live in $V$).

Using this $\omega$-sequence of lifted embeddings, you can form their inverse limit $J:V_\lambda^{V[G]}\prec V_{\bar{\lambda}}^{V[G]}$ and this is an $E_{n+1}$ embedding. Since there's a $E_n$ embedding $\hat{k}$ (the lift of $k$) on $V_{\bar{\lambda}}^{V[G]}$ there is one on $V_{\lambda}^{V[G]}$ as well.

I see no real obstacle to using the same argument (though I am cautious about such a claim): instead of $\omega_1$, collapse the first inaccessible to countable cofinality. Each critical point of any of the embeddings used is already measurable, so there are lots of inaccessibles for you to pick. If you don't want to damage the universe too much, perhaps you can even use Prikry forcing to get a $\lambda$ which is measurable in $V$ (or even a measurable limit of critical points of embeddings), but witnesses some $E_n$ in $V[G]$.

Laver's paper has lots of interesting ideas in it. If you haven't already looked at it, maybe check it out. Also, Scott Cramer has lots of good stuff on inverse limits and reflection properties of these gargantuan cardinals.

Perhaps you are familiar with all of this already. If so, have you come across some obstacle to repeating this argument in the case where your chosen $\lambda$ is an inaccessible or stronger?

• Everett Piper. I was not familiar with argument yet. I will need to read up on the papers and notes that you have referenced. – Joseph Van Name Nov 11 '15 at 17:16

If I understand the question correctly, then Woodin's Σ2 resurrection theorem gives examples. Assume there is a proper class of Woodin cardinals, and let φ be a Σ2 statement which holds in V. Let P be any poset. There is then a poset Q in VP such that φ holds in VP*Q. P could kill all rank-in-rank embeddings, but then Q would resurrect one.

• Ralf, it seems the problem is different. The question is not just whether forcing can create a rank-into-rank embedding $j\!:V_\lambda\to V_\lambda$ (which the argument you give shows it can) but whether we can do it starting with $\lambda$ an inaccessible cardinal in the ground model (which in the extension is necessarily singular of cofinality $\omega$). It is not clear in the argument you give that the witnessing $\lambda$ after forcing with your $\mathbb Q$ would be inaccessible in $V^{\mathbb P}$. – Andrés E. Caicedo Sep 17 '17 at 22:55
• Ah, okay, I missed that. – Ralf Schindler Sep 18 '17 at 7:28