# Can $Ord$ have nontrivial second-order elementary self-embeddings?

I forgot to mention originally: this was motivated by this old MSE question.

It's not hard to show in $$\mathsf{ZFC}$$ that there are nontrivial elementary embeddings from $$(Ord; \in)$$ to itself - or rather, it's not hard to write down specific formulas which $$\mathsf{ZFC}$$ proves define such maps. When we add common algebraic operations, things get messier but the picture seems to stay the same - e.g. the result continues to hold for $$Ord$$ equipped with ordinal addition, multiplication, and exponentiation in addition to $$\in$$. It seems like any "reasonably simple" expansion of $$Ord$$ has $$\mathsf{ZFC}$$-provable nontrivial elementary self-embeddings.

I'm interested in a particular precisiation of this intuition. Specifically, the following seems very close to the Kunen inconsistency but I can't quite get it there:

Is it consistent that there is a nontrivial second-order elementary self-embedding $$f:(Ord;\in)\rightarrow (Ord;\in)$$?

Note that the second-order quantifiers in this case range over subclasses of $$Ord$$ or its finite Cartesian powers, not just subsets (= bounded subclasses). So we really have to ask this within an appropriate class theory. Given that that may be a bit annoying to consider, here's a "set-ified" version of that question:

Is it consistent with $$\mathsf{ZFC}$$ that there is an inaccessible $$\kappa$$ and a second-order nontrivial elementary embedding $$f:(\kappa;\in)\rightarrow(\kappa;\in)$$?

(Here correspondingly the second-order quantifiers range over the $$\mathcal{P}(\kappa^{n})$$s for $$n\in\omega$$ as appropriate, so everything is nicely captured by $$V_{\kappa+2}$$: $$f$$ itself, or an equivalent coding thereof, lives in $$V_{\kappa+1}$$, the evaluation of each second-order formula over $$V_{\kappa}$$ is determined by $$V_{\kappa+1}$$, and $$V_{\kappa+2}$$ can express "is a second-order elementary embedding" directly.)

The difficulty I run into in getting a negative answer here by invoking the Kunen inconsistency is that even using second-order logic there doesn't seem to be a way to code $$V$$ into $$Ord$$ (or $$V_\kappa$$ into $$\kappa$$). Of course if $$V$$ is "sufficiently canonical" we can do this, e.g. the existence of such an $$f$$ (or such a $$\kappa$$ and $$f$$) clearly contradicts $$\mathsf{V=L}$$, but I don't see an argument for the general case.

• You can build $\mathbb{R}$ as the Dedekind (or Cauchy) completion of the field of fractions of the Grothendieck ring of $\omega$; does that count as a reasonably simple expansion? (the final step is second order) I believe that since the only automorphism of $\mathbb{R}$ is the identity, this would mean that there are no nontrivial second order elementary embeddings since they would preserve the final step in that construction and thusly yield a nontrivial automorphism of $\mathbb{R}$, correct? – Alec Rhea Dec 14 '20 at 20:44
• @AlecRhea No, that would only block an elementary embedding from moving "small" ordinals. Nothing about that prevents for example an elementary embedding which fixes all the ordinals below the first inaccessible, for example. – Noah Schweber Dec 14 '20 at 20:49
• @AlecRhea and actually it doesn't even do that: building the reals in that way does not amount to adding structure to $Ord$ directly. Rather, it gives an interpretation of $\mathbb{R}$ in $V_{\omega+1}$, and we still would need to find a way to code that directly into the ordinals - and it's not obvious how to do that. – Noah Schweber Dec 14 '20 at 20:53
• Always a pleasure to see someone giving a good answer to a nice question on MSE, and then making a nice question on MO as a result. That's quite the ideal way that the system should work, in my opinion. – Asaf Karagila Dec 14 '20 at 21:18
• @Noah: Maybe it's time you start using real large cardinals like super-Reinhardt cardinals. – Asaf Karagila Dec 15 '20 at 13:14

First, if choiceless cardinals are consistent, one cannot rule out the existence of a second-order elementary embedding. Corollary 4.8 in Usuba's "A note on Lowenheim-Skolem cardinals" states that if there is a proper class of Lowenheim-Skolem cardinals, one can force the Axiom of Choice by a homogeneous definable class forcing. (He doesn't say homogeneous, but it will be.) Therefore assume there is a $$j : V\to V$$ and there is a proper class of Lowenheim-Skolem cardinals. Let $$\mathbb P$$ be Usuba's forcing, and let $$G\subseteq \mathbb P$$ be a generic filter. Then $$j\restriction \text{Ord}$$ is second-order elementary in $$V[G]$$: if $$\varphi(\vec v)$$ is a second-order formula and $$V[G]\vDash \varphi(\vec \alpha)$$, then by homogeneity, $$1_\mathbb P\Vdash \varphi(\vec \alpha)$$, and so by the definability of $$\mathbb P$$ and the elementarity of $$j$$, $$1_\mathbb P\Vdash \varphi(j(\vec\alpha))$$, and hence $$V[G]\vDash \varphi(j(\vec\alpha))$$.
Second, assuming the HOD Conjecture and the existence of an extendible cardinal, one can rule out the existence of a nontrivial second-order elementary embedding of the ordinals. This follows from Theorem 3.47 of Woodin's paper "In search of Ultimate L." The reason is that any second-order elementary embedding $$j : \text{Ord}\to\text{Ord}$$ extends to an elementary embedding from the structure $$(\text{HOD},T)$$ to $$(\text{HOD},T)$$ where $$T$$ is the $$\Sigma_2$$ theory of $$V$$ with ordinal parameters. It is a prominent open question whether such an elementary embedding can ever exist. It is also open whether there can be an elementary embedding from $$\text{HOD}$$ to $$\text{HOD}$$ in the context of the HOD Conjecture.
• I mean, a super-Reinhardt implies there is a Reinhardt with a proper class of LS cardinals. No? In some sense an I3 is already "kinda" that. And if you want a 2nd order embedding, then Kunen cardinals (critical points of $j\colon V_{\lambda+2}\prec V_{\lambda+2}$, where $\kappa=\operatorname{crit}(j)$ and $\lambda=\sup j^n(\kappa)$) already kind of supply this situation to you, and the consistency of a Kunen cardinal is already bounded by I0. – Asaf Karagila Dec 15 '20 at 1:56
• @AsafKaragila If you want to get a model of NBG with one of these embeddings, a $j : V_{\lambda+2}\to V_{\lambda+2}$ is not going to work. Or an inaccessible $\kappa$ with a second order embedding $\kappa\to \kappa$. Feels like you're gonna need something like Reinhardt with proper class of Lowenheim-Skolems. (Of course it's even more reasonable to doubt the consistency of super-Reinhardt cardinals.) – Gabe Goldberg Dec 15 '20 at 2:07