On Oct 27, 2014, at 11:00 PM, Sy David Friedman wrote:
The Stability Predicate S is the important thing. V is generic over the Stable Core = (L[S],S). As far as I know, V may not be generic over HOD; but it is generic over (HOD,S).
V is always a symmetric extension of HOD but maybe you have something else in mind.
Let A be a V-generic class of ordinals (so A codes V). Then A is (HOD, P)-generic for a class partial order P which is definable in V. So if T is the -theory of the ordinals then P is definable in (HOD,T) and A is generic over (HOD,T).
But you did not answer my question. Are you just really conjecturing that if V is generic over N then there is no nontrivial ? Let me phrase this more precisely.
Suppose A is a V-generic class of ordinals, N is an inner model of V, P is a partial order which is amenable to N and that A is (N,P)-generic.
Are you conjecturing that there is no non-trivial ? Or that there is no nontrivial ? Or nothing along these general lines?
I would think that based on HP etc., you would actually conjecture that there _is_ a nontrivial j:HOD \to HOD.
No. This is the “reality check” that Peter and I discussed. Maximality suggests that V is as far from HOD as possible, but we have to acknowledge what is not possible.
So maximality considerations have no predictive content. It is an idea which has to be continually revised in the face of new results.
Yet you propose to deduce the non existence of large cardinals at some level based on maximality considerations. I would do the reverse, revise maximality.
I guess this is yet another point we just disagree on.
PS: With embarrassment and apologies to the group, I have to report that I found a bug in my argument that maximality kills supercompacts. I’ll try to fix it and let you know what happens. I am very sorry for the premature claim.
Suppose that there is an extendible and that the HOD Conjecture fails. Then:
1) Every regular cardinal above the least extendible cardinal is measurable in HOD (so HOD computes no successors correctly above the least extendible cardinal).
2) Suppose is an inaccessible cardinal which is a limit of extendible cardinals. Then there is a club such that every is a regular cardinal in HOD (and hence inaccessible in HOD).
So, if you fix the proof, you have proved the HOD Conjecture.