What interesting/nontrivial results in Algebraic geometry require the existence of universes? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T05:28:47Z http://mathoverflow.net/feeds/question/24552 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/24552/what-interesting-nontrivial-results-in-algebraic-geometry-require-the-existence-o What interesting/nontrivial results in Algebraic geometry require the existence of universes? Harry Gindi 2010-05-13T23:11:06Z 2010-07-31T16:59:29Z <p>Brian Conrad indicated a while ago that many of the results proven in AG using universes can be proven without them by being very careful (<a href="http://mathoverflow.net/questions/15897/in-what-topology-dm-stacks-are-stacks/15910#15910" rel="nofollow">link</a>). I'm wondering if there are any results in AG that actually depend on the existence of universes (and what some of the more interesting ones are).</p> <p>I'm of course aware of the result that as long as we require that the classes of objects and arrows are sets (this is the only valid approach from Bourbaki's perspective), for every category C, there exists a universe U such that the U-Yoneda lemma holds for U-Psh(C) (this relative approach makes proper classes pointless because every universe allows us to model a higher level of "largeness"), but this is really the only striking application of universes that I know of (and the only result I'm aware of where it's clear that they are necessary for the result). </p> http://mathoverflow.net/questions/24552/what-interesting-nontrivial-results-in-algebraic-geometry-require-the-existence-o/28913#28913 Answer by Joel David Hamkins for What interesting/nontrivial results in Algebraic geometry require the existence of universes? Joel David Hamkins 2010-06-21T04:25:53Z 2010-07-31T14:17:09Z <p>My belief is that no result in algebraic geometry that does not explicitly engage the universe concept will fully require the use of universes. Indeed, I shall advance an argument that no such results actually need anything beyond ZFC, and indeed, that they need much less than this. (But please note, I answer as a set theorist rather than an algebraic geometer.)</p> <p>My reason is that there are several hierarchies of weakened universe concepts, which appear to be sufficient to carry out all the arguments that I have heard using universes, but which are set-theoretically strictly weaker hypotheses.</p> <p>A universe, as you know, is known in set theory as $H_\kappa$, the set of all hereditarily-size-less-than-$\kappa$ sets, for some inaccessible cardinal $\kappa$. Every such universe also has the form $V_\kappa$, in the cumulative Levy hierarchy, because $H_\kappa=V_\kappa$ for any beth-fixed point, which includes all inaccessible cardinals. Thus, the Universe Axiom, asserting that every set is in a universe, is equivalent to the large cardinal assertion that there are unboundedly many inaccessible cardinals.</p> <p>This hypothesis is relatively low in the large cardinal hierarchy, and so from this perspective, it is relatively mild to just go ahead and use universes. In consistency strength, for example, it is strictly weaker than the existence of a single Mahlo cardinal, which is considered to be very weak large-cardinal theoretically, and much stronger hypotheses are routinely considered in set theory. Nevertheless, these hypotheses do definitely exceed ZFC in strength, unless ZFC is already inconsistent, and so your question is a good one. It follows the pattern in set theory of inquiring the exact large cardinal strength of a given hypothesis.</p> <p>The weaker universe concepts that I propose to use in replacement of universes are the following, where I take the liberty of introducing some new terminology.</p> <ul> <li>A <em>weak universe</em> is some $V_\alpha$ which models ZFC. The Weak Universe axiom is the assertion that every set is in a weak universe.</li> </ul> <p>This axiom is strictly weaker than the Universe Axiom, since in fact, every universe is already a model of it. Namely, if $\kappa$ is inaccessible, then there are unboundedly many $\alpha\lt\kappa$ with $V_\alpha$ elementary in $V_\kappa$, by the Lowenheim-Skolem theorem, and so $V_\kappa$ satisfies the Weak Universe Axiom by itself. From what I have seen, it appears that most of the applications of universes in algebraic geometry could be carried out with weak universes, if one is somewhat more careful about how one treats universes. The difference is that when using weak universes, one must pay attention to whether a given construction is definable inside the universe or not, in order to know whether the top level $\kappa$ of the weak universe, which may now be singular (and this is the difference), is reached.</p> <ul> <li>Let us say that a <em>very weak universe</em> is simply a transitive set model of ZFC. (In set theory, one would want just to call these <em>universes</em>, but here that word is taken to have the meaning above; so we could call them <em>set-theoretic universes</em>.) The Very Weak Universe Axiom (or Set-theoretic Universe Axiom) is the assertion that every set is an element of a very weak universe.</li> </ul> <p>The difference between a very weak universe and a weak universe is that a very weak universe $M$ may be wrong about power sets, even though it satisfies its own version of the Powerset Axiom. Set theorists are very attentive to such very weak universes, and pay attention in a set-theoretic construction to which model of set theory it is undertaken. If the algebraic geometers were to give similar attention to this point, thereby turning themselves into set theorists, I believe that all of their arguments using universes could be essentially replaced with very weak universes. Another important point is that while universes are always linearly ordered by inclusion, this is no longer true for very weak universes.</p> <p>Now, even the Very Weak Universe Axiom transcends ZFC in consistency strength, because it clearly implies Con(ZFC). So let me now describe how one might provide an even greater reduction in the strength of the hypothesis, and capture a use of universes within ZFC itself.</p> <p>The key is to realize that algebraic geometry does not really use the full force of ZFC. (Please take this with some skepticism, given my comparatively little exposure to algebraic geometry.) It seems to me unlikely, for example, that one needs the full Replacement Axiom in order to carry out the principal goal constructions of algebraic geometry. Let me suppose that these arguments can be carried out in some finite fragment $ZFC_0$ of $ZFC$, for example, $ZFC$ restricted to formulas of complexity $\Sigma_N$ for some definite number $N$, such as $100$ or so. In this case, let me define that a <em>good-enough-universe</em> is $V_\kappa$, provided that this satisfies $ZFC_0$. All such good-enough universes have $V_\kappa=H_\kappa$, just as with universes, since these will be beth-fixed points. The Good-enough Universe Axiom is the assertion that every set is a member of a good-enough universe.</p> <p>Now, my claim is first, that this Good-enough Universe Axiom is sufficient to carry out most or even all of the applications of universes in algebraic geometry, provided that one is sufficiently attentive to the set-theoretic issues, and second, that this axiom is simply a theorem of ZFC. Indeed, one can get more, that the various good-enough universes agree with each other on truth.</p> <p><b>Theorem.</b> There is a definable closed unbounded class of cardinals $\kappa$ such that every $V_\kappa$ is a good-enough universe and furthermore, whenever $\kappa\lt\lambda$ in $C$, then $\Sigma_N$-truth in $V_\kappa$ agrees with $\Sigma_N$-truth in $V_\lambda$, and moreover, agrees with $\Sigma_N$ truth in the full universe $V$.</p> <p>This theorem is exactly an instance of the Levy Reflection Theorem.</p> <p>OK, so if I am right, then the algebraic geometers can carry out their universe arguments by paying a lot of careful attention to the set-theoretic complexity of their constructions, and using good-enough universes instead of universes.</p> <p>But should they do this? For most purposes, I don't think so. The main purpose of universes is as a simplifying device of convenience to stratify the full universe by levels, which can be fruitfully compared by local notions of large and small. This makes for a very convenient theory, having numerous local concepts of large and small.</p> <p>I can imagine, however, a case where one has used the Universe theory to prove an elementary result, such as Fermat's Last Theorem, and one wants to know what are the optimal hypotheses for the proof. The question would be whether the extra universe hypotheses are required or not. The thrust of my answer here is that such a question will be answered by replacing the universe concepts that are used in the proof with any of the various weakened universe concepts that I have mentioned above, and thereby realizing the theorem as a theorem of ZFC or much less.</p> http://mathoverflow.net/questions/24552/what-interesting-nontrivial-results-in-algebraic-geometry-require-the-existence-o/34036#34036 Answer by Andreas Blass for What interesting/nontrivial results in Algebraic geometry require the existence of universes? Andreas Blass 2010-07-31T16:59:29Z 2010-07-31T16:59:29Z <p>There is a paper of Solomon Feferman from 1969 in which he proposes a conservative extension of ZFC adequate for formulating much of category theory. Since the need for universes in algebraic geometry seems to arise mainly via the use of category theory, I conjecture that Feferman's approach may be useful in algebraic geometry. The paper in question is "Set-theoretical foundations of category theory" (in "Reports of the Midwest Category Seminar III", Springer Lecture Notes in Math 106, edited by Saunders Mac Lane, pages 201-247). Feferman proposes to add to the language of ZFC a new constant symbol $\kappa$ and to add to the axioms the schema saying that <code>$V_\kappa$</code> is an elementary submodel of the universe $V$ of all sets, with respect to the original language of ZFC. That is, for each formula <code>$\phi(x_1,\dots,x_n)$</code> in the original language (i.e., not involving $\kappa$), there is an axiom saying that, for each <code>$x_1,\dots,x_n\in V_\kappa$</code>, the statement $\phi(x_1,\dots,x_n)$ is equivalent to the same statement with all quantified variables restricted to range over <code>$V_\kappa$</code>. The idea is that this $V_\kappa$ can play the role of a Grothendieck universe, even though it isn't really one. Furthermore, in situations where one would expect to use two or more universes, one can often get by with one, by a judicious use of the "elementary submodel" axioms. </p> <p>Note that one cannot express the notion of "elementary submodel of the universe" as a single formula, since there is no uniform way to express "truth in the universe" (Tarski's theorem). But one can express it one formula at a time, as an infinite axiom scheme, since there's no problem expressing truth of a single formula $\phi$ (just say $\phi$). And this is what Feferman does.</p> <p>Feferman's axiom system is conservative over ZFC. That is, if you can prove, in Feferman's system, a sentence that doesn't involve $\kappa$, then you can prove the same sentence in ZFC. (This follows from Levy's reflection principle plus the compactness theorem of first-order logic.)</p> <p>Many years ago, I taught a category theory course, using Feferman's axioms as the foundation. For the most part, this foundation worked well, but, if I remember correctly, I ran into a problem proving the existence of Kan extensions. I don't remember what I did to overcome the problem (nor do I even remember exactly what the problem was). </p> <p>I believe Feferman has done additional work improving this foundational system, but I have not yet absorbed that work.</p>