29
$\begingroup$

Forcing construction in set theory leads to a new understanding of the mathematical (multi)universe by providing a machinery through which one can construct new models of the universe from the existing ones in a fairly controlled and comprehensible way and connect them to one another through forcing extensions.

Since the very beginning, the true nature of this rather bizarre construction method has been a matter of debate. A widely accepted (?) analogy emphasizes on a degree of similarity between forcing method in set theory and field extensions in Galois theory where one constructs new fields using the existing ones by adding new elements. Besides some papers and textbooks, a good description of this analogy could be found in this MathOverflow good oldie: Forcing as a new chapter of Galois Theory?.

However, as Dorais mentioned in his answer, despite the undeniable similarity, this analogy is not as perfect as what it seems at the first glance. So, forcing is not a one hundred percent field extension-like construction anyway. This fact causes some confusion concerning the possible translation of Galois theory machinery into the language of forcing and set theory. It is often not immediately clear what the forcing analogy of a given notion in the field extensions would be. Also, sometimes there is more than one approach towards defining a corresponding Galois theoretic notion in forcing so that one needs to decide which the natural one is.

One such central concept in Galois theory is the notion of the degree of a field extension, that is the dimension of the extended field while viewed as a vector space over the ground field. The question that arises here is whether there is any corresponding well-defined, well-behaved and natural similar notion in set-theoretic forcing.

One may think if a forcing extension could be viewed as a kind of vector space-like structure over the ground model. In this sense, there might be a kind of basis associated with any such pair of models which itself may satisfy some uniqueness properties that give rise to the existence of a well-defined notion of (relative) dimension of a generic forcing extension with respect to its ground model. Our axiomatic expectations of the possible behavior of any such notion of basis/dimension is also an interesting topic to explore even before defining any such notion. Of course, we need them to behave as natural as possible and resemble their Galois theoretic counterparts fairly closely.

Question. What are examples of defining a notion of dimension or basis for forcing generic extensions in the set-theoretic literature (possibly close to the same fashion that exists for the field extensions and vector spaces)?

I couldn't find much along these lines in the literature except a short unpublished note of Golshani in which he takes a Galois theoretic approach towards dimension in forcing by dealing with mutually generic sequences and the chain of forcing extensions. It also shares many features with Hamkins' set-theoretic geology project. However, this notion of forcing dimension seems not to be complete enough to cover all aspects of a vector space view towards generic extensions but could be a really good starting point anyway.

Remark. As Peter stated in his comment, absoluteness would be an issue while defining a notion of (relative) dimension in the set-theoretic multiverse. One may ask, from whose perspective are you trying to calculate the relative dimension of two set-theoretic universes and why? In fact, due to the highly contradictory views of different models of $\sf ZFC$ towards anything beyond the narrow scope of universally absolute properties, there is not more than a little hope to obtain an absolute notion of forcing dimension after all. However, one may argue that absoluteness is not a crucial condition in the long list of our expectations from a possible natural definition of dimension for generic models. The same situation happens in the Hamkins' Well-foundedness Mirage Principle stating that "Every universe $V$ is non-well-founded from the perspective of another universe" and so there is no standard model of $\sf ZFC$.


Update. According to the Joel and Mohammad's answers, it turned out that there is more than one approach towards developing a dimension theory for forcing extensions; each with their own characteristics. While Joel's definition requires all forcing dimensions to be infinite, Mohammad's approach allows finite dimensions as well as infinite ones. Also, as Monroe pointed out both definitions fail to satisfy the so-called downwards closure property.

In general, Joel's approach sounds a little bit vector space-like to me while Mohammad's reminds me the Krull's dimension in commutative algebra. There is also a chance that these different dimensions be related or coincide with each other under certain circumstances.

The point is that in the absence of a general common sense about the expected behavior of a nice forcing dimension, it is not immediately clear whether the mentioned features are flaws or advantages of the presented definitions. Maybe one needs to fix some abstract list of required properties for any such dimension operator and then search for its existence in the forcing extensions. Anyway, if there is any suggestion for such an axiomatic approach towards forcing dimension, I will be so happy to hear about it.

$\endgroup$
  • 2
    $\begingroup$ Two thoughts: (1) The sheaf-theoretic/topos-theoretic view of forcing seems perhaps quite relevant. (2) One must of course be careful to distinguish between discussing a forcing extension from three viewpoints: the logic of the ground model, the logic of the extension, and a model-theoretic setting external to both. A priori, dimension seems maybe a quite extrinsic property of a model/extension; compare how in the topos-theoretic picture, every topos thinks that it is the topos of sheaves on the one-point space, even though externally it might be a non-trivial space. $\endgroup$ – Peter LeFanu Lumsdaine May 31 '18 at 9:30
  • $\begingroup$ @PeterLeFanuLumsdaine (+1) Nice point, Peter! I have added some explanations about what I consider the soul of the second part of your remark into the original post for better visibility. Thanks! $\endgroup$ – Morteza Azad May 31 '18 at 10:31
35
$\begingroup$

My co-authors and I introduced a notion of dimension for forcing extensions in the following paper:

Specifically, for any forcing extension $V\subset V[G]$, we defined the essential size of the extension to be the smallest cardinality in $V$ of a complete Boolean algebra $\mathbb{B}\in V$ such that $V[G]$ is realized as a forcing extension of $V$ using $\mathbb{B}$, so that $V[G]=V[H]$ for some $V$-generic $H\subset\mathbb{B}$. More recently, I have been inclined to call this the forcing dimension of $V[G]$ over $V$.

This can indeed be seen as a dimension, in light of the following (essentially lemma 23 of the paper above):

Theorem. If $V\subset V[G]$ has essential size $\delta$, then the essential size of any further extension $V[G][H]$ over $V$ is at least $\delta$.

Proof. By combining the forcing into an iteration, we may view $V[G][H]$ as a single-step forcing extension of $V$, and so it has some essential size over $V$. Since we have an intermediate model $V\subset V[G]\subset V[G][H]$, it follows by the intermediate model theorem that $V[G]$ can be realized as an extension by a complete subalgebra of that forcing notion. So the smallest size of a complete Boolean algebra realizing $V[G]$ is not larger than the smallest size of a compete Boolean algebra realizing $V[G][H]$ over $V$. $\Box$

We had used the fact that there is a definable dimension, in the forcing extensions over $L$, to show in general circumstances that the modal logic of $\Gamma$-forcing over $L$ is contained in S4.3, for a wide collection of forcing classes $\Gamma$.

Monroe Eskew points out in the comments that, contrary to my initial thoughts about this, the size of the smallest partial order giving rise to the extension will also serve as a forcing dimension. The reason is that the density of a complete subalgebra of a complete Boolean algebra is at most the density of the whole algebra, simply by projecting any dense set of the larger algebra to the subalgebra. It follows by the same argument as in the theorem above that the poset-based forcing dimension of any intermediate model in a forcing extension is bounded by the minimal size of a partial order giving rise to the whole extension.

I propose that we officially adopt the poset-based notion as the forcing dimension of a forcing extension $V\subset V[G]$, denoting this dimension by $\left[V[G]\mathrel{:}\strut V\right]$. We may now observe the following attractive identity, confirming the suggestion of Will Brian.

Theorem. For any successive forcing extensions $V\subset V[G]\subset V[G][H]$, we have $$\left[V[G][H]\mathrel{:}\strut V\right]=\left[V[G]\mathrel{:}\strut V\right]\cdot\left[V[G][H]\mathrel{:}\strut V[G]\right].$$

Proof. Suppose that $G\subset\mathbb{P}\in V$ and $H\subset\mathbb{Q}\in V[G]$, where these are the minimal-size partial orders realizing the extensions. Suppose that $\mathbb{Q}$ has size $\kappa$ in $V[G]$. So without loss there is $\mathbb{P}$-name for a relation $\dot\leq$ on $\kappa$, such that $\mathbb{Q}=\langle\kappa,{\dot\leq}_G\rangle$ in $V[G]$. We can now use the partial order $\{(p,\check\alpha)\mid p\in\mathbb{P},\alpha<\kappa\}$, which is dense inside $\mathbb{P}*\dot{\mathbb{Q}}$, to realize $V\subset V[G][H]$. This shows $\leq$ of the desired identity.

Conversely, if we can realize $V[G][H]$ as a forcing extension of $V$ by some partial order $\mathbb{R}$, then $V[G]$ arises as a subforcing notion, and $V[G][H]\supset V[G]$ arises as quotient forcing. The quotient forcing $\mathbb{R}/G$ can be thought of as the conditions in $\mathbb{R}$ that are compatible with every element of (the image of) $G$ in (the Boolean completion of) $\mathbb{R}$. So the smallest partial order giving rise to $V[G][H]$ over $V[G]$ is at most the size of the smallest partial order giving rise to $V[G][H]$ over $V$, as desired. $\Box$

$\endgroup$
  • 2
    $\begingroup$ Just curious -- if we write $[V[G]:V]$ for the essential size of $V[G]$ with respect to the ground model $V$, then is it true that $[V[G][H]:V] = [V[G]:V] \cdot [V[G][H]:V[G]]$? $\endgroup$ – Will Brian May 31 '18 at 12:31
  • 2
    $\begingroup$ Why can’t you use minimal size of a partial order? If B is a complete subalgebra of a cba C, and C contains a dense set of size kappa, then the canonical projection of this set to B is dense in B, so the density of B is at most kappa. $\endgroup$ – Monroe Eskew May 31 '18 at 12:32
  • $\begingroup$ @Monroe: Because then you can have dimension $\aleph_0$, and set theorists are generally countablists. $\endgroup$ – Asaf Karagila May 31 '18 at 12:33
  • $\begingroup$ Monroe, it seems that you are right! I have edited. $\endgroup$ – Joel David Hamkins May 31 '18 at 12:41
  • $\begingroup$ @WillBrian Yes, that identity works, and I have added a proof, using the poset-based dimension. $\endgroup$ – Joel David Hamkins May 31 '18 at 13:33
12
$\begingroup$

Let me first give the motivations of the definitions given in my paper.

So suppose $F \subseteq K$ be fields. Note that $[K:F],$ the dimension of $K$ over $F$ considered as a vector space is equal to $\kappa$ iff $\kappa$ is maximal length of a chain $F=F_0 \subset F_1 \dots F_\kappa=K$, where each $F_i$ is a vector space over $F$. My definition of dimension is essentially taken from this fact.

The note is that based on Hamkins definition, we don't have finite dimension, while in my case for example a one dimenional extension of a model of ZFC is just a minimal generic extension, which in my opinion seems reasonable.

Also note that if $[K:F]=\kappa$  and if $\lambda < \kappa,$ then there exists a vector space $G$ over $F$ with $F \subset G \subset K$ and $[G:F]=\lambda.$ However this is not true in general nor for the Hamkins definition nor for my definition.

Also my definition allows to consider pairs $V \subset W$, where $W$ is not necessarily a set forcing extension of $V$, and for this reason I have defined the upward generic dimension and the downward generic dimension, the latter being related to set theoretic geology.

Another important notion is the notion of independence.Note that $x, y \in K$ are independent over $F$ iff $F[x] \cap F[y]=F$ (where for $X \subseteq K,$ $F[X]$ is the vector space generated by $F \cup X$). Note that a subset $X$ of $K$ is independent over $F$ iff for any finite disjoint subsets $X_1, X_2$ of $X$ we have $F[X_1] \cap F[X_2]=F$. On the other hand, using a theorem of Solovay, it is natural  for $V \subset W$ and for $x, y \in W$ to say that $x$ and $y$ are independet over $V$ if they are mutually generic. This is  my motivation for $\aleph_0$-transendence degree between models of set theory, of course in this case we can define $\kappa$-transendence degree for any $\kappa \geq \aleph_0.$  

Note that as it is stated in the comments, in general we don't need for example $x$ and $y$ to be mutually generic over a model $V$ to have  $V[x] \cap V[y]=V$, however the assumption of they being mutually generic has some advantages, for example we can form $V[x, y]$ and it is again a model of ZFC, while in general this may not be the case.

Regarding your update, I don't see any essential similarities between these two concepts:

1) Suppose $\mathbb{P}$ is a forcing notion of size $\kappa$ which gives a minimal generic extension of the universe and let $\mathbb{Q}=Add(\omega, \kappa).$ Then:

(1-a) In the sense of Hamkins definition $[V^{\mathbb{P}}, V]=[V^{\mathbb{Q}}, V]=\kappa$.

(1-b) In the sense of my definition $[V^{\mathbb{P}}, V]=1$ while $[V^{\mathbb{Q}}, V]=\kappa^+$.

2) Suppose $\mathbb{P}$ and $\mathbb{Q}$ gives minimal generic extensions of the universe and are of size $\kappa$ and $\lambda$ respectively where $\kappa < \lambda$. Then

(2-a) In the sense of Hamkins definition $[V^{\mathbb{P}}, V]=\kappa$ and $[V^{\mathbb{Q}}, V]=\lambda$.

(2-b) In the sense of my definition $[V^{\mathbb{P}}, V]=[V^{\mathbb{Q}}, V]=1$.

$\endgroup$
  • 3
    $\begingroup$ I don't think it is true that your notion of dimension has the "downwards closure" property in your third paragraph. If $W$ is a generic extension of $V$ for adding one Cohen real, then any $V'$ such that $V \subseteq V' \subseteq W$ has dimension either $0$ or $\omega_1$ over $V$, since $V' = V$ or is another Cohen-real extension. $\endgroup$ – Monroe Eskew May 31 '18 at 20:05
  • 2
    $\begingroup$ Regarding your last paragraph, while I do think it is reasonable to define independence over $V$ as being mutually generic, it should be noted that Solovay's result is not an equivalence. For example, you can construct two Cohen reals $c,d$ over (a countable model) $V$ which are not mutually generic, but still satisfy $V[c]\cap V[d]=V$. $\endgroup$ – Miha Habič May 31 '18 at 20:39
  • 2
    $\begingroup$ (+1) Very nice, Mohammad! Thanks for sharing your view of forcing dimension here! It is interesting that we have at least two different approaches towards the concept of forcing dimension, namely the the field extension-like (i.e. yours) and the vector space-like (i.e. Joel's). It would be even more interesting if it turns out that these two are closely connected and are behaving as different types of dimension on the same mathematical space. Any thoughts along these lines? Is your dimension related to Joel's? Maybe the key is in the relation between modal logic of forcing and its geology. $\endgroup$ – Morteza Azad Jun 1 '18 at 3:40
  • 3
    $\begingroup$ I don't know anything about logic, and maybe I am missing something here, but unless you are using some non-standard notation, your characterisation of $[F:K]$ is wrong. It does not apply to any field extension of degree $3$, for example. The claim about $\lambda < \kappa$ is, accordingly, also wrong for degree $3$ extensions. $\endgroup$ – Alex B. Jun 1 '18 at 10:08
  • 1
    $\begingroup$ @AlexB. Maybe I have used bad notation, but I mean the dimension of $K$ over $F$ when considered $K$ as a vector space over $F$. $\endgroup$ – Mohammad Golshani Jun 2 '18 at 4:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.