When is a classification problem “wild”?

I hope someone can point me to a quick definition of the following terminology.

I keep coming across wild and tame in the context of classification problems, often adorned with quotes, leading me to believe that the terms are perhaps not being used in a formal sense. Yet I am sure that there is some formal definition.

For example, the classification problem for nilpotent Lie algebras is said to be wild in dimension $\geq 7$. All that happens is that in dimension $7$ and above there are moduli. In what sense is this wild?

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This was essentially asked at mathoverflow.net/questions/5895/… – Mariano Suárez-Alvarez Jan 2 2010 at 12:58
That's fair, though I think the question is different enough to not qualify as a duplicate. Certainly the answers there will be helpful, though... – Ben Webster Jan 2 2010 at 14:25
@Mariano: I had a look at the question before posting, but it was not obvious to me that the questions were related except in the terminology. It's still not totally clear. Note that I am not talking about a particular algebra being wild or tame, but of a classification problem being wild or tame. The reference below by mathphysicist seemed to contain exactly what I wanted and this was not mentioned in the earlier question. – José Figueroa-O'Farrill Jan 2 2010 at 14:41
Well, that why it isn't really a duplicate. For reference, calling a finite dimensional algebra wild if the problem of classifying it representations is wild is standard terminology. – Ben Webster Jan 3 2010 at 0:37

I am not an expert but in the algebra and representation theory the apparently standard definition is as follows (see also here and here): a problem is wild if it contains a subproblem which is equivalent to the problem of simultaneously reducing to canonical form two linear operators on a finite-dimensional space.

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Many thanks! However I do not at once that the classification problem of, say, 7-dimensional nilpotent Lie algebras is "wild" in this sense, but it has given me something to think about. – José Figueroa-O'Farrill Jan 2 2010 at 11:45
I am accepting the answer, because I think that this is precisely the definition that I was looking for. Thanks again! – José Figueroa-O'Farrill Jan 2 2010 at 11:49
As for the 7d nilpotent Lie algebras, maybe the thesis (etd.uwaterloo.ca/etd/mpgong1998.pdf) and references therein could be of some help. Also you may wish to look here: mathoverflow.net/questions/5895/…, and at this page in Kirillov's book: books.google.com/… – mathphysicist Jan 2 2010 at 13:04

In local arithmetic, one talks of wild and tame ramification. Let me explain a special case.

Let $K$ be a finite extension of $\mathbb{Q}_p$ ($p$ prime) and $L$ a finite extension of $K$. There are two integers $e$, $f$ attached to $L|K$, called the ramification index and the residual degree.

If we denote the valuation of $L$ by $w:L^\times\to\mathbb{Z}$, then $e$ is the index $(w(L^\times):w(K^\times))$. If we denote by $l,k$ the residue fields of $L,K$, then $f$ is the degree $[l:k]$. We have $ef=[L:K]$ always.

The extension $L|K$ is said to be unramified if $e=1$, tamely ramified if $p\not| e$, wildly ramified if $p|e$, and totally ramified if $f=1$.

Tamely ramified extensions $L|K$ are very easy to understand. They are all of the form $L=L_0(\root n\of\pi)$ for some unramified extension $L_0|K$, some uniformiser $\pi$ of $L_0$ and some $n>0$ (prime to $p$).

Wildly ramified extensions are not fully understood. The simplest example where we know them all is $p=2$, $[L:K]=2$. For $K=\mathbb{Q}_2$, there are seven quadratic extensions, obtained by adjoining $$\sqrt5,\sqrt3,\sqrt{15},\sqrt2,\sqrt{10},\sqrt6,\sqrt{30}.$$ All of these are (wildly) ramified, except $\mathbb{Q}_2(\sqrt5)$, which is unramified. This is to be contrasted with the fact that for $p\neq2$, $\mathbb{Q}_p$ had only three quadratic extensions, obtained by adjoining $$\sqrt u,\sqrt p, \sqrt{up},$$ where $u$ is a unit which is not a square. Of these $\mathbb{Q}_p(\sqrt u)$ is unramified, the other two are (tamely) ramified.

I hope this gives some idea of the difference between tame and wild in arithmetic.

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This is an a completely unrelated use of the words "tame" and "wild." – Ben Webster Jan 2 2010 at 14:23
There was nothing in the question to indicate that it was about some very specific classification problem. How I wish the question had been more specific... – Chandan Singh Dalawat Jan 3 2010 at 3:27
Really? He did talk specifically about the classification problem for nilpotent Lie algebras... – Ben Webster Jan 3 2010 at 17:16
The question as stated is actually quite general, with the case of Lie algebras given merely as an example. Although perhaps a more specific question was intended, it is this general quiestion that fascinates me. Mathematics is saturated with diverse classification problems that we want to say are simple or complicated, or "tame" or "wild". As I explained in my answer, Borel equivalence relation theory provides a sweeping general context for comparing them, and thereby answers this general question. I wonder where the classification problem for nilpotent Lie algebras fits into the hiearchy? – Joel David Hamkins Jan 4 2010 at 0:03

Although your specific request for terminology in the case of Lie algebras has now been answered, there is a very interesting broader question underlying your inquiry. Namely, how can we understand in a precise general way the idea that a given classification problem is complicated? How are we to compare the relative difficulty of two classification problems?

These questions form the central motivation for the emerging subject known as Borel equivalence relation theory (see Greg Hjorth's survey article). The main idea is that many of the most natural equivalence relations arising in many parts of mathematics turn out to be Borel relations on a standard Borel space. To give one example, the isomorphism problem on finitely generated groups, but of course, there are hundreds of other examples. A classification problem for an equivalence relation E is really the problem of finding a way to describe the E-equivalence classes, of finding an E-invariant function that distinguishes the classes.

Harvey Friedman defined that one equivalence relation E is Borel-reducible to another relation F if there is a Borel function f such that x E y if and only if f(x) F f(y). That is, the function f maps E classes to F classes in such a way that different E classes get mapped to different F classes. This provides a classification of the E classes by using the F classes. The concept of reducibility provides a precise, robust way to say that one relation F is at least as complex as another E. Two relations are Borel equivalent if they reduce to each other, and we are led to the hierarchy of equivalence relations under Borel reducibility. By placing an equivalence relation into this hierarchy, we come to understand how complex it is in comparision with other equivalence relations. In particular, we say that one equivalence relation E is strictly simpler than F, if E reduces to F but not conversely.

It sometimes happens that one has a classification problem E and is able to provide a classification by assigning to each structure a countable list of data, such that two structures are equivalent iff they have the same data. This amounts to a reduction of E to the equality relation =, for two structures are E equivalent iff their data is equal. Such relations that reduce to equality are called smooth, and lay near the bottom of the hierarchy of Borel equivalence relations. These are the simplest equivalence relations. Thus, one way of showing that a relation is comparatively simple, is to show that it is smooth, and to show it is comparatively hard, show that it is not smooth.

The subject of Borel equivalence relation theory, as now developed by A. Kechris, G. Hjorth, S. Thomas and many others, is focused on placing many of the natural classification problems of mathematics into this hierarchy. Some of the main early results are the following interesting dichotomies:

Theorem.(Silver dichotomy) Every Borel equivalence relation E either has only countably many equivalence classes or = reduces to E.

The relation E0 says that two binary sequences are equivalent iff they agree from some point onward. It is easy to see that = reduces to E0, and an elementary argument shows that E0 does not reduce to =. Thus, E0 is strictly harder than equality. Moreover, it is a kind of next-step up in the hiearchy, in light of the following.

Theorem.(Glimm-Effros dichotomy) Every Borel equivalence relation E either reduces to = or E0 reduces to E.

The subject continues with many interesting results that gradually illuminate more and more of the hierarchy of Borel equivalence relations. For example, the Feldman-Moore theorem shows that every Borel equivalence relation E having every equivalence class countable is the orbit equivalence of a countable group of Borel bijections of the space. The relation Eoo is the orbit equivalence of the left-translation action of the free group F2 on its power set. This relation is complete for the countable Borel equivalence relations, in the sense that every countable Borel equivalence relation reduces to it. It's great stuff!

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