Does "classification" of mathematical objects have a precise meaning?

Question

I've heard about things in math that have been "completely classified," like finite simple groups, and other things that are not completely classified, like topological spaces. Does "classification" in this sense have a generally accepted and precise meaning?

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By "precise," I'm hoping it could clarify questions like the following:

1. Does "complete classification" of some objects necessarily mean that we can write an algorithm that determines whether two input objects are isomorphic, where the algorithm completes after a finite number of steps? That's what this answer suggests, but this seems overly restrictive. Suppose I have a finite list of properties such that iff $A$ and $B$ agree on all properties, then $A$ and $B$ are isomorphic. Suppose further that there is no algorithm that can, in a finite number of steps, always determine whether these properties agree for any two input objects. Is this a still classification?

2. Suppose I have a finite list of properties such that iff $A$ and $B$ agree on all properties, then $A$ and $B$ are isomorphic. But suppose the "value" of one such property takes on a type of object that is not completely classified. For example, suppose I have such a list of properties for classifying manifolds, and that one of these properties is a certain Banach space associated with the manifold. I.e., if two manifolds have the same (or isomorphic) associated Banach space, and also agree in all of the other properties on my list, then the manifolds are isomorphic. Banach spaces, however, are not completely classified. Does this mean I would still not have completely classified manifolds?

3. Suppose I don't have a list of properties that uniquely determine each object (up to isomorphism), but instead have a list of properties that still somehow provide a nice organization scheme. Is this ever considered a classification?

Answer 1

I don't think it has a precise meaning, but usually to classify a set $X$ means to express $X$ as the disjoint union of the images of a some functions $$f_0 : P_0 \rightarrow X, \;\ldots, f_n : P_n \rightarrow X$$ For example, for the classification of closed surfaces, we let $X$ denote the groupoid of all connected closed surface, and we define:

1. $f_0 : 1 \rightarrow X$ returns the sphere
2. $f_1 : \{1,2,3,\ldots\} \rightarrow X$ maps $g$ to a connected sum of $g$ tori
3. $f_2 : \{1,2,3,\ldots\} \rightarrow X$ maps $k$ to a connected sum of $k$ real projective planes

One can then show that for each $x \in X$, there exists unique $i \in I$ such that for some $p \in \mathrm{dom}(f_i)$, we have $f_i(p) \cong x$. This fact, which amounts to saying that the images of our functions partition $X$, is called the classification theorem of closed surfaces.

This is different to being able to tell whether any two objects specified in any given language are equal/equivalent/isomorphic/whatever. Such problems are usually called "isomorphism problems" or "recognition problems" - consider, for example, the graph isomorphism problem from graph theory, or the recognition problem from knot theory.

So the answer to your questions are as follows:

1. Does "complete classification" of some objects necessarily mean that we can write an algorithm that determines whether two input objects are isomorphic, where the algorithm completes after a finite number of steps?

No. For example, there's a good classification of closed surfaces, but if you specify a surface in a sufficiently rich language, you're not going to be able to work out what standard-form object it's isomorphic to.

2. Suppose I have a finite list of properties such that iff A and B agree on all properties, then A and B are isomorphic. But suppose the "value" of one such property takes on a type of object that is not completely classified. [...] Does this mean I would still not have completely classified manifolds?

Reducing one isomorphism problem to another isomorphism problem does not necessarily solve the former unless the latter has already been solved. It's still important progress though.

3. Suppose I don't have a list of properties that uniquely determine each object (up to isomorphism), but instead have a list of properties that still somehow provide a nice organization scheme. Is this ever considered a classification?

As long as each class in our scheme can be realized as the image of a function that is considered "constructive" and "well-understood", it would be considered a classification.

I'll also remark that a lot of classification theorems aren't recognized as such even though, strictly speaking, they fit the above pattern. For example, the set $X = \{(x,y) : x^2 + y^2 = 1\}$ is classified by the function $f_0 : [0,2\pi) \rightarrow X$ given by $f_0(\theta) = (\cos \theta, \sin \theta)$. Yet few would call this a classification of the points on the unit circle. You sometimes here the word "parametrization" thrown around when only one $f_i$ is involved in the classification.

Answer 2

The notion classification in mathematics is used colloquially but it can be assigned more rigorous meaning, particularly if you change 'classificatoin' to 'classification of something by something'. Very often objects of interest form categories, so there is the category of $\mathbf {Set}$ of sets, $\mathbf{Grp}$ of groups, etc. Very often constructions in mathematics that turn one type of structure into another give rise to functors, so, for instance there is the free group functor $\mathbf {Set}\to \mathbf{Grp}$. Now, consider the category $\mathbf C$ of finite sets where the morphisms are the injective functions. Consider also the category $\mathbb N$ of natural numbers with a morphism $m\to n$ precisely when $m\le n$. The functor $\mathbf C \to \mathbb N$ that sends each finite set to its cardinality is a rigorous way of saying that the cardinality of a set classifies the set. The reason is that two sets are essentially the same, i.e., are isomorphic in the category $\mathbf C$ if, and only if, they have the same cardinality, i.e., the functor sends them to isomorphic (in this case identical) objects in $\mathbb N$. One can thus say that cardinality provides a complete classification of finite sets by means of natural numbers. More accurately, one can say that the functor is the classification. Note that the choice of the morphisms being the injections is crucial.

So, whenever you have a functor $\mathbf C\to \mathbf D$ you can think of it as a classification of things of type $\mathbf C$ by things of type $\mathbf D$. The functor will always map isomorphic things in $\mathbf C$ to isomorphic things in $\mathbf D$. If it also reflects isomorphisms, then you can think of it as a complete classification. Usually, you would like $\mathbf D$ to be a much simpler category than $\mathbf C$. Question of how difficult it is to actually compute the functor are relevant as well of course, as you indicate in your question, so you may impose further computability conditions on the functor.