# What does it mean for a space to be naturally identified with another?

1. Jan 30, 2012

### daveyp225

What does it mean for a space to be "naturally identified" with another?

Does it mean there is a bijection? Or maybe an isomorphism?

2. Jan 30, 2012

### micromass

Staff Emeritus
Re: What does it mean for a space to be "naturally identified" with another?

It means that there exists an isomorphism between the two spaces.

However, there are many kinds of isomorphisms: isomorphisms of groups, rings, metric spaces, topological spaces, etc.

So what it means actually depends on the context. For example, we can say that $A\times B$ is naturally isomorphic to $B\times A$.
If you're reading this in a set theory book, then it means that there exists a bijection between the two sets.
If you're reading this in a group theory book, then it means that there exists a group isomorphism between the two.
If you're reading this in a topology book, then it means that there is a homeomorphism between the two.
And so on.

3. Jan 31, 2012

### daveyp225

Re: What does it mean for a space to be "naturally identified" with another?

It is a question in a linear algebra book which states that linear functionals on a subspace of F^n (the subspace which consists of all vectors whose components sum to 0) can be identified with linear functionals whose coefficients sum to 0. I understand the "identification" part, it is obvious, but I wasn't sure what that implied about the set of linear functionals it is "identified" with. So to show that a "natural identification" in fact exists, one should show an isomorphism exists?

4. Jan 31, 2012

### Jamma

Re: What does it mean for a space to be "naturally identified" with another?

In this context it will mean "there exists an isomorphism of vector spaces between the two".

A way to think of isomorphisms (in categories that aren't too abstract) is that the two objects in question that the isomorphism links are "the same" up to a relabeling of the elements. The structure of the vector space is the same if you identify the elements of each using the isomorphism in question.

5. Jan 31, 2012

### SteveL27

Re: What does it mean for a space to be "naturally identified" with another?

I think there's an element of psychology in what we mean by a natural identification. It's not just an isomorphism.

For example, the natural numbers are naturally identified with the subset {0, 1, 2, 3, ...} of the real numbers. There's also a ring isomorphism between the naturals and the set of reals {0, 2, 4, 6, ...} but I don't think anyone would call that "natural." If you asked 100 mathematicians to tell you the natural embedding (or natural identification) of the naturals in the reals, they'd all give the first map but not the second.

I think that in this context the word natural is not purely a technical term. You know it when you see it, but there may not be a formal definition. I could be wrong.

(edit) I thought of another good example. Consider the two vector spaces:

* R^2, the set of ordered pairs (a,b) of real numbers with component-wise addition and scalar multiplication.

* The set of 1-degree polynomials with real coefficients; that is, polynomials of the form a + bx.

These are both 2-dimensional vector spaces so they are isomorphic. Can we think of a particular isomorphism? Well, how about mapping (1,0) to 5 + 3x and mapping (0,1) to 11 - 2x. Since the two target vectors are linearly independent, we have defined an isomorphism between the two vector spaces. (Verification by the devoted reader of course).

But is this a natural identification? I'd argue not. I would say that there is only one natural isomorphism between the two vector spaces, namely the one that sends (a,b) to a + bx.

So a natural identification between structures is

a) An isomorphism;

b) Is the isomorphism that everyone would instantly name if asked to give an obvious isomorphism;

c) And any other isomorphism seems forced or contrived.

As you can see, conditions (b) and (c) are psychological in nature. There are infinitely many isomorphisms between the two vector spaces, but only one is the obvious one. Only one is natural, or perhaps canonical.

Anyone agree? Disagree? Is there perhaps some clever category-theoretic way to define some isomorphisms as natural or canonical and others not?

Last edited: Jan 31, 2012
6. Feb 1, 2012

### Jamma

Re: What does it mean for a space to be "naturally identified" with another?

Actually, I don't totally agree with you here - often "natural" does have a very precise mathematical meaning (although, admittedly, not always and sometimes natural will be used in exactly the way you described above). I think that indeed in the OPs example, the isomorphism in question may turn out to be naturally isomorphic to the identity functor [edit: actually, rereading, I don't understand what we are trying to naturally identify].

A precise version of the word "natural" often turns up when dealing with functors, see here, for example:
http://en.wikipedia.org/wiki/Natural_transformation

Last edited: Feb 1, 2012
7. Feb 1, 2012

### Office_Shredder

Staff Emeritus
Re: What does it mean for a space to be "naturally identified" with another?

In the world of linear algebra, naturally identifying two spaces usually means giving an isomorphism that does not depend on a choice of basis.

For example, a vector space over R of dimension 2, and the set of all points of the form acos(x)+bsin(x) are isomorphic as vector spaces. One such map is to pick a basis u,v of our vector space, and send au+bv to acos(x)+bsin(x). But this isomorphism depends on how we chose u and v.

On the other hand, consider the isomorphism between V and V** (its double dual space) that sends a vector v to fv where fv(h) = h(v) for h in V*. This is an isomorphism (assuming V is finite dimensional) that does NOT depend on how you choose a basis, so you would describe it as a natural isomorphism

8. Feb 1, 2012

### Jamma

Re: What does it mean for a space to be "naturally identified" with another?

Yes, this makes sense. It also overlaps nicely with what I was saying in my previous comment (in fact, it talks about this in the wiki article I linked:)