Is there a relationship between isometry as defined on metric spaces and those on vector spaces?

Question

I am taking a course on linear algebra and another on real analysis.

• In linear algebra we defined that two vector spaces are isomorphic if there existed a

  bijective and linear map

  between the two vector spaces

• In real analyis we defined that two metric spaces are isometric if there existed a

  bijective, distance preserving map $d(x,y) = d'(fx, fy)$ between the two metric spaces

I looked up the definition of isometry online and many sources tell me it is a bijective "structure preserving" map.

Is there some commonality between these so called structures?

Or is it hopeless for me to guess what would be an isomorphism defined between topological spaces, Hilbert spaces or Banach spaces until I see the definitions?

Answer 1

Given two structures of $A,B$ of some kind, you first have to guess what a structure-preserving 'map', i.e. a (homo-) morphism $A \to B$ (from $A$ to $B$) is. For the sake of simplicity, we are going to assume, that a morphism is a function with some special properties (this need not be the case, but it very often is). If you have this, then $A$ is commonly called isomorphic to $B$, if there is an isomorphism $A\to B$.

An isomorphism $f : A\to B$ is a morphism $A\to B$ with an inverse, that is a morphism $g : B\to A$ with $g\circ f = \operatorname{id}_A$ and $f\circ g = \operatorname{id}_B$. Of course, in our case we then have $g = f^{-1}$. So, being an isomorphism implies being bijective. The converse is not generally true, because even if $f$ is a bijective morphism, $f^{-1}$ is not necessarily a morphism too (this is indeed the case for topological spaces and their morphisms), but it is true for example for linear maps.

How do you decide what a morphism should be, i.e. what kind of special properties a map between structures should have? Well, it depends. Sometimes its obvious (vector spaces), sometimes it is not (metric spaces). But in any case, the morphisms should preserve "all" the structure. So vector space morphisms (linear maps) should preserve addition and scalar multiplication and a metric space morphism should preserve the metric.

Unfortunately or maybe interestingly, there are many things you could consider "preserving the metric". One kind of possible morphisms are simply called "distance-preserving", but there are other possibilities (short maps, lipschitz-continuous maps, etc.). Because in this case, we have many notions of morphisms, someone decided to use the term "isometric" in place of "isomorphic".

The general theory dealing with "morphisms between structures" is called "category theory".

Answer 2

As celtschk mentioned, I think you are looking for the general concept of "isomorphism", which is a commonality shared by linear isomorphisms for vector spaces and isometries for metric spaces. Intuitively, an isomorphic things should have equivalent relevant structure, whether that's linear structure, or a metric, or whatever else.

The most general idea of isomorphism comes from Category Theory. Basically, a morphism is a relevant function or function-like thing, in that it plays nicely with the structure. For example, a vector space morphism is a linear transformation, because it plays nicely with addition and scalar multiplication. An isomorphism is a morphism with an inverse morphism. In some contexts, the definition of "isomorphism" is equivalent to "a bijective morphism", but this doesn't always work.

Isometry does not really make sense outside of the context of metric spaces, so extending to topological spaces does not make sense. In the context of general topological spaces, the morphisms are the continuous functions, and the isomorphisms are homeomorphisms.

Since metric spaces are topological spaces, you might ask whether homeomorphic (topologically "the same") metric spaces are isometric. Unfortunately, that's often not the case (see, for example, this question). As a consequence, when dealing with Banach spaces, there are different concepts of isomorphism that may be relevant: a linear isomorphism that's also a homeomorphism is good, but a linear isomorphism that's an isometry is better because it "preserves the metric/norm structure" as well. These two types of isomorphism are contrasted at the wikipedia page for Banach space.

Answer 3

Normed spaces (including Banach spaces) are in particular metric spaces: the metric is given by $d(x,y) = |x - y|$. (This should agree with your intuition for $\mathbb{R}^3$). The notion of isometry is then the same as what you would guess.

But an isomorphism doesn't have to be an isometry: for example, scaling by 2 is an isomorphism (of a real vector space), but it is not an isometry.

Generally one uses the expression "isometric isomorphism" to describe structure preserving maps between normed vector spaces. For example: Every infinite dimensional separable Hilbert space is isometrically isomorphic to the space of square summable sequences. (The space of square summable sequences is the space of sequences of real numbers $(a_i)$ with $\Sigma a_i^2 < \infty$. The norm is $\Sigma a_i^2$. Actually it's a Hilbert space, so it has an inner product. Can you see it? Exercise: this space is complete.)

Note that an isometric map is necessarily injective, so if the vector space is finite dimensional, then a linear isometry is also an isomorphism (of vector spaces). But in the infinite dimensional case this is not true: for example, consider the "shift" map on the (Hilbert) space of square summable sequences, sending the sequence $(a_0, \ldots)$ to $(0,a_0, \ldots)$.

In general a topological space does not have a metric.

The general notion of a structure preserving map varies depending on the objects involved, but the general idea is: if $f : X \to Y$ is BLAH-structure preserving, then any proof of a fact having to do with X's BLAH-structure can be translated into a proof of that same fact about $Y$'s BLAH-structure via $f$.

For example: If $V$ and $W$ are linear spaces, an $f$ is an isomorphism of vector spaces ...

1. Any basis in $V$ $v_i$ becomes a basis in $W$ under $f$. The proof that $v_i$ is a basis has two parts, linear independence and spanning. In addition to the linearity of $f$, one uses the surjectivity of $f$ to show that $f(v_i)$ still spans, and the injectivity to show that $f(v_i)$ are still independent. (Though there are other ways to do this.)

2. Then the space of linear functions $W \to \mathbb{R}$ can be (isomorphically) identified with the space of linear functions $V \to \mathbb{R}$ via $f$. How? (Exercise.)

Etc.

If $V$ has more structure, maybe a norm, then a linear isomorphism $f$ may not preserve statements about it. One has to go to infinite dimensions for this to be interesting, since all norms on $\mathbb{R}^n$ are equivalent (good exercise. Possible hint: Think about the unit ball, and the relationship between the unit ball and norms via the so called Minkowski functional.)

As an example: Any two vector spaces with the basis of the same cardinality are isomorphic. So take $V$ to be the Banach space (norm is sup norm) of continuous functions on $[0,1]$, called $C[0,1])$, and $W$ to be the normed space $V$ of $C^1$ functions (continuously differentiable) on $[0,1]$, again with the sup norm.

These spaces are isomorphic, having basis with the same cardinality (you can use Taylors theorem to write $V = \mathbb{R} \oplus C[0,1]$: $f = f(0) + \int_0^x f'(t) dt$, so the map sends $f$ to the pair $(f(0), f')$, that it is bijective and linear is a combination of some standard calculus theorems: every continuous function $g(x)$ is the derivative of the function $f(x) = \int_0^x g(t) dt$, the derivative of a constant is zero, the integral of zero is zero, integration and differentiation are linear, etc.)*, but not isometric because the former is complete as a metric space and the latter is not.

*Alternatively, you can use the functionals $e_a(f) = f(a)$, for $a \in \mathbb{Q}$ to embed each in the space $\mathbb{R}^{\mathbb{Q}}$. This gives an upper bound for the cardinality of their basis, which is $P(\mathbb{N})$. Then you need to show that neither has countable basis: since $V \subset C[0,1]$, you just need to show that $V$ cannot have a countable basis. So you need to build an uncountable linearly independent set: if you subdivide the interval into halves (and quarters, and eights, etc.), and use bump functions, you can build this uncountable family. (I am assuming that there is no cardinality between $|\mathbb{N}|$ and $|P(\mathbb{N})|$... I think this is right but I don't really know set theory too well. Am I assuming the continuum hypothesis?)

(Question: You can put a norm on the space of $C^k$ functions on the interval so that it becomes a Banach space - the $C^k$ norm, which is the sum of the sup norms of the first $k$ derivatives. Are these Banach spaces isometrically isomorphic, for various $k$, or maybe they are all distinct? $C^k([0,1]) \cong \mathbb{R}^k \oplus C^0([0,1])$ - as vector spaces, using Taylors theroem, but this is not an isometry I think. However, I am reasonably sure that it is a homeomorphism.)

It may be enlightening to look up the definition of an isomorphism in a category, and also the Yoneda lemma, but again maybe not. Possibly better to think about different objects that have many layers of structure, and what kind of things are preserved by different maps.