What are we allowed to do when we know that two metric spaces are isometrically isomorphic?

Question

What motivated the question

Let $E$ and $F$ be two normed vector spaces and let $f: \Omega \subset E \to F$, where $\Omega$ is open.

We denote by $\mathcal L(E,F)$ the set $\{L: E \to F, L$ is linear and continuous$\}$.

• Definition: $f$ is said to be Fréchet differentiable (henceforth abbreviated "F.D") at a point $x_0 \in \Omega$ if there exists $L \in \mathcal L(E,F)$, such that:

$$f(x_0 + h) - f(x_0) - L(h) = o (h)$$

In such a case $L$ is called the differential of $f$ at $x_0$ and denoted by $Df(x_0)$.

$f$ is said to be F.D on $\Omega$ if it's F.D at each $x \in \Omega$; that's, at each $x \in \Omega$, there is such a $Df(x) \in \mathcal L(E,F)$. In such a case, we define the differential map of $f$:

$Df: \Omega \subset E \to \mathcal L(E,F)$, $x \mapsto Df(x)$.

• Definition: $f$ is said to be twice F.D at $x_0 \in \Omega$ if $f$ is F.D in a neighbourhood $V$ of $x_0$, and the differential map $Df$ is F.D at $x_0$.

The $2$nd derivative of $f$ at $x_0$ is defined to be the differential at the point $x_0$ of the differential map of $f$. That's,

$$\text{The second derivative of $f$ at $x_0$} = D(Df)(x_0)$$

It's denoted by $D^2f(x_0)$.

So,

$$\large D^2f(x_0) \in \mathcal L(E, \mathcal L(E,F))$$

Or,

$D^2 f(x_0): E \to \mathcal L(E,F)$, $h \to [D^2f(x_0)](h): E \to F$, $k \mapsto ([D^2 f(x_0)](h))(k)$

Now, in a lecture, our professor stated that since $\mathcal L(E,\mathcal L(E,F)) \cong \mathcal L(E,E; F)$ isometrically, where the latter denotes the vector space of bilinear and continuous maps from $E \times E$ into $F$, we can identify the two and deal with $D^2f(x_0)$ as a bilinear map, i.e. that we will consider it as follows:

$D^2f(x_0): E \times E \to F$, $(h,k) \to D^2f(x_0)(h,k)$.

This confuses me; I know that there is mathematical rigour behind this consideration and what would follow from that, however, I was not exposed to that rigour.

What makes me quite uncomfortable is that the nature of $D^2 f(x_0)$ was completely changed. Rather than being a map of one argument, producing a linear map, it is now a mapping of two arguments, producing a vector.

Question:

In the example I considered, the setting was taking place in normed vector spaces, but I suppose that this is not special to such spaces because we can talk about isometric isomorphisms in metric spaces in general. So, here's the question:

What are we allowed to do when we know that two metric spaces are isometrically isomorphic? What does this tell us on an intuitive level? More importantly, what does this tell us on a rigorous level? Are there any good online references in which I can read about that?

Answer 1

Isometric isomorphisms are the correct notion of isomorphism between metric spaces. (It's a more subtle question what the correct notion of morphism between metric spaces is.) In the same way that two groups being isomorphic means that any "genuinely group-theoretic" question about them has the same answer, two metric spaces being isomorphic means that any "genuinely metric" question about them has the same answer. Here are some examples of "genuinely metric" questions:

• Is the metric space bounded?
• Does it contain two points which are a distance $r$ apart?
• Does some other metric space isometrically embed into it?
• Is the metric intrinsic?

That's an answer to your question as stated, but I think it doesn't address your motivation. There's a closely related but easier thing that happens already at the level of vector spaces: if $U, V, W$ are three vector spaces, $[U, V]$ denotes the vector space of linear maps $U \to V$, and $U \otimes V$ denotes the tensor product of vector spaces, then there's a natural isomorphism

$$[U, [V, W]] \cong [U \otimes V, W]$$

(the tensor-hom adjunction) which says that the following two kinds of gadgets are naturally (as a functor in $U, V$, and $W$) the same:

• A linear map from $U$ into the space of linear maps $V \to W$, and
• A bilinear map from $U \times V$ to $W$.

I'm stressing the naturality here because it's important in applications. This is essentially an example of currying, which is a very important and basic pattern in mathematics. It's good to get used to as soon as you can. A more fundamental example at the level of sets is that if $X, Y, Z$ are three sets, then the following two kinds of gadgets are naturally (as a functor in $X, Y, Z$) the same:

• A function from $X$ into the set of functions $Y \to Z$, and
• A function $X \times Y \to Z$.

The correspondence sends a function $f(-, -) : X \times Y \to Z$ to the function $f : x \mapsto (y \mapsto f(x, y))$.