# Isometry and isomorphism

1. Feb 3, 2016

### ericm1234

I have read a definition of isomorphism as bijective isometry. I was also showed a definition that isomorphism is a bijective map where both the map and its inverse are bounded (perhaps only for normed spaces??). This does not seem to be the same thing as an isometry.
For example, the poisson problem, from H^1_0 to H^-1 (dual space of H^1_0), is bijective by Lax Milgram, and I can show both maps (the original and the inverse) are bounded. But showing an isometry doesn't seem possible.

2. Feb 3, 2016

### andrewkirk

Isomorphism is an algebraic concept (bijection that preserves algebraic structure), whereas isometry is a concept that applies to metric spaces (bijection that preserves distances). For both concepts to be applicable, we'd need to be talking about a metric space that also has an algebraic structure, like with a Lie Group.
What is the context of your question? What sort of spaces are you asking about?

3. Feb 3, 2016

### Staff: Mentor

"iso" in general indicates something of equal value or structure. Such as isotopes are of the same atom, i.e. have the same number of protons.

You can see what an isometry is by taking it literally: A mapping $f : U → V$ between two normed spaces that respects the norm, the metric: $||f(u)||_V = ||u||_U$.

The same holds for isomorphisms. Only that the morphism part of the word is far more general. It basically means "structure", e.g. sets, groups, vector spaces. Bijections in a categorical sense are isomorphisms which means they translate groups into groups, vector spaces into vector spaces and so on. The iso part of the word here means "basically the same". So isomorphisms transport the structure 1:1.

4. Feb 3, 2016

### micromass

Staff Emeritus
Yes, that is correct. You indeed have two different categories with two different notions of isomorphisms. For a very neat book exploring analysis from the explicit point of view of the two categories, see https://www.amazon.com/Exercises-Functional-Translations-Mathematical-Monographs/dp/0821840983

Last edited by a moderator: May 7, 2017
5. Feb 3, 2016

### micromass

Staff Emeritus
Not necessarily true.

6. Feb 3, 2016

### ericm1234

The context of my question would be taking place in the Hilbert spaces for the example I gave, if someone can comment more explicitly on that;

it'd be great if someone can walk through why the "Poisson map", from H^1_0 to dual space of H^1_0, is an isomorphism.

7. Feb 3, 2016

### mathwonk

an morphism is a map thar preserves whatever structure is present. an isomorphism is a morphism that has an inverse map that is also a morphism. for eample in the theory of banach spaces a morphism is a linear map that is bounded and an isomorphism is a bounded linear map with a bounded linear inverse, which is equivalent to a linear map that is bounded both above and below.

8. Feb 4, 2016

### ericm1234

Thank you for that explanation; How come the definition I have from a functional analysis textbook states it not in terms of bounded maps, but as an isometry?

9. Feb 4, 2016

### Fredrik

Staff Emeritus
An isomorphism is a bijective homomorhpism. The term "homomorphism" is defined differently for different types of structures (groups, vector spaces, etc). A vector space homomorphism is just a linear map. A normed space homomorphism is a vector space homomorphism that also preserves the norm. An isometry is a map that preserves distances.

The two calculations below show that a linear map is an isometry if and only if it preserves the norm. So a normed space isomorphism can be defined as a bijective linear isometry.
\begin{align*}
&d(f(x),f(y))=\|f(x)-f(y)\|=\|f(x-y)\|=\|x-y\|=d(x,y),\\
&\|f(x)\|=\|f(x)-0\|=\|f(x)-f(0)\| =d(f(x),f(0))= d(x,0)=\|x-0\|=\|x\|.
\end{align*}

Edit: Hm, I see that my post contradicts mathwonk's. That's usually means that I did something wrong. Maybe my definition of "homomorphism" in the context of normed vector spaces is wrong (but still leads to the correct notion of "isomorphism"). I need to think for a minute.

OK, I have thought about it. I don't have access to books on functional analysis at my current location, so I can't check the book definitions. I did some googling, but didn't find a definition of "homomorphism" in the context of normed spaces in the time I was willing to spend on it. I did however find something that reminded me that what I like to call an isomorphism in the context (or category) of normed spaces, is called an "isometric isomorphism" by most sources.

So I think my terminology in this post is a bit non-standard. I think that my definition of homomorphism in this post makes the most sense if we think of them as structure-preserving maps. But a map doesn't have to preserve anything to be a "morphism" in the sense of category theory.

I think a lot of people simply prefer to define homomorphisms (in the context of normed spaces) as bounded linear maps instead of as norm-preserving or (equivalently) distance-preserving linear maps, because it leads to the same kind of isomorphisms...if we define an "isomorphism" in the following way: A homomorphism $f:X\to Y$ is said to be an isomorphism if there's a homomorphism $g:Y\to X$ such that $f\circ g$ is the identity map on Y, and $g\circ f$ is the identity map on X.

Last edited: Feb 4, 2016
10. Feb 4, 2016

### micromass

Staff Emeritus
Hi Fredrik, this is not necessarily true. It happens to be true in the case of groups and vector spaces and most other algebraic structures. But for example, as you know, it is false in the topological case.

Your post does not contradict mathwonk's post. The "contradiction" is because of what I said already: there are at least two interesting categories of normed spaces. One of this gives as isomorphisms the linear bijective isometries (called isometric isomorphism or linear isometries), the other gives the linear bijective bounded maps whose inverse is bounded too (called topological isomorphism or linear homeomorphism). Both are a very interesting choice of isomorphism.

As for what a "morphism" is. Again, we have two interesting cases. The first case is that the morphisms are all the linear bounded maps, the other interesting case is that the morphisms are all the contractions. This leads to two categories which are interesting to study.