# Isomorphism isometry

1. Dec 22, 2011

### jfy4

Hi,

I am wondering if all isomorphisms between hilbert spaces are also isometries, that is, norm preserving. In another sense, since all same dimensional hilbert spaces are isomorphic, are they all related by isometries also?

Thank you,

2. Dec 22, 2011

### micromass

The answer is yes. The isomorphic Hilbert spaces are also isometric. This follows from what is called the Riesz-Fisher theorem (but a lot of theorems have that name).

I've tried to find an online reference, but I couldn't find one. In any case, you should check out the book "Fundamentals of the theory of operator algebras" by Kadison and Ringrose. You will want to look at §2.2.

Basically: Hilbert spaces are isometric <=> they are isomorphic <=> they have an orthonormal basis of the same cardinality.

Last edited: Dec 22, 2011
3. Dec 22, 2011

### jfy4

Thanks for the reference and answer.

4. Dec 22, 2011

### Fredrik

Staff Emeritus
It's very easy to prove that a linear bijection $T:\mathcal H\to\mathcal K$ is an isomorphism if and only if it's a isometry. If T is an isomorphism, it satisfies $\langle Tx,Ty\rangle=\langle x,y\rangle$ for all $x,y\in\mathcal H$, and this clearly implies that $\|Tx\|=\|x\|$ for all $x\in\mathcal H$, so T is an isometry. Conversely, if T is an isometry, it satisfies $\langle Tx,Tx\rangle=\langle x,x\rangle$. Consider this equality first with x=y+z and then with x=y-z, where $y,z\in\mathcal H$ are arbitrary. When you write it all out, using linearity and the assumption that T is an isometry, you see that T is an isomorphism.

5. Dec 22, 2011

### micromass

Hmmm. With isomorphism I though he meant a linear homeomorphism. You seem to have a definition that relies on the inproduct...

6. Dec 22, 2011

### Fredrik

Staff Emeritus
Since a Hilbert space is a complete inner product space, shouldn't an "isomorphism" be a vector space isomorphism (i.e. a linear bijection) that preserves the inner product?

I just checked my Conway, and he defines it that way. Page 19.

Last edited: Dec 22, 2011
7. Dec 22, 2011

### micromass

Some sources indeed define isomorphism like that. But other sources mean something entirely different with it. If you read literature on Banach spaces, then "linear homeomorphism" seems to be a very popular word for isomorphism.

But whatever definition you pick, in Hilbert spaces they are all the same thing.

I guess it is up to the OP to clarify what he meant.

8. Dec 22, 2011

### mathwonk

this is of course dependent on your definition of isomorphism. it is obvious that a linear isomorphism, such as x-->2x is not an isometry. but if you define an isometry as a linear isomorphism that preserves dot products, then of course it is trivial, indeed a tautology, that it presrves distances.

9. Dec 22, 2011

### I like Serena

If you look it up, isomorphism has lots of different definitions (wikipedia disambiguates into 9 different definitions that do not even include vector space isomorphism or topological isomorphism).
But I believe they have 1 thing in common.
I think they all say that an isomorphism is a bijection that preserves the properties of whatever it is that you're talking about.

10. Dec 22, 2011

### mathwonk

i agree, so the answers in 2,4 are in sync with this. but i wonder if the OP understood this, since then there is nothing to do really.

11. Dec 22, 2011

### I like Serena

Yes, the definition doesn't say so, but I believe an isometry is a metric space isomorphism.
This would be implied by the more generic Hilbert space isomorphism.

Last edited: Dec 22, 2011
12. Dec 22, 2011

### micromass

No, you need linearity as well. Just asking that $d(f(x),f(y))=d(x,y)$ is not enough.

And also, "metric space isomorphism" is not a standard name. There are various candidates for that title. Although I suspect you mean a bijective function f such that $d(f(x),f(y))=d(x,y)$. But the usual name for that is an isometric isomorphism.

13. Dec 22, 2011

### I like Serena

I know that "metric space isomorphism" is not a standard name.
I just invented it myself, since it seems to me it fits the bill.
It fits since linearity is not defined for a metric space.

For a space that is metric and that is linear, obviously an isomorphism also has to be linear.

14. Dec 23, 2011

### Fredrik

Staff Emeritus
There are? The definition you stated seems like the only one that "preserves the structure". I know I saw somewhere that you can take maps f such that d(f(x),f(y))≤d(x,y) to be the morphisms of a category of metric spaces, so there seem to be several candidates for that job. (An equality sign would of course work just as well as the ≤). But if the choice of morphisms make the isomorphisms anything but "structure preserving", wouldn't that choice be considered very non-standard?

The way I see it, the only use for terms like "isometric isomorphism" is if you're writing a book that covers vector spaces with additional structure, and you insist on using the term "isomorphism" for linear bijections. Then you would could say "isometric isomorphism" when you mean what I would prefer to call a "normed space isomorphism", and "unitary operator" when you mean what I would prefer to call an "inner product space isomorphism". Similarly, you can say "homemorphism" when you mean what could be called a "topological space isomorphism". (Yes, I know that everyone uses the term "homeomorphism" instead of "topological space isomorphism", but I would argue that it makes at least as much sense to use the latter term. The only thing I like better about the former is that it's much shorter).

15. Dec 23, 2011

### micromass

Yes, it all depends on what the maps of the category are. You would like an isomorphism to be a morphism f such that $f$ is also a morphism. In some way, it make sense to make the morphisms all the continuous maps, since these are the maps we work with in metric spaces. But then an isomorphism is an homeomorphism. If we want the isomorphisms to be isometric, then we need to use the samewhat strange maps $d(f(x),f(y))\leq d(x,y)$ (strange in the sense that it is somewhat a limitation to only look at these maps).

It's the same problem with Hilbert spaces. The maps you usually like to work with are the bounded linear maps. But then the isomorphisms become the linear homeomorphisms, which might not be the isomorphisms you want.

So if you want to work with category theory in Hilbert spaces, then you need two categories: one which will give the right morphisms and one which will give the right isomorphisms. It is probably for this reason that categories aren't really popular in Hilbert spaces.

Yes, isometric isomorphism is only really used when there are multiple candidates for isomorphisms, thus when there is some kind of linear structure on the space. But still: I never really heard many people talk about a "metric space" isomorphism or "normed space isomorphism". Don't know why these terms are not so popular.

16. Dec 24, 2011

### jfy4

Its nice that the thread got so many responses, since I was partly thrown a bit by the number of different isomorphisms I was encountering in my research, and trying to settle on what it really meant to be an isomorphism. I have been working with a category of Hilbert spaces whos morphisms are isomorphisms, which was something I was trying to figure out in a two-fold manner. First in category theory, isomorphism are defined to simply be maps that have an inverse that when composed go to the identity of either of the starting objects, but my morphisms that I chose to be the morphisms were also isomorphisms, and I guess I was just confused about exactly what I had on my hands with the category I was talking about in the first place...

The isometry question came from my Hoffman linear algebra book, where it gave a theorem saying that all same dimensional inner product spaces were isometric, which was equivalent to being isomorphic. But they defined the term isomorphic in that theorem as an isometry, and so I wanted to know if it was the same isomorphism I had learned in abstract algebra, a bijective homomorphism.

If it doesn't bother anyone, would it be okay to ask some questions about what possible geometric structures I will have on my hands when considering a category of hilbert spaces connected by isomorphisms, and also possibly considering the underlying morphisms of linear operators on (the objects of) each hilbert space?

Thanks again,

17. Dec 25, 2011

### jfy4

Given that all same dimensional hilbert spaces are isomorphic, and all isomorphisms are norm preserving, it seems to imply a rather curious consequence.

To me this seems to be saying that for physical systems with the same number of degrees of freedom, the descriptions are the same, regardless of the physical system they represent. That is, with the great success of quantum mechanics to both predict and accommodate many phenomena, physics has been reduced to calculating probability amplitudes for hilbert spaces of varying dimensions in one's favorite functional form; since, if need be, one can change to the appropriate functional form through isomorphism (guaranteed to exist). But this says nothing (...or everything...) about the physical situation being analyzed.

Is my take on this sound?

Thanks,

18. Dec 28, 2011

### Hawkeye18

For Banach spaces two types of isomorphism make sense.
Two normed spaces $X$ and $Y$ are called isomorphic if there exists an invertible linear transformation $A:X \to Y$ (both $A$ and $A^{-1}$ are assumed to be bounded). Note that for Banach spaces any bounded bijection is invertible (i.e. $A^{-1}$ is bounded), so one can use the words "bounded bijection" in the definition in the case of Banach spaces.

The spaces are called isometrically isomorphic if one can find and invertible (linear) isometry (i.e. norm preserving map) $U:X \to Y$.

Both definition make sense for Hilbert spaces (because they are a particular case of Banach spaces). However any 2 isomorphic Hilbert spaces are isometrically isomorphic. Namely, if $A:X \to Y$ is an invertible linear transformation, then $U=A (A^*A)^{-1/2}$ is an invertible isometry (unitary transformation) $U:X \to Y$.

Note, that for Banach spaces there are isomorphic spaces that are not isometrically isomorphic. For example, any 2 Banach spaces of the same dimension are isomorphic, and it is easy to see that $\mathbb R^n$ with the Eucledian norm is not isometrically isomorphic to $\mathbb R^n$ with the $\ell^p$ norm for $p\ne 2$ (parallelogram identity fails for $p\ne 2$). In fact, $\mathbb R^n$ with the $\ell^p$ norm is not isometrically isomorphic to $\mathbb R^n$ with the $\ell^q$ norm for $p\ne q$, but that is a bit harder to prove.