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1. • CommentRowNumber1.
   • CommentAuthorAndrew Stacey
   • CommentTimeMay 20th 2010
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   Author: Andrew Stacey
   Format: MarkdownItexLooking at the entry [[Banach spaces]], I found the following in the introduction: > So every $n$-dimensional real Banach space may be described (up to linear isometry, the usual sort of [[isomorphism]]) as the [[Cartesian space]] $\mathbb{R}^n$ equipped with the $p$-norm for $1 \leq p \leq \infty$ which seems to imply that every norm on a finite dimensional Banach space is a $p$-norm for some $p$. That feels to me like a load of dingo's kidneys. To define a norm on some $\mathbb{R}^n$ I just need a nice convex set, and there's lots more of these than the balls of $p$-norms, surely. Am I missing something?

   Looking at the entry Banach spaces, I found the following in the introduction:

   So every $n$-dimensional real Banach space may be described (up to linear isometry, the usual sort of isomorphism) as the Cartesian space $\mathbb{R}^n$ equipped with the $p$-norm for $1 \leq p \leq \infty$

   which seems to imply that every norm on a finite dimensional Banach space is a $p$-norm for some $p$. That feels to me like a load of dingo’s kidneys. To define a norm on some $\mathbb{R}^n$ I just need a nice convex set, and there’s lots more of these than the balls of $p$-norms, surely.

   Am I missing something?

2. • CommentRowNumber2.
   • CommentAuthorTim_van_Beek
   • CommentTimeMay 20th 2010
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   Author: Tim_van_Beek
   Format: MarkdownItexHunch: The difference may lie in the understanding of "isomorphism": All norms on $\mathbb{R}^n$ are equivalent, and thus induce the same topology, but they are not isometric isomorph.

   Hunch: The difference may lie in the understanding of “isomorphism”: All norms on $\mathbb{R}^n$ are equivalent, and thus induce the same topology, but they are not isometric isomorph.

3. • CommentRowNumber3.
   • CommentAuthorAndrew Stacey
   • CommentTimeMay 20th 2010
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   Author: Andrew Stacey
   Format: MarkdownItexThe quote from the page has the word "isometry" in it, so I assumed that it meant "isometric isomorphism". Up to just isomorphism, there's no need to list all the norms for different $p$s, as well. So either the list is overspecified, since just one would do, or is underspecified, since there are probably uncountably many suitable convex sets.

   The quote from the page has the word “isometry” in it, so I assumed that it meant “isometric isomorphism”. Up to just isomorphism, there’s no need to list all the norms for different $p$s, as well. So either the list is overspecified, since just one would do, or is underspecified, since there are probably uncountably many suitable convex sets.

4. • CommentRowNumber4.
   • CommentAuthorTodd_Trimble
   • CommentTimeMay 20th 2010
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   Author: Todd_Trimble
   Format: MarkdownItexGood catch, Andrew. Definitely wrong. (Gee, hope it wasn't me who wrote that steaming pile!) Continuous linear isomorphism is obviously what should have been said for that statement.

   Good catch, Andrew. Definitely wrong. (Gee, hope it wasn’t me who wrote that steaming pile!) Continuous linear isomorphism is obviously what should have been said for that statement.

5. • CommentRowNumber5.
   • CommentAuthorAndrew Stacey
   • CommentTimeMay 20th 2010
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   Author: Andrew Stacey
   Format: MarkdownItexGiven that this was in the "Idea" section, I tried to keep it fairly simple. This is particularly tricky since the question of what is an isomorphism of Banach spaces is an important issue later on!

   Given that this was in the “Idea” section, I tried to keep it fairly simple. This is particularly tricky since the question of what is an isomorphism of Banach spaces is an important issue later on!

6. • CommentRowNumber6.
   • CommentAuthorTodd_Trimble
   • CommentTimeMay 20th 2010
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   Author: Todd_Trimble
   Format: MarkdownItexWhile we're on [[Banach space]], I've long had a niggling doubt about something I wrote under the section Categorical Operations. It was about tensor products, where we want a left adjoint $- \otimes X$ to the standard internal hom functor $X \Rightarrow -$. I had invoked the uniform boundedness principle, remembering the fact that functions $X \times Y \to Z$ that are separately continuous and linear in each of the arguments are in fact jointly continuous. That invocation was, to be honest, a case of hedging bets, and I don't think that fact and the uniform boundedness principle (which ultimately depend on Dependent Choice) need enter the discussion at all to construct the tensor product -- the tensor product should be completely straightforward without it. Is that right?

   While we’re on Banach space, I’ve long had a niggling doubt about something I wrote under the section Categorical Operations. It was about tensor products, where we want a left adjoint $- \otimes X$ to the standard internal hom functor $X \Rightarrow -$. I had invoked the uniform boundedness principle, remembering the fact that functions $X \times Y \to Z$ that are separately continuous and linear in each of the arguments are in fact jointly continuous. That invocation was, to be honest, a case of hedging bets, and I don’t think that fact and the uniform boundedness principle (which ultimately depend on Dependent Choice) need enter the discussion at all to construct the tensor product – the tensor product should be completely straightforward without it. Is that right?

7. • CommentRowNumber7.
   • CommentAuthorAndrew Stacey
   • CommentTimeMay 20th 2010
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   Author: Andrew Stacey
   Format: MarkdownItexI think that you're right. In so far as the construction of the tensor product is concerned, the formula that you wrote down works just fine. Where one might need uniform boundedness is in proving that this is adjoint to the internal hom functor - I'd need to think about that. I'm not convinced that it's necessary, though, since one can do this construction for arbitrary LCTVS so invoking uniform boundedness seems a little over the top. (We really ought to transfer our discussion on monoidal structures in LCTVS to the lab from where it's mouldering in Cafe comments)

   I think that you’re right. In so far as the construction of the tensor product is concerned, the formula that you wrote down works just fine. Where one might need uniform boundedness is in proving that this is adjoint to the internal hom functor - I’d need to think about that. I’m not convinced that it’s necessary, though, since one can do this construction for arbitrary LCTVS so invoking uniform boundedness seems a little over the top.

   (We really ought to transfer our discussion on monoidal structures in LCTVS to the lab from where it’s mouldering in Cafe comments)

8. • CommentRowNumber8.
   • CommentAuthorTodd_Trimble
   • CommentTimeMay 20th 2010
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   Author: Todd_Trimble
   Format: MarkdownItexYes, you're right -- I didn't mean *just* construction, I meant adjointness too, that none of this should depend on the UBP. In fact, I think it's basically obvious that the natural isomorphism $\hom(X \otimes Y, Z) \cong \hom(X, Y \Rightarrow Z)$ will be an isometry (yes, an isometry in this case). But I may not have much time to write this in today.

   Yes, you’re right – I didn’t mean just construction, I meant adjointness too, that none of this should depend on the UBP. In fact, I think it’s basically obvious that the natural isomorphism $\hom(X \otimes Y, Z) \cong \hom(X, Y \Rightarrow Z)$ will be an isometry (yes, an isometry in this case). But I may not have much time to write this in today.

9. • CommentRowNumber9.
   • CommentAuthorTim_van_Beek
   • CommentTimeMay 20th 2010
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   Author: Tim_van_Beek
   Format: MarkdownItex<blockquote> <p> The quote from the page has the word "isometry" in it, so I assumed that it meant "isometric isomorphism". </p> </blockquote> Yes, sorry, I completly ignored that, which is a good example of <a href="http://en.wikipedia.org/wiki/Cognitive_dissonance">cognitive dissonance</a>.

   The quote from the page has the word “isometry” in it, so I assumed that it meant “isometric isomorphism”.

   Yes, sorry, I completly ignored that, which is a good example of cognitive dissonance.

10. • CommentRowNumber10.
    • CommentAuthorHarry Gindi
    • CommentTimeMay 20th 2010
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    Author: Harry Gindi
    Format: MarkdownHmm, Andrew, perhaps a proper isomorphism of Banach spaces is a uniform linear isomorphism?

    Hmm, Andrew, perhaps a proper isomorphism of Banach spaces is a uniform linear isomorphism?

11. • CommentRowNumber11.
    • CommentAuthorAndrew Stacey
    • CommentTimeMay 20th 2010
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    Author: Andrew Stacey
    Format: MarkdownItexIsn't every continuous linear isomorphism also uniform? The functor from TVS to Top factors through uniform spaces, so that should be correct.

    Isn’t every continuous linear isomorphism also uniform? The functor from TVS to Top factors through uniform spaces, so that should be correct.

12. • CommentRowNumber12.
    • CommentAuthorTodd_Trimble
    • CommentTimeMay 20th 2010
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    Author: Todd_Trimble
    Format: MarkdownItexIf I recall correctly, the following holds true for topological groups $G$, $H$ (with their standard uniform structures): a group homomorphism $f: G \to H$ is uniformly continuous iff it is continuous iff it is continuous at a single point of $G$. Assuming that's correct, I could Labbify this (later). Wouldn't this be in Bourbaki?

    If I recall correctly, the following holds true for topological groups $G$, $H$ (with their standard uniform structures): a group homomorphism $f: G \to H$ is uniformly continuous iff it is continuous iff it is continuous at a single point of $G$. Assuming that’s correct, I could Labbify this (later).

    Wouldn’t this be in Bourbaki?