On similar concepts in mathematics whose similarity is a non-trivial fact. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T11:16:13Z http://mathoverflow.net/feeds/question/116649 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116649/on-similar-concepts-in-mathematics-whose-similarity-is-a-non-trivial-fact On similar concepts in mathematics whose similarity is a non-trivial fact. Leonard 2012-12-17T21:43:26Z 2012-12-19T20:27:47Z <p>Recently, while undertaking a study of commutative algebra, I learned three concepts: (i) a local ring, (ii) a regular local ring and (iii) a regular ring.</p> <p>At the end, I found myself asking this seemingly naïve question: Are regular local rings the same objects as local rings that are regular? At first, I thought, "My mind must be acting stupid again." However, upon further analysis, it turned out that the answer to my question was non-trivial after all.</p> <p>One direction, namely proving that a local ring that is regular is actually a regular local ring, is not very hard to establish. Indeed, it can be assigned as a homework problem in an undergraduate abstract algebra course. The key observation is that for a local ring $(R,{\frak{m}})$, upon localization at ${\frak{m}}$, we obtain $R_{\frak{m}} = R$. This is because $R \setminus {\frak{m}}$ is precisely the set of units of $R$. Hence, by the definition of <em>regular ring</em>, we see that $(R,{\frak{m}})$ is a regular local ring.</p> <p>The other direction is a well-known (in my opinion, highly) non-trivial result in homological algebra, which states that the localization of a regular local ring at any prime ideal is still a regular local ring. By the definition of <em>regular ring</em> once again, regular local rings are therefore local rings that are regular.</p> <p>I am wondering, are there any pairs of concepts in other areas of mathematics that look so similar that their similarity may be mistaken for tautology but, in reality, can only be established by a hard proof?</p> http://mathoverflow.net/questions/116649/on-similar-concepts-in-mathematics-whose-similarity-is-a-non-trivial-fact/116657#116657 Answer by Alexandre Eremenko for On similar concepts in mathematics whose similarity is a non-trivial fact. Alexandre Eremenko 2012-12-17T22:13:08Z 2012-12-18T14:52:59Z <p>In my example, the similarity did not require a hard proof but it was not seen for many years for the reasons which I would call "social".</p> <p>In 1928 Weil (and simultaneously Siegel) defined and studied heights in algebraic number theory. In 1933 Henry Cartan introduced the Nevanlinna characteristic of a holomorphic curve in projective space. From certain view point these two things are the same:-)</p> <p>Only in 1987, Paul Vojta pursued this analogy quite far. This became famous as "Vojta's analogy", and many new results were proved inspired by this observation. S. Lang, was very much excited and widely popularized this discovery. Since then, Vojta's analogy led to substantial progress in both areas. The story is described in Lang's book Introduction to Complex hyperbolic spaces, on page 185. </p> <p>Why I called the reasons of this almost 60 years gap social? </p> <p>Because on my opinion the reason is that complex analysts and algebraic geometers do not communicate sufficiently with each other.</p> <p>Actually I noticed the analogy in 1982 (I am sure that I was not alone) and I told about my observation to a famous algebraic geometer in. He was not excited. But when the news of the Vojta's analogy reached him few years later, he run into my office, and said: "Alex, can you quickly tell me what's Nevanlinna theory about?"</p> http://mathoverflow.net/questions/116649/on-similar-concepts-in-mathematics-whose-similarity-is-a-non-trivial-fact/116729#116729 Answer by Peter Michor for On similar concepts in mathematics whose similarity is a non-trivial fact. Peter Michor 2012-12-18T19:09:52Z 2012-12-19T19:06:48Z <ol> <li><p>$f:\mathbb R^2\to \mathbb R$ is $C^\infty$.</p></li> <li><p>$f:\mathbb R^2\to \mathbb R$ is $C^\infty$ along each $C^\infty$-curve $c:\mathbb R\to \mathbb R^2$; i.e., $f\circ c$ is $C^\infty$ for each such $c$. </p></li> </ol> <p>Equivalence was proved only in 1979 by Jan Boman. EDIT: It was 1967, sorry for being careless.</p> <p>EDIT: Using "general abstract nonsense" and functional analysis, one push this result from $\mathbb R^2$ to Frechet spaces. Beyond Frechet spaces, the notions start to divergence. Analysis based on (2) is called convenient analysis, since it leads to a diffeomorphism $$C^\infty(U,C^\infty(V,W)) \cong C^\infty(U\times V, W)$$ and a monoidally closed category.</p> <p>See:</p> <p>A.Fr&ouml;licher and A.Kriegl: Linear spaces and differentiation theory. John Wiley &amp; Sons Ltd., Chichester, 1988.</p> <p>Andreas Kriegl, Peter W. Michor: The Convenient Setting of Global Analysis. Mathematical Surveys and Monographs, Volume: 53, American Mathematical Society, Providence, 1997. <a href="http://www.mat.univie.ac.at/~michor/apbookh-ams.pdf" rel="nofollow">(pdf)</a></p> http://mathoverflow.net/questions/116649/on-similar-concepts-in-mathematics-whose-similarity-is-a-non-trivial-fact/116776#116776 Answer by Peter Michor for On similar concepts in mathematics whose similarity is a non-trivial fact. Peter Michor 2012-12-19T10:36:37Z 2012-12-19T19:12:13Z <p>(1) $f:\mathbb R\to \mathbb R$ is $C^\infty$.</p> <p>(2) $f^2:\mathbb R\to \mathbb R$ is $C^\infty$. <BR> EDIT: The right formulation is: If $f\ge 0$ is $C^\infty$, then one can choose a square root of $f$ which is twice differentiable, but not better in general. </p> <p>(3) $f^2$ and $f^3:\mathbb R\to \mathbb R$ are both $C^\infty$.</p> <p>(4) $f^p$ and $f^q:\mathbb R\to \mathbb R$ are both $C^\infty$. EDIT: Where $p$, $q$ are relatively prime. </p> <p>Obviously, (1) $\implies$ (3), (4).</p> <p>(2) See:<BR> Dmitri Alekseevky, Andreas Kriegl, Mark Losik, Peter W. Michor: Choosing roots of polynomials smoothly, Israel J. Math 105 (1998), p. 203-233.<a href="http://www.mat.univie.ac.at/~michor/roots.pdf" rel="nofollow">(pdf)</a></p> <p>(3) $\implies$ (1). See: <BR> MR0682456 Joris, Henri: Une $C^\infty$-application non-immersive qui possède la propriété universelle des immersions. Arch. Math. (Basel) 39 (1982), no. 3, 269–277. <BR> and: <BR> MR2179865 Myers, Robert: An elementary proof of Joris's theorem. Amer. Math. Monthly 112 (2005), no. 9, 829–831. </p> <p>(4) $\implies$ (1). See:<BR> MR0833407 Duncan, John; Krantz, Steven G.; Parks, Harold R.: Nonlinear conditions for differentiability of functions. J. Analyse Math. 45 (1985), 46–68.</p> http://mathoverflow.net/questions/116649/on-similar-concepts-in-mathematics-whose-similarity-is-a-non-trivial-fact/116817#116817 Answer by Yemon Choi for On similar concepts in mathematics whose similarity is a non-trivial fact. Yemon Choi 2012-12-19T20:27:47Z 2012-12-19T20:27:47Z <p>Here is a possible example that I have been speculating about, on and off, since around 2006: it might not fit the OP's question, and I may be talking rubbish, so I'd welcome corrections from the true connoisseurs (as opposed to this dilletant).</p> <p>Call a short exact sequence of Banach spaces and bounded linear maps <em>naively exact</em> if it is exact as a s.e.s. of vector spaces (this is just to get round the fact that category-theoretic cokernels and epimorphisms in Ban don't behave as they do in Vect). Call a Banach space $E$ <em>homologically flat</em> if $E\widehat{\otimes}\underline{\quad}$ preserves the short naively exact sequences of Banach spaces, where the tensor is the projective tensor product a.k.a. the "right one from the POV of closed monoidal categories".</p> <p>Now if I understand the literature correctly, the homologically flat Banach spaces are known: they are precisely the ${\mathcal L}^1$-spaces of Lindenstrauss and Pelczynski, which are roughly speaking those where each fin-dim subspace $E$ is contained in a fin-dim subspace $F$ which is uniformly isomorphic to $\ell_1^{|F|}$.</p> <p>As $\ell_1^n$ is the "free Banach space" on an $n$-element set, this result could be airily if imprecisely captured by saying "the flat Banach spaces are those built locally from finitely generated free Banach spaces". But now this sounds awfully like the Govorov-Lazard theorem from ring theory: <em>a module over a fixed commutative ring $R$ is flat iff it is a colimit of finitely generated free $R$-modules</em>.</p> <p>As far as I know the Banach-space result is proved using duality (duals of flat spaces are injective, and the injective Banach spaces admit a description as ${\mathcal L}^\infty$-spaces) and doesn't seem to have a proof along the lines of the GL-theorem. But one might hope to find a proof that made use of a meaningful analogy between the category of Banach spaces and bounded linear maps, and categories of modules...</p> <p>(Of course the module-categories are abelian, while Ban is not, but there may be some way to exploit the "embedding" of Ban into an abelian envelope, cf. work of No&euml;l, Waelbroeck and in greater generality Schneiders. Calling Theo B&uuml;hler... calling Theo B&uuml;hler...)</p>