User theo buehler - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T17:27:15Z http://mathoverflow.net/feeds/user/11081 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47747/freyd-mitchells-embedding-theorem/47762#47762 Answer by Theo Buehler for Freyd-Mitchell's embedding theorem Theo Buehler 2010-11-30T07:05:38Z 2012-12-26T16:43:09Z <p>$\DeclareMathOperator{\Hom}{Hom}\newcommand{\amod}{\mathscr{A}\text{-}{\bf Mod}}\newcommand{\scrA}{\mathscr{A}}\newcommand{\scrE}{\mathscr{E}}\newcommand{\Ab}{\mathbf{Ab}}\DeclareMathOperator{\Lex}{\mathbf{Lex}}\DeclareMathOperator{\coker}{Coker}$I only know one proof of the embedding theorem&mdash;the expositions differ heavily in terminology but the approaches all are equivalent, as far as I can tell. I think the proof in Swan's book on K-theory makes the relation between the Freyd-Mitchell approach and Gabriel's approach pretty clear. Let me just say that there is no cheap way of getting the Freyd-Mitchell embedding theorem since there is a considerable amount of work you need to invest in order to get all the details straight. On the other hand, if you manage to feel comfortable with the details, you will have learned quite a good amount of standard tools of homological algebra, so I think it's well worth the effort.</p> <p>Think of the category of functors $\scrA \to \Ab$ as $\scrA$-modules, that's why the notation $\amod$ is fairly common. The Yoneda embedding $A \mapsto \Hom(A,{-})$ even yields a fully faithful contravariant functor $y\colon \scrA\to\amod$.</p> <p>The category $\amod$ inherits a bunch of nice properties of the category $\Ab$ of abelian groups:</p> <ul> <li>It is abelian.</li> <li>It is complete and cocomplete ((co-)limits can be computed pointwise on objects)</li> <li>The functors $\Hom(A,{-})$ are injective and $\prod_{A \in\scrA} \Hom{(A,{-})}$ is an injective cogenerator.</li> <li>Since there is an injective cogenerator, the category $\amod$ is well-powered.</li> <li>etc.</li> </ul> <p>However, the Yoneda embedding is <em>not</em> exact: If $0 \to A' \to A \to A'' \to 0$ is a short exact sequence, we only have an exact sequence</p> <p>$$0 \to \Hom(A',{-}) \to \Hom(A,{-}) \to \Hom(A'',{-})$$</p> <p>in $\amod$.</p> <p>It turns out that the functor $Q = \coker{(\Hom(A,{-}) \to \Hom(A'',{-}))}$ is &ldquo;weakly effaceable&rdquo;, so we would want it to be zero in order to get an exact functor. How can we achieve this? Well, just <em>force</em> them to be zero: say a morphism $f\colon F \to G$ in $\amod$ is an isomorphism if both its kernel and its cokernel are weakly effaceable. If this works then a weakly effaceable functor $E$ is isomorphic to zero because $E \to 0$ has $E$ as kernel. Now the full subcategory $\scrE$ of weakly effaceable functors is a Serre subcategory, so we may form the Gabriel quotient $\amod/\scrE$. By its construction, isomorphisms in the Gabriel quotient have precisely the description above.</p> <p>On the other hand, the category $\Lex(\scrA,\Ab)$ of left exact functors $\scrA \to \Ab$ is abelian. This is far from obvious when you start from the definitions. However, $\Lex(\scrA,\Ab)$ sits comfortably inside the abelian category $\amod$. The inclusion has an <em>exact</em> left adjoint (!) (= &ldquo;sheafification&rdquo;), so again ${\bf Lex}({\scr A}, \Ab)$ inherits many useful properties from $\amod$. Moreover, the kernel of the left adjoint can be identified with the weakly effaceable functors, and that's why $\Lex{(\scrA, \Ab)} = \amod/\scrE$.</p> <p>All this work shows that $A \mapsto \Hom{(A,{-})}$ is a fully faithful and <em>exact</em> embedding of $\scrA$ into ${\Lex}{(\scrA, \Ab)}$, so it remains to show that the latter can be embedded into a category of modules. This is well described in Weibel's or Swan's books, so I won't elaborate on that point and content myself by saying that you simply need to look at the endomorphism ring of an injective cogenerator.</p> <p>As for references, I think you can't do much better than Freyd's book. Don't be too intimidated by Swan's exposition in his K-theory book. If you're really interested in understanding this proof, I think it's worth reading the two expositions (first Freyd, then Swan). There also is a proof in volume 2 of Borceux's Handbook of categorical algebra with a more &ldquo;hands on&rdquo; approach.</p> http://mathoverflow.net/questions/112679/does-there-exist-an-isometry-between-lp-and-lp/112776#112776 Answer by Theo Buehler for Does there exist an isometry between $L^p$ and $l^p$? Theo Buehler 2012-11-18T17:57:19Z 2012-11-25T01:54:18Z <p>Variants of this question show up often enough here on MO and over at <a href="http://math.stackexchange.com" rel="nofollow">math.SE</a> that it seems worthwhile to collect some facts and links. I say <em>isomorphic</em> for <em>linearly homeomorphic</em> and <em>isometric</em> for <em>isometrically isomorphic</em>.</p> <p>One main upshot is:</p> <blockquote> <p>The family of Banach spaces $L^p, \ell^q$ for $1 \leq p, q \leq \infty$ consists of pairwise non-isomorphic spaces, except for two cases:</p> <ol> <li>There is the obvious isometry between $L^2$ and $\ell^2$ and</li> <li>There is a non-obvious isomorphism between $L^\infty$ and $\ell^\infty$, due to Pe&#322;czy&#324;ski. However, $L^\infty$ and $\ell^\infty$ are not isometric.</li> </ol> </blockquote> <p>Plenty of references can be found in the following threads on MO and math.SE:</p> <ol> <li><p>Bill Johnson explains in <a href="http://mathoverflow.net/questions/79713/" rel="nofollow">$L^p$ vs. $L^q$</a> how one can tell these spaces apart for $1 \lt p \neq q \lt \infty$ using type and cotype considerations, and in the comments there is also some discussion of existence of embeddings.</p></li> <li><p>A detailed explanation of the non-existence of an isomorphism between the spaces $L^p$ and $\ell^q$ for $1 \leq p,q \lt \infty$ (modulo Bill Johnson's answer in 1.) is in the answer to <a href="http://math.stackexchange.com/q/97126" rel="nofollow">If $1\leq p &lt; \infty$ then show that $L^p([0,1])$ and $\ell_p$ are not topologically isomorphic</a>. The main ingredient in that answer is Pitt's theorem stating that every operator $\ell^q \to \ell^p$ is compact for $1 \leq p \lt q \lt \infty$, the existence of an embedding $L^2 \to L^p$ plus a little bit of duality theory. See also</p> <ul> <li><a href="http://math.stackexchange.com/q/99086/" rel="nofollow">How do you prove that $\ell^p$ is not isomorphic to $\ell^q$?</a> for a discussion how Pitt's theorem implies non-isomorphism of $\ell^p$ and $\ell^q$ and </li> <li><a href="http://math.stackexchange.com/q/170743" rel="nofollow">Inclusion of $L^p$-spaces, reloaded</a> for a discussion of embeddings of $L^2$ into $L^p$ via Rademacher functions or Gaussians.</li> </ul></li> <li><p>Pe&#322;czy&#324;ski's isomorphism between $L^\infty$ and $\ell^\infty$ is discussed in <a href="http://math.stackexchange.com/q/110438" rel="nofollow">Is there an explicit isomorphism between $L^\infty[0,1]$ and $\ell^\infty$?</a> and non-existence of an isometry is discussed in <a href="http://math.stackexchange.com/q/207435" rel="nofollow">Isometry between $L^\infty$ and $\ell^\infty$</a>. </p></li> </ol> <p>Let me finish by recommending the very nice book by Albiac and Kalton, <a href="http://books.google.com/books?id=vjOBcgAACAAJ" rel="nofollow">Topics in Banach Space Theory</a>, as an alternative to Lindenstrauss-Tzafriri. It contains a gentle introduction to the above ideas and much more.</p> <hr> <p><strong>Edit:</strong> Further links to related topics:</p> <ul> <li><p>One crucial point used in establishing Pe&#322;czy&#324;ski's isomorphism is the injectivity of $\ell^\infty$ and $L^\infty$. For $\ell^\infty$ this is a standard exercise in applying Hahn-Banach (coordinatewise) and for $L^\infty$ Bill Johnson gives a quick proof in <a href="http://mathoverflow.net/questions/110461/" rel="nofollow">Direct proof of injectivity of $L^\infty$</a>. See also <a href="http://mathoverflow.net/questions/111190/" rel="nofollow">Direct proof of "K is projective iff C(K) has the Hahn-Banach property"?</a> for a generalization and related results.</p></li> <li><p><a href="http://mathoverflow.net/questions/111882/" rel="nofollow">Complemented subspaces of $\ell_p(I)$ for uncountable $I$</a></p></li> <li><p><a href="http://mathoverflow.net/questions/111039/" rel="nofollow">Isometric embeddings of $\ell_q^m$ into $\ell_p$ and $L_p$ for $p,q\in[1,+\infty]$</a></p></li> </ul> http://mathoverflow.net/questions/64477/universal-sets-in-metric-spaces/107900#107900 Answer by Theo Buehler for Universal sets in metric spaces Theo Buehler 2012-09-23T12:45:36Z 2012-09-23T12:45:36Z <p>While idly browsing around I stumbled over the follwing paper and remembered this question:</p> <p>A.W. Miller, <em><a href="http://www.math.wisc.edu/~miller/res/hrus.pdf" rel="nofollow">Uniquely Universal Sets</a></em>, Topology and its Applications&nbsp;<strong>159</strong> (2012), pp. 3033&ndash;3041. It's available in various formats <a href="http://www.math.wisc.edu/~miller/res/index.html" rel="nofollow">here</a>.</p> <p>Let me quote the abstract (to avoid confusion: Miller's terminology reverses the r&ocirc;les of $X$ and $Y$ in your question):</p> <blockquote> <p>We say that $X \times Y$ satisfies the Uniquely Universal property (UU) iff there exists an open set $U \subseteq X \times Y$ such that for every open set $W \subseteq Y$ there is a unique cross section of $U$ with $U_x=W$. Michael Hru&#353;&aacute;k raised the question of when does $X \times Y$ satisfy UU and noted that if $Y$ is compact, then $X$ must have an isolated point. We consider the problem when the parameter space $X$ is either the Cantor space $2^\omega$ or the Baire space $\omega^\omega$. We prove the following:</p> <ol> <li><p>If $Y$ is a locally compact zero dimensional Polish space which is not compact, then $2^\omega\times Y$ has UU.</p></li> <li><p>If $Y$ is Polish, then $\omega^\omega \times Y$ has UU iff $Y$ is not compact.</p></li> <li><p>If $Y$ is a $\sigma$-compact subset of a Polish space which is not compact, then $\omega^\omega \times Y$ has UU.</p></li> </ol> </blockquote> <p>His results are mostly positive: &ldquo;a certain space or family of spaces has UU&rdquo; and various permanence properties. One nice &ldquo;negative&rdquo; result:</p> <blockquote> <p>Proposition 30: There exists a partition $X\cup Y=2^\omega$ into Bernstein sets $X$ and $Y$ such that for every Polish space $Z$ neither $Z\times X$ nor $Z\times Y$ has UU.</p> </blockquote> <p>He also raises a few questions, e.g.:</p> <ul> <li>Question&nbsp;4: Does $(2^\omega\oplus 1) \times [0,1]$ have UU?</li> <li>Question&nbsp;6: Does either $\mathbb{R} \times \omega$ or $[0,1]\times \omega$ have UU? Or more generally, is there any example of UU for a connected parameter space?</li> <li>Question&nbsp;11: Is the converse of Corollary&nbsp;10 false? That is: Does there exist $Y$ such that $\omega^\omega \times Y$ has UU but $2^\omega\times Y$ does not have UU?</li> </ul> http://mathoverflow.net/questions/104965/non-regular-connected-hausdorff-banach-manifold/105595#105595 Answer by Theo Buehler for Non-regular Connected Hausdorff Banach Manifold Theo Buehler 2012-08-27T02:34:31Z 2012-08-27T20:37:15Z <p>Apparently the answer is <strong>no</strong>, not every connected Hausdorff Banach manifold is regular, not even when it is modeled on a separable Hilbert space.</p> <p>I quote (verbatim) from J.&nbsp;Margalef-Roig, E.&nbsp;Outerelo-Dominguez, <em>Differential Topology</em>, North Holland Mathematics Studies 173, 1992, page&nbsp;44f.</p> <blockquote> <p>It is well known the result of General Topology that every Hausdorff locally compact topological space satisfies the Tychonoff axiom [M-O-P, V.2, pg. 231]. By this and the Riesz's theorem every Hausdorff locally finite dimensional differentiable manifold satisfies the Tychonoff axiom. This last affirmation is not true for arbitrary Hausdorff differentiable manifolds. In [M.O.1] there is an example of a Hausdorff connected differentiable manifold $X$ of class $\infty$, such that $\partial X = \emptyset$, $X$ is not regular and $X$ admits an atlas whose charts are modelled over an infinite dimensional separable real Hilbert space.</p> </blockquote> <p>They continue to add the regularity hypothesis in their results whenever it is needed.</p> <p>The cited references are:</p> <ul> <li><p>[M.O.P.] MARGALEF, J.-OUTERELO, E.-PINILLA, J.L.: Topologia, I-V. Alhambra, Madrid 1975, 79, 79, 80 and 1982.</p></li> <li><p>[M.O.1] MARGALEF, J.-OUTERELO, E.: Una variedad diferenciable de dimension infinita, separada y no regular. Rev. Mat. Hisp.-Am, IV, V.42, 1982, 51-55. <a href="http://tinyurl.com/8kb3m45" rel="nofollow">(QuickView link)</a>.</p></li> </ul> <hr> <p><strong>Edit:</strong> As was pointed out by Benjamin Dickman in the comments, the example also appears in English in A. Kriegl, P.W.&nbsp;Michor, <em><a href="http://books.google.com/books?id=s7fPYRqhXEUC" rel="nofollow">The convenient setting of global analysis</a></em>, AMS (1997), <a href="http://books.google.com/books?id=s7fPYRqhXEUC&amp;pg=PA266" rel="nofollow"><strong>27.6</strong> Non-regular manifold</a>, page&nbsp;266. The book is available as <a href="http://www.mat.univie.ac.at/~kriegl/Skripten/apbook.pdf" rel="nofollow">a pdf file</a> on Kriegl's homepage.</p> http://mathoverflow.net/questions/100245/no-injective-groups-with-more-than-one-element/100258#100258 Answer by Theo Buehler for No injective groups with more than one element? Theo Buehler 2012-06-21T16:21:05Z 2012-06-21T17:04:02Z <p>The argument in <a href="http://zimmer.csufresno.edu/~mnogin/talks/regAMSapril2004.pdf" rel="nofollow">the slides</a> by <a href="http://zimmer.csufresno.edu/~mnogin/" rel="nofollow">Maria Nogin</a> (née Voloshina) linked to by a-fortiori in a comment was published as:</p> <p>Maria Nogin, <em><a href="http://dx.doi.org/10.2478/s11533-006-0040-7" rel="nofollow">A short proof of Eilenberg and Moore’s theorem</a></em>, Central European Journal Of Mathematics Volume&nbsp;5, Number&nbsp;1 (2007), 201&ndash;204. <a href="http://mail.csufresno.edu/~mnogin/short-proof-published.pdf" rel="nofollow">Also available on her homepage</a>.</p> <hr> <p><strong>Added:</strong> The above paper was also mentioned in <a href="http://math.stackexchange.com/q/102820/5363" rel="nofollow">Jonas Meyer's answer</a> to the same question on <a href="http://math.stackexchange.com" rel="nofollow">math.SE</a>. As Steve D. points out in a comment there, the result appears as <a href="http://books.google.com/books?id=L_E6AAAAIAAJ&amp;pg=PA9" rel="nofollow">Exercise&nbsp;7 on page&nbsp;9</a> of D.L.&nbsp;Johnson's <em><a href="http://books.google.com/books?id=L_E6AAAAIAAJ" rel="nofollow">Topics in the Theory of Group Presentations</a></em>, Cambridge University Press&nbsp;1980:</p> <p><img src="http://i.stack.imgur.com/1xMQC.png" alt="Exercise 7"></p> <hr> <p>The argument by Eilenberg and Moore appears on pages&nbsp;21/22 of <em>Foundations of Relative Homological Algebra</em>, Memoirs of the AMS, Volume&nbsp;<strong>55</strong> (1965). Here's a scan of the relevant passage for the convenience of the readers:</p> <p><img src="http://i.stack.imgur.com/5Kns9.png" alt="Original argument part 1"> <img src="http://i.stack.imgur.com/xmsoS.png" alt="Original argument part 2"></p> http://mathoverflow.net/questions/73550/ergodic-decomposition-of-quasi-invariant-measure/73576#73576 Answer by Theo Buehler for Ergodic decomposition of quasi-invariant measure Theo Buehler 2011-08-24T15:31:40Z 2011-08-25T05:46:27Z <p>This is a bit too long for a comment, hence I post it as an answer.</p> <p>I honestly don't know where you can find a group theoretic version of ergodic decomposition proved via Choquet theory (and I'm not convinced that it exists in the setting you're interested in).</p> <p>However, the exact result you quote from Zimmer is proved as Theorem&nbsp;1.1 in the carefully written paper</p> <p>G.&nbsp;Greschonig, K.&nbsp;Schmidt, <em><a href="http://matwbn.icm.edu.pl/ksiazki/cm/cm84/cm84217.pdf" rel="nofollow">Ergodic Decomposition of quasi-invariant probability measures</a></em>, <a href="http://journals.impan.gov.pl/cm/" rel="nofollow">Colloq. Math.</a>&nbsp;<strong>84/85</strong>&nbsp;(2000), part&nbsp;2, 495&ndash;514, <a href="http://www.ams.org/mathscinet-getitem?mr=1784210" rel="nofollow">MR1784210</a>.</p> <p>You'll find many interesting references in there.</p> <p>For a lot of extremely helpful results that are used in and around Zimmer's work, I recommend Section&nbsp;2 of</p> <p>David Fisher, Dave Witte Morris, and Kevin Whyte, <em><a href="http://nyjm.albany.edu/j/2004/10-15.html" rel="nofollow">Nonergodic actions, cocycles and superrigidity</a></em>, New York Journal of Mathematics, Volume <strong>10</strong> (2004) 249&ndash;269, <a href="http://www.ams.org/mathscinet-getitem?mr=2114789" rel="nofollow">MR2114789</a>.</p> <p>I can't say anything on your final question.</p> http://mathoverflow.net/questions/51494/why-the-name-separable-space/51501#51501 Answer by Theo Buehler for Why the name 'separable' space? Theo Buehler 2011-01-08T22:35:47Z 2011-08-09T01:23:30Z <p>As far as I know the word <em>separable</em> was introduced by M. Fréchet in <em>Sur quelques points du calcul fonctionnel</em>, Rend. Circ. Mat. Palermo <strong>22</strong> (1906), 1-74. The paper can be obtained via <a href="http://dx.doi.org/10.1007/BF03018603" rel="nofollow">this link</a> (Springer). It's the famous paper in which he introduced metric spaces. He considers first slightly more general objects which he calls <em>classes (V)</em>: where (V) stands for <em>voisinage</em> &mdash; neighborhood.</p> <p><strong>Remark:</strong> Metrics are introduced under the name <em>&eacute;cart</em> in n<sup>o</sup>&nbsp;49 on page&nbsp;30. It is peculiar that the symmetry condition is <em>not</em> explicitly mentioned but it seems to be understood as Fr&eacute;chet immediately mentions that metric spaces generalize <em>classes (V)</em> cf.&nbsp;n<sup>o</sup>&nbsp;27 on page&nbsp;17f. However, I couldn't find an instance where he actually uses it, he is always careful to respect the order&nbsp;&mdash;&nbsp;I may have missed something since I haven't read the paper in detail.</p> <p>I quote the relevant passage [from n<sup>o</sup>&nbsp;37 on page 23f]:</p> <blockquote> <p>Nous appellerons ensuite <em>classe s&eacute;parable</em> une classe qui puisse &ecirc;tre consid&eacute;r&eacute;e d'au moins une fa&ccedil;on comme l'ensemble d&eacute;riv&eacute; d'un ensemble d&eacute;nombrable de ses propres &eacute;l&eacute;ments.</p> <p>[...]</p> <p>Ceci &eacute;tant, nous nous bornerons maintenant &agrave; l'&eacute;tude des <em>classes (V)</em> NORMALES, <em>c'est-&agrave;-dire parfaites, s&eacute;parables et admettant une g&eacute;n&eacute;ralisation du th&eacute;or&egrave;me de</em> CAUCHY. Cette limitation n'a du reste rien d'artificiel, elle provient directement de la comparaison des classes (V) avec les ensembles lin&eacute;aires [...]</p> <p>[...]</p> <p>Passons maintenant aux classes s&eacute;parables. On peut qualifier ainsi les ensembles lin&eacute;aires en consid&eacute;rant la droite ind&eacute;finie comme l'ensemble d&eacute;riv&eacute; de l'ensemble des points d'abscisses rationnelles. Mais il n'en est pas de m&ecirc;me pour toute classe parfaite (V).</p> </blockquote> <p>I am unable to translate this in a reasonable way (but see Amit's comment below for a translation). Very roughly: Fr&eacute;chet defines separable spaces as we do it today and says that in the following he will restrict attention to complete, perfect and separable metric spaces. The last quoted paragraph indeed confirms Qiaochu's comment.</p> http://mathoverflow.net/questions/60255/information-about-publishing-and-citations/60256#60256 Answer by Theo Buehler for Information about publishing and citations Theo Buehler 2011-04-01T04:30:51Z 2011-07-17T17:26:39Z <p>On Mariano's request I'm adding my comment on meta.MO as an answer here. This only concerns (part of) the first bullet point in the question.</p> <p>There was an article <a href="http://www.ams.org/notices/201011/rtx101101421p.pdf" rel="nofollow"><em>Topical Bias in Generalist Mathematics Journals</em></a> by Joseph F. Grcar in the <a href="http://www.ams.org/notices/201011/index.html" rel="nofollow">december 2010</a> issue of the Notices of the AMS. According to the text, the statistics are based on 854,547 entries from the 2000-2009 period of the Zentralblatt database. Unfortunately the article remains silent on exactly how the data was gathered, but it might be a starting point for your own investigations.</p> <p>For the convenience of the readers I take the liberty of reproducing the statistics most relevant for the present question from that article:</p> <p><img src="http://i.stack.imgur.com/hIAhN.png" alt="" title="Percentage of all Mathematics Papers 2000-2009" /></p> <p>For more detailed information please follow <a href="http://www.ams.org/notices/201011/rtx101101421p.pdf" rel="nofollow">this link</a> or the ones provided above.</p> http://mathoverflow.net/questions/69317/extensions-of-banach-spaces/69321#69321 Answer by Theo Buehler for Extensions of Banach spaces Theo Buehler 2011-07-02T09:53:38Z 2011-07-02T20:48:29Z <p>I don't understand the questions for the following reason: If the image of $X$ is complemented in $E_1$ then the extension is split. Indeed, if $P$ is a projection of $E_1$ onto the image of $X$ then $1-P$ is a projection onto an isomorph of $Y$ by the open mapping theorem (see e.g. <a href="http://dx.doi.org/10.1007/b80626" rel="nofollow">Nicolas Monod's thesis</a> Corollary&nbsp;4.2.4 for a detailed proof).</p> <p>First question: If you're asking about a pair of extensions of $Y$ by $X$ with $E_1$ split and $E_2$ non-split, take $X = c_0$ and $Y = \ell^{\infty}/c_0$. Then $E_2: 0 \to X \to \ell^{\infty} \to Y \to 0$ is not split by Phillips' lemma (see <a href="http://www.jstor.org/stable/2315346" rel="nofollow">Whitley's note</a> in the Monthly for a simple proof), and $E_1: 0 \to X \to X \oplus Y \to Y \to 0$ is split by definition.</p> <p>Second question: Yes, $(x, y) \mapsto \left(\frac{1}{2}(x+y), \frac{1}{2}(x+y)\right)$ is a projection of $X \oplus X$ onto $\Delta$. I recommend you to prove that this sequence is isomorphic to the obvious extension $0 \to X \to X \oplus X \to X \to 0$ (inclusion into the first summand, projection onto the second).</p> <p>Two final remarks:</p> <ul> <li><p>A very interesting procedure for producing non-split extensions of Banach spaces is the <em>twisted sum construction</em> due to <a href="http://www.ams.org/mathscinet-getitem?mr=542869" rel="nofollow">Kalton-Peck</a> (I recently learned about this from Bill Johnson in <a href="http://mathoverflow.net/questions/66345/example-of-a-compact-set-that-isnt-the-spectrum-of-an-operator" rel="nofollow">this thread</a>).</p></li> <li><p>Basically, you're asking about the <a href="http://en.wikipedia.org/wiki/Ext_functor#Ext_and_extensions" rel="nofollow">Yoneda Exts</a> in the <a href="http://en.wikipedia.org/wiki/Exact_category" rel="nofollow">exact category</a> of Banach spaces with the exact structure consisting of all kernel-cokernel pairs. If you're interested in such abstract nonsense, please allow me a bit of <a href="http://dx.doi.org/10.1016/j.exmath.2009.04.004" rel="nofollow">self-advertisement</a>.</p></li> </ul> http://mathoverflow.net/questions/66345/example-of-a-compact-set-that-isnt-the-spectrum-of-an-operator Example of a compact set that isn't the spectrum of an operator Theo Buehler 2011-05-29T08:27:39Z 2011-05-29T09:42:42Z <p>This question is somewhat ill-posed (due to the word easy) and is triggered by idle curiosity:</p> <blockquote> <p>Is there an <em>easy</em> example of a (separable, infinite-dimensional) Banach space $X$ and a nonempty compact set $K \subset \mathbb{C}$ such that $K$ is <em>not</em> the spectrum of a bounded linear operator $T: X \to X$?</p> </blockquote> <p>As I'm not at all knowledgeable beyond the the very first basics in the geometry of Banach spaces I apologize if the following notes completely miss the point. I add them to give some background and in order to clarify what I consider "easy":</p> <p><strong>Notes:</strong></p> <ul> <li><p>If $X$ is a Hilbert space (or more generally, if $X$ admits an unconditional basis $\{e_{n}\}$), it is easy to construct a diagonal operator with spectrum $K$ by choosing a countable dense subset $\{\lambda_{n}\} \subset K$ and letting $T$ be the diagonal operator sending $\sum x_n e_n$ to $\sum (\lambda_n x_n)e_n$.</p></li> <li><p>Standard examples for spaces without an unconditional basis are $L^1[0,1]$ and $C[0,1]$. I think I convinced myself that in both cases every non-empty compact set of $\mathbb{C}$ arises as the spectrum of an operator, so these obvious candidates don't seem to answer my question. (If this should be wrong, please tell me!)</p></li> <li><p>Variants of the <a href="http://arxiv.org/abs/math/9205204" rel="nofollow">Gowers-Maurey</a> space and the <a href="http://arxiv.org/abs/0903.3921" rel="nofollow">Argyros-Haydon</a> space afford examples such that the spectrum $K$ must be countable with at most one accumulation point. See <a href="http://gowers.wordpress.com/2009/02/07/a-remarkable-recent-result-in-banach-space-theory/" rel="nofollow">Gowers's blog</a> for background on that. For the Argyros-Haydon space this is easy to see by the very motivation for its construction: It has the remarkable property that a bounded linear operator is of the form $\lambda \cdot \operatorname{id} + C$, where $C$ is compact (thus solving the long-standing scalar-plus-compact problem).</p></li> <li><p>I asked a <a href="http://math.stackexchange.com/q/35623/" rel="nofollow">version of this question</a> a few weeks ago on <a href="http://math.stackexchange.com" rel="nofollow">math.SE</a> but with no answers so far. In view of the illuminating comments by Robert Israel and Jonas Meyer I got there I updated it a bit.</p></li> <li><p>The present question is related to Pietro Majer's question <a href="http://mathoverflow.net/questions/30669/" rel="nofollow">Banach spaces with few linear operators ?</a> here on MO. I looked at Maurey's chapter <em><a href="http://dx.doi.org/10.1016/S1874-5849%2803%2980036-0" rel="nofollow">Banach spaces with few operators</a></em> in the <em><a href="http://www.sciencedirect.com/science/handbooks/18745849" rel="nofollow">Handbook of the Geometry of Banach spaces</a></em> Vol. 2, Elsevier 2003, (Johnson, Lindenstrauss, eds) but the examples discussed there are <em>way beyond</em> what I would count as easy.</p></li> <li><p>It may well be (as I'm rather ignorant on this topic) that the level of difficulty of an example must be comparable to the one of the construction of the Gowers-Maurey space or even the Argyros-Haydon space, so if there's a compelling reason pointing in this direction, please let me know.</p></li> </ul> http://mathoverflow.net/questions/62380/when-do-0-preserving-isometries-have-to-be-linear/62382#62382 Answer by Theo Buehler for When do 0-preserving isometries have to be linear? Theo Buehler 2011-04-20T07:16:01Z 2011-04-20T07:16:01Z <p>If you assume $f$ to be surjective then $f$ has to be linear without any assumptions on $V$ by the <a href="http://en.wikipedia.org/wiki/Mazur%2DUlam_theorem" rel="nofollow">Mazur-Ulam theorem</a>. Wikipedia doesn't offer much more information than a link to the beautiful recent <a href="http://www.helsinki.fi/~jvaisala/mazurulam.pdf" rel="nofollow">proof</a> by J.&nbsp;V&auml;is&auml;l&auml;.</p> http://mathoverflow.net/questions/60050/projective-banach-spaces/60052#60052 Answer by Theo Buehler for Projective Banach spaces Theo Buehler 2011-03-30T10:28:34Z 2011-03-30T10:28:34Z <p>You essentially answered your first question yourself: the ground field is a (contractive) retract of any nonzero Banach space by Hahn-Banach. If there were a non-zero projective Banach space in your sense then the ground field would be projective as well.</p> <p>On the other hand: It is a theorem due to K&ouml;the and Pe&#322;czy&#324;ski that every projective Banach space (lifting over surjective maps in the additive category of Banach spaces) is isomorphic to $\ell^1{(S)}$ for some $S$. I don't know how well the norm of the isomorphism is controlled, but as all these spaces already satisfy your relaxed condition, you won't find any others.</p> <p>The references for the second paragraph are:</p> <ul> <li>A. Pe&#322;czy&#324;ski, <em>Projections in certain Banach spaces</em>, Studia Math. <strong>19</strong> (1960), 209–228. MR0126145.</li> <li>G. K&ouml;the, <em>Hebbare lokalkonvexe R&auml;ume</em>, Math. Ann. <strong>165</strong> (1966), 181–195. MR0196464.</li> </ul> http://mathoverflow.net/questions/57616/automatic-continuity-measurable-homomorphisms-of-c-c-or-c-c/57623#57623 Answer by Theo Buehler for [automatic continuity] measurable homomorphisms of (C,+)-->(C,+) or (C,+)-->(C,*) are continuous and admit an explicit description ? Theo Buehler 2011-03-07T00:44:27Z 2011-03-07T00:44:27Z <p>On page 23 of his 1932 book <a href="http://matwbn.icm.edu.pl/kstresc.php?wyd=10&amp;tom=1&amp;jez=en" rel="nofollow"><em>Sur la th&eacute;orie des op&eacute;rations lin&eacute;aires</em></a> Banach proves:</p> <p><strong>Theorem.</strong> A Baire measurable homomorphism between Polish groups is continuous.</p> <p>Note that Banach writes $+$ for the composition but neither does he assume nor use that a group is abelian. He proves the result first for Borel measurable homomorphisms and remarks immediately afterwards that the same argument even shows that a Baire measurable homomorphism between Polish groups is continuous, a fact usually attributed to Pettis for reasons that are unclear to me.</p> <p>Recall that a topological group is <em>Polish</em> if its topology is second countable and metrizable with respect to a complete metric. The Baire $\sigma$-algebra is the $\sigma$-algebra generated by the Borel sets and the meager sets (beware that there are other uses of the term "Baire measurable" in the literature -- there are even published papers whose authors fell for this trap).</p> <p>In particular, the specific question you ask about the real and complex numbers has a positive answer in all cases.</p> <p>There is an entire industry, called <em>automatic continuity</em>, asking the question whether homomorphisms/derivations etc are continuous only due to their algebraic property. The easiest result in that direction states that a $*$-homomorphism between $C^{\ast}$-algebras is a linear contraction. One of the major players of that topic, H.G. Dales, has recently written a voluminous book of the same title, containing many results of that spirit and many historical remarks.</p> http://mathoverflow.net/questions/56651/on-a-decomposition-of-l1g/56791#56791 Answer by Theo Buehler for On a decomposition of L^1(G) Theo Buehler 2011-02-27T05:27:41Z 2011-02-28T13:06:41Z <p>I'm answering Yemon's version of the question.</p> <p>The answer is trivially yes for discrete $G$ since $\ell^1(G) \subset \ell^2(G)$, so let me focus on the non-discrete case.</p> <p>The first observation to make is that $B(G)$ is contained in the bounded (and uniformly continuous) functions of $G$. So the question asks in particular if every integrable function on $G$ is the sum of a bounded function and a square-integrable function.</p> <p>This is clearly false for compact infinite $G$: For such $G$ we have the strict inclusions $L^\infty \subsetneqq L^2 \subsetneqq L^1$ so $L^\infty + L^2 \subset L^2$, and hence every function in $L^1 \smallsetminus L^2$ provides a counterexample to the question.</p> <p>Since the question asks for a counterexample in $\mathbb{R}^{n}$, I'll give one for $\mathbb{R}$ which is easily adapted to the higher-dimensional case and with a little care <s>should</s> also gives a counterexample for any <s>non-compact and</s> non-discrete locally compact abelian group.</p> <p>Take $f = \sum_{n=1}^{\infty} n \cdot [n,n+\frac{1}{n^{3}}]$. This is a function in $L^1 \smallsetminus L^2$. For a bounded function $h$ we have for all $n \geq \Vert h \Vert_{\infty}$ and all $x \in [n,n+\frac{1}{n^3}]$ that $|f(x) - h(x)| \geq n- \Vert h \Vert_{\infty}$, which implies that $g = f - h \notin L^2(\mathbb{R})$ by a straightforward estimate.</p> http://mathoverflow.net/questions/56718/when-do-isometric-actions-exist/56734#56734 Answer by Theo Buehler for When do isometric actions exist? Theo Buehler 2011-02-26T13:20:56Z 2011-02-26T18:26:09Z <p>This is at best a partial answer but rather too long for a comment (I only adress the last paragraph of the question).</p> <hr> <p>Indeed, if $X$ is locally compact second countable and the action of $G$ is proper then there exists a $G$-invariant metric compatible with the topology. As you suspect, this can be done by integrating the metric multiplied by a (generalized) <em>Bruhat function</em> (see Bourbaki, <em>Int&eacute;gration</em>, VII, &sect; 2, No. 4).</p> <p>More precisely: Let $G$ be a locally compact group and fix a left Haar measure on $G$. The action of $G$ on a locally compact space $X$ (such that $X/G$ is paracompact) is proper if and only if there exists a continuous function $\beta: X \to [0,\infty)$ such that</p> <ul> <li>For every compact set $K \subset X$ the set $\operatorname{supp}\beta \cap GK$ is compact.</li> <li>For all $x \in X$ we have $\int_{G} \beta(g^{-1}x)\,dg = 1$.</li> </ul> <p>This fact is folklore but it is difficult to locate a simple proof in the literature. Therefore I've given a short one in Appendix E of my thesis, available <a href="http://dx.doi.org/10.3929/ethz-a-005561107" rel="nofollow">here</a>. Sometimes these functions are called cut-off functions but I don't like the name.</p> <p>Using this, the existence of an invariant metric compatible with the topology on a locally compact second countable proper $G$-space $X$ is an easy exercise in integration theory: Pick any metric $d_{X}$ compatible with the topology and replace it by $\frac{d_{X}}{1+d_{X}}$ in case it is unbounded. Then put $\delta_{X}(x,y) = \iint_{G \times G} \beta(g^{-1}x)\, \beta(h^{-1} y)\, d_X (g^{-1}x, h^{-1}y)\,dh\,dg$ and verify that $\delta_{X}:X \times X \to [0,1)$ is an invariant metric compatible with the topology.</p> <p>A similar and detailed argument can be found in one of the first few sections of Koszul's <em>Lectures on groups of transformations</em> (I think it's in the the third section of the first chapter but I can't verify this at the moment).</p> <hr> <p>Finally, you're asking about the relation to amenability, here I also have at best some comments. Of course, proper actions are known to be topologically amenable (for instance because there is a Bruhat function). In the other direction, I think there is no hope. There are plenty of actions of amenable groups that can't be made into isometric actions (and any continuous action of an amenable group is amenable). Since Vaughn has given some nice examples, I can end this long post now.</p> http://mathoverflow.net/questions/56497/morphisms-of-banach-spaces/56505#56505 Answer by Theo Buehler for Morphisms of Banach spaces Theo Buehler 2011-02-24T08:34:48Z 2011-02-24T12:18:24Z <p>I don't know if there's a universally accepted name for this notion. Of course, if the kernel and the complement of the image are finite-dimensional, such a map is a Fredholm operator. I would suggest to call them pseudo-invertible (in reference to the Moore-Penrose pseudo-inverse), because $f:E \to F$ has complemented kernel and image if and only if there is a $g:F \to E$ such that $fgf = f$ (and $gfg = g$).</p> <p>Since the question is tagged homological algebra, let me point out that these morphisms are precisely the morphisms factoring as $E \twoheadrightarrow I \rightarrowtail F$, where $\twoheadrightarrow$ stands for a split epimorphism and $\rightarrowtail$ for a split monomorphism. These are the <em>admissible monomorphisms</em> and <em>admissible epimorphisms</em> for the <em>split exact structure</em> (in the sense of Quillen) on the additive category of Banach spaces and bounded linear maps, see <a href="http://en.wikipedia.org/wiki/Exact_category" rel="nofollow">here</a> for more on this. However, there are several exact structures on the category of Banach spaces (at least three of interest), so admissible monomorphism/epimorphism alone is not good enough from this point of view.</p> <p>I've seen several names for the maps you're asking about in the literature:</p> <ul> <li><p>In Borel-Wallach, <em>Continuous cohomology, discrete subgroups and representations of reductive groups</em>, Annals of Math. Studies <strong>94</strong>, Princeton University Press (1980), IX 1.5, they are called $s$-<em>morphisms</em> (actually for Hausdorff locally convex spaces). </p></li> <li><p>Guichardet, <em>Cohomologie des groupes topologiques et des alg&egrave;bres de Lie</em>, Textes Mathématiques&nbsp;<strong>2</strong>, Fernand Nathan, Paris (1980), Appendice D, Def. D.1 calls them <em>strong</em> (again in the setting of Hausdorff locally convex spaces).</p></li> <li><p>Monod, <em>Continuous Bounded Cohomology of Locally Compact Groups</em>, Springer Lecture Notes in Mathematics, <strong>1758</strong> (2001), Definition 4.2.2, calls them <em>weakly admissible</em> (the weakly refers to the fact that his notion of admissibility requires that if $f$ is of norm $1$, there is a pseudo-inverse $g$ of norm $1$, at least for monomorphisms).</p></li> </ul> http://mathoverflow.net/questions/55726/properly-discontinuous-action/56490#56490 Answer by Theo Buehler for Properly Discontinuous Action Theo Buehler 2011-02-24T06:10:07Z 2011-02-24T06:10:07Z <p>Below <em>locally compact</em> spaces are assumed to be <em>Hausdorff</em>. The following is essentially a distillate of results from Bourbaki's <em>Topologie G&eacute;n&eacute;rale</em>, Chapitres II and III.</p> <hr> <p><strong>Definition.</strong> A continuous function $f: X \to Y$ is called <em>proper</em> if $f$ maps closed sets to closed sets and $f^{-1}(K)$ is compact for all compact $K \subset Y$.</p> <p><strong>Remark.</strong> If $X$ is Hausdorff and $Y$ is locally compact then a continuous function $f: X \to Y$ is proper if and only if $f^{-1}(K)$ is compact for all compact $K \subset Y$. Moreover, $X$ must be locally compact.</p> <p>To see this, cover $Y$ with open and relatively compact sets $U_{\alpha}$. Then $f^{-1}(U_{\alpha})$ is an open covering of $X$ by relatively compact sets, hence $X$ is locally compact. If $F \subset X$ is closed then $f(F)$ is closed. Indeed, if $(y_{n}) \subset f(F)$ is a net converging to $y$, then we may assume that all $y_{n}$ are in a compact neighborhood $K$ of $y$. Pick a pre-image $x_{n}$ of each $y_{n} \in f^{-1}(K)$, which is compact by assumption. If $x_{i} \to x \in f^{-1}(K)$ is a convergent subnet of $(x_{n})$ then $(f(x_{i}))$ is a subnet of $(y_{n})$, hence $f(x) = y$ by continuity and thus $y \in f(F)$. </p> <p><strong>Remark.</strong> In the definition of properness it would suffice to require that $f$ is closed and $f^{-1}(y)$ is compact for all $y \in Y$, but the definition above is good enough for the present purposes.</p> <hr> <p><strong>Definition.</strong> Let $G$ be a topological group acting continuously on a topological space $X$. The action is called <em>proper</em> if the map $\rho: G \times X \to X \times X$ given by $(g,x) \mapsto (x,gx)$ is proper.</p> <p><strong>Proposition.</strong> If $G$ acts properly on $X$ then $X/G$ is Hausdorff. In particular, each orbit $Gx$ is closed. The stabilizer $G_{x}$ of each point is compact and the map $G/G_{x} \to Gx$ is a homeomorphism. Moreover, if $G$ is Hausdorff then so is $X$.</p> <p><em>Proof.</em> Indeed, the orbit equivalence relation is the image of $\rho$, hence it is closed. Since the projection $X \to X/G$ is open, this implies that $X/G$ is Hausdorff. Since the pre-image of the point $[x]$ in $X/G$ is its orbit $Gx$, we have that orbits are closed. The stabilizer $G_{x}$ of a point $x$ is the projection of $\rho^{-1}(x,x)$ to $G$, hence it is compact. The map $G/G_{x} \to Gx$ is proper and $1$-to-$1$, hence a homeomorphism. Finally, if $G$ is Hausdorff, then $\{e\} \times X \subset G \times X$ is closed and therefore the diagonal $\Delta_{X} = \rho(\{e\} \times X)$ of $X \times X$ is closed, hence $X$ is Hausdorff.</p> <p><strong>Exercise.</strong> Let $G$ be a Hausdorff topological group acting properly on a locally compact space $X$. Then $G$ and $X/G$ are both locally compact. If $X$ is compact Hausdorff then so are $G$ and $X/G$.</p> <hr> <p>Replace <em>finite</em> by <em>compact</em> in Type A and Type B. Then we have the following implications for a continuous action:</p> <p>Proper $\Longrightarrow$ Type A, the converse holds if both $G$ and $X$ are locally compact.</p> <p>Type A $\Longrightarrow$ Type B.</p> <p>Let $K \subset X$ be compact. Then $K \times K \subset X \times X$ is compact. Thus, if the action is of type A, then $\rho^{-1}(K \times K) = \{(g,x) \in G \times X\,:\,(x,gx) \in K \times K\} \subset G \times X$ is compact. The projection of this set to $G$ is compact and consists precisely of the $g \in G$ for which $K \cap gK \neq \emptyset$.</p> <p>Type B $\Longrightarrow$ Type A if $X$ is Hausdorff.</p> <p>We have to show that $\rho^{-1}(L)$ is compact for every compact $L \subset X \times X$. Let $K$ be the union of the two projections of $L$. Then $(g,x) \in \rho^{-1}(K \times K)$ is equivalent to $x \in K \cap gK$. Since $\rho^{-1}(K \times K)$ is compact and $\rho^{-1}(L)$ is a closed subset of $\rho^{-1}(K \times K)$, we have that $\rho^{-1}(L)$ is compact.</p> <p><strong>Corollary.</strong> If $G$ and $X$ are locally compact, properness, Type A and Type B are all equivalent.</p> <hr> <p>Let me now show that in the locally compact setting properness is equivalent to a refinement of Type C:</p> <p><strong>Proposition.</strong> Let $G$ and $X$ be locally compact and assume that $G$ acts continuously on $X$. The following are equivalent:</p> <ol> <li>The action is proper.</li> <li>For all $x,y \in X$ there are open neighborhoods $U_{x}, U_{y} \subset X$ of $x$ and $y$ such that $C = \{g \in G\,:\,gU_x \cap U_{y} \neq \emptyset \}$ is relatively compact.</li> </ol> <p><em>Proof.</em> $1.$ implies $2.$ Let $K_{x}$ and $K_{y}$ be compact neighborhoods of $x$ and $y$. Then the set $\rho^{-1}(K_{x} \times K_{y})$ is compact and its projection to $G$ contains $C$ and is compact. Now let $U_{x}$ and $U_{y}$ be the interiors of $K_{x}$ and $K_{y}$.</p> <p>$2.$ implies $1$. Let $K \subset X \times X$ be compact. We want to show that $\rho^{-1}(K)$ is compact as well. Let $(g_{n},x_{n})$ be a universal net in $\rho^{-1}(K)$. Then $(x_{n},g_{n}x_{n})$ is a universal net in $K$ and hence converges to some $(x,y) \in K$. Let $U_{x}, U_{y}$ and $C$ be as in $2.$. Then $(x_{n},g_{n}x_{n}) \in U_{x} \times U_{y}$ eventually and thus also $(g_{n}) \subset C$ eventually. Since $(g_{n})$ is universal and $C$ is relatively compact, $(g_{n})$ converges to some $g \in G$. Hence $(g_{n},x_{n})$ converges to $(g,x) \in \rho^{-1}(K)$.</p> <p><strong>Example.</strong> To see that Type C is weaker than properness, consider $A = \begin{pmatrix} 2 &amp; 0 \\ 0 &amp; 2^{-1} \end{pmatrix}$ and the action of $\mathbb{Z}$ on $\mathbb{R}^{2} \smallsetminus \{0\}$ given by $n \cdot x = A^{n} x$. For instance for $x = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $y = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ and all neighborhoods $U_{x} \ni x$ and $U_{y} \ni y$ the set $\{n \in \mathbb{Z}\,:\, U_{x} \cap n \cdot U_{y} \neq \emptyset \}$ is infinite. Thus this action isn't proper. On the other hand, it is easy to see that it is of Type C.</p> <p><strong>Remark.</strong> The previous example shows that properness of an action is <em>not</em> a local property.</p> <p><strong>Exercise.</strong> If the action of a locally compact group $G$ on a locally compact space $X$ is of type C and $X/G$ is Hausdorff then it is proper.</p> <hr> <p>To finish this discussion, it is evident that an action of type C is also of type E, hence type E is also weaker than properness. Finally, a trivial action is of type D, hence this property has nothing to do with properness.</p> <hr> <p>Here are some references:</p> <p>I've followed Bourbaki, Topologie G&eacute;n&eacute;rale, Ch. III, in terminology, and the proofs I've given are variants of Bourbaki's. I happen to like Koszul's <em>Lectures on groups of transformations</em>. If you're looking for a more pedestrian approach, you can find the most important facts in Lee's <em>Introduction to topological manifolds</em>.</p> http://mathoverflow.net/questions/48970/why-are-the-integers-with-the-cofinite-topology-not-path-connected Why are the integers with the cofinite topology not path-connected? Theo Buehler 2010-12-10T20:22:31Z 2011-02-15T23:12:22Z <p>An apparently elementary question that bugs me for quite some time:</p> <blockquote> <p><strong>(1)</strong> Why are the integers with the cofinite topology not path-connected?</p> </blockquote> <p>Recall that the open sets in the cofinite topology on a set are the subsets whose complement is finite or the entire space.</p> <p>Obviously, the integers are connected in the cofinite topology, but to prove that they are not path-connected is much more subtle. I admit that this looks like the next best homework problem (and was dismissed as such in <a href="http://mathoverflow.net/questions/41016/finite-complement-topology-closed" rel="nofollow">this thread</a>), but if you think about it, it does not seem to be obvious at all.</p> <p>An equivalent reformulation of <strong>(1)</strong> is:</p> <blockquote> <p><strong>(2)</strong> The unit interval $[0,1] \subset \mathbb{R}$ cannot be written as a countable union of pairwise disjoint non-empty closed sets.</p> </blockquote> <p>I <em>can</em> prove this, but I'm not really satisfied with my argument, see below.</p> <p>My questions are:</p> <blockquote> <ol> <li>Does anybody know a reference for a proof of <strong>(1)</strong>, <strong>(2)</strong> or an equivalent statement, and if so, do you happen to know who has proved this originally?</li> <li>Do you have an easier or slicker proof than mine?</li> </ol> </blockquote> <hr> <p>Here's an outline of my rather clumsy proof of <strong>(2)</strong>:</p> <p>Let $[0,1] = \bigcup_{n=1}^{\infty} F_{n}$ with $F_{n}$ closed, non-empty and $F_{i} \cap F_{j} = \emptyset$ for $i \neq j$.</p> <p>The idea is to construct by induction a decreasing family $I_{1} \supset I_{2} \supset \cdots$ of non-empty closed intervals such that $I_{n} \cap F_{n} = \emptyset$. Then $I = \bigcap_{n=1}^{\infty} I_{n}$ is non-empty. On the other hand, since every $x \in I$ lies in exactly one $F_{n}$, and since $x \in I \subset I_{n}$ and $I_{n} \cap F_{n} = \emptyset$, we see that $I$ must be empty, a contradiction.</p> <p>In order to construct the decreasing sequence of intervals, we proceed as follows:</p> <p>Since $F_{1}$ and $F_{2}$ are closed and disjoint, there are open sets $U_{1} \supset F_{1}$ and $U_{2} \supset F_{2}$ such that $U_{1} \cap U_{2} = \emptyset$. Let $I_{1} = [a,b]$ be a connected component of $[0,1] \smallsetminus U_{1}$ such that $I_{1} \cap F_{2} \neq \emptyset$. By construction, $I_{1}$ is not contained in $F_{2}$, so by connectedness of $I_{1}$ there must be infinitely many $F_{n}$'s such that $F_{n} \cap I_{1} \neq \emptyset$. </p> <p>Replacing $[0,1]$ by $I_{1}$ and the $F_{n}$'s by a (monotone) enumeration of those $F_{n}$ with non-empty intersection with $I_{1}$, we can repeat the argument of the previous paragraph and get $I_{2}$.</p> <p>[In case we have thrown away $F_{3}, F_{4}, \ldots, F_{m}$ in the induction step (i.e, their intersection with $I_{1}$ is empty but $F_{m+1} \cap I_{1} \neq \emptyset$), we put $I_{3}, \ldots, I_{m}$ to be equal to $I_{2}$ and so on.]</p> <hr> <p><strong>Added:</strong> Feb 15, 2011</p> <p>I was informed that a proof of <strong>(2)</strong> appears in C. Kuratowski, <em>Topologie II</em>, &sect;42, III, 6 on p.113 of the 1950 French edition, with essentially the same argument as I gave above. There it is attributed to W.&nbsp;Sierpi&#324;ski, <em>Un th&eacute;or&egrave;me sur les continus</em>, T&ocirc;hoku Mathematical Journal <strong>13</strong> (1918), p. 300-303.</p> http://mathoverflow.net/questions/54111/separability-of-a-certain-space-of-continuous-functions-ii/54114#54114 Answer by Theo Buehler for separability of a certain space of continuous functions, II Theo Buehler 2011-02-02T17:59:57Z 2011-02-02T19:30:24Z <p><strong>No.</strong> Ady's construction still works, I think. Here's another, easier one:</p> <p>Choose an orthonormal basis $\{e_{n}\}$ of $H$. Since $\|e_{i} - e_{j}\| = \sqrt{2}$, the continuous functions $f_{n}(x) = \max\{0, 1 - 2 \cdot \|e_{n} - x\|\}$ have disjoint support and sup-norm $1$. This gives an isometric embedding $\ell^{\infty} \to C_{b}(B)$ by sending a bounded sequence $(a_{n})$ to $\sum a_{n} f_{n}$.</p> <p>A simple modification of this argument shows that $C_{b}(X)$ contains a copy of $\ell^{\infty}$ (and thus isn't separable) whenever the completely regular space $X$ has a countable discrete subset.</p> http://mathoverflow.net/questions/53767/constructions-unique-up-to-non-unique-isomorphism/53784#53784 Answer by Theo Buehler for Constructions unique up to non-unique isomorphism Theo Buehler 2011-01-30T14:41:43Z 2011-01-30T16:01:44Z <p>Here are some examples that are less of an algebraic nature (but all seem to be subsumed by Qiaochu's observation in that they are "weakly initial" or "weakly terminal" objects in appropriate categories):</p> <p>Consider the categories of metric spaces or complete metric spaces and $1$-Lipschitz maps. Isbell has shown that in these categories there are injective hulls, unique up to non-unique isomorphism. A metric space $I$ is injective if for every isometric embedding $A \to B$ and every $1$-Lipschitz map $A \to I$ there exists a $1$-Lipschitz extension $B \to I$. The automorphism groups of the injective hull of a space seems exceedingly hard to determine (even for finite spaces) but there's one case I find interesting. If $M$ happens to be a (real) Banach space and $I(M)$ is its injective hull then $I(M)$ is a Banach space, uniquely determined up to <em>unique</em> linear isometry, and it is of the form $C(K)$ where $K$ is an extremally disconnected Hausdorff space. H. Elton Lacey and co-authors have given a complete (finite!) list of possible injective hulls of separable Banach spaces.</p> <p>Closely related are projective covers in the category of compact Hausdorff spaces and continuous maps. There, the projectives are precisely the extremally disconnected spaces (Gleason).</p> http://mathoverflow.net/questions/53029/smooth-cohomology-of-lie-groups/53034#53034 Answer by Theo Buehler for smooth cohomology of Lie groups Theo Buehler 2011-01-24T10:03:31Z 2011-01-24T10:35:12Z <p>This is indeed very well studied. The standard references are Borel-Wallach, <em>Continuous cohomology, discrete subgroups and representations of reductive groups</em>, Annals of Math. Studies <strong>94</strong>, Princeton University Press (1980) and the more gentle book by A. Guichardet, <em>Cohomologie des groupes topologiques et des alg&egrave;bres de Lie</em> Textes Math&eacute;matiques <strong>2</strong> Fernand Nathan, Paris (1980). But there has been a lot of progress since. For instance, there is some recent work by Crainic extending the theory to Lie groupoids and algebroids.</p> <p>The continuous cohomology is usually attributed to Hochschild-Mostow and coincides with the smooth cohomology under reasonably weak hypotheses.</p> <p>The interesting fact is that there is an intimate interplay between the continuous cohomology of, say a semi-simple Lie group and the cohomology of its Lie algebra and the de Rham cohomology of the associated symmetric space. One of the most important results is the <em>van Est-isomorphism:</em></p> <blockquote> <p>Let $G$ be a semi-simple Lie group with finite center and no compact factors. Let $M = G/K$ be the associated symmetric space. Then the de Rham complex $\mathbb{R} \to \Omega^{0}(M) \to \Omega^{1}(M) \to \cdots$ is an injective resolution of $\mathbb{R}$. In particular, since a $G$-invariant differential form on $M$ is automatically closed, there is a natural isomorphism $H_{c}^{\ast}(G,\mathbb{R}) \cong \Omega^{*}(M)^{G}$ where the right hand side are the $G$-invariant differential forms on $M$. If $A$ is a sufficiently nice smooth $G$-module then the $A$-valued differential forms need not be closed, but one still has $H_{c}^{\ast}(G,A) \cong H^{\ast}(\Omega^{\ast}(M,A)^{G})$ Moreover, one may identify this with <em>relative Lie algebra cohomology</em> $H^{\ast}(\mathfrak{g},\mathfrak{k};A)$.</p> </blockquote> <p>This immediately gives us an interpretation of $H_{c}^{2}(G,\mathbb{R}) = \Omega^{2}(M)^{G}$. Namely, if $G$ is simple, then $\Omega^{2}(M)^{G}$ is one-dimensional if and only if $G$ is Hermitian (when it is generated by the K&auml;hler form on $M$), and otherwise it is zero. So the dimension of $H_{c}^{2}(G,\mathbb{R})$ corresponds to the number of of Hermitian simple factors of $G$.</p> <p>Of course, one may also bring discrete subgroups into play, together with all their rich and beautiful connections to geometry and number theory. There are far too many things to mention here, so I'd better stop now. </p> <p>Before I forget: I don't know of a direct interpretation of $H_{c}^{2}(G,A)$ as equivalence classes of suitable extensions of $G$. One problem is that in the topological category, it is not clear at all that there should be a smooth (or continuous) section of non-trivial extension.</p> http://mathoverflow.net/questions/52881/is-anything-known-about-the-closure-of-an-additive-category-by-adding-all-the-i/52895#52895 Answer by Theo Buehler for Is anything known about the "closure" of an additive category by adding all the images and kernels? Theo Buehler 2011-01-23T01:12:54Z 2011-01-23T03:23:39Z <p><strong>Note:</strong> I'm only addressing David's question how one one can "add cokernels" to an additive category and how one can use this in order to embed an additive category into an abelian one, even in a universal way.</p> <p>Concerning the request on minimality in the question, it is unclear to me what exactly it is, Lev wants to achieve. What happens in the mentioned example of sheaves that one identifies an interesting category (vector bundles = finitely generated projective $\mathcal{O}_{X}$-modules) and ends up with the category coherent sheaves. In this case, the category constructed below embeds into the category of sheaves just because one started out with a class of <em>projective</em> objects in an abelian category already.</p> <hr> <p>The idea is to embed the additive category $\mathcal{A}$ into the category of morphisms $\mathcal{A}^{\to}$. Note that a morphism in $\mathcal{A}^{\to}$ corresponds to a commutative square. More precisely, the idea is that a morphism $(A \to B)$ of $\mathcal{A}$ <em>should</em> represent its cokernel.</p> <p>Let's pretend this works. Since the cokernel of $0 \to A$ in $\mathcal{A}$ is $A$, we see that we must embed $\mathcal{A}$ via $A \mapsto (0 \to A)$. This embedding of $\mathcal{A} \to \mathcal{A}^{\to}$ is clearly fully faithful. But we're not quite there, yet. There are nonzero morphisms of morphisms $(A^{-1} \to A^{0}) \to (B^{-1} \to B^{0})$ that <em>should</em> induce the zero map on the (putative) cokernels, namely precisely those for which the map $A^{0} \to B^{0}$ factors over $B^{-1}$:</p> <p><img src="http://i.imgur.com/1JWSb.png" alt="" title="Hosted by imgur.com" /></p> <p>It is easy to see that these morphisms form an <em>ideal</em> $\mathcal{J}$ in $\mathcal{A}^{\to}$ (they are closed under composition and sums), so we may factor this ideal out. In other words, a morphism of $\mathcal{A}^{\to}$ is identified with zero if and only if it lies in $\mathcal{J}$. The resulting category $\text{fp}(\mathcal{A}) = \mathcal{A}^{\to}/\mathcal{J}$ is what we want because we have:</p> <blockquote> <p><strong>Theorem</strong> (Freyd, 1965; Beligiannis, 2000)</p> <ol> <li>The category $\text{fp}(\mathcal{A})$ has cokernels. The functor $A \mapsto (0 \to A)$ is fully faithful and universal among functors to additive categories with cokernels: more precisely, every functor $F: \mathcal{A} \to \mathcal{C}$ to a category with cokernels extends uniquely to a cokernel-preserving functor $\text{fp} (\mathcal{A}) \to \mathcal{C}$.<br></li> <li>Moreover, $\text{fp}{(\mathcal{A})}$ is abelian if and only if $\mathcal{A}$ has weak kernels (a <em>weak kernel</em> has the factorization property of a kernel but uniqueness of the factorization is not required).</li> </ol> </blockquote> <p>So if $\mathcal{A}$ has weak kernels, we're already done. If not, we may play the same game again, using this theorem. We first embed the category with kernels $(\text{fp} \mathcal{A})^{\text{op}}$ into the abelian category $\text{fp}\left( (\text{fp} \mathcal{A})^{\text{op}} \right)$ and then pass to the opposite category $\mathcal{F}{(\mathcal{A})} = \left(\text{fp}\left( (\text{fp} \mathcal{A})^{\text{op}} \right)\right)^{\text{op}}$. The diagrams get a bit unwieldy, but one may check that:</p> <blockquote> <p><strong>Theorem</strong> (Adelman, 1971) The embedding $\mathcal{A} \to \mathcal{F}(\mathcal{A})$ is fully faithful and universal among all functors to abelian categories (every functor to an abelian category extends to an <em>exact</em> functor on $\mathcal{F}(\mathcal{A})$).</p> </blockquote> <p>The category $\text{fp}(\mathcal{A})$ is sometimes used in connection with the derived category/triangulated categories (the inclusion $\mathcal{T} \to \text{fp}(\mathcal{T})$ is the universal homological functor on the triangulated category $\mathcal{T}$. It already appears in Verdier's thesis - attributed to Freyd). Closely related are also the hearts of $t$-structures (perverse sheaves).</p> <p>The most important references are:</p> <ol> <li><p>Peter Freyd, Representations in abelian categories, Proc. Conf. Categorical Algebra (La Jolla, Calif., 1965), Springer, New York, 1966, pp. 95–120.</p></li> <li><p>A. Beligiannis: ''On the Freyd Categories of an Additive Category'', Homology, Homotopy and Applications Vol. 2, No.11 (2000), pp. 147-185.</p></li> </ol> http://mathoverflow.net/questions/52864/colimit-of-locally-finitely-presented-quasi-coherent-modules/52876#52876 Answer by Theo Buehler for Colimit of locally finitely presented quasi-coherent modules Theo Buehler 2011-01-22T22:08:38Z 2011-01-22T22:08:38Z <p>The answer is <strong>yes</strong>, at least if you believe Thomason-Trobaugh, <em>Higher algebraic $K$-theory of schemes and of derived categories</em>, which David Ben-Zvi already mentioned.</p> <p>I quote from Appendix B.3 (p. 409f):</p> <blockquote> <p>B.3. If $X$ is a quasi-compact and quasi-separated scheme, every sheaf in $\text{Qcoh}(X)$ is a direct colimit of its sub-<code>$\mathcal{O}_{X}$</code>-modules of finite type. Also, every sheaf in $\text{Qcoh}(X)$ is a filtering colimit of finitely presented <code>$\mathcal{O}_{X}$</code>- modules. ([EGA] I 6.9.9, 6.9.12.) In this case, the set of finitely presented $\mathcal{O}_{X}$-modules forms a set of generators for $\text{Qcoh}(X)$, which is then a Grothendieck abelian category and has enough injectives (cf. B.12.).</p> </blockquote> http://mathoverflow.net/questions/52734/on-locally-convex-and-compactly-generated-topological-vector-spaces/52740#52740 Answer by Theo Buehler for On locally convex (and compactly generated) topological vector spaces Theo Buehler 2011-01-21T07:29:21Z 2011-01-22T00:00:04Z <p>Part 1: The "cheeky" answer is: <strong>huge</strong>. There is a left adjoint to the forgetful functor $LCTVS \to Vect$ (in particular there is a left adjoint to the forgetful functor $LCTVS \to Sets$): Equip a vector space $V$ with the locally convex topology induced by <em>all</em> linear functionals on $V$ (or as Pietro Majer put it: the topology given by all semi-norms).</p> <hr> <p><strong>Edit 2:</strong></p> <p>Every linear map $f: V \to W$ is continuous: every semi-norm $|\,\cdot\,|$ on $W$ gives rise to a semi-norm on $V$ by $v \mapsto |f(v)|$. For every net $v_{i} \to v$ we have $|f(v_{i} - v)| \to 0$, hence $f(v_{i}) \to f(v)$ and thus $f$ is continuous.</p> <hr> <p><strong>Edit:</strong> The following summarizes what has transpired from Bill's, Neil's and my answers/comments:</p> <p>Part 2: If $S$ is any set then the space $\ell^{2}(S) = \{\lambda = \sum_{s \in S} \lambda_{s} s\,|\,\sum |\lambda_{s}|^{2} \lt \infty \}$ is a Hilbert space with respect to the scalar product $\langle \lambda, \mu \rangle = \sum_{s \in S} \lambda_{s} \overline{\mu}_{s}$ and it contains the free vector space on $S$. Since metrizable spaces are compactly generated and weakly Hausdorff (see <a href="http://neil-strickland.staff.shef.ac.uk/courses/homotopy/cgwh.pdf" rel="nofollow">N. Strickland's notes</a>, Propositions 1.6 and 1.2), and since the cardinality of $S$ determines the isomorphism type of $\ell^{2}{(S)}$ (see <a href="http://en.wikipedia.org/wiki/Hilbert_space#Hilbert_dimension" rel="nofollow">here</a>), the category of compactly generated locally convex topological vector spaces cannot be essentially small.</p> http://mathoverflow.net/questions/51842/does-a-compact-semilocally-simply-connected-geodesic-space-have-the-homotopy-type/51867#51867 Answer by Theo Buehler for Does a compact semilocally simply connected geodesic space have the homotopy type of a CW complex? Theo Buehler 2011-01-12T19:42:10Z 2011-01-12T19:56:16Z <p>I hope I have not goofed, but I think the answer to your modified question is <strong>yes</strong>:</p> <blockquote> <p>The fundamental group of a semi-locally simply connected, compact and geodesic space is finitely presented.</p> </blockquote> <p>Here are the ingredients - all numbers in parentheses refer to Bridson-Haefliger, <em>Metric spaces of non-positive curvature</em>, Springer Grundlehren, 1999, Part I.</p> <ul> <li>The universal covering space $\widetilde{X}$ equipped with the length metric induced by the covering projection is a length space (3.25).</li> <li>The fundamental group $\pi_{1}(X)$ acts on $\widetilde{X}$ properly and cocompactly by isometries (8.3 (2)), see also (8.5).</li> <li>If a length space $\widetilde{X}$ admits a proper and cocompact action by isometries then it is locally compact (8.4 (1)) and hence proper and geodesic (3.7).</li> <li>A group is finitely presented if and only if it acts properly and cocompactly by isometries on a simply connected geodesic space (8.11).</li> </ul> <p>All this taken together yields that $\widetilde{X}$ is a simply connected geodesic metric space and $\pi_{1}{(X)}$ acts properly and cocompactly by isometries, hence $\pi_{1}{(X)}$ must be finitely presented.</p> http://mathoverflow.net/questions/51225/direct-construction-of-cocontinuous-functors-on-moda/51237#51237 Answer by Theo Buehler for Direct construction of cocontinuous functors on Mod(A) Theo Buehler 2011-01-05T19:47:16Z 2011-01-05T19:55:48Z <p>This is too long for a comment only, therefore I write it into an answer (all categories are assumed to be additive). This will not answer your question but maybe it will provide some insight.</p> <p>Let me first show you how to construct $F$ in the specific situation of $\text{Mod}(A)$ -- this is essentially the proof of the comparison theorem for projective resolutions: You've already explained how to get $g_{2}: M_{2} \to N_{2}$ from a given $f: M \to N$.</p> <p>Let $ZM$ be the image of $M_{1} \to M_{2}$ and let $ZN$ be the image of $N_{1} \to N_{2}$. Since the morphism $ZM \to N$ is zero, there is a unique morphism $h: ZM \to ZN$ such that the diagram</p> <pre><code>M_1 ---&gt;&gt; ZM &gt;--&gt; M_2 ----&gt;&gt; M | | | h g_2 f | | | v v v N_1 ---&gt;&gt; ZN &gt;--&gt; N_2 ----&gt;&gt; N </code></pre> <p>commutes. Now use projectivity of $M_{1}$ to lift $M_1 \to ZN$ up to $g_{1}: M_{1} \to N_{1}$. Note that $g_{2}$ and $g_{1}$ are far from unique. If $g_{2}$ and $g_{2}'$ are two lifts of $f$ then their difference will factor through a map $M_{2} \to ZN$ and you may lift this map to a morphism $M_{2} \to N_{1}$ by using projectivity of $M_{2}$ again. The non-uniqueness of $g_{1}$ need not concern us.</p> <p>We are thus led to the following category $\text{mod}(\mathscr{F}(A))$: Objects are morphisms $(M_{1} \to M_{2})$ with $M_{1}$ and $M_{2}$ free modules in $\text{Mod}(A)$. Two morphisms of arrows (commutative squares) are identified if $g_{2} - g_{2}'$ factors through a morphism $M_{2} \to N_{1}$. Think of an object $(M_{1} \to M_{2})$ as representing its cokernel in $\text{Mod}{(A)}$, two morphisms are identified if and only if they induce the same morphism on the cokernel.</p> <p>By construction, a functor from the category of free modules $\mathscr{F}(A)$ to a category with cokernels $C$ will extend uniquely (up to unique isomorphism) to a functor $\text{mod}(\mathscr{F}(A)) \to C$. Moreover, it is not difficult to check that the inclusion $\mathscr{F}(A) \to \text{Mod}(A)$ extends to an equivalence $\text{mod}(\mathscr{F}(A)) \to \text{Mod}(A)$. So this should be more or less a complete answer for the question of how to get $F$ and its functoriality.</p> <p>The category $\text{mod}(\mathscr{A})$ described above can be defined for every additive category. The notation is motivated by thinking of it as the category of finitely presented functors on $\mathscr{A} \to \text{Mod}{(\mathbb{Z})}$ - that is to say of functors of the form $\text{Coker}(\text{Hom}(-,M_{1}) \to \text{Hom}(-,M_{2}))$. It is abelian if and only if $\mathscr{A}$ has weak kernels (weak kernels are defined as kernels but without requiring uniqueness). A discussion of all this can be found in Freyd, <em>Representations in abelian categories</em>, in the La Jolla proceedings 1966. A more recent exposition is in Beligiannis, <em>On the Freyd categories of an additive category</em>, which you can download from <a href="http://users.uoi.gr/abeligia/hha.pdf" rel="nofollow">his homepage</a>.</p> <p>I do not know how to do this without using that free modules are projective and whether it is helpful for your original motivation.</p> http://mathoverflow.net/questions/50541/quotient-sigma-algebra-on-quotient-space-of-locally-compact-groups/50546#50546 Answer by Theo Buehler for quotient sigma-algebra on quotient space of locally compact groups Theo Buehler 2010-12-28T07:00:31Z 2010-12-28T12:39:26Z <p><strong>Yes</strong>. $\DeclareMathOperator{\calB}{\mathcal{B}}$</p> <p>Since the projection $\pi: G \to G/H$ is continuous, pre-images of Borel sets are Borel, so we have that $\pi^{-1}{(\mathcal{A})}$ is contained in the $\sigma$-algebra $\calB_{H}$ of right $H$-invariant Borel-sets on $G$.</p> <p>To see the reverse inclusion $\calB_{H} \subset \pi^{-1}(\mathcal{A})$, it suffices to show that $\pi(B) \in \mathcal{A}$ for every $B \in \mathcal{B}_{H}$ since for such $B$ we have $\pi^{-1}(\pi(B)) = B$.</p> <p><sub>It is <em>not</em> true that $\pi$ maps Borel sets to Borel sets. The image of a Borel set under $\pi$ will only be <em>analytic</em> in general. Historically, this remark can be seen as the starting point of descriptive set theory: Lebesgue famously assumed in an argument that the projection of a Borel set in $\mathbb{R}^{2}$ to one of its coordinates is Borel in $\mathbb{R}$. Sierpinski gave an example showing that this is wrong.</sub></p> <p>It is a basic but non-trivial result in descriptive set theory that there is a Borel transversal for the right $H$-action on $G$, that is to say, there is a Borel set $T \subset G$ meeting every $H$-coset exactly once $-$ see Kechris, <em>Classical descriptive set theory</em>, Springer GTM vol. 156, Theorem 12.17. Observe that the restriction $\pi: T \to G/H$ is a Borel measurable bijection, hence it maps Borel subsets of $T$ to Borel sets in $G/H$ (see Kechris, Corollary 15.2). In other words, $\pi$ yields a Borel isomorphism $T \to G/H$. For every $B \in \calB_{H}$ we have $\pi(B \cap T) = \pi (B)$ and because $B \cap T$ is Borel we have $\pi(B) = \pi(B \cap T) \in \mathcal{A}$.</p> http://mathoverflow.net/questions/50277/sections-of-topological-group-extensions/50279#50279 Answer by Theo Buehler for Sections of topological group extensions Theo Buehler 2010-12-24T09:54:21Z 2010-12-24T12:07:44Z <p>A very easy example (inspired by an answer by Laurent Moret-Bailly that he removed immediately after posting): Take $0 \to 3\mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/3\mathbb{Z} \to 0$ and define a section $s: \mathbb{Z}/3\mathbb{Z} \to \mathbb{Z}$ by $s(0) = 0$, $s(1) = 1$ and $s(2) = -1$.</p> <p>Here's an example from functional analysis that I happen to like, but it certainly is rather involved:</p> <p>Take a non-complemented subspace $F$ of a Banach space $E$. The quotient map $p:E \to E/F$ admits a continuous and homogeneous right inverse $\sigma$ by a theorem of Bartle-Graves and Michael, see E. Michael "Continuous selections, I", Ann. Math. Vol. 63, No. 2 (Mar., 1956), pp. 361-382, Proposition 7.2. Homogeneity yields in particular $\sigma(-x) = -\sigma(x)$, but $\sigma$ cannot be linear because $F$ is not complemented in $E$.</p> <p>You can take for instance $F = c_{0}$, the space of sequences converging to zero, inside the Banach space $E = \ell^{\infty}$ of bounded sequences.</p> http://mathoverflow.net/questions/49942/killing-cohomology-of-a-complex/49950#49950 Answer by Theo Buehler for Killing cohomology of a complex Theo Buehler 2010-12-20T10:47:56Z 2010-12-20T10:47:56Z <p>Let $\tau^{\geq 1}M$ be the complex $\cdots \to 0 \to M^{0} / \ker{d^{0}} \to M^{1} \to M^{2} \to \ldots$. The obvious map $f: M \to \tau^{\geq 1}M$ yields a triangle $M \to \tau^{\geq 1}M \to C(f) \to M[1]$ and shifting this back by $[-1]$ we get a map $C(f)[-1] \to M$ with the desired properties.</p> <hr> <p>Superfluous remark: If $f: A \to B$, its shifted cone $C(f)[-1]$ could be called the <em>homotopy fiber</em> of $f$, as we get a triangle $A[-1] \to C(f)[-1] \to A \to B$. Now write $\Omega A = A[-1]$...</p> http://mathoverflow.net/questions/49418/nondifferentiability-set-of-an-arbitrary-real-function/49487#49487 Answer by Theo Buehler for Nondifferentiability set of an arbitrary real function Theo Buehler 2010-12-15T05:51:10Z 2010-12-15T05:51:10Z <p>Apparently, continuity is not essential.</p> <p>According to A. Brudno, <em>Continuity and differentiability</em> (Russian), Rec. Math [Mat. Sbornik], N.S. 13 (<strong>55</strong>), (1943), 119–134 (<a href="http://www.ams.org/mathscinet-getitem?mr=12321" rel="nofollow">MathSciNet review here</a>, <a href="http://www.mathnet.ru/php/archive.phtml?wshow=paper&amp;jrnid=sm&amp;paperid=6175&amp;option_lang=eng" rel="nofollow">online article here</a>), the set of non-differentiability of any real function is the union of a $G_{\delta}$-set with a $G_{\delta\sigma}$-set of measure zero.</p> <p>I quote from Brudno's English summary:</p> <blockquote> <p>In the present paper we investigate the structure of the set of points, in which the function of one real variable is not differentiable. All the functions are only supposed to be finite in every point.</p> <p>(...)</p> <p>Theorem IV. In order that the set $Q$ should be the totality of points, in which a function f(x) does not possess a derivative, it is necessary and sufficient that $Q = G_{\delta} + G_{\delta\sigma}$ $(\mathrm{mes}\,G_{\delta\sigma} = 0)$.</p> </blockquote> <p>Since I do not read Russian, I cannot tell you anything about the methods of proof apart from the fact that it involves an investigation of the sets where the Dini derivatives are infinite or distinct.</p> http://mathoverflow.net/questions/118442/reference-for-x-compact-c-bx-separable-x-metric-space Comment by Theo Buehler Theo Buehler 2013-01-09T12:29:52Z 2013-01-09T12:29:52Z I don't have a reference, but I'd suggest this argument: if $X$ is not compact, there is an infinite closed discrete subset $D$ of $X$. For every $A \subset D$ choose a continuous function $f_A \colon X \to [0,1]$ such that $f_A|_A = 1$ and $f_A|_{D \setminus A} = 0$ (by Urysohn). This gives an uncountable family of continuous bounded functions such that $\|f_A - f_B\| = 1$ whenever $A \neq B$. Alternatively, embed $\ell^\infty$ using a similar trick. If $X$ is compact then $C(X)$ is separable by Stone-Weierstrass. http://mathoverflow.net/questions/12169/why-are-abelian-groups-amenable/118434#118434 Comment by Theo Buehler Theo Buehler 2013-01-09T11:37:53Z 2013-01-09T11:37:53Z In the definition of $K$ you should replace $\lVert \Phi\rVert \leq 1$ by $\Phi(\chi_G) = 1$ (which together with $\Phi(F) \geq 0$ implies $\lVert \Phi \rVert = 1$). The way you define it, $0 \in K$ is always a fixed-point, whether $G$ is amenable or not. http://mathoverflow.net/questions/12169/why-are-abelian-groups-amenable/12172#12172 Comment by Theo Buehler Theo Buehler 2013-01-09T10:39:15Z 2013-01-09T10:39:15Z (and the $F$ and $F_n$ must have positive measure, of course). http://mathoverflow.net/questions/12169/why-are-abelian-groups-amenable/12172#12172 Comment by Theo Buehler Theo Buehler 2013-01-09T10:34:29Z 2013-01-09T10:34:29Z I'm not sure what exactly the deleted comments you responded to said, but it is not true that F&#248;lner limits you to discrete groups. A locally compact group $G$ with Haar measure $\mu$ is amenable if and only if for every compact $K$ and every $\varepsilon \gt 0$ there is a compact set $F$ such that $\mu(F \mathbin{\Delta} kF) \leq \varepsilon \mu(F)$ for all $k \in K$. If $G$ happens to be $\sigma$-compact then you can arrange $F$ to be from an exhaustive sequence $F_n$. http://mathoverflow.net/questions/109505/about-the-definition-of-borel-and-radon-measures/117693#117693 Comment by Theo Buehler Theo Buehler 2013-01-04T16:42:38Z 2013-01-04T16:42:38Z Thanks! Meanwhile, I remembered an old construction due to Oxtoby <a href="http://dx.doi.org/10.1090/S0002-9947-1946-0018188-5" rel="nofollow">dx.doi.org/10.1090/S0002-9947-1946-0018188-5</a> which produces a non-trivial (invariant and inner regular) Borel measure on every separable completely metrizable group. If the group is not locally compact then every open set has infinite measure, but there are always sets of finite measure, thus Borel regularity fails. This construction can be adapted to give a non-Borel regular Borel measure even on $\mathbb{R}$, by working on the set of irrationals and using that they are homeomorphic to $\Bbb{Z^N}$. http://mathoverflow.net/questions/117705/the-practical-impact-of-set-theoretic-axioms-on-measure-theory Comment by Theo Buehler Theo Buehler 2013-01-01T23:12:08Z 2013-01-01T23:12:08Z A standard application of Martin's axiom <a href="http://en.wikipedia.org/wiki/Martin%27s_axiom" rel="nofollow">en.wikipedia.org/wiki/Martin%27s_axiom</a> is the existence of Banach limits satisfying some measurability conditions (medial means in the sense of Mokobodzki). See my answer to <a href="http://math.stackexchange.com/q/54554" rel="nofollow">math.stackexchange.com/q/54554</a> for some details and references and part 4 of that answer for a basic sample application that might illustrate their power. http://mathoverflow.net/questions/109505/about-the-definition-of-borel-and-radon-measures/117693#117693 Comment by Theo Buehler Theo Buehler 2013-01-01T22:51:40Z 2013-01-01T22:51:40Z What would be a good example of a non-Borel regular Borel measure on a second countable metric space? All the standard procedures for constructing measures I'm aware of seem to yield measures satisfying this condition. The examples I was able to produce either live on large (i.e. non-separable) spaces or fail to measure all Borel sets. http://mathoverflow.net/questions/117650/continuum-hypothesis/117653#117653 Comment by Theo Buehler Theo Buehler 2012-12-30T19:08:14Z 2012-12-30T19:08:14Z The point of &quot;naturality&quot; is exactly what I meant by emphasizing &quot;algebraic&quot;. Thanks a lot for all the additional pointers and all these interesting links in your answer. http://mathoverflow.net/questions/117602/what-do-i-need-to-understand-wiles-proof-of-fermats-last-theorem Comment by Theo Buehler Theo Buehler 2012-12-30T18:19:24Z 2012-12-30T18:19:24Z I believe this would be the relevant meta thread. <a href="http://meta.mathoverflow.net/discussion/1327/closing-questions-as-spam/" rel="nofollow">meta.mathoverflow.net/discussion/1327/&hellip;</a> http://mathoverflow.net/questions/117650/continuum-hypothesis/117653#117653 Comment by Theo Buehler Theo Buehler 2012-12-30T18:02:15Z 2012-12-30T18:02:15Z Hi Andres. Some minor points: 1) The question asked especially about <i>algebraic</i> equivalents of CH. Do you happen to know one beyond the one mentioned by Mariano in another thread? 2) I think the original article by Erdős is very readable <a href="http://projecteuclid.org/euclid.mmj/1028999028" rel="nofollow">projecteuclid.org/euclid.mmj/1028999028</a> and crystal clear. 3) I would add G&#246;del's Monthly article <a href="http://www.jstor.org/stable/2304666" rel="nofollow">jstor.org/stable/2304666</a> to recommended reading. 4) Since you brought up MA: As a non-set theorist who happened to need to understand it I can recommend Fremlin's book <a href="http://books.google.com/books?id=tXVrPwAACAAJ" rel="nofollow">books.google.com/books?id=tXVrPwAACAAJ</a> Happy New Year! http://mathoverflow.net/questions/117415/old-books-still-used/117421#117421 Comment by Theo Buehler Theo Buehler 2012-12-28T17:20:04Z 2012-12-28T17:20:04Z Maybe Dunford-Schwartz (1957) could be added here? http://mathoverflow.net/questions/117295/automorphisms-of-higher-genus-riemann-surfaces-act-nontrivially-on-homology-refe/117298#117298 Comment by Theo Buehler Theo Buehler 2012-12-27T17:30:16Z 2012-12-27T17:30:16Z Hurwitz's paper is freely available from the G&#246;ttinger Digitalisierungszentrum here: <a href="http://resolver.sub.uni-goettingen.de/purl?GDZPPN002253941" rel="nofollow">resolver.sub.uni-goettingen.de/&hellip;</a> http://mathoverflow.net/questions/117292/why-is-a-ring-called-a-ring/117314#117314 Comment by Theo Buehler Theo Buehler 2012-12-27T12:07:31Z 2012-12-27T12:07:31Z Hi quid. It seems to me that the sources I give in the answer to <i>where does the term “integral domain” come from?</i> are somewhat relevant to your answer (bridging the gap between Hilbert and <i>Moderne Algebra</i>): <a href="http://math.stackexchange.com/q/45945" rel="nofollow">math.stackexchange.com/q/45945</a> The history of the <i>concept</i> of rings (as opposed to the etymology) was also discussed in this thread: <a href="http://math.stackexchange.com/q/362" rel="nofollow">math.stackexchange.com/q/362</a> http://mathoverflow.net/questions/117292/why-is-a-ring-called-a-ring Comment by Theo Buehler Theo Buehler 2012-12-27T00:41:17Z 2012-12-27T00:41:17Z Several lengthy discussion of parts of this question can be found on math.SE, e.g. in this thread: <a href="http://math.stackexchange.com/q/61497" rel="nofollow">math.stackexchange.com/q/61497</a> and the threads mentioned there. http://mathoverflow.net/questions/47747/freyd-mitchells-embedding-theorem/47762#47762 Comment by Theo Buehler Theo Buehler 2012-12-26T16:45:22Z 2012-12-26T16:45:22Z @Axel: thanks again, I finally found the time for doing it.