User wildildildlife - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T06:46:56Z http://mathoverflow.net/feeds/user/10729 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99492/is-there-a-different-construction-of-the-tensor-product-of-two-modules/102069#102069 Answer by wildildildlife for Is there a different construction of "the" tensor product of two modules? wildildildlife 2012-07-12T20:26:01Z 2012-07-12T20:26:01Z <p>In my first encounter with the tensor product of modules (in a course on representation theory by prof. Lenstra), it was done in the following spirit:</p> <p>First, the tensor product $M\otimes _RE$ is defined using the universal property. Next, we prove the following (and other) elementary properties (here I am concentrating on the object and leaving out the morphism):</p> <ul> <li>$M\otimes _RR$ exists, and equals $M$.</li> <li>'$\otimes$ commutes with $\oplus$': if $(M\otimes E_i)_i$ exist, then $M\otimes \oplus_iE_i$ exists and equals their direct sum.</li> <li>'$\otimes$ commutes with $coker$ (right-exactness of $M\otimes_R-$)': Let $f:E\to F$ be $R$-linear, and assume $M\otimes_RE$ and $M\otimes_RF$ exist. Then $M\otimes_R coker f$ exists and equals $coker(M\otimes_RE\xrightarrow{1\otimes f}to M\otimes_RF)$</li> </ul> <p><strong>Theorem</strong>: The tensor product $M\otimes_RE$ exists.</p> <p><strong>Proof</strong>: Take a generating set S of E, i.e. the natural map $f:R^{(S)}\to E$ is surjective. Next pick a generating set T of ker(f), so the natural map $h:R^{(T)}\to R^{(S)}$ has image ker(f). Now $coker(h)=R^{(S)}/\ker f$ is (isomorphic via f to) $E$. </p> <p>By property 1 and 2, $M\otimes_RR^{(T)}$ and $M\otimes_RR^{(S)}$ exist. By property 3, we conclude that $M\otimes_RE$ exists. </p> <p><em>Remark</em>: I guess a didactical merit of this approach (compared to the standard construction as the free abelian group on the product modulo bilinear relations) is that it forces you to think and reason in terms of the universal property and exact sequences. I am not sure if this is also the reason my teachers had in mind. </p> http://mathoverflow.net/questions/99506/blackbox-theorems/99699#99699 Answer by wildildildlife for Blackbox Theorems wildildildlife 2012-06-15T12:23:27Z 2012-06-15T12:23:27Z <p>Someone mentioned existence and uniqueness of Haar measure on a locally compact topological group. But if one uses the Riesz representation theorem and Tychonoff, the standard proof is not so long or hard, and may even be considered conceptual. For example a clear proof is in Bourbaki's Integration, and in Principles of Harmonic Analysis [by Deitmar and Echterhoff]. </p> <p>I think </p> <p>the <a href="http://en.wikipedia.org/wiki/Riesz_representation_theorem" rel="nofollow">Riesz representation theorem</a> (about the dual of $C_c(X)$)</p> <p>is more often used as a Blackbox theorem. Of course this is a main result in analysis, and many standard books (Rudin, Folland, Appendix of Conway's Functional analysis) have a proof, but they are all long and technical, and in my opinion very difficult to remember. See also Remark 4 in these <a href="http://terrytao.wordpress.com/2009/03/02/245b-notes-12-continuous-functions-on-locally-compact-hausdorff-spaces/" rel="nofollow">wonderful notes</a> by Terry Tao.</p> http://mathoverflow.net/questions/99239/non-identity-operator-on-a-separable-hilbert-space/99263#99263 Answer by wildildildlife for non-Identity operator on a separable Hilbert space wildildildlife 2012-06-10T22:42:48Z 2012-06-10T22:42:48Z <p>Nothing new compared to Andreas's answer, just wanted to stress the polarization idea:</p> <p><strong>Notation</strong>: For $H$ a Hilbert space, and $A\in B(H)$ (bounded linear operator), write $q_A$ for the quadratic form $x\mapsto \langle Ax,x\rangle$. </p> <p><strong>Lemma</strong> ('polarization'): If $H$ is a complex Hilbert space, $q_A=q_B\Leftrightarrow A=B$.</p> <p><strong>Proof</strong>: We may assume $B=0$ [replacing $A$ by $A-B$]. If $\langle Ax,x\rangle=0$ for all $x$, then $0=\langle A( x+y),x+y\rangle$ implies $\langle Ax,y\rangle+\langle Ay,x\rangle=0$. But then [replace x by ix] also $\langle Ax,y\rangle-\langle Ay,x\rangle=0$.</p> <p><strong>Answer to question</strong>: Yes, and separability is not needed. Proof by contraposition:</p> <p>If $\lambda:=q_A(x)=q_A(y)$ for all $x,y$ of norm 1, then $q_A(h)=\lambda\|h\|^2=q_{\lambda I}(h)$ for all $h\in H$. Hence $q_A=q_{\lambda I}$, and the lemma implies $A=\lambda I$.</p> http://mathoverflow.net/questions/83881/a-book-in-topology/83895#83895 Answer by wildildildlife for A book in topology wildildildlife 2011-12-19T20:22:25Z 2011-12-19T20:22:25Z <p>Willard's <a href="http://www.amazon.com/General-Topology-Dover-Books-Mathematics/dp/0486434796" rel="nofollow">General Topology</a> is my favourite book on point-set topology (together with Bourbaki, but the latter is not suited as course text for several reasons). It also defines the fundamental group, but doesn't really do anything with it.</p> <p>More geometric is Lee's <a href="http://www.amazon.com/Introduction-Topological-Manifolds-Graduate-Mathematics/dp/0387950265" rel="nofollow">Introduction to Topological Manifolds</a>, it is also very student friendly.</p> http://mathoverflow.net/questions/72318/textbooks-for-pde-between-strauss-and-folland/79379#79379 Answer by wildildildlife for Textbooks for PDE between Strauss and Folland wildildildlife 2011-10-28T12:15:15Z 2011-10-28T12:15:15Z <p>While Evans (which I think fits your requirement) is probably the 'standard' text, I slightly prefer <a href="http://www.amazon.com/Introduction-Partial-Differential-Equations/dp/0387004440" rel="nofollow">Renardy and Rogers' book</a>, which covers roughly the same material.</p> http://mathoverflow.net/questions/65340/paracompact-hausdorff-but-not-compactly-generated/65380#65380 Answer by wildildildlife for Paracompact Hausdorff but not compactly generated? wildildildlife 2011-05-18T22:33:31Z 2011-05-18T22:33:31Z <p>(This should be a comment, but my rep is too low.)</p> <p>It seems that it's certainly Hausdorff, as the topology of $k(X)$ is finer (if $U$ is open in $X$ then $U\cap K$ is open in $K$ for all compacta $K$, by definition of the subspace topology.) So the two separating sets that worked for $X$ still work for $k(X)$. </p> http://mathoverflow.net/questions/31623/an-advanced-exposition-of-galois-theory/50200#50200 Answer by wildildildlife for An advanced exposition of Galois theory wildildildlife 2010-12-22T22:17:44Z 2010-12-22T22:17:44Z <p>Another great set of notes by H.W.Lenstra which discusses algebra in general, and in particular Field and Galois theory at the end, can be found <a href="http://www.math.leidenuniv.nl/~jbrakenh/topics/total.pdf" rel="nofollow">here (pdf)</a>. Two quotes about the approach:</p> <blockquote> <p>We indulge next in a casual and motivational comparison of the classical and modern approaches to Galois theory. In all current textbooks, Galois theory is studied using ﬁnite separable ﬁeld extensions L of a given base ﬁeld K. Our approach follows that of the Grothendieck formulation, in which the objects under consideration are ﬁnite étale K-algebras A. We now consider the relation between the two perspectives.</p> </blockquote> <p>and </p> <blockquote> <p>One can track Galois theory through the years from being a discussion of polynomials, to an exploration of splitting ﬁelds, and ﬁnally to the Grothendieck formulation that we have used in this unit.</p> </blockquote> http://mathoverflow.net/questions/20314/good-books-on-theory-of-distributions/47582#47582 Answer by wildildildlife for Good books on theory of distributions wildildildlife 2010-11-28T12:42:55Z 2010-11-28T12:42:55Z <p>I'd like to point out a recent (Birkhäuser Cornerstones) <a href="http://www.springer.com/birkhauser/mathematics/book/978-0-8176-4672-1?cm_mmc=Google-_-Book%2520Search-_-Springer-_-0" rel="nofollow">textbook</a> on Distribution Theory by Duistermaat and Kolk. </p> <blockquote> <p>The present text has evolved from a set of notes for courses taught at Utrecht University over the last twenty years, mainly to bachelor-degree students in their third year of theoretical physics and/or mathematics. </p> </blockquote> <p>(I have followed this course, which was quite fun.)</p> <p>For a more advanced exposition, Knapp's <a href="http://www.springer.com/birkhauser/mathematics/book/978-0-8176-4382-9" rel="nofollow">Advanced Real Analysis</a> is great.</p> <p>Very complete and advanced (and dry) is Hörmander's <a href="http://www.springer.com/mathematics/analysis/book/978-3-540-00662-6" rel="nofollow">The Analysis of Linear Partial Differential Operators I: Distribution Theory and Fourier Analysis</a>, which has already been mentioned.</p> http://mathoverflow.net/questions/43408/is-this-formulation-of-the-singular-value-decomposition-standard/45573#45573 Answer by wildildildlife for Is this formulation of the Singular Value Decomposition standard? wildildildlife 2010-11-10T17:21:21Z 2010-11-10T17:21:21Z <p>It's in my favourite linear algebra book <a href="http://www.amazon.com/Advanced-Linear-Algebra-Graduate-Mathematics/dp/0387728287" rel="nofollow">Advanced Linear Algebra by Steven Roman</a>, chapter 17 (called Singular Values and the Moore-Penrose inverse).</p> http://mathoverflow.net/questions/103082/recommendation-textbooks-on-d-module Comment by wildildildlife wildildildlife 2012-07-25T12:25:05Z 2012-07-25T12:25:05Z also asked at <a href="http://math.stackexchange.com/questions/174975/recommendation-textbooks-on-d-module" rel="nofollow" title="recommendation textbooks on d module">math.stackexchange.com/questions/174975/&hellip;</a> http://mathoverflow.net/questions/102161/proving-the-existence-of-good-covers Comment by wildildildlife wildildildlife 2012-07-13T19:56:39Z 2012-07-13T19:56:39Z Interesting question. Small comment: the Riemannian argument does not use compactness, but of course it guarantees existence of a <i>finite</i> good cover (in Bott&amp;Tu's terminolgy).