User alekzander - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T05:09:21Z http://mathoverflow.net/feeds/user/707 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/2533/homological-algebra-texts Homological Algebra texts alekzander 2009-10-25T23:37:37Z 2012-05-02T13:12:39Z <p>I would like to hear the communities' ideas on good Homological Algebra textbooks / references. The standard example is of course Weibel (which I'll leave for someone else to describe).</p> <p>As usual, the rule is one reference per post. Please include some description which distinguishes it from other texts.</p> http://mathoverflow.net/questions/63857/interesting-examples-of-adjoint-functors Interesting examples of adjoint functors? alekzander 2011-05-03T22:27:10Z 2011-05-03T22:27:10Z <p>It would be useful to a beginner of category theory to see some examples of adjoint pairs which are not equivalences, and which are not forgetful/inclusion types. Of course, a first is the standard Hom-tensor duality. </p> http://mathoverflow.net/questions/4782/splitting-in-triangulated-categories Splitting in triangulated categories alekzander 2009-11-09T22:35:41Z 2009-11-10T00:25:22Z <p>Using the <a href="http://en.wikipedia.org/wiki/Triangulated%5Fcategory#Definition" rel="nofollow">axioms</a> for a triangulated category, is it possible to prove the following:</p> <blockquote> <p>$A\stackrel{0}{\to}B\to A\oplus B\to$ is a distinguished triangle.</p> </blockquote> <p>From the first axiom, the map <code>0:A--&gt;B</code> extends to its cone, but there is no guarantee I see that the direct sum fits into a triangle. If it does, however, clearly they are (should be?) isomorphic.</p> <p>I have tried using the universal properties of the direct sum, in that finite coproducts and finite products coincide in additive categories, so that I have two diagrams, and extended each of these diagrams into triangles in every way I can imagine, but I think I'm just getting lost in the plethora of sequences. <code>0:A--&gt;B</code> extends to a triangle <code>A--&gt;B--&gt;X--&gt;</code> and so I can get things like $X\to A\oplus B\to \Sigma^{-1}X\cong X$, but by moving away from triangles, I have lost notions of exactness (so that this sequence of maps merely commutes..)</p> <p>I ask this because the proof of Lemma 3.3(2) of <a href="http://www2.math.uni-paderborn.de/fileadmin/Mathematik/AG-Krause/publications%5Fkrause/support.pdf" rel="nofollow">this paper</a> seems to use this without reference.</p> http://mathoverflow.net/questions/4762/in-compatible-gradings-of-an-associative-algebra-tell-us (in-)compatible gradings of an associative algebra tell us...? alekzander 2009-11-09T20:06:16Z 2009-11-09T20:18:33Z <p>If an associative algebra <code>A</code> is $\mathbb{Z}$-graded, then it is automatically $\mathbb{Z}_2$ (aka $\mathbb{Z}/2\mathbb{Z}$) graded by defining $A_{\bar{0}}$ to be the direct sum over the even graded elements of <code>A</code>, and $A_{\bar{1}}$ to be the direct sum over the odds. Conversely, what may be said to distinguish those algebras <code>A</code> for which a $\mathbb{Z}_2$-grading exists, but no compatible $\mathbb{Z}$-grading exists? (Of course, compatible in the sense that the induced grading just described matches the given one.)</p> <p>My motivation is the study of the <code>B(0,n)</code> (aka <code>osp(1|2n)</code>) series of Lie superalgebras, which I have been told cannot be $\mathbb{Z}$-graded, thus making their study a bit different from several of the other classes of Lie superalgebras.</p> <p>The question can also be generalized greatly from above, and I think this is the right generalization. Say $\pi:R\to S$ is a surjection of abelian groups, so that an associative algebra <code>A</code> graded over <code>R</code> is automatically graded over <code>S</code>. What properties does an algebra have if it has an <code>S</code> grading, but no compatible <code>R</code> grading?</p> <p>For a boring example, the associative algebra $\mathbb{Z}_2$ (with itself as base field) is graded over itself as an abelian group, but clearly cannot be $\mathbb{Z}$-graded. In fact, any finite associative algebra with nontrivial grading over $\mathbb{Z}_2$ cannot be given a compatible $\mathbb{Z}$ grading.</p> http://mathoverflow.net/questions/2931/localizations-of-categories Localization(s) of Categories alekzander 2009-10-27T23:09:42Z 2009-10-28T16:57:04Z <p>I've been trying to read a <a href="http://www2.math.uni-paderborn.de/fileadmin/Mathematik/AG-Krause/publications%5Fkrause/support.pdf" rel="nofollow">paper</a> by Krause and came across a strange (to me, of course) notion of localization. After looking around for a long time, and then finding <a href="http://www2.math.uni-paderborn.de/fileadmin/Mathematik/AG-Krause/publications%5Fkrause/local.pdf" rel="nofollow">this</a> on his site, I see that there are two notions for localization, both with significant usage online. These are namely <strong>Verdier localization</strong> and <strong>Bousfield localization</strong>. <em>Is there a strong motivation to use one over the other?</em> A little bit of context:</p> <p>I see that Bousfield localization is defined for model categories, and this includes the notion of modules over a ring, among many many others. I don't see a similar restriction for the Verdier localization.</p> <p>Verdier localization uses the (standard for ''localization'') idea of a multiplicative set <code>S</code> of maps which are formally inverted by a functor <code>Q</code> from a category <code>T</code> to a new category denoted <code>T/S</code>. Hartshorne's Residues and Duality is a reference for this. (BTW, where does the assumption that the pullback of a multiplicative map is multiplicative come from?)</p> <p>Bousfield localization is stated in several places (such as the Krause reference above) as a Verdier localization composed with a right adjoint for <code>Q</code>, which I understand to mean a functorial way of choosing objects in the isomorphism classes, and maps in the multiplicative subsets of each <code>Hom(A,B)</code>. It is also stated in the generality of model categories as needing three distinguished collections of morphisms: namely quasi-isomorphisms and (co)fibrations. What bothers me more is the definition as given in Krause: an <strong>exact</strong> functor <code>L</code> from a <strong>triangulated category</strong> <code>T</code> to itself for which there exists a natural transformation &eta;<code>:Id--&gt;L</code> which commutes with <code>L</code> (&eta;<code>L=L</code>&eta;) and for which &eta;<code>L</code> is invertible. As a second, smaller, question, what is encoded by the commutative condition (what would be lost without it?)? I can come up with contrived examples (using the automorphisms of the objects <code>LX</code>) of course, but in what precise way does &eta; really just encode <code>L</code> as a natural transformation?</p> http://mathoverflow.net/questions/2530/elementary-ext1-intuition elementary Ext^1 intuition alekzander 2009-10-25T23:30:44Z 2009-10-27T22:46:20Z <p>I am wondering what sort of basic basic intuitive meaning <code>Ext</code><sup><code>1</code></sup><code>(M,N)</code> has.<br /> As a base case: if <code>M</code> and <code>N</code> are say, (finite-dimensional) vector spaces (with a compatible group/algebra action), and <code>M</code> and <code>N</code> are indecomposable inequivalent (so <code>Hom(M,N)={0}</code>), can I somehow conclude that <code>Ext</code><sup><code>1</code></sup><code>(M,N)</code> is zero? </p> <p>All I can get from Weibel/Wikipedia is that <code>Ext</code><sup><code>1</code></sup><code>(M,N)</code> is a group under the Baer sum operation, and is in bijection with the set of solutions <code>{X}</code> to the short exact sequence <code>0 --&gt; N --&gt; X --&gt; M --&gt; 0</code>. I don't know how to use this second meaning, but it seems the most hands-on.</p> <p>Full disclosure: If this sounds like a homework exercise, it (almost) <strong>was</strong> -- past tense. Although, the problem/text/instructor had no desire for use of <code>Ext</code> or <code>Hom</code>, I just want to know how to use these functors (better). I could give character references (even from some past Berkeley grads) to allay fears....</p> <p>I would be happy to have a good reference to look this up myself. I hear Rotman's first book was good, but I've only read negative responses to the new edition (and the old one isn't for sale anywhere I've seen), and Weibel is apparently too abstract for me, in some way. I'll post a separate question for that, in fact.</p> http://mathoverflow.net/questions/9037/how-is-it-that-you-can-guess-if-one-of-a-pair-of-random-numbers-is-larger-with-p/9164#9164 Comment by alekzander alekzander 2009-12-19T06:28:15Z 2009-12-19T06:28:15Z Your argument on information bothers me a little bit. I agree that probabilities calculated vary with information available, of course, but is the information given in this problem <i>relevant</i> to the problem at hand? E.g.: the Monty Hall problem tells us &quot;not that one&quot;, which has application to what we were considering in the first place. http://mathoverflow.net/questions/9163/whats-the-nearest-algebraic-theory-to-inner-product-spaces Comment by alekzander alekzander 2009-12-19T06:06:12Z 2009-12-19T06:06:12Z Then a field can be stated (up to isomorphism) as a one-dimensional object in the category of vector spaces over that field? What I mean is, we can single out a field as an object in this category of algebraic objects.. thus making it algebraic? I'm still missing this distinction. If a vector space does not encode the constraints on its scalars, how is it not just a module? http://mathoverflow.net/questions/9163/whats-the-nearest-algebraic-theory-to-inner-product-spaces Comment by alekzander alekzander 2009-12-18T05:59:59Z 2009-12-18T05:59:59Z Maybe I'm missing something obvious, but I don't see why vector spaces are algebraic if fields aren't, since we require the field axioms as a subset. http://mathoverflow.net/questions/8056/what-are-good-non-english-languages-for-mathematicians-to-know/8813#8813 Comment by alekzander alekzander 2009-12-14T04:00:44Z 2009-12-14T04:00:44Z It may be a shame that my motivation to learn Chinese (Mandarin, I assume) for this reason will probably always stay low. Without knowing some of the language, and with institutions seeing low demand to subscribe to anything academic in Chinese, my own opportunity to come face-to-face with the language and its usefulness in math will probably remain identically zero. http://mathoverflow.net/questions/8290/why-is-the-decomposition-theorem-awesome/8294#8294 Comment by alekzander alekzander 2009-12-09T02:49:10Z 2009-12-09T02:49:10Z This shouldn't be an answer, but a comment to the question above. http://mathoverflow.net/questions/8189/cardinality-of-equivalence-classes-of-cauchy-sequences/8265#8265 Comment by alekzander alekzander 2009-12-08T23:50:19Z 2009-12-08T23:50:19Z Violin, Cello, Bass?! http://mathoverflow.net/questions/8056/what-are-good-non-english-languages-for-mathematicians-to-know/8094#8094 Comment by alekzander alekzander 2009-12-07T16:13:22Z 2009-12-07T16:13:22Z I have to upvote the second half. I think I've accepted the fact that learning a new language for reading is excessive. (Opportunity cost high, likelihood of forgetting high, frequency of usefulness low.) Instead, I tend to make fairly quick progress whenever I try to read, and a (paper!) dictionary quickly finishes the process. Reading math is slow, anyway. http://mathoverflow.net/questions/8056/what-are-good-non-english-languages-for-mathematicians-to-know/8058#8058 Comment by alekzander alekzander 2009-12-07T07:28:52Z 2009-12-07T07:28:52Z Jantzen's Moduln Mit Einem Hochsten Gewicht is a '79 text which I would still love to read, and which I think has no (English) translation available. http://mathoverflow.net/questions/7389/what-are-the-most-overloaded-words-in-mathematics/7408#7408 Comment by alekzander alekzander 2009-12-01T21:03:53Z 2009-12-01T21:03:53Z It seems to me that Abel(ian) gets thrown around as an adjective much more than Riemannian or Eulerian -- so much so that I often forget that it's a name and really do think of it as an adjective all on its own. http://mathoverflow.net/questions/5499/which-mathematicians-have-influenced-you-the-most/5550#5550 Comment by alekzander alekzander 2009-11-26T06:28:11Z 2009-11-26T06:28:11Z Might you want to say what he has contributed, for those of us who might not recognize the name? (Is he the same as in the Peirce Decomposition for a ring in terms of its idempotents?) http://mathoverflow.net/questions/6675/periods-and-commas-in-mathematical-writing/6703#6703 Comment by alekzander alekzander 2009-11-24T23:55:00Z 2009-11-24T23:55:00Z I'm surprised to not see more comments in this direction: I tend to recast sentences so that my formulae need no punctuation IF it appears confusing when I first type it; the only exception is that of Simon's last example -- namely, I feel it forgivable to end a sentence with no punctuation in the context of an offset formula following a colon. http://mathoverflow.net/questions/6675/periods-and-commas-in-mathematical-writing/6739#6739 Comment by alekzander alekzander 2009-11-24T23:52:16Z 2009-11-24T23:52:16Z I can be sure that there would be very few disagreements that it is bad form to start a sentence with a mathematical statement. http://mathoverflow.net/questions/6376/why-forgetful-functors-usually-have-left-adjoint/6406#6406 Comment by alekzander alekzander 2009-11-23T09:22:24Z 2009-11-23T09:22:24Z If you're just talking about the adjectives left/right, well .. that's terminology, sure, since you're talking about .. terms. There was some usage historically of &quot;adjoint&quot;/&quot;coadjoint&quot;, but it's unnatural AFAIAC to say that one is in the &quot;correct&quot; (I guess I have to avoid the word &quot;right&quot; here) direction. However, we do have &quot;unit&quot;/&quot;counit&quot;, so it wouldn't be a stretch to associate &quot;adjoint&quot; with &quot;unit&quot;, and co-. I think this isn't done mostly because it was historically not used consistently this way (or that's what I gleaned from a comment in MacLane). http://mathoverflow.net/questions/6376/why-forgetful-functors-usually-have-left-adjoint/6380#6380 Comment by alekzander alekzander 2009-11-23T01:47:15Z 2009-11-23T01:47:15Z I don't think I've ever seen it defined as such, but it has surely been strongly hinted at. http://mathoverflow.net/questions/6376/why-forgetful-functors-usually-have-left-adjoint/6406#6406 Comment by alekzander alekzander 2009-11-23T01:46:43Z 2009-11-23T01:46:43Z In what way do you mean &quot;mostly just terminology&quot;? As Mark Hovey's topological example (an exercise in MacLane, also) points out, one may have both a left and a right adjoint which are <i>clearly</i> distinct.