most recent 30 from http://mathoverflow.net 2013-05-22T05:01:58Z http://mathoverflow.net/feeds/user/182 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/2115/what-does-a-projective-resolution-mean-geometrically Justin DeVries 2009-10-23T16:52:26Z 2011-01-29T04:15:19Z

<p>For R a commutative ring and M an R-module, we can always find a projective resolution of M which replaces M by a sequence of projective R-modules. But as R is commutative, we can consider the affine variety X=Spec R and the sheaf of modules associated to M. What is the projective resolution doing geometrically to this sheaf?</p> <p>Projectives are locally free sheaves, so if M itself is not projective then it must have some sort of "sharp twisting" or "pinching". In some way a projective resolution is "un-pinching" M. Geometrically, is this the same "un-pinching" that happens in a resolution of a singularity of a variety? Is there an example in low dimensions where one can actually draw a picture of this happening for modules?</p>

http://mathoverflow.net/questions/35388/a-chain-homotopy-that-does-not-arise-from-a-homotopy-of-spaces/35395#35395 Justin DeVries 2010-08-12T20:48:37Z 2010-08-12T20:48:37Z

<p>A bounded below complex of free $R$-modules is acyclic if and only if it is contractible (in the sense that the identity map is chain homotopic to zero). Since the singular chain complex of a space is constructed out of free $\mathbb{Z}$-modules, any space with no homology would have to contract to a point, which is not the case (check out the <a href="http://en.wikipedia.org/wiki/Acyclic_space" rel="nofollow">wikipedia page</a> for some examples/references).</p> <p>Thus the chain homotopy exhibiting contractibility of the identity does not come from a topological homotopy.</p>

http://mathoverflow.net/questions/652/homological-algebra-and-calculus-as-in-newton/663#663 Justin DeVries 2009-10-15T21:22:02Z 2010-02-01T02:43:40Z

<p>A while ago I worked on the question of what we can say if $d^n=0$, but I got distracted by more concrete problems. A few people have certainly thought about this question. One place to start looking is "$d^N=0$: generalized homology" or "Generalized homologies for $d^N=0$ and graded $q$-differential algebras" both by Michel Dubois-Violette.</p> <p>(Sorry for the lack of links; I'm off-campus so I can't actually get to the MathSciNet entries right now.)</p>

http://mathoverflow.net/questions/6704/how-to-think-about-cm-rings/6720#6720 Justin DeVries 2009-11-24T19:25:27Z 2009-11-24T19:25:27Z

<p>Geometrically, depth is measuring "dimension" via hypersurfaces. The set of zero-divisors is the union of the associated primes, so to say that an element x is a non-zero-divisor is to say that it is not contained in any associated prime. Thus the hypersurface $V(x)$ does not intersect $M$ in a component.</p> <p>The other condition on a regular sequence is that $xM\neq M$, which amounts to saying that the hypersurface $V(x)$ must intersect $M$ somewhere.</p> <p>So basically, you're cutting down $M$ by hypersurfaces. Since they can't intersect in a component, they actually do cut it down by some amount, and since they must intersect somewhere they aren't throwing everything away at once. This is a very loose description, but I don't have time right now to make it more precise.</p> <p>If you look at quotients of polynomial rings you can actually see this at work. Here you can compute depth by drawing pictures: take an ideal $I$ in $k[x,y,z]$ say, and look at $V(I)$. Find some hypersurface (a plane in this case) that intersects $V(I)$ but not in a component. Then repeat on this intersection. Using this you should be able to find the classic example of a regular sequence that does not stay regular under permutation. You can also convince yourself that in a local ring, all permutations <em>are</em> regular.</p> <p>In this interpretation, CM rings are exactly those for which the dimension can be measured by using hypersurfaces in this way.</p> <p>I'm somewhat rushing to catch a flight (bad time to look at mathoverflow!), so there may be some mistakes but the idea is sound.</p>

http://mathoverflow.net/questions/3951/memorizing-theorems/3955#3955 Justin DeVries 2009-11-03T16:47:36Z 2009-11-03T16:47:36Z

<p>Don't memorize theorems. That said, if you want to remember what a theorem is saying then there are a few things I find helpful:</p> <ul> <li><p>Try it out in a computable example. If it's a classification theorem, pick some object and follow the steps of the proof on your chosen object.</p></li> <li><p>Build examples and counter-examples. The theorem likely has some conditions where it applies and doesn't apply. Try to figure out what examples force the hypotheses of the theorem.</p></li> <li><p>Try to remove hypotheses. Maybe you can't find counter-examples for the hypotheses of the theorem because there aren't any! See if you can tweak the proof a little to remove a hypothesis.</p></li> </ul> <p>After you've gone through a few of these you'll find yourself much more familiar with the theorem and its proof, and (hopefully) you'll find it easy to remember it.</p>

http://mathoverflow.net/questions/1750/can-anyone-give-me-a-good-example-of-two-interestingly-different-ordinary-cohomol/1754#1754 Justin DeVries 2009-10-21T22:39:25Z 2009-10-21T22:39:25Z

<p>An example I recently came across while learning about shape theory is the <a href="http://en.wikipedia.org/wiki/Topologist%27s%5Fsine%5Fcurve" rel="nofollow">closed topologist's sine curve</a>. The zero-th singular cohomology (integer coefficients) is Z^2, while the zero-th Cech cohomology (again integer coefficients) is Z.</p> <p>I can't give a great reason why this is so as I'm just learning about the technicalities myself, but it seems to be a standard example.</p>

http://mathoverflow.net/questions/46541/how-to-introduce-notions-of-flat-projective-and-free-modules/46544#46544 Justin DeVries 2010-11-18T23:42:52Z 2010-11-18T23:42:52Z

@Pete: In my mind, the exactness of the tensor product is relatively easy to motivate through change of rings. Say you've got your sequence over $R$ and you've got an ideal $I$. Is it still exact when we think of everything as $R/I$ modules? Well, that's what the exactness of the tensor product will tell you. Of course this becomes very useful once you have localization and completions to play with. For $Hom$, it seems easiest to motivate $Hom(-,R)$ by analogy with vector space duals, and you can build up from there to $Hom(-,M)$ and $Hom(M,-)$.

http://mathoverflow.net/questions/10255/effective-teaching/10263#10263 Justin DeVries 2009-12-31T17:01:20Z 2009-12-31T17:01:20Z

There is a recent book from the MAA titled &quot;The Moore method : a pathway to learner-centered instruction&quot; that describes five teachers' variations on the Moore method. It may be helpful to look through it for ideas.

http://mathoverflow.net/questions/5357/theorems-for-nothing-and-the-proofs-for-free/5388#5388 Justin DeVries 2009-11-13T16:59:11Z 2009-11-13T16:59:11Z

That's quite surprising! What are some good references for this?

http://mathoverflow.net/questions/1827/what-representative-examples-of-modules-should-i-keep-in-mind/1846#1846 Justin DeVries 2009-11-06T18:24:31Z 2009-11-06T18:24:31Z

@Dinakar: the top part of the picture is a module M. The labels M_P and M_Q denote the localizations at primes. The bottom part of the picture is the ring R, with primes P and Q labeled. The curves drawn in R represent the support of M. Presumably the thicker lines are indicating embedded primes (non-minimal associated primes), and eta is a generic point (non-maximal prime).

http://mathoverflow.net/questions/2115/what-does-a-projective-resolution-mean-geometrically/2163#2163 Justin DeVries 2009-10-23T20:21:51Z 2009-10-23T20:21:51Z

In my mind, the easiest way for a sheaf to not be locally free is for some point p to lack a local trivialization. For vector bundles I can visualize this happening when the stalks do not vary nicely: at p they change directions suddenly (what I meant by &quot;sharply twisting&quot;) or the dimension of the stalks change (&quot;pinching&quot; of some sort). This might be confusing smoothness with local free-ness; I'm not really sure what's going on geometrically - hence my question. I feel like the maps in the resolution can be described geometrically rather than algebraically in some way. I just don't know how.