User arnav tripathy - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T01:11:24Z http://mathoverflow.net/feeds/user/7690 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60166/rational-map-to-a-non-uniruled-manifold/60208#60208 Answer by Arnav Tripathy for Rational map to a non-uniruled manifold Arnav Tripathy 2011-03-31T17:08:15Z 2011-03-31T17:08:15Z <p>I think you have already answered this yourself, for the most part. We have a birational morphism $X' \to X$ with some exceptional locus $E$ inside $X'$, which is some finite union of ruled components $\mathbb{P}^1 \times Z_i$. We also have the (proper) map $f': X' \to Y$, and the restriction of the map to any $\mathbb{P}^1 \times Z_i \to Y$ does not factor through $\mathbb{P}^1 \times Z_i \to Z_i$, so the image inside $Y$ will be uniruled. As such, the image of $E$ inside $Y$ is some closed set $V$ which by hypothesis cannot be all of $Y$. Restricting to the complement $U \subset Y$, the map $f'^{-1}(U) \to U$ has domain away from the exceptional locus inside $X'$, which is what you wanted, yes?</p> <p>Also, the analysis you gave in your answer almost works, I think, except there is a slight problem in that you can only take $p$ to be very general (in the complement of a countable union of closed proper subsets) rather than merely general. For example, a very general K3 surface has infinitely many rational curves despite not being uniruled, so the locus in $Y$ to which your analysis applies would be precisely the complement of these infinitely many rational curves.</p> http://mathoverflow.net/questions/23643/books-about-history-of-recent-mathematics/55900#55900 Answer by Arnav Tripathy for Books about history of recent mathematics Arnav Tripathy 2011-02-18T19:51:06Z 2011-02-18T19:51:06Z <p>The degree to which you like this answer is very dependent on the degree to which you find physics even in its own right of interest to mathematicians, but Abraham Pais is a wonderful historian of (relatively) recent physics; I'm most familiar with his biography of Einstein (Subtle is the Lord) and his history of particle physics (Inward Bound), but I understand he has many more books as well.</p> http://mathoverflow.net/questions/52665/survey-article-on-intersection-theory/52705#52705 Answer by Arnav Tripathy for Survey article on Intersection Theory Arnav Tripathy 2011-01-21T01:51:53Z 2011-01-21T01:51:53Z <p>Eisenbud and Harris are coming out with a book on intersection theory, "3264 and all that", and if you know Harris's style at all, you'll know it's chock full of down-to-earth examples that should be right along the lines of what you're looking for. (Sorry to recommend a book that's not strictly speaking published yet, but it does sound like exactly what you're asking!)</p> http://mathoverflow.net/questions/52191/is-a-fibration-in-algebraic-geometry-a-fibre-bundle Is a fibration in algebraic geometry a fibre bundle? Arnav Tripathy 2011-01-15T22:50:01Z 2011-01-16T01:11:04Z <p>Let $X \to B$ be a smooth, proper, dominant map of schemes over $\text{Spec }k$ an algebraically closed field of characteristic zero with $B$ integral. We have the generic fibre $\overline{F}$ defined over $\text{Spec }\overline{K(B)}$ and by base-changing along $\text{Spec }\overline{K(B)} \to \text{Spec }k$, we obtain a map $\overline{X} \to \overline{B}$ such that we can now write down the sequence $\overline{F} \to \overline{X} \to \overline{B}$. To what extent is this a fibre bundle? To ask a definite question, is there some etale map $\overline{B}' \to \overline{B}$ such that further pulling back will yield an isomorphism $\overline{X}' \simeq \overline{F}' \times \overline{B}'$? </p> <p>This question is closely related to <a href="http://mathoverflow.net/questions/20184/flatness-in-algebraic-geometry-vs-fibration-in-topology" rel="nofollow">http://mathoverflow.net/questions/20184/flatness-in-algebraic-geometry-vs-fibration-in-topology</a> and <a href="http://mathoverflow.net/questions/28162/is-an-algebraic-geometers-fibration-also-an-algebraic-topologists-fibration" rel="nofollow">http://mathoverflow.net/questions/28162/is-an-algebraic-geometers-fibration-also-an-algebraic-topologists-fibration</a>. In particular, it is motivated by (1) Ehresmann's theorem that locally analytically such a morphism should be a (topological) fibre bundle and (2) the fuzzy thinking that "locally analytically" should mean "after an etale base change", but I feel like the answer to the question I posed it above is probably in the negative. For example, it seems unlikely to me that two smooth hypersurfaces of degree $d$ in $\mathbb{P}^n$ which are (automatically) diffeomorphic but not isomorphic should suddenly become isomorphic after an etale base change. However, I don't know of any weaker way to algebro-geometrically state the condition that some map be a fibre bundle, however -- is there anything then that we can say algebro-geometrically with respect to the above maps, or do we have to be content with the differential-geometric statement that it's a fibre bundle in that category?</p> http://mathoverflow.net/questions/22927/why-worry-about-the-axiom-of-choice/22938#22938 Comment by Arnav Tripathy Arnav Tripathy 2011-07-28T15:21:29Z 2011-07-28T15:21:29Z Could you sketch what you mean by anomalies' ? Is the idea simply that we have more interesting behaviour in objects internal to these categories than in the similarly defined objects internal to simpler' categories, or is there some more precise notion being hit upon (in particular, I would love to know if you're making reference to the anomalies in quantum theory !) ? http://mathoverflow.net/questions/58500/is-every-abelian-variety-a-direct-summand-of-the-jacobian-of-a-curve/58502#58502 Comment by Arnav Tripathy Arnav Tripathy 2011-03-15T05:17:06Z 2011-03-15T05:17:06Z Ah yes, everything I said just holds working in abelian varieties with principal polarisations, and I am echoing Ben Wieland's comment. http://mathoverflow.net/questions/54769/is-there-evidence-whether-undergraduate-math-courses-improve-problem-solving Comment by Arnav Tripathy Arnav Tripathy 2011-02-08T15:11:32Z 2011-02-08T15:11:32Z ...for, at least at the very basic level that we're talking about when we discuss &quot;courses required for graduating&quot;, is being able to easily see flaws in reasoning, gaps in arguments, things that follow 'tautologically', and so forth. http://mathoverflow.net/questions/54769/is-there-evidence-whether-undergraduate-math-courses-improve-problem-solving Comment by Arnav Tripathy Arnav Tripathy 2011-02-08T15:10:33Z 2011-02-08T15:10:33Z I'm surprised to hear that cited as the justification -- in my humble opinion, learning mathematics, especially at an undergraduate level, is less about developing problem-solving skills and more about honing logical acuity. For example, I am not infrequently asked which courses exactly one should take to develop skill at the Putnam (presumably one of the most elementary examples of mathematical problem-solving), to which I hem and haw and then say to just take some standard classes so you know the material and then go work some problems instead. What I think a mathematics education IS good... http://mathoverflow.net/questions/54215/group-theory-problem Comment by Arnav Tripathy Arnav Tripathy 2011-02-03T18:02:47Z 2011-02-03T18:02:47Z also, the only finite subgroup of the group of complex numbers is trivial ;-) http://mathoverflow.net/questions/53700/is-there-a-way-to-graphically-imagine-smash-product-of-two-topological-spaces Comment by Arnav Tripathy Arnav Tripathy 2011-01-29T15:34:34Z 2011-01-29T15:34:34Z What do you mean by a pure geometric proof? The obvious geometric intuition seems to work -- imagine $S^n$ as the one-point compactification of $\mathbb{R}^n$ (with the point at infinity as the base-point), so that when you smash two of these gadgets together, you get your larger Euclidean space $\mathbb{R}^{n+m}$ with a lot of stuff at $\infty$ that all gets contracted together. Of course it is hard to visualise except for $n + m \le 3$, but the idea is clear from those cases. http://mathoverflow.net/questions/52191/is-a-fibration-in-algebraic-geometry-a-fibre-bundle/52195#52195 Comment by Arnav Tripathy Arnav Tripathy 2011-01-16T13:56:04Z 2011-01-16T13:56:04Z Thanks for your answer! Yes, I was fairly sure asking the moduli to simply collapse after an etale base change would be quite nonsense, but that was really just the only example I knew of to indicate what an algebro-geometric fibre bundle might be like; I am interested in whether there is any possible weaker algebro-geometric statement along those lines.