User daniel litt - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T13:57:42Z http://mathoverflow.net/feeds/user/6950 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130127/etale-cohomology-of-punctured-spectra-of-local-rings Etale Cohomology of Punctured Spectra of Local Rings Daniel Litt 2013-05-08T23:13:27Z 2013-05-09T12:00:05Z <p>Let $R=\mathbb{C}[[x,y]]$ be a power series ring in two variables (or maybe more generally a strictly Henselian local ring) with maximal ideal $\mathfrak{m}$. </p> <blockquote> <p>What is <code>$H^*_{et}(\operatorname{Spec}(R)\setminus\{\mathfrak{m}\}, \mathbb{G}_m)$</code>?</p> </blockquote> <p>My motivation is this: I'm trying to understand the extent to which etale cohomology of $\mathbb{G}_m$ resembles the cohomology of $\mathcal{O}_X^*$ in the complex-analytic setting. For example, one might expect that in the case above, $H^2$ ought to be $\mathbb{Z}$.</p> <p>I'd be satisfied with an answer that computes $H^2$.</p> http://mathoverflow.net/questions/122509/two-quetions-on-complex-geometry/122514#122514 Answer by Daniel Litt for Two quetions on complex geometry. Daniel Litt 2013-02-21T07:57:28Z 2013-02-21T09:05:33Z <p>(1) Why is the existence of an almost-complex structure a topological question? Suppose $M$ is a $2n$-manifold. The tangent bundle is classified by some map $M\to BGL_{2n}(\mathbb{R})$; $M$ admits an almost-complex structure if and only if this map admits a lift to $BGL_{n}(\mathbb{C})$ (that is, an almost-complex structure is the same as endowing the tangent bundle with the structure of a complex vector bundle). The existence of such a lift depends only on the homotopy type of the map $M\to BGL_{2n}(\mathbb{R})$, and is thus a topological question. As such, it can be analyzed via standard methods in obstruction theory, which I will leave for you to google.</p> <p>(2) Suppose we a projective space $\mathbb{P}V$, where $V$ is some vector space of dimension $n+1$. Then a map $X\to\mathbb{P}V$ is the same as a line bundle $\mathcal{L}$ on $X$ and a surjective map $V\otimes \mathcal{O}_X\to \mathcal{L}$. The identity map on $\mathbb{P}V$ classifies such data---namely, a map <code>$V\otimes \mathcal{O}_{\mathcal{P}V}\to \mathcal{O}(1)$</code>. Explicitly, this map is given by identifying the global sections of $\mathcal{O}(1)$ with $V$, via the standard computation of the cohomology of line bundles on projective space.</p> <p>Now your map is dual of this map. (Picking a basis $(x_i)$ of $V=\Gamma(\mathbb{P}V, \mathcal{O}(1))$, this map is given by multiplication by $(x_i)$.) </p> <p>By the way, the Euler exact sequence admits a very geometric interpretation. Namely, there is a sequence of maps $\mathbb{A}^{n+1}\setminus{0}\to \mathbb{P}^n\to \operatorname{pt}$, where the first map is the usual quotient by $\mathbb{C}^*$. This induces a short exact sequence of cotangent bundles <code>$$0\to \pi^*\Omega^1_{\mathbb{P}^n}\to \Omega^1_{\mathbb{A}^{n+1}\setminus\{0\}}\to \Omega^1_{(\mathbb{A}^{n+1}\setminus\{0\})/\mathbb{P}^n}\to 0.$$</code></p> <p>Each of these sheaves admits an (equivariant) action of $\mathbb{C}^*$, and so descend to sheaves on $\mathbb{P}^n$, giving exactly the Euler exact sequence.</p> http://mathoverflow.net/questions/121136/nice-algebraic-approximations-of-classifying-spaces/121146#121146 Answer by Daniel Litt for Nice algebraic approximations of classifying spaces Daniel Litt 2013-02-08T03:21:26Z 2013-02-08T03:21:26Z <p>This <a href="http://arxiv.org/pdf/math/9802097.pdf" rel="nofollow">beautiful paper of Totaro</a> does much of what you ask. Namely, choose a representation $V$ of $G$ so that $G$ acts freely on $V$ outside of a codimension $\geq s$ subset $S$. Then Totaro shows that the Chow ring of $X=(V-S)/G$ (the geometric quotient) is independent of $V$ and $S$, in degree less than $s$. Furthermore, he shows that such pairs $(V, S)$ exist with $s$ arbitrarily large. So the set of such $(V-S)/G$ form a reasonable approximation to $BG$ (and in particular, give a reasonable definition of the Chow ring of $BG$)!</p> <p>Let's see how these varieties stack up against your criteria. </p> <p>1) They don't necessarily admit pavings by affines (unless $G$ is a so-called "special group"), but they do in some cases, e.g. if $G=GL_n$ or $SL_n$. Moreover, their "motives" admit a paving by affines, in the sense that $S$ can be chosen so that $V-S$ admits a paving by affines, as does $G$ in many cases (e.g. if it is split, affine, ...), so $(V-S)/G$ is morally $[X]/[G]$. (This can be made precise in the Grothendieck ring of varieties, tensored with $\mathbb{Q}$; the issue is that this quotient isn't necessarily a Zariski-fiber bundle. The question of whether such a quotient is even rational is, I think, pretty hard, and possibly open(?).)</p> <p>(2+3) I have no idea about this (largely because I don't know many of the words you use). </p> <p>That said, these have one additional virtue, mentioned by Totaro. Namely, in every case he computes, the Chow ring of $BG$ (defined in degree $&lt; s$ by choosing $(V, S)$ as above with $S$ of codimension at least $s$), actually equals the ring $MU^*BG\otimes_{MU^*} \mathbb{Z}$. (This only makes sense over $\mathbb{C}$ of course.) These examples include $GL(n), Sp(2n), O(n), SO(2n+1),$ and $SO(4)$. So these spaces "really are" algebraic approximations to $BG$!</p> <p>Hopefully this is in a similar spirit to your question! </p> http://mathoverflow.net/questions/121057/from-topological-to-smooth-and-holomorphic-vector-bundles/121098#121098 Answer by Daniel Litt for From Topological to Smooth and Holomorphic Vector Bundles Daniel Litt 2013-02-07T17:40:21Z 2013-02-07T17:40:21Z <p>(A) As Michael Murray remarks, the answer is no (though some care is needed in interpreting the remark that "any two such choices [of maps into the classifying space] are smoothly homotopic") since in this case the classifying space is an infinite-dimensional manifold (the Grassmannian of $n$-planes in $\mathbb{R}^\infty$).</p> <p>(B) An elliptic curve $X$ over $\mathbb{C}$ will admit infinitely many distinct holomorphic structures on the (topologically) trivial line bundle! Choose two (distinct) points $x_0, x_1$; then $\mathcal{O}(x_0-x_1)$ is an example (this is the ideal sheaf of $x_1$ tensored with the dual of the ideal sheaf of $x_0$). One can see that this is distinct from e.g. $\mathcal{O}_X$ because it has no non-zero global sections, whereas $\mathcal{O}_X$ does (constant functions).</p> <p>(C) You need a flat connection to define this complex, as Liviu Nicolaescu remarks. In this case, let $\mathcal{E}$ be the sheaf of flat sections to $E$. Then the de Rham complex is <code>$$\mathcal{E}\otimes_{\mathbb{C}} \Omega^\bullet(B).$$</code> The differential is defined to be a derivation, which kills $e\otimes 1$ for any section $e$ to $\mathcal{E}$ (this uniquely determines a differential).</p> <p>(D) Perhaps I'm confused, but I think Alex's answer is misleading--one can define a Dolbeault complex for any holomorphic vector bundle; this doesn't require a Hermitian metric. If $\mathcal{E}$ is the sheaf of holomorphic sections to your vector bundle, this complex is defined as $$\mathcal{E}\otimes_{\mathcal{O}_X} A^{0, \bullet}(B).$$ Again, the differential is a derivation which kills sections to $\mathcal{E}$. Here $A^{0, \bullet}$ is the complex of smooth $(0, \bullet)$ forms; the complex above should compute the sheaf cohomology of $\mathcal{E}$.</p> http://mathoverflow.net/questions/121005/the-picard-group-over-artin-ring/121013#121013 Answer by Daniel Litt for The Picard Group over artin ring Daniel Litt 2013-02-06T20:02:17Z 2013-02-06T22:56:15Z <p>In FGA, no.232, Thm 3.1, Grothendieck shows that if $f: X\to S$ is flat, projective and finitely presented, with reduced and irreducible geometric fibers, then $\operatorname{Pic}_{X/S}$ is representable by a separated $S$-scheme, locally of finite presentation over $S$. </p> <p>As for smoothness, unfortunately the "well-known" result you cite is actually not true in general--in particular, Igusa <a href="http://www2.iag.uni-hannover.de/~kass/files/IgusaExample.pdf" rel="nofollow">constructed a smooth projective surface</a> (<a href="http://www.jstor.org/stable/89189" rel="nofollow">here</a> is the original paper, written in somewhat archaic language) in characteristic $2$ with non-reduced Picard scheme; the example is the quotient of a particular Abelian surface by a fixed-point-free involution. The smoothness result you claim is true in characteristic zero.</p> <p>Luckily, Theorem 5 of Section 8.4 of Bosch-Lütkebohmert-Raynaud's Neron Models [BLR] answers your question when $X$ is an Abelian $S$-scheme; in this case the Picard functor is representable by another Abelian $S$-scheme, and is in particular smooth.</p> <p>Just a warning: in the general case (e.g. if $f: X\to S$ has no section, for example), the Picard functor represented might not be what you think it is, but rather the etale or fppf sheafification of the usual Picard functor. (This can happen e.g. if $X$ has no rational points and $S$ is $\operatorname{Spec}(k)$.) Of course an Abelian scheme always admits a section, so you're OK in this situation.</p> <p>BTW, Chapter 8 of BLR is a great place to learn about this stuff.</p> http://mathoverflow.net/questions/120253/research-trends-in-geometry-of-numbers/120266#120266 Answer by Daniel Litt for Research trends in geometry of numbers? Daniel Litt 2013-01-29T22:25:42Z 2013-01-29T22:25:42Z <p>There has indeed been exciting recent work in this area, by Bhargava and Shankar (see <a href="http://www-math.mit.edu/~poonen/papers/Exp1049.pdf" rel="nofollow">this Bourbaki expose by Poonen</a>) and also by <a href="http://www.math.harvard.edu/~gross/preprints/stable18.pdf" rel="nofollow">Bhargava and Gross</a>. Briefly, the work of Bhargava and Shankar bounds the average rank of the group of rational points of elliptic curves over $\mathbb{Q}$, while the Bhargava and Gross paper does the same for Jacobians of hyperelliptic curves.</p> <p>Section 4 of the (quite readable) write-up by Poonen explains why I refer to these results as recent advances in the geometry of numbers: both of these results boil down to (subtle!) computations of adelic volumes! It's worth noting that the <a href="http://arxiv.org/abs/1006.1002" rel="nofollow">work of Bhargava and Shankar</a> does not use adelic language, and so is more obviously related to the "classical" geometry of numbers.</p> http://mathoverflow.net/questions/120081/line-bundles-and-vector-bundles-on-mathbb-p1-times-mathbb-p1/120087#120087 Answer by Daniel Litt for Line bundles and vector bundles on $\mathbb P^1 \times \mathbb P^1$ Daniel Litt 2013-01-28T08:14:03Z 2013-01-28T08:14:03Z <p>As Will Sawin remarks in the comments, the answer to your question (2) is "yes," either via the exponential exact sequence or by an easy application of the "<a href="http://staff.science.uva.nl/~bmoonen/boek/LineBund.pdf" rel="nofollow">theorem of the square</a>," (which works e.g. over a general field).</p> <p>The answer to your question (1) is rather more interesting--while there is no classification of vector bundles on $\mathbb{P}^1\times \mathbb{P}^1$ (that I know of), there <em>is</em> a splitting theorem analogous to Grothendieck's result. First, it's worth noting that there is a result of Horrocks generalizing Grothendieck's theorem to projective spaces of arbitrary dimension:</p> <blockquote> <p><strong>Theorem.</strong> (Horrocks) Let $E$ be a vector bundle on $\mathbb{P}^n$. Then $E$ is a direct sum of line bundles if and only if $H^i(\mathbb{P}^n, E(r))=0$ for <code>$0&lt;i&lt;n$</code> and all $r$.</p> </blockquote> <p>Note that if $n=1$, this is precisely Grothendieck's splitting theorem--a good place to learn about this result and its cousins is <a href="http://books.google.com/books/about/Vector_Bundles_on_Complex_Projective_Spa.html?id=OG348syTqC8C" rel="nofollow">Vector Bundles on Complex Projective Spaces</a> by Okonek et al. It is this theorem which generalizes to $\mathbb{P}^1\times \mathbb{P}^1$. </p> <p>Indeed, a literally identical theorem to Horrocks' result above works for a quadric surface $\mathbb{P}^1\times \mathbb{P}^1$ in $\mathbb{P}^3$ (a vector bundle splits if and only if the middle cohomology of all of its twists $E(n)$ vanishes). One reference is <a href="https://tspace.library.utoronto.ca/bitstream/1807/9843/1/cohen_macaulay_hypersurface_1987.pdf" rel="nofollow">this paper</a> by Buchweitz, Greuel, and Schreyer; the relevant comments are Conjecture B and Remark 2 on page 169. Conjecture B gives a (conjectural) generalization to smooth hypersurfaces; I don't know too much about this subject, so I have no idea what the status of this conjecture is. </p> <p>Of course, this is far from a classification of vector bundles on $\mathbb{P}^1\times \mathbb{P}^1$--that said, I think that such a classification (at anywhere near the level of completeness of the classification on $\mathbb{P}^1$) is well beyond current technology. Vector bundles on $\mathbb{P}^2$ are already quite interesting and complicated.</p> http://mathoverflow.net/questions/119953/definition-of-area/119954#119954 Answer by Daniel Litt for Definition of area Daniel Litt 2013-01-26T18:40:21Z 2013-01-26T18:40:21Z <p>There is an intuitive approach to area, based on the fact that polygons $P, P'$ have the same area if and only if they are equidecomposable (that is, one may be cut into pieces and reassembled to form the other).</p> <p>The first three pages of <a href="http://arxiv.org/pdf/math/0312229.pdf" rel="nofollow">this note</a> sketch a "motivic" approach to the definition of area, for polygons. Namely, one defines $K(\text{Poly})$ to be the free Abelian group generated by plane polygons $P$, subject to the following two relations:</p> <ol> <li>$[P]=[P']$ if $P$ is congruent to $P'$</li> <li>$[P]=[P_1]+[P_2]$ if $P$ may be cut into polygons $P_1$ and $P_2$.</li> </ol> <p>An easy exercise (sketched in the note I link to) shows that $[P]=[P']$ if and only if $P$ and $P'$ have the same area, so $K(Poly)\simeq \mathbb{R}$. But even better, one can <em>define</em> the area of a polygon $P$ as its class $[P]$ in $K(\text{Poly})$. </p> <p>Indeed, for many reasonable classes of subsets of the plane, one may extend this definition to assign to such a set a class in $K(\text{Poly})$. For example, say a sequence of classes $[P_i]$ in $K(\text{Poly})$ converges to $[P]$ if there exists a representative $[A]-[B]$ for $[P-P_i]$ with both $A$ and $B$ contained in $[0, \epsilon_i]\times [0, \epsilon_i]$, with $\epsilon_i\to 0$. </p> <p>Suppose $X$ is a subset of the plane so that there is a sequence of polygons $P_i$, such that the symmetric difference $(X\cup P_i)-(X\cap P_i)$ is contained in a polygon $Q_i$. Suppose further that $[Q_i]=[Q'_i]$ and $Q'_i\subset [0, \epsilon_i]\times [0,\epsilon_i]$, with $\epsilon_i\to 0$. Then we assign to $X$ the class $$\lim_{i\to \infty} [P_i]$$ if it exists. It's not hard to check (geometrically!) that this assignment is well-defined.</p> <p>NB: This approach does not work to define volume in $\mathbb{R}^n, n>2$. Indeed, Dehn showed that there are many polyhedra with the same volume that are not equidecomposable.</p> http://mathoverflow.net/questions/119712/geometrizing-the-third-cohomology-of-a-complex-lie-group Geometrizing the Third Cohomology of a Complex Lie Group Daniel Litt 2013-01-24T00:26:55Z 2013-01-24T00:26:55Z <p>If $G_\mathbb{C}$ is a simply-connected simple complex Lie group, theorem 5.4.10 of Brylinski's "Loop Spaces, Characteristic Classes, and Geometric Quantization" claims that there is a natural <code>$\mathcal{O}_{G_\mathbb{C}}^*$</code>-banded gerbe on $G_\mathbb{C}$, essentially corresponding to a generator of $H^3(G_\mathbb{C}, \mathbb{Z})=\mathbb{Z}$; the gerbe comes with some extra "connective" structure. However, he cryptically says on p. 227 of my edition that</p> <blockquote> <p>an explicit construction of [this gerbe] would involve, in one way or another, algebraic $K$-theory.</p> </blockquote> <p>He goes on to give an explicit (ad hoc) construction in the case of $SL(2, \mathbb{C})$, but says no more about the algebraic $K$-theory construction. I can't figure out the relationship to algebraic $K$-theory at all, so my question is:</p> <blockquote> <p>How does one construct this gerbe (with connective structure) via algebraic $K$-theory? I'd also be happy with other constructions, even if they are special to the case of $SL(n, \mathbb{C})$, say.</p> </blockquote> <p>My interest in this question comes from the fact that it gives a (reasonably) concrete geometric interpretation of the $2$nd Chern class of a principal $G_\mathbb{C}$-bundle (say, $G_\mathbb{C}=SL(n, \mathbb{C})$) as follows: the second Chern class of $P$ may be viewed as the obstruction to building a gerbe with connective structure on the total space of $P$ which is fiberwise equivalent to the canonical gerbe on $G_\mathbb{C}$. Actually, this even gives a geometric interpretation of the "refined" $2$nd Chern class valued in the Deligne cohomology of the base.</p> http://mathoverflow.net/questions/114081/explicit-computations-of-the-etale-homotopy-type/118334#118334 Answer by Daniel Litt for Explicit computations of the étale homotopy type? Daniel Litt 2013-01-08T05:37:37Z 2013-01-08T05:37:37Z <p>Here's an example which is, in my opinion, illuminating. It is also quite easy, which I view as a plus.</p> <p>Namely, consider the étale homotopy type of $\text{Spec}~\mathbb{R}$. By your comparision theorem, this is (pro)-equivalent to $B(\mathbb{Z}/2\mathbb{Z})$. But in fact the (pro)-simplicial set one gets is precisely the bar construction for <code>$G=\text{Gal}(\mathbb{C}/\mathbb{R})=\mathbb{Z}/2\mathbb{Z}$</code>(!!!), showing that computing the étale cohomology of $\text{Spec}~\mathbb{R}$ is "the same" as computing the group cohomology of $G$. This is a good, and not hard, exercise.</p> <p>In general, if $k$ is a field with finite Galois group $G$, its étale homotopy type will equal(!) the bar construction of $BG$ for $G=\text{Gal}(k^s/k)$; with an appropriate version of $BG$ for $G$ profinite, this will be true for any field.</p> http://mathoverflow.net/questions/117904/elementary-examples-of-the-weil-conjectures/117908#117908 Answer by Daniel Litt for Elementary examples of the Weil conjectures Daniel Litt 2013-01-02T21:42:26Z 2013-01-02T21:42:26Z <p>Weil himself <a href="http://www.ams.org/journals/bull/1949-55-05/S0002-9904-1949-09219-4/S0002-9904-1949-09219-4.pdf" rel="nofollow">verifies the conjectures</a> "by hand" for diagonal hypersurfaces, that is, hypersurfaces defined by an equation of the form </p> <p>$$a_0x_0^{n_0}+a_1x_1^{n_1}+\cdots+a_kx_k^{n_k}=b.$$</p> <p>The argument is pretty elementary--it essentially uses only character theory. It seems to me likely that this argument heavily influenced Dwork's original proof of rationality.</p> <p>The paper is quite readable; I learned of it from Akshay Venkatesh.</p> http://mathoverflow.net/questions/116896/liouvilles-theorem-with-your-bare-hands/116908#116908 Answer by Daniel Litt for Liouville's theorem with your bare hands Daniel Litt 2012-12-20T20:33:04Z 2012-12-22T00:05:51Z <p>Here's a proof via character theory; in some ways it's a recasting of the Cauchy proof, but it has the advantage of being purely algebraic. <strong>In particular, this proof will work over any algebraically closed complete valued field</strong>, for example (some mild care must be taken over a field of positive characteristic when one takes the limit $k\to\infty$).</p> <p>Let $$f(z)=\sum_{n=0}^\infty a_nz^n$$ be a bounded entire function; we wish to show that $a_n=0$ for $n>0$. We use the fact that $$\sum_{\zeta \text{ a k-th rooth of unity}} \zeta^n$$ equals $k$ if $k$ divides $n$ and equals $0$ otherwise. So in particular setting $g_n(z)=f(z)/z^n$ we have for $k>n$ that</p> <p>$$\frac{1}{k}\sum_{\zeta \text{ a k-th rooth of unity}} g_n(r\zeta)=a_n+\sum_{m=1}^\infty a_{km+n}r^{km}.$$ The sum on the right tends to zero with $k$, as the $a_{km+n}$ decrease very rapidly by the entireness of $f$. So we have $$a_n=\lim_{k\to\infty}\frac{1}{k}\sum_{\zeta \text{ a k-th rooth of unity}} g_n(r\zeta).$$ But taking the limit as $|r|\to \infty$ and using the boundedness of $f$, this is zero for $n>0$.</p> <p>I say that this is a recasting of the Cauchy integral proof because the character sum we use is essentially a Riemann sum for the integral of $g_n$ around a circular contour. Indeed, one may prove the Cauchy integral formula for circular contours this way.</p> http://mathoverflow.net/questions/116913/on-local-parameters-at-the-origin-in-an-algebraic-group/116922#116922 Answer by Daniel Litt for On local parameters at the origin in an algebraic group Daniel Litt 2012-12-20T21:52:55Z 2012-12-20T21:52:55Z <p>This <em>never</em> happens for a reductive group $G$ of dimension $n>0$. The conditions you give induce a surjective finite group homomorphism $G\to \mathbb{A}^n$, for some mysterious group structure on $\mathbb{A}^n$. In particular, $G$ would act transitively on $\mathbb{A}^n$, which a reductive group cannot do.</p> <p>On the other hand, there are many unipotent groups whose underlying schemes are affine spaces, e.g. unipotent upper triangular matrices. Any extension of such a group by a finite group will satisfy the conditions of your question.</p> http://mathoverflow.net/questions/116885/infinitesimal-deformations-and-moving-cycles/116920#116920 Answer by Daniel Litt for Infinitesimal deformations and moving cycles Daniel Litt 2012-12-20T21:38:45Z 2012-12-20T21:38:45Z <p>I'd like to expand a bit on Francesco Polizzi's excellent answer, with a couple of examples. He's right of course that deformations may be obstructed, but there's another issue which prevents infinitesimal deformations from being realized as families of rationally equivalent cycles.</p> <p>As Francesco observes, the existence of obstructed deformations of a subscheme $Z\subset X$ is equivalent to the non-smoothness of the Hilbert $\operatorname{Hilb}_X^{p(Z)}$ at the point corresponding to $Z$. Now suppose one has some unobstructed deformation of $Z$, say a flat family $\mathcal{Z}\to k[t]/t^n$ with closed fiber $Z$. If $X$ is projective and $[Z]\in\operatorname{Hilb}_X^{p(Z)}$ is a smooth point, then by the projectivity of $\operatorname{Hilb}^{p(Z)}_X$ we can slice $\operatorname{Hilb}^{p(Z)}_X$ by hypersurfaces tangent to the family $\mathcal{Z}$ and thus realize $\mathcal{Z}$ inside a family over some curve.</p> <p>This will give a family of subschemes <em>algebraically</em> equivalent to $Z$--but unless the curve is rational, they may not be <em>rationally</em> equivalent to $Z$. As a simple example, consider deforming points in an elliptic curve. Such deformations are always unobstructed, since elliptic curves are smooth by definition. But no two distinct points in an elliptic curve are rationally equivalent (because e.g. any map from a rational curve to an elliptic curve is constant, or because distinct points on an elliptic curve give distinct line bundles).</p> <p>The question you are asking is thus not a <em>local</em> question about the Hilbert scheme, but rather one about its global geometry---namely, is there a <em>rational curve</em> in the Hilbert scheme realizing some given germ of a deformation. This is very hard in general, even for zero cycles on surfaces--see e.g. Bloch's conjecture, and the related beautiful paper by Mumford <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.kjm/1250523940" rel="nofollow">here</a>. </p> http://mathoverflow.net/questions/111464/self-intersection-and-the-normal-bundle/116834#116834 Answer by Daniel Litt for Self-intersection and the normal bundle Daniel Litt 2012-12-20T02:05:55Z 2012-12-20T02:05:55Z <p>I'd like to expand a bit on the excellent comments of Charles Staats and Donu Arapura. They both suggest understanding the self-intersection number of a curve as the number of fixed points of an infinitesimal deformation of the curve, which is manifestly the degree of the normal bundle when such a deformation exists. Here's a slightly more pedestrian route, which I think has the benefit of being rigorous and almost as intuitive.</p> <p>Suppose we have two curves in a surface: $$\iota_C: C\hookrightarrow X, \iota_D: D\hookrightarrow X.$$ If $C\cap D$ has dimension zero, the intersection number should manifestly be <code>$$C\cdot D:=\dim\Gamma(C\cap D, \mathcal{O}_{C\cap D})=\dim\Gamma(C, \iota_C^*\mathcal{O}_D).$$</code> We'd like to write this as an Euler characteristic, to make it constant if we vary $C$ or $D$ in a flat family. But this is easy; since $\mathcal{O}_{C\cap D}$ has zero-dimensional support, it has no higher cohomology, so its Euler characteristic equals $C\cdot D$ as defined above. Line bundles are nice (and more importantly, are acyclic with respect to restriction), so we use the short exact sequence <code>$$0\to \mathcal{O}_X\to \mathcal{O}_X(D)\to \mathcal{O}_D\to 0$$</code> to rewrite this Euler characteristic as <code>$$\chi(\mathcal{O}_X(D)|_C)-\chi(\mathcal{O}_C)=\operatorname{deg}(\mathcal{O}_X(D)|_C).$$</code></p> <p>I think this is a reasonably intuitive motivation for the definition of the intersection number. So to fully answer your question, one should give an intuitive reason for why <code>$\mathcal{O}_X(D)|_D$</code> is <code>$\mathcal{N}_{D/X}.$</code> Of course, this is just the definition of the normal bundle, but let's motivate the definition. First, why is the conormal bundle is $I/I^2$, for $I$ the ideal sheaf of a closed subvariety $V\subset X$? Well, an element of $I/I^2$ is precisely a function on $X$ vanishing at $V$, but ignoring higher-order terms. A section to the normal bundle precisely takes functions $f$ defined in a neighborhood of $Y$ and differentiates them--but the partial derivative should depend only on the first-order part of $f$. So the normal bundle should be precisely $(I/I^2)^\vee$. This is another name for $\mathcal{O}_X(D)|_D.$</p> <p>I hope that was some reasonable intuition/motivation.</p> http://mathoverflow.net/questions/77611/serres-analogue-of-the-weil-conjectures-for-non-compact-kahler-manifolds/116832#116832 Answer by Daniel Litt for Serre's Analogue of the Weil Conjectures for Non-Compact Kahler Manifolds Daniel Litt 2012-12-20T01:01:23Z 2012-12-20T01:01:23Z <p>Since the link in the comments is broken, here is a link to Serre's paper: <a href="http://www.jstor.org/stable/1970088?seq=2" rel="nofollow">Analogues Kahleriennes de Certaines Conjectures de Weil</a>. There seems to be some misunderstanding about what Serre actually proves implicit in the question, so I'll clarify it a bit, explain an easy analogue for some non-compact varieties, and indicate what should be true in general.</p> <p>First of all, Serre's result is <em>not</em> for arbitrary compact Kahler manifolds. Rather, he starts with a smooth complex projective variety $X$, and an endomorphism $f: X\to X$, with an ample divisor $E$ so that $f^{-1}(E)$ is algebraically equivalent to $qE$ for some integer <code>$q&gt;0$</code>. Then he shows that the eigenvalues of $f^*$ acting on $H^r(X, \mathbb{C})$ have absolute value $q^{r/2}$. Contrary to the statement of the question, the algebraicity of $X$ is built into the very result, since it requires the existence of an ample divisor.</p> <p>Serre doesn't explicitly define the zeta function of $(X, f)$, but by analogy to the Weil conjectures, one may define <code>$$Z(X, f, t)=\prod_i \det(1-f^*t\mid H^i(X, \mathbb{C}))^{(-1)^{i+1}}.$$</code> Then this zeta function satisfies the desired "Riemann hypothesis," by the result above.</p> <p>When looking for a non-compact analogue, the motto one should have in mind is that zeta functions should behave well under "cutting-and-pasting." For example, if $Z(X, t)$ is the zeta function from the Weil conjectures, where <code>$X/\mathbb{F}_q$</code> is an arbitrary variety, and <code>$Y\subset X$</code> is a closed subscheme, then <code>$$Z(X, t)=Z(X\setminus Y, t) \cdot Z(Y, t).$$</code></p> <p>So here's an analogue of this fact in the setting of smooth quasi-projective varieties. Suppose $X$ is a smooth projective variety over <code>$\mathbb{C}$</code>, and $f: X\to X$ is a self-map satisfying the conditions of Serre's result. Suppose <code>$Y\subset X$</code> is a smooth closed subvariety, so that <code>$f(Y)\subset Y$</code>. Then <code>$Y, f|_Y$</code> also satisfy the conditions of the theorem, and so <code>$Z(X, f, t)$</code> and <code>$Z(Y, f|_Y, t)$</code> satisfy the "Riemann Hypothesis." Now suppose <code>$f(X\setminus Y)\subset X\setminus Y$</code> as well, and <code>$f|_{X\setminus Y}$</code> is proper. Then we may define <code>$$Z(X\setminus Y, f|_{X\setminus Y}, t)=\prod_i \det(1-f|_{X\setminus Y}^*t\mid H^i_c(X\setminus Y, \mathbb{C}))^{(-1)^{i+1}}.$$</code></p> <p>Here $H^i_c(X, \mathbb{C})$ denotes the cohomology of $X$ with compact support. Does this zeta function satisfy some kind of Riemann hypothesis? Well, from the long exact sequence relating the compactly supported cohomology of $X\setminus Y$ to that of $X$ and $Y$, we have that $$Z(X\setminus Y, f|_{X\setminus Y}, t)=Z(X, f, t)/Z(Y, f|_Y, t).$$</p> <p>So certainly all of the poles and zeroes of $Z(X\setminus Y, f|_{X\setminus Y}, t)$ have absolute value $q^{-r/2}$ for some integer $r$. This gives a sort of "Riemann hypothesis" for quasi-projective varieties admitting a particularly nice compactification. What you'll notice immediately though is that the $r$ does not match up with the cohomological degree, as it does in the compact case---there is some "slippage" coming from the boundary map in the long exact sequence. Rather, the $r$ is related to the "weight filtration" on the cohomology of $X\setminus Y$.</p> <p>Now suppose $U$ is an arbitrary quasiprojective variety, and $f: U\to U$ is a proper map. $U$ might not have the ridiculously nice compactification we need to run the above argument, but the zeta function defined using cohomology with compact support still makes sense. I doubt that it will in general satisfy a "Riemann hypothesis" of the type you're looking for, since the condition on ample divisors may not makes sense. What is true, though, is that if $U$ admits a smooth compactification $X$ so that $f$ extends to a map $g: X\to X$ satisfying the conditions of Serre's result, so that $g(X\setminus U)\subset X\setminus U$, (where $X\setminus U$ is not necessarily smooth!) this zeta function will satisfy a "Riemann hypothesis." Namely, the zeroes and poles of $Z(U, f, t)$ will be of the form $q^{-r/2}$, with the $r$ coming from the weight filtration on the compactly supported cohomology of $U$.</p> <p>To see this, one may imitate Serre's argument using the mixed Hodge structure on the cohomology of $X\setminus Z$. I don't immediately see how to give an analogue of Serre's condition on the existence of the ample divisor $E$ for $U$ without reference to a compactification, though there should be a way to do so; then one should be able to imitate Serre's argument, just working with the mixed Hodge structure on the compactly supported cohomology of $U$.</p> <p><code></code></p> http://mathoverflow.net/questions/116639/hodge-numbers-of-reduction-mod-p/116658#116658 Answer by Daniel Litt for Hodge numbers of reduction mod $p$ Daniel Litt 2012-12-17T22:23:30Z 2012-12-19T21:29:26Z <p>Since people have addressed (2-4), I'll address (1). There is indeed a fast argument. </p> <p>Namely, you have a flat family of curves $X\to \operatorname{Spec}(\mathcal{O}_{K, p})$. You are interested in comparing the geometric genus of the generic fiber $X_K$ to that of the special fiber <code>$X_{\mathbb{F}_q}$</code>. Note that by constancy of Euler characteristic in flat families, <code>$$1-p_a({X_K})=\chi(\mathcal{O}_{X_K})=\chi(\mathcal{O}_{X_{\mathbb{F}_q}})=1-p_a(X_{\mathbb{F}_q}).$$</code> where $p_a$ denotes arithmetic genus. So the arithmetic genera of the fibers are equal. Since you've required that the fibers $X_K$ and $X_{\mathbb{F}_q}$ are smooth, their geometric genera equal their arithmetic genera, which completes the argument.</p> <hr> <p>ADDED (12/19/2012): I'd also like to add a comment about the Hodge decomposition for curves, to complement Christian's excellent answer. Namely, one can see that the Hodge-theoretic predictions one would make from the situation over $\mathbb{C}$ are true for <strong><em>curves over any base</em></strong>. Here's what I mean by this:</p> <p>Let $\pi: C\to S$ be a smooth projective relative curve, with geometrically connected fibers. Let $\Omega^{dR}_{C/S}$ be the relative de Rham complex <code>$$\Omega^{dR}_{C/S}: 0\to \mathcal{O}_C\overset{d}{\to} \Omega^1_{C/S}\to 0.$$</code> Then <code>$$R^2\pi_*(\Omega^{dR}_{C/S})=R^1\pi_*(\Omega^1_{C/S})$$</code> <code>$$R^0\pi_*(\Omega^{dR}_{C/S})=R^0\pi_*(\mathcal{O}_C)$$</code> and there is a natural short exact sequence <code>$$0\to R^0\pi_*(\Omega^1_{C/S})\to R^1\pi_*(\Omega^{dR}_{C/S})\to R^1\pi_*(\mathcal{O}_C)\to 0.$$</code></p> <p>These statements are equivalent to the claim that the Frolicher spectral sequence degenerates at $E_2$. But this is clear, since all of the differentials have $0$ as either their target or source (if you write out the spectral sequence, you'll see that all the non-zero entries are in a little $2\times 2$ box, but all the differentials have bi-degree $(1, 2)$). </p> <p>So one cannot hope to have any "pathological" behavior in $h^{p,q}$'s for smooth relative curves. One can see this from Christian's answer (since the cohomology of a curve has no torsion); I wanted to observe that this is true over any base for essentially formal reasons. </p> http://mathoverflow.net/questions/116037/a-question-about-free-resolutions/116038#116038 Answer by Daniel Litt for A Question About Free Resolutions Daniel Litt 2012-12-11T02:54:55Z 2012-12-11T02:54:55Z <p>No. Consider $\mathfrak{m}:=(x,y,z)\subset k[x,y,z]_{(x,y,z)}=:R$. Then the kernel of the map $$R^3\to \mathfrak{m}$$ defined by the minimal generating set $x,y,z$ is minimally generated by $$k_1:=(y, -x, 0), k_2:=(z, 0, -x), k_3:=(0, z, -y).$$ But $$zk_1-yk_2-xk_3=0$$ so the submodule of $R^3$ that these generate is not free. Rather, we have a free resolution $$0\to R\to R^3\to R^3\to \mathfrak{m}\to 0.$$ where the middle map is defined by the matrix $(k_1 ~k_2 ~k_3)$ and the first map sends a generator of $R$ to</p> <p><code>$$\begin{pmatrix} z \\ -y\\ -x \end{pmatrix}.$$</code></p> <p>(I should mention--one can know this example will work by "pure thought" using local cohomology; essentially there is a natural way of identifying the local cohomology of $(R, \mathfrak{m})$, with the coherent cohomology of $\mathbb{P}^2$. But this does not always vanish in degree $2$, so there cannot be a length $2$ resolution...this argument also shows that there's no way of, say, choosing different generators to get a shorter resolution.)</p> http://mathoverflow.net/questions/32968/slick-ways-to-make-annoying-verifications/32971#32971 Answer by Daniel Litt for Slick ways to make annoying verifications Daniel Litt 2010-07-22T16:31:06Z 2012-10-22T23:29:36Z <ul> <li>The Yoneda Lemma (used everywhere!)</li> <li>In algebraic geometry, the vanishing locus of a map of vector bundles on a scheme $X$ is a closed subscheme of $X$.</li> </ul> <p>I'll give an example where we use both. Say we want to show that there is such a thing as a scheme of "flags" on a vector space $V$, and that it is a closed subscheme of projective space. Such a flag scheme $\text{Fl}_{i_1, ..., i_k}$ represents the functor sending a scheme $S$ to the set of $M_1\hookrightarrow ...\hookrightarrow M_k\hookrightarrow V\otimes \mathcal{O}_S$ where $M_j$ is a vector bundle on $S$ of rank $i_j$ and where all the maps are injective bundle maps. We could do this by working with coordinates, but that would be a huge pain, so let's use our lemmas.</p> <p>It suffices to do this for $k=2$, as larger flag schemes can be written as fiber products of $\text{Fl}_{i_1, i_2}$. We embed $\text{Fl}_{i_1, i_2}$ as a closed subscheme of the scheme $T:=\text{Gr}(i_1, V)\times \text{Gr}(i_2, V)$. We know that this is a closed subscheme of $\mathbb{P}(\Lambda^{i_1}(V)\otimes \Lambda^{i_2}(V))$ so this suffices.</p> <p>But indeed, letting $M_j$ be the canonical rank $i_j$ bundle over $\text{Gr}(i_j, V)$ (induced by the identity map through the universal property of the Grassmannian and Yoneda), we have that the flag bundle $\text{Fl}_{i_1, i_2}$ is the vanishing locus of the map $p_1^*M_1\to \text{coker}(p_2^*M_2\to V\otimes \mathcal{O}_T)$. It's easy to check that this represents the desired functor.</p> http://mathoverflow.net/questions/29949/what-is-the-shortest-program-for-which-halting-is-unknown What is the shortest program for which halting is unknown? Daniel Litt 2010-06-29T18:31:54Z 2012-10-04T16:06:46Z <p>In short, my question is:</p> <blockquote> <p>What is the shortest computer program for which it is not known whether or not the program halts?</p> </blockquote> <p>Of course, this depends on the description language; I also have the following vague question:</p> <blockquote> <p>To what extent does this depend on the description language?</p> </blockquote> <p>Here's my motivation, which I am sure is known but I think is a particularly striking possibility for an application to mathematics:</p> <p>Let $P(n)$ be a statement about the natural numbers such that there exists a Turing machine $T$ which can decide whether $P(n)$ is true or false. (That is, this Turing machine halts on every natural number $n$, printing "True" if $P(n)$ is true and "False" otherwise.) Then the smallest $n$ such that $P(n)$ is false has low <a href="http://en.wikipedia.org/wiki/Kolmogorov_complexity" rel="nofollow">Kolmogorov complexity</a>, as it will be printed by a program that tests $P(1)$, then $P(2)$, and so on until it reaches $n$ with $P(n)$ false, and prints this $n$. Thus the Kolmogorov complexity of the smallest counterexample to $P$ is bounded above by $|T|+c$ for some (effective) constant $c$. </p> <p>Let $L$ be the length of the shortest computer program for which the halting problem is not known. Then if $|T|+c &lt; L$, we may prove the statement $\forall n, P(n)$ simply by executing all halting programs of length less than or equal to $|T|+c$, and running $T$ on their output. If $T$ outputs "True" for these finitely many numbers, then $P$ is true.</p> <p>Of course, the Halting problem places limits on the power of this method.</p> <p>Essentially, this question boils down to: What is the most succinctly stateable open conjecture?</p> <p>EDIT: By the way, an amazing implication of the argument I give is that to prove any theorem about the natural numbers, it suffices to prove it for finitely many values (those with low Kolmogorov complexity). However, because of the Halting problem it is impossible to know which values! If anyone knows a reference for this sort of thing I would also appreciate that.</p> http://mathoverflow.net/questions/103636/modern-proof-of-the-theorem-of-the-base Modern Proof of the Theorem of the Base Daniel Litt 2012-07-31T21:14:58Z 2012-07-31T21:58:23Z <p>I am looking for a modern proof of the so-called "Theorem of the Base"--that the Neron-Severi rank of a smooth projective variety is finite. One can prove this for varieties over $\mathbb{C}$ easily via transcendental methods---however, <a href="http://books.google.com/books?id=vRhtM0UMUh4C&amp;pg=RA1-PA221&amp;lpg=RA1-PA221&amp;dq=%2522theorem+of+the+base%2522+lang&amp;source=bl&amp;ots=YmQSU9ZBZp&amp;sig=Jwm8WbbILU26tVN6bLvIgzdSA9M&amp;hl=en&amp;sa=X&amp;ei=-EcYUIPmCsbL0QH3goD4DQ&amp;ved=0CFUQ6AEwAg#v=onepage&amp;q=%2522theorem%2520of%2520the%2520base%2522%2520lang&amp;f=false" rel="nofollow">the most recent proof</a> I can find which works e.g. in positive characteristic is due to Lang and Neron and is written in the language of Weil's foundations. (Lang includes a similar proof in his book "Diophantine Geometry.") </p> <blockquote> <p>Does anyone know of a proof written in the language of schemes?</p> </blockquote> <p>My motivation is that I'd like to have a brief reading seminar on this theorem with some graduate students at my institution---while it looks like the Lang/Neron paper is translatable, I think that this sort of translation is a serious burden for seminar participants. So it would be nice to have a more modern reference.</p> http://mathoverflow.net/questions/93330/why-is-the-fundamental-group-of-a-compact-riemann-surface-not-free/93340#93340 Answer by Daniel Litt for Why is the fundamental group of a compact Riemann surface not free ? Daniel Litt 2012-04-06T16:51:53Z 2012-04-06T16:51:53Z <p>As per Theo's request, I'm posting this as an answer, though it's largely an expansion on Vitali's comment. Let $F_n$ be the free group on $n$ letters; $K(F_n, 1)$ is a wedge of $n$ circles and so has vanishing cohomology in degrees $>1$. On the other hand, if $X$ is a compact Riemann surface of genus $g>1$, $X$ is a $K(\pi_1(X), 1)$ as its universal cover is the upper-half plane, which is contractible. But then $$H^2(\pi_1(X), \mathbb{Z})=H^2_{sing}(X, \mathbb{Z})=\mathbb{Z},$$ which is non-zero. In particular, $\pi_1(X)$ has non-vanishing cohomology in degree $2$ and so is not free, as Vitali says.</p> http://mathoverflow.net/questions/76541/what-do-the-stable-homotopy-groups-of-spheres-say-about-the-combinatorics-of-fini What do the stable homotopy groups of spheres say about the combinatorics of finite sets? Daniel Litt 2011-09-27T18:18:27Z 2012-02-28T10:26:53Z <p>The Barratt-Priddy-Quillen(-Segal) theorem says that the following spaces are homotopy equivalent in an (essentially) canonical way:</p> <ol> <li><p>$\Omega^\infty S^\infty:=\varinjlim~ \Omega^nS^n$</p></li> <li><p><code>$\mathbb{Z}\times ({B\Sigma_\infty})_+$</code>, where $\Sigma_\infty$ is the group of automorphisms of a countable set which have finite support, and $+$ is the Quillen plus-construction.</p></li> <li><p>The group completion of $B\left(\bigsqcup_n \Sigma_n\right)$, where $\Sigma_n$ is the symmetric group on $n$ letters, and $B(\sqcup_n \Sigma_n)$ is given the structure of a topological monoid via the block addition map $\Sigma_n\times \Sigma_m\to \Sigma_{n+m}$.</p></li> <li><p>$\Omega|S^\bullet.\operatorname{FinSet}|$, where $S^\bullet$ is the Waldhausen $S$-construction, and $\operatorname{FinSet}$ is the category of pointed finite sets, given the structure of a Waldhausen category by declaring cofibrations to be injections and weak equivalences to be isomorphisms. I don't want to define this since it's complicated (for a reference, see Chapter IV of Weibel's K Book), but it should be thought as a homotopical version of the Grothendieck ring of finite sets, where addition is given by disjoint union and multiplication is given by the cartesian product. Clark Barwick's answer <a href="http://mathoverflow.net/questions/11421/is-higher-k-functor-the-derived-functor-of-k0/11434#11434" rel="nofollow">here</a> makes this more precise.</p></li> </ol> <p>Now, the homotopy groups of the first space are manifestly the stable homotopy groups of spheres; on the other hand, the last two spaces clearly encode some information about the combinatorics of finite sets. So my question is:</p> <blockquote> <p>Is there a concrete combinatorial interpretation of the higher stable homotopy groups of spheres in terms of the combinatorics of finite sets or symmetric groups?</p> </blockquote> <p>For example it is easy to see via (3) or (4) that $\pi_{0+k}(S^k)=\mathbb{Z}$ corresponds to the Grothendieck ring of finite sets. Similarly, (2), or with some theory (4), make it clear that $\pi_{1+k}(S^k)=\mathbb{Z}/2\mathbb{Z}$ corresponds to the abelianization of $\Sigma_n$ (via the sign homomorphism). I am interested in concrete interpretations of the higher stable homotopy groups in this style.</p> <p>A good answer would be, for example, a direct combinatorial interpretation of $\pi_{2+k}(S^k)=\mathbb{Z}/2\mathbb{Z}$ and $\pi_{3+k}(S^k)=\mathbb{Z}/24\mathbb{Z}$; a not-so-good answer would be a statement like "the sphere spectrum is a categorification of the integers," which is not the sort of concrete thing I'm looking for. </p> <p>EDIT: So with the exception of Jacob Lurie's comment on $\pi_{2+k}(S^k)$ below (interpreting it as the Schur multiplier $H_2(\Sigma_\infty, \mathbb{Z})$ of $\Sigma_\infty$), it seems like it might be too much to hope for any reasonably complete combinatorial interpretation of the stable homotopy groups. So I'd settle for something like the following: namely, a sequence of groups $G_n$ defined in some combinatorial way, and maps $f_n: G_n\to \pi_{n+k}(S^k)$ or $g_n: \pi_{n+k}(S^k)\to G_n$ such that </p> <ol> <li><p>$f_n$ or $g_n$ are nontrivial for infinitely many $n$,</p></li> <li><p>The maps are related in some way to the constructions 2-4 above, and</p></li> <li><p>The $G_n$ are combinatorially interesting.</p></li> </ol> <p>One such example is $G_n:=H_n(\Sigma_\infty, \mathbb{Z})$ with $g_n$ the Hurewicz map (whence the interpretation of $\pi_{2+k}(S^k)$). But even in this case, the combinatorial meaning is sort of mysterious (to me at least) for large $n$. </p> http://mathoverflow.net/questions/89069/should-the-formula-for-the-inverse-of-a-2x2-matrix-be-obvious/89074#89074 Answer by Daniel Litt for Should the formula for the inverse of a 2x2 matrix be obvious? Daniel Litt 2012-02-21T02:58:23Z 2012-02-21T02:58:23Z <p>My favorite way to remember this is to think of $SL_2(\mathbb{R})$ as a circle bundle over the upper half-plane, where $SL_2(\mathbb{R})$ acts on the upper half-plane via fractional linear transformations; then the fiber over a point is the stabilizer of that point.</p> <p>This naturally gives the Iwasawa decomposition of $SL_2(\mathbb{R})$ as $$SL_2(\mathbb{R})=NAK$$ where </p> <p><code>$$K=\left\{\begin{pmatrix} \cos(\theta) &amp; -\sin(\theta) \\ \sin(\theta) &amp; \cos(\theta) \end{pmatrix} , ~0\leq\theta&lt;2\pi \right\}$$</code></p> <p><code>$$A=\left\{\begin{pmatrix} r &amp; 0\\ 0 &amp;1/r\end{pmatrix},~ r\in \mathbb{R}\setminus\{0\}\right\}$$</code></p> <p><code>$$N=\left\{\begin{pmatrix} 1 &amp; x \\ 0 &amp; 1\end{pmatrix},~ x\in \mathbb{R}\right\}$$</code></p> <p>Here $K$ is the stabilizer of $i$ in the upper half-plane picture; viewed as acting on the plane via the usual action of $SL_2(\mathbb{R})$ on $\mathbb{R}^2$ it is just rotation by $\theta$ (and likewise if we view the upper half plane as the unit disk, sending $i$ to $0$ via a fractional linear transformation). $A$ is just scaling by $r^2$, in the upper half-plane picture, and is stretching in the $\mathbb{R}^2$ picture. $N$ is translation by $x$ in the upper half-plane picture, and is a skew transformation in the $\mathbb{R}^2$ picture.</p> <p>In each case, the inverse is geometrically obvious: for $K$, replace $\theta$ with $-\theta$; for $A$ replace $r$ with $1/r$, and for $N$, replace $x$ with $-x$. Since $$SL_2(\mathbb{R})=NAK$$ this lets us invert every $2\times 2$ matrix by "pure thought", at least if you remember the Iwasawa decomposition (which is easy from the geometric picture, I think). Of course this easily extends to $GL_2$; if $A$ has determinant $d$, then $A^{-1}$ had better have determinant $d^{-1}$.</p> <p>If you'd like to derive the formula you've written down by "pure thought" it suffices to look at any one of these cases if you remember the general form of the inverse; or you can simply put them all together to give a rigorous derivation.</p> http://mathoverflow.net/questions/84182/shape-of-snowflakes/84243#84243 Answer by Daniel Litt for Shape of snowflakes Daniel Litt 2011-12-24T22:31:29Z 2011-12-25T19:14:48Z <p>I was curious about the OP's second question, which I now think is actually rather difficult. Namely</p> <blockquote> <p>Why is it [the shape of a snowflake] not round?</p> </blockquote> <p>Various physics-y sources (e.g. <a href="http://www.its.caltech.edu/~atomic/publist/rpp5_4_R03.pdf" rel="nofollow">this paper</a>) suggest the following "explanation" for the shape of a snowflake, which I mention in my comment on the original question--namely, that water crystallizes in a hexagonal lattice, so small snowflakes are just hexagons; new water molecules are more likely to attach at corners than edges or faces (for complicated reasons I don't really understand) so vertices grow faster than edges. Thus the hexagon will become a non-convex 6-pointed star; then the edges of this figure will split similarly, and so on. This interpretation is born out by e.g. the picture on pg. 884 of the paper above.</p> <p>This inspired the following simple model, which comes in both deterministic and random flavors. We'll build a snowflake on the standard hexagonal lattice in $\mathbb{R}^2$, spanned by e.g. $(1, 0)$ and $(1/2, \sqrt{3}/2)$. Start with a single regular hexagon of side length $1$ centered at the origin, with vertices the six shortest lattice vectors. </p> <p>In the deterministic version of the model, at each positive integer time $t$ we add a regular lattice hexagon with side length $1$ centered at each lattice point which is the vertex of exactly one hexagon. In the random version, at each positive integer time $t$, we add a hexagon centered at a random lattice point which is the vertex of exactly one hexagon with uniform probability over such lattice points.</p> <p>I had some time this morning, so I coded up both models in the language "Processing." Here is a typical pair of snowflakes from the deterministic model: </p> <p><img src="http://img607.imageshack.us/img607/323/deterministic1.png" alt="alt text"></p> <p><img src="http://img40.imageshack.us/img40/1539/deterministic2.png" alt="alt text"></p> <p>This model has the following interesting properties, none of which are particularly difficult to prove. </p> <blockquote> <p>1) By the envelope of a snowflake I mean the smallest simply-connected polygon containing it. Let $S_n$ be the envelope of the snowflake at time $n$. Consider the sequence $S_n$ in the space of plane polygons metrized by Hausdorff distance, modulo homothety (two polygons are homothetic if one is congruent to a rescaling of the other). Then $S_n$ is recurrent (that is, any homothety class visited by $S_n$ is approached arbitrarily closely infinitely many times). However, the only homothety class taken infinitely many times by the $S_n$ is that of a regular hexagon. (Thus in this setting the adage that no two snowflakes are alike is pretty far off.)</p> <p>2) Let $H_n$ be the smallest regular hexagon containing $S_n$. Then $$\frac{\text{area}(S_n)}{\text{area}(H_n)}$$ is bounded above by $1$ and below by, say $1/2$ (though one can do better). By virtue of the recurrence of the $S_n$, however, this ratio does not attain a limit.</p> <p>3) Certain interior triangles are never filled in, and as is visible in the pictures above these follow a beautiful regular pattern which I haven't bothered to work out.</p> </blockquote> <p>Now let's look at the random model. Here are two typical snowflakes:</p> <p><img src="http://img197.imageshack.us/img197/9476/random1bq.png" alt="alt text"></p> <p><img src="http://img689.imageshack.us/img689/6692/random2oh.png" alt="alt text"></p> <p>As you can see, these are quite round, so they might be better called snowballs. I understand this model much less well than the deterministic one above. However, the following conjectures are natural given the pictures.</p> <blockquote> <p>4) (Conjecture) In the space of homothety classes of plane polygons, metrized by Hausdorff distance, as in (1) above, the envelopes of these shapes tend towards the homothety class of a circle with probability $1$.</p> <p>5) (Conjecture) The ratio $$\frac{\text{perimiter}(S_n)}{\sqrt{\text{area}(S_n)}}$$ tends to infinity with probability $1$.</p> </blockquote> <p>In other words--the random model I implicitly suggested in my comment on the original question seems to give round snowflakes! So I at least think the physics question as to why snowflakes aren't round is still pretty interesting.</p> <hr> <p>In the comments, Rebecca Bellovin suggests another random model--namely, fix a probability $0\leq p\leq 1$ and at each time $t$, and each valid lattice point (namely, each lattice point which is the vertex of exactly one hexagon) add a hexagon centered at that point with probability $p$. At least for small $t$ (e.g. $t&lt;10000$), this seems to interpolate between the two models I give here, and certainly if one scales $p$ in proportion to the number of valid lattice points (so that for example the probability is negligible that more than one hexagon will be added, or that no valid points will be missed), these models will behave exactly like the ones I give. On the other hand, for middling $p$, something interesting happens--namely, the snowflakes look like rounded hexagons. At Rebecca's request, I am posting a picture for $p=0.7$, below:</p> <p><img src="http://img267.imageshack.us/img267/8533/rebeccasmodel07.png" alt="alt text"></p> <p>I have no real explanation for this phenomenon; only unconvincing heuristics.</p> http://mathoverflow.net/questions/81342/elementary-results-with-p-adic-numbers/81361#81361 Answer by Daniel Litt for Elementary results with p-adic numbers Daniel Litt 2011-11-19T17:22:47Z 2011-11-19T17:22:47Z <p>Here is a beautiful and essentially elementary result using the $p$-adics: the Skolem-Mahler-Lech theorem.</p> <blockquote> <p><strong>Theorem</strong>. (<a href="http://en.wikipedia.org/wiki/Skolem%25E2%2580%2593Mahler%25E2%2580%2593Lech_theorem" rel="nofollow">Skolem-Mahler-Lech</a>) Let $(a_i)$ be a sequence defined by an integer linear recurrence. Then the set of $i$ such that $a_i=0$ is the union of a finite set with finitely many arithmetic progressions.</p> </blockquote> <p>A quick proof may be found on Terry Tao's blog, <a href="http://terrytao.wordpress.com/2007/05/25/open-question-effective-skolem-mahler-lech-theorem/#more-34" rel="nofollow">here</a>. Essentially, the $p$-adic step of the proof works by defining a $p$-adic analytic function with infinitely many zeros, and then concluding that this function is identically zero--by the definition of this function, this gives some congruence information about the structure of the zero set of the linear recurrence, as desired. The proof is quite elementary and beautiful, and I think accessible to people seeing the $p$-adics for the first time.</p> http://mathoverflow.net/questions/69035/the-category-of-l-adic-sheaves/80743#80743 Answer by Daniel Litt for The category of l-adic sheaves Daniel Litt 2011-11-12T09:23:38Z 2011-11-12T09:23:38Z <p>Section 1.4 in <a href="http://math.unice.fr/~dehon/CohEtale-09/Elencj_Etale/CONRAD%2520Etale%2520Cohomology.pdf" rel="nofollow">these notes</a> of Brian Conrad is as nice as one could hope for, given the dryness of the adic formalism. I don't think the material differs substantially from Frietag-Kiehl, except in that the presentation is much cleaner. For the derived category stuff, the notes refer to Behrend's paper <a href="http://www.math.ubc.ca/~behrend/ladic.pdf" rel="nofollow">"Derived $\ell$-adic Categories for Algebraic Stacks"</a> which I haven't looked at really, but a brief skim suggests it contains everything you might desire (constructions in <em>extremely</em> general situations), but nonetheless includes examples (!).</p> http://mathoverflow.net/questions/78423/clean-proofs-of-properties-of-projective-space/78496#78496 Answer by Daniel Litt for Clean Proofs of Properties of Projective Space Daniel Litt 2011-10-18T22:19:58Z 2011-10-18T22:19:58Z <p>I agree with Anton that it would be too much to hope for to get serious results (e.g. cohomology of line bundles) from the "nice" universal property of projective space, but one can indeed prove that there are no non-constant regular functions on $\mathbb{P}^n$ using only the universal property. </p> <p>Namely, it suffices to check that $\mathbb{P}^n$ is proper and connected. For properness, one may use the valuative criterion. Namely, let $R$ be a valuation ring and $K$ its fraction field. Then a map from $\operatorname{Spec}(K)\to \mathbb{P}^n$ is a surjection $K^{n+1}\to K$; then the image of $R^n\hookrightarrow K^n\to K$, where the inclusion is the obvious one, is isomorphic to $R$. In particular, the map $K^{n+1}\to K$ lifts to a surjective map $R^{n+1}\to R$, which is a map $\operatorname{Spec}(R)\to \mathbb{P}^n$ fullfilling the valuative criterion (its uniqueness up to automorphisms of the given diagram is also clear).</p> <p>As for connectedness, it suffices to check that $\mathbb{P}^n$ is path-connected in the following sense--for any two geometric points, represented by surjections $x_0: \bar k^{n+1}\to\bar k, x_1: \bar k^{n+1}\to\bar k$, there is a path" connecting them; namely a map $f: \mathbb{A}^1\to \mathbb{P}^n$ with $f(0)=x_0, f(1)=x_1$. This is a surjection $\bar k[t]^{n+1}\to \bar k[t]$ such that reducing mod $(t), (t-1)$ gives the desired maps. Translating, we must choose polynomials $f_1, ..., f_{n+1}$ such that $f_i(0)=x_0(e_i), f_i(1)=x_1(e_i)$, where $e_i$ are the standard basis of $\bar k^{n+1}$, and where all the $f_i$ do not vanish simultaneously.</p> <p>But one can do this by Lagrange interpolation; choose any $f_1$ with the desired values at $0,1$, then any $f_2$ with the desired values at $0, 1$ and not vanishing at the other zeros of $f_1$, then any $f_3$ analogously, and so on. </p> http://mathoverflow.net/questions/77992/is-mathbbcx-y-isomorphic-to-mathbbcx-otimes-mathbbr-mathbbcy/77997#77997 Answer by Daniel Litt for Is $\mathbb{C}[x,y]$ isomorphic to $\mathbb{C}[x]\otimes_{\mathbb{R}}\mathbb{C}[y]$ as rings? Daniel Litt 2011-10-13T04:14:33Z 2011-10-13T04:14:33Z <p>Here's an answer to your second question. There is a cartesian square</p> <p>$$X\underset{Z}{\times} Y \to X\underset{W}{\times} Y$$ $$\downarrow~~~~~~~~~~~~\downarrow$$ $$Z ~~\to ~~Z\underset{W}{\times} Z$$</p> <p>where the bottom arrow is the diagonal map. Then the top arrow is an isomorphism if (but not only if) the bottom arrow is. Of course, if you are asking when the top two objects are abstractly isomorphic, rather than isomorphic through this particular map, there's unlikely to be a reasonable answer.</p> http://mathoverflow.net/questions/73571/f-g-f-p-groups-with-exactly-n-normal-subgroups/73574#73574 Answer by Daniel Litt for (F.g., f.p.) groups with exactly $n$ normal subgroups Daniel Litt 2011-08-24T15:16:18Z 2011-08-24T15:16:18Z <p>In general, $\mathbb{Z}/2^k\mathbb{Z}$ has $k+1$ normal subgroups, namely $\mathbb{Z}/2^j\mathbb{Z}$ for $0\leq j\leq k$. So the answer to your question is "yes."</p> http://mathoverflow.net/questions/122509/two-quetions-on-complex-geometry/122514#122514 Comment by Daniel Litt Daniel Litt 2013-02-26T21:45:02Z 2013-02-26T21:45:02Z @Jude: Real vector bundles of rank $n$ over a finite CW complex $X$ are naturally in bijection with homotopy classes of maps from $X\to BGL_n(\mathbb{R})$, e.g. the Grassmannian of $n$-planes in $\mathbb{R}^\infty$. This is a well-known fact from algebraic topology; see e.g. <a href="http://en.wikipedia.org/wiki/Classifying_space" rel="nofollow">en.wikipedia.org/wiki/Classifying_space</a> http://mathoverflow.net/questions/122645/injective-morphism-from-curves-to-mathbb-cp2 Comment by Daniel Litt Daniel Litt 2013-02-22T19:08:22Z 2013-02-22T19:08:22Z @J&#233;r&#233;my: I assume that aglearner is allowing unibranch singularities. http://mathoverflow.net/questions/122509/two-quetions-on-complex-geometry/122514#122514 Comment by Daniel Litt Daniel Litt 2013-02-22T05:11:24Z 2013-02-22T05:11:24Z I'm happy I could help, Jude--is there any specific point on which you'd like clarification? http://mathoverflow.net/questions/122499/topologically-embedding-curves-in-jacobian/122501#122501 Comment by Daniel Litt Daniel Litt 2013-02-21T04:58:45Z 2013-02-21T04:58:45Z This is really nice! It's essentially a version of the Abel-Jacobi map, where instead of integrating holomorphic differentials, one integrates differentials supported on (or near) each of the tori you've glued together. http://mathoverflow.net/questions/122442/functional-equations-of-zeta-functions-over-global-fields Comment by Daniel Litt Daniel Litt 2013-02-20T19:57:36Z 2013-02-20T19:57:36Z Tate's thesis... http://mathoverflow.net/questions/121638/does-every-commutative-monoid-admit-a-translation-invariant-measure Comment by Daniel Litt Daniel Litt 2013-02-12T19:48:57Z 2013-02-12T19:48:57Z If $T$ is a group, this is not a very interesting topology... http://mathoverflow.net/questions/121379/what-is-an-infty-1-topos-and-why-is-this-a-good-setting-for-doing-different Comment by Daniel Litt Daniel Litt 2013-02-12T19:23:06Z 2013-02-12T19:23:06Z @Urs: Thanks--the links you give are an excellent summary of the current state of applications of &quot;higher&quot; techniques to differential geometry. I just wanted to say, since I think what I'd written may have been harsher than I intended--I have a huge amount of respect for everyone working on this stuff! I was worried that the original wording of the question was based on a misapprehension about diff. geom., which ignored the centuries of difficult analytic work that has gone into proving beautiful theorems. The new and exciting &quot;higher&quot; techniques are as yet a small part of diff. geom... http://mathoverflow.net/questions/121379/what-is-an-infty-1-topos-and-why-is-this-a-good-setting-for-doing-different/121436#121436 Comment by Daniel Litt Daniel Litt 2013-02-11T18:43:36Z 2013-02-11T18:43:36Z Ah, OK. That's awesome--I'll keep an eye out for the preprint! And thanks for the references! http://mathoverflow.net/questions/121379/what-is-an-infty-1-topos-and-why-is-this-a-good-setting-for-doing-different/121436#121436 Comment by Daniel Litt Daniel Litt 2013-02-11T05:19:23Z 2013-02-11T05:19:23Z I am amazed by the possibility of a &quot;purely formal&quot; proof of GRR--is there a write-up somewhere? The Borel-Serre writeup of Grothendieck's proof is quite natural but far from formal--does the work they do move somewhere else in setting up the &quot;derived&quot; theory? http://mathoverflow.net/questions/121379/what-is-an-infty-1-topos-and-why-is-this-a-good-setting-for-doing-different Comment by Daniel Litt Daniel Litt 2013-02-11T00:42:06Z 2013-02-11T00:42:06Z (cont.) it is a setting for differential geometry at all! I hope and believe that your work and that of others will rectify this lack of &quot;external triumphs&quot; (or perhaps there have already been such triumphs, of which I am not aware). But until these techniques produce a &quot;gem,&quot; I think it is reasonable to maintain a position of (hopeful) skepticism. http://mathoverflow.net/questions/121379/what-is-an-infty-1-topos-and-why-is-this-a-good-setting-for-doing-different Comment by Daniel Litt Daniel Litt 2013-02-11T00:37:35Z 2013-02-11T00:37:35Z @Urs Schreiber: My objection is not to the idea that one might find higher-categorical techniques useful--I've actually been thinking about related things recently. But if one measures the impact of these techniques by their &quot;external triumphs,&quot; I think it is clear that they have not yet begun to penetrate differential geometry (or algebraic geometry, which has never been averse to abstraction!) to the extent which they have started to penetrate homotopy theory. So it is at best misguided to call $(\infty, 1)$-stuff the proper setting for differential geometry--few mathematicians would say http://mathoverflow.net/questions/121379/what-is-an-infty-1-topos-and-why-is-this-a-good-setting-for-doing-different Comment by Daniel Litt Daniel Litt 2013-02-10T07:54:27Z 2013-02-10T07:54:27Z The idea that this is &quot;the appropriate setting for the study of principal bundles, i.e., doing differential geometry&quot; is ridiculous. The $(\infty, 1)$-language gives (supposedly) concrete interpretations of higher cohomology classes, which can be useful--but it's not as if the preponderance of interesting theorems in differential geometry require higher-categorical techniques to prove. http://mathoverflow.net/questions/121364/fixed-point-theorem-on-graphs Comment by Daniel Litt Daniel Litt 2013-02-09T22:50:41Z 2013-02-09T22:50:41Z This process doesn't converge for the graph consisting of two vertices and a single edge. Or any path. Or any cycle. (With appropriate choice of weights.) It might be worth adding some hypotheses. http://mathoverflow.net/questions/121057/from-topological-to-smooth-and-holomorphic-vector-bundles/121150#121150 Comment by Daniel Litt Daniel Litt 2013-02-08T06:14:22Z 2013-02-08T06:14:22Z (cont.) $\mathbb{R}^4$ has many smooth structures. But in your language, any continuous (but not smooth) automorphism of $\mathbb{R}$ will induce a distinct smooth structure on $\mathbb{R}$. I think that very few people would consider this standard usage. But in any case, it's a distinction worth drawing. BTW, I think we're both at Stanford--hi! http://mathoverflow.net/questions/121057/from-topological-to-smooth-and-holomorphic-vector-bundles/121150#121150 Comment by Daniel Litt Daniel Litt 2013-02-08T06:11:52Z 2013-02-08T06:11:52Z Of course, I think we agree on all the mathematical content--and thanks, I appreciate the change. As I interpret the question, it's about <i>isomorphism classes of smooth vector bundles over $B$</i> (where the isomorphism may be any isomorphism over $B$); if I understand correctly, you interpret the question as requiring that isomorphism to be the identity. This is just a question of usage--I'd argue that &quot;differential structure&quot; or &quot;smooth structure&quot; is almost always used to mean isomorphism class...for example, it is common to say that $\mathbb{R}$ has only one smooth structure, but...