User acl - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T16:50:20Z http://mathoverflow.net/feeds/user/10696 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/129703/the-etale-fundamental-group-in-the-non-normal-case/129711#129711 Answer by ACL for The étale fundamental group in the non-normal case ACL 2013-05-05T11:07:27Z 2013-05-05T11:07:27Z <p>Take a nodal cubic $C$, that is a projective line of which you identify two points $0$ and $\infty$. This curve has connected étale covers of any degree: take $n$ copies of the projective line, numbered circularly, and identify the $\infty$ of each of them with the $0$ of the next one. This is a $\mathbf Z/n\mathbf Z$-Galois cover of $C$.</p> <p>(Taken from Hartsorne's <em>Algebraic Geometry</em>, page 276.)</p> http://mathoverflow.net/questions/128874/tangent-space-in-algebraic-geometry-and-differential-geometry/128983#128983 Answer by ACL for Tangent space in Algebraic geometry and Differential geometry ACL 2013-04-28T07:33:12Z 2013-04-28T07:33:12Z <p>I would say that it is easier to view the things the other way round and to transport in differential geometry the algebraic-geometry definition of differentials. This is the one which is alluded to by Martin Brandenburg when he writes the tangent space $T_xX$ as the dual of $\mathfrak m_x/\mathfrak m_x^2$. </p> <p>What the Taylor expansion of $f$ says is that a function $f$ is equal to its value at $x$ plus its differential plus something which has order $2$. Moreover, a function which has order $2$ at $x$ can be written as a linear combination of products of two functions vanishing at $x$. </p> <p>This leads to the definition: the space of differentials at $x$ is the quotient of the ideal $\mathfrak m_x$ of the ring of $C^\infty$-functions vanishing at $x$ modulo its square. Then the differential of $f$ (satisfying $f(x)=0$) is just $f$ up to second order terms''.</p> http://mathoverflow.net/questions/114555/does-physics-need-non-analytic-smooth-functions/128686#128686 Answer by ACL for Does Physics need non-analytic smooth functions? ACL 2013-04-25T06:54:05Z 2013-04-25T06:54:05Z <p>A beautiful <a href="http://en.wikipedia.org/wiki/Bertrand%27s_theorem" rel="nofollow">19th century theorem of mechanics</a> due to Joseph Bertrand says the following. <em>Consider the motion of a particle which is driven by a central force potential $V$</em>, that is, the potential $V$ is a function from the distance of the particle to some fixed origin and the force exerted on the particle is given by $\vec F=-\mathop{\rm grad}(V)$. <em>If (almost) all trajectories are periodic, then either $V(r)\propto 1/r$ (celestial mechanics), or $V(r)\propto r^2$ (harmonic oscillator).</em></p> <p>At some point of the proof one knows that all periods of all trajectories are rational multiples of a common period, and one needs to conclude that there is a common period. That part of the proof is always incorrect in the litterature I know. (In particular, Wikipedia's argument is not complete.)</p> <p>It is in fact easy to construct smooth real families of periodic functions with non-constant rational period. The only way I can correct the proof assumes that the potential is real analytic.</p> http://mathoverflow.net/questions/1420/whats-the-best-proof-of-quadratic-reciprocity/128198#128198 Answer by ACL for What's the "best" proof of quadratic reciprocity? ACL 2013-04-20T22:10:46Z 2013-04-20T22:10:46Z <p>I learned the following proof from Jean-François Mestre, it is a variant of Zolotarev's.</p> <p>For every prime number $p>2$, let $T_p\in\mathbf Z[X]$ be the monic polynomial such that $T_p(X+1/X)X^{(p-1)/2}=(X^p-1)/(X-1)=1+X+\dots+X^{p-1}$. The complex roots of $T_p$ are $x+1/x$, where $x$ is a primitive $p$th root of unity. The same holds in any field of characteristic $\neq p$. In a field of characteristic $p$, the only root of $T_p$ is $2$, with multiplicity $(p-1)/2$.</p> <p>Let $p,q$ be two odd prime numbers, with $p\neq q$.</p> <p>The resultant $\mathop{\rm Res}(T_p,T_q)$ of $T_p$ and $T_q$ is an integer. Since these polynomials have no common root, this integer is non-zero. </p> <p>Since these polynomials have no common root in every field, in particular modulo every prime number, this integer is $\pm1$.</p> <p>Compute this resultant modulo $p$. One gets $\mathop{\rm Res}(T_p,T_q)\equiv (-1)^{(p-1)(q-1)/4} T_q(2)^{(p-1)/2}\equiv (-1)^{(p-1)(q-1)/2} q^{(p-1)/2}\pmod p$. Consequently, $\mathop{\rm Res}(T_p,T_q)=\epsilon \left(\frac qp\right)$, with $\epsilon=(-1)^{(p-1)(q-1)/4}$.</p> <p>Similarly, $\mathop{\rm Res}(T_q,T_p)=\epsilon \left(\frac pq\right)$.</p> <p>Now, $\mathop{\rm Res}(T_p,T_q) = (-1)^{\deg(T_p)\deg(T_q)} \mathop{\rm Res}(T_q,T_p),$ hence the quadratic reciprocity law.</p> http://mathoverflow.net/questions/123942/how-many-flat-connections-has-a-line-bundle-in-algebraic-geometry/123945#123945 Answer by ACL for How many flat connections has a line bundle in algebraic geometry? ACL 2013-03-08T06:52:17Z 2013-03-09T00:51:55Z <p>I presume that your variety $X$ is smooth.</p> <p>Consider the additive map $\mathrm d\log \colon \mathscr O_X^*\to \Omega^1_X$ that sends $f$ to $\mathrm df/f$. It induces a map $c_1$ in cohomology from $H^1(X,\mathscr O_X^*)$ to $H^1(X,\Omega^1_X)$ — a coherent avatar of the first Chern class. By Hodge Theory, $H^1(X,\Omega^1_X)$ is a subspace of $H^2(X,\mathbf C)$ and the two notions of first Chern class coincide. </p> <p>A line bundle $\mathscr L$ has a connection if and only if its first Chern class $c_1(\mathscr L)\in H^1(X,\Omega^1_X)$ vanishes. The proof is straightforward: take an open cover $(U_i)$ of $X$, an invertible section $s_i$ of $\mathscr L$ on $U_i$ and the associated cocycle $(f_{ij})$ representing your line bundle in $H^1(X,\mathscr O_X^*)$. A connection $\nabla$ maps $s_i$ to $s_i\otimes\omega_i$, for some 1-form $\omega_i\in H^0(U_i,\Omega^1_X)$. The condition that these $s_i\otimes\omega_i$ come from a global connection on $X$ is exactly the vanishing of $c_1(\mathscr L)$.</p> <p>It is a non-trivial fact that if $\mathscr L$ has an algebraic connection, then it is automatically flat. Torsten Ekedahl gave an algebraic proof on <a href="http://mathoverflow.net/questions/80280/line-bundles-with-integrable-connection-on-abelian-varieties" rel="nofollow">this thread of MO</a> (Ekedahl also observes that $p$th power of line bundles in characteristic $p$ have an integrable connection), but an analytic proof seems easy. The algebraic connexion $\nabla$ gives rise to a connexion $\nabla+\bar\partial$ on the associated holomorphic line bundle. One checks that the curvature of this connection is a $(2,0)$-form, while it should be a $(1,1)$-form. Consequently, it vanishes. </p> <p>When non empty, the set of flat connections on a vector bundle $\mathscr E$ is an affine space under $H^0(X,\Omega^1_X\otimes\mathscr E\mathit{nd}(\mathscr E))$, a finite dimensional vector space. In our case, $\mathscr L$ is a line bundle, hence $\mathscr E\mathit{nd}(\mathscr L)$ is the trivial line bundle so that we get $H^0(X,\Omega^1_X)$.</p> <p>NB. Following the comment of Ben McKay, I edited the last paragraph.</p> http://mathoverflow.net/questions/123948/kodaira-dimension-of-symmetric-products-of-curves Kodaira dimension of symmetric products of curves ACL 2013-03-08T07:33:32Z 2013-03-08T17:26:00Z <p>What is the Kodaira dimension of symmetric products of curves? That is, given a projective smooth, connected complex curve $C$, what is the Kodaira dimension of $C^{(d)}=C^d/\mathfrak S_d$?</p> <p>When $d> g$, the genus of $C$, then $C^{(d)}$ is a bundle in projective spaces over the Jacobian of $C$, hence all pluriforms on $C^{(d)}$ vanish on the fibers of this fibration and $\kappa(C^{(d)})=0$ in this case. </p> <p><em>Is something known for $2\leq g\leq g-1$?</em> (This question is prompted by <a href="http://mathoverflow.net/questions/123862/rational-points-on-surfaces-of-general-type" rel="nofollow">this other post</a>.) I suspect that the answer will strongly depend on fine properties of the curve $X$ (gonality, Brill-Noether properties) and there might not be a general and neat answer.</p> <p>Perhaps surprisingly, the analogous question in higher dimensions is quite different since if $X$ is a projective smooth connected variety of dimension $>1$, the Kodaira dimension of (any desingularization) of $X^{(d)}$ is equal to $d \kappa(X)$, where $\kappa(X)$ is the Kodaira dimension of $X$ (D. Arapura, S. Archava, <em><a href="http://www.ams.org/journals/proc/2003-131-05/S0002-9939-02-06797-7/S0002-9939-02-06797-7.pdf" rel="nofollow">Kodaira dimension of symmetric powers</a></em>, Proc. AMS, 2003).</p> http://mathoverflow.net/questions/123760/topological-characterization-of-the-closed-interval-0-1/123798#123798 Answer by ACL for Topological characterization of the closed interval $[0,1]$. ACL 2013-03-06T18:46:04Z 2013-03-06T18:46:04Z <p>The following topological characterization is close to that of the real line which is indicated in <a href="http://mathoverflow.net/questions/76134/topological-characterisation-of-the-real-line" rel="nofollow">another MO thread</a>, but does not seem to have been pointed out here.</p> <p>Let $X$ be a topological space, let $a,b$ be points of $X$ such that $a\neq b$. Assume that $X$ is compact, connected and separable. The following conditions are equivalent:</p> <ol> <li>There exists a homeomorphism $f\colon [0,1]\to X$ such that $f(0)=a$ and $f(1)=b$.</li> <li>Every connected subset of $X$ containing $\lbrace a,b\rbrace$ is equal to $X$.</li> <li>The space $X$ is locally connected, and any connected and compact subset of $X$ which contains $\lbrace a,b\rbrace$ is equal to $X$.</li> <li>For every $x\in X\setminus\lbrace a,b\rbrace$, the space $X\setminus\lbrace x\rbrace$ is not connected.</li> </ol> http://mathoverflow.net/questions/123513/the-role-of-anr-in-modern-topology The role of ANR in modern topology ACL 2013-03-04T09:23:57Z 2013-03-05T04:50:09Z <p>Absolute neighborhood retracts (ANRs) are topological spaces $X$ which, whenever $i\colon X\to Y$ is an embedding into a normal topological space $Y$, there exists a neighborhood $U$ of $i(X)$ in $Y$ and a retraction of $U$ onto $i(X)$. They were invented by Borsuk in 1932 (Über eine Klasse von lokal zusammenhängenden Räumen, Fundamenta Mathematicae <strong>19</strong> (1), p. 220-242) and have been the object of a lot of developments from 1930 to the 60s (Hu's monograph on the subject dates from 1965), being a central subject in combinatorial topology.</p> <p>The discovery that these spaces had good topological (local connectedness), homological (finiteness in the compact case) and even homotopical properties must have been a strong impetus for the developement of the theory. Also, they probably played some role in the discovery of the homotopy extension property (it is easy to extend homotopies whose source is a normal space and target an ANR) and of cofibrations.</p> <p>I have the impression that this more or less gradually stopped being so in the 70s: a basic MathScinet search does not refer that many recent papers, although they seem to be used as an important tool in some recent works (a colleague pointed to me those of Steve Ferry).</p> <p>My question (which does not want to be subjective nor argumentative) is the following: what is the importance of this notion in modern developments of algebraic topology?</p> http://mathoverflow.net/questions/123096/is-the-canonical-height-of-a-totally-p-adic-point-on-an-abelian-variety-bounded-a/123166#123166 Answer by ACL for Is the canonical height of a totally p-adic point on an abelian variety bounded away from zero? ACL 2013-02-27T23:32:21Z 2013-02-27T23:32:21Z <p>In my paper <em>Mesures et équidistribution sur les espaces de Berkovich</em>, Crelle, 2006, I had proved an equidistribution theorem in the good reduction case. The limit measure is a Dirac mass at a single point of the Berkovich space (the one whose reduction is the generic point of the special fiber). Since the Galois orbit of a totally $p$-adic point is contained in a compact subset disjoint from that point, this implies that there exist a dense Zariski open subset $U$ and a strictly positive real number $c$ such that the Néron-Tate height of any totally $p$-adic point of $U$ is at least $c$.</p> <p>For related results, see also the paper of Baker-Hsia (Crelle, 2005).</p> http://mathoverflow.net/questions/122390/volumes-of-families-of-semialgebraic-sets/122392#122392 Answer by ACL for Volumes of families of semialgebraic sets ACL 2013-02-20T12:43:14Z 2013-02-20T12:43:14Z <p>A theorem by Jean-Philippe Rolin and Jean-Marie Lion asserts a similar property in the analytic category. See <em><a href="https://eudml.org/doc/75301" rel="nofollow">Intégration des fonctions sous-analytiques et volumes des sous-ensembles sous-analytiques</a></em>, Annales de l'institut Fourier (1998) <strong>48</strong> (3), p.755-767. </p> <p>I quote their abstract : "Let $f(x,y)$ be a positive subanalytic function defined on $\mathbf R^n×\mathbf R^m$. We prove that the integral $\int_{\mathbf R^m}f(x,y)\,dy$ is a log-analytic function of $x$. Let $Y_x$ be a subanalytic family of global subanalytic subsets of the euclidean space $\mathbf R^m$. We deduce from the previous result that the $k$-dimensional volume of $Y_x$ is a log-analytic function of $x$. A corollary is the log-analytic behaviour of the $k$-dimensional density of a $k$-dimensional subanalytic set at any point of its topological closure."</p> http://mathoverflow.net/questions/19644/what-is-the-definition-of-canonical/121991#121991 Answer by ACL for What is the definition of "canonical" ? ACL 2013-02-16T15:02:42Z 2013-02-16T15:02:42Z <p>I have two competing interpretations of the word canonical.</p> <p>One, apparently the one used by Bourbaki, is mathematically informal. In various contexts, some objects (maps, modules, etc.) are defined unambiguously and <em>called</em> canonical. For example, the canonical basis of the free module $A^{(I)}$ over a set $I$, the canonical surjection from a set $X$ to its set $X/R$ of equivalence classes with respect to some equivalence relation $R$, the canonical bilinear map from a product of two modules to their tensor product, etc.</p> <p>The other interpretation is categorical. The given context defines (often implicitly) some categories and the canonical object is functorial with respect to <em>isomorphisms</em>. It is more or less what Kevin Buzzard says in its answer, when he defines canonical by the property that he and a colleague, when asked to define the object, would agree on the same object.</p> http://mathoverflow.net/questions/15438/a-slick-proof-of-the-bruhat-decomposition-for-gl-nk/121931#121931 Answer by ACL for A slick proof of the Bruhat Decomposition for GL_n(k)? ACL 2013-02-15T18:41:21Z 2013-02-15T18:41:21Z <p>One can prove the Bruhat decomposition by applying the theorem of Jordan-Hölder. This theorem shows that two chains of submodules of a module of finite length whose successive quotients are simple have the same lengths, and that the same quotients appear. But it is slightly more precise, because it gives a precise recipe for a bijection between the two lists. Here we apply it for modules of finite length over a field, aka finitely dimensional vector spaces.</p> <p>Let $E=(e_1,\dots,e_n)$ and $F=(f_1,\dots,f_n)$ be two bases of a vector space $M$ over a field $K$.</p> <p>For $0\leq i\leq n$, define $M_i=\langle e_1,\dots,e_i\rangle$ and $N_i=\langle f_1,\dots,f_i\rangle$. The proof of the theorem of Jordan-Hölder furnishes a (unique) permutation $\sigma$ of ${1,\dots,n}$ such that $M_{i-1}+M_i\cap N_{\sigma(i)-1}=M_{i-1}$ and $M_{i-1}+M_i\cap N_{\sigma(i)}=M_i$, for every $i\in{1,\dots,n}$. </p> <p>For any $i$, let $x_i$ be a vector belonging to $M_{i}\cap N_{\sigma(i)}$ but not to $M_{i-1}$. For every $i$, one has $\langle x_1,\dots,x_i\rangle =M_i$; it follows that $X=(x_1,\dots,x_n)$ is a basis of $M$; moreover, there exists a matrix $B_1$, in upper triangular form, such that $X=E B_1$.</p> <p>Set $\tau=\sigma^{-1}$. Similarly, one has $\langle x_{\tau(1)},\dots,x_{\tau(i)}\rangle=N_i$ for every $i$. Consequently, there exists a matrix $B_2$, still in upper-triangular form, such that $(x_{\tau(1)},\dots,x_{\tau(n)})=F B_2$.</p> <p>Let $P_\tau$ be the permutation matrix associated to $\tau$, we have $(x_{\tau(1)},\dots,x_{\tau(n)})=(x_1,\dots,x_n)P_\tau$. This implies that $FB_2=EB_1P_\tau$, hence $F=E B_1 P_{\tau} B_2^{-1}$. Therefore, the matrix $A=B_1P_{\tau}P_2^{-1}$ that expresses the coordinates of the vectors of $F$ in the basis $E$ is the product of an upper-triangular matrix, a permutation matrix and another upper-triangular matrix.</p> <p>In the group $\mathop{\rm GL}(n,K)$, let $B$ be the subgroup consisting of upper-triangular matrices, and let $W$ be the subgroup consisting of permutation matrices. We have proved that $\mathop{\rm GL}(n,K)=BWB$: this is precisely the Bruhat decomposition.</p> http://mathoverflow.net/questions/121824/what-sets-of-primes-can-we-pick-out-with-first-order-statements/121847#121847 Answer by ACL for What sets of primes can we pick out with first-order statements? ACL 2013-02-14T22:45:36Z 2013-02-14T22:45:36Z <p>You guessed the correct answer. This is explained in the paper of Mike Fried and George Sacerdote, <em><a href="http://www.math.uci.edu/~mfried/paplist-ff/annals76.pdf" rel="nofollow">Solving Diophantine Problems Over All Residue Class Fields of a Number Field and All Finite Fields</a></em>, The Annals of Mathematics, 2nd Ser., Vol. 104, No. 2. (Sep., 1976), pp. 203-233.</p> <p>Since the theory of fields lacks elimination of quantifiers in the language of rings (the formula $\exists y,\ x=y^2$ which says that $x$ is a square cannot be expressed directly as a polynomial condition on $x$), the authors introduce a richer language, using the concept of Galois stratifications, which allows for elimination of quantifiers. Geometrically, this basically means that one can eliminate quantifiers up to the level of finite extensions of fields.</p> <p>See also Chapters 30 and 31 on Galois stratifications in the book <em>Field arithmetic</em> by Mike Fried and Moshe Jarden.</p> http://mathoverflow.net/questions/121571/connected-sum-of-topological-manifolds Connected sum of topological manifolds ACL 2013-02-12T08:59:55Z 2013-02-13T21:06:57Z <p>A definition of the connected sum of two $n$-manifolds $M$ and $M'$ begins by considering two $n$-balls $B$ in $M$, $B'$ in $M'$, and glueing the varieties $M\setminus \mathring B$ and $M'\setminus \mathring B'$ along their boundary (an $(n-1)$-sphere) by an orientation-reversing homeomorphism. The construction depends a priori on these various choices, but it is asserted at many places of the litterature (Lee's book on topological manifolds for example, as well as <a href="http://en.wikipedia.org/wiki/Connected_sum" rel="nofollow">Wikipedia</a>) that the result does not depend on these choices.</p> <p>In the differentiable case, a reference is given to a theorem of Palais (<em>Natural operations on differential forms</em>, Thm. 5.5) which asserts — roughly — that two embedding of $n$-balls differ by a global diffeomorphism which is isotopic to identity.</p> <p>Are the details of this independence written somewhere in the litterature, both in the continuous and the smooth case?</p> http://mathoverflow.net/questions/121602/two-different-analytic-curves-cannot-intersect-in-infinitely-many-points/121686#121686 Answer by ACL for Two different analytic curves cannot intersect in infinitely many points ACL 2013-02-13T08:31:07Z 2013-02-13T08:31:07Z <p>Let $f\colon [0,1]\to \mathbf R^n$ and $g\colon[0,1]\to\mathbf R^n$ parametrize your curves. The set $C$ of $t\in[0,1]$ such that there exists $s\in[0,1]$ with $f(t)=g(s)$ is a closed subset of $[0,1]$. More importantly, it is <em>definable</em> in the sense of model theory in the language — called $\mathbf R_{\text{an}}$ — consisting of polynomial functions and so called restricted analytic functions, namely analytic functions defined over compact subsets of $\mathbf R^m$.</p> <p>It is an important theorem (Gabrielov; Denef and van den Dries) that $\mathbf R_{\text{an}}$ is $o$-minimal. This means that definable subsets of the real line are finite unions of intervals. </p> <p>From there, one should be able to analyse the situation further when the coincidence set is infinite. I would conjecture that it can only be the union of one or two intervals. Ramiro gave an example with one interval, though more complicated examples are possible, e.g., of the form $s\mapsto F(u(s))$, $s\mapsto F(v(s))$, where $u,v\colon[0,1]\to\mathbf R$ are analytic and $F\colon\mathbf R\to\mathbf R^n$ is a fixed curve. In particular, the set is a union of two intervals if $F$ is a parameterization of a circle and $u$ and $v$ are suitably chosen so as to draw two arcs which overlap twice.</p> http://mathoverflow.net/questions/84674/algebraization-theorems/120994#120994 Answer by ACL for algebraization theorems ACL 2013-02-06T17:40:22Z 2013-02-06T17:40:22Z <p>There are algebraizations theorems in Diophantine Geometry of an apparently different nature. In fact, Jean-Benoît Bost has explained how to think of them as variants of Grothendieck's existence theorem over a compactification of $\mathop{\rm Spec} (\mathbf Z)$ and has developed this point of view in many papers.</p> <p>Examples are: </p> <ul> <li><p>Theorems of Chudnovsky, André, Bost, according to which some formal subgroups of Abelian varieties defined over number fields are algebraic. This works more generally for leaves of appropriate foliations in algebraic varieties. See Bost's paper, <a href="http://www.numdam.org/item?id=PMIHES_2001__93__161_0" rel="nofollow">Algebraic leaves of algebraic foliations over number fields</a>. Publications Mathématiques de l'IHÉS, 93 (2001), p. 161-221. See also my Bourbaki talk on the subject, <a href="http://www.numdam.org/item?id=SB_2000-2001__43__175_0" rel="nofollow">Théorèmes d'algébricité en géométrie diophantienne.</a> Séminaire Bourbaki, 43 (2000-2001), Exposé No. 886.</p></li> <li><p>A Lefschetz type theorem, also due to Bost, proving that fundamental groups of some algebraic surfaces are trivial (come from the base), see <a href="http://www.numdam.org/item?id=ASENS_1999_4_32_2_241_0" rel="nofollow">Potential theory and Lefschetz theorems for arithmetic surfaces.</a> Annales scientifiques de l'École Normale Supérieure, Sér. 4, 32 no. 2 (1999), p. 241-312 </p></li> <li><p>The theorem of Borel-Dwork-Polya-Bertrandias asserting that some power series with rational coefficients are rational is also of this kind, see my paper with Bost, <a href="http://www.ams.org/mathscinet-getitem?mr=2641171" rel="nofollow">Analytic curves in algebraic varieties over number fields</a>. In: Algebra, arithmetic, and geometry: in honor of Yu. I. Manin. Vol. I, 69–124, Progr. Math., 269, Birkhäuser Boston, Inc., Boston, MA, 2009. <a href="http://arxiv.org/abs/math/0702593" rel="nofollow">(http://arxiv.org/abs/math/0702593)</a></p></li> </ul> http://mathoverflow.net/questions/120536/is-the-empty-graph-a-tree/120842#120842 Answer by ACL for Is the empty graph a tree? ACL 2013-02-05T08:05:44Z 2013-02-05T08:05:44Z <p>The correct formulation of Kirchoff's Theorem is the following result, which does not require any connectedness assumption on the graph $G$, and does not lead to false expectations when the graph is empty.</p> <p>Let $\Delta\colon L^2(G)\to L^2(G)$ be the Laplacian on a finite graph $G$. Then $\ker(\Delta)$ is the space of locally constant functions on $G$; its dimension is the number of connected components of $G$. Let $L^2_0(G)$ be the orthogonal of $\ker(\Delta)$, a subspace stable under $\Delta$. Then $\det(\Delta|L^2_0(G))$ is the number of maximal forests in $G$.</p> http://mathoverflow.net/questions/120442/are-rational-varieties-simply-connected/120454#120454 Answer by ACL for Are rational varieties simply connected? ACL 2013-01-31T19:59:36Z 2013-02-03T20:27:55Z <p>In positive characteristic, the (étale) fundamental group of a rationally connected variety is finite (Kollár, Inventiones Math., 1993). </p> <p>And this happens: Let us assume that the base field has characteristic $p$, where $p\neq 5$ and $p\not\equiv 1\pmod 5$. Then, the hypersurface with equation $X_0^5+\dots+X_3^5=0$ in $\mathbf P^3$ is unirational; so is its quotient by the obvious action of $\mu_5$ (a surface of general type known as the Godeaux surface), which is therefore an unirational variety with fundamental group $\mathbf Z/5$.</p> <p>However, Ekedahl has proved that this group is prime to the characteristic. I discussed that in my Bourbaki Seminar talk, « <a href="http://front.math.ucdavis.edu/math.AG/0303052" rel="nofollow">Points rationnels et groupes fondamentaux, applications de la cohomologie $p$-adique</a> », <em>Astérisque</em> <strong>294</strong>, p. 125-146.</p> http://mathoverflow.net/questions/120184/how-to-write-down-explictly-the-isomorphism-of-two-finite-dimensional-representat/120201#120201 Answer by ACL for How to write down explictly the isomorphism of two finite dimensional representation of compact groups? ACL 2013-01-29T12:10:20Z 2013-01-29T12:10:20Z <p>I find easier to construct "the" isomorphism between the direct sums, for variable $n$, of the pair of spaces you have.</p> <p>We consider the algebraic De Rham complex $C$ of the affine space~$V$, that is: for every $p$, $C^p$ is the space of differential forms of degree $p$ with polynomial coefficients and $C$ is the direct sum of the spaces $C^p$. I let $C^+$ be the direct sum of the $C^p$, for $p>0$. The exterior derivation $d$ maps $C^p$ to $C^{p+1}$; let $Z^p$ be its kernel (space of closed forms of degree $p$). It is known (we assume that the characteristic of $K$ zero) that the cohomology of the complex $(C,d)$ is zero in strictly positive degrees. In other words, $d$ induces isomorphisms $C^p/Z^p\simeq Z^{p+1}$. Moreover, $C^0$ is the ring of polynomials and $Z^0=K$ (constant polynomials). In differential topology, the obvious contraction from the affine space to the origin furnishes canonical antiderivatives to closed forms. One can check that it preserves forms with polynomial coefficients. We thus get linear maps $I\colon Z^{p+1}\to C^p$ such that $d\circ I = \mathrm{id}$ and $C^p=Z^p\oplus I(Z^{p+1})$ for every $p\geq 0$.</p> <p>Let $C_e$ and $C_o$ be the subspace of $C$ consisting of differential forms whose coefficients are even, respectively odd, polynomials. One checks that $d(C_e)\subset C_o$, $d(C_o)\subset C_e$, $I(C_e^+)\subset C_o$ and $I(C_o^+)\subset C_e$. Consequently, $$C_e = Z_e \oplus I(Z_o^+) = K \oplus Z_e^+ \oplus I(Z_o^+)$$ and $$C_o = Z_o\oplus I(Z_e^+)=Z_o^+\oplus I(Z_e^+).$$</p> <p>Let $u\colon C_e\to C_o$ be the map given by $0$ on $K$, $I$ on $Z_e^+$ and $d$ on $I(Z_o^+)$, let $v\colon C_o\to C_e$ be the map given by $I$ on $Z_o^+$ and $d$ on $I(Z_e^+)$. Then $u\circ v$ is the identity and $v\circ u$ is the projection from $C_e$ to $Z_e^+\oplus I(Z_o^+)$ with kernel $K$. They induce the isomorphisms you are looking for.</p> http://mathoverflow.net/questions/118254/usage-of-set-theory-in-undergraduate-studies Usage of set theory in undergraduate studies ACL 2013-01-07T09:09:17Z 2013-01-09T14:50:39Z <p>I would like to ask my colleagues their thought on good practices concerning set theorical framework in undergraduate studies. For example, have there been any attempt to use another mathematical formalism, such as ETCS? (for research issues, see <a href="http://mathoverflow.net/questions/116701/how-would-set-theory-research-be-affected-by-using-etcs-instead-of-zfc" rel="nofollow">this question</a>).</p> <p>While most, if not all, of our mathematics are thought, done, written using set theory, our younger students seem to struggle with these concepts. Some can well put a $4\times 6$ matrix in row reduced echelon form but plainly do not understand the meaning of a question like "If $A,B$ are two square matrices of size~$n$, prove that $\ker(AB)$ contains $\ker(B)$." The difficulties with $\varepsilon,\delta$ definition of limits may be of a similar nature.</p> <p>In fact, one may argue that all set theoretical concepts presently are more or less eliminated from the lower levels of mathematical education. One may even argue that it should be so. I remember that each one of the first years of middle school (from 6th grade on, the French and US systems coincide here!) taught me one new definition in set theory; sets and mappings at the age of 11, then equivalence relations, then sets of equivalence classes (to define vectors)... And a few years later, students are taught quotient groups like $\mathbf Z/n\mathbf Z$ as sets of equivalence classes, a definition which they of course take litteraly. </p> <p>While Set theory is very useful to formalize things, at least once you're used to it, it is true that it allows stupid questions, requires abuses of notations (so that one does not distinguish between the $1$ of $\mathbf Z$ with the $1$ of $\mathbf R$, not forgetting thoses of $\mathbf Q$ and $\mathbf C$). In some sense, modern mathematicians, especially algebraists, speak sets but think categories. This may be related with the fact that the precise definition of the axioms of set theory (ZFC, say) are not so well known among mathematicians, and even not really taught (for example, no mention of the replacement axiom in my own mathematical education). In contrast, a more recent book like Terence Tao's <em>Analysis</em> begins with a precise exposition of these axioms, up to this replacement axiom.</p> <p>I can't really make my mind between one attitude and the opposite. So what do you think?</p> http://mathoverflow.net/questions/15611/to-prove-the-nullstellensatz-how-can-the-general-case-of-an-arbitrary-algebraica/118210#118210 Answer by ACL for To prove the Nullstellensatz, how can the general case of an arbitrary algebraically closed field be reduced to the easily-proved case of an uncountable algebraically closed field? ACL 2013-01-06T16:34:09Z 2013-01-06T18:08:19Z <p>The easiest way to reduce to the uncountable case may be as follows. Let $I$ be an ideal of $k[X_1,...,X_d]$ which does not contain $1$. Let $P_1,\dots,P_r$ be a generating family of $I$.</p> <p>Let $A=k^{\mathbf N}$ and let $m$ be a maximal ideal of $A$ which contains the ideal $N=k^{(\mathbf N)}$ of $A$. Then $K=A/m$ is an algebraically closed field which is has at least the power of the continuum. (Alternative description: let $K$ be an ultrapower of $k$, with respect to a non-principal ultrafilter.)</p> <p><strong>Lemma</strong>. *For $i\in{1,\dots,r}$, let $a_i=(a_{i,n})\in A$. Assume that $(\bar a_1,\dots,\bar a_r)=0$ in $K^r$. Then the set of $n\in\mathbf N$ such that $(a_{1,n},\dots,a_{r,n})=0$ is infinite.*</p> <p>Proof. Assume otherwise. For every $n$ such that $(a_{1,n},...,a_{r,n}) \neq 0$, choose $(b_{1,n},\dots,b_{r,n})$ such that $\sum a_{i,n}b_{i,n}=1$, and let $b_i=(b_{i,n})_n\in A$. Then $\sum a_i b_i - 1$ belongs to $N^r$, hence $\sum \bar a_i \bar b_i=1$. Contradiction.</p> <p>Thanks to the lemma, one proves easily that the ideal $I_K$ of $K[X_1,...,X_d]$ generated by $I$ does not contain $1$. By the uncountable case, there exists $x=(x_1,...,x_d)\in K^d$ such that $P_j(x_1,...,x_d)=0$ for every $j$. For every $i$, let $a=(a_{n})\in A^d$ be such that $\bar a=x$. By the lemma again the set of integers $n$ such that $P_j(a_n) \neq 0$ for some $j$ is finite. In particular, there exists a point $y\in k^d$ such that $P_j(y)=0$ for every $j$.</p> http://mathoverflow.net/questions/55701/are-there-only-finite-many-maximal-left-ideals-for-a-left-artinian-ring/117621#117621 Answer by ACL for Are there only finite many maximal left ideals for a left Artinian ring? ACL 2012-12-30T12:44:37Z 2012-12-30T13:00:03Z <p>As indicated by Pete Clark, an Artinian commutative ring has only finitely many maximal ideals.</p> <p>However, the correct generalization is not the one you seem to expect but the following: <em>If $A$ is right-Artinian, the set of isomorphism classes of simple right-$A$-modules is finite.</em></p> <p>This implies the commutative case. Assume that $A$ is commutative. Then, if $S$ is a simple $A$-module, there exists a maximal ideal $\mathfrak m$ of $A$ such that $S\simeq A/\mathfrak m$; moreover, $\mathfrak m$ is the annihilator of $S$ so that there is a bijection between isomorphism classes of simple $A$-modules and maximal ideals of $A$.</p> http://mathoverflow.net/questions/117241/arithmetic-dynamics-and-dynamics-on-moduli-spaces/117257#117257 Answer by ACL for Arithmetic dynamics and dynamics on moduli spaces ACL 2012-12-26T14:07:10Z 2012-12-26T14:07:10Z <p>Concerning dynamics on moduli spaces (and work of Avila, Eskin, Forni, Gouëzel, Hubert, Kontsevich, McMullen, Yoccoz, Eskin,...), I would first study three Bourbaki Seminars talks:</p> <ul> <li>Raphaël Krikorian, <em>Déviations de moyennes ergodiques</em>, Exp. 927, 2003-2004</li> <li>Jean-Christophe Yoccoz, <em><a href="http://www.college-de-france.fr/media/jean-christophe-yoccoz/UPL26933_Exp.996.JC.Yoccoz.pdf" rel="nofollow">Échanges d'intervalles et surfaces de translation</a></em>, Exp. 996, 2008-2009.</li> <li>Julien Grivaux &amp; Pascal Hubert, <em><a href="http://www.cmi.univ-mrs.fr/~jgrivaux/articles/Bourbaki.pdf" rel="nofollow">Exposants de Lyapunov du ﬂot de Teichmüller</a></em>, 2012.</li> </ul> http://mathoverflow.net/questions/117238/how-to-calculate-zeroth-crystalline-cohomology/117254#117254 Answer by ACL for How to calculate zeroth crystalline cohomology ACL 2012-12-26T13:46:09Z 2012-12-26T13:46:09Z <p>In this generality, this is unfortunately not so easy because it requires to compute universal PD-envelopes. The case where $X/S$ is embedded in a smooth scheme $Y/S$ is explained in the book by Pierre Berthelot, <em>Cohomologie cristalline des schémas de caractéristique $p>0$</em>, Lecture notes in mathematics 407, 1974 (see chapter V). </p> <p>You may also look at a paper by Jean-Marc Fontaine (« Cohomologie de de Rham, cohomologie cristalline et représentations $p$-adiques », in <em>Algebraic geometry Kyoto-Tokyo</em>, Lecture notes in mathematics 1016, 1983, p. 86-108). There, he proves that the 0th crystalline cohomology of $O_{\bar K}$ ($K$ local field) is the ring $B_{\rm{cris}}$ he had defined earlier.</p> http://mathoverflow.net/questions/117027/how-we-obtain-information-about-a-variety-from-an-algebraic-group-acting-on-it/117080#117080 Answer by ACL for How we obtain information about a variety from an algebraic group acting on it ACL 2012-12-23T10:42:04Z 2012-12-23T10:42:04Z <p>As indicated by the other answers, you question is not specific enough.</p> <p>In the particular case where $G$ is the multiplicative group, a theorem of Białynicki-Birula (On fixed point schemes of actions of multiplicative and additive groups. <em>Topology</em> <strong>12</strong> (1973), 99–103, <a href="http://www.ams.org/mathscinet-getitem?mr=313261" rel="nofollow">MR:313261</a>), furnishes a decomposition of $V$ into locally closed subsets $V_i$, each of them being stable under the action of $G$ and a trivial fibration over the fixed point set $V_i^G$.</p> http://mathoverflow.net/questions/116956/numerable-covers-from-the-point-of-view-of-grothendieck-topologies Numerable covers from the point of view of Grothendieck topologies ACL 2012-12-21T08:46:59Z 2012-12-21T11:12:33Z <p>Let $G$ be a topological group. Recall that its classifying space $BG$ is a CW-complex which is the base of a locally trivial principal bundle of group $G$, with contractible total space $EG$. It classifies principal $G$-bundles in the sense that for any <em>paracompact</em> space $B$, isomorphism classes of locally trivial principal $G$-bundles are in natural bijection with homotopy classes of maps from $B$ to $BG$. </p> <p>More generally, if $B$ is any topological space, homotopy classes of maps from $B$ to $BG$ are in bijection with isomorphism classes of principal $G$-bundles for which there exists a <em>numerable open cover</em> $(U_i)_i$ of $B$ such that the bundle is trivial on each $U_i$. This means that there exists a partition of unity subordinate to that cover. I believe that this is due do A. Dold (Partitions of unity in the theory of fibrations, Annals of math., 1965).</p> <p>Now observe that the family numerable covers defines a <em><a href="http://en.wikipedia.org/wiki/Grothendieck_topology" rel="nofollow">Grothendieck topology</a></em> on a topological space. Hence, this result of Dold falls within the circle of thoughts (development of étale cohomology on algebraic varieties, sites, toposes,...) that was actively developed by Grothendieck at the time Dold wrote his paper. </p> <p>My question is whether there has been any kind of connections between these two groups of mathematicians or, in the contrary, whether this is a mere coïncidence.</p> http://mathoverflow.net/questions/42512/awfully-sophisticated-proof-for-simple-facts/116957#116957 Answer by ACL for Awfully sophisticated proof for simple facts ACL 2012-12-21T09:02:02Z 2012-12-21T09:02:02Z <p>Liouville remarked that the fundamental theorem of algebra could be derived from his theorem that elliptic functions (doubly periodic meromorphic functions of one complex variable) must have poles. The proof goes by substituting the inverse of a polynomial as the argument of, say, Weierstrass $\wp$-function with large enough periods, and observing that it has no poles.</p> <p>Of course, the proof of Liouville's theorem on elliptic functions requires the same kind of arguments used for proving the famous Liouville theorem (due to Cauchy) that bounded holomorphic functions are bounded and, apparently, already used before by Cauchy for algebraic functions.</p> <p>But Liouville's observation is really more complicated than the present proof. What it simplifies, however, is the compactness argument. For elliptic functions, or for algebraic functions, one has at hand a compact Riemann surface on which some holomorphic function is bounded, hence achieves its supremum, etc. This may be the reason why the general form of Liouville theorem came only after the case of algebraic or elliptic functions.</p> http://mathoverflow.net/questions/116755/can-liouvilles-number-be-expressed-as-a-physical-ratio-in-the-sense-that-pi-i/116767#116767 Answer by ACL for Can Liouville's number be expressed as a physical ratio in the sense that $\pi$ is? ACL 2012-12-19T08:08:39Z 2012-12-19T08:25:39Z <p>The answer is probably <em>no</em>.</p> <p>In his paper <a href="http://www.math.jussieu.fr/~miw/articles/pdf/TranscendencePeriods.pdf" rel="nofollow">Transcendence of Periods: the State of the Art</a> (Pure and Applied Mathematics Quarterly, Volume 2, Number 2, p. 435-463, 2006), Michel Waldschmidt conjectures that no period is a Liouville number (see questions 2 &amp; 3 in the introduction). This expectation is supported by the main advances of transcendental number theory in 20th century, notably Baker's theory of linear forms in logarithms and its extensions to commutative algebraic groups.</p> <p>Here, the word <em>period</em> has to be understood in the sense of Kontsevich and Zagier: a complex number whose real and imaginary parts are values of absolutely convergent integral of rational functions with rational coeﬃcients, over domains in $\mathbf R$ given by polynomial inequalities with rational coeﬃcients. (The preceding definition is quoted from the paper <a href="http://arxiv.org/abs/0805.0349" rel="nofollow">Periods and elementary real numbers</a> by Masahiko Yoshinaga, who was apparently the first to prove that periods belong to the field of elementary complex numbers, those whose real and imaginary parts can be effectively approximated by Cauchy sequences of rationals.)</p> http://mathoverflow.net/questions/115567/covering-maps-in-real-life-that-can-be-demonstrated-to-students/115593#115593 Answer by ACL for Covering maps in real life that can be demonstrated to students ACL 2012-12-06T09:02:10Z 2012-12-06T09:02:10Z <p>The fact that the fundamental group SO(3) is Z/2 is sometimes shown by a funny movement of the arm, twisting it once while keeping the hand palm up, and detwisting it by twisting it once more. See also <a href="http://math.stackexchange.com/questions/17856/visualizing-the-fundamental-group-of-so3" rel="nofollow">this thread</a> on Stackexchange.</p> http://mathoverflow.net/questions/115560/primitive-kth-root-of-unity-in-a-finite-field-mathbbf-p/115561#115561 Answer by ACL for Primitive $k$th root of unity in a finite field $\mathbb{F}_p$ ACL 2012-12-05T23:52:48Z 2012-12-05T23:52:48Z <p>Presumably, you are assuming that $k$ divides $p-1$, so that there is effectively a primitive $k$th root of unity in ${\bf F}_p$, even $\phi(k)$ of them ($\phi$ is Euler's totient function). The simplest method I know to get your hand on one is as follows.</p> <p>A. Factor $k=\ell_1^{n_1}\dots \ell_r^{n_r}$ as a product of distinct prime numbers with exponents. </p> <p>B. For every $i=1,\dots,r$, do the following: Take a random element $x$ in ${\bf F}_p$ and compute $x^{(p-1)/\ell_i}$ in $F_p$, until the result is different from $1$. Then set $a_i=x^{(p-1)/\ell_i^{n_i}}$.</p> <p>C. Set $a=a_1 a_2\cdots a_r$. This is a primitive $k$th root of unity in ${\bf F}_p$.</p> <p>In practice, $k$ should be a power of $2$, $k=2^n$, so that $r=1$ and you only have to repeat step B once.</p> http://mathoverflow.net/questions/130062/concrete-examples-of-noncongruence-arithmetic-subgroups-of-sl2-r/130063#130063 Comment by ACL ACL 2013-05-08T20:18:31Z 2013-05-08T20:18:31Z To construct non-congruence subgroups of $PSL(2,\mathbf Z)$, it is easier to start from the congruence subgroup $\Gamma(2)$ — matrices congruent to identity modulo $2$, for this group is free on two generators. Since the symmetric group $S_k$ is generated by a transposition and a $k$-cycle, it is a quotient of $\Gamma(2)$. http://mathoverflow.net/questions/129086/borels-paris-lectures/129089#129089 Comment by ACL ACL 2013-04-29T13:49:23Z 2013-04-29T13:49:23Z Some (illegal but useful) websites have a copy of it in DJVU format. http://mathoverflow.net/questions/114555/does-physics-need-non-analytic-smooth-functions/128686#128686 Comment by ACL ACL 2013-04-25T14:45:57Z 2013-04-25T14:45:57Z @Willie. Indeed, thanks! http://mathoverflow.net/questions/127643/algebraic-machinery-for-algebraic-geometry/127662#127662 Comment by ACL ACL 2013-04-19T06:54:53Z 2013-04-19T06:54:53Z @Filippo: Unfortunately, not for cohomology. http://mathoverflow.net/questions/125189/have-we-ever-proved-any-non-solvable-case-of-reciprocity-without-the-langlands-pr/125807#125807 Comment by ACL ACL 2013-03-28T23:27:03Z 2013-03-28T23:27:03Z @Jo&#235;l. Something can be deduced by transcendental number theoretical methods (Chudnovski, Andr&#233;, Bost) applied to the Grothendieck-Katz conjecture on differential equations, although this is certainly much weaker than a reciprocity. For example, one can prove without any Galois theory that an irreducible polynomial of degree $&gt;1$ with integer coefficients can't have a root modulo almost all prime numbers, even modulo too high a (weak form of) density of prime numbers. (One proves that the power series $(1+x)^\alpha$ would be an algebraic function, hence $\alpha\in\mathbf Q$.) http://mathoverflow.net/questions/119825/constructible-topology-on-schemes Comment by ACL ACL 2013-03-25T18:11:02Z 2013-03-25T18:11:02Z Together with its constructible topology, the spectrum of $k[X_1,...,X_n]$ is homeomorphic to the space of $n$-types with parameters in $k$ (in the sense of model theory) for the first-order theory ACF. (If $a\in K^n$, where $K$ is an overfield of~$a$, send its type $\mathrm{tp}(a)$ to the prime ideal of polynomials vanishing at $a$). Two apparently (but only apparently) distinct worlds in which the same object is defined... http://mathoverflow.net/questions/16991/what-are-the-connections-between-pi-and-prime-numbers/19502#19502 Comment by ACL ACL 2013-03-15T17:20:49Z 2013-03-15T17:20:49Z What is funny is that they consider 1 to be a prime number... (although this makes no difference for the product). http://mathoverflow.net/questions/40632/what-is-the-inverse-image-sheaf-necessary-for-in-algebraic-geometry/40706#40706 Comment by ACL ACL 2013-03-14T21:55:45Z 2013-03-14T21:55:45Z I would also add : 6. Sheaves of groups (such as the multiplicative group whose first cohomology group is the Picard group). http://mathoverflow.net/questions/124529/volume-of-pgl2-f-pgl2-a Comment by ACL ACL 2013-03-14T21:50:49Z 2013-03-14T21:50:49Z @Paul Garrett: I am confused. Up to a rational number, the measure of the arithmetic quotient $SL_n(\mathbf R)/SL_n(\mathbf Z)$ is a product of zeta values for every reasonable normalization. What is equal to 1, is the measure of the adelic quotient $SL_n(\mathbf A)/SL_n(\mathbf Q)$ - once the Haar measure is suitably normalized. http://mathoverflow.net/questions/124529/volume-of-pgl2-f-pgl2-a Comment by ACL ACL 2013-03-14T21:47:22Z 2013-03-14T21:47:22Z @Agol: The Tamagawa number is 1 for the special linear group. For the projective linear group, which is isogeneous to the special linear group, it is some rational number (whom I do not remember). http://mathoverflow.net/questions/124529/volume-of-pgl2-f-pgl2-a Comment by ACL ACL 2013-03-14T15:57:01Z 2013-03-14T15:57:01Z This should be in Weil's <i>Adeles and algebraic groups</i>, chapter 3. http://mathoverflow.net/questions/124505/the-work-of-pierre-deligne Comment by ACL ACL 2013-03-14T15:06:11Z 2013-03-14T15:06:11Z The proof that the Weil conjectures implies the Ramanujan conjecture on coefficients of modular forms only exists as a Bourbaki Seminar. Serre wrote a few times (as a footnote in his Course in Arithmetic, or in his Complete Works) that he regrets this. Around 10 years ago, Brian Conrad had undertaken to write a book on the topic, but I don't think that the book has appeared (yet). http://mathoverflow.net/questions/124484/stokes-theorem-for-manifolds-without-orientation Comment by ACL ACL 2013-03-14T09:22:03Z 2013-03-14T09:22:03Z You need orientability to define the integral, so the form $\omega$ would have to be a twisted differential form -&#160;with values in some kind of orientation bundle so that $\omega$ has an integral over $\partial M$ and $d\omega$ (whose existence now requires a connexion on this bundle) has an integral over $M$. Presumably, if the definitions are correct, harvesting the formula will be easy. http://mathoverflow.net/questions/124414/proving-1-p-infty-f-in-lp-mathbb-rn-g-in-lp-mathbb-rn Comment by ACL ACL 2013-03-13T21:10:22Z 2013-03-13T21:10:22Z .. and some definition of $p'$... http://mathoverflow.net/questions/108169/modern-proof-of-serres-open-image-theorem/108237#108237 Comment by ACL ACL 2013-03-13T09:09:22Z 2013-03-13T09:09:22Z @Adam Harris: The image of the $2g$-dimensional Galois representation is contained in the $\ell$-adic Mumford-Tate group which is a subgroup of the general symplectic group $GSp_{2g}(\mathbf Z_\ell)$. The Mumford-Tate conjecture asserts that the image is open in the M-T group. Hall's paper provides a criterion which asserts that the M-T group is the full symplectic group and the Galois image is open.