User damian r&#246;ssler - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T04:13:09Z http://mathoverflow.net/feeds/user/17308 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/133059/belyis-theorem-for-function-fields/133065#133065 Answer by Damian Rössler for Belyi's theorem for function fields Damian Rössler 2013-06-07T13:12:50Z 2013-06-07T13:12:50Z <p>Much stronger results are available in positive characteristic. See </p> <p>Kedlaya, Kiran S.</p> <p>More étale covers of affine spaces in positive characteristic. J. Algebraic Geom. 14 (2005), no. 1, 187–192. </p> http://mathoverflow.net/questions/111980/reference-request-for-the-theory-of-heights-over-function-fields Reference request for the theory of heights over function fields Damian Rössler 2012-11-10T11:17:25Z 2013-01-21T20:40:32Z <p>I am looking for an article or book where the theory of heights over function fields (in any characteristic) is treated. I am especially interested in Northcott-type statements. For instance, over a function field $K$ over $\bf Q$, say, a subvariety $X$ of ${\bf P}^n_{K}$, which has a dense subset of $K$-points with bounded height, should have a model over (possibly a finite extension of) $\bf Q$. Where can I find the proof of such a statement ? When $K$ is a function field over a finite field, then one can use Hilbert schemes to get Northcott-type finiteness statements but in general, it seems that one should combine the theory of Hilbert schemes with some descent arguments. I would be grateful for any suggestions.</p> http://mathoverflow.net/questions/118989/the-dual-abelian-scheme/118999#118999 Answer by Damian Rössler for the dual abelian scheme Damian Rössler 2013-01-15T16:56:44Z 2013-01-15T16:56:44Z <p>See Bosch, Luetkebohmert, Raynaud, "Néron Models", chap. 8, 8.4, p. 234.</p> http://mathoverflow.net/questions/115483/vanishing-of-local-cohomology-h2-x-y-left-frac-bbb-zx-y5x4y-right/115603#115603 Answer by Damian Rössler for vanishing of local cohomology $H^2_{(x,y)}\left(\frac{\Bbb Z[x,y]}{(5x+4y)}\right)=0$ Damian Rössler 2012-12-06T11:31:54Z 2012-12-11T08:21:52Z <p>I wasn't able to use the method you suggest. Here is a different proof. Let $M$ be any ${\bf Z}[x,y]$-module. We have (see Brodmann and Sharp, Th. 1.3.8), $$ H^2_{(x,y)}(M)=\varinjlim_n \operatorname{Ext}^2({\bf Z}[x,y]/(x,y)^n,M) $$ where the limit is an inductive limit. Now we have $$ \varinjlim_n \operatorname{Ext}^2({\bf Z}[x,y]/(x,y)^n,M)=\varinjlim_n \operatorname{Ext}^2({\bf Z}[x,y]/(x^n,y^n),M) $$ because for any $n$, we have $(x^n,y^n)\subseteq (x,y)^n$ and $(x,y)^{2n}\subseteq (x^n,y^n)$. Now the ideal $(x^n,y^n)$ defines a regular closed immersion into $\operatorname{Spec} {\bf Z}[x,y]$, so that we have the "fundamental local isomorphism" (see Hartshorne, Residues and duality, Prop. 7.2): ... EDIT: the end of the original argument was wrong. I don't know how to make it work. </p> http://mathoverflow.net/questions/114415/global-section-of-vector-bundle-and-reduction/114498#114498 Answer by Damian Rössler for global section of vector bundle and reduction Damian Rössler 2012-11-26T07:55:56Z 2012-11-26T07:55:56Z <p>Here is a counterexample to (2).</p> <p>Let $H_1$ be the lifting of the trivial line bundle on $C_0$ and suppose that $H_1\not\simeq{\cal O}_{C_1}$. Examples of such line bundles $H_1$ may be produced using the Picard scheme of $C_1$ over $W_2(k)$. I contend that the morphism $H^0(C_1,H_1)\to H^0(C_0,H_0)$ vanishes. To see this, let $\sigma_1\in H^0(C_1,H_1)$. This corresponds to a morphism of sheaves $\sigma_1:{\cal O}_{C_1}\to H_1$. Let $K_1$ be the kernel of $\sigma_1$ and ${\rm CK}_1$ be the cokernel of $\sigma_1$. Let let $K_0$ (resp. ${\rm CK}_0$) be the reduction mod. $p$ of $K_1$ (resp. ${\rm CK}_1$). The reduction mod. $p$ of $\sigma_1$ gives a morphism $\sigma_0:{\cal O}_{C_0}\to H_0\simeq{\cal O}_{C_0}$. We want to show that $\sigma_0=0$. To get a contradiction, suppose that $\sigma_0\not=0$. Then $\sigma_0$ is an isomorphism, since $C_0$ is proper over $k$ and the source and target of $\sigma_0$ are trivial. Since the tensor product is right-exact, we deduce that ${\rm CK}_0$ vanishes; but this implies that ${\rm CK}_1$ vanishes. Now using the fact that $H_1$ is locally free, we deduce likewise that $K_0$ vanishes and hence that $K_1$ vanishes. This shows that $\sigma_1$ is an isomorphism, which contradicts the assumption on $H_1$. Hence $\sigma_0=0$, which is what we wanted. </p> http://mathoverflow.net/questions/104436/ti-functors-are-isomorphic-for-analytically-isomorphic-isolated-singular-points/104605#104605 Answer by Damian Rössler for T^i functors are isomorphic for analytically isomorphic isolated singular points Damian Rössler 2012-08-13T09:43:36Z 2012-08-13T09:43:36Z <p>Here is a stab at an answer, but it is incomplete. </p> <p>Let $S:={\rm Spec}\, B$ and let $\widehat{S}:={\rm Spec}\,\widehat{B}$ be the completion of $B$ along its maximal ideal $m$. Let $\phi:\widehat{S}\to S$ be the natural morphism (which is faithfully flat). The composition of morphisms $\widehat{S}\to S\to{\rm Spec}\,k$ gives rise to a triangle of cotangent complexes and hence to an exact sequence $$ \dots\to T^1_{\phi}\to T^1_{\widehat{S}/k}\to \phi^*T^1_{S/k}\to T^2_{\phi}\to T^2_{\widehat{S}/k}\to \phi^*T^2_{S/k}\to T^3_\phi\to\dots {\rm (*)} $$ so what you need to show is that $T^i_\phi$ vanishes for $i=1,2,3$. You would then get isomorphisms $T^i_{\widehat{S}/k}\to \phi^*T^i_{S/k}$ and since you can repeat this for $B'$ instead of $B$, you would get the required isomorphism. </p> <p>Let $L_\phi$ be the cotangent complex of $\widehat{S}/S$. This complex $L_\phi$ is concentrated in degree $0$. A quick way to see this is to notice that </p> <ul> <li><p>the formation of the cotangent complex is compatible with direct limits of rings (see Quillen, "Cohomology of commutative rings", eq. (4.11))</p></li> <li><p>the ring $B$ is excellent because it is essentially of finite type over a field (Grothendieck) and thus the fibres of $\phi$ are geometrically regular;</p></li> <li><p>a (deep...) result of Popescu (see for instance Th. 1.3 in "Approximations of versal deformations", by B. Conrad and AJ de Jong) then implies that $\widehat{B}$ is a direct limit of smooth $B$-algebras, and for the latter the cotangent complex is clearly concentrated at $0$. </p></li> </ul> <p>By considering the sequence analogous to (*) for the modules $T_i$ instead of $T^i$, this proves the analog of your assertion for the $T_i$. Furthermore, we see that $T^i_\phi={\rm Ext}^i(L_\phi,\widehat{B})={\rm Ext}^i(\Omega_\phi,\widehat{B})$.</p> <p>So the issue is to show that ${\rm Ext}^i(\Omega_\phi,\widehat{B})=0$ for $i>0$. Maybe your hypothesis on the fact that the singularity of $S$ is isolated at the closed point plays a role here.</p> http://mathoverflow.net/questions/103848/mordell-weil-group-of-the-universal-abelian-scheme Mordell-Weil group of the universal abelian scheme Damian Rössler 2012-08-03T09:56:50Z 2012-08-03T11:27:07Z <p>Let $n>2$ and let $k$ be either $\bf Q$ or a finite field whose characteristic is prime to $n$. Let $A_{g,n}$ be the moduli scheme, which represents the functor, which with every $k$-scheme $S$ associates the set of principally polarized abelian schemes $\cal A$ over $S$, together with a symplectic isomorphism <code>$({\bf Z}/n{\bf Z})^{2g}_S\simeq A[n]$.</code> This scheme is geometrically irreducible by Chai-Faltings. Let <code>${\widetilde A}_{g,n}\to A_{g,n}$</code> be the universal family and let $K$ be the function field of $A_{g,n}$.</p> <p>My ${\bf question}$ is: is anything known about <code>${\widetilde A}_{g,n}(K)$</code> ? $(\ast)$ </p> <p>A guess would be that <code>${\widetilde A}_{g,n}(K)\simeq {\widetilde A}_{g,n}[n](K)$.</code> </p> <p>Note that part of the difficulty of the question $(\ast)$ lies in the fact that I am asking for the structure of <code>${\widetilde A}_{g,n}(K)$</code> and not for the structure of its subset <code>${\widetilde A}_{g,n}({A}_{g,n})$.</code> </p> <p>In the case $g=1$ (elliptic curves), these two sets coincide and the question should be easier to answer.</p> <p>A final remark is that question $(\ast)$ is maybe not "the right one". It might make more sense to ask for the structure of the group of rational sections of the universal abelian scheme over the moduli stack of all abelian varieties (forgetting level structures and even polarizations) - but this group is not the Mordell-Weil group of a concrete abelian variety so I prefer to focus on the more down-to-earth question $(\ast)$. </p> http://mathoverflow.net/questions/103695/does-the-self-product-of-a-g-dimensional-abelian-variety-contain-an-abelian-var/103697#103697 Answer by Damian Rössler for Does the self-product of a $g$-dimensional abelian variety contain an abelian variety of dimension smaller than $g$ at some point Damian Rössler 2012-08-01T15:40:38Z 2012-08-01T15:40:38Z <p>No (I suppose that $k$ is algebraically closed). This is because Poincaré's complete reducibility theorem contains a unicity statement for the intervening factors (up to isogeny). See Mumford, Abelian varieties, p. 173-174.</p> http://mathoverflow.net/questions/103657/degree-of-a-variety-is-well-defined/103673#103673 Answer by Damian Rössler for Degree of a variety is well-defined Damian Rössler 2012-08-01T09:08:36Z 2012-08-01T09:08:36Z <p>See Fulton, "Intersection Theory", Ex. 8.4.12, p. 149.</p> http://mathoverflow.net/questions/86221/fibre-cardinality-of-an-unramified-morphism/86315#86315 Answer by Damian Rössler for Fibre cardinality of an unramified morphism Damian Rössler 2012-01-21T16:35:19Z 2012-01-21T16:40:27Z <p>After I wrote the comments above, I found the following reference : </p> <p>Formula (12.6.2), p. 329 in Görtz-Wedhorn, Algebraic Geometry I, Viehweg &amp; Teubner Verlag</p> <p>for (a generalisation of) the equality you are looking for, when $\phi$ is assumed flat (which is true if you assume that $X$ and $Y$ are non-singular, as pointed out in the comments of K. M. Pera and S. Kovacs). </p> http://mathoverflow.net/questions/81338/coherent-sheaf-with-connection-is-locally-free/81340#81340 Answer by Damian Rössler for Coherent sheaf with connection is locally free? Damian Rössler 2011-11-19T13:04:00Z 2011-11-19T13:04:00Z <p>See Prop. 8.8, p. 206 in N. Katz, "Nilpotent connections and the monodromy...", Publications Mathématiques de l'IHES, 39 (1970), p. 175-232.</p> http://mathoverflow.net/questions/80951/algebraic-proof-of-atiyah-bott-fixed-point-formula/80964#80964 Answer by Damian Rössler for algebraic proof of Atiyah-Bott fixed point formula? Damian Rössler 2011-11-15T08:27:16Z 2011-11-15T08:27:16Z <p>Notice that you must assume that the graph of $f$ intersects the diagonal tranversally (otherwise some determinants in the formula might vanish). This transversality condition is automatic if $f$ has finite order. With that assumption, the above formula is a special case of the "Woods hole" formula, which is proven using Grothendieck duality in SGA 5 (Springer Lecture Notes in mathematics 589), Appendix to Exp. III, Cor. 6.12, p. 131.</p> http://mathoverflow.net/questions/80898/references-for-theta-characteristic/80909#80909 Answer by Damian Rössler for references for theta characteristic Damian Rössler 2011-11-14T17:07:42Z 2011-11-14T17:07:42Z <p>In general, <em>if $A$ is principally polarised</em>, there is a canonical isomorphism $(\Omega_A^{g})^{\otimes 4}\simeq{\cal O}_A(\Theta)^{\otimes 8}$ (if you choose to divide by $4$, it is not canonical anymore). The justification for this isomorphism is the Grothendieck-Riemann-Roch theorem, applied to ${\cal O}(\Theta)$. For this, see the book by Moret-Bailly, "Pinceaux des variétés abéliennes" (Astérisque 129), whose principal aim is the investigation of (a generalization of) this canonical isomorphism. See also his article "Sur l'équation fonctionelle..." (Compositio Math. 75). Finally see his article "La formule de Noether sur les surfaces arithmétiques" (Invent. Math. 98), par. 2.3 for the case of jacobians that you are interested in. Other references are the book by Chai-Faltings, "Degeneration of abelian varieties", chap. I, Th. 5.1 and the article by V. Maillot and myself, "On the determinant bundles of abelian schemes" (Compositio 144). </p> http://mathoverflow.net/questions/7330/which-math-paper-maximizes-the-ratio-importance-length/79483#79483 Answer by Damian Rössler for Which math paper maximizes the ratio (importance)/(length)? Damian Rössler 2011-10-29T22:28:30Z 2011-10-29T22:28:30Z <p>I know that this question was posted almost two years ago but I cannot resist suggesting</p> <p>Zagier, D. Newman's short proof of the prime number theorem. Amer. Math. Monthly 104 (1997), no. 8, 705–708. </p> <p>which is difficult to beat, I think.</p> http://mathoverflow.net/questions/78597/c-infty-function-f-bf-c-mapsto-bf-c-such-that-fz-in-overline-bf-q $C^\infty$ function $f:{\bf C}\mapsto {\bf C}$ such that $f(z)\in\overline{{\bf Q}(z)}$ for all $z\in {\bf C}$ Damian Rössler 2011-10-19T17:38:23Z 2011-10-19T20:54:05Z <p>Suppose that $f:{\bf C}\mapsto {\bf C}$ is a $C^\infty$ function such that $f(z)\in\overline{{\bf Q}(z)}$ for all $z\in {\bf C}$, ie $f(z)$ is algebraic over the field ${\bf Q}(z)$ generated by $z$ over ${\bf Q}$, for all $z\in{\bf C}$. </p> <p>Is it true that $f(z)\in\bar{\bf Q}[z]$, in other words, that $f(z)$ is a polynomial function in $z$, with coefficients in $\bar{\bf Q}$ ?</p> <p>One of my colleagues sketched an argument showing that this is true, which uses Baire's category theorem. I would like to know if anybody knows a reference for this result in the mathematical literature (or possibly a counterexample - I haven't checked his argument in detail) . </p> <p>This question is related to the question <a href="http://mathoverflow.net/questions/78526" rel="nofollow">http://mathoverflow.net/questions/78526</a> but the properties requested from $f$ are stronger here.</p> http://mathoverflow.net/questions/78252/etale-morphism-and-direct-image/78254#78254 Answer by Damian Rössler for etale morphism and direct image Damian Rössler 2011-10-16T07:29:02Z 2011-10-16T07:29:02Z <p>Since $Y$ and $X$ are proper over the base field, the morphism $f$ is finite (by Zariski's main theorem). Hence by the semi-continuity theorem (see Hartshorne p. 288, Cor. 12.9), $f_*V$ is locally free, because $f$ is finite and flat. So the answer is yes.</p> http://mathoverflow.net/questions/76928/grothendieck-riemann-roch-interpretation-of-a-calculation/76944#76944 Answer by Damian Rössler for Grothendieck-Riemann-Roch interpretation of a calculation Damian Rössler 2011-10-01T21:01:20Z 2011-10-01T21:01:20Z <p>Let $T\phi$ be the relative tangent bundle. So we have an exact sequence <code>$0\to T\phi\to TY\to \phi^*TB\to 0$</code>. Now GRR and the projection formula gives <code>$$ \phi_*{\rm Td}(Y)=\phi_*({\rm Td}(T\phi){\rm Td}(\phi^*TB))= {\rm Td}(TB)\phi_*({\rm Td}(T\phi))={\rm ch}(1-R^1\pi_*({\cal O}_Y)){\rm Td}(TB) $$</code> which suggests that <code>${\cal L}=R^1\pi_*({\cal O}_Y)^\vee=\pi_*(\Omega_\phi):=\pi_*(T\phi^\vee)$</code> (by Grothendieck duality). I will take this for granted; up to $\otimes$ by a torsion bundle, it is forced upon you by the equation; if $\cal L$ and $\pi_*(\Omega_\phi)$ differ by a torsion line bundle, the calculations below still work. </p> <p>Furthermore, applying GRR and the projection formula again, we may compute <code>$$ \phi_*{\cal H}_1(Y)=\phi_*({\rm ch}(\Omega_Y){\rm Td}(TY))= \phi_*(\ [\phi^*{\rm ch}(\Omega_B)+{\rm ch}(\Omega_\phi)]{\rm Td}(T\phi)\phi^*{\rm Td}(TB)\ )= $$ $$ {\rm Td}(TB){\rm ch}(\Omega_B)\phi_*({\rm Td}(T\phi))+{\rm Td}(TB)\phi_*({\rm Td}(T\phi) {\rm ch}(\Omega_\phi))= $$ $$ {\rm Td}(TB){\rm ch}(\Omega_B)\phi_*({\rm Td}(T\phi))+{\rm Td}(TB){\rm ch}({\cal L}- R^1\pi_*(\Omega_\phi))= $$ $$ {\rm Td}(TB){\rm ch}(\Omega_B)\phi_*({\rm Td}(T\phi))+{\rm Td}(TB){\rm ch}({\cal L}-1)= $$ $$ {\rm Td}(TB){\rm ch}(\Omega_B)(1-{\rm ch}({\cal L}^\vee))+{\rm Td}(TB){\rm ch}({\cal L}-1)= (1-e^{-L}){\cal H}_1(B)+(e^{L}-1){\rm Td}(TB)\,\,\, (*) $$</code> Now use the fact that $\cal L$ is actually a torsion bundle, because the discriminant modular form will trivialise ${\cal L}^{\otimes 12}$ (or possibly a higher power, if one needs to introduce level structures). This last fact is also a consequence of GRR, since <code>$$ \phi_*({\rm Td}(T\phi))=\pi_*({\rm ch}(1))\phi^*\phi_*({\rm Td}(T\phi))=0={\rm ch}(1-R^1\pi_*({\cal O}_Y)) $$</code> (because <code>$T\phi=\pi^*\pi_* T\phi)$</code>. Hence, one gets, all in all, that <code>$$ \phi_*{\rm Td}(TY)=0 $$</code> and in view of (*), that <code>$$ \phi_*{\cal H}_1(Y)=0 $$</code> which is equivalent to the two equations you are considering, since ${\cal L}$ is a torsion line bundle (observe that the degree $0$ part of $−4−e^{-L}+3e^{−2L}+2e^{−3L}$ vanishes).</p> http://mathoverflow.net/questions/12347/algebraic-versus-analytic-brauer-group/75830#75830 Answer by Damian Rössler for Algebraic versus Analytic Brauer Group Damian Rössler 2011-09-19T08:47:27Z 2011-09-19T10:43:33Z <p>I think the article by B. Toen "Derived Azumaya algebras and generators for twisted derived categories", arXiv:1002.2599, gives a pointer to a possible answer to your question.</p> <p>EDIT I have weakened the assertion...</p> http://mathoverflow.net/questions/75410/flatness-of-sheaf-of-relative-kahler-differentials/75584#75584 Answer by Damian Rössler for Flatness of sheaf of relative Kahler differentials Damian Rössler 2011-09-16T09:04:56Z 2011-09-18T06:49:38Z <p>Suppose that $Y$ is the spectrum of a smooth curve over perfect field $k$ and let $D\subseteq Y$ be a finite set of closed points. Let $f:X\to Y$ be a proper morphism and let $D\subseteq X$ be a normal crossings divisor. Suppose that $f$ is semi-stable relatively to $E,D$ and $k$, in the sense of Illusie in par. 1.4 of "Réduction semi-stable et décomposition...", Duke Math. J. 60 (1990). </p> <p>The morphism $f$ is then flat and lci and its fibres are reduced normal crossings divisors. There is a relative residue sequence $$ 0\to \Omega_{X/Y}\to\Omega_{X/Y}({\rm log})\to F\to 0\ \ \ \ (*) $$ where $F$ is supported on the singular locus of the singular fibres of $f$, and $\Omega_{X/Y}({\rm log})$ is the locally free sheaf of differentials with (relative) logarithmic singularities along $D$. See for instance p. 23 in "Une conjecture sur la torsion..." by V. Maillot and D. Rössler (Publ. Res. Inst. Math. Sci. 46, no. 4 (2011) - for lack of a canonical reference (?)). </p> <p>Now let $M$ be any quasi-coherent ${\cal O}_Y$-module. The tor-sequence corresponding to $\otimes_Y M$ when applied to (*) gives $$ \dots\to {\rm Tor}^1_Y(\Omega_{X/Y},M)\to{\rm Tor}^1_Y(\Omega_{X/Y}({\rm log}),M)\to{\rm Tor}^1_Y(F,M) $$ $$ \to \Omega_{X/Y}\otimes_Y M\to\Omega_{X/Y}({\rm log})\otimes_Y M\to F\otimes_Y M\to 0 $$ and since ${\rm Tor}^l_Y(\Omega_{X/Y}({\rm log}),M)=0$ for all $l>0$ (because $\Omega_{X/Y}({\rm log})$ is locally free and $f$ is flat) and ${\rm Tor}^l_Y(N,K)=0$ for any $l>1$ and any quasi-coherent ${\cal O}_Y$-modules $N,K$ (that is because $Y$ is the spectrum of a Dedekind domain and any finitely generated quasi-coherent ${\cal O}_Y$-module has a two-step projective resolution; the general case follows from compatibility of Tor with direct limits), we see that ${\rm Tor}^l_Y(\Omega_{X/Y},M)=0$, for all $l>0$, ie $\Omega_{X/Y}$ is flat over $Y$. </p> <p>EDIT As remarked by Liu below, the sheaf $\Omega_{X/Y}$ can also be seen to be flat simply because it is the subsheaf of a torsion free sheaf.</p> http://mathoverflow.net/questions/74009/functoriality-of-the-blow-up/74042#74042 Answer by Damian Rössler for Functoriality of the Blow-Up Damian Rössler 2011-08-30T07:30:21Z 2011-08-30T07:30:21Z <p>Suppose <code>$f={\rm Id}_X$</code>, <code>$X={\bf A}^3_{\bf C}$</code> (affine space of dimension $3$ over the complex numbers). Suppose that $\cal I$ is the sheaf of ideals of a smooth curve going through $0$ and that $\cal K$ is the sheaf of ideals of the point $0$ in ${\bf A}^3_{\bf C}$. Then the pull-back of ${\cal J}={\cal I}$ to $\widetilde{W}$ defines a subscheme $Z$ of $\widetilde{W}$ and the dimension of the intersection of $Z$ with the complement of the exceptional divisor of $\widetilde{W}$ is of dimension $1$ and thus $Z$ is not a Cartier divisor (ie $(\rho^{-1}J){\cal O}_{\widetilde{W}}$ is not an invertible sheaf). </p> http://mathoverflow.net/questions/73396/etale-endomorphisms-of-abelian-varieties-in-positive-characteristic Etale endomorphisms of abelian varieties in positive characteristic Damian Rössler 2011-08-22T12:07:56Z 2011-08-28T01:45:12Z <p>Let $K$ be the function field of a smooth curve over the algebraic closure $k$ of the finite field ${\bf F}_p$ (where $p>0$ is a prime number). </p> <p>My question is : does there exist an abelian variety $A$ over $K$, with the following properties : </p> <p>(a) the $K|k$-image of $A$ is trivial;</p> <p>(b) there exists an étale $K$-endomorphism of $A$, whose degree is a power of $p$ ?</p> <p>The condition on the $K|k$-image can be rephrased as : there are no non-zero $K$-homomorphisms<br> $A\to C_K$, where $C$ is an abelian variety over $k$. </p> <p>For examples of abelian varieties satisfying condition (b) only, look at abelian varieties $C_K$, where $C$ is an ordinary abelian variety over ${\bf F}_p$. The abelian variety $C$ is endowed with the étale endomorphism given by the Verschiebung morphism. </p> <p>Also, notice that if there is an abelian variety over $K$ satisfying (a) and (b), then the dimension of $A$ is larger than one (ie it is not an elliptic curve). Indeed, if an elliptic curve $E$ satisfies the above conditions, then $E$ has an endomorphism, which is not a multiplication by a scalar and thus it has complex multiplications; this implies that it is isogenous to an elliptic curve defined over $k$, by a theorem of Grothendieck (or by more direct arguments). </p> <p>Finally, I would like to point out that if $A$ is an ordinary abelian variety over $K$, which has maximal Kodaira-Spencer rank, then $A[p]$$(K^{\rm sep})=0$ by a theorem of J-F Voloch (see p. 1093 in "Diophantine Approximation on Abelian Varieties in Characteristic $p$", Amer. J. Math., Vol. 117, No. 4., pp. 1089-1095); this shows that such an abelian variety cannot have an endomorphism as in (b). </p> http://mathoverflow.net/questions/73867/torsion-points-of-abelian-varieties-in-the-perfect-closure-of-a-function-field Torsion points of abelian varieties in the perfect closure of a function field Damian Rössler 2011-08-27T20:56:01Z 2011-08-27T20:56:01Z <p>The following is a problem, which was recently brought to my attention by H. Esnault and A. Langer. </p> <p>Let $K$ be the function field of a smooth curve over the algebraic closure $k$ of the finite field ${\bf F}_p$ (where $p>0$ is a prime number).</p> <p>Let $A$ be an abelian variety over $K$ and suppose that the $K|k$-image of $A$ is trivial (ie there are no non-vanishing $K$-homomorphisms from $A$ to an abelian variety over $K$, which has a model over $k$). </p> <p><strong>Question</strong> : is it true that <code>$\#{\rm Tor}(A(K^{\rm perf}))&lt;\infty$</code> ? (*)</p> <p>Here $K^{\rm perf}$ is the maximal purely inseparable extension of $K$ and ${\rm Tor}(A(K^{\rm perf}))$ is the subgroup of $A(K^{\rm perf})$ consisting of elements of finite order.</p> <p>To put things in context, recall that by the Lang-Néron theorem, we have <code>$\#{\rm Tor}(A(K))&lt;\infty$</code>. </p> <p>Furthermore, one can show using a specialization argument that <code>$\#{\rm Tor}(A(K^{\rm perf}))&lt;\infty$</code> if $k$ is replaced by a finite extension of ${\bf F}_p$; in this case, the assumption on the $K|k$-image can actually be dropped. </p> <p>Notice also that the inequality in question (*) is actually equivalent to the inequality <code>$\#{\rm Tor}_p(A(K^{\rm perf}))&lt;\infty$</code>, where ${\rm Tor}_p(A(K^{\rm perf}))$ is the subgroup of $A(K^{\rm perf})$ consisting of the elements, whose order is a power of $p$. This follows from the fact the multiplication by $n$ morphism is étale if $p\not|n$.</p> <p>Question (*) has a positive answer if $A$ is an elliptic curve by the work of M. Levin, who proves a much stronger result (see "On the group of rational points...", Amer. J. Math. 90 (1968)). </p> <p>The question (*) is in part complementary to the following other question in MO : </p> <p><a href="http://mathoverflow.net/questions/73396/etale-endomorphisms-of-abelian-varieties-in-positive-characteristic" rel="nofollow">http://mathoverflow.net/questions/73396/etale-endomorphisms-of-abelian-varieties-in-positive-characteristic</a></p> http://mathoverflow.net/questions/73634/subscheme-structure-of-support/73644#73644 Answer by Damian Rössler for subscheme structure of support Damian Rössler 2011-08-25T09:11:06Z 2011-08-25T09:11:06Z <p>No it isn't the reduced induced closed subscheme structure in general. For example, let $A={\bf Z}$, $M={\bf Z}/4{\bf Z}$. Then ${\rm Ann}(M)=4{\bf Z}$ and the prime ideal defining $S$ (with its reduced structure) is $2{\bf Z}={\rm rad}({\rm Ann}(M))$. So if $S$ is endowed with the reduced structure, it is isomorphic to ${\rm Spec}({\bf Z}/2{\bf Z})$ and if it is endowed with the structure given by the annihilator then it is isomorphic to ${\rm Spec}({\bf Z}/4{\bf Z})$.</p> http://mathoverflow.net/questions/24110/locally-free-modules/73610#73610 Answer by Damian Rössler for Locally free modules Damian Rössler 2011-08-24T21:25:51Z 2011-08-24T21:25:51Z <p>I think the theory you are looking for can be found in Th. 8.5.3, chap. 8 (p. 210) of the book by B. Fantechi et al., "FGA explained". </p> http://mathoverflow.net/questions/72829/which-curves-have-stable-faltings-height-greater-or-equal-to-1/73594#73594 Answer by Damian Rössler for Which curves have stable Faltings height greater or equal to 1 Damian Rössler 2011-08-24T18:12:57Z 2011-08-24T18:12:57Z <p>Dear Ariyan, the elliptic curve with equation $$y^2=x^3+6$$ has Faltings height $$-(3/2)\log(\Gamma(1/3)/\Gamma(2/3))+(1/4)\log(3)=-0.748752...;$$ the curve of genus $2$ with equation $$y^2+y=x^5$$ has Faltings height $$ h_{\rm Fal}(C_{\bar{\bf Q}})=2\log(2\pi)- {1\over 2}\log\big(\Gamma(1/5)^5\Gamma(2/5)^3\Gamma(3/5)\Gamma(4/5)^{-1}\big) $$ $$ \approx -1.452509239645644650317707042; $$ For the first example, see Deligne, "Preuve des conjectures de Tate et Shafarevich", Séminaire Bourbaki. For the second one, see Bost, Mestre, Moret-Bailly, "Sur le calcul explicite des 'classes de Chern' des surfaces arithmétiques de genre $2$", Séminaire sur les Pinceaux de Courbes Elliptiques (Paris, 1988). Astérisque No. 183 (1990), 69–105. </p> <p>Another explicit formula that should allow you to produce elliptic curves of arbitrarily large Faltings height is the inequality $$ |h(j_E)-12h_{\rm Fal}(E)|\leqslant 6\log(1+h(j_E))+47.15 $$ See paragraph 5. of the article "Serre's uniformity..." by Bilu and Parent for references.</p> <p>Something else you can do is make numerical experiments with formula in Conj. 3 of the article of Colmez, "Hauteur de Faltings..." (Compositio), which is true (without $\log(2)$ factor !, see A. Obus, arXiv:1107.0684) if the CM field is abelian over $\bf Q$. In that case, the Artin $L$-functions become Dirichlet $L$-functions and can be computed explicitly in terms of values of the Gamma function using the Hurwitz formula.</p> <p>This is not a complete answer but I hope that it helps.</p> http://mathoverflow.net/questions/73450/top-chern-class-euler-characteristic/73474#73474 Answer by Damian Rössler for Top Chern Class = Euler Characteristic Damian Rössler 2011-08-23T08:12:01Z 2011-08-23T08:12:01Z <p>As an alternative to R. Budney's answer, one might also notice that the Gauss-Bonnet formula (the one you mention - mind that you must assume that $X$ is projective, otherwise the integral might not even make sense) is a consequence of the Hirzebruch-Riemann-Roch theorem. Indeed, the HRR theorem says $$ \chi(V)=\int_{X}{\rm Td}({\rm T}X){\rm ch}(V) $$ where $$\chi(V):=\sum_{l}{(-1)}^l{\rm rk}(H^l(X,V))$$ is the Euler characteristic of coherent sheaves. Now there is an universal identity of Chern classes $$ {\rm ch}(\sum_{r}(-1)^r\Omega_X^r){\rm Td}(\Omega^\vee_X)=c^{\rm top}(\Omega^\vee_X) $$ (called the Borel-Serre identity). Here $\Omega_X$ is the sheaf of differential of $X$ and thus $\Omega^\vee_X={\rm T}X$. Plugging the element $\sum_{r}(-1)^r\Omega_{X}^r$ into the HRR theorem, one gets $$ \sum_{k,l}(-1)^{l+k}{\rm rk}(H^k(X,\Omega^l))=\int_{X}c^{\rm top}(TX) $$ and by the Hodge decomposition theorem $$ \sum_{k,l}(-1)^{l+k}{\rm rk}(H^k(X,\Omega^l))=\sum_{r}{(-1)}^r{\rm rk}(H^r(X({\bf C}),{\bf C})) $$ where $H^r(X({\bf C}),{\bf C})$ is the $r$-th singular cohomology group. The quantity $\sum_{r}{(-1)}^r{\rm rk}(H^r(X({\bf C}),{\bf C}))$ is the topological Euler characteristic, so this proves what you want. The HRR theorem is proved in chap. 15 of Fulton's book (or in Hirzebruch's book "Topological methods...") and the Borel-Serre identity is Ex. 3.2.5, p. 57 of the same book.</p> http://mathoverflow.net/questions/133059/belyis-theorem-for-function-fields Comment by Damian Rössler Damian Rössler 2013-06-08T20:22:02Z 2013-06-08T20:22:02Z (answering your comment below). I think that it is unlikely that this will work if you want the morphism to be tamely ramified. Galois coverings of the affine line in char. p, which are tamely ramified along $\infty$, are trivial. http://mathoverflow.net/questions/132912/what-analysis-should-i-know-for-studying-arakelov-theory Comment by Damian Rössler Damian Rössler 2013-06-07T18:37:12Z 2013-06-07T18:37:12Z Global Arakelov theory requires a lot of index theory. Some of that material is covered in the book Heat kernels and Dirac operators, by N. Berline, E. Getzler and M. Vergne, Grundlehren Math. Wiss., vol. 298, Springer-Verlag, New York, 1992. The difficult part of local index theory, which is needed in the proof of the arithmetic Riemann-Roch theorem is only described in the articles of Bismmut and his coworkers (the most self-contained one is his article with Lebeau). http://mathoverflow.net/questions/132912/what-analysis-should-i-know-for-studying-arakelov-theory Comment by Damian Rössler Damian Rössler 2013-06-07T14:06:39Z 2013-06-07T14:06:39Z You could have a look at C. Soul&#233; book &quot;Lectures on Arakelov Geometry&quot; (Cambridge Univ. Press). http://mathoverflow.net/questions/132673/duality-and-hirzebruch-riemann-roch Comment by Damian Rössler Damian Rössler 2013-06-04T19:06:49Z 2013-06-04T19:06:49Z There is a link between duality for coherent sheaves and Grothendieck-Riemann-Roch. Consider the Lefschetz-Verdier formula for coherent sheaves (see SGA 5, Exp. III, Appendix) and (with the terminology of the appendix) let $X_1=X_2=X$, $Y_1=Y_2=Y$ and $C'=C''=$diagonal, $D'=D''$=diagonal. Then the formula gives (although this is difficult to show...) Grothendieck-Riemann-Roch with values in Hodge cohomology, ie $\oplus_{p}H^p(Y,\Omega^p)$. The Lefschetz-Verdier formula depends only on duality. http://mathoverflow.net/questions/131063/connectedness-of-hyperplane-sections-reference-request Comment by Damian Rössler Damian Rössler 2013-05-19T06:35:47Z 2013-05-19T06:35:47Z ( Hartshorne, Algebraic Geometry, Springer, GTM 52, Cor. 7.9, p. 244 ) http://mathoverflow.net/questions/130804/identity-on-topological-space-but-not-on-scheme Comment by Damian Rössler Damian Rössler 2013-05-16T11:19:03Z 2013-05-16T11:19:03Z Notice that an automorphism of a field will also do the trick. http://mathoverflow.net/questions/130804/identity-on-topological-space-but-not-on-scheme Comment by Damian Rössler Damian Rössler 2013-05-16T09:18:18Z 2013-05-16T09:18:18Z ($k$ is a field in my last comment). http://mathoverflow.net/questions/130804/identity-on-topological-space-but-not-on-scheme Comment by Damian Rössler Damian Rössler 2013-05-16T09:17:25Z 2013-05-16T09:17:25Z There is the composition $k[\epsilon]/\epsilon^2\to k\to k[\epsilon]/\epsilon^2$, where the first arrow is the morphism of $k$-algebras sending $\epsilon$ to $0$ et the second one is the morphism making $k[\epsilon]/\epsilon^2$ into a $k$-algebra. http://mathoverflow.net/questions/130270/what-is-a-good-reference-preferably-thorough-for-the-derived-category-of-a-sche Comment by Damian Rössler Damian Rössler 2013-05-10T16:44:50Z 2013-05-10T16:44:50Z There is the first chapter of the book &quot;Sheaves on manifolds&quot; by Kashiwara-Shapira. http://mathoverflow.net/questions/129993/shafarevichs-theorem-for-elliptic-curves-defined-over-function-field-of-algebrai Comment by Damian Rössler Damian Rössler 2013-05-09T07:40:46Z 2013-05-09T07:40:46Z I think the following problem is lurking in the background here: the action of the automorphism group of $K$ on $j$. In order to get finiteness statements, this group should be seen to be finite (eg when $K$ is the function field of a curve over a finite field). http://mathoverflow.net/questions/129814/lefschetz-hyperplane-section-theorem-for-connections Comment by Damian Rössler Damian Rössler 2013-05-06T14:23:46Z 2013-05-06T14:23:46Z @ulrich: I see - you are right of course. http://mathoverflow.net/questions/129814/lefschetz-hyperplane-section-theorem-for-connections Comment by Damian Rössler Damian Rössler 2013-05-06T10:39:31Z 2013-05-06T10:39:31Z ... what I wrote above does not provide a counterexample but it might suggest how to construct one (or provide a proof !). http://mathoverflow.net/questions/129814/lefschetz-hyperplane-section-theorem-for-connections Comment by Damian Rössler Damian Rössler 2013-05-06T10:38:02Z 2013-05-06T10:38:02Z By Cor. II.6.11 in Deligne's &quot;Equations diff&#233;rentielles...&quot; (LNM 163), the cohomology groups you want to compute are the cohomology groups of the corresponding local systems (for the ordinary topology). Now notice that the topological proof of Lefschetz's theorem is based on the result that a non-singular $k$-dim. affine variety has the homotopy type of $k$-dim. CW complex. So you are led to the question: does the cohomology of a local systems on a $k$-dim. CW complex vanish above $2k$ ? This is unlikely to be true in general (classifying spaces probably provide counterexamples). http://mathoverflow.net/questions/128456/reference-request-for-the-finiteness-of-cuspidal-subgroup-of-x-0n Comment by Damian Rössler Damian Rössler 2013-05-02T10:30:11Z 2013-05-02T10:30:11Z See also R. Elkik, &quot;Le th&#233;or&#232;me de Manin-Drinfeld.&quot; Ast&#233;risque, pages 59-67, 1990. S&#233;minaire sur les Pinceaux de Courbes Elliptiques (Paris, 1988). The proof given there follows an idea of Deligne and uses mixed Hodge structures. http://mathoverflow.net/questions/129093/zariskis-main-theorem-in-the-form-of-grothendieck-universal-properties Comment by Damian Rössler Damian Rössler 2013-04-29T14:00:45Z 2013-04-29T14:00:45Z I don't think that (3) holds. This would imply that every ${\bf Z}/n$-torsor on $W^\circ$ can be extended to a ${\bf Z}/n$-torsor over $W$. There is an obstruction for this to be possible, which lies in a $H^2$ group.