User torsten ekedahl - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T03:09:26Z http://mathoverflow.net/feeds/user/4008 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80697/top-chern-class-in-positive-characteristic/80702#80702 Answer by Torsten Ekedahl for Top chern class in positive characteristic Torsten Ekedahl 2011-11-11T18:57:32Z 2012-11-10T22:02:29Z <p>The same thing is true in positive characteristic, the degree of $c_n$ is equal to the Euler characteristic (except if you consider de Rham cohomology where it only is the Euler characteristic mod $p$). The proof of course cannot use the standard proof in the complex case, using Hopf's theorem that says that the degree of the Euler class is the Euler characteristic and the identification of the Euler class with the top Chern class). One can instead use the Riemann-Roch theorem and the identification of the Euler characteristic with the de Rham Euler characteristic. Given the latter the rest is just to verify that the seemingly complicated expresssion of the Riemann-Roch simplifies by a calculation using the splitting principle to just $c_n$. Alternatively, if I remember correctly one can use a Lefschetz pencil and induction over the dimension.</p> <p><b>Addendum</b>: It's all coming back to me, a third possibility is to use that the self-intersection of the diagonal is on the one hand the Euler characteristic (as it is gives the trace of the identitity map), on the other hand that self-intersection is given by the top Chern class of the normal bundle of the diagonal which is exactly the cotangent bundle.</p> http://mathoverflow.net/questions/25020/what-is-the-universal-enveloping-algebra/25046#25046 Answer by Torsten Ekedahl for What is the universal enveloping algebra? Torsten Ekedahl 2010-05-17T19:06:37Z 2012-08-02T16:12:20Z <p>I have now understood the situation better so my previous post has been replaced by this. (The only thing that was in the original but will not be here are some explicit formulas but Theo has given a reference for that.)</p> <p>As I understand the question the poser wanted a construction of the enveloping algebra of a Lie algebra in a symmetric monoidal pseudoabelian (i.e., idempotents have kernels) $K$-category $\mathcal C$ with arbitrary sums over a field $K$ of characteristic zero. This means that for any $K[\Sigma_n]$-module $M$ and any object $V\in\mathcal C$ we can define $M\bigotimes_{\Sigma_n}V^{\otimes n}$ and for any $\Sigma$-module $M_\bullet$ (i.e., a collection $(M_n)$ of $K[\Sigma_n]$-modules) we can define $M(V):=\bigoplus_nM_n\bigotimes_{\Sigma_n}V^{\otimes n}$ which is an endofunctor of $\mathcal C$. Furthermore, a map $M \to N$ of $\Sigma$-modules gives a natural transformation of functors $M(V) \to N(V)$. In the particular case when $\mathcal C$ is the category of $K$-vector spaces such a natural transformation comes from a unique map of $\Sigma$-modules. The idea is to do what we know to do for $K$-vector spaces, interpret it as a set of natural transformations, get the corresponding maps of $\Sigma$-modules and use them to induce natural transformations for a general $\mathcal C$.</p> <p>Let thus $S(V)$ be the symmetric algebra on the $K$-vector space $V$, $T(V)$ the tensor algebra on $V$ and $L(V)$ the free Lie algebra on $V$. Symmetrisation gives an isomorphism $S(L(V)) \to U(L(V))=T(V)$, where $U(-)$ is the enveloping algebra of Lie algebras. As $T(V)$ is a $T$-algebra (i.e., an associative algebra) and we can use this isomorphism to give $S(L(V))$ a $T$-algebra structure (i.e., a natural transformation $T(S(L(V))) \to S(L(V))$ fulfilling the appropriate conditions with respect to the monad structure on $T(-)$). Furthermore, if $\mathfrak g$ is a Lie algebra, then the $T$-algebra structure on $S(\mathfrak g)$ induced by the isomorphism $S(\mathfrak g) \to U(\mathfrak g)$ is given as the composite of $T(S(\mathfrak g)) \to T(S(L(\mathfrak g)))$ induced by the inclusion $\mathfrak g \to L(\mathfrak g)$, the map $T(S(L(\mathfrak g))) \to S(L(\mathfrak g))$ given by the $T$-module structure on $S(L(V))$ above and the map $S(L(\mathfrak g)) S(\mathfrak g)$ induced by the structure map $L(\mathfrak g) \to \mathfrak g$</p> <p>Now, the functors $S(-)$, $L(-)$ and $T(-)$ are associated to $\Sigma$-modules which will be denoted by the same letters (instead of the standard $Com$, $Lie$ and $Ass$). Furthermore, composition of functors correspond to the plethysm $\circ$. Hence we get that $S\circ L$ is a $T$-module, i.e., we have a map $T\circ S\circ L \to S\circ L$ compatible with the operad structure on $T$. Consider now the case of a general $\mathcal C$. Each of $S$, $L$ and $T$ give endofunctors on $\mathcal C$ and $\circ$ again corresponds to composition. Let $\mathfrak g$ be a Lie algebra in $\mathcal C$ and define a $T$-algebra structure (i.e., the structure of associative algebra) on $S(\mathfrak g)$ as the composite $$T(S(\mathfrak g)) \to T(S(L(\mathfrak g))) \to S(L(\mathfrak g)) \to S(\mathfrak g)$$ as above. The verification that this does indeed give a $T$-algebra structure is just a question of unwinding the definitions. The fact that $S$ is an operad gives us a natural transformation $V \to S(V)$ which applied to $\mathfrak g$ gives a morphism $\mathfrak g \to S(\mathfrak g)$ which we now want to show is a Lie algebra homomorphism. Here the Lie algebra structure on $S(\mathfrak g)$ is induced by its $T$-algebra structure and the operad map $L \to T$. Again unwinding definitions shows that it is indeed a Lie algebra morphism.</p> <p>Finally assuming that $\mathfrak g \to A$ is a Lie algebra homomorphism where $A$ is an associative algebra with $L \to T$ inducing its Lie algebra structure. Note that we have an isomorphism (now going back to vector spaces) $S(L(V))\to T(V)$ and hence an isomorphism is $\Sigma$-modules $S\circ L=T$. This gives us a map $S(\mathfrak g)\to S(L(\mathfrak g))=T(\mathfrak g) \to T(A) \to A$ and it is easy to see that this is an algebra morphism.</p> http://mathoverflow.net/questions/80280/line-bundles-with-integrable-connection-on-abelian-varieties/80281#80281 Answer by Torsten Ekedahl for Line bundles with integrable connection on abelian varieties Torsten Ekedahl 2011-11-07T08:49:58Z 2011-11-13T10:27:41Z <p>Yes, it is true, though an algebraic proof seems (there may be a simpler proof however) somewhat tricky.</p> <ul> <li> Such a line bundle lies in $\mathrm{Pic}^\tau(X)$. This is a general fact as a line bundle lies in $\mathrm{Pic}^\tau(X)$ if its rational Chern classes are trivial (this follows from Riemann-Roch) and the Chern classes of a line bundle with integrable connection are torsion (an algebraic proof is given in Dix exposés sur la cohomologie des schémas). This works without the assumption of $X$ being abelian. <li>It is then a further fact that for abelian varieties $X$ we have that $\mathrm{Pic}^\tau(X)= \mathrm{Pic}^0(X)$ (this I think is an Mumford's abelian varieties somewhere). </ul> <p>This extends to immediately to families by checking fibre by fibre.</p> <p><b>Addendum</b>: Sorry forgot to say that this all requires characteristic zero. In characteristic $p$ every $p$'th power line bundle has an integrable connection (even of $p$-curvature $0$) but will in general not lie in $\mathrm{Pic}^0(X)$.</p> <p><b>Addendum 1</b>: Lars (in a comment) makes an interesting point about the positive characteristic situation. A module structure over the ring of differential operators (aka a stratification) implies in particular that the line bundle is a $p^n$'th power for each $n$ and as $\mathrm{Pic}(X)/\mathrm{Pic}^\tau(X)$ is a finitely generated group this implies that the line bundle lies in $\mathrm{Pic}^\tau(X)$. The same idea could be applied to the characteristic zero situation if the $p$-curvature of the reduction modulo $p$ for an infinite number of primes $p$ were zero. However that should be true only if the line bundle has finite order so it doesn't give very much.</p> <p><b>Addendum 2</b>: Veen is asking about the equality $\mathrm{Pic}^0(A)=\mathrm{Pic}^\tau(A)$ particularly in a family (when $A$ is abelian). The easiest way to answer all of these questions simultaneously is to assume that there is an ample line bundle $\mathcal L$ on $A$ (which is true locally on the base) and then consider the map $A\to \mathrm{Pic}(A)$ given by $a\mapsto \mathcal L_a\bigotimes\mathcal L^{-1}$. To get all equalities needed it is enough to show that the image is $\mathrm{Pic}^\tau(A)$. This is something that can be checked fibrewise and then it can be extracted from Mumford.</p> http://mathoverflow.net/questions/22943/reflexive-modules-over-a-2-dimensional-regular-local-ring/80737#80737 Answer by Torsten Ekedahl for Reflexive modules over a 2-dimensional regular local ring Torsten Ekedahl 2011-11-12T07:02:33Z 2011-11-12T07:02:33Z <p>If you accept the fact that a $2$-dimensional (local) ring has global dimension $2$, the following is a (somewhat) alternative proof. Choose a free f.g. presentation $F_1 \to F_0 \to M^\ast \to 0$ and then dualise giving an exact sequence $0\to M^{\ast\ast}\to F_0^\ast\to F_1^\ast\to K \to 0$, where $K$ is defined as the cokernel. As the $F^\ast_i$ are free and $\mathrm{pd} K\le 2$ we get that $M^{\ast\ast}$ is projective (and then free if the ring is local).</p> http://mathoverflow.net/questions/80110/structure-of-the-variety-of-normal-matrices/80124#80124 Answer by Torsten Ekedahl for structure of the variety of normal matrices Torsten Ekedahl 2011-11-05T08:17:05Z 2011-11-05T09:05:22Z <p>You have to be careful with what you mean here. As your equations involve complex conjugation they do not define a <em>complex</em> variety. They do define a real algebraic variety. However, then you have to be careful because there may be reducible components that "are not seen" easily in the real picture, they may not have real points, only complex ones (to be precise all their real points may lie on othere irreducible components). This is what happens in your example.</p> <p>The first step in understanding a real algebraic variety is usually to extend scalars to the complex numbers. To understand what happens in your case it pays to describe a little bit more abstractly. Hence considered over the reals what we have is a finite dimensional $\mathbb R$-algebra with an $\mathbb R$-involution, namely $\mathbb C$. We then consider $n\times n$-matrices over $\mathbb C$ which inherits an involution from the involution of $\mathbb C$, $M\mapsto M^\dagger$. As it is $\mathbb R$-linear the condition $MM^\dagger=M^\dagger M$ is given by polynomials with real coefficients and hence defines a real algebraic subvariety of $M_n(\mathbb C)$ which is seen as the affine space <code>$\mathbb A^{2n^2}_{\mathbb R}$</code> over the reals. When we extend scalars we get exactly the same description but replacing the involutive $\mathbb R$-algebra $\mathbb C$ with its scalar extension $\mathbb C\bigotimes_{\mathbb R}\mathbb C$. This algebra is isomorphic to $\mathbb C\times\mathbb C$ with the involution permuting the two factors. Hence $M$ is now described by two complex matrices $(A,B^t)$ and $(A,B^t)^\dagger=(B,A^t)$ and $MM^\dagger=M^\dagger M$ is turned into $(AB,B^tA^t)=(BA,A^tB^t)$ which just describes pairs $(A,B)$ of commuting matrices. Hence the complex scalar extension is the variety of pairs of commuting matrices with a non-standard real structure where complex conjugation takes $(A,B)$ to $(\overline B,\overline A^t)$ (instead of $(A,B)\to (\overline A,\overline B)$ which is the more standard one).</p> <p>Now, it is well-known that the variety of pairs of commuting matrices is <em>very</em> complicated. There is a more easily understood subset consisting of the pairs $(A,B)$ where both $A$ and $B$ are semi-simple. From an algebraic point of view it is not as it a subvariety (it is a constructible but not locally closed subset). It does contain the open algebraic subset of pairs $(A,B)$ where all the roots of the characteristic polynomials of $A$ (say) are distinct. That subset is Zariski dense in the set of all semi-simple matrices and hence the latter lie in the closure of the former. All the real points will then lie in this closure $\overline S$. However, it is well-known that unless $n$ is <em>very</em> small the closure is not equal to the whole variety (there are pairs that can not be deformed into semi-simple ones). This means that while $\overline S$ is an irreducible component (that is defined over the real numbers) there are (many) other irreducible components. They will be exactly of the type described above, all their real points will also lie in $\overline S$.</p> <p><b>Addendum</b>: As to the question of dimension, of course the closure of semi-simple matrices has the right dimension but I think (but don't quite remember) that there are other components of larger dimension.</p> http://mathoverflow.net/questions/80105/why-do-the-definition-of-deck-transformations-requires-homeomorphism/80120#80120 Answer by Torsten Ekedahl for Why do the definition of deck transformations requires homeomorphism Torsten Ekedahl 2011-11-05T06:45:06Z 2011-11-05T08:28:29Z <p>Assume given a group $G$, a subgroup $H$ and a $g\in G$ such that $gHg^{-1}$ is <em>properly</em> contained in $H$. Let now $Y$ be a topological space with an action of $G$ such that $Y\rightarrow Y/G=:X$ is a covering map (such $Y$'s are plentiful). Let $\tilde X:=Y/H$ so that $p\colon\tilde X\rightarrow X$ is another covering map. Now, $g\cdot\colon Y\rightarrow Y$ maps $H$-orbits to $H$-orbits and hence induces a map $f\colon\tilde X\rightarrow\tilde X$ which fulfils $p\circ f=p$. It is not injective however as the fibres look like cosets $H/gHg^{-1}$ which by assumption contains more than one element. (Using some covering space theory it is easy to see that possibly - I don't think this possibility exists - excluding some very strange examples any example must appear in this way.)</p> <p>It remains to show that such $(G,H,g)$ exist. One can actually start with any $H$ and an injective non-surjective endomorphism of it but it is easier to give a concrete example (the general construction is very similar). Hence we let $H':=\mathbb Z[1/2]$, the group of rational numbers whose denominators are powers of $2$. We have an action of $\mathbb Z$ on $H'$ where $1\in\mathbb Z$ acts by multiplication by $2$. We then let $G$ be the semi-direct product of $H'$ and $\mathbb Z$, $H:=\mathbb Z\subseteq H'$ and $g=(1,0)$. Note that this group is finitely generated (by $(1,0)$ and $(0,1)$) but it seems not to be finitely presented. However, it is easy enough to modify it to be finitely presented: Just take the group generated by $g$ and $h$ and relation $ghg^{-1}=h^2$ with $H$ generated by $h$. It maps to the semi-direct product which shows that $gHg^{-1}$ is indeed properly contained in $H$. Hence we can get examples where $X$ is a finite CW-complex.</p> <p><b>Addendum</b>: I just saw Benjamin's answer when I posted this. He makes use of the claim that any $G$-endomorphism of a transitive $G$-set is an automorphism. However, this claim seems to be false and under my assumptions we get an example of an endomorphism of $G/H$ that is not bijective. Curiously enough my first reaction was the same as Benjamin's. Maybe both of us encountered such statements when learning of finite groups where of course it is true.</p> <p><b>Addendum 1</b>: In a comment to Benjamin's answer (that may disappear as Benjamin has deleted his answer) Georges Elencwajg makes a reference to page 179 of Lima's <em>Fundamental Groups and Covering Spaces</em> where a counterexample to bijectivity is given where $X$ is the figure eight with a $2$-cell adjoined. Let me further add that my last example with the group generated by $g$ and $h$ is most likely to be the same example. We can indeed construct a CW-complex with my example as its fundamental group by starting with the figure eight and then adding the relation $ghg^{-1}=h^2$ by adjoining the appropriate 2-cell.</p> http://mathoverflow.net/questions/79872/chern-numbers-of-primitve-classes-in-bu/79873#79873 Answer by Torsten Ekedahl for Chern numbers of primitve classes in BU Torsten Ekedahl 2011-11-02T23:01:00Z 2011-11-03T05:11:48Z <p>We have that if $f\colon S^{2k}\to BU$ is an actual map of topological spaces (it is a little bit unclear from your formulation if you assume this) then $\langle c_k,[f]\rangle>$ is a multiple of $(k-1)!$ and all multiple are possible. See for instance Husemoller: Fibre bundles, Cor 18.9.8, GTM 20, Springer Verlag.</p> http://mathoverflow.net/questions/79356/regular-singularities-and-logarithmic-singularities/79362#79362 Answer by Torsten Ekedahl for regular singularities and logarithmic singularities Torsten Ekedahl 2011-10-28T08:04:20Z 2011-10-28T08:04:20Z <p>My guess is that the OP is thinking of singularities of integrable connections rather than singularities of varieties. If so logarithmic singularities usually refer to solutions of $y'=1/x$ whose (many-valued) solution is exactly $\log x$. It may possibly also refer to higher order equations whose solutions are powers of $\log x$. In any case $y'=\alpha/x y$ with solution $x^\alpha$ also has regular singularities. In one dimension the general connection with regular singularities is locally a combination of these two cases. In higher dimension there are even more complicated connections with regular singularities (even when the divisor of singularities of the connection only has normal crossing singularities). However, there is the notion of log-connections and by a result (of Deligne I believe) when the divisor of singularities has normal crossings any connection with regular singularities has an extension across the divisor which is a log-connection.</p> http://mathoverflow.net/questions/78508/brauer-group-of-complete-dvr/78531#78531 Answer by Torsten Ekedahl for Brauer group of complete DVR Torsten Ekedahl 2011-10-19T04:35:09Z 2011-10-19T04:35:09Z <p>This is more trivial (in the sense that we have hidden all the non-trivial parts among general preliminaries...) than the identification of $\text{Br}(K)$ that Alex is talking about. We have that $\text{Spec } A_{nr}\to \text{Spec } A$ is an algebraic universal covering map and $\text{Spec } A_{nr}$ is acyclic so that $H^\ast(\text{Spec} A,\mathbb G_m)=H^\ast(\pi,A^\ast_{nr})$, where $\pi$ is the Galois group of $\text{Spec } A_{nr}\to \text{Spec } A$ (which is also the Galois group of $K_{nr}/K$). Hence $H^2(\text{Gal}(K_{nr}/K),A^\ast_{nr})$ is the cohomological Brauer group of $A$ which in this case is easily seen to be equal to the Brauer group.</p> <p>If one looks at what actually goes into this proof one has the following:</p> <ul> <li>The Brauer group of $A_{nr}$ is trivial. This comes from the fact that the special fibre of an Azumaya algebra is trivial as the residue field is separably closed and as $A_{nr}$ is Henselian idempotents lift. <li>This immediately gives an embedding $\text{Br}(A)\subseteq H^2(\pi,A^\ast_{nr})$. The surjectivity is done in the same way as for fields, one explicitly constructs the Azumaya algebra associated to a $2$-cocycle. </ul> http://mathoverflow.net/questions/78218/brauer-group-of-a-field-of-power-series-in-two-variables/78478#78478 Answer by Torsten Ekedahl for Brauer group of a field of power series in two variables. Torsten Ekedahl 2011-10-18T19:28:03Z 2011-10-18T19:28:03Z <p>You can get a fairly good picture of the elements of order $2$ of the Brauer group in the following way. There is no reason to fixate on characteristic $2$ so I assume that we are dealing with $K:=\mathbb F_p((X,Y))$ and in fact the only reason to stick to the prime field is notational convenience as the Frobenius map on $\mathbb F_p$ is the identity so the relative Frobenius is equal to the absolute one which means that I won't have to distinguish between $Z$ and $Z^{(p)}$. Technically it may be that proper references would require $Z$ to be of finite type over the base but everything I say will be clearly true also for $Z=\text{Spec}\mathbb F_p((X,Y))$. (On the other hand the only thing I will need of $Z$ except for a smoothness/regularity assumption is that it is affine and with trivial Picard group).</p> <p>We define the sheaf (on the small étale site of $Z$) $\nu$ by the exact sequence $$0\rightarrow\mathcal O_Z^\ast\xrightarrow{p}O_Z^\ast\rightarrow\nu\rightarrow0.$$ We have a map $\text{dlog}\colon\mathcal O_Z^\ast\rightarrow\Omega^1_Z$, where $\text{dlog}(f):=df/f$ and it factors to give an injection $\nu\subseteq \Omega^1_Z$. More precisely, it lands in the subsheaf $Z^1$ of closed forms and we have an exact sequence (again on the small étale site): $$0\rightarrow\nu\rightarrow Z^1\xrightarrow{C-\iota}\Omega^1_Z\rightarrow0,$$ where $C\colon Z^1/B^1\rightarrow\Omega^1_Z$, is the Cartier isomorphism ($B^1$ being the exact $1$-forms) and $\iota\colon Z^1\subseteq \Omega^1_Z$ is the inclusion.</p> <p>Now, the first sequence (and the fact that $\text{Pic}(Z)=0$) gives that $H^1(Z,\nu)$ is the kernel of multiplication by $p$ on the Brauer group. The fact that $Z$ is affine gives that $H^1(Z,Z^1)=0$ and hence the second sequence gives that $H^1(Z,\nu)$ is the cokernel of $C-\iota\colon H^0(Z,Z^1)\rightarrow H^0(Z,\Omega^1_Z)$. This cokernel can be made very explicit (and to make it very explicit we temporaritly assume $p=2$):</p> <p>$H^0(Z,Z^1)$ is a module over $K$, where scalar multiplication is given by the square map $f\cdot\omega=f^2\omega$, and has a basis given by $dX$, $dX/X$, $d(XY)$, $dY$ and $dY/Y$. We have that $C$ is $0$ on $dX$, $d(XY)$ and $dY$ and $C(dX/X)=dX/X$ and $C(dY/Y)=dY/Y$. Furthermore, $C$ is linear in the sense that $C(f^2\omega)=fC(\omega)$. This implies that the relations in the cokernel are given by $f^2dX=0$, $f^2dY=0$, $f^2XY(dX/X+dY/Y)=0$, $(f^2-f)dX/X=0$ and $(f^2-f)dY/Y=0$ (where $dX$, $dY$, $dX/X$, $dY/Y$, $XdY$ and $YdX$ is a $K$-basis for $H^0(Z,\Omega^1_Z)$). This allows for a fairly transparent normal form for elements in $H^1(Z,\nu)$.</p> <p>If one wants a direct description of the central simple algebra associated to an element $\omega\in H^0(Z,\Omega^1_Z)$ one can apply the What Else Can It Be-principle (a very useful though somewhat dangerous principle, in this case it is probably OK). Recall that we have the algebra $\mathcal D$ of differential operators of order $<p>We can then twist this by$\omega$by replacing the last relation with$D^p=D^{[p]}+\omega(D)^p$. This still gives a central simple algebra and it should (as I said according to the WECIB-principle) be the associated element of${}_p\text{Br}(K)$.</p> <p>There is however a different way of getting explicit representatives. For this we realise instead${}_p\text{Br}(K)$as$H^2(Z,\mu_p)$(now in the flat topology). We then have the usual cup product map$H^1(Z,\mathbb Z/p)\bigotimes H^1(Z,\mu_p)\rightarrow H^2(Z,\mu_p)$. We can represent elements of$H^1(Z,\mathbb Z/p)$by Artin-Schreier extensions$b^p-b=a$, where$a\in K$and elements of$H^1(Z,\mu_p)$by$p$'th root extensions$g^p=f$, where$f\in K^\ast$. The central simple algebra associated to the cup product of these two classes is the algebra generated by$K(b)$and$g$and relations$gbg^{-1}=b+1$and$g^p=f$. On the other hand a straightforward computation shows that the class of the cup product in$H^1(Z,\nu)$is the residue of$adf/f\in H^0(Z,\Omega^1_Z)$. As$H^0(Z,\Omega^1_Z)$is generated as a group by such elements we get a different description of the class of${}_p\text{Br}(K)$associated to elements of$H^0(Z,\Omega^1_Z)$. Note however, that they will not be isomorphic as algebras, the algebra associated by the first procedure to$adf/f$has$K$-dimension$p^4$whereas the second construction has dimension$p^2$.</p> http://mathoverflow.net/questions/77812/uniqueness-of-splitting-field-for-linear-representations-of-finite-groups/77824#77824 Answer by Torsten Ekedahl for Uniqueness of splitting field for linear representations of finite groups Torsten Ekedahl 2011-10-11T15:08:30Z 2011-10-12T06:10:22Z <p>You need two conditions for a field to be a splitting field for a specific irreducible representation (in characteristic zero to begin with): It must contain the character values of the representation. For this there is of course a minimal field, the field generated by those values. However, a splitting field must also split a division algebra and for that there is no unique minimal field. </p> <p>As a specific example we can take the quaternion group of order$8$over the rational numbers. The component of the group algebra corresponding to the only faithful representation is a quaternion algebra over$\mathbb Q$ramified at$\infty$and$2$and hence is split by any quadratic imaginary field for which$2$is either ramified or non-split.</p> <p>To be more concrete about the quaternion representation, the ordinary real quaternion algebra makes sense over the rationals, denoted$\mathbb H_{\mathbb Q}$. It has$\mathbb Q$-basis 1,i,j,k and the usual multiplication table. Then <code>$\pm\{1,i,j,k\}$</code> is a multiplicative group that is a copy of the quaternion group$Q$and therefore we have an algebra map$\mathbb Q[Q]\to\mathbb H_{\mathbb Q}$. It is evidently surjective so that$\mathbb H_{\mathbb Q}$is one of the factors in the group algebra (more precisely$\mathbb Q[Q]=\mathbb Q^4\times\mathbb H_{\mathbb Q}$, where the first four factors correspond to the four one-dimensional representations). Now, for a field$K$of characteristic zero$\mathbb H_K$has a two-dimensional irreducible representation exactly when it is split, i.e., when it is isomorphic to the algebra of$2\times2$-matrices. That is thus exactly the condition for a two-dimensional irreducible representation to exist over$K$. Now, it is well-known that the algebra is split precisely when the reduced norm form ($N(\alpha)=\alpha\overline{\alpha}$) restricted to the purely imaginary quaternions has a non-trivial zero. As$N(ai+bj+ck)=a^2+b^2+c^2$we get the condition that David mentions in the comments.</p> <p>The situation in positive characteristic is different. As the group algebra can be defined over the prime field and the Brauer group of finite fields is trivial, one only needs for (the mod$p$) character values to be in the field so in that case there is a minimal field.</p> <p><b>Addendum</b>: The question of a minimal field for all irreducible representations has essentially the same answer. First the field has to contain all character values, then there are still division algebras to split (in characteristic$0$). The example of the quaternion group still illustrates the problem, all characters are rational-valued and one must still kill the quaternion algebra for which there is no unique minimal field.</p> http://mathoverflow.net/questions/77233/fields-obtained-by-adjoining-x-coordinates-of-torsion-points-on-elliptic-curves/77266#77266 Answer by Torsten Ekedahl for Fields obtained by adjoining x coordinates of torsion points on elliptic curves Torsten Ekedahl 2011-10-05T17:31:15Z 2011-10-05T17:31:15Z <p>I think this works: Take two non-isogenous (over$\overline{K}$) curves$E$and$E'$with$K(E[\ell])=K(E'[\ell])=K$. Replace$E'$, say, by a quadratic twist. Then$K(E[\ell])=K(x(E[\ell]))=K(x(E'[\ell]))=K$but$K(E'[\ell])\ne K$.</p> http://mathoverflow.net/questions/65653/invariant-differential-forms-on-commutative-group-schemes-are-closed/75141#75141 Answer by Torsten Ekedahl for Invariant differential forms on commutative group schemes are closed!? Torsten Ekedahl 2011-09-11T14:03:45Z 2011-09-11T14:03:45Z <p>I would be a little bit nervous about things when the group scheme is not smooth (there may not be any problems though) but you are interested in a smooth case anyway. To me it seems that the most natural way of proving this is to look at the relation between commutators of vector fields and the differential of forms. Thus if$X$and$Y$are vector fields and$\omega$is a$1$-form we have$d\omega(X,Y)=X(\omega(Y))-Y(\omega(X))+\omega([X,Y])$. Applying this to the case when$X$,$Y$and$\omega$are (left say) translation invariant gives$d\omega(X,Y)=\omega([X,Y])$as$\omega(X)$and$\omega(Y)$are constant functions. This shows that all translation invariant$1$-forms are closed precisely when the Lie algebra of the group is commutative. Of course the Lie algebra is commutative if the group is (I guess the converse does not hold in positive characteristics though I cannot offhand come up with an example).</p> http://mathoverflow.net/questions/75071/i-was-wondering-if-the-set-of-singular-loops-is-a-somewhere-submanifold-of-loop/75077#75077 Answer by Torsten Ekedahl for I was wondering if the set of singular loops is a (somewhere) submanifold of loop space? Torsten Ekedahl 2011-09-10T09:10:22Z 2011-09-10T09:10:22Z <p>The following argument I think should work to prove that the answer is no (though I haven't checked whether it really fits in with the proper definition): Consider a loop where three points of$S^1$maps to the same point in$M$but with linearly independent (to be on the safe side) tangents. Then you keep within the singular loops by moving two of the three points in the same way but the third in any which way. To first order that should correspond to a$TM$-valued vector field on$S^1$which is the same at the two points but arbitrary at the third. However, there are three possibilities for the choice of the two points that should travel together and all the vector fields that you get from varying the choices will not form a linear subspace. Hence what you get is rather three submanifolds (corresponding to the three choices) coming together meeting transversally.</p> http://mathoverflow.net/questions/74611/is-the-composition-of-two-bundle-projections-necessarily-a-bundle-projection/74722#74722 Answer by Torsten Ekedahl for Is the composition of two bundle projections necessarily a bundle projection? Torsten Ekedahl 2011-09-07T07:18:04Z 2011-09-07T07:18:04Z <p>I think that the following works: Let$X\to Y$and$Y\to Z$be locally trivial fibrations with all spaces paracompact and$Z$locally contractible (I do not assume that a fibration implies that all fibres are homeo- or diffeomorphic). We want to show that$X\to Z$is locally trivial. We may then reduce to the case when$Z$is contractible and$Y=Z\times F$. Under the paracompactness assumption locally trivial fibrations are homotopy invariant (see for instance Husemoller: Fibre bundles GTM 20, Springer, Thm 9.8 plus a simple reduction to the principal bundle case). Hence there is a locally trivial fibre bundle$X'\to F$such that$X\to Y$is isomorphic to$Z\times X'$which is the local trivialtity.</p> http://mathoverflow.net/questions/73989/the-jacobson-radical-of-an-infinite-dimensional-algebra/74331#74331 Answer by Torsten Ekedahl for The Jacobson radical of an infinite dimensional algebra Torsten Ekedahl 2011-09-02T05:49:01Z 2011-09-02T07:58:32Z <p>As there seem to be some differing opinions in the comments as to whether all irreducible representations are finite-dimensional let me give the argument I had in mind. A module over the path algebra is the same thing as a representation of the quiver, which in this case means two vector spaces$U$and$V$and linear maps$e\colon U\rightarrow V$and$f\colon V\rightarrow U$. We introduce also$S:=fe\colon U\rightarrow U$and$T:=ef\colon V\rightarrow V$. Assume now that$(U,V,e,f)$is irreducible and pick a non-zero$u\in U$and let$W$be the subrepresentation generated by$u$. It is clear that the$U$-part of$W$is the$k[S]$-submodule generated by$u$so by irreducibility we have that$U=k[S]u$. Now, do the same argument for$Su$giving us also that$U=k[S]Su$. In particular there is a polynomial$p(S)$such that$u=p(S)Su$, i.e.,$q(S)u=0$where$q(S)=p(S)S-1$which in particular is non-zero so that$U=k[S]u$is finite dimensional. By symmetry the same argument applies to$V$so the module is finite dimensional.</p> <p>Note, that slightly extending this also gives us a classification of the finite dimensional modules. In particular there are enough of them to make the Jacobson radical be equal to$0$.</p> <p><b>Addendum</b>: Rather than mixing together several steps it is probably better to divide it up: First show that if$(U,V,e,f)$is irreducible then$U$is irreducible as$k[S]$-module and then use the classification of simple$k[S]$-module.</p> <p>Also I feel that the path algebra is something of a red herring. Representations of a quiver are clearly modules over a ringoid (aka ring with several objects). A ringoid with a finite number of objects is Morita equivalent to a ring (as each object gives rise to a compact projective which collectively are faithful, giving a compact faithful projective module as their sum). However, passing to the endomorphism ring of that projective just hides some extra structure that you started with which seems silly.</p> http://mathoverflow.net/questions/74113/every-involution-of-an-enriques-surface-is/74225#74225 Answer by Torsten Ekedahl for every involution of an Enriques surface is Torsten Ekedahl 2011-09-01T04:12:57Z 2011-09-01T08:04:13Z <p>Let me try an argument different from Christian's:$\sigma$does not act freely as$\chi(\mathcal O_X)=1$and hence not divisible by$2$. At a fixed point$x$,$\sigma$acts by$\pm1$on the fibre of$\omega_X$and hence acts by$1$on the fibre of$\omega_X^{\otimes2}$. It also acts by a scalar on a global non-zero section of$\omega_X^{\otimes2}$but as that section is non-zero at$x$this scalar must be$1$.</p> <p><b>Addendum</b>: It seems that it is essential that there are fixed points. If we look at a bielliptic example;$E\times F$the product of two elliptic curves with$\tau$acting by an automorphism of order$4$on$E$(assumed to have one) and translation by an element of order$4$on$F$then if we divide by$\tau^2$we have that$\tau$induces an involution which acts by multiplication by$-1$on global sections of$\omega^{\otimes 2}$.</p> http://mathoverflow.net/questions/73889/sq1-cohomology-of-spaces/73891#73891 Answer by Torsten Ekedahl for$Sq^1$cohomology of spaces Torsten Ekedahl 2011-08-28T09:43:43Z 2011-08-29T17:49:36Z <p>I think the easiest way to understand the Bockstein spectral sequence is through the exact couple coming from the long exact sequence of cohomology associated to$0\to\mathbb Z\to\mathbb Z\to \mathbb Z/2\to0$. This shows first that indeed the first differential is$Sq^1$and tells you that the next page is the direct sum of the cokernel and kernel (shifted one step) of multiplication by$2$on$2H^\ast(X,\mathbb Z)$. Hence it is like what you would get from applying the universal coefficient formula to$2H^\ast(X,\mathbb Z)$(instead of$H^\ast(X,\mathbb Z)$). When each cohomology group$H^\ast(X,\mathbb Z)$is finitely generated this means concretely that you "keep" each$\mathbb Z$-factor (as well as odd torsion) and downgrade each$\mathbb Z/2^n$to$\mathbb Z/2^{n-1}$.</p> <p>In particular the difference between the dimension of$H^n(X,\mathbb Z/2)$and that of the$Sq^1$-cohomology is equal to the number of$\mathbb Z/2$-factors in$H^n(X,\mathbb Z)$and$H^{n+1}(X,\mathbb Z)$.</p> <p>I found a reference to Q2. In Madsen, Milgram: The classifying spaces for surgery and cobordism of manifolds, Ann of Math Studies 92 where they refer to Browder: Torsion in H-spaces, Ann of Math 74 for the Bockstein s.s. of$K(\mathbb Z_{(2)},n)$and$K(\mathbb Z/2,n)$. The Madsen-Milgram book also contains other examples of computations with the Bss.</p> http://mathoverflow.net/questions/73879/canonical-liftings-of-endomorphisms-of-ordinary-abelian-varieties/73882#73882 Answer by Torsten Ekedahl for Canonical liftings of endomorphisms of ordinary abelian varieties Torsten Ekedahl 2011-08-28T05:00:48Z 2011-08-29T13:01:08Z <p>An almost reference (some assembly required) is N. Katz: Serre-Tate local moduli, Springer Lecture notes in Mathematics 868. By Lemma 1.1.2 an endomorphism lifts if and only if it lifts on the$p$-divisible group and for the canonical lift the endomorphism lifts trivially. I haven't looked at Drinfeld's original article to see if that is more suitable.</p> <p><b>Addendum</b>: There is no denying that my reference suffers in comparison with the two others given. I would like to point out however that Drinfeld's argument is very beautiful and arguably the slickest approach to canonical liftings (and Serre-Tate coordinates in general).</p> http://mathoverflow.net/questions/73760/connected-extensions-of-finite-by-connected-algebraic-groups/73769#73769 Answer by Torsten Ekedahl for Connected extensions of finite by connected algebraic groups Torsten Ekedahl 2011-08-26T13:10:08Z 2011-08-26T17:31:02Z <p>In Groupes algébriques et corps de classes Serre classifies the$2$-dimensional commutative unipotent connected algebraic groups$G$(VII:11). With the exception of the product of the additive group with itself they are all isogenous to the Witt vector group$W_2$so that there is an exact sequence as per above with$E=W_2$. For some of them Serre notes that the isogeny can be chosen to be separable so that$H$is a finite group (i.e., étale group scheme) yet$G$is not necessarily isomorphic to$W_2$.</p> <p><b>Addendum</b>: An algebraic group is a vector group precisely when it is commutative, connected and killed by$p$(again in Serre's book). Clearly$G$is killed by$p$if$E$is. Conversely, if$G$is killed by$p$then multiplication by$p$induces a map$G\to H/pH$which is constant as$G$is connected and$H$is finite. Thus$G$is a vector group precisely when$E$is. Also vector groups are determined by their dimension and$E$and$G$have the same dimension.</p> http://mathoverflow.net/questions/73545/composition-of-covering-map-and-bundle-projection/73546#73546 Answer by Torsten Ekedahl for composition of covering map and bundle projection Torsten Ekedahl 2011-08-24T06:32:08Z 2011-08-25T11:11:02Z <p>It is better to just assume that$f$is a covering space. By shrinking$Z$we may assume that that$Z$is a ball and that$Y=Z\times F$. As$X\to Y$is a covering space and$Z$is simply-connected there is a covering space$X'\to F$such that$X\to Y$is isomorphic to$Z\times X'\to Z\times F$which gives what you want.</p> <p><b>Addendum</b>: The reason that$X'$exists is that if$h\colon T\to T'$is a homotopy equivalence (and possibly$T$and$T'$fulfil some local niceness conditions which certainly are fulfilled in the case at hand), then pullback along$h$induces an equivalence between the category of covering spaces of$T$and that of$T'$. This is true irregardless on whether$T$and$T'$are connected or not. </p> http://mathoverflow.net/questions/73447/central-extensions-of-group-schemes/73470#73470 Answer by Torsten Ekedahl for Central extensions of group schemes Torsten Ekedahl 2011-08-23T05:50:34Z 2011-08-24T04:28:11Z <p>If we have a central extension of group schemes$1\rightarrow B \rightarrow C\rightarrow A\rightarrow1$with$A$abelian, then we get a commutator mapping$\Lambda^2A\rightarrow B$(of sheaves as$\Lambda^2A$in general is not a group scheme) and the extension is abelian precisely when this map is zero. Hence for an non-abelian extension to exist there must be a non-zero map$\Lambda^2A\rightarrow B$. Let us now assume that$A=\mu_n$and consider first the case when$n=p$, the characteristic of the field$k$(which we may assume is algebraically closed). A non-zero map$\Lambda^2A\rightarrow B$would give a non-zero map$A\rightarrow\mathrm{Hom}(A,B)$, where the right hand side is the sheaf of group homomorphisms. As the Frobenius map is zero on$\mu_p$we may replace$B$by its Frobenius kernel so we may assume that$B$is either$\mu_p$or the Cartier dual of$\alpha_{p^m}$. Now, as sheaves$\mathrm{Hom}(A,B)$is isomorphic to$\mathrm{Hom}(D(B),D(A))$, where$D(-)$denotes the Cartier dual. However$D(\mu_p)=\mathbb Z/p$so when$A=\mu_p$we get that$\mathrm{Hom}(D(B),D(A))=\mathbb Z/p$and there is only the zero map from$A=\mu_p$into it. In the other case$D(A)=\alpha_{p^m}$and$\mathrm{Hom}(\alpha_{p^m},\mathbb Z/p)$is zero. If instead$n=p^k$, the argument is the same. The case when$n=\ell^k$is even simpler so in all cases all possible commutator maps are zero and the extension is commutative.</p> <p>(When$A=B=\mathbb G_a$then there are candidates for commutator maps and in fact$(a,b)(a',b')=(a+a',b+b'+a^pa')$gives a non-commutative central extension which I imagine is the fake Heisenberg group.)</p> <p><b>Addendum</b>: A general comment is that it is more convenient to work with sheaves (in the fppf topology say) as that means that we essentially can pretend that we work with set-theoretic groups. It is however also necessary if we want to see the commutator map as a map$\Lambda^2A\to B$as the sheaf$\Lambda^2A$(of$A$considered as an abelian sheaf) is in general not reprsentable. The$\langle-,-\rangle\colon\Lambda^2A\to B$view point is convenient as it allows us to do what one usually does when having a pairing: We get for instance a map$A\to\mathrm{Hom}(A,B)$given by$a\mapsto (a'\mapsto \langle a,a'\rangle)$just from the fact that$\langle-,-\rangle$is biadditive.</p> <p>I have implicitly assumed that$B$is of finite type (as I claim that its Frobenius kernel is finite) even though it may not be necessary (a limit argument anyone?).</p> http://mathoverflow.net/questions/72173/coderivations-of-sv-correspond-to-linear-maps-sv-v-only-over-characterist/72176#72176 Answer by Torsten Ekedahl for Coderivations of S(V) correspond to linear maps S(V) -> V. Only over characteristic 0? Torsten Ekedahl 2011-08-05T12:59:00Z 2011-08-05T12:59:00Z <p>Let us assume$k$has characteristic$p$. The problem is (to me at least) easier to understand by dualising (assume that we are only looking at homgeneous derivations so that we can take the graded dual and have no problems at least if$V$is finite-dimensional, things will go wrong even here). Then the statement would say that any linear map$V\to V\subset S(V^\ast)^\ast$has a unique extension to a derivation of$S(V^\ast)^\ast$. However,$S(V^\ast)^\ast$is the divided power algebra on$V$which is not generated as an algebra by its degree$1$elements so the uniqueness doesn't follow as for the symmetric algebra. In fact for$V=kx$we have that the divided power algebra is generated as a commutative algebra by$x_n:=\gamma_{p^n}(x)$and relations$x_n^p=0$. Hence one can arbitrarily choose the value of a derivation on the$x_n$which makes it very clear that the derivation is not determined by its value on$x_1$.</p> http://mathoverflow.net/questions/71664/obstructions-to-formally-integrating-vector-fields-in-characteristic-p/71670#71670 Answer by Torsten Ekedahl for Obstructions to formally integrating vector fields in characteristic p? Torsten Ekedahl 2011-07-30T18:21:04Z 2011-08-01T12:10:30Z <p>This is not an answer to the questions but some general comments. One should be aware that the relation between vector fields and Hasse derivations in characteristic$p$is not at all analogous to the characteristic$0$. It is true that a Hasse derivation in all characteristics is the same thing as an acction of the formal additive group. The difference is whereas in characteristic$0$the formal additive group is the only$1$-dimensional formal group whereas in positive characteristic there are more, for instance the formal multiplicative group. In positive characteristic the derivation part of a Hasse derivation corresponds to an action of the group scheme that is the kernel of the Frobenius map which is$\alpha_p$. Again there are several group schemes of order$p$(such as$\mu_p$which is the kernel of Frobenius on the formal multiplicative group). On the vector field side there thus is a first obstruction on the vector field itself, that it should give rise to an$\alpha_p$-action. Concretely that means that the$p$-th power of the derivation should be$0$.</p> <p>Note that that means that there may not even be a vector field to start with even though there might be many vector fields on$M$there may not be any with that property (for instance a smooth proper toric variety with no automorphisms outside of the torus).</p> <p>In any case if one wants an obstruction theory one should note that given$D_1$,$D_n$for$n=1,\ldots,p-1$is just$D_1^n$so the first undetermined one would be$D_p$. If one has local liftings one should compare two such liftings$D_p$and$D'_p$and it follows that their difference$D'_p-D_p$is a derivation and one gets a torsor of the tangent sheaf as a first obstruction. Unfortonately if$D$is a derivation$D'_p=D_p+D$may not fulfil the further condition for being a part of a Hasse derivation namely that <em>its</em>$p$'th power should be zero. One can expand its$p$'th power using the Jacobson formula but it leads to a (seemingly) nasty non-linear problem. Anyway if solvable one can continuer with$D_{p^2}$which would be the next undetermined term and continue in the same manner but it looks like a nightmare (the fact that I don't think that local liftings may exist makes it even less palatable).</p> <p>However, it is clear that you do not have unique extensions: Take some$M$on which an action of$\hat G_a\times \hat G_a$is given. Then the actions of$\hat G_a$given by the inclusions$t\mapsto (t,0)$and$t\mapsto (t,t^p)$have the same first order action.</p> <p><b>Addendum</b>: I think that all in all Hasse-Schmidt derivations are better than Hasse derivations (recall that a Hasse-Schmidt derivation is just a map$D_\infty\times M\to M$, i.e., not necessarily a formal group action). There the liftings of an order$n$HS-derivation to order$n+1$is just a pseudo-torsor over the tangent sheaf with local liftings existing so that the obstruction for extension is just an$H^1$. Of course you will have many more of them but as you don't have uniqueness for lifting vector fields to Hasse derivations anyway it seems to matter less.</p> <p><b>Addendum (formal multiplicative group)</b>: Whereas actions of the formal additive group are quite messy, the formal multiplicative group (as always with tori) is much nicer. As$\mu_p$is linearly reductive an action of it is nothing but a$\mathbb Z/p$-grading on$M$. This is the best description though in terms of vector fields it corresponds to a derivation$D$with$D^p=D$. Similarly a$\mu_{p^n}$-action is nothing but a$\mathbb Z/p^n$-grading. Hence, lifting from$\mu_{p^n}$to$\mu_{p^{n+1}}$corresponds to a refinement of a$\mathbb Z/p^n$-grading to a$\mathbb Z/p^{n+1}$-grading. Hence an action of the formal multiplicative group is the same thing as a$\mathbb Z_p$-grading. </p> <p><b>NB</b>: As a topological group that is; a collection of compatible$\mathbb Z/p^n$-gradings for each$n$. In particular the part of degree$a\in \mathbb Z_p$is an infinite intersection and could very well be$0$for all$a$. On the other hand an action of the actual multiplicative group is the same thing as a$\mathbb Z$-grading and the associated$\mathbb Z_p$-grading is just induced by$\mathbb Z\subset \mathbb Z_p$.</p> <p>However things can be more complicated yet very close to such an action. If we have a derivation$D$with$D^p=fD$where$f$is an invertible function then we don't have a$\mu_p$-action unless$f=1$and we may not even be able to get a$\mu_p$-action by scaling$D$. However, on the étale cover where we extract a$p-1$'th root of$f$we do get such a modification. So what we have is a kind of twisted$\mu_p$-action and a sheaf of gradings which is an abelian sheaf of$M$isomorphic to$\mathbb Z/p$over the above étale cover (in fact it is the twist of$\mathbb Z/p$by exactly this cover). This extends without problem to twisted actions of the formal multiplicative group.</p> <p>Such thingies do occur in practice. My favourite example is the following: Look at the moduli space (i.e., stack if you want to quibble) of <em>ordinary</em> elliptic curves. Formally at each point this space can be identified using Serre-Tate coordinates (which I guess in this$1$-dimensional case is earlier than Serre-Tate) with the formal multiplicative group. We can now consider the (pro)-étale covering (with structure group$\mathbb Z_p^\ast$) trivialising the group of$p$-torsion points of the universal elliptic curve. By Igusa this is connected and on it we have an action of the formal multiplicative group realising formally at each point the Serre-Tate coordinates.</p> http://mathoverflow.net/questions/70920/simplest-example-of-jumping-of-cohomology-of-structure-sheaf-in-smooth-families/71197#71197 Answer by Torsten Ekedahl for Simplest example of jumping of cohomology of structure sheaf in smooth families? Torsten Ekedahl 2011-07-25T07:38:39Z 2011-07-25T07:38:39Z <p>This is an attempt to realise Sándor's program of getting an example based on Kodaira vanishing or non-vanishing varying in a family. It will be done by keeping the surface fixed but varying the line bundle.</p> <p>I shall start not with the examples of Raynaud but rather a variation given by Raynad-Szpiro (for details on it see Szpiro's article in Astérisque 64 or Flexors exposé in Astérisque 86). Recall that a Tango-Raynaud structure on a relative smooth and proper curve$f\colon X\to C$consists of a line bundle$\mathcal L$on$X^{(p)}$together with a map$\mathcal L\to B^1_{X/C}$, where$B^1_{X/C}\subseteq \alpha_\ast\Omega^1_{X^{(p)}/C}$is the image of$d\colon\alpha_\ast\mathcal O_{X}\rightarrow\alpha_\ast\Omega^1_{X/C}$(and$\alpha\colon X\to X^{(p)}$is the relative Frobenius), such that the adjoint map$\alpha^\ast\mathcal L\rightarrow \Omega^1_{X/C}$is an isomorpism. If$C$is a proper smooth curve and the fibration is non-isotrivial, then such an$\mathcal L$is ample as$\Omega^1_{X/C}$is by Szpiro (and$\alpha$is finite) and on the other hand, by looking at the exact sequence$0\rightarrow\mathcal O_X\rightarrow\alpha_\ast\mathcal O_{X^{(p)}}\rightarrow B^1_{X/C}\rightarrow0$tensored with$\mathcal L^{-1}$we get an embedding$H^0(X,B^1\bigotimes\mathcal L^{-1})\hookrightarrow H^1(X^{(p)},\mathcal L^{-1})$and thus the given map$\mathcal L\to B^1_{X/C}$gives a non-zero element of$H^1(X^{(p)},\mathcal L^{-1})$. The map$X\rightarrow C$is constructed by the Kodaira procedure (using that if the second starting curve$D$is Tango-Raynaud, then so is$f$).</p> <p>The aim is now to show that there exists an$\mathcal{M}\in\mathrm{Pic}^0(X^{(p)})$such that$H^1(X^{(p)},\mathcal L\bigotimes \mathcal{M})=0$. Of course, for every$\mathcal{M}$,$\mathcal L\bigotimes \mathcal{M}$is ample and$\mathrm{Pic}^0(X^{(p)}$is connected so we would be finished if we could do this. Note that$\mathrm{Pic}^0(X^{(p)})$is positive dimensional (as it contains a subgroup isogenous to$\mathrm{Pic}^0(C)\times \mathrm{Pic}^0(D)$) and I shall in fact show that the vanishing is true for all but a finite number of$\mathcal{M}$'s.</p> <p>I shall use$\mathcal L'$to denote a general element of the form$\mathcal L\bigotimes\mathcal{M}$, where$\mathcal{M}\in\mathrm{Pic}^0(X^{(p)})$. </p> <ul> <li> We have that$H^1(X^{(p)},\mathcal L'^{-p^n})=0$for$n>0$. This is proven by descending induction on$n$, the statement being true for large$n$by Serre and ampleness of$\mathcal L'$. It is therefore enough to show that the$p$'th power map$H^1(X^{(p)},\mathcal L'^{-p^n})\rightarrow H^1(X^{(p)},\mathcal L'^{-p^{n+1}})$is injective. This is the map induced by adjunction$\mathcal L''\rightarrow F_\ast F^\ast\mathcal L''$, where$F\colon X\rightarrow X$is the absolute Frobenius (on both$X$and$X^{(p)}$) and$\mathcal L'':=\mathcal L'^{-p^n}$. We prove this by factoring$F$as$\alpha\circ\beta$, where$\beta\colon X^{(p)}\rightarrow X$is the base change of the Frobenius on$C$. Hence it will be enough to show that$H^1(X^{(p)},\mathcal L'')\rightarrow H^1(X,\alpha^\ast\mathcal L'')$and$H^1(X,\alpha^\ast\mathcal L'')\rightarrow H^1(X,\beta^\ast\alpha^\ast\mathcal L'')$are both injective. For the first we have that its kernel is equal to$\mathrm{Hom}(\mathcal L'',B^1_{X/C})$and an element of it gives (by inclusion and adjunction as above) rise to a map$\alpha^\ast\mathcal L''\rightarrow \Omega^1_{X/C}\cong \alpha^\ast\mathcal L$. Such a map is zero as because of the ampleness of$\mathcal L$,$\mathcal L''$is more positive than$\mathcal L$. As for the injectivity of$H^1(X,\alpha^\ast\mathcal L'')\rightarrow H^1(X,\beta^\ast\alpha^\ast\mathcal L'')$we get similarly that the kernel is equal to$\mathrm{Hom}(\alpha^\ast\mathcal L'',f^{\ast}B^1_{C})$. However,$\alpha^\ast\mathcal L''$has strictly positive degree on the fibres of$f$so all such maps are zero. <li> Hence to show that$H^1(X^{(p)},\mathcal L'^{-1})=0$for general$\mathcal{M}$it would be enough to show that for such$\mathcal{M}$the$p$'th power map$H^1(X^{(p)},\mathcal L'^{-1})\rightarrow H^1(X^{(p)},\mathcal L'^{-p})$is injective. Again we can factor it and the second part of the argument works as before so it will be enough to show that there are only a finite number of$\mathcal{M}$'s for which$\mathrm{Hom}(\mathcal L',B^1_{X/C})$could be non-zero. As for$\mathcal L$a non-zero such map would give a non-zero map$\alpha^\ast\mathcal L'\rightarrow \Omega^1_{X/C}\cong \alpha^\ast\mathcal L$. However,$\mathcal L'$is numerically equivalent to$\mathcal L$and hence the map would have to be an isomorphism. This implies that$\mathcal{M}$would have to lie in the kernel of$\alpha^\ast$which is finite as$\alpha$is finite flat (and surjective). </ul> http://mathoverflow.net/questions/71087/presentation-of-the-dual-of-a-locally-free-sheaf/71115#71115 Answer by Torsten Ekedahl for Presentation of the dual of a locally free sheaf Torsten Ekedahl 2011-07-24T09:12:25Z 2011-07-24T09:12:25Z <p>We have that$\mathcal F^\ast$is, by the pairing induced by the exterior algebra, canonically isomorphic to$\Lambda^{d-1}\mathcal F\bigotimes(\Lambda^d\mathcal F)^{-1}$. Now, in general if$\mathcal H\to\mathcal G\to \mathcal F\to 0$is exact then the kernel of the surjective map$\Lambda^\ast \mathcal G\to\Lambda^\ast\mathcal F$is the ideal generated by the image of$\mathcal H\to\mathcal G$. Hence we get an exact sequence$\Lambda^{i-1}\mathcal G\bigotimes \mathcal H \to \Lambda^i \mathcal H\to\Lambda^i\mathcal F\to0$. Applpy, this your second exact sequence and$i=d-1$gives a presentation of the desired type for$\Lambda^{d-1}\mathcal F$and then twist it by$(\Lambda^d\mathcal F)^{-1}$.</p> http://mathoverflow.net/questions/70920/simplest-example-of-jumping-of-cohomology-of-structure-sheaf-in-smooth-families/70963#70963 Answer by Torsten Ekedahl for Simplest example of jumping of cohomology of structure sheaf in smooth families? Torsten Ekedahl 2011-07-22T04:45:47Z 2011-07-24T06:53:05Z <p>One example is given by Enriques surfaces in characteristic$2$. There are three types depending on the value of$\mathrm{Pic}^\tau$(as a group scheme) which can be either$\mathbb Z/2$,$\mu_2$or$\alpha_2$. In the first case$\omega_X$is the generator so in particular it is non-trivial and$H^2(X,\mathcal O_X)=0$(using of course Serre duality). In the two other cases$\omega_X$is trivial (as it is numerically trivial and all numerically trivial line bundles are trivial) so that$h^2(X,\mathcal O_X)=1$. Now,$\alpha_2$can be deformed to both$\mathbb Z/2$and$\mu_2$and such deformations can be lifted to deformations of Enriques surfaces (in fact$\mathrm{Pic}^\tau$is flat in families of Enriques surfaces and the functor from deformations of the surfaces to those of$\mathrm{Pic}^\tau$is formally smooth - Liedtke: arXiv:1007.0787, Ekedahl-Shepherd-Barron: unpublished). If we pick a connected family of Enriques surfaces with some special value being an$\alpha_2$-surface and generically$\mathbb Z/2$, then we get an example.</p> <p>Such an example can be constructed (very) explicitly without deformation theory. Here is a semi-explicit construction which works in any positive characteristic. Fixa a a group scheme$A$of order$p$over$\mathbb A^1$localised at$0$(say) which is$\alpha_p$at$0$and$\mathbb Z/p$elsewhere. By the Godeaux construction (which Raynaud - Prop. 4.2.3, p-torsion du schema de Picard, Astérisque 64 - showed works for such families) there is a free action of$A$on a flat complete intersection$Y$(of any dimension, which we assume is$\ge 2$) such that$X=Y/A$is smooth (note that contrary to the case of an étale group scheme$Y$will <em>not</em> be smooth). As$Y$is a complete intersection we have that$\mathrm{Pic}^\tau_Y=0$and from that it follows that$\mathrm{Pic}^\tau_X=A$. Now,$H^1(-,\mathcal O_-)$is the tangent space of$\mathrm{Pic}^\tau$so it is zero outside of$0$and$1$-dimensional at$0$. (By being careful one can get Enriques surfaces for$p=2$, this I guess was the inspiration for the Godeaux construction).</p> http://mathoverflow.net/questions/14944/have-people-successfully-worked-with-the-full-ring-of-diferential-operators-in-ch/70876#70876 Answer by Torsten Ekedahl for Have people successfully worked with the full ring of diferential operators in characteristic p? Torsten Ekedahl 2011-07-21T04:51:33Z 2011-07-21T04:51:33Z <p>Certainly the fact that the ring of differential operators is non-Noetherian is an inconvenience but it is not clear if it is more than that. For instance one can define the notion of holonomic module. It is not a direct translation of the characteristic zero definition (and this is certainly related to this inconvenience) but once given it seems to work as well as in characteristic zero:</p> <p>MR1918185 (2003h:14030) Bögvad, Rikard(S-STOC) An analogue of holonomic D-modules on smooth varieties in positive characteristics. (English summary) The Roos Festschrift volume, 1. Homology Homotopy Appl. 4 (2002), no. 2, part 1, 83–116. 14F10 (16S32 32C38)</p> http://mathoverflow.net/questions/70813/the-number-of-orbits-of-a-permutation-action/70817#70817 Answer by Torsten Ekedahl for The number of orbits of a permutation action Torsten Ekedahl 2011-07-20T14:38:12Z 2011-07-20T14:38:12Z <p>For$r=2$a$k$-subset can be thought of as a graph with vertices$[n]$and$k$edges. Hence the number of orbits is equal to the number of isomorphism classes of graphs on$n$vertices and$k$edges. Counting them seems like a fairly intractable problem.</p> http://mathoverflow.net/questions/70716/characterization-of-locally-free-modules-via-exterior-powers/70731#70731 Answer by Torsten Ekedahl for Characterization of locally free modules via exterior powers Torsten Ekedahl 2011-07-19T12:36:41Z 2011-07-19T15:23:13Z <p>I think that$\mathcal F$is indeed locally free of rank$n$:</p> <p>Pick a point$x\in X$. It will be enough to show that there is a neighbourhood of$x$on which$\mathcal F$is free of rank$n$. Now, the exterior power commutes with pullbacks (aka scalar extensions) so that in particular the fibre (in the sense of pullback to$\mathrm{Spec}k(x)$) of$\Lambda^m\mathcal F$at$x$equals <code>$\Lambda^m\mathcal{F}_x$</code>. This shows that <code>$\mathcal{F}_x$</code> is an$n$-dimensional vector space. After possibly shrinking$X$we may assume that there is a an <code>$\mathcal{O}_X$</code>-map <code>$f\colon \mathcal{O}_X^n\to \mathcal F$</code> which induces an isomorphism on fibres at$x$. Thus$\Lambda^nf$is a map between locally free modules (of rank$1$) that gives an isomorphism on fibres at$x$and hence is an isomorphism in a neighbourhood of$x$so that we may assume that it is a global isomorphism. The wedge product induces pairings$\mathcal{F}\times\Lambda^{n-1}\mathcal{F}\to \Lambda^{n}\mathcal{F}$and <code>$\mathcal{O}_X^n\times\Lambda^{n-1}\mathcal{O}_X^n\to \Lambda^{n}\mathcal{O}_X^n$</code>, the latter being a perfect pairing. Composing the second with$\Lambda^nf$gives a pairing <code>$\mathcal{O}_X^n\times\Lambda^{n-1}\mathcal{O}_X^n\to \Lambda^{n}\mathcal{F}$</code>. As$\Lambda^\ast f$is multiplicative we get that the composite <code>$$\mathcal{O}_X^n\xrightarrow{f}\mathcal{F}\to \mathrm{Hom}(\Lambda^{n-1}F,\Lambda^{n}\mathcal{F})\xrightarrow{\Lambda^{n-1}f^*}\mathrm{Hom}(\Lambda^{n-1}\mathcal{O}_X^n,\Lambda^{n}\mathcal{F})$$</code> equals the map induced by the pairing for <code>$\mathcal{O}_X^n$</code>. This is an isomorphism (as <code>$\mathcal{O}_X^n$</code> is free of rank$n$and$\Lambda^nf$is an isomorphism) so we get that$f$is split injective and we may write$\mathcal{F}$as <code>\mathcal{O}_X^n\bigoplus \mathcal G$</code> for some quasi-coherent sheaf $\mathcal{G}$. Now, <code>$\Lambda^n(\mathcal{O}_X^n\bigoplus \mathcal{G})$</code> splits up as $$\bigoplus_{i+j=n}\Lambda^i\mathcal{O}_X^n\bigotimes \Lambda^j\mathcal{G}$$ and $\Lambda^nf$ is the inclusion into the $j=0$ factor. As that inclusion is an isomorphism, the other factors are zero but $\Lambda^{n-1}\mathcal{O}_X^n\bigotimes \Lambda^1\mathcal{G}$ has $\mathcal{G}$ as a direct factor and hence $\mathcal{G}=0$.</p> http://mathoverflow.net/questions/81350/simple-proof-of-that-kxg-cohen-macaualy-g-finite Comment by Torsten Ekedahl Torsten Ekedahl 2011-11-19T18:47:38Z 2011-11-19T18:47:38Z @Steve: Of course (lots of head smacking). http://mathoverflow.net/questions/81350/simple-proof-of-that-kxg-cohen-macaualy-g-finite Comment by Torsten Ekedahl Torsten Ekedahl 2011-11-19T17:26:00Z 2011-11-19T17:26:00Z It would be enough to (locally) find a regular sequence consisting of invariants as then $k[X]$ would be free over the subring generated by the sequence and so would $k[X]^G$ being a direct factor of it. It seems that for a polynomial ring the coefficients of the characteristic polynomial of a general linear polynomial might do the trick. http://mathoverflow.net/questions/80983/basic-question-about-picard-functor Comment by Torsten Ekedahl Torsten Ekedahl 2011-11-16T05:17:15Z 2011-11-16T05:17:15Z @Veen: Notice that I said <i>invertible scalar</i>, the scalars referring to the base $S$. I.e., the functor associates to $S$ the isomorphism classes of rigidified line bundles on $X\times S$. http://mathoverflow.net/questions/81035/coincidences-amongst-classifying-spaces-and-eilenberg-mac-lane-spaces Comment by Torsten Ekedahl Torsten Ekedahl 2011-11-16T05:13:14Z 2011-11-16T05:13:14Z In any case, disregarding the $K$-notation, the result is true and for the same reason the others are; $\mathbb{H}P^\infty$ is the quotient of $S^\infty$ by $Sp(1)$. The action is free and $S^\infty$ is contractible so (more or less) by definition the quotient is a $BSp(1)$. http://mathoverflow.net/questions/81037/homotopy-type-of-the-simplicial-action-groupoid/81039#81039 Comment by Torsten Ekedahl Torsten Ekedahl 2011-11-16T05:05:34Z 2011-11-16T05:05:34Z From this you also see that it is just the Borel construction (associated if you prefer to geometric realisations). http://mathoverflow.net/questions/80983/basic-question-about-picard-functor Comment by Torsten Ekedahl Torsten Ekedahl 2011-11-15T17:23:44Z 2011-11-15T17:23:44Z If you change the rigidification by an invertible scalar you get an isomorphic rigidification by letting the scalar act on $L$. http://mathoverflow.net/questions/80717/why-should-the-anabelian-geometry-conjectures-be-true/80745#80745 Comment by Torsten Ekedahl Torsten Ekedahl 2011-11-14T10:50:55Z 2011-11-14T10:50:55Z Note that in the genus $1$ case you do not recover the elliptic curve from its Tate modules. If you twist a (CM-)curve by a rank $1$ projective module over its endomorphism ring then the result will have the same Tate modules but only be isomorphic to the curve if the module is free. The same in some sense holds true in higher genus only that the twists may not be Jacobians. Hence it is crucial to look at the fundamental group and not its abelianisation. http://mathoverflow.net/questions/80803/what-does-the-tannakian-formalism-reconstruct-when-fed-the-category-of-chain-comp Comment by Torsten Ekedahl Torsten Ekedahl 2011-11-14T08:43:09Z 2011-11-14T08:43:09Z @Theo: You are absolutely right, I am glad for the qualifications I put in... I leave my comment in case someone else makes the same mistake or more precisely &quot;enom till straff, androm till varnagel&quot; (&quot;punishment for one, warning to others&quot; a phrase used in old Swedish laws). http://mathoverflow.net/questions/80803/what-does-the-tannakian-formalism-reconstruct-when-fed-the-category-of-chain-comp Comment by Torsten Ekedahl Torsten Ekedahl 2011-11-13T09:37:29Z 2011-11-13T09:37:29Z To me it seems that the $G$-representations in $\mathrm{SVect}$ consist of a graded supervector space together with a differential of superdegree $1$ and internal degree $1$. Such a graded vector spaces can be thought of as a graded vector space where the even vectors of degree $n$ correspond to new degree $2n$ and the odd vectors of degree correspond to new degree $2n+1$. In the new degree the differential is of degree $1$. This looks like the category of complexes (with the Koszul rule built into the monoidal structure as it should). http://mathoverflow.net/questions/80280/line-bundles-with-integrable-connection-on-abelian-varieties/80281#80281 Comment by Torsten Ekedahl Torsten Ekedahl 2011-11-13T05:19:28Z 2011-11-13T05:19:28Z @Veen: $\mathrm{Pic}^\tau$ is defined by the condition that the line bundle lie in the $\mathrm{Pic}^\tau$ of each fibre and the same i strue for $\mathrm{Pic}^0$ (though not by definition). http://mathoverflow.net/questions/80753/automorphisms-of-local-fields Comment by Torsten Ekedahl Torsten Ekedahl 2011-11-12T14:43:03Z 2011-11-12T14:43:03Z It seems to me that the same argument as you gives that all automorphisms for finite extensions of $\mathbb Q_p$ are continuous in the $p$-adic topology (and hence are the identity on $\mathbb Q_p$. http://mathoverflow.net/questions/80717/why-should-the-anabelian-geometry-conjectures-be-true Comment by Torsten Ekedahl Torsten Ekedahl 2011-11-12T05:18:19Z 2011-11-12T05:18:19Z I agree that that seems a tenuous argument. Any convincing argument has to be able to make a distinction between the genus $1$ and higher genus cases. http://mathoverflow.net/questions/80697/top-chern-class-in-positive-characteristic/80702#80702 Comment by Torsten Ekedahl Torsten Ekedahl 2011-11-12T05:11:17Z 2011-11-12T05:11:17Z That's right, though you could also use crystalline cohomology which could be defined as the cohomology of a complex of sheaves in the Zariski topology. http://mathoverflow.net/questions/80110/structure-of-the-variety-of-normal-matrices/80265#80265 Comment by Torsten Ekedahl Torsten Ekedahl 2011-11-08T07:58:59Z 2011-11-08T07:58:59Z Just to emphasise the reality of possible confusions, Ikramov uses irreducibility in the sense of the Zariski topology induced on the real points of a real algebra variety and not in the sense of irreducibility of the whole variety. This is quite common in real algebraic geometry. http://mathoverflow.net/questions/80280/line-bundles-with-integrable-connection-on-abelian-varieties/80281#80281 Comment by Torsten Ekedahl Torsten Ekedahl 2011-11-07T09:17:14Z 2011-11-07T09:17:14Z Line bundles numerically equivalent to the trivial line bundle.