User marc nieper-wi&#223;kirchen - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T12:51:10Z http://mathoverflow.net/feeds/user/1841 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/121792/scalar-curvature-notion-for-cartan-connections Scalar curvature notion for Cartan connections Marc Nieper-Wißkirchen 2013-02-14T12:15:28Z 2013-02-17T15:11:33Z <p>In Riemannian geometry, there is a well-known notion of the scalar curvature on a Riemannian manifold $M$, which is a function on $M$ given by a suitable contraction the Riemannian curvature tensor. The scalar curvature gives a rough measure how much the geometry of $M$ differs locally from the flat Euclidean case, namely how much the volume (or surface) of a small ball with a given radius differs from the usual value in flat space.</p> <p>I am interested in whether (or in which cases) this notion generalises to arbitrary Cartan geometries, given by a model Klein geometry $(G, H)$, an $H$-principal bundle $P$ and a one-form $\omega$ on $P$ with values in the Lie algebra $\mathfrak g$ of $G$, the Cartan connection. For what follows, you may assume that $H$ is the trivial subgroup.</p> <p>The curvature is given by the $\mathfrak g$-valued two-form $\Omega = d \omega + \frac 1 2 [\omega, \omega]$. If and only if $\Omega$ vanishes, $P$ looks locally like $G$ in that there is a local diffeomorphism between $P$ and $G$ such that $\omega$ becomes the Maurer--Cartan form of $G$. In this sense, the curvature is completely analogous to the Riemannian curvature.</p> <p>Now my question is whether one can form out of $\Omega$, $\omega$ and maybe some scalar product on $\mathfrak g$ (e.g. the Killing form) a scalar valued function (or maybe density) on $P$ which gives a rougher measure for how much the geometry differs from the Lie group $G$ (just as the scalar curvature does for Riemannian curvature)? For example, does it make sense to compare the volume of a small ball around the identity of $G$ with the "same" (in terms of the connection $\omega$) ball around a point in $P$, or is there some other well-established notion?</p> <p>Of course, I would like to see whether one gets back the scalar curvature of Riemannian geometry (maybe up to some term depending on the torsion) in the case that $G$ is the group of Euclidean motions and $H$ the subgroup of rotations.</p> http://mathoverflow.net/questions/116702/the-quillen-model-structure-on-simplicial-sets-as-a-bousfield-localization The Quillen model structure on simplicial sets as a Bousfield localization Marc Nieper-Wißkirchen 2012-12-18T13:52:36Z 2013-02-12T07:59:42Z <p>Starting with the trivial model structure on the category of simplicial sets (that is the weak equivalences are exactly the isomorphisms and the cofibrations and fibrations are arbitrary maps), is it possible to get the usual Quillen model structure on simplicial sets by performing a number of explicit left and right Bousfield localizations (e.g. by localizing along the inclusions of horns into simplices)?</p> http://mathoverflow.net/questions/85323/topos-associated-to-a-category Topos associated to a category Marc Nieper-Wißkirchen 2012-01-10T09:03:04Z 2012-01-11T00:49:45Z <p>For each topos $\mathbb E$ let $\mathcal O(\mathbb E)$ be the locally presentable category of objects in $\mathbb E$. We can make $\mathcal O$ into a contravariant functor to the category of locally presentable categories (with morphisms being cocontinuous) by assigning to each geometric morphism $(f^*, f_*)$ the functor $f^*$. By [Mac Lane, Moerdijk: Sheaves in Geometry and Logic] this functor is representable, that is there is a topos $\mathbb A$, called the <em>object classifier</em>, such that there is a natural equivalence $$\mathrm{Hom}(\mathbb E, \mathbb A) \to \mathcal O(\mathbb E).$$ Now I wonder whether $\mathcal O$ has a right adjoint, which I want to call $\operatorname{Spec}$ due to the analogy with algebraic geometry, that is whether there exists a contravariant functor $\operatorname{Spec}$ from the category of locally presentable categories to the category of topoi (with geometric morphisms) such that there is a natural equivalence $$\mathrm{Hom}(\mathbb E, \operatorname{Spec}\mathcal C) \to \mathrm{Hom}(\mathcal C, \mathcal O(\mathbb E))$$ of categories.</p> <p>(Here, <em>topos</em> shall mean <em>Grothendieck topos</em>.)</p> http://mathoverflow.net/questions/9269/category-of-categories-as-a-foundation-of-mathematics Category of categories as a foundation of mathematics Marc Nieper-Wißkirchen 2009-12-18T13:08:20Z 2011-02-15T21:07:54Z <p>In</p> <blockquote> <p>Lawvere, F. W., 1966, “The Category of Categories as a Foundation for Mathematics”, Proceedings of the Conference on Categorical Algebra, La Jolla, New York: Springer-Verlag, 1–21.</p> </blockquote> <p>Lawvere proposed an elementary theory of the category of categories which can serve as a foundation for mathematics.</p> <p>So far I have heard from several sources that there are some flaws with this theory so that it does not completely work as proposed.</p> <p>So my question is whether there is currently any (accepted) elementary theory of the category of categories that is rich enough so that one can formulate, say, the following things in the theory:</p> <ul> <li>The category of sets.</li> <li>Basic notions of category theory (functor categories, adjoints, Kan extensions, etc.).</li> <li>Other important categories (like the category of rings or the category of schemes).</li> </ul> <p>The elementary theory I am looking for should allow me to identify what should be called a category of commutative rings (at best I would like to see this category defined by a universal 2-categorical property) or how to work with this category. I am not interested in defining groups, rings, etc. as special categories as this seems to be better done in an elementary theory of sets.</p> <p>P.S.: The same question has an analogue one level higher. Assume that we have constructed an object in the category of categories (=: CAT) which can serve as a, say, category C of spaces. Classically, we can associate to each space X in C the sheaf topos over it. In the picture I have in mind, one should ask whether there is a similar elementary theory of the category of 2-categories (=: 2-CAT). Then one should be able to lift the object C from CAT to 2-CAT (as one is able to form the discrete category from a set), define an object T in 2-CAT that serves as the 2-category of toposes, and a functor C -> T in 2-CAT.</p> http://mathoverflow.net/questions/47945/a-infinity-tensor-categories A-infinity tensor categories Marc Nieper-Wißkirchen 2010-12-01T20:47:05Z 2011-01-07T03:44:15Z <p>My question is rather simple:</p> <p>What is the correct notion of a monoidal A-infinity category C?</p> <p>Or is there any reference where such a notion is explained?</p> http://mathoverflow.net/questions/49633/how-to-define-the-orientation-of-a-vector-space-over-an-arbitrary-field How to define the orientation of a vector space over an arbitrary field? Marc Nieper-Wißkirchen 2010-12-16T11:24:28Z 2010-12-17T03:50:39Z <p>I know the construction of the Hodge star operator in the context of (pseudo-)euclidean real vector spaces. Apart from the scalar product it involves a orientation of the vector space, which one has to choose (at least if one is not willing to deal with forms of odd parity). </p> <p>While I was wondering how to extend this definition to vector spaces over arbitrary fields, I realised that I even do not know what an orientation of a vector space $V$ over a general field $k$ is. In case there is such a notion: How could one use the definition in order to define a Hodge star operator?) Or is there a better definition?</p> <p>A slight extension of this question I am also interested in is the case of a finitely presented flat module over a local ring (which includes the case where the module is the germ of sections of a vector bundle over a locally ringed space at a point).</p> http://mathoverflow.net/questions/731/points-in-algebraic-geometry-why-shift-from-m-spec-to-spec/49684#49684 Answer by Marc Nieper-Wißkirchen for "Points" in algebraic geometry: Why shift from m-Spec to Spec? Marc Nieper-Wißkirchen 2010-12-16T20:59:07Z 2010-12-16T20:59:07Z <p>The reason why $\operatorname{Spec} A$ is an important notion is because it solves the following problem for a commutative ring $A$:</p> <p>Find a local ring $\mathcal O$ together with a localisation morphism $A \to \mathcal O$ such that every other localisation morphism $A \to B$ to a local ring $B$ factors as a local morphism over $A \to \mathcal O$, i.e. one looks for a kind of universal localisation of $A$.</p> <p>Stated as above, this problem has no solution at least as long one is not willing to leave the world of rings in the category of sets. However it has a solution in the following more general setting:</p> <p>There is a topos $X$ endowed with a local ring object $\mathcal O$ and a localisation morphism $A \to \mathcal O$ such that for every other topos $Y$ together with a local ring object $B$ and a localisation morphism $A \to B$ there is a pair of a geometric morphism $f\colon Y \to X$ and a morphism $f^* \mathcal O \to B$ of local rings, which is unique up to a unique natural isomorphism, such that $A \to B$ is given by the composition of $f^* \mathcal O \to B$ and $A \to f^* \mathcal O$.</p> <p>In fact, the solution to this problem is the topos $X$ of sheaves on $\operatorname{Spec} A$ together with the structure sheaf $\mathcal O_X$ as a local ring object. Now if you replace $\operatorname{Spec} A$ by the max-spectrum, the locally ringed sheaf topos you get will not solve the universal localisation problem in general.</p> <p>This means that the usual definition of $\operatorname{Spec} A$ with prime ideals is <em>a</em> correct one (as long as one is working in classical logic with the axiom of choice) but it does not mean that it is the only correct definition: You can, for example, replace $\operatorname{Spec}$ by any other topological space or, more generally, by any other site such that the sheaf topos over it is still equivalent to $X$.</p> http://mathoverflow.net/questions/13616/a-book-on-locally-ringed-spaces/49632#49632 Answer by Marc Nieper-Wißkirchen for A book on locally ringed spaces? Marc Nieper-Wißkirchen 2010-12-16T11:15:28Z 2010-12-16T14:49:15Z <p>A locally ringed space is nothing but a local ring object (in the internal sense) in a category of sheaves over a topological space, which happens to be an example of a topos.</p> <p>So in order to study general properties of locally ringed spaces, one could proceed by studying properties of local ring objects in arbitrary topoi. As any proposition (in the internal language) on a local ring (a commutative ring with $0 \neq 1$ and $s + t = 1 \implies s \in R^\times \lor t \in R^\times$) is true for any topos if and only if it can be derived intuitionistically (which is more or less the same as constructively), the original questions seems to boil down to:</p> <p>"What are the constructively valid properties and constructions for a local ring?"</p> <p>For example, the construction of Kähler differentials makes also sense constructively, which immediately implies (by the above reasoning) that every morphism $X \to Y$ of locally ringed spaces has an associated module $\Omega_{X/Y}$ of Kähler differentials.</p> <p>And there is a lot of literature on constructive algebra. The book of Mines, Richman and Ruitenburg as well as many of the preprints on Fred Richman's homepage are a start. Some material can also be found in "Sheaves in Geometry and Logic" by Mac Lane and Moerdijk.</p> http://mathoverflow.net/questions/46754/what-do-you-call-an-algebraic-element-with-the-property-that-the-generated-field What do you call an algebraic element with the property that the generated field extension is normal? Marc Nieper-Wißkirchen 2010-11-20T17:30:38Z 2010-11-22T03:14:25Z <p>Let $L/K$ be a field extension. Let $\alpha \in L$ be algebraic over $K$. Is there an established terminology for the property of $\alpha$ that $K(\alpha)/K$ is a normal field extension? Would you call $\alpha$ <em>normal over $K$</em> in case $K(\alpha)/K$ is normal?</p> http://mathoverflow.net/questions/46615/what-is-classified-by-the-big-crystalline-topos What is classified by the (big) crystalline topos? Marc Nieper-Wißkirchen 2010-11-19T11:29:01Z 2010-11-20T15:38:17Z <p>In his paper "Generic Galois Theory of Local Rings", G.C. Wraith states on p. 743 that the (big) crystalline topos "can be conveniently described in terms of the theory it describes". What exactly is the theory it describes? I.e., what type of objects in a topos E correspond to geometric morphisms from E to the big crystalline topos (over some scheme)?</p> <p>P.S.: One paragraph later, G.C. Wraith adds that he conjectures that the fppf-topos classifies algebraically closed local rings. Has this been established somewhere?</p> http://mathoverflow.net/questions/22161/what-is-the-german-translation-of-catenary-ring What is the German translation of "catenary ring"? Marc Nieper-Wißkirchen 2010-04-22T07:58:30Z 2010-04-22T10:32:51Z <p>I am looking for the correct technical term in German for the notion of <i>catenary ring</i> in commutative algebra.</p> <p>Does anyone know?</p> <blockquote> <p>For those who don't know what a catenary ring is but would like to: A Noetherian commutative ring A is called <i>catenary</i> if the following codimension formula holds for irreducible closed subsets T &sube; Y &sube; Z of Spec A:</p> <blockquote> <p>codim(T, Z) = codim(T, Y) + codim (Y, Z).</p> </blockquote> </blockquote> http://mathoverflow.net/questions/20622/construction-of-the-petit-zariski-topos-out-of-the-gros-topos-of-a-scheme Construction of the petit Zariski topos out of the gros topos of a scheme Marc Nieper-Wißkirchen 2010-04-07T11:43:52Z 2010-04-12T20:49:11Z <p>Let S be a scheme. Let (Sch/S) be a small category of schemes over S (including essentially all finitely presented schemes affine over S). Let E = (Sch/S)<sub>zar</sub> denote the gros Zariski topos with its local ring object <b>A</b><sup>1</sup>.</p> <p>Is there a nice way to construct the petit Zariski topos X = S<sub>zar</sub> out of the locally ringed topos E? (By nice I mean, for example, that there is a universal property that the locally ringed topos X possesses with respect to E.)</p> <p>There are variations of this question in which I am also interested: For example, one can replace E by the gros étale (or fppf or fpqc) topos (Sch/S)<sub>ét</sub> and ask for the construction of S<sub>zar</sub> out of (Sch/S)<sub>ét</sub>. Or one can replace X by the petit étale (or fppf or fpqc) topos S<sub>ét</sub> and ask for the construction of it out of E = (Sch/S)<sub>zar</sub>.</p> http://mathoverflow.net/questions/19719/category-of-copresheaves-over-commutative-monoids Category of copresheaves over commutative monoids Marc Nieper-Wißkirchen 2010-03-29T10:54:10Z 2010-04-07T09:37:48Z <p>Let C be a symmetric monoidal category. Let Comm(C) be the category of commutative monoids in C. Consider the topos X = CoPSh(Comm(C)) of covariant functors from Comm(C) to the category Set of sets.</p> <p>Which extra data do we have to specify on the topos X such that we can recover (up to some notion of equivalence) the essential structure of the underlying site? For example, every representable copresheaf F = Hom(A, _) has a category of modules attached to it, which is a kind of extra data. [I know I am a bit unprecise here with what I mean by essential, so making this precise could also be part of the answer to my question.]</p> <p>One could also rephrase this question as follows: Given a topos X that satisfies Giraud's axioms, one can extract a site such that X is the Grothendieck topos over this site. Which extra data do we need to impose on X such that we can recover X as a Grothendieck topos over a site that is (the dual) to commutative monoids in a symmetric monoidal category.</p> <p>When I write down this question, I have the following example in mind: Let C be the category of abelian groups. Then X is the topos of presheaves on the category of affine schemes, which gives rise to algebraic geometry. X possesses a commutative ring object, namely the affine line A<sup>1</sup> and one has the stack of categories of quasi-coherent sheaves over objects of X. Taking the idea of (Grothendieck) topoi seriously, one should be able to forget about C and just consider the topos X (i.e. without a fixed base site). Of course, one has to remember (at least) A<sup>1</sup>. This allows to recover the stack of categories of quasi-coherent modules. </p> <p>Added for clarification:</p> <p>But what if C is not the category of abelian groups? In this case, X = CoPSh(Comm(C)) also carries a stack QCoh of categories of quasi-coherent modules as follows: Let F be an object of X, i.e. F is a covariant functor from Comm(C) to Set. An object M of the category QCoh(F) maps a morphism a: Hom(A, _) -> F to an A-module M(a) in C together with natural isomorphisms M(b) = M(a) &otimes;<sub>A</sub> B for all morphisms a -> b in Comm(C). [It is here where the category C itself comes in.]</p> <p>What is the minimal amount of data we need on X so that X is equivalent to sheaves on a site S such that the dual of S is of the form Comm(C') with C' giving rise to a somewhat equivalent stack of categories of quasi-coherent modules.</p> http://mathoverflow.net/questions/6840/does-a-universal-frobenius-map-exist Does a universal Frobenius map exist? Marc Nieper-Wißkirchen 2009-11-25T20:30:12Z 2009-12-15T11:44:43Z <p>For any prime p, one has the Frobenius homomorphism F<sub>p</sub> defined on rings of characteristic p.</p> <p>Is there any kind of object, say U, with a "universal Frobenius map" F such that for any prime p and any ring R of characteristic p we can view the Frobenius F<sub>p</sub> over R as "the" base change of F from U to R?</p> <p>I have the following picture in mind: In some sense it should be possible to view the category of Z-algebras as a sheaf of categories over Spec Z such that the fibre over Spec F<sub>p</sub> is just the category of F_p-algebras. A natural transformation f of the identity functor on the category of Z-algebras should restrict to a natural transformation f<sub>p</sub> of the identity functor on the category of F<sub>p</sub>-algebras. In this naive picture one cannot expect the existence of an f such that f<sub>p</sub> is the Frobenius on F<sub>p</sub>-algebras for all primes p. But is there way to make this picture work?</p> <p>Another possible way to answer my question could be the following: Is there a classifying topos of, say, algebras with a Frobenius action? By this I mean the following: Is there a topos E with a fixed ring object R and an algebra A over it and an R-linear endomorphism f of A such that for any other topos E' with similar data R', A' there is a unique morphism of topoi E' -> E that pulls back R, A to R', A' and such that f is pulled back to the Frobenius f<sub>p</sub> of A' in case R' is of prime characteristic.</p> <p>(Feel free to modify my two pictures to make them work.)</p> http://mathoverflow.net/questions/7080/definition-of-the-symmetric-algebra-in-arbitrary-characteristic-for-graded-vector Definition of the symmetric algebra in arbitrary characteristic for graded vector spaces Marc Nieper-Wißkirchen 2009-11-28T22:11:44Z 2009-11-29T22:41:57Z <p>What is the right definition of the symmetric algebra over a graded vector space V over a field k?</p> <p>More generally: What is the right definition of the symmetric algebra over an object in a symmetric monoidal category (which is suitably (co-)complete)?</p> <p>Two possible definitions come to my mind:</p> <p>1) Take the tensor algebra over V and identify those tensors which differ only by an element of the symmetric group, i.e. take the coinvariants wrt. the symmetric group. The resulting algebra A is then the universal algebra together with a map V -> A such that the product of elements in V is commutative.</p> <p>2) Take the tensor algebra over V and divide out the ideal generated by antisymmetric two-tensors. In this case, the resulting algebra A is the universal algebra together with a map V -> A such that the product of A vanishes on all antisymmetric two-tensors (one could say that all commutators of A vanish).</p> <p>The definition 1) looks more natural and gives, for example, the polynomial ring in case V is of degree 0.</p> <p>The definition 2) applied a vector space shifted by degree 1 gives (up to degree shift) the exterior algebra over the unshifted vector space. However, in characteristic 2 for example, one doesn't get the polynomial ring if one starts with a vector space of degree 0.</p> <p>Finally, both definitions have a shortcoming in that they don't commute well with base change.</p> http://mathoverflow.net/questions/6789/why-are-flat-morphisms-flat/6818#6818 Answer by Marc Nieper-Wißkirchen for Why are flat morphisms "flat?" Marc Nieper-Wißkirchen 2009-11-25T14:53:21Z 2009-11-25T14:53:21Z <p>As others have stated above, flatness of a family should mean that the fibres of the family vary somehow continuously. Let state this in terms of a module M over a ring R. Here a fibre of M over a prime P of R is M(P), the k(P)-vector space M<sub>P</sub>/PA<sub>P</sub>, where k(P) denotes the quotient field of R/P. If the fibres vary continuously, it should be possible to extend a basis of M(P) to nearby fibres, i.e. that the lift of a k(P)-basis wrt. the canonical map M<sub>P</sub> -> M(P) should yield a basis of M<sub>P</sub> over A<sub>P</sub>, i.e. that the stalk M<sub>P</sub> is a free module.</p> <p>And in fact: If M is a finitely presented R-module it is flat if and only if M is locally free, i.e. that stalks are free.</p> <p>(And that a notion may become less geometric when we turn to non finitely presented modules is something which one may expect anyway.)</p> http://mathoverflow.net/questions/4157/universal-definition-of-tangent-spaces-for-schemes-and-manifolds/6734#6734 Answer by Marc Nieper-Wißkirchen for Universal definition of tangent spaces (for schemes and manifolds) Marc Nieper-Wißkirchen 2009-11-24T21:20:40Z 2009-11-24T21:20:40Z <p>Tangent vectors in a C^r-manifold are defined by mappings of an open interval into the manifold. We would like to do a similar thing in the algebraic context.</p> <p>Thus we want to consider maps from the line A^1 into a scheme <em>X</em>. As we are considering tangent vectors at a fixed point, we may assume that the map <em>F</em>: A^1 -> <em>X</em> maps the origin to the given point of $X$.</p> <p>Of course, this is not quite the right thing as it should be enough to map an open neighborhood of the origin in A^1 to $X$. The Zariski topology is (contrary to the analytic topology of manifolds) rather coarse, so let us consider étale open neighborhoods <em>T</em> of the origin A^1, which we map to $X$. A tangent vector to <em>X</em> is then an equivalence class of morphisms from pointed étale neighborhoods <em>T</em> of the origin of A^1 to <em>X</em> where we define the equivalence of two such morphisms as in the C^r-case, namely when (after restricting to a common étale neighborhood $U$) the two morphisms $f, g\colon U \to X$ have the property that <em>df</em> and <em>dg</em> coincide in the base point of $U$.</p> <p>This gives the right notion at least in the case of smooth schemes of finite type over a field (non-smooth schemes definitely compare badly to manifolds): Every such scheme possesses (at least Zariski-locally) an étale map to the affine space A^n_k. This allows one to show that the above definition gives the right tangent space, namely an <em>n</em>-dimensional one.</p> <p>P.S.: Trying to go the other way round by using synthetic differential geometry in order to imitate the algebraic definition in the context of differential geometry does not seem to be a solution as functions in SDG are always of class $\mathcal C^\infty$.</p> http://mathoverflow.net/questions/1684/why-is-the-exterior-algebra-so-ubiquitous/6728#6728 Answer by Marc Nieper-Wißkirchen for Why is the exterior algebra so ubiquitous? Marc Nieper-Wißkirchen 2009-11-24T20:18:03Z 2009-11-24T20:18:03Z <p>If the exterior algebra was just the symmetric algebra (up to a degree shift), it wouldn't be a useful notion by itself.</p> <p>But to me it seems it isn't: Consider an ordinary vector space <em>V</em> over a field <em>k</em> of characteristic 2. For simplicity lets assume that <em>V</em> is one-dimensional with generator $x$. As $x \wedge x = 0$, the Graßmann algebra of <em>V</em> is <em>k</em> in weight 0 and 1, and trivial in all other weights.</p> <p>Now change the parity of the vector space <em>V</em>, i.e. consider <em>x</em> to be of odd degree. Let us call the resulting odd vector space $W$. What is the symmetric algebra over $W$? Neglecting the parity, it is simply the polynomial algebra <em>k</em> in one variable, which is <em>k</em> in all positive weights. The discrepancy comes from the fact that in characteristic 2, the symmetric algebra does not see the difference between even and odd elements.</p> <p>As to the ubiquity of the exterior algebra: Given an operator <em>A</em> acting on any space, it is quite natural to ask whether $A \circ A = 0$ (e.g. any would-be differential in homological algebra). Whenever I have a vector space of operators that has this property, the exterior algebra shows up. And such a situation looks quite common.</p> <p>Furthermore, the exterior algebra enjoys a property the symmetric algebra (of the shifted space) does not have: If V is free of rank n, the natural pairing $\Lambda^p V \otimes \Lambda^{n - p} V \to \Lambda^n V$ is a perfect pairing.</p> http://mathoverflow.net/questions/5691/atiyah-macdonald-exercise-5-29-local-ring-of-a-valuation-ring Atiyah-MacDonald: exercise 5.29 - "local ring of a valuation ring" Marc Nieper-Wißkirchen 2009-11-16T11:51:58Z 2009-11-16T18:41:14Z <p>The exercise is the following:</p> <blockquote> <p>Let $A$ be a valuation ring, $K$ its field of fractions. Show that every subring of $K$ which contains $A$ is a local ring of $A$.</p> </blockquote> <p>Does anyone know what is meant by "to be a local ring of a valuation ring"?</p> http://mathoverflow.net/questions/121792/scalar-curvature-notion-for-cartan-connections/121795#121795 Comment by Marc Nieper-Wißkirchen Marc Nieper-Wißkirchen 2013-02-25T14:36:29Z 2013-02-25T14:36:29Z @Robert: Please excuse my late silence — I have caught a serious cold and had no head for maths. Will think about your new input during the next days. http://mathoverflow.net/questions/121792/scalar-curvature-notion-for-cartan-connections/121795#121795 Comment by Marc Nieper-Wißkirchen Marc Nieper-Wißkirchen 2013-02-14T14:51:54Z 2013-02-14T14:51:54Z Thanks for these thoughts so far; expressing all tensors using the natural coframe makes things much clearer. I have one question, though: In case of $H$ being non-trivial, why would I want to look at $\mathfrak h^* \otimes \Lambda^2 (\mathfrak g/\mathfrak h)$? I specifically do not want to restrict to torsion-free geometries, so I would need something like $\mathfrak g^* \otimes \Lambda^2 (\mathfrak g/\mathfrak h)$, wouldn't I? http://mathoverflow.net/questions/116702/the-quillen-model-structure-on-simplicial-sets-as-a-bousfield-localization/121566#121566 Comment by Marc Nieper-Wißkirchen Marc Nieper-Wißkirchen 2013-02-14T11:45:31Z 2013-02-14T11:45:31Z Thanks for this clear answer and the reference to the paper of Freyd, which (unfortunately) I didn't know of. http://mathoverflow.net/questions/85323/topos-associated-to-a-category/85379#85379 Comment by Marc Nieper-Wißkirchen Marc Nieper-Wißkirchen 2012-01-11T09:10:48Z 2012-01-11T09:10:48Z Thanks! That's exactly the kind of answer I hoped for! http://mathoverflow.net/questions/85323/topos-associated-to-a-category Comment by Marc Nieper-Wißkirchen Marc Nieper-Wißkirchen 2012-01-10T21:44:08Z 2012-01-10T21:44:08Z @Martin: For the moment, the analogy is just formal to me. For the topos that I write $\mathbb{A}$ Mac Lane and Moerdijk write $S[U]$ in analogy to a polynomial ring because the set of morphisms from the polynomial algebra over a ground ring to another algebra is just the set of elements of that other algebra as the category of morphisms from a topos to $S[U]$ is the category of objects of that topos. I doesn't like that notation too much as one has to turn arrows around so that one should introduce, at least formally, somewhere a $\operatorname{Spec}$. http://mathoverflow.net/questions/85323/topos-associated-to-a-category Comment by Marc Nieper-Wißkirchen Marc Nieper-Wißkirchen 2012-01-10T21:32:59Z 2012-01-10T21:32:59Z @Sergio: If you have a topos, which is a category of a certain form, and you forget about that extra information, you are left with just a category, which I call the category of objects of the topos. It happens to be a locally presentable category [Francis Borceux: Handbook of Categorical Algebra: Categories of Sheaves (proposition 3.4.16)]. http://mathoverflow.net/questions/85323/topos-associated-to-a-category Comment by Marc Nieper-Wißkirchen Marc Nieper-Wißkirchen 2012-01-10T11:53:28Z 2012-01-10T11:53:28Z @Tom: You are right; in my question I am a bit sloppy when it comes to size issues. @Andrej: Do you possibly mean cocontinuous? I will changed my question to address both comments in a manner that is hopefully helpful. http://mathoverflow.net/questions/42508/repairing-the-lie-operad-in-characterstic-2/43723#43723 Comment by Marc Nieper-Wißkirchen Marc Nieper-Wißkirchen 2011-12-09T15:11:18Z 2011-12-09T15:11:18Z Thanks for the great references! After a first look at the paper, I haven't found the statement about $\Lambda(P)$, though. (Your paper references &quot;On the Homotopy of Simplicial Algebras over an Operad&quot;, where I couldn't find the fact. Could you enlighten me? http://mathoverflow.net/questions/13616/a-book-on-locally-ringed-spaces/49632#49632 Comment by Marc Nieper-Wißkirchen Marc Nieper-Wißkirchen 2010-12-16T12:38:30Z 2010-12-16T12:38:30Z @Peter. Did so. http://mathoverflow.net/questions/49633/how-to-define-the-orientation-of-a-vector-space-over-an-arbitrary-field Comment by Marc Nieper-Wißkirchen Marc Nieper-Wißkirchen 2010-12-16T12:34:48Z 2010-12-16T12:34:48Z Thanks for your comment on $\mathrm{SL}(V)$ being the wrong group!(When I wrote this I was still thinking about the Hodge star where we have a metric and an $\mathrm{SO}(V)$.) I am correcting my question. http://mathoverflow.net/questions/46754/what-do-you-call-an-algebraic-element-with-the-property-that-the-generated-field Comment by Marc Nieper-Wißkirchen Marc Nieper-Wißkirchen 2010-11-30T07:17:46Z 2010-11-30T07:17:46Z @Laurent: I like the terminology you suggested although it is not established. Given the discussion above and the fact that no one could provide a usual terminology it seems pretty clear that the answer to my original question is No. http://mathoverflow.net/questions/46615/what-is-classified-by-the-big-crystalline-topos Comment by Marc Nieper-Wißkirchen Marc Nieper-Wißkirchen 2010-11-30T07:10:12Z 2010-11-30T07:10:12Z @Harry: Thanks for providing the link to your related question. http://mathoverflow.net/questions/46754/what-do-you-call-an-algebraic-element-with-the-property-that-the-generated-field Comment by Marc Nieper-Wißkirchen Marc Nieper-Wißkirchen 2010-11-20T21:14:56Z 2010-11-20T21:14:56Z @Martin: You can find a link to the script of the course on my webpage. During the next weeks I will put further chapters online. As to your other question: I haven't made up my mind yet how I will present Galois theory in characteristic $p$ when more than one non-separable element is adjoined. It has also something to do with that I don't have factorisation methods for non-separable polynomials over non-separable extensions in general. (This is important as I want my reasoning to be constructively valid. See also here: <a href="http://rmmc.asu.edu/TO%20DOUGLAS/RMJ/vol12/vol12-1/min.pdf" rel="nofollow">rmmc.asu.edu/TO%20DOUGLAS/RMJ/vol12/vol12-1/&hellip;</a>.) http://mathoverflow.net/questions/46754/what-do-you-call-an-algebraic-element-with-the-property-that-the-generated-field Comment by Marc Nieper-Wißkirchen Marc Nieper-Wißkirchen 2010-11-20T20:26:53Z 2010-11-20T20:26:53Z @Jim: I agree with you that Galois theory should be treated purely algebraically without constructing the field of complex numbers or proving the so-called fundamental theorem of algebra. However, it is due to didactical reasons that I make use of this theorem as it allows me to develop Galois' theory without having to talk about the abstract concept of a field. Furthermore, I need the complex numbers when talking about the various impossibility theorems of circle and ruler constructions. After the first part of the course I will mention Kronecker's construction of abstract splitting fields. http://mathoverflow.net/questions/46754/what-do-you-call-an-algebraic-element-with-the-property-that-the-generated-field Comment by Marc Nieper-Wißkirchen Marc Nieper-Wißkirchen 2010-11-20T18:31:47Z 2010-11-20T18:31:47Z @Jim: In the first half of the course, everything is inside $\mathbf C$ so there is no need for abstract field extensions. The Galois group of an algebraic number $\alpha$ over an algebraic number $\beta$ (where $\alpha$ is rational in $\beta$) is being be defined as those permutations of the roots of the minimal polynomial of $\alpha$ over $\beta$ that respect all algebraic relations over $\beta$ between the roots. So I don't have to talk about field automorphisms.