Get e-book Algebraic Groups

Free download. Book file PDF easily for everyone and every device. You can download and read online Algebraic Groups file PDF Book only if you are registered here. And also you can download or read online all Book PDF file that related with Algebraic Groups book. Happy reading Algebraic Groups Bookeveryone. Download file Free Book PDF Algebraic Groups at Complete PDF Library. This Book have some digital formats such us :paperbook, ebook, kindle, epub, fb2 and another formats. Here is The CompletePDF Book Library. It's free to register here to get Book file PDF Algebraic Groups Pocket Guide.
In mathematics, a linear algebraic group is a subgroup of the group of invertible n ×n matrices (under matrix multiplication) that is defined by polynomial.
Table of contents

A rigidity result will be explained and I explain how to generalize known classification results for spherical varieties over algebraically closed fields to spherical spaces over more general bases.

Bibliographic Information

Stefan Wewers Ulm , Thursday, Computing semistable reduction, and non-archimedian structure of covers of curves. The motivating problem for my talk is the explicit computation of semistable reduction of curves over a p -adic number field K. The formulation of this solution benefits from the language of nonarchimedian analytic geometry, in the following way.

This affinoid can be described in terms of explicit inequalities. The main step in computing the semistable reduction of Y is then achieved by explictly determining the Shilov boundary of X et. The resulting algorithm is being implemented in Sage this is joint work with Julian Rueth.

Lie algebra in physics

Massey products in the Galois cohomology of an arbitrary field are conjectured to always vanish. This conjecture is known to hold for triple Massey products. Map centered at Hotel West. To get to the hotel or the lecture room from Frankfurt Main Station you can either walk min , take the tram 16, or take the subway U4. With both tram or U4 you get off at station Bockenheimer Warte. Time table at RMV as well as information about tickets. There are plenty of places to eat nearby, e.

Moreover, on Thursday there is the weekly market at Bockenheimer Warte. Amaury Thuillier Lyon , Friday, Tropicalization of analytic spaces Let k be a non-archimedean field and let X be a closed analytic subspace of the n-dimensional unit ball B n over k, i. Torsten Wedhorn Darmstadt , Friday, Spherical Spaces Spherical varieties form an important class of varieties that includes flag varieties, toric varieties, and symmetric spaces.

Stefan Wewers Ulm , Thursday, Computing semistable reduction, and non-archimedian structure of covers of curves The motivating problem for my talk is the explicit computation of semistable reduction of curves over a p -adic number field K. Jakob Stix stix[at]math. Annette Werner werner[at]math.

Torsten Wedhorn Darmstadt. Wewers Computing semistable reduction, and non-archimedian structure of covers of curves. An important family of examples of Lie groups are the algebraic groups — algebraic varieties with a group law given by algebraic maps. Given that one can always automatically upgrade the smooth structure on a Lie group to analytic structure by using the Baker-Campbell-Hausdorff formula , it is natural to ask whether one can upgrade the structure further to an algebraic structure.

Unfortunately, this is not always the case. A prototypical example of this is given by the one-parameter subgroup.

This is a Lie group for any exponent , but if is irrational, then the curve that traces out is not an algebraic subset of as one can see by playing around with Puiseux series. This is not a true counterexample to the claim that every Lie group can be given the structure of an algebraic group, because one can give a different algebraic structure than one inherited from the ambient group.

Indeed, is clearly isomorphic to the additive group , which is of course an algebraic group. However, a modification of the above construction works:. Proposition 2 There exists a Lie group that cannot be given the structure of an algebraic group. We consider the subgroup. This is a three-dimensional metabelian Lie group, whose Lie algebra is spanned by the elements. As such, we see that if we use the basis to identify to , then adjoint representation of is the identity map.

If is an algebraic group, it is easy to see that the adjoint representation is also algebraic, and so is algebraic in. Specialising to our specific example, in which adjoint representation is the identity, we conclude that if has any algebraic structure, then it must also be an algebraic subgroup of ; but projects to the group 1 which is not algebraic, a contradiction.

A slight modification of the same argument also shows that not every Lie algebra is algebraic , in the sense that it is isomorphic to a Lie algebra of an algebraic group. However, there are important classes of Lie algebras that are automatically algebraic, such as nilpotent or semisimple Lie algebras. The main theorem in that paper asserted that given any semisimple algebraic group over an uncountable algebraically closed field , there existed a free subgroup which was strongly dense in the sense that any non-abelian subgroup of was Zariski dense in.

DSpace/Manakin Repository

This type of result is useful for establishing expansion in finite simple groups of Lie type, as we will discuss in a subsequent paper. An essentially equivalent formulation of the main result is that if are two non-commuting elements of the free group on two generators, and is a generic pair of elements in , then and are not contained in any proper closed algebraic subgroup of. In most cases, one expects that if are generically drawn from , then will also be generically drawn from , but this is not always the case, which is a key source of difficulty in the paper.

For instance, if is conjugate to in , then and must be conjugate in and so the pair lie in a proper subvariety of. It is currently an open question to determine all the pairs of words for which is not generic for generic or equivalently, the double word map is not dominant. The main strategy of proof was as follows.

It is not difficult to reduce to the case when is simple. Suppose for contradiction that we could find two non-commuting words such that were generically trapped in a proper closed algebraic subgroup. As it turns out, there are only finitely many conjugacy classes of such groups, and so one can assume that were generically trapped in a conjugate of a fixed proper closed algebraic subgroup.

Acta Mathematica

One can show that , , and are generically regular semisimple, which implies that is a maximal rank semisimple subgroup. The key step was then to find another proper semisimple subgroup of which was not a degeneration of , by which we mean that there did not exist a pair in the Zariski closure of the products of conjugates of , such that generated a Zariski-dense subgroup of. This is enough to establish the theorem, because we could use an induction hypothesis to find in and hence in such that generated a Zariski-dense subgroup of , which contradicts the hypothesis that was trapped in for generic and hence in for all.

To illustrate the concept of a degeneration, take and let be the stabiliser of a non-degenerate -space in. All other stabilisers of non-degenerate -spaces are conjugate to. However, stabilisers of degenerate -spaces are not conjugate to , but are still degenerations of. For instance, the stabiliser of a totally singular -space which is isomorphic to the affine group on , extended by is a degeneration of.

This required using the standard classification of algebraic groups via Dynkin diagrams, and knowledge of the various semisimple subgroups of these groups and their representations as we used the latter as obstructions to degeneration, for instance one can show that a reducible representation cannot degenerate to an irreducible one.

During the refereeing process for this paper, we discovered that there was precisely one family of simple algebraic groups for which this strategy did not actually work, namely the group or the group that is double-covered by this group in characteristic. This group which has Dynkin diagram , as discussed in this previous post has one maximal rank proper semisimple subgroup up to conjugacy, namely , which is the stabiliser of a line in.

To find a proper semisimple group that is not a degeneration of this group, we basically need to find a subgroup that does not stabilise any line in. In characteristic larger than three or characteristic zero , one can proceed by using the action of on the five-dimensional space of homogeneous degree four polynomials on , which preserves a non-degenerate symmetric form the four-fold tensor power of the area form on and thus embeds into ; as no polynomial is fixed by all of , we see that this copy of is not a degeneration of.

Unfortunately, in characteristics two and three, the symmetric form on degenerates, and this embedding is lost.

Group Definition (expanded) - Abstract Algebra

In the characteristic two case, one can proceed by using the characteristic fact that is isomorphic to because in characteristic two, the space of null vectors is a hyperplane, and the symmetric form becomes symplectic on this hyperplane , and thus has an additional maximal rank proper semisimple subgroup which is not conjugate to the subgroup.

But in characteristic three, it turns out that there are no further semisimple subgroups of that are not already contained in a conjugate of the. This is not a difficulty for larger groups such as or , where there are plenty of other semisimple groups to utilise; it is only this smallish group that has the misfortune of having exactly one maximal rank proper semisimple group to play with, and not enough other semisimples lying around in characteristic three.

As a consequence of this issue, our argument does not actually work in the case when the characteristic is three and the semisimple group contains a copy of or , and we have had to modify our paper to delete this case from our results. We believe that such groups still do contain strongly dense free subgroups, but this appears to be just out of reach of our current method. One thing that this experience has taught me is that algebraic groups behave somewhat pathologically in low characteristic; in particular, intuition coming from the characteristic zero case can become unreliable in characteristic two or three.

GR , paper Tags: algebraic groups , dominant maps , free groups , generic points , zariski density by Terence Tao 4 comments. This paper was originally motivated by and provides a key technical tool for another forthcoming paper of ours, on expander Cayley graphs in finite simple groups of Lie type , but also has some independent interest due to connections with other topics, such as the Banach-Tarski paradox. Recall that one of the basic facts underlying the Banach-Tarski paradox is that the rotation group contains a copy of the free non-abelian group on two generators; thus there exists such that obey no nontrivial word identities.

In particular, using Haar measure on , almost every pair has this property. In fact one has a stronger property, given any non-trivial word , the associated word map from to is a dominant map, which means that its image is Zariski -dense. More succinctly, if is generic, then is generic also. In contrast, if one were working in a solvable, nilpotent, or abelian group such as , then this property would not hold, since every subgroup of a solvable group is still solvable and thus not free and similarly for nilpotent or abelian groups.

This already goes a long way to explain why the Banach-Tarski paradox holds in three or more dimensions, but not in two or fewer. On the other hand, a famous result of Borel asserts that for any semisimple Lie group over an algebraically closed field , and any nontrivial word , the word map is dominant, thus generalising the preceding discussion for. There is also the even more famous Tits alternative , that asserts that any linear group that is not virtually solvable will contain a copy of the free group ; as pointed out to me by Michael Cowling, this already shows that generic pairs of generators will generate a free group, and with a little more effort one can even show that it generates a Zariski-dense free group.

Now suppose we take two words , and look at the double word map on a semisimple Lie group. For instance, if the two words commute, then must also commute and so the image of the double word map is not Zariski-dense. It is still not clear exactly what pairs of words have the dominance property. However, we are able to establish that all pairs of non-commuting words have a weaker property than dominance:.

enpaconcoa.tkaic geometry - Elementary reference for algebraic groups - MathOverflow

Let be non-commuting words, and let be generic elements of a semisimple Lie group over an algebraically closed field. Then generate a Zariski-dense subgroup of. To put it another way, not only contains free subgroups, but contains what we call strongly dense free subgroups: free subgroups such that any two non-commuting elements generate a Zariski-dense subgroup. Our initial motivation for this theorem is its implications for finite simple groups of Lie type. It also has implications for results of Banach-Tarski type; it shows that for any semisimple Lie group G, and for generic , one can use to create Banach-Tarski paradoxical decompositions for all homogeneous spaces of.

In particular there is one pair of that gives paradoxical decompositions for all homogeneous spaces simultaneously. Let be generic elements of , and suppose for contradiction that generically generated a group whose algebraic closure was conjugate to a proper algebraic subgroup of. If we were in the model case , then we would already be done, as there are no such maximal rank semisimple subgroups; but in the other groups, such proper maximal semisimple groups unfortunately exist.

Fortunately, they have been completely classified, and we take advantage of this classification in our argument. The degeneration argument comes in as follows. Let be a non -generic pair in. Then lies in the Zariski closure of the generic pairs, which means that lies in the Zariski closure of the set formed by and its conjugates. In particular, if the non-generic pair is such that generates a group that is dense in some proper algebraic subgroup , then is in the Zariski closure of the union of the conjugates of.

When this happens, we say that is a degeneration of. For instance, could be the stabiliser of some non-degenerate quadratic form, and could be the stabiliser of a degenerate limit of that form. The key fact we need that relies on the classification, and a certain amount of representation theory is:.