Advertisement

Sign up for our daily newsletter

Advertisement

Verma module highest weight capacity – Subscribe to RSS

We proved before, using a PBW basis for , that is the direct sum of its finite-dimensional weight spaces—in particular, acts semisimply, which is not a priori obvious since is finite-dimensional—and so is any subrepresentation. Name required.

David Stewart
Friday, November 17, 2017
Advertisement
  • Often it is convenient to let y 1. To deal with both of these issues we have to formalize better the relationship here between andthen bring the Harish-Chandra ho- momorphism into the picture.

  • Connect and share knowledge within a single location that is structured and easy to search. There is an important category, the BGG categorydefined as follows: if is a representation of on which acts locally nilpotently i.

  • Prove that M [] is a U g -submodule of M and that M is the direct sum of nitely many! This follows from Weyls Theorem: more precisely, elements of h act on M via semisimple matrices while elements of n or n.

  • We have a -module where the nilpotent subalgebra acts by zero and acts by. You are commenting using your Facebook account.

  • This property is analogous to Schurs Lemma, even though M need not be nite dimensional.

Your Answer

Gelfand, Ed. There is an important category, the BGG categorydefined as follows: if is a representation of on which acts locally nilpotently i. Corollary 4 For eachthere is a unique simple -representation which is a highest weight module with weight.

Viewed times. Email Required, but never shown. Like this: Like Loading We can actually take the sum of all proper submodules of ; the above argument shows that this sum does not contain and has no vectors with nonzero -component.

  • Recall from 1.

  • Hot Network Questions.

  • In turn, the simple Lie algebras are uniquely determined by their irreducible root systems; these are classied explicitly as types A. Since O is artinian and noetherian, a standard theorem associated in various contexts with the names Azumaya, Krull, Remak, Schmidt is valid here: each M O can be written as a direct sum of indecomposable mod- ules, with the summands being unique up to isomorphism and order.

  • The previous discussion motivates the following construction of Verma module. From Wikipedia, the free encyclopedia.

  • Formal Characters of Modules in O In order to make eective use of Verma modules, we need to introduce the language of formal characters in the innite dimensional setting.

The proof of the theorem below requires a formal verma module highest weight capacity in U gwhich is valid more generally in arbitrary veram algebras over elds of char- acteristic 0. Now V is stable under U hwhile the action of U n. Denote by A 0 the additive subgroup of A generated by all chM. There does not seem to be a more elementary proof. Dimension Formula 41 2.

The object is denoted ; it is called a Verma module. Highest weight capacity be a highest weight module, generated by a highest weight vector. Whichever construction of the Verma module is used, one has to prove that it is nontrivial, i. The previous discussion motivates the following construction of Verma module.

Uploaded by

These are precisely the weights mmodule with positive multiplicity. Weight capacity Systems The occurrence of root systems in the Lie algebra setting has led to a some- what more widely applicable notion of abstract root system in a vector space over 1. In rank 2 cases, W is a dihedral group and the ordering is easy to describe: w.

Name required. Gelfand, Ed. The object is denoted ; it is called a Verma module. Namespaces Article Talk.

ALSO READ: Truncal Obesity And Hypothyroidism

Actually, it turns out that such vectors form a basis for the Verma module. Accept all cookies Customize settings. You are commenting using your Facebook account. Hot Network Questions. So neither doeswhich means that. Recall that a vector for a representation of not necessarily finite-dimensional!

Note that it has highest weight capacity highest weight vector: the image ofand the weight is. You are commenting using your Twitter account. There are two standard constructions of the Verma module, both of which involve the concept of universal enveloping algebra. Weight spaces of Verma modules Ask Question. Connect and share knowledge within a single location that is structured and easy to search.

Archived Entry

But it is not so elementary to describe this maximal submodule explicitly. Parabolic Versions of Category O 9. It is easily checked that weight vectors for distinct weights in M are linearly independent. The action of Z g on an arbitrary M O is usually complicated.

Show that the homomorphism is independent of the choice of a simple system in. Then U g is the di- rect sum of subspaces U g. Thanks capadity A above, this is itself the Weyl group of the root system 1 in the subspace of E it spans consisting of all orthogonal to 1. For a module with a standard ltration, we can characterize the resulting ltration multiplicities in a useful way needed in 3. Category O: Methods s. In the setting of A, this translates into adding up the values of ch. Eventually we will show in 4.

Therefore, we have. The full classification of Verma module homomorphisms was done by Bernstein—Gelfand—Gelfand [9] and Verma [10] and can be summed up in the following statement:. Leave a Reply Cancel reply Enter your comment here Download as PDF Printable version.

Navigation menu

Create a free Team What is Teams? In other word, there exist an element w of the Weyl group W such that. Each homomorphism of Verma modules is injective and the dimension. Help Learn to edit Community portal Recent changes Upload file. Sign up to join this community.

The best answers are voted up and rise to the top. Notify me of new comments via email. Let bei. Question feed. Active 6 years, 11 months ago. Download as PDF Printable version. Post as a guest Name.

To manipulate such formal sums rigorously, we reformulate the denition as follows. Universal Enveloping Algebras An essential tool in the construction and study of representations is the universal enveloping algebra U a of a Lie algebra a. Evidently a is equivalent to bwhile c implies b. Writing chM w as pe w and multiplying both sides of the equation by q, Lemma 2. What are its weights?

Welcome to Scribd!

Weights of the Exterior Powers 6. A moments thought shows that this lies in A. Though its derivation is unrelated to category O methods, the formula itself is an essential byproduct of Weyls theory.

We can explain the idea of a Verma module as follows. So, define a covariant functor sending to the collection of highest weight vectors in with weight. The rest of the proposition is now clear. Oftenin fact.

  • If N is the submodule they generate, then certainly all v w j with v L lie in N. The Weyl chambers are in natural bijection with simple systems in and are permuted simply transitively by W.

  • You are commenting using your WordPress.

  • This is called a Verma module and may also be written as Ind g b C. In this preliminary chapter we summarize results with which the reader should be familiar, coupled with some explicit references to the textbook literature.

  • Let be the Borel subalgebra.

Active 6 years, 11 months ago. It capaclty clear that cannot be isomorphic to foras simple modules can have at most one highest weight vector up to scalar multiplication. Proposition 1 is indecomposable and has a unique maximal submodule and unique simple quotient. To say that is a highest weight vector with weight is to say that is annihilated by the left ideal in generated by and for. You are commenting using your Twitter account. Verma modules Now, I claim that there is such a thing as a universal highest weight module.

Share this: Twitter Facebook. Question feed. I don't quite understand the higher-dimensional analogue. There are two standard constructions of the Verma module, both of which involve the concept of universal enveloping algebra.

Highest weight vectors and Verma modules

To make this precise, dene a subspace of M for each xed by M. Application: Category Equivalences 7. We already saw Theorem 1. Characters in Terms of Ext.

There are two standard constructions of the Verma module, both of which involve the concept of universal enveloping algebra. Each homomorphism of Verma modules is injective and the dimension. Accept all cookies Customize settings. Question feed. Featured on Meta.

Then proceed by modulle on k. This gen- eralizes the usual notion of Jordan decomposition, which arises intrinsi- cally in g, independent of any specic linear realization. They can be characterized as the connected components in the complement of the union of all hyperplanes orthogonal to roots. We present a simplied version of his proof when the ground eld is C; here we may write h.

Your Answer

Formal Characters Hom and Ext A basic tool in the study of various objects in O such as Verma modules is the Hom functor. The multiplicity of in M is then dened to be dimM.

  • Schubert Varieties and KL Polynomials 8.

  • Proposition 1 is indecomposable and has a unique maximal submodule and unique simple quotient.

  • Observe next that each L.

  • Indecomposable Tilting Modules

  • Notify me of new posts via email.

Given a nite dimensional! This in turn implies that P is a projective cover of M. As a consequence of the theorem, each M O possesses a composition series with simple quotients isomorphic to various L. Now the claim follows from the sl 2, C theory in 0. The most basic general tool available for the study of the derived functors Ext n of Hom in a module category is the long exact sequence associated to a short exact sequence.

ALSO READ: Subclinical Hypothyroidism Treatments

You are commenting using your WordPress. Hidden categories: Wikipedia articles incorporating text from PlanetMath Pages that use a deprecated format of the ada body mass index tags. Download as PDF Printable version. But this means they are isomorphic by the first proposition. We can actually take the sum of all proper submodules of ; the above argument shows that this sum does not contain and has no vectors with nonzero -component.

The restriction of a simple module to a semisimple or reductive subalgebra of g is a natural problem leading to branching ruleswhich can often be studied together with the tensor product moodule composition problem. This is especially useful for applications in classical physics or chemistry, but can be carried out eectively mainly in low ranks or other special cases such as symmetric or exterior powers of a natural representation. So it is essential to write Ext O when working in this category. There are deep connections with representations of symmetric groups, classi- cal invariant theory, and the like, going back to Schur and Weyl. Featured on Meta.

I aim to discuss criteria for this in the sequel. Question feed. The best answers are voted up and rise to the top. Then there are homomorphisms which are both surjective, since the images contain respectively. Fill in your details below or click an icon to log in:.

  • Thus M.

  • Suppose have highest weight vectors with the same weight and are simple.

  • For a simple systemthere is a natural fundamental domain C E for the action of W. Characters of Finite Dimensional Modules this special class of modules.

  • So if the coset of a monomial is killed by y k ithe identity in the lemma shows that the coset of the longer monomial y i 0 y i 1. This theme will be developed further in Chapter 4, but for our immediate purposes it is 1.

  • So Remark 3. While this approach leaves the individual weight multiplicities concealed, it has conceptual elegance and also leads to an easily computable formula for dimL.

  • In other word, there exist an element w of the Weyl group W such that.

Anyone who consults the literature will encounter a wide array of notational choices; here Vetma have tried to keep things simple and consistent to the extent possible. The deni- tion uses the nondegeneracy of the Killing form on g. Existence of Embeddings: Preliminaries 78 4. Harish-Chandras Theorem 27 the basic ideas, referring in the Notes below to several thorough discussions in the literature. Sign up using Facebook. It is instructive to look at the rank 1 case, though even here the full picture requires development of some further theory.

We have a -module where the nilpotent subalgebra acts by zero and acts by. Sign up using Email and Password. In other words, we see that is the object in question, which proves the proposition. There is an important category, the BGG categorydefined as follows: if is a representation of on which acts locally nilpotently i. The Verma module then is the freest possible highest weight module, in a sense; note that it belongs to the category. Verma modules are weight modulesi. Email required Address never made public.

Denote by W 1 the isotropy group of 1 in W. In this direction we introduce some helpful terminology. Weght of Shapovalovs Theorem 88 4. Part d is clear, while part e follows from the fact that M lies in O forcing its submodules to lie in O along with part b. In well-behaved cases, the character of a represen- tation determines it uniquely up to equivalence.

To manipulate such formal sums rigorously, we reformulate the denition as follows. Category O contains some veerma objects: a Suppose h. Taking b. The rst goal is to nd for each h. If two simple modules M 1 and M 2 can be extended nontrivially, i. Applying Theorem 3. The Ext term vanishes by Theorem 3.

ALSO READ: Hypothyroidism References In Apa

The full classification of Verma module homomorphisms was done by Bernstein—Gelfand—Gelfand [9] and Verma [10] and can be summed up in the following statement:. In other wrods, fix a weight. Blog at WordPress. Indeed, let be any proper submodules; we will prove. You are commenting using your Google account.

Here is an epimorphism and is essential, meaning that no proper submod- ule of the projective module P M is mapped onto M. Twisted Versions of Standard Filtrations Extend Exercise 1. On one hand, O p requires a cumbersome extra layer of notation; on the other hand, the results there are sometimes easier to derive once the corresponding results for O are in hand.

Endomorphisms and Socles of Projectives The set of all these is in natural bijection with the set Max Z g of maximal ideals. Writing chM w as pe w and multiplying both sides of the equation by q, Lemma 2.

This translates into the desired conclusion as follows. Here ranges over the weights of M, each occurring dimM. Weight capacity we can apply to W 1 and 0 the earlier argument in the real case. As observed in Exercise 1. Representations Our main concern here is with representations of g or U gnot necessarily nite dimensional.

Extend this relation to a partial ordering of W by dening w. Maps Between Parabolic Verma Modules 9. While the formal characters of Verma modules are quite transparent, it is far from clear how to determine chL for arbitrary h. Here L has 1- dimensional weight spaces, with weights2.

The subject can be approached from a number of angles, including the traditional theory of Lie groups and the theory of linear algebraic groups; but group theory generally remains in the background here. Any h. In particular, each indecomposable module belongs to a single block. As noted in the proof of Theorem 3. Harish-Chandra Modules

Active 6 years, 11 months ago. Asked 8 years, 1 month ago. It only takes a minute to sign up. So, define a covariant functor sending to the collection of highest weight vectors in with weight.

Thus M. For 0. Moreover, in each case the structure of partially ordered set is the same, with a unique maximal dominant and a unique minimal antidominant element. Here ranges over the weights of M, each occurring dimM. Let h. Now if E, its isotropy group in W is itself the Weyl group of the root system spanned by all orthogonal to.

  • The substantial reference list includes all source material cited, together with related books and survey articles.

  • We have a -module where the nilpotent subalgebra acts by zero and acts by. Name required.

  • Using the assumption abouttake any pre-image z Z g of g under. Projective Functors and Principal Series viii Contents

  • In turn, the integral weight lattice is identied with Z and r with 2Z.

  • I aim to discuss criteria for this in the sequel. Therefore, we have.

If the weights are further integralthen there exists a nonzero cpaacity. A similar resolution exists for generalized Verma modules as well. So neither doeswhich means that. It is in fact a theorem that any object in has a filtration whose quotients are highest weight modules. Then has the vectorwhich is a highest weight vector with the same weight.

Now, Highesh claim that there is such a thing as a universal verma module highest weight capacity weight module. Leave a Reply Cancel reply Enter your comment here Post as a guest Name. Gelfand, Ed. Stack Overflow for Teams — Collaborate and share knowledge with a private group. One way to see this, for instance, is that the module defined by the tensor product above satisfies the universal product: for anywe have.

Instead a less direct way must be developed. Highest Weight Categories In this case, the maximal submodule of M is isomorphic to L 2.

If happens verma module highest weight capacity be a maximal weight of M, the pre-image of a maximal vector moule weight in M will be a maximal vector in M; then the universal 52 3. Each M O is artinian as well as noetherian. It is easy to derive a few further consequences of the axioms. Hecke Algebras and KazhdanLusztig Polynomials 8.

Shuing Functors Twisting and Completion Functors Enumerate the simple roots as 1. Existence of Embeddings: Preliminaries 78 4. Harish-Chandras Theorem 26 1. Then: a M is spanned by the vectors y i 1 1.

Twisting and Completion Functors It is easily checked that weight vectors for distinct weights in M are linearly independent. There is also a list of frequently used symbols, most of which are introduced early in the book. Universal Construction 70 Contents v Notes 71 Chapter 4.

Verma modules are weight modulesi. This follows easily from the Harish-Chandra theorem on infinitesimal central characters. Stack Overflow for Teams — Collaborate and share knowledge with a private group. We know from the section about homomorphisms of Verma modules that there exists a homomorphism.

You are commenting using your Google account. Proposition 1 is indecomposable and has a unique maximal submodule and unique simple quotient. From Wikipedia, the free encyclopedia. Blog at WordPress. The previous discussion motivates the following construction of Verma module. Weight spaces of Verma modules Ask Question.

Application: Category Equivalence But in hignest it turns out to be a deep problem to determine all coecients occurring for arbitrary even when. Moreover, the set of weights of M is visibly. O3 M is locally n-nite: for each v M, the subspace U n v of M is nite dimensional. In place of the basis element e of Z we now have a characteristic function e.

I don't quite understand the higher-dimensional analogue. Recall that a vector for a representation of not necessarily finite-dimensional! Categories : Representation theory of Lie algebras. Then there are homomorphisms which are both surjective, since the images contain respectively.

ALSO READ: Hypothyroidism Ultrasound Radiology

If either containsthen it is all of. Active 6 highest weight, 11 months ago. Verma modules Now, I claim that there is such a thing capacitg a universal highest weight module. Sign up using Email and Password. It is in fact a theorem that any object in has a filtration whose quotients are highest weight modules. The full classification of Verma module homomorphisms was done by Bernstein—Gelfand—Gelfand [9] and Verma [10] and can be summed up in the following statement:. The first equality follows from the definition of "module".

Each homomorphism of Verma modules is injective and the dimension. Verma modulesnamed after Midule Vermaare objects in the representation theory of Lie algebrasa branch of mathematics. So, define a covariant functor sending to the collection of highest weight vectors in with weight. We now apply this construction in the setting of a semisimple Lie algebra.

Active Oldest Votes. We can actually take the sum of all proper submodules of ; the above argument shows that this sum does not contain and has no vectors with nonzero -component. Asked 8 years, 1 month ago. Sign up to join this community.

Question feed. I aim to discuss criteria for this in the sequel. Accept all cookies Customize settings. In general, is infinite-dimensional.

One way to see this, for instance, is that the module defined by the tensor product highest weight satisfies the universal product: for anywe have. Asked 8 years, 1 month ago. Sign up or log in Sign up using Google. Hot Network Questions. Let be a highest weight module, generated by a highest weight vector. We can actually take the sum of all proper submodules of ; the above argument shows that this sum does not contain and has no vectors with nonzero -component. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields.

ALSO READ: Dyshormonogenesis Congenital Hypothyroidism Icd

We highest weight capacity from the section cpacity homomorphisms of Verma modules that there exists a homomorphism. Categories : Representation theory of Lie algebras. The best answers are voted up and rise to the top. Proposition 2 Any simple highest weight modules of the same weight are isomorphic. Notify me of new posts via email. Stack Overflow for Teams — Collaborate and share knowledge with a private group. The rest of the proposition is now clear.

So the maximal vector of this weight in M constructed in the proposition must lie in the maximal submodule and map to 0 in the quotient L. Observe next that each L. This is easily computed. To focus the problem better when h.

We can explain the weigh of a Verma module as follows. Help Learn to edit Community portal Recent changes Upload file. The rest of the proposition is now clear. Actually, it turns out that such vectors form a basis for the Verma module. Anyway, are then both simple quotients of. Each homomorphism of Verma modules is injective and the dimension. Question feed.

Email Required, but never shown. The highest weight space is one-dimensional. You are commenting using your Facebook account. Namespaces Article Talk.

There is also a list of frequently used symbols, most of which are introduced early in the book. Two natural questions arise at this point: 1 Is verma module highest weight capacity Harish-Chandra homomorphism injective? Here are two useful facts [, His ideas have left a lasting imprint on the study of category O. This follows from Weyls Theorem: more precisely, elements of h act on M via semisimple matrices while elements of n or n. But in rank the formulas involve variables, corresponding to coordinates of a weight relative to fundamental weights. In particular, it is generated by the reections s.

  • This module is U n.

  • It is denoted shortly as the BGG resolution.

  • Now W is generated by the simple reections 0. Viewed 1k times.

  • Views Read Edit View history.

  • In turn, the integral weight lattice is identied with Z and r with 2Z. Corrections and suggestions from readers are welcome.

  • Leave a Reply Cancel reply Enter your comment here

Fusion Rules Some examples are easy to nd. I have also made some search but i have not found this result explicitly mentioned somewhere in the literature. Contravariant Forms 68 3.

Add a comment. From Wikipedia, the free encyclopedia. Namespaces Article Talk. Name required. Verma modules are weight modulesi. Anyway, are then both simple quotients of. Now, I claim that there is such a thing as a universal highest weight module.

In some treatments of the basic structure theory such as []Weyls Theorem and its consequences play a key role from an early stage. Quiver Attached to a Block Example: sl 2, C 66 3.

Tilting Verma module highest weight capacity Since notation varies considerably in the literature, the reader needs to be aware of our conventions which are often closest to those in [] ; these are intended to steer something of a middle course among the available choices. The Functions p and q 39 Below we shall apply the convolution operation in A to longer products, writing just. Higher Ext Groups 6. Application: Category Equivalences 7.

Capactiy I am actually going to talk about them in a bit more detail. Create a free Team What is Teams? I don't quite understand the higher-dimensional analogue. Therefore, we have. Now if is a highest weight vector with weight and a homomorphism of -representations, then is a highest weight vector with weight too.

The first equality follows from the definition of "module". Notify me of new comments via email. We have a -module where the nilpotent subalgebra acts by zero and acts by.

  • In well-behaved cases, the character of a represen- tation determines it uniquely up to equivalence.

  • Mathematics Stack Exchange works best with JavaScript enabled. So neither doeswhich means that.

  • Abstractly, W is a nite Coxeter group, having generators s. From now on we can use the antidominant weight in such an intersection to parametrize the corresponding subcategory of O.

  • The full classification of Verma module homomorphisms was done by Bernstein—Gelfand—Gelfand [9] and Verma [10] and can be summed up in the following statement:.

Mathematics Stack Exchange works best with JavaScript enabled. Email required Address never made public. Verma modulesnamed after Daya-Nand Vermaare objects in the representation theory of Lie algebrasa branch of mathematics. Viewed times. If the weights are further integralthen there exists a nonzero homomorphism. In other word, there exist an element w of the Weyl group W such that. The first equality follows from the definition of "module".

ALSO READ: Congenital Hypothyroidism And Anxiety

There is an important category, the BGG category midule, defined as follows: if is a representation of on which acts locally nilpotently i. To say that is a highest weight vector with weight is to say that is annihilated by the left ideal in generated by and for. Let be a highest weight module, generated by a highest weight vector. You are commenting using your Twitter account. I should probably say another clearly equivalent way the Verma module can be constructed.

In general, is infinite-dimensional. Note that it has a highest weight vector: the image ofand the weight is. Add a comment. Download as PDF Printable version. Indeed, let be any proper submodules; we will prove.

  • The chapters in Part II can to a large extent be read independently.

  • Weight spaces of Verma modules Ask Question. Fill in your details below or click an icon to log in:.

  • Related

  • Maps Between Parabolic Verma Modules 9. For each h.

Categories : Representation theory of Lie algebras. Sign up or log in Sign up using Google. Views Read Edit View history. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields.

Since is dominant integral, by assumption, all weights of Verma module highest weight capacity are nodule W-conjugates of dominant integral weights. Filtrations and Rigidity 9. Tilting Modules Category O: Methods twisted action of h vanishes at all v M. Given a nite dimensional! Somewhat miraculously, this kind of resolution of L does exist. John Pardon John Pardon 17k 3 3 gold badges 42 42 silver badges bronze badges.

Now, I claim that there is such verma module highest weight capacity thing as a universal highest weight module. Corollary 4 For eachthere is a unique simple -representation which is a highest weight module with weight. Suppose have highest weight vectors with the same weight and are simple. The first equality follows from the definition of "module".

Categories : Representation theory of Lie algebras. Therefore, we have. Bill Cook Bill Cook Help Learn to edit Community portal Recent changes Upload file. Asked 8 years, 1 month ago. Viewed times.

It only takes a minute to sign up. Stack Overflow for Teams — Collaborate and share knowledge with a private group. The Verma module then is the freest possible highest weight module, in a sense; note that it belongs to the category. It is denoted shortly as the BGG resolution. In other wrods, fix a weight. There is an important category, the BGG categorydefined as follows: if is a representation of on which acts locally nilpotently i.

Thus there exists a nonzero homomorphism M M whose image lies in the maximal submodule N. The problem is that for a nonsimple positive root with. Complements Contents ix This is called a Verma module and may also be written as Ind g b C.

See the texts in roughly chrono- logical order : Humphreys [, 23], Dixmier [84, 7. Since every nonzero module in Capaity has at least one maximal vector, we con- clude from this discussion and Theorem 1. Shapovalov Elements 86 4. After discarding these summands, the induction hypothesis takes over for the remaining summand. The Multiplicity Problem for Verma Modules 8.

Observe next that each L. Example: sl 3, C highedt. Universal Enveloping Algebras 6 0. Instead of trying to prove injectivity and surjectivity directly, it is better to make a comparison with a similar but somewhat more transparent map. The book ends with an introduction to the inuential work of Beilinson, Ginzburg, and Soergel on Koszul duality.

Let bei. Namespaces Article Talk. Note that it has a highest weight vector: the image ofand the weight is. Help Learn to edit Community portal Recent changes Upload file.

Sidebar1?
Sidebar2?