User george melvin - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T10:25:41Z http://mathoverflow.net/feeds/user/9970 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130371/semi-small-resolutions-of-peterson-varieties (semi-)Small resolutions of Peterson varieties George Melvin 2013-05-11T21:02:25Z 2013-05-11T21:02:25Z <p>Peterson varieties (in type A) can be described as the subvarieties of the full flag variety </p> <p>$$\{(F_{i})\;|\; F_{i}\subset \mathbb{C}^{n}, \; \dim F_{i} =i,\; N(F_{i})\subset F_{i+1}\}$$</p> <p>where $N$ is a regular nilpotent endomorphism of $\mathbb{C}^{n}$. Peterson introduced these types of varieties in the 90s in his work on quantum cohomology of flag varieties and Kostant showed they are singular and gave partial descriptions of their singularities; this work has been recently extended by Insko and Yong <a href="http://arxiv.org/pdf/1101.3255v2.pdf" rel="nofollow">here</a>.</p> <p>Question: do there exist (semi)small resolutions of these varieties?</p> http://mathoverflow.net/questions/129988/connectedness-of-springer-fibers/129998#129998 Answer by George Melvin for Connectedness of Springer Fibers George Melvin 2013-05-07T17:57:05Z 2013-05-07T18:03:11Z <p>Yes: this is discussed in Chriss &amp; Ginzburg 'Representation Theory and Complex Geometry', p.161 Remark 3.3.26. In short, the nilpotent cone is normal and we can apply Zariski's Main Theorem to deduce connectedness of Springer fibres. (This works for reductive $G$, and does not depend on working over $\mathbb{C}$; I'm not sure in what generality the normality condition holds though). This was originally proved by Spaltenstein via a different method, however - EDIT: see Jim Humphreys's answer for this.</p> http://mathoverflow.net/questions/129927/computing-the-grothendieck-springer-resolution-for-g-sl-2/129928#129928 Answer by George Melvin for Computing the Grothendieck-Springer resolution for $G = SL_2$ George Melvin 2013-05-07T05:43:54Z 2013-05-07T06:00:07Z <p>Hi Vinoth, here are my thoughts, hopefully they're correct and what you're after: </p> <p>You can think of $\mathfrak{\tilde{g}}$ as the set of pairs</p> <p>$$\{(X,L)\in \mathfrak{g}\times \mathbb{P}^{1}\;|\; X(L)\subset L\}$$</p> <p>(identify $\mathfrak{g}$ and $\mathfrak{g}^{\ast}$ via the Killing form). Thus, the fibre over $L\in \mathbb{P}^{1}$ is the the set of $X\in \mathfrak{sl}_{2}$ that have $L$ as an 'eigenline'. Since the projection to $G/B$ is $G$-equivariant it suffices to consider a particular line <code>$L_{0}$</code>; take <code>$L_{0}=span\{e_{1}\}$</code>, to see that the fibre is the standard Borel $\mathfrak{b}$, so that the fibre over <code>$g\cdot L_{0}$</code> is $g\cdot\mathfrak{b}$.</p> <p>EDIT: providing the answer in terms of the dual $\mathfrak{g}^{\ast}$, as the question states, we see that the fibre is the span of the dual basis (via the Killing form) of the standard basis $\{F,H\}$ of the lower triangular Borel.</p> http://mathoverflow.net/questions/129451/generalised-coinvariant-algebras 'Generalised' coinvariant algebras George Melvin 2013-05-02T19:47:39Z 2013-05-04T14:27:33Z <p>Let $\mathfrak{g}$ be a simple complex Lie algebra, and $\mathfrak{h}\subset\mathfrak{g}$ a Cartan subalgebra with Weyl group $W$. Consider the fibre product $\mathfrak{h}\times_{\mathfrak{g}} N$, where $N\subset \mathfrak{g}$ is the nilpotent cone. It is a zero-dimensional affine $\mathbb{C}$-scheme with coordinate ring $S/(S_{+}^{W})$, where $S=\mathbb{C}[\mathfrak{h}^{\ast}]$ is the coordinate ring of $\mathfrak{h}^{\ast}$ and $(S_{+}^{W})$ is the ideal generated by the $W$-invariants in $\mathbb{C}[\mathfrak{h}^{\ast}]$ with no constant term: the ring $S/(S_{+}^{W})$ is known as the <em>algebra of coinvariants</em>; it is a finite dimensional $W$-module. In fact, it is isomorphic (as a graded $W$-module) to the cohomology of the flag variety.</p> <p>Now, let $\lambda$ be a dominant weight lying in the root lattice. Then, the corresponding finite dimensional $\mathfrak{g}$-module $L(\lambda)$ admits a zero weight space $L(\lambda)[0]$; moreover, since $W$ permutes weight spaces we have that this zero weight space is $W$-invariant, hence a $W$-module. Now set $S=\text{Sym}(L(\lambda)[0])$, the symmetric algebra of $L(\lambda)[0]$ and define $S_{+}^{W}$ to be the set of $W$-invariants without constant term. Then, $R_{\lambda} \stackrel{def}{=}S/(S_{+}^{W})$ is a finite dimensional $W$-module (since this is the coordinate ring of the scheme theoretic fibre of the image of $0$ in the quotient $L(\lambda)[0]^{\ast}//W$). Hence, to each dominant root lattice element $\lambda$ of $\mathfrak{g}$ we have associated a finite dimensional $W$-module $R_{\lambda}$.</p> <p><strong>Question(s)</strong>: What is known about these $W$-modules $R_{\lambda}$? Do they arise 'naturally' anywhere? (eg, if $\lambda$ is the highest root of $\mathfrak{g}$ then $R_{\lambda}$ is the cohomology of the flag variety).</p> <p>Any references or pointers to what's 'really' going on would be much appreciated.</p> <p>I should also point out that in type $A_{n}$ it is known that when </p> <p>$$\lambda=(\mu_{1}-\mu_{2})\omega_{1}+ (\mu_{2}-\mu_{3})\omega_{2}+\ldots + \mu_{n}\omega_{n}$$</p> <p>and $\mu_{1}+\ldots + \mu_{n}=n+1$ then $L(\lambda)[0]$ is the simple $S_{n+1}$-module $V^{\mu'}$, where $\mu'$ is the dual partition of $\mu: \mu_{1}\geq \mu_{2} \geq \ldots \mu_{n}\geq 0$.</p> http://mathoverflow.net/questions/128678/motivating-geometric-representation-theory/128766#128766 Answer by George Melvin for motivating geometric representation theory George Melvin 2013-04-25T21:13:57Z 2013-04-25T21:42:53Z <p>I'm struggling to see what the actual question is but here is an example of a non-trivial use of geometry to prove a simple statement in representation theory; moreover, it is the only known way to obtain the result (apologies if this is not what you're after): the $n!$ conjecture states that the dimension of a certain bigraded $S_{n}$-module is $n!$ (in fact, something stronger is true, the bigraded module is the left regular representation). This statement is equivalent to a certain morphism being Gorenstein and Cohen-Macaulay, namely the morphism $\rho: X_{n}\to H_{n}$, where $H_n$ is the Hilbert scheme of n points in $\mathbb{C}^2$ and $X_n$ is the isospectral Hilbert scheme. Mark Haiman gave a proof of the geometric statement in 2000 (math.AG/0010246).</p> <p>The bigraded module (call it $D_{\mu}$) considered here is the span of partial derivatives (with respect to $x$'s and $y$'s) of a bihomogeneous polynomial $\Delta_{\mu}(x_1,\ldots,x_n;y_1,\ldots,y_n)$, where $\mu$ is a partition of $n$. Furthermore, the 'bigraded multiplicity' of the simple $S_n$-module $V^{\lambda}$ in $D_{\mu}$ gives the coefficients of the Macdonald-Kostka polynomials, thereby proving their positivity (Macdonald's conjecture). </p> <p>Again, I'm not sure this is an example of 'geometric representation theory' as most people see it, but it's a nice example of using geometry to solve a 'classical' representation theoretic problem. (I suppose my struggle to see the question is more to do with understanding what comes under 'geometric representation theory')</p> http://mathoverflow.net/questions/127409/springer-action-on-centre-of-parabolic-category-o-after-brundan Springer Action on Centre of Parabolic Category O (after Brundan) George Melvin 2013-04-12T19:54:33Z 2013-04-13T19:16:46Z <p>I recently learned of a result of Brundan describing the centre of the regular block of parabolic category $\mathcal{O}$ for $\mathfrak{gl}_{n}$ as the cohomology of a corresponding Springer fibre (available <a href="http://darkwing.uoregon.edu/~brundan/papers/springernew.pdf" rel="nofollow">here</a>); this is an isomorphism at the level of $\mathbb{C}$-algebras. However, we also have an action of the symmetric group on these cohomology algebras.</p> <p><em>Question</em>: Is there a way to define an $S_{n}$-action on the centre of (parabolic) category $\mathcal{O}$ for <code>$\mathfrak{gl}_{n}$</code>, and in such a way that Brundan's isomorphism becomes a morphism of graded $S_{n}$-modules? </p> <p>Of course, I can simply transport the $S_{n}$-action in the obvious way but I was looking for a more intrinsic action on the centre that arises from (parabolic) category $\mathcal{O}$ itself. Also, I've had a look at Brundan's paper but it doesn't appear to be discussed there (if I'm being blind then please let me know!)</p> <p>Thanks in advance.</p> http://mathoverflow.net/questions/127475/endomorphisms-in-category-o-and-schubert-classes Endomorphisms in Category O and Schubert Classes George Melvin 2013-04-13T16:41:55Z 2013-04-13T19:04:56Z <p>Let $\mathfrak{g}\supset \mathfrak{b}\supset \mathfrak{h}$ be a complex semisimple Lie algebra, with choice of Borel and Cartan subalgebras, $W$ the Weyl group. </p> <p>W. Soergel's 'Endomorphismensatz' allows for the identification of <code>$End_{\mathcal{O}_{0}}(P(w_{0}))$</code> with the algebra of coinvariants <code>$\mathbb{C}[\mathfrak{h}^{\ast}]/\mathbb{C}[\mathfrak{h}^{\ast}]^{W}_{+}$</code>, a finite dimensional quotient of a polynomial algebra, equipped with a $W$-action. Here $\mathcal{O}_{0}$ is the block of the BGG category $\mathcal{O}$ corresponding to the trivial central character.</p> <p>Moreover, it is a classical result (due to Borel?) that we can identify $\mathbb{C}[\mathfrak{h}^{\ast}]/\mathbb{C}[\mathfrak{h}^{\ast}]^{W}_{+}$ with the cohomology algebra of the flag variety of $\mathfrak{g}$, and that there is a basis of this cohomology algebra given by the <em>Schubert basis</em> <code>$\lbrace S_{w}\rbrace_{w\in W}$</code>, where $S_{w}$ is the class of the corresponding Schubert cell.</p> <p><em>Question(s)</em>: 1) does anyone know to which morphisms in category $\mathcal{O}$ the Schubert classes correspond to under the above identifications? </p> <ul> <li><p>If yes; is there a 'nice' intrinsic (in terms of category $\mathcal{O}$) description of these morphisms that would give a 'canonical' description of the Schubert classes?</p></li> <li><p>(rubbish, vague question) if no; would this be an interesting/valuable thing to know? (ie, are there any immediate applications?)</p></li> </ul> <p>2) Is there a way to see the $W$-action on the endomorphism algebra in category $\mathcal{O}$?</p> <p>Also, replace 'Schubert class' by 'first Chern class of tautological bundles' in the above questions; is anything known in this case?</p> <p>If this is standard material then my apologies; any references/directions would be much appreciated. In particular, any references for Soergel's work (in English/French) would be particularly appreciated.</p> <p>Cheers, George</p> http://mathoverflow.net/questions/127478/flag-varieties-via-quiver-varieties Flag Varieties via Quiver Varieties George Melvin 2013-04-13T17:22:44Z 2013-04-13T17:35:52Z <p>In type $A$ it is possible to realise the flag variety $\mathcal{F}$ of $\text{SL}_{n}(\mathbb{C})$ via Nakajima's quiver varieties: consider the vectors $v=(1,2,\ldots, n-1), w=(0,\ldots,0,n)$. Then, we can identify the Nakajima quiver variety $M(v,w) = T^{\ast}\mathcal{F}$, the cotangent bundle of $\mathcal{F}$. Hence, we can realise the flag variety as the zero section.</p> <p><em>Question</em>: Is it possible to realise the flag varieties of other symmetric Kac-Moody algebras via quiver varieties? To make things easier, let's restrict to finite type; and, if needs be, even further, to $D_{4}$.</p> <p>I'm aware that there is not a reasonable description of most (all?) Nakajima quiver varieties outside of type $A$, and was wondering whether this is just type $A$ phenomena (similar in spirit, perhaps, to Ginzburg's Lagrangian construction of representations of $U(\mathfrak{sl}_{n})$, which doesn't work outside of type $A$ for explicit reasons).</p> <p>Cheers, George</p> http://mathoverflow.net/questions/107621/status-of-a-conjectural-definition-of-h-nakajima Status of a conjectural definition of H. Nakajima George Melvin 2012-09-19T21:06:44Z 2013-03-19T12:53:45Z <p>In his paper '$t$-analogue of $q$-characters of finite dimensional representations of quantum affine algebras' - <a href="http://arxiv.org/abs/math/0009231" rel="nofollow">http://arxiv.org/abs/math/0009231</a> - H. Nakajima states a conjectural definition of the $t$-analogue of the $q$-character of a standard module $M_{P}$ (with weight $P$) of a quantum affine algebra at level $0$ - this appears as Conjecture 3.1.1 on p. 5 of the arXived preprint above. In summary, Nakajima conjectures that the $q,t$'-character of a standard module $M_{p}$ can be determined using certain filtrations on individual weight spaces.</p> <p>I was hoping that someone can let me know of the status of this conjecture - is it true that we can describe the '$q,t$'-character as conjectured?</p> <p>I imagine that this has been resolved since the paper is a bit more mature now. If this is the case, can someone point me in the direction of a resolution? </p> <p>If this conjecture has not been resolved, does anyone know of any progress towards its resolution/any problems that have arisen in resolving this conjecture?</p> <p>Thanks in advance.</p> http://mathoverflow.net/questions/123009/simply-connected-simple-algebraic-groups/123011#123011 Answer by George Melvin for Simply connected simple algebraic groups George Melvin 2013-02-26T18:41:06Z 2013-02-26T18:55:43Z <p>The almost simple algebraic groups $G$ are classified by their <em>root datum</em> (defined <a href="http://en.wikipedia.org/wiki/Root_datum" rel="nofollow">here</a>): in short, this is a quadruple $(X,X^{\ast},R,R^{\ast})$ with $X,X^{\ast}$ dual finitely generated free abelian groups and $R\subset X, R^{\ast}\subset X^{\ast}$ finite subsets (satisfying some conditions). Think of $X$ as the character group of some maximal torus in $G$ and $X$ a collection of <em>roots</em> (in particular, $R$ will define a root system once we tensor $X$ with $\mathbb{R}$). Then, $R^{\ast}\subset X^{\ast}$ are the corresponding <em>co</em>-objects (cocharacters, coroots). This datum characterises almost simple algebraic groups in the following sense: the <em>type of $G$</em> is determined by the root system determined by $R\subset X$ (type ABCDEFG, as you've mentioned) and the extra information supplied by knowing the co-objects determines the weight lattice of this root system. The <em>simply connected</em> groups are those groups for which the weight lattice of the root system of $G$ is equal to $X$; this is the same as those groups of each type with the largest' (finite) centre. The <em>simple</em> groups are those groups for which $\mathbb{Z}R = X$, that is, when the root lattice equals the character lattice. Hence, the simple simply connected groups are those $G$ whose weight lattice is equal to its root lattice.</p> <p>Some nice, introductory notes are available <a href="http://pages.uoregon.edu/brundan/math681fall07/notes9.pdf" rel="nofollow">here</a>. Also, see Chevalley's Collected Works <a href="http://books.google.com/books/about/Classification_des_Groupes_Alg%25C3%25A9briques.html?id=LAXvAAAAMAAJ" rel="nofollow">here</a>.</p> <p>I suppose this does not completely answer your question (I haven't given a list of the simply connected simple groups) but hopefully this will give you an idea of the general framework.</p> <p>EDIT: I appear to have been too slow! As I was off looking for the required list @Jay Taylor has given a complete answer. </p> http://mathoverflow.net/questions/122843/reductive-groups-and-hilbert-schemes-reference "Reductive Groups and Hilbert Schemes" - Reference George Melvin 2013-02-24T23:29:56Z 2013-02-24T23:29:56Z <p>Bezrukavnikov and Ginzburg have unpublished notes, 'Hilbert Schemes and Reductive Groups' (referenced <a href="http://arxiv.org/abs/math/9907151" rel="nofollow">here</a>, for example): does anyone know what became of these notes? Did Bezrukavnikov-Ginzburg publish anything based on these notes? I am interested in finding them as it is claimed (in the reference linked to above) that Bezrukavnikov-Ginzburg obtain an isomorphism between equivariant K-theory of $X^{n}$ ($X$ a quasi-projective smooth surface) and cohomology of $X^{[n]}$ (Hilbert scheme of $n$ points on $X$); this should(?) allow an interpretation of the Nakajima-Grojnowski construction of the Heisenberg algebra via equivariant K-theory.</p> <p>Thanks in advance.</p> http://mathoverflow.net/questions/121996/applications-for-intersection-cohomology-and-for-the-decomposition-theorem-for/122036#122036 Answer by George Melvin for Applications for intersection (co)homology and for the Decomposition Theorem for students? George Melvin 2013-02-17T01:46:22Z 2013-02-17T01:55:17Z <p>A nice example is given by Borho-Macpherson's construction of Weyl group representations (<em>Representations des groupes de Weyl et homologie d'intersection pour les varietes nilpotentes</em>, Comptes rendus. Acad. Sci. Paris, p.707–710 (1981)).</p> <p>Here's a short description: consider the Springer resolution $p:\tilde{N}\rightarrow N$, where $N\subset \mathfrak{g}$ is the nilpotent cone of some reductive finite dimensional Lie algebra $\mathfrak{g}$ (over $\mathbb{C}$, say). Then, $Rp_{\ast}\mathbb{C}$ carries an action of the Weyl group $W$ of $\mathfrak{g}$, providing a representation of $\mathbb{C}[W]\rightarrow End(Rp_{\ast}\mathbb{C})$; this was an idea of Lusztig, I think. Borho-Macpherson show that this is an isomorphism and the Decomposition Theorem furnishes an identification between irreducible representations of $W$ and irreducible summands of $Rp_{\ast}\mathbb{C}$. For the case <code>$\mathfrak{g}=\mathfrak{gl}_{n}$</code>, this provides a natural bijection between irreducible representations of $S_{n}$ and partitions of $n$; to an irreducible representation $V$ the corresponding summand of $Rp_{\ast}\mathbb{C}$ is supported on a (closure of?) nilpotent orbit associated to some partition.</p> <p>For <code>$\mathfrak{g}=\mathfrak{gl}_{n}$</code>, you can state Borho-Macpherson's Theorem and give a good idea of the proof with nothing more than some formal properties of IC (fibre squares, for example), the notion of a covering space and some linear algebra. For <code>$\mathfrak{gl}_{n}$</code>, with $n$ small, you can compute examples using some linear algebra; you can also show that the representation corresponding to the subregular nilpotent orbit is the standard irreducible representation $V\subset \mathbb{C}^{n}$, where $\mathbb{C}^{n}$ is the permutation representation (the Springer fibre is a union of $n-1$ projective lines, whose intersection configuration is given by the $A_{n-1}$ Dynkin diagram). </p> http://mathoverflow.net/questions/117415/old-books-still-used/118501#118501 Answer by George Melvin for Old books still used George Melvin 2013-01-10T05:01:03Z 2013-01-10T05:01:03Z <p>My first thought was Atiyah &amp; Macdonald's 'An Introduction to Commutative Algebra' - which has already been mentioned - and 'anything by J.P. Serre' (that's old enough, of course!). It appears that not quite everything in this latter category has been mentioned; notably, 'Algebres de Lie Semi-simple Complexes', first published in 1966. There is also a later English translation, 'Complex Semisimple Lie Algebras' published in 1987. </p> <p>While not quite an introduction, I find myself referring back to this text often for its streamlined, beautiful exposition (a hallmark of Serre). It also has the best exposition of root systems I've encountered. </p> <p>Furthermore, another classic text on semisimple Lie algebras (J. Humphreys - 'Introduction to Lie Algebras &amp; Representation Theory') is a 'fleshing out' of Serre's notes. Actually, Humphreys's textbook was first published in 1972 so might squeeze onto this list too?</p> http://mathoverflow.net/questions/114919/non-symmetric-quiver-varieties Non-Symmetric Quiver Varieties George Melvin 2012-11-29T19:52:13Z 2012-12-08T08:37:17Z <p>Given a symmetric Cartan datum $(I,\cdot)$, H. Nakajima has defined a family of varieties - known as quiver varieties - and has used them to give geometric constructions of the representation theory of the corresponding Kac-Moody algebra associated to $(I,\cdot)$ using (Borel-Moore) homology and the convolution product (discussed, for example, in 'Representation Theory and Complex Geometry', by Chriss &amp; Ginzburg). Furthermore, Nakajima has given constructions of representations of quantum affine algebras using the equivariant K-theory of said varieties leading him to prove several results on the $q$-characters of representations of these quantum algebras. These results are excellently expounded in Nakajima's original papers (which involve the term 'quiver variety' in their title).</p> <p>For a non-symmetric Cartan datum (eg, the Cartan datum associated to non-simply laced finite dimensional Lie algebras of type $B,C,F,G$) the approach of Nakajima does not extend in a simple manner to provide non-symmetric quiver varieties (to me, doubling the quiver seems to destroy any information we have on root lengths, but I don't know if this is the correct way to think about this). </p> <p><strong>Question</strong>: Has there been any progress on defining/constructing quiver varieties in the non-symmetric case? Can anyone provide a reference to this material or know of anyone working on this? Or, is there a reason why such a construction cannot be realised? I am interested in this as I am wondering if there are any connections with any 'Langlands'-y things (eg, Frenkel-Hernandez's papers on Langlands duality).</p> <p>F. Xu and A. Savage have have managed to realise the crystal structure of representations in the non-simply laced case using the notion of admissable quiver automorphisms. Their approach is based on an observation by Lusztig (section 14, 'Introduction to Quantum Groups') that every Cartan datum in the affine/finite-type case can be constructed as a 'quotient' of a type A,D,E quiver with an admissable automorphism. To the ADE quiver we <strong>can</strong> define a quiver variety and they show that one can realise a crystal basis using irreducible components of quiver varieties that are invariant under the admissable automorphism. Savage explicitly constructs the crystal structure using the ADE crystal structure. Both authors mention that it seems plausible that there should be a geometric construction of this crystal structure (ie, a geometric construction of the representations of the corresponding Kac-Moody algebras). However, after a cursory search of the literature (ie, typing 'quiver varieties non symmetric' into Google), this seems to be most of what is known. </p> <p>Furthermore, Nakajima (<a href="http://www.kurims.kyoto-u.ac.jp/~nakajima/TeX/kika03.pdf" rel="nofollow">here</a>, Problem 2.1) also states a desire to give a construction of quiver varieties in the non-symmetric case so it seems hopeful that such a construction should/might exist.</p> http://mathoverflow.net/questions/115231/expository-papers-related-to-quantum-groups/115233#115233 Answer by George Melvin for expository papers related to quantum groups George Melvin 2012-12-03T02:42:40Z 2012-12-03T02:42:40Z <p><a href="http://www.mate.uncor.edu/~ggarcia/encuentros/notas-curso-qg-ha.pdf" rel="nofollow">These</a> are a nice set of introductory notes that I like discussing the example of quantum $SL_{2},\mathfrak{sl}_{2}$.</p> <p>Also, Kashiwara's original papers on the 'crystals' and 'crystal bases' in quantised universal enveloping algebras are very readable and discuss the relationship between the representation theory of these objects and Kac-Moody algebras. Essentially, for generic $q$ the representation theories are the same. Jantzen's AMS book goes into further detail on this story and has a whole chapter devoted to several examples highlighting some of the main features.</p> http://mathoverflow.net/questions/111837/what-does-multiplying-a-matrix-by-its-transpose-have-to-do-with-spectral-theorem/111841#111841 Answer by George Melvin for What does multiplying a matrix by its transpose have to do with spectral theorem? George Melvin 2012-11-08T18:16:55Z 2012-11-08T18:16:55Z <p>In the finite dimensional case the Spectral Theorem says that we can decompose a self-adjoint operator into a sum of projection operators: if $A$ is self-adjoint then we can write</p> <p>$A=\lambda_{1}P_{1}+\cdots +\lambda_{r}P_{r}$</p> <p>where the $\lambda$'s are the (necessarily real) eigenvalues of $A$ and the $P$'s are orthogonal projection onto the corresponding eigenspaces. If we choose an orthonormal basis of each eigenspace and let $U_{i}$ be the matrix whose columns are this eigenbasis, then we have $P_{i}=U_{i}U_{i}^{\ast}$. </p> <p>As mentioned in the answer by Robert Israel, given any $A$ we always have $AA^{\ast}$ is self-adjoint so the discussion above holds. In particular, we can determine an orthonormal basis for which the operator determined by $A$ is a diagonal matrix real entries on the diagonal (ie, $A$ is (orthogonally) diagonalisable). </p> <p>This material can be found in any good linear algebra textbook (my personal favourite is by Hoffmann &amp; Kunze). </p> http://mathoverflow.net/questions/111232/triviality-of-associated-bundles Triviality of Associated Bundles George Melvin 2012-11-02T00:47:56Z 2012-11-02T09:49:12Z <p>Let $P\rightarrow M$ be a principal (right) $G$-bundle, where $G$ is a Lie group. Given a finite-dimensional representation of $G$, $V$ say, we can define the associated bundle $P\times_{G}V\rightarrow M$. This is a vector bundle over $M$ defined as the quotient of the (free, right) action of $G$ on $P\times V$ - $(p,v)\cdot g =(p\cdot g, g^{-1}v)$.</p> <p>Hence, for a given representation $V$ of $G$ we can associate to a principal $G$-bundle $P\rightarrow M$ a vector bundle $P\times_{G} V\rightarrow M$. Moreover, this assignment is functorial and so induces a map from isomorphism classes of principal $G$-bundles to $K_{0}(M)$, the Grothendieck group of vector bundles on $M$. Call this functor (and, by abuse of notation, the map it induces) $\theta_{V}$. Furthermore, it seems (there may be problems here?) that we obtain a functor </p> <p>$\theta: Rep_{G}\rightarrow Fun(Prin_{G}(M),Vec(M))$</p> <p>where the left hand side is the category of (finite dimensional) representations of $G$ and the right hand side is the category of functors from $Prin_{G}(M)$ to $Vec(M)$, the categories of principal $G$-bundles on $M$ and vector bundles on $M$ (respectively).</p> <p>Question 1: Which representations induce the trivial map on iso-classes? For example, the trivial representation $T$ will always give </p> <p>$\theta_{T}(P\rightarrow M)=M\times T$</p> <p>since we can choose linearly independent generating sections of $P\times_{G} T$ using triviality of $T$. My question is, are there other representations of $G$ which afford this property?</p> <p>Question 2: What am I <em>really</em> discussing here? Is there a name for $\theta$? Do these ideas arise in some 'deeper' (or more natural) framework?</p> <p>Question 3: Is this formulation useful? Are there any interesting results related to this construction? </p> <p>I have come to these conclusions as a result of thinking about associated bundles based on knowing the basic definition only and any references/comments would be appreciated. My apologies if this is standard material to topologists, or well-known to experts - I am neither.</p> http://mathoverflow.net/questions/110471/classification-of-platonic-solids/110473#110473 Answer by George Melvin for Classification of Platonic solids George Melvin 2012-10-23T20:54:13Z 2012-10-23T20:54:13Z <p>I believe Hermann Weyl's classic book 'Symmetry' discusses this question but I can't recall what his approach is (ie, whether he makes use of the Euler formula). </p> http://mathoverflow.net/questions/109310/a-line-bundle-that-does-not-admit-a-g-linearisation A line bundle that does not admit a G-linearisation George Melvin 2012-10-10T16:43:11Z 2012-10-10T19:35:46Z <p>I have been thinking about quotients lately and pondered the following:</p> <p>Let $G$ be a connected linear algebraic group and $X$ a $G$-variety acting via the morphism $\sigma:G\times X\rightarrow X$. Let $p:L\rightarrow X$ be a line bundle on $X$. </p> <p>A $G$-linearisation of $L$ is an action of $G$ on $L$ such that $p(g\cdot l)= g\cdot p(l)$, for $l\in L, g\in G$, and which restricts to a linear isomorphism <code>$L_{x}\rightarrow L_{g\cdot x}$</code> on the fibres. This last condition can be expressed as saying that there is an isomorphism $L\rightarrow g^{\ast}L$, for each $g\in G$ (here $g^{\ast}L$ is the pullback bundle by the automorphism $g$ of $X$). In fact, since $G$ is connected, a $G$-linearisation of $L$ exists if and only if there is an isomorphism <code>$p_{2}^{\ast}L\rightarrow \sigma^{\ast}L$</code> of bundles on $G\times X$, with $p_{2}$ the projection to $X$.</p> <p>It is known (Corollary 7.2, p.109, 'Lectures on Invariant Theory' - Dolgachev) that if $X$ is normal then for any $L$ there is some power of $L$ that admits a $G$-linearisation.</p> <p>Question 1: Can someone provide an example of a non-normal $G$-variety $X$ and a line bundle $L$ for which no power $L^{n}$ admits a $G$-linearisation? . </p> <p>Question 1': If no such example can exist can someone point me towards the literature (if any) where this question is addressed?</p> <p>The existence result for normal $X$ relies on that fact that there is an exact sequence</p> <p>$0\rightarrow K \rightarrow Pic^{G}(X)\rightarrow Pic(X) \rightarrow Pic(G)$</p> <p>and that $Pic(G)$ is finite. Here $K$ is the group of rational characters of $G$ and $Pic^{G}(X)$ is the group of line bundles admitting a $G$-linearisation (or line $G$-bundles in Dolgachev's terminology).</p> <p>Question 2: Can we extend the exact sequence</p> <p>$0\rightarrow K \rightarrow Pic^{G}(X) \rightarrow Pic(X)$</p> <p>to the right for arbitrary $X$ and in a 'canonical' manner? (ie, is this exact sequence the tail of a canonical long exact sequence for any $G$-variety X?) </p> <p>Question 2': If so, what groups appear? Do they have any 'down-to-earth' interpretations? (eg, we have $Pic(G)$ appearing for normal $X$).</p> <p>Thanks in advance and apologies if this is standard material in GIT - I only have a copy of Dolgachev's notes at hand and these questions are not addressed.</p> http://mathoverflow.net/questions/108820/springer-resolution-over-wedge3-mathbbc6/108887#108887 Answer by George Melvin for springer resolution over $\wedge^3 \mathbb{C}^6$ George Melvin 2012-10-05T03:26:00Z 2012-10-05T03:49:12Z <p>The dimension of the nilpotent orbits of $\mathfrak{gl}_{6}$ can be described using the corresponding partition <code>$\pi: d_{1}+d_{2}+\ldots + d_{k} =6$</code> associated to a nilpotent orbit - so $\pi$ is the partition corresponding to the Jordan representative of the nilpotent orbit - as follows: </p> <p>consider the 'dual partition of $\pi$', <code>$\pi': e_{1}+\ldots +e_{l}=6$</code>. If we consider the Young diagram of $\pi$ then $\pi'$ is the partition of 6 corresponding to the transpose Young diagram. Then, the dimension of a nilpotent orbit is <code>$6^{2} - \sum_{i=1}^{l}e_{i}^{2}$</code>. A quick check of the 11 partitions of 6 shows that there can't exist nilpotent orbits of dimension 9,14 or 19 (furthermore, orbits are always even dimensional). Perhaps you could give more information as to why these orbits given in the question are expected to be nilpotent orbits? </p> <p>There are two orbits of dimension 18 (corresponding to the partitions $3+1+1+1$ and $2+2+2$, with dual partitions $4+1+1$ and $3+3$, respectively). In this case, more information as required.</p> http://mathoverflow.net/questions/106481/a-remark-in-jantzens-lectures-on-quantum-groups A remark in Jantzen's 'Lectures on Quantum Groups' George Melvin 2012-09-06T06:32:30Z 2012-09-06T15:10:37Z <p>In Jantzen's AMS text 'Lectures on Quantum Groups' he makes the following remark (p.187, preface to Chapter 9):</p> <p>"For general (complex semisimple f.d. Lie algebra) $\frak{g}$ we can consider for each simple root $\alpha$... a Lie subalgebra (of $\frak{g}$) isomorphic to $\frak{sl}_{2}$ (ie, the Lie subalgebra <code>$\frak{s}_{\alpha}$</code> generated by suitable <code>$X_{\alpha}\in \frak{g}_{\alpha}$</code>, <code>$Y_{\alpha}\in\frak{g}_{-\alpha}$</code>). So, if $M$ is a f.d. $\frak{g}$-module, then one can find (for fixed $\alpha$) a basis <code>$v_{1},\ldots,v_{n}$</code> such that <code>$Y_{\alpha}v_{i}$</code> is either $0$ or a nonzero multiple of another <code>$v_{j}$</code> and such that also each <code>$X_{\alpha}v_{h}$</code> is either $0$ or a nonzero multiple of another <code>$v_{l}$</code>. However, in general, there does not exist a basis that works simultaneously for all simple $\alpha$. (There are exceptions, such as the adjoint representations...)..."</p> <p>The first part of this statement is standard (we are applying complete reducibility of $M$ as an <code>$\frak{s}_{\alpha}$</code>-module). However, I hope that someone can illuminate the last sentence on 'exceptions' - is this a typo/mis-statement? Or am I missing something here? </p> <p>It is not possible to obtain a simultaneous basis for the adjoint representation of $\frak{sl}_{3}$: indeed, if <code>$\alpha_{1},\alpha_{2}$</code> are the simple roots and <code>$\frak{s}_{1},\frak{s}_{2}$</code> the corresponding <code>$\frak{sl}_{2}$</code>-triples then we can decompose <code>$\frak{sl}_{3}$</code> as </p> <p>$\frak{sl}_{3}$$\cong L(1)\oplus L(2)\oplus L(0)\oplus L(1)</p> <p>when we consider <code>\frak{sl}_{3}</code> as either a <code>\frak{s}_{1}-</code> or <code>\frak{s}_{2}-</code>module. Here, L(n) is the irreducible \frak{sl}_{2}-module of dimension n+1. Also, in both decompositions we have L(0) appears as a subspace of \frak{h} (the 0-weight space).</p> <p>If we were to have a simultaneous basis as described above we would need a basis vector u\in\frak{h}\subset\frak{sl}_{3} corresponding to the copy of L(0) appearing in the <code>\frak{s}_{1}</code>- and <code>\frak{s}_{2}</code>-decompositions of <code>\frak{sl}_{3}</code>. This would imply that <code>\ker ad \;X_{\alpha_{1}}\cap \ker ad \; X_{\alpha_{2}}\cap \frak{h}</code> is nonzero, which is impossible (as can be seen by a basic calculation) since we are in characteristic 0.</p> <p>Thanks in advance for your comments.</p> http://mathoverflow.net/questions/103522/nilpotent-lie-algebras/104875#104875 Answer by George Melvin for Nilpotent Lie Algebras George Melvin 2012-08-16T22:26:12Z 2012-08-16T22:26:12Z <p>Whenever ad_{\xi} is an endomorphism of \mathfrak{g} whose corresponding partition \pi: 1^{s_{1}}2^{s_{2}} \cdots \; of \dim \mathfrak{g} is such that s_{1} =0, then we have im\; ad_{\xi} \cap Z(\mathfrak{g}) \neq \{0\}. </p> <p>As \mathfrak{g} is nilpotent, ad_{\xi} acts as a nilpotent endomomorphism of \mathfrak{g}, for every \xi \in \mathfrak{g}, and we can associate to ad_{\xi} a partition \pi(\xi) of n=\dim \mathfrak{g} as follows: the Jordan form of ad_{\xi} is </p> <p>$$J=\begin{bmatrix} J_{1}\\ &amp; \ddots \\ &amp;&amp; J_{r}\end{bmatrix}$$</p> <p>where J_{i} is an n_{i}\times n_{i} 0-Jordan block and n_{1}\geq \ldots \geq n_{r} >0. Then, n_{1} +\ldots +n_{r}=n is a partition of n, which we will denote \pi(\xi): 1^{s_{1}}2^{s_{2}} \cdots , so that$$1s_{1}+2s_{2}+\ldots =n$$is the partition of n determined by J.</p> <p>Let \xi \in \mathfrak{g} and let \pi(\xi) : 1^{s_{1}}2^{s_{2}} \cdots be the partition associated to ad_{\xi} and suppose that (x_{1},\ldots,x_{n}) is a Jordan basis of ad_{\xi}, so that the matrix of ad_{\xi} is in Jordan form. Thus, x_{1},x_{n_{1}+1},\ldots, x_{n_{r-1}+1} is a basis of$$Z_{\mathfrak{g}}(\xi)=\ker ad_{\xi}.$$</p> <p>As \mathfrak{g} is nilpotent we must have Z(\mathfrak{g}) \neq \{0\}. Let w \in Z(\mathfrak{g}) be nonzero. Then, w \in Z_{\mathfrak{g}}(\xi) and so w= c_{1}x_{1}+\ldots + c_{r} x_{n_{r-1}+1}.</p> <p>Assume that s_{1}=0 in the partition \pi (so that \pi contains no parts equal to 1). Thus, x_{2},\ldots, x_{n_{r-1}+2} \notin Z_{\mathfrak{g}}(\xi) and x=c_{1}x_{2}+\ldots +c_{r}x_{n_{r-1}+2} is such that [\xi,x] =w\in Z(\mathfrak{g}).</p> <p>While this condition is sufficient it is not necessary: for example, consider the nilpotent Lie algebra \mathfrak{n}_{3} of strictly upper triangular 3\times 3 complex matrices. Then,</p> <p>$$\xi = \begin{bmatrix} 0&amp;1&amp;0\\0&amp;0&amp;1\\0&amp;0&amp;0\end{bmatrix}$$</p> <p>is such that$im \; ad_{\xi}=Z(\mathfrak{n}_{3})$, while$\pi(\xi): 12$.</p> http://mathoverflow.net/questions/103751/about-the-intrinsic-definition-of-the-weyl-group-of-complex-semisimple-lie-algebr/103753#103753 Answer by George Melvin for About the intrinsic definition of the Weyl group of complex semisimple Lie algebras George Melvin 2012-08-02T02:52:50Z 2012-08-02T02:52:50Z <p>Yes: this is the approach to defining the 'abstract Weyl group' introduced in "Representation Theory and Complex Geometry" by Chriss/Ginzburg on p. 135 (2nd Edition, Birkhauser).</p> http://mathoverflow.net/questions/129988/connectedness-of-springer-fibers/129997#129997 Comment by George Melvin George Melvin 2013-05-07T17:58:42Z 2013-05-07T17:58:42Z It appears I was too slow with my answer by 32 seconds...! http://mathoverflow.net/questions/129451/generalised-coinvariant-algebras/129466#129466 Comment by George Melvin George Melvin 2013-05-03T00:51:23Z 2013-05-03T00:51:23Z Thanks for the clarification and references, Prof. Humphreys. I had discovered Reeder's work (and the more recent work of Achar-Henderson-Riche) and it seems quite surprising that progress on this problem is recent; after all, it seems like a reasonably straightforward question to consider. http://mathoverflow.net/questions/129451/generalised-coinvariant-algebras Comment by George Melvin George Melvin 2013-05-03T00:48:36Z 2013-05-03T00:48:36Z Yes, of course. Thanks for pointing that out! Cheers, George http://mathoverflow.net/questions/127475/endomorphisms-in-category-o-and-schubert-classes/127486#127486 Comment by George Melvin George Melvin 2013-04-13T19:41:16Z 2013-04-13T19:41:16Z Excellent, thanks for your comments. I'll have a think about this. I take it that it will suffice to have a look in Humphreys's 'Category O' for the tilting module material? http://mathoverflow.net/questions/127409/springer-action-on-centre-of-parabolic-category-o-after-brundan/127488#127488 Comment by George Melvin George Melvin 2013-04-13T19:38:28Z 2013-04-13T19:38:28Z Thanks for the reference, Ben. It's very much appreciated. http://mathoverflow.net/questions/127475/endomorphisms-in-category-o-and-schubert-classes Comment by George Melvin George Melvin 2013-04-13T18:13:03Z 2013-04-13T18:13:03Z Hi Jos\'{e}! Thanks for the help. I may be around Edinburgh in the summer, if so then I'll drop by and say 'hello'. Hope you're well. Cheers, George http://mathoverflow.net/questions/127475/endomorphisms-in-category-o-and-schubert-classes Comment by George Melvin George Melvin 2013-04-13T17:06:16Z 2013-04-13T17:06:16Z Great, thanks for pointing this out. http://mathoverflow.net/questions/127409/springer-action-on-centre-of-parabolic-category-o-after-brundan Comment by George Melvin George Melvin 2013-04-13T04:44:49Z 2013-04-13T04:44:49Z Great, thanks for your comment. I'll have a think about this and see where I get. http://mathoverflow.net/questions/107621/status-of-a-conjectural-definition-of-h-nakajima/124956#124956 Comment by George Melvin George Melvin 2013-03-19T19:26:03Z 2013-03-19T19:26:03Z Thanks for your response and reference; I will take a look! Cheers, George http://mathoverflow.net/questions/123429/is-fx-tanx-a-diffeomorphism-why-or-why-not/123432#123432 Comment by George Melvin George Melvin 2013-03-02T18:31:32Z 2013-03-02T18:31:32Z @Todd: Yes, I was moderately aware of this agreement but think my morning coffee spurred me towards a 'answer before think of the community' response. Noted for the future and thanks for the clarification. http://mathoverflow.net/questions/123135/modern-developments-in-finite-dimensional-linear-algebra/123151#123151 Comment by George Melvin George Melvin 2013-02-28T03:53:28Z 2013-02-28T03:53:28Z Subspace problems of the form 'classify all ways to embed n subspaces in a vector space' can be studied using quivers. The four subspace problem is studied in a nice paper of Gelfand and Ponomarev 'Problems of linear algebra and classification of quadruples of subspaces...'. http://mathoverflow.net/questions/123152/exceptional-schur-weyl-duality Comment by George Melvin George Melvin 2013-02-27T22:29:31Z 2013-02-27T22:29:31Z ... Here's the link <a href="http://www.math.ethz.ch/~khorosh/teaching/sym_functions/list.html" rel="nofollow">math.ethz.ch/~khorosh/teaching/sym_functions/&hellip;</a> http://mathoverflow.net/questions/123152/exceptional-schur-weyl-duality Comment by George Melvin George Melvin 2013-02-27T22:27:38Z 2013-02-27T22:27:38Z Schur-Weyl duality can be obtained via the more general notion of Howe duality of reductive pairs (in this case, GLn and GLm). It might be useful to look at R. Howe's 'Schur Lectures' where this framework is discussed; these notes should be available at a good library and there is also a copy floating around online somewhere. http://mathoverflow.net/questions/122973/homogenous-polynomials-as-sum-or-differences-of-squares-and-symmetric-polynomials Comment by George Melvin George Melvin 2013-02-26T22:26:50Z 2013-02-26T22:26:50Z At the risk of stating the even more obvious, the degree of$Q^{2}$is twice the degree of$Q\$... http://mathoverflow.net/questions/122843/reductive-groups-and-hilbert-schemes-reference Comment by George Melvin George Melvin 2013-02-25T02:33:07Z 2013-02-25T02:33:07Z Thanks for the reference.