User mts - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T11:27:48Z http://mathoverflow.net/feeds/user/703 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/129037/criterion-for-nilradical-of-a-maximal-parabolic-subalgebra-to-be-abelian/129043#129043 Answer by MTS for Criterion for nilradical of a maximal parabolic subalgebra to be abelian? MTS 2013-04-28T23:53:15Z 2013-04-28T23:53:15Z <p>Denote by $\mathfrak{l}$ the Levi factor of the parabolic, so that $\mathfrak{p} = \mathfrak{l} \oplus \mathfrak{n}$, and note that this is a splitting as $\mathfrak{l}$-modules. Also denote by $\mathfrak{n}_-$ the nilradical of the opposite parabolic subalgebra; this is the dual of $\mathfrak{n}$ via the Killing form of $\mathfrak{g}$.</p> <p>Here are the equivalent conditions that I know:</p> <ol> <li>$\mathfrak{n}$ is abelian.</li> <li>$\mathfrak{n}_-$ is abelian.</li> <li>$\mathfrak{n}$ is an irreducible representation of $\mathfrak{l}$.</li> <li>$\mathfrak{n}_-$ is an irreducible representation of $\mathfrak{l}$.</li> <li>$\mathfrak{p}$ is maximal and the simple root of $\mathfrak{g}$ that is removed from $\mathfrak{l}$ has coefficient 1 in the highest root of $\mathfrak{g}$.</li> <li>$[\mathfrak{n},\mathfrak{n}] \subseteq \mathfrak{l}$.</li> <li>$(\mathfrak{g},\mathfrak{l})$ is a symmetric pair, i.e. there is an involutive Lie algebra automorphism of $\mathfrak{g}$ whose fixed-point subalgebra is precisely $\mathfrak{l}$.</li> </ol> <p>Clearly condition 5 is the easiest way to check this, assuming you have handy a table of highest roots of the simple Lie algebras. One can be found in, e.g. Table 2 in the exercises of Chapter 12 of <em>Introduction to Lie Algebras and Representation Theory</em>, by... well, you.</p> <p>I have not seen this entire collection of equivalent criteria written up in one place, although many of them are discussed in Lemma 7.3.1 of <em>Multiplicity-free Theorems of the Restrictions of Unitary Highest Weight Modules with respect to Reductive Symmetric Pairs</em>, by Toshi Kobayashi.</p> http://mathoverflow.net/questions/128254/possible-directions-in-noncommutative-geometry/128284#128284 Answer by MTS for Possible directions in noncommutative geometry MTS 2013-04-21T20:40:55Z 2013-04-21T20:40:55Z <p>Connes' book is pretty tough to get through as a beginner. I would suggest as an alternative the book <em>Elements of Noncommutative Geometry</em> by Gracia-Bondia, Varilly, and Figueroa. Or for a more concise, but less thorough, introduction, I like Varilly's book <em>An Introduction to Noncommutative Geometry</em>, which is also a little more recent.</p> http://mathoverflow.net/questions/19014/finding-questions-between-functional-analysis-and-set-theory/128027#128027 Answer by MTS for Finding questions between functional analysis and set theory MTS 2013-04-18T23:33:01Z 2013-04-18T23:33:01Z <p>Chris Phillips and Nik Weaver wrote a paper called <em>The Calkin Algebra has Outer Automorphisms</em>, where they showed that the Continuum Hypothesis implies that the Calkin algebra $\mathcal{B(H)/K(H)}$ has outer automorphisms. See also <a href="http://arxiv.org/abs/1211.4134" rel="nofollow">this paper</a> of Farah, McKenney, and Schimmerling, and references therein; they show that it is relatively consistent with ZFC that the Calkin algebra has only inner automorphisms, and hence the question of existence of outer automorphisms is independent of ZFC.</p> http://mathoverflow.net/questions/126462/generators-of-the-quantum-coordinate-algebras-and-quantized-enveloping-algebra-re/126475#126475 Answer by MTS for Generators of the Quantum Coordinate Algebras and Quantized Enveloping Algebra Representations MTS 2013-04-04T05:06:58Z 2013-04-04T05:06:58Z <p>For your first question, the answer is yes, as Casteels pointed out in the comments. The reason is that, for $\mathfrak{sl}_N$, every finite-dimensional irreducible representation appears as a subrepresentation of some tensor power of the natural representation. Since the finite-dimensional representation theory of the quantized enveloping algebra behaves the same as in the classical case (i.e. tensor products decompose in the same way), the same is true in the quantum setting.</p> <p>I don't know off the top of my head an expression in the generators corresponding to the second exterior power of the natural representation. But you can find one as follows: take $V$ to be the natural representation of $U_q(\mathfrak{sl}_N)$, and decompose $V \otimes V$ explicitly into irreducible components. There will be two: one of them will correspond to $\mathrm{Sym}^2(V)$ and the other to $\Lambda^2(V)$. The generators $u_j^i$ are the coordinate functions for $V$, so this decomposition will give you expressions for the coordinate functions of the two subrepresentations of $V\otimes V$ in terms of the $u_j^i$.</p> http://mathoverflow.net/questions/126101/domain-of-the-wedge-product-in-little-spivak/126136#126136 Answer by MTS for Domain of the wedge product in Little Spivak MTS 2013-03-31T22:05:36Z 2013-03-31T22:05:36Z <p>Without having the book handy, I can't say for sure what Spivak means by this. As KConrad points out in his comment, this is usually applied to tensors that are already alternating, and it turns the vector space $$\Lambda(V) := \bigoplus_{k=0}^\infty \Lambda^k(V)$$ into an associative algebra, called the exterior algebra of $V$.</p> <p>I would argue that this is sort of an unnatural way to view the exterior algebra. See <a href="http://mathoverflow.net/questions/54343/is-there-a-preferable-convention-for-defining-the-wedge-product/54375#54375" rel="nofollow">this question</a>, where this viewpoint is discussed. The gist of it is the following: the exterior algebra is most naturally viewed as the <em>quotient</em> algebra $T(V)/\langle x \otimes y + y \otimes x \mid x,y \in V\rangle$, where $T(V)$ is the tensor algebra of $V$. The description above of $\Lambda(V)$ as the direct sum of the antisymmetric tensors means that we are embedding the exterior algebra as a subspace of the tensor algebra $T(V)$ rather than a quotient. It is not a sub-<em>algebra</em>, because the concatenation $\omega \otimes \eta$ of two antisymmetric tensors is not antisymmetric. So the definition you/Spivak give of the wedge product is the way to express the natural multiplication of the exterior algebra in terms of operations in the tensor algebra. Applying $\mathrm{Alt}$ makes it antisymmetric again, and the factorials are a fudge factor required to make it associative.</p> http://mathoverflow.net/questions/126037/almost-lie-algebras/126040#126040 Answer by MTS for Almost-Lie Algebras? MTS 2013-03-30T18:53:00Z 2013-03-30T18:53:00Z <p>See Section 2.3 of the lecture notes called <em>Geometric Models for Noncommutative Algebras</em> by Ana Cannas da Silva and Alan Weinstein. There they define an "almost Lie algebra" to be something with an antisymmetric bracket but which does not necessarily satisfy Jacobi. In Section 3.2 they connect this to the notion of "almost Poisson manifold", which is just a manifold equipped with a bivector field. The bivector field defines a skew-symmetric bracket on smooth functions which may or may not satisfy Jacobi.</p> http://mathoverflow.net/questions/124513/non-drinfeld-jimbo-deformations-and-finite-quantum-groups/124585#124585 Answer by MTS for Non-Drinfeld--Jimbo Deformations and Finite Quantum Groups MTS 2013-03-15T03:04:33Z 2013-03-15T03:04:33Z <p>I do not know of a general method for quantizing the group algebra of a finite group. However, there is a way to do it for Coxeter groups (finite or not): the result is called an <a href="http://en.wikipedia.org/wiki/Iwahori-Hecke_algebra" rel="nofollow">Iwahori-Hecke algebra</a>. These are closely related to Drinfeld-Jimbo quantum groups, at least when the Coxeter group is the Weyl group of a finite-dimensional semisimple Lie algebra (and probably this extends to Kac-Moody algebras, but I don't know enough about that to make a definitive statement).</p> <p>One thing to think about is that if $G$ is a finite group, the group algebra $kG$ is semisimple as long as the characteristic of the field doesn't divide the order of $G$. And semisimple algebras are somewhat resistant to deformation. See <a href="http://mathoverflow.net/questions/81005/a-matrix-algebra-has-no-deformations" rel="nofollow">this question</a> for some more details on that story.</p> http://mathoverflow.net/questions/122918/real-forms-of-drinfeld-jimbo-quantum-groups Real forms of Drinfeld-Jimbo quantum groups MTS 2013-02-25T19:24:01Z 2013-02-26T18:12:10Z <p>A real form of a Hopf algebra $H$ over $\mathbb{C}$ is defined to be a $\ast$-structure on $H$ which is compatible with the coproduct. Compatibility of the $\ast$-structure with the counit and antipode then follows.</p> <p>The real forms of the Drinfeld-Jimbo quantum groups $U_q(\mathfrak{g})$ have been classified. This result is stated as Theorem 20 in Chapter 6 of <em>Quantum Groups and their Representations</em>, by Klimyk and Schmudgen, and also Proposition 9.4.2 of <em>A Guide to Quantum Groups</em>, by Chari and Pressley.</p> <p>For a given $\mathfrak{g}$, there are $\ast$-structures when $q \in \mathbb{R}$ or when $|q|=1$. These $\ast$-structures depend on a diagram automorphism of the Dynkin diagram of $\mathfrak{g}$, plus some extra parameters, and it is understood which sets of parameters give equivalent $\ast$-structures.</p> <p>There is also an exceptional case, but the two sources I have cited differ on what the exceptional case is. Klimyk and Schmudgen say that the exceptional case is when <code>$\mathfrak{g} = \mathfrak{sp}_{2n}$</code> and $q \in i \mathbb{R}$, while Chari and Pressley say that the exceptional case is $\mathfrak{g} = \mathfrak{so}_{2n+1}$ and $q \in i \mathbb{R}$.</p> <p>Neither book contains a proof, nor cites a source, although Chari-Pressley gives a sketch of the idea of the proof. So I would be interested in knowing the following things:</p> <ol> <li>Which is correct?</li> <li>What is the original reference for the classification of the real forms of $U_q(\mathfrak{g})$?</li> </ol> <hr> <h3>To set the record straight</h3> <p>It appears that the paper <em>Real forms of <code>$U_q(\mathfrak{g})$</code></em>, by Eric Twietmeyer, is the original reference. According to that paper, it is $\mathfrak{sp}_{2n}$ that has real forms for $q \in i \mathbb{R}$.</p> <p>Many thanks to Uwe Franz for digging up that reference!</p> http://mathoverflow.net/questions/122748/what-is-a-spinor-structure/122750#122750 Answer by MTS for what is a spinor structure? MTS 2013-02-23T19:55:01Z 2013-02-23T19:55:01Z <p>Chapter 9 of <em>Elements of Noncommutative Geometry,</em> by Gracia-Bondia, Varilly, and Figueroa, has this perspective on spin$^c$ and spin structures. </p> <p>The way to think about this algebraically is that the module of (continuous, say) sections of a spinor bundle over a (compact, Riemannian) manifold $M$ is Morita equivalence bimodule for the algebras $C(M)$ and $Cl(M)$, where $C(M)$ is the algebra of continuous functions and $Cl(M)$ is the algebra of continuous sections of the Clifford bundle (formed using the Riemannian metric). You can replace "continuous" with "smooth" here with no problems.</p> http://mathoverflow.net/questions/120452/does-the-braid-group-act-faithfully-on-the-quantized-enveloping-algebra Does the braid group act faithfully on the quantized enveloping algebra? MTS 2013-01-31T19:51:44Z 2013-01-31T19:51:44Z <p>Let $\mathfrak{g}$ be a finite-dimensional simple Lie algebra over $\mathbb{C}$, and let $U_q(\mathfrak{g})$ be some incarnation of the quantized universal enveloping algebra of $\mathfrak{g}$, where $q \in \mathbb{C}^\times$ is not a root of unity.</p> <p>Lusztig showed that the braid group $B_{\mathfrak{g}}$ (an infinite cover of the Weyl group of $\mathfrak{g}$ obtained by dropping the relations that the generators are involutive) acts by algebra automorphisms on $U_q(\mathfrak{g})$. (In fact there are several different ways that $B_{\mathfrak{g}}$ can act, but they are closely related and it doesn't matter which one we pick for the purposes of this question.)</p> <p>Has anybody seen anything addressing the question of whether this action is faithful? I would much appreciate an explanation or a reference that resolves the question either way. Thank you!</p> http://mathoverflow.net/questions/120127/representation-theory-for-the-exterior-algebra/120149#120149 Answer by MTS for Representation theory for the exterior algebra MTS 2013-01-28T21:49:21Z 2013-01-28T21:49:21Z <p>I'm not sure if this answers your question, but if $V$ is finite-dimensional and you are just asking for representations of the algebra structure, then the only irreducible representation of $\Lambda(V)$ is the trivial one-dimensional representation, in which $V$ acts as $0$.</p> <p>Indeed, let $W$ be an irreducible representation of $\Lambda(V)$. Let $0 \neq x \in V$, and consider $\mathrm{ker}(x) \subseteq W$ and $\mathrm{im}(x) \subseteq W$. Since $x$ anticommutes with all generators of $\Lambda(V)$, both of these subspaces of $W$ are actually $\Lambda(V)$-submodules.</p> <p>We are trying to show that $x$ acts as zero. If not, then since $W$ is irreducible, we must have that $\mathrm{ker}(x) = (0)$. Since the image is nonzero, again by irreducibility we conclude that $\mathrm{im}(x) = W$, so $x$ is an automorphism on $W$. But then $x \wedge x = 0$ in $\Lambda(V)$, so the square of this automorphism of $W$ is zero, which is a contradiction.</p> http://mathoverflow.net/questions/119114/clifford-lie-algebras/119132#119132 Answer by MTS for Clifford Lie Algebras MTS 2013-01-17T03:41:06Z 2013-01-17T03:41:06Z <p>A little bit of what you want can be found in Chapter 5 of Gracia-Bondia, Varilly, and Figueroa's book <em>Elements of Noncommutative Geometry.</em> They don't say much about subalgebras, I think, but they do prove (Lemma 5.7, page 182) the result that bivectors in the Clifford algebra $Cl(V)$ are closed under taking commutators, and that the adjoint action of bivectors on vectors in $V$ induces an isomorphism of the Lie algebra of bivectors with $\mathfrak{so}(V)$. That might be a good place to start.</p> http://mathoverflow.net/questions/116133/explicit-computations-of-examples-in-spin-geometry/116144#116144 Answer by MTS for Explicit Computations of Examples in Spin Geometry MTS 2012-12-12T05:36:14Z 2012-12-12T05:36:14Z <p>Appendix A to Chapter 9 of the book <em>Elements of Noncommutative Geometry</em> by Gracia-Bondia, Varilly, and Figueroa is titled "Spin geometry of the Riemann sphere". It is 15 pages long and goes into quite some detail. (Some might call that level of detail excruciating, but YMMV.)</p> <p>As Paul Siegel notes, computations on homogeneous spaces can be done quite effectively using representation theory. Some years ago, in the course of learning about that approach, I wrote up an account of the construction of the spinor bundle, Dirac operator, etc on $S^2$, viewed as the homogeneous space $SU(2)/U(1)$. If you're interested, email me (you can find my email address at my website, linked in my profile) and I can send it to you.</p> http://mathoverflow.net/questions/82917/r-matrices-crystal-bases-and-the-limit-as-q-1 R-matrices, crystal bases, and the limit as q -> 1 MTS 2011-12-07T23:16:31Z 2012-11-26T21:21:08Z <p>I am seeking references for precise statements and rigorous proofs of some facts about the actions of quantum root vectors and $R$-matrices on crystal bases for finite-dimensional representations of quantum groups. I am very new to crystal bases, so I would also appreciate corrections if my questions are not well-formulated. I am putting the questions first, followed by the motivation for those who are curious.</p> <h2>Questions</h2> <p>Do you know references (with proofs) for the following statements:</p> <ol> <li><p>Let $V$ be a finite-dimensional $U_q(\mathfrak{g})$-module. Then the divided powers $E_{\beta}^{(t)}, F_\beta^{(t)}$ of the quantum root vectors have matrix coefficients given by Laurent polynomials in $q$, with respect to the global crystal basis for $V$.</p></li> <li><p>Let $V,W$ be finite-dimensional $U_q(\mathfrak{g})$-modules. Then the matrix coefficients of the $R$-matrix $R_{V,W}$ are Laurent polynomials in $q$, with respect to the tensor product of the global crystal bases for $V,W$.</p></li> </ol> <p>I believe I have proofs for these statements, but it would be nice to just reference something definitive instead of writing the proofs out myself.</p> <h2>Background</h2> <p>Let $U_q(\mathfrak{g})$ be the quantized enveloping algebra of $\mathfrak{g}$ for $q$ not a root of unity, with generators $E_i,F_i,K_i$ corresponding to the simple roots of $\mathfrak{g}$. Using an action of the braid group of $\mathfrak{g}$ on $U_q(\mathfrak{g})$ one can define quantum root vectors $E_\beta,F_\beta$ for all positive roots $\beta$. (This depends on a choice of decomposition of the longest word of the Weyl group, so assume that we have fixed such a decomposition.)</p> <p>Let $R_{U,V}$ be the action of the $R$-matrix on $U \otimes V$, (as in Chari-Pressley or Klimyk-Schmudgen, say) so $\tau \circ R_{U,V}$ is the braiding. I would like to make sense of the statement that <code>$R_{V,W} \to \mathrm{id}_{V \otimes W}$</code> as $q \to 1$ (and hence the braiding tends to the flip as $q \to 1$). This is not trivial because for different $q$'s, the operators $R_{V,W}$ are really operators on different vector spaces. This is where the crystal bases come in. </p> <p>As I understand it, a crystal basis for a module has the property that the matrix coefficients of the generators $E_i,F_i$ of $U_q(\mathfrak{g})$ (and divided powers of the generators) are given by universal Laurent polynomials in $q$ whose coefficients are independent of $q$. Using this basis we can think of all of the algebras $U_q(\mathfrak{g})$ for various $q$'s acting on the same vector space. The point is that Laurent polynomials are continuous and well-defined at $q=1$, i.e. they are specializable to $q=1$.</p> <p>Taking the tensor product of the crystal bases for $V$ and $W$, we can think of all of the $R$-matrices for various $q$'s acting on the same space as well, and it makes sense to ask if this family of $R$-matrices is continuous in $q$, and if so, whether it can be extended to $q=1$.</p> <p>The formula for the action of the $R$-matrix is a big sum of products of operators of the form</p> <p><code>$$\frac{1}{[t]_{q_\beta}!} E_\beta^t \otimes F_\beta^t$$</code> </p> <p>with coefficients given by Laurent polynomials in $q$. Putting the $q$-factorial under, say, the $E_\beta^t$ term gives the divided power $E_\beta^{(t)}$. If the quantum root vectors and their divided powers act by Laurent polynomials, then the $R$-matrix does as well, and hence everything in sight is continuous in $q$, can be specialized to $q=1$, and it is clear that at $q=1$ the $R$-matrix is just the identity.</p> http://mathoverflow.net/questions/82917/r-matrices-crystal-bases-and-the-limit-as-q-1/114584#114584 Answer by MTS for R-matrices, crystal bases, and the limit as q -> 1 MTS 2012-11-26T21:21:08Z 2012-11-26T21:21:08Z <p>I never found a precise reference for the statement about the R-matrix, so I ended up writing it up myself. The precise statements and proofs can be found in $\S 4.1$ of my paper with Alex Chirvasitu, <em>Remarks on quantum symmetric algebras</em>, available <a href="http://arxiv.org/abs/1206.1614" rel="nofollow">here</a>.</p> http://mathoverflow.net/questions/113868/a-definition-of-non-commutative-metrisable-space/113876#113876 Answer by MTS for A definition of non-commutative metrisable space MTS 2012-11-19T21:54:25Z 2012-11-19T21:54:25Z <p>I would suggest that you look at the work of Rieffel on compact quantum metric spaces. His point of view is not to directly generalize the metric by understanding it as an element of the tensor square $A \otimes A$ (NB: you have not specified which tensor product you use here), but rather he generalizes the Lipschitz seminorm associated to the metric.</p> <p>As I understand it from Rieffel, it was known already to Kantorovich that the metric on a compact space $X$ is determined by the Lipschitz seminorm on $C(X)$: $$L(f) = \sup_{x \neq y} \{\frac{|f(x)-f(y)|}{d(x,y)}\}$$ via the identity $$d(x,y) = \sup\{ |f(x) - f(y)| : L(f) \le 1\}.$$</p> <p>Anyway, this doesn't directly answer your question, but this has been a fruitful line of inquiry, and I suggest you look at Rieffel's papers to see if they have anything useful for you. I think "Metrics on State Spaces" is a good one to start with.</p> http://mathoverflow.net/questions/93148/unitary-representations-of-quantum-groups Unitary representations of Quantum Groups MTS 2012-04-04T17:37:26Z 2012-09-22T21:13:23Z <p>Let $\mathfrak{g}$ be a finite-dimensional complex simple Lie algebra and let $U_q(\mathfrak{g})$ be some incarnation of the quantized universal enveloping algebra of $\mathfrak{g}$; here I am assuming that $q \in \mathbb{C}^\times$ is not a root of unity. I am interested in unitarizability of representations of $U_q(\mathfrak{g})$ with respect to various $*$-structures. </p> <p>The $*$-structures on $U_q(\mathfrak{g})$ have been classified; this can be found, for example, in section 9.4 of A Guide to Quantum Groups by Chari and Pressley, for example. For the $*$-structure known as the compact real form of $U_q(\mathfrak{g})$, it is known that each finite-dimensional irreducible representation $V_\lambda$ admits an invariant inner product, i.e. a positive-definite Hermitian form such that $$\langle av, w \rangle = \langle v, a^* w \rangle$$ for all $v,w \in V_\lambda$ and all $a \in U_q(\mathfrak{g})$.</p> <h3>Question: for an arbitrary $*$-structure on $U_q(\mathfrak{g})$, has anybody classified the set of dominant integral weights $\lambda$ for which $V_\lambda$ admits an invariant inner product?</h3> <p>Chari and Pressley mention in passing in their book that this is an open question, but that was in 1995 and I thought I'd see if anybody had resolved it in the meantime. I would expect that for an arbitrary $*$-structure, most irreps do not have such an invariant inner product, since in the classical situation the corresponding real form of the group is not compact, so you can't just average over Haar measure as you do in the compact case.</p> http://mathoverflow.net/questions/105493/generators-of-associated-graded-algebra Generators of associated graded algebra MTS 2012-08-26T02:21:34Z 2012-08-26T03:09:28Z <p>Suppose that $A = \bigcup_{n=0}^{\infty} A_n$ is a filtered algebra over a field $k$. The associated graded algebra is $\mathrm{gr} A = \bigoplus_{n=0}^{\infty} A_n/A_{n-1}$, where we define $A_{-1} = (0)$. There is no canonical algebra map from $A$ to $\mathrm{gr} A$, but there is a well-defined <em>function</em> $\gamma : A \to \mathrm{gr} A$ given by $$\gamma(x) = x + A_{n-1} \in A_n/A_{n-1},$$ where $n$ is the unique natural number (or 0) such that $x \in A_n$ but $x \notin A_{n-1}$. (To forestall nitpicking, let's say that $\gamma(0) = 0 \in A_0$.) Of course, this map fails to be even additive, but it does exist.</p> <h3>Question:</h3> <p>Given a set of generators $\{x_i\}$ for $A$, when is it the case that the set $\{\gamma(x_i)\}$ generates $\mathrm{gr} A$?</p> <p>Here is an easy example where this fails to happen. Let $\mathfrak{h}$ be the 3-dimensional Heisenberg Lie algebra (over $\mathbb{C}$, say), spanned by three elements $X,Y,Z$ with $[X,Y] = Z$ and $Z$ central. Let $A = U(\mathfrak{h})$ be its universal enveloping algebra with the usual filtration. Since $XY - YX = Z$, it follows that $A$ can be generated just by $X$ and $Y$. But Poincare-Birkhoff-Witt tells us that $\mathrm{gr} A \cong \mathbb{C}[X,Y,Z]$, which is certainly not generated just by $X$ and $Y$.</p> <p>The problem here is with the relation $XY-YX=Z$: since $Z$ has lower degree than $XY$ and $YX$, it drops out of the relation in the associated graded.</p> <p>Can anything be said about this, in general? Are there any nice criteria on the filtration and the generating set (and the relations, obviously) that ensure things don't go wrong in this way? Also I am amenable to making assumptions on the algebra $A$, for example that it is finitely generated or Noetherian (or ...?), if that helps. I do <em>not</em> want to increase the size of the generating set.</p> http://mathoverflow.net/questions/105103/reference-request-or-otherwise-adjoint-action/105125#105125 Answer by MTS for Reference request (or otherwise): Adjoint action MTS 2012-08-20T22:08:53Z 2012-08-20T22:08:53Z <p>I hope this clears up some of your confusion:</p> <ol> <li><p>The group $U(A)$ acts on $A$ itself by conjugation: $\mathrm{Ad}_u(a) = uau^\ast$.</p></li> <li><p>If $A$ acts on a Hilbert space $H$ then the group $U(A)$ acts on $H$ also just by restricting the action of $A$ on $H$. What you have written, $\mathrm{Ad}_u(\xi) = u \xi u^\ast$, doesn't make sense. What does it mean to multiply a vector on the right by an operator?</p></li> <li><p>What is true is that there is a certain intertwining relation between the conjugation action of $U(A)$ on $A$ and the action of $U(A)$ on $H$. Specifically: $$u (a \xi) = u a u^\ast u \xi = \mathrm{Ad}_u(a)(u \xi)$$</p></li> <li><p>The other adjoint action that you have written, $\mathrm{ad}_B(\xi) = B\xi - \xi B$, also does not make sense as stated, for the same reason as above: it doesn't make sense to multiply a vector on the right by an operator. This type of adjoint action is usually associated with Lie algebras. Since $A$ is an associative algebra, it is a Lie algebra with the bracket $[a,b] = ab - ba$. For any Lie algebra, the adjoint action is defined as $\mathrm{ad}_X(Y) = [X,Y]$.</p></li> <li><p>For Lie groups and Lie algebras, the relation between these two notions of adjoint is standard, and not really at the appropriate level for this site. But it works like this: let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. For $g \in G$, define $C_g : G \to G$ by $C_g(x) = gxg^{-1}$. This is a smooth map which takes the identity to the identity. The differential of this map at the identity is then a linear map from $\mathfrak{g}$ to $\mathfrak{g}$, which is called $\mathrm{Ad}_g$. Then the map $Ad : g \mapsto Ad_g$ is a smooth group homomorphism from $G$ to $GL(\mathfrak{g})$; the differential of this map at the identity is a linear map $ad : \mathfrak{g} \to \mathfrak{gl(g)}$. You can check that this agrees with the notion of adjoint defined in item 4 above.</p></li> </ol> <p>I would suggest thinking about these different notions of adjoint for a while and then coming back to ask a more focused question.</p> http://mathoverflow.net/questions/104348/which-concept-of-dimension-of-a-ring-of-functions-on-a-manifold-gives-the-dimens/104579#104579 Answer by MTS for Which concept of dimension of a ring of functions on a manifold, gives the dimension of the manifold? MTS 2012-08-12T21:11:50Z 2012-08-12T21:11:50Z <p>There is an approach to this question through a smooth version of the Hochschild-Kostant-Rosenberg (HKR) Theorem, which is related to Robert Bryant's answer. The smooth version is due to Alain Connes in its original form.</p> <p>First, the original HKR theorem says that for a regular affine $k$-algebra $R$ (think of the algebra of polynomial functions on a smooth affine variety), there are isomorphisms <code>$$\Lambda^\bullet (\Omega_{R/k}) \cong \mathrm{Tor}_\bullet^{R^e}(R,R) \overset{\mathrm{def}}{=} HH_\bullet(R)$$</code> and <code>$$\Lambda^\bullet (\mathrm{Der}(R)) \cong \mathrm{Ext}_{R^e}^\bullet (R,R) \overset{\mathrm{def}}{=} HH^\bullet(R).$$</code> Here $\Omega_{R/k}$ is the $R$-module of Kahler differentials and $\mathrm{Der}(R)$ is the $R$-module of derivations of $R$, which are algebraic analogues of 1-forms and vector fields on $R$, respectively. As Donu Arapura points out in his comment, Kahler differentials are the predual to derivations, rather than the other way around, as one normally defines forms to be dual to vector fields in the differential-geometric setting. Also $R^e = R \otimes R$, and $HH_\bullet(R)$ and $HH^\bullet(R)$ are the Hochschild homology and cohomology of $R$, respectively.</p> <p>The upshot is that when $R$ is the coordinate ring of a smooth affine variety, we can find the dimension of that variety by looking at the highest degree in which the Hochschild homology or cohomology does not vanish.</p> <p>Moving to the smooth case one needs to be quite careful, however. For a smooth manifold $M$, the Kahler differentials of $C^\infty(M)$ are not the same as the module of smooth 1-forms, as can be seen in <a href="http://mathoverflow.net/questions/6074/kahler-differentials-and-ordinary-differentials" rel="nofollow">this question</a>.</p> <p>The extra structure that $C^\infty(M)$ possesses is that of a Frechet algebra, where the seminorms are given by sup-norms of partial derivatives on compact subsets of $M$. Hochschild homology and cohomology can be adapted to the setting of certain topological algebras, and then there is an analogue of the HKR theorem. One formulation of it says that for $A = C^\infty(M)$, there are isomorphisms <code>$$_cHH_\bullet(A,A) \cong \Omega^\bullet(M)$$</code> and <code>$$_cHH^\bullet(A,A) \cong \Gamma^\infty(\Lambda^\bullet(TM)),$$</code> where $\Omega^\bullet(M)$ is the algebra of differential forms, and $\Gamma^\infty(\Lambda^\bullet(TM))$ is the space of smooth polyvector fields on $M$, and the subscript $c$ indicates the continuous Hochschild homology and cohomology. So again, the dimension of $M$ can be determined as the top degree in which the continuous Hochschild homology and cohomology do not vanish.</p> <p>This was proved by Connes (see Chapter 3, Section 2 of his book <em>Noncommutative Geometry</em>, plus references therein) for compact manifolds. For noncompact manifolds it is written up concisely in the paper <em>On Continuous Hochschild Homology and Cohomology Groups</em>, by Markus Pflaum. Connes' version uses the language of de Rham currents, which are dual to differential forms.</p> <p>Another source to look at is Chapter 8 of the book <em>Elements of Noncommutative Geometry</em> by Gracia-Bondia, Varilly, and Figueroa.</p> http://mathoverflow.net/questions/104034/a-good-primer-for-geometric-quantization/104037#104037 Answer by MTS for A good primer for geometric quantization. MTS 2012-08-05T18:51:20Z 2012-08-05T18:51:20Z <p>There is a set of lecture notes based on a course given by Alan Weinstein called Lectures on the Geometry of Quantization. It is part of the Berkeley Mathematics Lecture Notes series published by the AMS. You can also find the pdf freely available at Weinstein's <a href="http://math.berkeley.edu/~alanw/" rel="nofollow">homepage</a>.</p> http://mathoverflow.net/questions/103651/global-dimensions-of-non-commutative-rings/103727#103727 Answer by MTS for Global dimensions of non-commutative rings MTS 2012-08-01T20:58:52Z 2012-08-01T20:58:52Z <p>There is another way to see this than constructing an explicit resolution. This involves viewing $R$ as an iterated skew polynomial ring. I am assuming that you want $a_{ii} = 1$; otherwise $x_i^2 = 0$ for all $i$. Also, since this works over any field, I am just going to denote the base field by $k$.</p> <p>Start with $R_1 = k[x_1]$. Then let $\sigma_1$ be the $k$-algebra automorphism of $R_1$ defined by $\sigma_1(x_1) = a_{21} x_1$, and let $R_2$ be the skew-polynomial ring $$R_2 = R_1[x_2; \sigma_1].$$ Thus $R_2$ is generated by $x_1$ and $x_2$ with the relation $$x_2 x_1 = \sigma_1(x_1)x_2 = a_{21} x_1 x_2.$$ Then we continue this game. Having constructed $R_i$, define $$R_{i+1} = R_i[x_{i+1}, \sigma_i],$$ where $\sigma_i \in \mathrm{Aut}(R_i)$ is defined by $\sigma_i(x_j) = a_{i+1,j} x_j$ for $1 \le j \le i$. Then for $j &lt; i$ we have the relations $$x_i x_j = \sigma_{i-1}(x_j) x_i = a_{ij} x_j x_i,$$ and hence your ring $R$ coincides with $R_n$.</p> <p>This gives us two things. First, there is an analogue of the Hilbert Basis Theorem for skew polynomial rings; if $A$ is left Noetherian then the skew polynomial ring $A[x;\sigma]$ is left Noetherian for any automorphism $\sigma$ of $A$. You can find this in Section 1.2.9 of McConnell-Robson or Theorem 1.14 of Goodearl-Warfield.</p> <p>The other fact is that there is an analogue of the (generalized) Hilbert Syzygy Theorem for skew polynomial rings over Noetherian rings. This is in Section 7.9.10 of McConnell-Robson. Explicitly, it says the following: if $A$ is left Noetherian with $\mathrm{l.gl.dim} \, A = n &lt; \infty$, then $\mathrm{l.gl.dim} \, A[x;\sigma] = n+1$ for any automorphism $\sigma$ of $A$.</p> <p>Starting with $\mathrm{l.gl.dim}k[x] = 1$ and iterating shows that each $R_i$ is both left and right Noetherian and has $$\mathrm{l.gl.dim} R_i = \mathrm{r.gl.dim} R_i = i.$$</p> http://mathoverflow.net/questions/103426/why-do-we-distinguish-the-continuous-spectrum-and-the-residual-spectrum/103460#103460 Answer by MTS for Why do we distinguish the continuous spectrum and the residual spectrum? MTS 2012-07-29T17:47:33Z 2012-07-29T17:58:12Z <p>I'm afraid that this is more or less a reformulation of what you asked, but: a bounded operator $T$ on a Banach space $X$ is invertible if and only if it is bounded below (i.e. there is some constant $C>0$ such that $||Tx|| \geq C||x||$ for all $x\in X$) and has dense range. Bounded below implies injective, and it also implies that the range is closed; the range is closed and dense, hence is everything.</p> <p>So these two different types of spectrum distinguish different ways that the operator can fail to be invertible.</p> <p>Why is this distinction not made for Banach algebras in general? Well, it doesn't make sense to ask this unless the Banach algebra is presented as a subalgebra of operators on some Banach space. I suppose that you could always consider the left-regular representation of the algebra on itself and then interpret those two conditions in that setting. If $A$ is a Banach algebra, let $a \mapsto L_a$ denote the left-regular representation $L_ab = ab$.</p> <p>If $L_a$ does not have dense range then $\overline{aA}$ is not all of $A$, which condition I don't think can be reduced to anything simpler.</p> <p>If $L_a$ is not bounded below, that means that there is a sequence $(x_n)$ in $A$ with $||x_n||=1$ such that $ax_n \to 0$, i.e. $a$ is what is called a (left?) topological zero divisor.</p> http://mathoverflow.net/questions/33681/is-there-a-quantum-hermite-reciprocity/103366#103366 Answer by MTS for Is there a quantum Hermite reciprocity? MTS 2012-07-28T08:02:11Z 2012-07-28T08:02:11Z <p>There is in fact a reasonable way to define quantum analogues of symmetric and exterior powers of a finite-dimensional representation of $U_q(\mathfrak{g})$. Let $V$ be such a representation, and let $\hat{R} : V \otimes V \to V \otimes V$ be the braiding of $V$ coming from the universal R-matrix.</p> <p>It is a fact (see, for instance, Proposition 22 and Corollary 23 in Chapter 8 of the book Quantum Groups and Their Representations, by Klimyk and Schmudgen) that the eigenvalues of $\hat{R}$ are all of the form $\pm q^{t_i}$, where $t_i \in \mathbb{Q}$. Call the eigenvalues of the form $+q^{t_i}$ positive, and those of the form $-q^{t_i}$ negative (this notion is well-defined if $q$ is not a root of unity). Then call eigenvectors for $\hat{R}$ positive or negative, respectively, if their eigenvalues are positive or negative. The idea is that positive eigenvectors are $q$-symmetric, while negative eigenvectors are $q$-antisymmetric.</p> <p>Then define $$S_q^2 V = \mathrm{span} \{ \text{positive eigenvectors} \}$$ and $$\Lambda_q^2 V = \mathrm{span} \{ \text{negative eigenvectors} \}.$$ Since $\hat{R}$ is diagonalizable, $V \otimes V = S_q^2V \oplus \Lambda^2_q V$. For example, when $V$ is the 2-dimensional representation of $U_q(\mathfrak{sl}_2)$ with weight basis $x,y$, where $x$ is the highest weight vector, we have $$S_q^2 V = \mathrm{span} \{ x \otimes x, y \otimes x - q x \otimes y, y \otimes y \},$$ and $$\Lambda_q^2 V = \mathrm{span} \{ y \otimes x + q^{-1} x \otimes y \}.$$</p> <p>Finally, you can define higher quantum symmetric powers $S^n_qV$ to be the submodules of $V^{\otimes n}$ created by intersecting the submodules of tensors that are $q$-symmetric in all $n-1$ consecutive pairs of entries: $$S^n_q V = (S^2_q V \otimes V^{\otimes n -2}) \cap \dots \cap (V^{\otimes n -2} \otimes S_q^2 V).$$ There is also a closely related notion of quantum symmetric algebra, which is a graded $U_q(\mathfrak{g})$-module algebra whose homogeneous components are isomorphic to the quantum symmetric algebra defined above.</p> <p>Anyway, that's the good news; there is a not-too-bad definition of quantum symmetric powers. The bad news is that it doesn't always give you the classical result. The quantum symmetric powers of a module are no larger than their classical counterparts, and the module is called flat (in a different sense than the usual homological one) if all of its q-symmetric powers (or equivalently, just the q-symmetric cube) are the right size.</p> <p>The flat simple modules $V_\lambda$ have been classified by Sebastian Zwicknagl in his paper R-Matrix Poisson Algebras and their Deformations. For each semisimple Lie algebra there are only finitely many flat simple modules.</p> <p>In the paper Braided Symmetric Algebras of $U_q(\mathfrak{sl_2})$-modules and Their Geometry, he computes all of the quantum symmetric powers of simple $U_q(\mathfrak{sl_2})$-modules. It turns out that if $V$ is the 2-dimensional simple module, then its symmetric powers are the right size, i.e. $$S^n_q V \cong V_n,$$ where $V_n$ is the $(n+1)$-dimensional simple module. So your question about Hermite Reciprocity boils down to the question: are $S^m_q V_n$ and $S^n_q V_m$ isomorphic for any $m$ and $n$?</p> <p>The answer is that they are not. The first example is $m=3$ and $n = 4$, which follows from the computation in Theorem 3.1 of the second paper I referenced. The decompositions into simple modules are: $$S^4_q(V_3) \cong V_{12} \oplus V_8,$$ while $$S^3_q(V_4) \cong V_{12} \oplus V_{8} \oplus V_{4} \oplus V_{0}.$$ Of course, this doesn't rule out the possibility of a better definition which does satisfy Hermite Reciprocity, but nobody has come up with one yet. And if you want everything to be $U_q(\mathfrak{g})$-equivariant, then your choices are pretty rigid. But perhaps if you let go of that requirement then something more is possible.</p> http://mathoverflow.net/questions/63321/what-are-some-examples-of-chimeras-in-mathematics/103097#103097 Answer by MTS for What are some examples of "chimeras" in mathematics? MTS 2012-07-25T12:41:24Z 2012-07-25T12:41:24Z <p>Index theory for subfactors. Given an inclusion of subfactors $N \subseteq M$, there is an index $[M:N]$, which a priori is just a positive real number. Vaughan Jones showed that the index is constrained in the values it can take: it can be any real number $\geq 4$, or it can be of the form $4\cos^2(\pi/n)$ for some $n \ge 3$.</p> http://mathoverflow.net/questions/102517/how-can-i-tell-whether-a-manifold-is-homogeneous/102617#102617 Answer by MTS for How can I tell whether a manifold is homogeneous? MTS 2012-07-19T05:36:32Z 2012-07-19T06:08:04Z <p>Not really an answer, but too long to be a comment.</p> <p>I have thought about this question before, without much progress. It seems difficult. Even just to recognize if a manifold has the structure of a Lie group seems to be a difficult problem. And a manifold can be a Lie group in different ways.</p> <p>For instance, the exponential map from the Lie algebra $\mathfrak{n}$ of strictly upper triangular $n \times n$ matrices is a diffeomorphism with the group $N$ of unipotent upper triangular matrices, so $N$ has its natural Lie group structure as well as the abelian Lie group structure obtained by transporting the addition from $\mathfrak{n}$ through the exponential map.</p> <p>Thinking about it algebraically, asking for a manifold to be a Lie group is the same as asking for its function algebra to be a Hopf algebra (with some topology, perhaps), and the responses to <a href="http://mathoverflow.net/questions/90548/how-to-recognize-a-hopf-algebra" rel="nofollow">this question</a> seem to indicate that it is not very easy to recognize a Hopf algebra from the algebra structure alone.</p> <p>If you want a homogeneous space rather than a Lie group, you are then asking for the function algebra to be a (right or left) coideal subalgebra of a Hopf algebra. That seems even more difficult than looking for a Hopf algebra structure. And again, there can be no canonical construction because the same manifold can be a homogeneous space of different Lie groups: for example, $S^2$ is a homogeneous space of $SU(2)$, $SL(2,\mathbb{C})$, $O(3)$, and $SO(3)$.</p> <p>So I am pessimistic about there being a nice characterization, or even about the existence of simply stated sufficient conditions. But I'll be watching this thread and hoping that somebody comes up with something!</p> <p>Edit: Oh yeah! I forgot to mention something. Thinking further about the problem of identifying a Lie group, given only the manifold structure: the multiplication map on a Lie group $G$ turns the cohomology ring of $G$ into a graded Hopf algebra. I don't know the details, but my understanding is that (perhaps with some assumption on the cohomology groups being finitely generated free modules over the coefficient ring?) that if the cohomology ring of a manifold has the structure of a graded Hopf algebra, then the manifold is an <a href="http://en.wikipedia.org/wiki/H-space" rel="nofollow">$H$-space</a>, i.e. it has a multiplication and a unit map, but it is not necessarily associative. For instance $S^7$ is an $H$-space, but not a Lie group. So cohomology gets you a certain distance, but not all the way.</p> http://mathoverflow.net/questions/102180/representation-theory-of-antiself-dual-tensors/102407#102407 Answer by MTS for Representation theory of (anti)self-dual tensors MTS 2012-07-16T23:55:12Z 2012-07-17T20:47:59Z <p>I think what is going on is the following. I find it a bit difficult to wade through your notation, so I am going to use my own (or rather, the notation of Chapter 5 of Elements of Noncommutative Geometry, where I copied most of this from).</p> <p>Edit: this got kind of long, so let me say what the point is. You start with a real vector space $V$ with a symmetric bilinear form $g$ and construct the corresponding Clifford algebra $\mathrm{Cl}(V,g)$. The construction that I detail below constructs an isomorphism of $\mathfrak{so}(V,g)$ with a certain subspace of "degree 2 elements" in $\mathrm{Cl}(V,g)$ (quotes because the Clifford algebra is not a graded algebra). I believe that this is the source of the connection that you are trying to find.</p> <p>Let $V$ be a finite-dimensional vector space over $\mathbb{R}$ equipped with a symmetric bilinear form $g$ (of any signature). Edit: $g$ should be nondegenerate. It is standard (see for instance <a href="http://mathoverflow.net/questions/68378/clifford-algebra-non-zero" rel="nofollow">this question</a>) that the Clifford algebra $\mathrm{Cl}(V,g)$ is isomorphic as a vector space to the exterior algebra $\Lambda(V)$. In particular, looking just at the degree 2 part, there is an injective linear map $Q : \Lambda^2(V) \to \mathrm{Cl}(V,g)$ given by $$Q(v \wedge w) = \frac{1}{2} (vw - wv).$$</p> <p>For brevity, let's denote $b = Q(v \wedge w) = \frac12(vw - wv) \in \mathrm{Cl}(V,g)$, and consider the operator $\mathrm{ad}_b = [b,-]$ on $\mathrm{Cl}(V,g)$. In particular, consider what this operator does to a vector $x \in V \subseteq \mathrm{Cl}(V,g)$. First, note that in the Clifford algebra we have $wv = - vw + 2g(v,w)$, so $$\mathrm{ad}_b(x) = \left[\frac12(vw - wv), x\right] = [vw, x]$$ since the scalar term commutes with $x$.</p> <p>Then using the relation $[ab,c] = a(bc + cb) - (ac + ca)b$, (which holds trivially in any associative algebra), we get \begin{align*} \mathrm{ad}_b(x) &amp; = [vw,x] \\ &amp; = v(wx + xw) - (vx + xv)w \\ &amp; = 2g(w,x)v - 2g(v,x)w. \end{align*} In other words, the operator $\mathrm{ad}_b$ preserves the subspace $V \subseteq \mathrm{Cl}(V,g)$. Next, we check that this operator is actually in $\mathfrak{so}(V,g)$. From our calculations above, we have for any $x,y \in V$: $$g(y,[b,x]) = 2g(w,x)g(y,v) - 2g(v,x)(y,w) = -g([b,y],x).$$ Now the map $b \mapsto \mathrm{ad}_b$ is injective, so for dimension reasons this gives that $$\{ \mathrm{ad}_b|_V : b \in Q(\Lambda^2(V)) \} = \mathfrak{so}(V,g).$$</p> <p>This was all independent of signature, so it works for any symmetric bilinear form. Sorry I couldn't dovetail this all a little better with your notation!</p> <p>Edit: injectivity of the map $b \mapsto \mathrm{ad}_b$ is not totally trivial, and may only hold for certain signatures. It requires you to analyze the intersection of the even subalgebra with the center of $\mathrm{Cl}(V,g)$.</p> <p>Here is how the argument should go: If $\mathrm{ad}_b$ is zero, then $b$ is even and central. Then you need to be able to conclude that $b$ is scalar, and then that $b = 0$. It requires a case-by-case analysis (probably there is a better way, but I don't see it right now) to determine when the even part of the center consists just of scalars. Then you should be able to use the trace on the Clifford algebra to conclude that $b=0$.</p> <p>Further edit: I think my the argument that $b \mapsto \mathrm{ad}_b$ is injective actually does work in all signatures.</p> <p>As I noted above, if $\mathrm{ad}_b = 0$ then $b$ is an even, central element of the Clifford algebra. We want to conclude that $b$ must be scalar. In most signatures, the Clifford algebra is simple, so the center consists just of scalars, so we're done. But in the two case where the Clifford algebra is not simple, the even subalgebra is simple, so a central even element must be scalar.</p> <p>Then we want to show that if $\mathrm{ad}_b = 0$ then we must have $b=0$ (only for $b$ which is a linear combination of terms of the form $\frac12 (vw - wv)$). There is a unique trace $\tau$ on $\mathrm{Cl}(V,g)$ which vanishes on the odd part and satisfies $\tau(1)=1$; this is given by taking the coefficient of the identity element in a basis for the Clifford algebra consisting of increasing products of elements of a basis for $V$ (and this is independent of the basis). But the trace vanishes on our element $b$, since it is given to us as a sum of commutators. Thus $b=0$.</p> http://mathoverflow.net/questions/101644/fiction-books-about-mathematicians/101733#101733 Answer by MTS for Fiction books about mathematicians? MTS 2012-07-09T03:50:26Z 2012-07-09T03:50:26Z <p>There is a graphic novel called <a href="http://en.wikipedia.org/wiki/Logicomix" rel="nofollow">Logicomix</a>, by Apostolos Doxiadis and Christos Papadimitriou, which is about Bertrand Russell and the search for the foundations of mathematics. I know you were asking for fiction, but as with <em>Kepler</em>, this is sort of a fictionalized version of actual events, rather than an academic history book. Plus the pictures are great! And it's self-referential; the authors themselves appear in the book.</p> <p>I liked this one a lot.</p> http://mathoverflow.net/questions/100976/lie-algebras-over-non-algebraically-closed-fields/100978#100978 Answer by MTS for Lie algebras over non-algebraically closed fields MTS 2012-06-29T21:57:45Z 2012-06-29T21:57:45Z <p>Well, certainly things get more complicated when the field is not algebraically closed, as you can see from the classification of finite-dimensional simple Lie algebras over $\mathbb{R}$. But there are many cases where one just needs to be more careful with hypotheses.</p> <ol> <li>In the proof of Lie's Theorem: for a solvable Lie algebra $\mathfrak{g}$ and a finite-dimensional representation $\pi : \mathfrak{g} \to \mathfrak{gl}(V)$, you can get the theorem as long as you assume that all of the eigenvalues of all of the endomorphisms $\pi(X)$ lie in the field you are working over. I don't know how often this happens in practice when the field is not algebraically closed, but the proof does go through.</li> <li>Engel's theorem goes through without a problem over any field, I believe.</li> <li>The Poincare-Birkhoff-Witt Theorem works over any field (and in fact, for Lie algebras over any commutative ring $R$ where the underlying $R$-module of the Lie algebra is free).</li> <li>Cartan's Criterion for Semisimplicity works over any subfield of $\mathbb{C}$, essentially because $\mathfrak{g}$ is semisimple if and only if $\mathfrak{g} \otimes \mathbb{K}$ is semisimple for any extension field $\mathbb{K}$ (note that this is not true for simplicity; if $\mathfrak{g}$ is a simple real Lie algebra which is not obtained from a simple complex Lie algebra by restriction of scalars, then the complexification $\mathfrak{g} \otimes \mathbb{C}$ is not simple.)</li> </ol> <p>Hmmm... I'm sure there's a lot more to say. Let's just wait until Professor Humphreys shows up.</p> http://mathoverflow.net/questions/100353/how-can-one-find-generators-of-basic-differential-forms-on-homogeneous-spaces/100411#100411 Answer by MTS for How can one find generators of basic differential forms on homogeneous spaces? MTS 2012-06-23T00:15:01Z 2012-06-23T15:53:14Z <p>This actually can be done in much greater generality. Let $G$ be a compact group and $K \subseteq G$ a closed subgroup. Then for any finite-dimensional representation $(V,\pi)$ of $K$ you can form the associated bundle $G \times_K V$ over $G/K$. Sections of this bundle are given by functions $f : G \to V$ satisfying the equivariance condition $$f(xs) = \pi(s)^{-1} f(x)$$ for all $x \in G$ and $s \in K$. You are now asking for a collection of sections that generates the module of all sections of this bundle. One way to do this is via Frobenius reciprocity.</p> <p>Frobenius reciprocity implies that there is a representation $(W,\sigma)$ of $G$ such that $V \subseteq W$ and $\sigma(s)v = \pi(s)v$ for $v \in V$. Since $G$ is compact, there is a $G$-invariant inner product on $W$. Let $P$ be the orthogonal projection of $W$ onto $V$ with respect to this inner product.</p> <p>Then choose an orthonormal basis $(w_i)$ for $W$. For each $i$, define a function $\eta_i : G \to V$ by $$\eta_i(x) = P \sigma(x)^{-1}w_i.$$ It is not too hard to show that each $\eta_i$ is actually a section of the associated bundle.</p> <p>To show that these sections generate the module, define a $C^\infty(G/K)$-valued pairing on the module $\Gamma = \Gamma^\infty(G/K, G \times_K V)$ by $$\langle \zeta, \xi \rangle(x) = \langle \zeta(x), \xi(x) \rangle,$$ for sections $\zeta, \xi$, where $\langle , \rangle$ was the invariant inner product on $W$ that we chose above. Then it turns out that for any $\xi \in \Gamma$, we have $$\xi = \sum_i \langle \xi, \eta_i \rangle \eta_i.$$ When verifying this you need to use the fact that since the inner product on $W$ is $G$-invariant, any element $x$ of $G$ takes the orthonormal basis $(w_i)$ to another orthonormal basis.</p> <p>Anyway, this is only as canonical as choosing an orthonormal basis of $W$, so it may not be what you want. But it is at least a way of getting a nice generating set.</p> <p>Edit: I should explain what this is really doing. Note that a homogeneous vector bundle bundle $G \times_K V$ over $G/K$ will be a trivial bundle if and only if the representation $(V,\pi)$ of $K$ is actually the restriction of a representation of $G$ on $V$. To see this, note that if this is the case, then in the construction above we can take $W = V$, the projection $P$ is just the identity operator, and the global sections $\eta_i$ vanish nowhere and form a global frame for the bundle.</p> <p>In the case when we have to take $W$ to be strictly larger than $V$, what we are doing is embedding the homogeneous bundle $G \times_K V$ into the trivial bundle $G \times_K W$. The projection $P$ tells us how to cut down fiberwise from the trivial bundle to the nontrivial one.</p> <p>I should add that in your case, you can take the larger vector space $W$ to be $\mathcal{G}^\ast$ itself with the coadjoint action of $G$, and you can take the invariant inner product to be the one induced by the Killing form on $\mathcal{G}$.</p> http://mathoverflow.net/questions/130914/is-there-an-analogue-of-spin-oscillator-representation-for-the-general-linear-lie Comment by MTS MTS 2013-05-17T13:36:07Z 2013-05-17T13:36:07Z Steven, what properties of the spin/oscillator representations are you trying to generalize? And what properties of the Clifford/Weyl algebras? http://mathoverflow.net/questions/130945/c-c-infty0-tv-is-dense-in-c-c10-tv Comment by MTS MTS 2013-05-17T13:16:57Z 2013-05-17T13:16:57Z Isn't it superfluous to require compact support, given that the domain is a compact interval? http://mathoverflow.net/questions/129902/let-r-be-a-ring-and-e-e-idempotents Comment by MTS MTS 2013-05-06T22:28:28Z 2013-05-06T22:28:28Z This problem is off-topic here, as it appears to be homework. Try at <a href="http://math.stackexchange.com" rel="nofollow">math.stackexchange.com</a>. Voting to close. http://mathoverflow.net/questions/129870/in-order-to-factor-we-must-first-find-its-zeros Comment by MTS MTS 2013-05-06T18:39:45Z 2013-05-06T18:39:45Z Hi Anthony, your question is well-phrased, but unfortunately this is not the right forum for it. This site is for research-level mathematics. Your question is better suited for <a href="http://math.stackexchange.com" rel="nofollow">math.stackexchange.com</a>. Voting to close. http://mathoverflow.net/questions/129551/weak-convergence-on-a-hilbert-space Comment by MTS MTS 2013-05-03T17:34:39Z 2013-05-03T17:34:39Z Stephane, I think this question is better suited to <a href="http://math.stackexchange.com" rel="nofollow">math.stackexchange.com</a>, as this is not really research-level. Voting to close. http://mathoverflow.net/questions/129465/number-of-linear-maps-is-less-or-equal-than-the-dimension-of-the-vector-space Comment by MTS MTS 2013-05-02T23:01:14Z 2013-05-02T23:01:14Z Examples include most of the other problems on this site. Your question is better suited for <a href="http://math.stackexchange.com" rel="nofollow">math.stackexchange.com</a> http://mathoverflow.net/questions/129161/quotient-of-lie-rings-and-quotient-of-lie-groups Comment by MTS MTS 2013-04-30T02:18:08Z 2013-04-30T02:18:08Z I think this still needs to be clarified. Take $g$ to be a Lie algebra, $I \subseteq g$ an ideal. Are you asking whether $U(I)$ (the universal enveloping algebra of $I$) is an ideal in the universal enveloping algebra $U(g)$? http://mathoverflow.net/questions/129161/quotient-of-lie-rings-and-quotient-of-lie-groups Comment by MTS MTS 2013-04-30T01:51:28Z 2013-04-30T01:51:28Z I don't understand your question. What quotient ring do you mean? What does a Lie group have to do with anything? http://mathoverflow.net/questions/129160/existence-of-a-continuous-and-unbounded-map-f-with-ffxx Comment by MTS MTS 2013-04-30T01:41:49Z 2013-04-30T01:41:49Z Continuous for which topology? http://mathoverflow.net/questions/129155/bayes-classification-rule-in-matlab-histogram-command Comment by MTS MTS 2013-04-29T23:25:52Z 2013-04-29T23:25:52Z This site is for research level math questions, which your question is not. See <a href="http://mathoverflow.net/faq#whatquestions" rel="nofollow">mathoverflow.net/faq#whatquestions</a>. Voting to close. http://mathoverflow.net/questions/129151/stones-theorem-on-one-parameter-unitary-groups Comment by MTS MTS 2013-04-29T22:30:05Z 2013-04-29T22:30:05Z This is the first and only example given on the Wikipedia page for Stone's Theorem: see <a href="http://en.wikipedia.org/wiki/Stone" rel="nofollow">en.wikipedia.org/wiki/Stone</a>'s_theorem_on_one-parameter_unitary_groups#Example http://mathoverflow.net/questions/129037/criterion-for-nilradical-of-a-maximal-parabolic-subalgebra-to-be-abelian/129043#129043 Comment by MTS MTS 2013-04-29T01:06:46Z 2013-04-29T01:06:46Z @Jim, check your email. http://mathoverflow.net/questions/129037/criterion-for-nilradical-of-a-maximal-parabolic-subalgebra-to-be-abelian/129043#129043 Comment by MTS MTS 2013-04-28T23:57:10Z 2013-04-28T23:57:10Z I should say that some people refer to this property by saying that $\mathfrak{p}$ is of <i>cominuscule type</i>. It is related to the property of minusculity, but is not quite the same. See the beginning of Chapter 9 of Billey and Lakshmibai, <i>Singular Loci of Schubert Varieties</i> for a little more information on that connection. http://mathoverflow.net/questions/128869/incidences-of-quadratic-forms-and-points Comment by MTS MTS 2013-04-27T01:20:40Z 2013-04-27T01:20:40Z -1. This question needs much more explanation to be clear and useful. See <a href="http://mathoverflow.net/howtoask" rel="nofollow">mathoverflow.net/howtoask</a>. http://mathoverflow.net/questions/128701/existence-of-a-projection-operator-onto-a-classical-set-of-density-matrices Comment by MTS MTS 2013-04-26T18:55:33Z 2013-04-26T18:55:33Z FYI, you should use \langle and \rangle rather than &lt; and &gt;, as they are interpreted differently by the formatting engine.