User peter tingley - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T18:06:24Z http://mathoverflow.net/feeds/user/1799 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/5545/where-can-i-find-questions-motivating-important-ideas-in-math Where can I find questions motivating important ideas in math? Peter Tingley 2009-11-14T18:32:44Z 2012-10-19T04:14:58Z <p>I would like questions that demonstrate why a mathematical tool or technique is useful, and which can be used to introduce that idea. Ideally, this would be a compilation of problems organized by the idea they motivate. Does anyone know of such a list? I am thinking of middle school and high school math, but I think such a list would be useful at any level. </p> http://mathoverflow.net/questions/43638/a-distinct-parts-odd-parts-identity-for-standard-young-tableaux/43837#43837 Answer by Peter Tingley for A Distinct parts/Odd parts identity for standard Young tableaux Peter Tingley 2010-10-27T17:15:28Z 2010-10-27T17:15:28Z <p>A possibly related result says that the number of partitions on n into distinct parts is equal to the number of partitions of n into odd parts. There is a bijective proof, I think due to Sylvester. I think a simpler version of the original bijection can be found in Kim and Yee's paper <a href="http://docs.google.com/viewer?a=v&amp;q=cache:0jVPrNaVJKcJ:citeseerx.ist.psu.edu/viewdoc/download%3Fdoi%3D10.1.1.46.4811%26rep%3Drep1%26type%3Dpdf+%22distinct+parts%22+%22odd+parts%22+partition&amp;hl=en&amp;gl=us&amp;pid=bl&amp;srcid=ADGEESjCkRyxeXcrIDJZ4y-W_cG1fFc5l5wB5mPXZWlj0laoA7HGVUNB4X_U966FtdZ8U7b5WvbvckOZbfWD0S7VYWapNu5fL5LOQQTelBhEA6-SCoPh4fV7HFTV5SJoFCi9t7GNCyzL&amp;sig=AHIEtbSk5WDMQfdsle3LxjqBvsVrUOplbg" rel="nofollow"> A Note on Partitions into Distinct Parts and Odd Parts</a>. There are also some refinements of this statement out there. </p> http://mathoverflow.net/questions/41033/realizing-higher-level-fock-spaces/41078#41078 Answer by Peter Tingley for Realizing higher level Fock spaces Peter Tingley 2010-10-04T22:51:31Z 2010-10-04T22:51:31Z <p>Higher level Fock spaces have been studied in the context of the quantum affine algebra $U_q(\widehat{sl}_n)$. There is a "higher level Fock space" representation for this algebra whose underlying space looks like semi-infinite wedge space. I believe the original reference is Jimbo, Miwa, Misra and Okado "Combinatorics of representations of $U_q(\widehat{sl}_n)$ at $q=0$"</p> <p><a href="http://www.ams.org/mathscinet/search/publdoc.html?arg3=&amp;co4=AND&amp;co5=AND&amp;co6=AND&amp;co7=AND&amp;dr=all&amp;pg4=AUCN&amp;pg5=AUCN&amp;pg6=AUCN&amp;pg7=ALLF&amp;pg8=ET&amp;r=1&amp;review_format=html&amp;s4=Jimbo&amp;s5=Miwa&amp;s6=Misra&amp;s7=&amp;s8=All&amp;vfpref=html&amp;yearRangeFirst=&amp;yearRangeSecond=&amp;yrop=eq" rel="nofollow">http://www.ams.org/mathscinet/search/publdoc.html?arg3=&amp;co4=AND&amp;co5=AND&amp;co6=AND&amp;co7=AND&amp;dr=all&amp;pg4=AUCN&amp;pg5=AUCN&amp;pg6=AUCN&amp;pg7=ALLF&amp;pg8=ET&amp;r=1&amp;review_format=html&amp;s4=Jimbo&amp;s5=Miwa&amp;s6=Misra&amp;s7=&amp;s8=All&amp;vfpref=html&amp;yearRangeFirst=&amp;yearRangeSecond=&amp;yrop=eq</a>,</p> <p>although there the wedge space structure is not clear. That is explained in Uglov's paper "Canonical bases of higher level $q$-deformed Fock space and Kahzdan Lusztig polynomials"</p> <p><a href="http://arxiv.org/abs/math/9905196" rel="nofollow">http://arxiv.org/abs/math/9905196</a>.</p> <p>Higher level Fock space is more complicated then the level 1 case. For instance many different irreducible representation occur as direct summands of Fock space. In order to get a realization of a single irreducible highest weight representation, you need to pick off the irreducible subrepresentation generated by a certain overall highest weight vector. On the level of representations, this is difficult. However, in the "crystal limit" (i.e. at $q=0$), this can be done quite easily. The basis of the resulting representation is naturally indexed by $\ell$ tuples of partitions (where $\ell$ is the level), satisfying a couple conditions. This fact has been useful in studying crystal bases of these higher level representations. </p> http://mathoverflow.net/questions/25592/what-role-does-the-dual-coxeter-number-play-in-lie-theory-and-should-it-be-cal/25729#25729 Answer by Peter Tingley for What role does the "dual Coxeter number" play in Lie theory (and should it be called the "Kac number")? Peter Tingley 2010-05-23T22:33:30Z 2010-05-23T22:33:30Z <p>I think part of the question here is "why is this thing called the dual Coxeter number? It looks pretty different, so why don't we just give it a different name?" I think the case is made in Kac's book that dual Coxeter number is the right name. </p> <p>The Coxeter number for $\mathfrak{g}$ is the sum of the labels in the Dynkin diagram for the untwisted affine algebra corresponding to $\mathfrak{g}$. These labels are the coefficients of a minimal integer linear dependence among the columns of the affine Cartan matrix, which seems fairly intrinsic, so I think this is a reasonable definition. I won't try to explain why it is equivalent to more standard definitions. The dual Coxeter number is then the sum of the labels in the dual affine Dynkin diagram. See Kac, section 6.1 for these definitons.</p> <p>I think what is confusing is that "dual" and "affine" do not commute. For instance, the dual of the affine diagram of type $B_\ell^{(1)}$ is the twisted affine Dynkin diagram of type $A_{2\ell-1}^{(2)}$.</p> http://mathoverflow.net/questions/19063/reference-for-the-existence-of-a-shapovalov-type-form-on-the-tensor-product-of-in/19145#19145 Answer by Peter Tingley for Reference for the existence of a Shapovalov-type form on the tensor product of integrable modules Peter Tingley 2010-03-23T20:58:59Z 2010-03-24T16:06:25Z <p>I know a couple of ways to get a Shapovalov type form on a tensor product. The details of what I say depends on the exact conventions you use for quantum groups. I will follow Chari and Pressley's book. </p> <p>The first method is to alter the adjoint slightly. If you choose a * involution that is also a coalgebra automorphism, you can just take the form on a tensor product to be the product of the form on each factor, and the result is contravariant with respect to *. There is a unique such involution up to some fairly trivial modifications (like multiplying $E_i$ by $z$ and $F_i$ by $z^{-1}$). It is given by: $$*E_i = F_i K_i, \quad *F_i=K_i^{-1}E_i, \quad *K_i=K_i,$$ The resulting forms are Hermitian if $q$ is taken to be real, and will certainly satisfy your conditions 1) ad 3). Since the $K_i$s only act on weight vectors as powers of $q$, it almost satisfies 2).</p> <p>The second method is in case you really want * to interchange $E_i$ with exactly $F_i$. This is roughly contained in this <a href="http://www.ams.org/mathscinet-getitem?mr=1470857" rel="nofollow">http://www.ams.org/mathscinet-getitem?mr=1470857</a> paper by Wenzl, which I actually originally looked at when it was suggested in an answer to one of your previous questions.</p> <p>It is absolutely essential that a * involution be an algebra-antiautomorphism. However, if it is a coalgebra anti-automorphism instead of a coalgebra automorphism there is a work around to get a form on a tensor product. There is again an essentially unique such involution, given by</p> <p>$$*E_i=F_i, \quad *F_i=E_i, \quad *K_i=K_i^{-1}, \quad *q=q^{-1}.$$</p> <p>Note that $q$ is inverted, so for this form one should think of $q$ as being a complex number of the unit circle. By the same argument as you use to get the Shapovalov form, then is a unique sesquilinear *-contravariant form on each irreducible representation $V_\lambda$, up to overall rescaling. </p> <p>To get a form on $V_\lambda \otimes V_\mu$, one should define $$(v_1 \otimes w_1, v_2 \otimes w_2)$$ to be the product of the form on each factor applied to $v_1 \otimes w_1$ and $R( v_2 \otimes w_2)$, where $R$ is the universal $R$ matrix. It is then straightforward to see that the result is *-contravariant, using the fact that $R \Delta(a) R^{-1} =\Delta^{op}(a).$</p> <p>If you want to work with a larger tensor product, I believe you replace $R$ by the unique endomorphism $E$ on $\otimes_k V_{\lambda_k}$ such that $w_0 \circ E$ is the braid group element $T_{w_0}$ which reverses the order of the tensor factors, using the minimal possible number of positive crossings. Here $w_0$ is the symmetric group element that reverses the order of the the tensor factors. </p> <p>The resulting form is *-contravariant, but is not Hermitian. In Wenzl's paper he discusses how to fix this. </p> <p>Now 1) and 2) on your wish list hold. As for 3): It is clear from standard formulas for the $R$-matrix (e.g. Chari-Pressley Theorem 8.3.9) that $R$ acts on a vector of the form $b_\lambda \otimes c \in V_\lambda \otimes V_\mu$ as multiplication by $q^{(\lambda, wt(c))}$. Thus if you embed $V_\mu$ into $V_\lambda \otimes V_\mu$ as $w \rightarrow b_\lambda \otimes w$, the result is isometric up to an overall scaling by a power of $q$. This extends to the type of embedding you want (up to scaling by powers of $q$), only with the order reversed. I don't seem to understand what happen when you embed $V_\lambda$ is $V_\lambda \otimes V_\mu$, which confuses me, and I don't see your exact embeddings.</p> http://mathoverflow.net/questions/18197/restriction-of-a-representation-of-gln-to-gln-1/18200#18200 Answer by Peter Tingley for restriction of a representation of GL(n) to GL(n-1) Peter Tingley 2010-03-14T19:32:07Z 2010-03-14T19:32:07Z <p>If you use complex numbers instead of real numbers, it is true that the restriction is multiplicity free. This is an important fact, and is used to construct the Gelfand-Zetlin basis for V. This is discussed is Fulton and Harris' book "representation theory", section 25.3. They also discuss generalizations to other classical algebras</p> http://mathoverflow.net/questions/16570/are-there-positive-formulae-for-the-inner-product-between-elements-of-a-lie-algeb/16982#16982 Answer by Peter Tingley for Are there positive formulae for the inner product between elements of a Lie algebra representation in the Shapovalov form? Peter Tingley 2010-03-03T17:00:04Z 2010-03-03T17:00:04Z <p>This may not be exactly what you want, but I'd suggest you look at Kashiwara's paper On crystal bass of the q-analogue of the universal enveloping algebra" (see MR1115118 ). </p> <p>In section 2.5 Kashiwara discusses the quantum version of the Shapovalov form. More relevant to what I want to say is Proposition 3.4.4, which defines/proves existence of a modification of the Shapovalov form defined on $U_q^-$. Roughly what he is doing is the following: Consider the pairing of, for example, $F_1F_2F_1v_\lambda$ and $F_1^2 F_2v_\lambda$, using the Shapovalov form on $M(\lambda)$, and allow $\lambda$ to vary. What you get is $$P(\lambda)/(q-q^{-1})^3$$ where $P$ is some Laurent polynomial in the $K_i$ and $q$, and evaluation at $\lambda$ means setting $K_i$ equal to $q^{(\alpha_i, \lambda)}.$ $P$ has a well defined highest order term in the $K_i$, and the coefficient of this term is a Laurent polynomial in $q$. Let $(F_1F_2F_1, F_1^2 F_2)$ be that leading coefficient. This will be Kashiwara's inner product on $U_q^-$, up to a power of $q$. In general, when pairing two monomials in the $F_i$ applied to $v_\lambda$, the denominator in the above equation has a factor of $q_i-q_i^{-1}$ for each $F_i$ in the first monomial. </p> <p>The way Kashiwara sets things up, it is clear that the inner product of two monomials in the $F_i$ is a Laurent polynomial in $q$ with positive coefficients. For monomials $m_1$ and $m_2$, the inner product will be zero unless $m_1$ and $m_2$ have the same weight (i.e. they are both the products of the same number of each $F_i$, but possibly in a different order). Furthermore, the sum of the coefficients is the number of ways of matching each $F_i$ in $m_1$ with an $F_i$ in $m_2$ for all $i$. This all follows from Equations (3.3.1) and (3.4.6) in Kashiwara's paper. I believe you can find the power of $q$ associated to a given matching by arranging the monomials correctly, drawing a line between each matching pair, and counting a contribution for each crossing in the resulting picture. So there should be a completely combinatorial formula. </p> <p>As I said, this may well not be what you want. For instance, this construction does not depend on $\lambda$. But maybe it is related. </p> <p>In case you are wondering about the connection with crystal bases (i.e. the title of Kashiwara's paper), Kashiwara shows that the inner product of any two elements in the crystal lattice $L(\infty)$ is regular at $q=0$, and a crystal basis is an orthonormal basis for the evaluation $(\cdot, \cdot)_0$ of the inner product at $q=0$. Of course for this to be true you need to get the powers of $q$ right, which I have not done here. </p> http://mathoverflow.net/questions/16786/is-an-algebraic-bijection-from-a-projective-variety-to-itself-necessarily-an-isom Is an algebraic bijection from a projective variety to itself necessarily an isomorphism? Peter Tingley 2010-03-01T21:35:28Z 2010-03-02T18:56:20Z <p>Let $X$ be a projective variety. Assume there is an algebraic map $f: X \rightarrow X$ that is a bijection. I am thinking of $X$ as a variety, not a scheme, so by a bijection I mean a bijection on closed points. Most likely I am working over the complex numbers, so if you like I mean a bijection on complex points. Can you conclude that $f$ has an algebraic inverse? </p> <p>I think this is not immediately obvious, since it is not true that any algebraic bijection between two projective varieties is an isomorphism. For instance, there is an algebraic bijection from ${\Bbb P}^1$ to a cuspidal cubic in ${\Bbb P}^2$ given by $[x,y] \rightarrow [x^3, x^2y, y^3]$. So if this is true one must use the fact that the map is from $X$ to itself.</p> <p>I am interested in cases where $X$ is both singular and reducible (although is of pure dimension, if that helps), so a complete answer would cover any such case. Alternatively, if it is not true that such a map has an algebraic inverse, I would like an explicit counter example. </p> http://mathoverflow.net/questions/14251/asymptotics-of-symmetry-types-of-tensors/14295#14295 Answer by Peter Tingley for Asymptotics of symmetry types of tensors Peter Tingley 2010-02-05T17:01:47Z 2010-02-05T17:01:47Z <p>This is not a complete answer, but perhaps will help. The probability distribution on tuples that shows up in your Theorem 1 is well known: It is the joint probability density function for the eigenvalues of a random unitary matrix, in the standard Gaussian unitary ensemble". See equation (9) in the paper of Terry Tao and Van Vue</p> <p><a href="http://arxiv.org/abs/0906.0510v9" rel="nofollow">http://arxiv.org/abs/0906.0510v9</a></p> <p>I have unfortunately not read that closely, but the title, "random matrices: universality of local eigenvalue statistics" suggests it might have something to say about why this distribution would appear in other places. I've looked more closely at the following paper of Okounkov. </p> <p><a href="http://arxiv.org/abs/math-ph/0309015" rel="nofollow">http://arxiv.org/abs/math-ph/0309015</a></p> <p>There he explains how a similar distribution does show up in a system of random partitions. See especially Section 1.4.2. There Okounkov uses the distribution coming from the Plancherel measure", which is slightly different then the distribution you describe: the probability of observing $\lambda$ is propositional to $dim S(\lambda)^2$, where $S(\lambda)$ is the representation of the symmetric group corresponding to $\lambda$. You seem to have chosen the distribution where the probability of observing $\lambda$ is proportional to $dim S(\lambda) dim V(\lambda)$, where $V(\lambda)$ is an irreducible representation of $U(n)$. Also, he has $-1/2$ where you have $-m/2$ in the exponential part. But perhaps it is still related.</p> <p>Anyway, if you are interested in this type of question about partitions, I highly recommend looking at these papers.</p> http://mathoverflow.net/questions/9308/are-the-identity-object-axioms-in-the-definition-of-a-braided-monoidal-category Are the “identity object axioms” in the definition of a braided monoidal category needed? (Answered: No) Peter Tingley 2009-12-18T20:03:57Z 2009-12-19T14:45:54Z <p>I am asking because the literature seems to contain some inconsistencies as to the definition of a braided monoidal category, and I'd like to get it straight. According Chari and Pressley's book A guide to quantum groups," a braided monoidal category is a monoidal category $\mathcal{C}$ along with a natural system of isomorphisms $\sigma_{U,V}: U \otimes V \rightarrow V \otimes U$ for all pairs of objects $U$ and $V$, such that</p> <p>(i) The Hexagon" axioms (two commutative diagrams) hold. </p> <p>(ii) The identity object" axioms: $\rho_V= \lambda_V \circ \sigma_{{\bf 1},V}: {\bf 1} \otimes V \rightarrow V$ and $\lambda_V= \rho_V \circ \sigma_{V, {\bf 1}}: {V} \otimes {\bf 1} \rightarrow V$, where $\lambda_V$ and $\rho_V$ are the isomorphisms of $V \otimes {\bf 1}$ and ${\bf 1} \otimes V$ with $V$ that are part of the definition of monoidal category. See Chari-Pressley Definitions 5.2.1 and 5.2.4. They use the term "quasitensor category," but note on p153 that the term "braided monoidal category" is equivalent.</p> <p>However, in some references (ii) seems to have been dropped. I am thinking in particular of Definition 3.1 is this <a href="http://arxiv.org/abs/0804.4688" rel="nofollow">expository paper</a>, and the <a href="http://en.wikipedia.org/w/index.php?title=Braided_monoidal_category&amp;oldid=309075821" rel="nofollow">wikipedia article</a>. The wikipedia article goes further, and suggests that (ii) somehow follows from (i) and the axioms of a monoidal category. So, my questions are.</p> <p>1) Is (ii) needed? That is if we do not impose (ii), does it follow from (i) and the axioms of a monoidal category?</p> <p>2) If (ii) is needed, can someone provide an example demonstrating why? That is, provide an example of a monoidal category $\mathcal{C}$ along with maps $\sigma_{U,V}$ such that (i) holds but (ii) fails. Alternatively, if (ii) is not needed, I'd like a proof (or reference to a proof) that it follows from other axioms.</p> http://mathoverflow.net/questions/5103/solutions-of-the-quantum-yang-baxter-equation/5715#5715 Answer by Peter Tingley for Solutions of the Quantum Yang-Baxter Equation Peter Tingley 2009-11-16T16:46:56Z 2009-11-17T13:23:31Z <p>The form of the Yang Baxter equation you are studying developed to a large extent in the context of solvable lattice modes in statistical mechanics. There the R matrix is used to construct a transition function in a certain statistical system. At least in some cases, the R matrix coming from statistical mechanics agrees with the R matrix coming from a quantum affine algebra. The standard example is the relationship between the six-vertex model" and $U_q(\widehat{sl}_2)$. A good reference is Jimbo and Miwa's book "algebraic analysis of solvable lattice models" (<a href="http://www.ams.org/mathscinet/search/publdoc.html?arg3=&amp;co4=AND&amp;co5=AND&amp;co6=AND&amp;co7=AND&amp;dr=all&amp;pg4=AUCN&amp;pg5=ICN&amp;pg6=TI&amp;pg7=ALLF&amp;pg8=ET&amp;review_format=html&amp;s4=jimbo&amp;s5=miwa&amp;s6=solvable&amp;s7=&amp;s8=All&amp;vfpref=html&amp;yearRangeFirst=&amp;yearRangeSecond=&amp;yrop=eq&amp;r=1&amp;mx-pid=1308712" rel="nofollow">mathscinet review</a>). At least for the R matrix coming from $U_q(\widehat{sl}_2)$, they do give explicit formulas. Unfortunately, I don't know a good online reference. </p> http://mathoverflow.net/questions/5497/what-are-the-qualities-of-a-good-math-teacher/5543#5543 Answer by Peter Tingley for What are the qualities of a good (math) teacher? Peter Tingley 2009-11-14T18:27:23Z 2009-11-14T18:27:23Z <p>I have answered this question with what I hope are some qualities a teacher can learn in order to improve their class. This is as opposed to listing qualities of a truly exceptional math teacher. I wrote this thinking of middle and high school level math, but most of it could be applied at other levels as well. </p> <p>1) A good math teacher should motivate the math and engage the students. Take the example of solving linear equation. One can start by telling students the formal rules for how to manipulate an equation, but I think students will find this very dry, and won't understand why they are doing what they are doing. It becomes and exercise in memorization. Instead, one can start with problems that can be solved with such equation. One can first get students to solve them with other techniques (e.g. guessing and checking or using some sort of graph). After a while one realizes there should be an easier way, which turns out to be solving a linear equation. This way the students understand why the formal math was developed, understand how to apply it, and see how it is related with other ideas (like graphs). Right now you might ask where one can find good problems to use in this way. I think I will ask that as a separate question...</p> <p>2) A good math teacher makes their students do math. I think it is crucial that every student, in every math class, every day, solve some math problems. Some of these should be easy (i.e. just practice solving equations, once they have been introduced), and some should require more creativity. It is of course a mistake to drill students with boring problems until they hate the subject, but it is also a mistake to let them do "interesting" or "discovery based" math all the time, and not make them practice the techniques they discover. </p> <p>3) A good math teacher should convey the beauty of the subject. One of the other answers said infectious enthusiasm" was needed. That would be great, but in reality not all math teachers can be that charismatic. Even without a great deal of charisma, I believe it is possible to show students the wonder of extracting a simple answer from a seemingly difficult question, and the beauty of the tools that help one do this. Often it is enough that students see that their teacher believes this. So in particular, I do not think it is a good idea to say things like "I hated math when I was your age too, but we'll get through this". </p> http://mathoverflow.net/questions/55512/classification-of-generalized-cartan-matrices-gcms Comment by Peter Tingley Peter Tingley 2011-02-15T13:48:25Z 2011-02-15T13:48:25Z Finite Cartan matrices are not all symmetric. Take e.g. $B_2$. They are all symmetrizable. Also, it is quite easy to see that the number of non-affine symmetric Cartan matrices is infinite. http://mathoverflow.net/questions/54926/longest-element-of-weyl-groups Comment by Peter Tingley Peter Tingley 2011-02-10T20:31:37Z 2011-02-10T20:31:37Z Just for the action of the longest element on the root system, it will always act as $-\sigma$ for a diagram automorphism $\sigma$. $\sigma$ always has order 2, and in types $A_n, E_6$ and $D_n$ for $n$ odd it is the unique non-trivial such diagram automorphism. http://mathoverflow.net/questions/51077/modular-forms-and-root-systems Comment by Peter Tingley Peter Tingley 2011-01-04T14:08:30Z 2011-01-04T14:08:30Z Connections between modular forms and affine Lie algebras are discussed in chapters 12 and 13 of Victor Kac's book &quot;Infinite dimensional Lie algebras&quot; http://mathoverflow.net/questions/41922/group-which-resembles-the-free-product-of-a-cyclic-group-of-order-two-and-a-cyc Comment by Peter Tingley Peter Tingley 2010-10-12T18:27:03Z 2010-10-12T18:27:03Z @Richard Are you missing at $=1$&quot; in your example? Otherwise I don't think the example you give is a quotient of the group described in the question. http://mathoverflow.net/questions/41033/realizing-higher-level-fock-spaces/41078#41078 Comment by Peter Tingley Peter Tingley 2010-10-05T01:58:50Z 2010-10-05T01:58:50Z The papers I mentioned deal with generic $q$. I think one should be able to make sense of the construction at $q=1$ though. This is just because the formulas for the actions of $E_i$ and $F_i$ on the standard basis of Fock space make sense at $q=1$. See Theorem 2.1 in Uglov's paper. These seem to make sense at other roots of unity as well...although certainly the structure of the representation would be much much complicated in those cases. http://mathoverflow.net/questions/28143/180-vs-360-twists-in-string-diagrams-for-ribbon-categories/29370#29370 Comment by Peter Tingley Peter Tingley 2010-06-25T15:19:39Z 2010-06-25T15:19:39Z @Peter I'm actually very interested to see the case where the dark side functor is duality worked out in detail. Among the things that confuse me about that case is that, the way Noah and I set things up, the dark side functor and the duality functor don't actually commute, so cannot be equal. See our Comment 4.12. They do commmute up to isomorphism, and one can set things up so that they commute exactly. But then other things get worse. In particular, our proposition 4.18 becomes messy (i.e. the half-twist on a tensor product is hard to describe). Do you see such issues in your work? http://mathoverflow.net/questions/28143/180-vs-360-twists-in-string-diagrams-for-ribbon-categories/29370#29370 Comment by Peter Tingley Peter Tingley 2010-06-24T15:35:05Z 2010-06-24T15:35:05Z Much of this argument is written up in Section 4 of my paper with Noah (cited above). See especially Prop 4.15 which argues that there is a unique functor incorporating the half twist. This argument actually requires almost no assumptions on the half twist (only that it is invertible), but does not imply all the properties one would want (for instance, it does not imply all caps/cups are what you expect). These extra properties are checked later with additional assumptions on t. We work with the quantum group case, but I think the argument is unchanged for more general ribbon categories. http://mathoverflow.net/questions/28143/180-vs-360-twists-in-string-diagrams-for-ribbon-categories/29261#29261 Comment by Peter Tingley Peter Tingley 2010-06-23T19:38:17Z 2010-06-23T19:38:17Z I actually once started working out this story in the case of representations of quantum groups. There $V$ need not necessarily isomorphic to $V^*$, but one can define natural inner products on each irreducible $V_\lambda$. Equivalently, one has a chosen system of isomorphism of vector spaces $V_\lambda \rightarrow V_\lambda^*$ (at least up to scaling). These behave quite well with respect to the representation structure, but are not morphisms). I wanted to use this as the half-twist...but never worked out the details. I've very glad to see that other people are thinking these things through! http://mathoverflow.net/questions/28143/180-vs-360-twists-in-string-diagrams-for-ribbon-categories/29261#29261 Comment by Peter Tingley Peter Tingley 2010-06-23T19:30:30Z 2010-06-23T19:30:30Z Thanks, I was not aware of this work. In Noah's language, perhaps the most obvious guess for a &quot;dark side functor&quot; would be duality. A half twist should then be a natural transformation from the identity functor $C \rightarrow C'$ and the duality functor, and it seems Selinger is exactly writing down the conditions to make this work. http://mathoverflow.net/questions/21875/multiplication-of-extreme-vector Comment by Peter Tingley Peter Tingley 2010-04-21T17:18:52Z 2010-04-21T17:18:52Z I believe this is just the observation that the $w(\lambda)+w(\nu)$ weight space in $V_\lambda \otimes V_\nu$ is 1-dimensional. Since both $e_{w\lambda}\otimes e_{w\nu}$ and $e_{w(\lambda +\nu)}$ (however you want to precisely define that) both lie in this weight space, one is a scalar multiple of the other. To see that this weight space is 1-dimensional, it suffices to notice that the $\lambda+\nu$ weight space is 1-dimensional and to recall that weight-multiplicities are invariant under weyl group reflections. http://mathoverflow.net/questions/16786/is-an-algebraic-bijection-from-a-projective-variety-to-itself-necessarily-an-isom Comment by Peter Tingley Peter Tingley 2010-03-23T22:02:51Z 2010-03-23T22:02:51Z I don't think this counter example is correct. The w be a primitive third root of unity. Then $(x,y)= (-1,1), (-w^2, w), (-w,w^2)$ are all sent to $(0,1)$ by your map. So it is not bijective. http://mathoverflow.net/questions/16786/is-an-algebraic-bijection-from-a-projective-variety-to-itself-necessarily-an-isom/16896#16896 Comment by Peter Tingley Peter Tingley 2010-03-02T20:38:18Z 2010-03-02T20:38:18Z Yes, not requiring irreducibility is important. Frank's argument seems to be more or less the same as the one in Kaliman's paper, but perhaps you are right that their varieties are irreducible.. http://mathoverflow.net/questions/15817/when-is-a-generalized-cartan-matrix-invertible/15846#15846 Comment by Peter Tingley Peter Tingley 2010-02-19T23:49:35Z 2010-02-19T23:49:35Z Something is not right. First, you can certainly have a block matrix where one block is affine and the other is indefinite, and this satisfies the condition of the question. But I think there is a more serious problem as well. Consider for instance the matrix 2,-4,0,0 \\ -4,2,-6,0 \\ 0,-6,2,-4 \\ 0,0,-4,2 I think you will find this has determinant zero, and it is certainly not affine http://mathoverflow.net/questions/13198/does-the-canonical-basis-of-a-tensor-product-of-quantum-group-representations-spa Comment by Peter Tingley Peter Tingley 2010-02-07T16:31:25Z 2010-02-07T16:31:25Z I think part of the confusion here is that there is more then one possible notion of canonical basis for the tensor product. One can consider the tensor product of the canonical basis of each factor. This does not restrict to a basis of each isotypic component. On the other hand there is a notion of a canonical basis of the tensor product itself, which I think does have this property. The distinction is explained for the simplest case in <a href="http://arxiv.org/pdf/math/0511467" rel="nofollow">arxiv.org/pdf/math/0511467</a> section 5.2, but I think it has been defined in greater generality (probably by Lusztig and/or Kashiwara) http://mathoverflow.net/questions/9308/are-the-identity-object-axioms-in-the-definition-of-a-braided-monoidal-category/9320#9320 Comment by Peter Tingley Peter Tingley 2009-12-19T19:10:14Z 2009-12-19T19:10:14Z The condition (ii) above is different. Notice that it only involves one object other than the identity. There are examples of monoidal categories with commutativity constraints that do not satisfy (i) but not (ii), like the coboundary categories I mentioned earlier.