User sammy black - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T11:59:47Z http://mathoverflow.net/feeds/user/813 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104126/knot-theory-without-planar-diagrams/119995#119995 Answer by Sammy Black for Knot theory without planar diagrams? Sammy Black 2013-01-27T07:28:41Z 2013-01-27T07:28:41Z <p>Here's a recent example.</p> <p><a href="http://arxiv.org/pdf/0911.2518v1.pdf" rel="nofollow">http://arxiv.org/pdf/0911.2518v1.pdf</a></p> <p>Adam McDougall constructs a "diagramless" homology theory that ends up being essentially equivalent to Khovanov homology.</p> http://mathoverflow.net/questions/20020/who-thought-that-the-alexander-polynomial-was-the-only-knot-invariant-of-its-kind Who thought that the Alexander polynomial was the only knot invariant of its kind? Sammy Black 2010-04-01T00:36:06Z 2011-06-30T02:35:55Z <p>I apologize that this is vague, but I'm trying to understand a little bit of the historical context in which the zoo of quantum invariants emerged.</p> <p>For some reason, I have in my head the folklore:</p> <p>The discovery in the 80s by Jones of his new knot polynomial was a shock because people thought that the Alexander polynomial was the only knot invariant of its kind (involving a skein relation, taking values in a polynomial ring, ??). Before Jones, there were independent discoveries of invariants that each boiled down to the Alexander polynomial, possibly after some normalization.</p> <p>Is there any truth to this? Where is this written?</p> http://mathoverflow.net/questions/1465/can-we-categorify-the-equation-1-t1-t-t2-1 Can we categorify the equation (1 - t)(1 + t + t^2 + ...) = 1? Sammy Black 2009-10-20T18:21:52Z 2011-02-18T07:10:40Z <p>Polynomials in &#8484;[t] are categorified by considering Euler characteristics of complexes of finite-dimensional graded vector spaces. Now, given a rational function that has a power series expansion with integer coefficients, it seems natural to consider complexes of (locally finite-dimensional) graded vector spaces.</p> <p>Are there nice examples of this in nature?</p> http://mathoverflow.net/questions/4764/does-some-version-of-u-qgl11-have-a-basis-like-lusztigs-basis-for-dotusl Does some version of U_q(gl(1|1)) have a basis like Lusztig's basis for \dot{U(sl_2)}? Sammy Black 2009-11-09T20:07:52Z 2011-01-08T12:22:14Z <p>There's a non-unital algebra $\dot{U}$ formed from $U_q (sl_2)$ by including a system of mutually orthogonal idempotents $1_n$, indexed by the weight lattice. You can think of this as a category with objects $\mathbb{Z}$ if you prefer.</p> <p>Lusztig's basis $\mathbb{\dot{B}}$ for $\dot{U}$ has nice positivity properties: structure coefficients are in $\mathbb{Z}[q,q^{-1}]$.</p> <p>Has anyone tried to write down a similar type of basis for the algebra associated to $U_q (gl_{1|1})$?</p> http://mathoverflow.net/questions/42236/is-a-positive-link-the-closure-of-a-positive-braid Is a positive link the closure of a positive braid? Sammy Black 2010-10-15T00:22:33Z 2010-10-15T05:17:54Z <p>Alexander's Theorem guarantees that every oriented link is the closure of some braid. In other words, the map</p> <p>$$\displaystyle \coprod_n \mathcal B_n\longrightarrow \{\text{ oriented links }\}$$</p> <p>is surjective. One algorithm (I'm actually not sure that it's Alexander's original demonstration) involves choosing a basepoint in the complement of a diagram for the link and applying Reidemeister moves until the resulting diagram winds around the point in a consistent direction, at which point the diagram is manifestly the closure of a link.</p> <blockquote> <p>If we begin with a positive diagram for a link, then do we necessarily obtain a positive braid?</p> </blockquote> http://mathoverflow.net/questions/33724/idempotency-of-the-q-antisymmetrizer/33739#33739 Answer by Sammy Black for Idempotency of the q-antisymmetrizer Sammy Black 2010-07-29T00:15:45Z 2010-07-29T00:15:45Z <p><em>Group Theory: Birdtracks, Lie's, and Exceptional Groups</em> by Predrag Cvitanovic has a graphical way of presenting the idempotents. His book is available online here:</p> <p><a href="http://birdtracks.eu/" rel="nofollow">http://birdtracks.eu/</a></p> http://mathoverflow.net/questions/31036/can-you-characterize-the-group-of-transformations-of-knot-diagrams-which-preserve/31917#31917 Answer by Sammy Black for Can you characterize the group of transformations of knot diagrams which preserve the knot embedding? Sammy Black 2010-07-14T22:48:39Z 2010-07-14T22:48:39Z <p>I remember attending a talk by Barbara Jablonska at Knots in Washington (2009) in which she studied a knot in a geometric, rigid fashion. As the direction of projection varies over $S^2$, she obtained interesting surfaces by looking at the locus of a particular crossing (if memory serves). Here's an abstract of the talk, but I cannot find anything published.</p> <p><a href="http://atlas-conferences.com/c/a/x/q/18.htm" rel="nofollow">http://atlas-conferences.com/c/a/x/q/18.htm</a></p> http://mathoverflow.net/questions/2692/matrices-into-path-algebras Matrices into path algebras Sammy Black 2009-10-26T22:16:44Z 2010-07-07T01:57:56Z <p>I was thinking about quivers recently, and the following idea came to me.</p> <p>Let e<sub>i,j</sub> denote the matrix unit in M<sub>n</sub> for 1 &le; i,j &le; n. Let &Gamma; denote the complete quiver on vertices {1, &hellip;, n}: one directed edge E<sub>i,j</sub> for each ordered pair (i, j), including self-loops E<sub>i,i</sub>.</p> <p>M<sub>n</sub>(k) is then the quotient of the path algebra P&Gamma; by a (rather large) ideal generated by "2-faces" of the simplex: E<sub>i,j</sub>E<sub>k,l</sub> = &delta;<sub>j,k</sub>E<sub>i,l</sub>.</p> <p>In this language, for example, the Borel of upper triangular matrices corresponds to the ordered simplex inside &Gamma;.</p> <ul> <li>Is this correspondence <strong><em>interesting</em></strong>?</li> <li>Can we transport Lie theoretic ideas about gl<sub>n</sub>(k) to the quiver language? Should we?</li> <li>What happens if we quotient by a smaller ideal? Say, only reduce paths of length at least 3 (E<sub>i,j</sub>E<sub>k,l</sub>E<sub>p,q</sub> = &delta;<sub>j,k</sub>&delta;<sub>l,p</sub>E<sub>i,q</sub>).</li> </ul> <p>My apologies in advance for these questions being vague.</p> http://mathoverflow.net/questions/26420/beginner-group-theory-question/26424#26424 Answer by Sammy Black for Beginner group theory question Sammy Black 2010-05-30T05:38:30Z 2010-05-30T05:38:30Z <p>Write $|G| = 2n-1$, so $x^{2n-1} = 1$ for all $x$ by Lagrange's Theorem. Now,</p> <blockquote> <p>$\varphi^n(x) = x^{2n} = x$ for all $x \in G$.</p> </blockquote> <p>In other words, $\varphi^{-1} = \varphi^{n-1}$.</p> http://mathoverflow.net/questions/26420/beginner-group-theory-question/26423#26423 Answer by Sammy Black for Beginner group theory question Sammy Black 2010-05-30T05:37:55Z 2010-05-30T05:37:55Z <p>Write $|G| = 2n-1$, so $x^{2n-1} = 1$ for all $x$ by Lagrange's Theorem. Now,</p> <blockquote> <p>$\varphi^n(x) = x^{2n} = x$ for all $x \in G$.</p> </blockquote> <p>In other words, $\varphi^{-1} = \varphi^{n-1}$.</p> http://mathoverflow.net/questions/22658/why-are-inverse-images-more-important-than-images-in-mathematics/22663#22663 Answer by Sammy Black for Why are inverse images more important than images in mathematics? Sammy Black 2010-04-27T01:19:35Z 2010-04-27T01:19:35Z <p><strong>1</strong> Many important properties of topological spaces are preserved by continuous maps (but not necessarily open maps): connectedness and compactness come to mind immediately. But more importantly, the most familiar, natural maps that we can define are continuous, but not necessarily open: polynomials $\mathbb{R}^n \to \mathbb{R}^m$.</p> <p><strong>2</strong> Inverse image of a subgroup under a homomorphism is a subgroup.</p> <p><strong>3</strong> There is a contravariant functor from the category Set to itself, mapping a set $X$ to its power set $\mathcal{P}(X)$ and sending the morphism $f:X \to Y$ to the inverse image $f^{-1}:\mathcal{P}(Y) \to \mathcal{P}(X)$. A "purely symmetric" function would be a symmetric relation on $X \times Y$, no? Functions are, after all, relations with an extra property that deliberately breaks the symmetry!</p> http://mathoverflow.net/questions/22252/is-there-any-analogs-of-vassiliev-invariants-in-higher-dimensions/22258#22258 Answer by Sammy Black for Is there any analogs of Vassiliev invariants in higher dimensions? Sammy Black 2010-04-22T21:08:41Z 2010-04-22T21:08:41Z <p>Gauss defined the linking number of two closed curves $\gamma_i: S^1 \to \mathbb{R}^3$ with an integral.</p> <blockquote> <p>$L(\gamma_1, \gamma_2) = \frac{1}{4\pi} \oint_{\gamma_1}\oint_{\gamma_2} \frac{\mathbf{r}_1 - \mathbf{r}_2}{|\mathbf{r}_1 - \mathbf{r}_2|^3} \cdot (d\mathbf{r}_1 \times d\mathbf{r}_2)$.</p> </blockquote> <p>In modern terms, this measures the degree of the Gauss map $\Gamma(\gamma_1,\gamma_2): S^1 \times S^1 \to S^2$.</p> <p>See <a href="http://en.wikipedia.org/wiki/Linking_number" rel="nofollow">http://en.wikipedia.org/wiki/Linking_number</a>.</p> http://mathoverflow.net/questions/21424/how-to-draw-knots-with-latex/21431#21431 Answer by Sammy Black for How to draw knots with Latex? Sammy Black 2010-04-15T07:58:00Z 2010-04-15T07:58:00Z <p>Aaron Lauda has a nice description using the package XY-pic <a href="http://www.math.columbia.edu/~lauda/xy/" rel="nofollow">here</a>. There are commands that generate pieces of knots (such as crossings in various orientations), although I prefer just using the \crv "curve" command to make splines.</p> http://mathoverflow.net/questions/19957/how-do-you-approach-your-childs-math-education/19978#19978 Answer by Sammy Black for How do you approach your child's math education? Sammy Black 2010-03-31T17:13:01Z 2010-03-31T21:51:29Z <p>Here are some activities that my son (almost 3 years) has enjoyed. They are all motivated by the idea: <strong>make mathematics visceral</strong> (especially for the young ones).</p> <p>As Kevin says, count, count, count things. Count backwards. Count by twos. Do it while you're moving.</p> <p>Rather than show them the symbols $3 \times 2 = 6$, take 6 bottle caps and arrange them into a rectangle. Can you do the same with 7 bottle caps?</p> <p>Let them play with a nice length of rope. Show them the "trick" of a slip-knot. Do it repeatedly (you've taught them to crochet!)</p> <p>Take off your T-shirt while keeping your sweater on. Or put on a shirt that is upside-down and inside out so that it comes out right. </p> <p>Draw big shapes with chalk on the sidewalk. A perennial request from my son: "Draw it bigger!"</p> <p>When you do get to the stage of learning the strange code called "alphabet," keep it tactile. Cut out big letters with scissors. Recognition of symmetry seems to be a pretty natural phenomenon when you can hold the object in your hands. "What happens when you turn M upside down, flip over the b?"</p> <p>Most importantly, don't push it. If their interest wanders elsewhere, then let it go.</p> http://mathoverflow.net/questions/3299/does-the-super-temperley-lieb-algebra-have-a-z-form Does the super Temperley-Lieb algebra have a Z-form? Sammy Black 2009-10-29T17:46:09Z 2010-03-28T18:34:45Z <p><strong>Background</strong> Let V denote the standard (2-dimensional) module for the Lie algebra sl<sub>2</sub>(C), or equivalently for the universal envelope U = U(sl<sub>2</sub>(C)). The Temperley-Lieb algebra TL<sub>d</sub> is the algebra of intertwiners of the d-fold tensor power of V.</p> <blockquote> <p>TL<sub>d</sub> = End<sub>U</sub>(V&otimes;&hellip;&otimes;V)</p> </blockquote> <p>Now, let the symmetric group, and hence its group algebra CS<sub>d</sub>, act on the right of V&otimes;&hellip;&otimes;V by permuting tensor factors. According to Schur-Weyl duality, V&otimes;&hellip;&otimes;V is a (U,CS<sub>d</sub>)-bimodule, with the image of each algebra inside End<sub>C</sub>(V&otimes;&hellip;&otimes;V) being the centralizer of the other.</p> <p>In other words, TL<sub>d</sub> is a quotient of CS<sub>d</sub>. The kernel is easy to describe. First decompose the group algebra into its Wedderburn components, one matrix algebra for each irrep of S<sub>d</sub>. These are in bijection with partitions of d, which we should picture as Young diagrams. The representation is faithful on any component indexed by a diagram with at most 2 rows and it annihilates all other components.</p> <p>So far, I have deliberately avoided the description of the Temperley-Lieb algebra as a diagram algebra in the sense that Kauffman describes it. Here's the rub: by changing variables in S<sub>d</sub> to u<sub>i</sub> = s<sub>i</sub> + 1, where s<sub>i</sub> = (i i+1), the structure coefficients in TL<sub>d</sub> are all integers so that one can define a &#8484;-form TL<sub>d</sub>(&#8484;) by these formulas.</p> <blockquote> <p>TL<sub>d</sub> = C &otimes; TL<sub>d</sub>(&#8484;)</p> </blockquote> <p>As product of matrix algebras (as in the Wedderburn decomposition), TL<sub>d</sub> has a &#8484;-form, as well: namely, matrices of the same dimensions over &#8484;. These two rings are very different, the latter being rather trivial from the point of view of knot theory. They only become isomorphic after a base change to C.</p> <p><hr /></p> <p>There is a super-analog of this whole story. Let U = U(gl<sub>1|1</sub>(C)), let V be the standard (1|1)-dimensional module, and let the symmetric group act by signed permutations (when two odd vectors cross, a sign pops up). An analogous Schur-Weyl duality statement holds, and so, by analogy, I call the algebra of intertwiners the super-Temperley-Lieb algebra, or STL<sub>d</sub>.</p> <p>Over the complex numbers, STL<sub>d</sub> is a product of matrix algebras corresponding to the irreps of S<sub>d</sub> indexed by hook partitions. Young diagrams are confined to one row and one column (super-row!). In that sense, STL<sub>d</sub> is understood. However, idempotents involved in projecting onto these Wedderburn components are nasty things that cannot be defined over &#8484;</p> <p><hr /></p> <p><strong>Question 1:</strong> Does STL<sub>d</sub> have a &#8484;-form that is compatible with the standard basis for CS<sub>d</sub>?</p> <p><strong>Question 2:</strong> I am pessimistic about Q1; hence, the follow up: why not? I suspect that this has something to do with cellularity.</p> <p><strong>Question 3:</strong> I care about q-deformations of everything mentioned: U<sub>q</sub> and the Hecke algebra, respectively. What about here? I am looking for a presentation of STL<sub>d,q</sub> defined over &#8484;[q,q<sup>-1</sup>].</p> http://mathoverflow.net/questions/19589/do-jones-wenzl-idempotents-lift-to-anything-interesting-in-the-hecke-algebra Do Jones-Wenzl idempotents lift to anything interesting in the Hecke algebra? Sammy Black 2010-03-28T05:20:45Z 2010-03-28T07:07:35Z <h2>Background</h2> <p>Inside the Temperley-Lieb algebra $TL_n$ (with loop value $\delta=-[2]$ and standard generators $e_1,\ldots,e_{n-1}$), the Jones-Wenzl idempotent is the unique non-zero element $f^{(n)}$ satisfying $$f^{(n)}f^{(n)} = f^{(n)} \quad \textrm{and} \quad e_i\;f^{(n)} = 0 = f^{(n)}e_i \quad \textrm{for each } i.$$ Consider the Iwahori-Hecke algebra $\mathcal{H}_n$, $n\ge3$, normalized so that $(T_i-q)(T_i+q^{-1})=0$, where $q$ is generic. Let $\mathcal{I}$ be the two-sided cellular ideal generated by canonical basis element $$C_{121} = T_1T_2T_1-qT_1T_2-qT_2T_1+q^2T_1+q^2T_2-q^3.$$ The assignment $\mathcal{H}_n \rightarrow TL_n$ given by $T_i \mapsto e_i + q$ is a surjective $\mathbb{C}(q)$-algebra homomorphism with kernel $\mathcal{I}$.</p> <p>We can lift the generators $e_i$ in $TL_n$ to the Kazhdan-Lusztig elements $C_i=T_i-q \in \mathcal{H}_n$. In fact, we have $C_{121} = C_1C_2C_1 - C_1$, hence the relation down below. Rescaling a bit, $E=-\frac{1}{[3]!}C_{121}$ is an idempotent, corresponding to the partition $(1,1,1)$. Actually, all of the primitive idempotents in $\mathcal{H}_n$ that correspond to Young diagrams with more than two rows live in the ideal $\mathcal{I}$.</p> <p>Now, any preimage of $f^{(n)}$ in the Hecke algebra (call it $F^{(n)}$) satisfies $$F^{(n)}F^{(n)} \equiv F^{(n)} \quad \textrm{and} \quad C_iF^{(n)} \equiv 0 \equiv F^{(n)}C_i \quad (\operatorname{mod} \mathcal{I})$$</p> <h2>Question</h2> <blockquote> <p>Can we choose $F^{(n)}$ to be an idempotent in $\mathcal{H}_n$?</p> </blockquote> <p>When $n=2$, the map is an isomorphism and we have no choice. $$F^{(2)} = \frac{1}{[2]}(T_1+q^{-1}),$$ which projects onto the $q$-eigenspace for $T_1$. In other words, it is the idempotent corresponding to the partition $(2)$.</p> http://mathoverflow.net/questions/13899/signed-and-unsigned-hecke-algebra-canonical-basis/19376#19376 Answer by Sammy Black for Signed and unsigned Hecke algebra canonical basis Sammy Black 2010-03-26T01:20:59Z 2010-03-26T01:20:59Z <p>You probably know all of this already, but here goes...</p> <p>Write $C'_w = T_w + \sum_{x &lt; w} p_{x,w} T_x$ where $p_{x,w} \in u\mathbb{Z}[u]$. Now, the other basis can be defined by applying the involutive automorphism $b: \mathcal{H}_n \to \mathcal{H}_n$, given by $b(T_w)=T_w$ and $b(u)=-u^{-1}$.</p> <blockquote> <p>Define $C_w := b(C'_w)$.</p> </blockquote> <p>Since, $b$ commutes with the bar involution, this basis is bar invariant as well.</p> <p>Explicitly, $C_w = T_w + \sum_{x &lt; w} (-1)^{\ell(w)+\ell(x)} \bar p_{x,w} T_x$.</p> <p>So $C_w = \bar{P}^{-1} P C'_w$ which seems hard to compute in general.</p> http://mathoverflow.net/questions/18522/how-should-i-visualise-rpn/18532#18532 Answer by Sammy Black for How should I visualise RP^n? Sammy Black 2010-03-17T23:28:27Z 2010-03-18T00:05:58Z <p>For $\mathbb{R}P^2$, I like <a href="http://en.wikipedia.org/wiki/Boy%27s_surface" rel="nofollow">Boy's surface</a>, which is a particularly symmetric immersion of the projective plane into $\mathbb{R}^3$.</p> <p>Also, see <a href="http://www.physics.unc.edu/~amellnik/surfaces/boy_jv.html" rel="nofollow">this</a> Java-based model.</p> <p>You can build a piecewise-linear version of one of these out of paper. If you cut out a disk-shaped window (to see inside to the triple point), what you have is a model of the Mobius band for which the boundary circle is really a round circle!</p> <p>Of course, this doesn't really help with all of the rest of them $(n \ge 3)$!</p> http://mathoverflow.net/questions/17197/how-does-this-relationship-between-the-catalan-numbers-and-su2-generalize/17219#17219 Answer by Sammy Black for How does this relationship between the Catalan numbers and SU(2) generalize? Sammy Black 2010-03-05T20:47:07Z 2010-03-05T20:47:07Z <p>Caveat lector: This does not address your questions directly, but I think that it's interesting and somewhat relevant.</p> <p>The $n$th Catalan number counts the number of walks of length $2n$, beginning and ending at the origin, in the positive cone of the root lattice of type $A_1$. This counts the dimension of the invariant subspace of endomorphisms of $2n$-fold tensor power of the standard irrep $V$, relating back to Scott's answer.</p> <p>This phenomenon generalizes to other root systems.</p> http://mathoverflow.net/questions/15727/group-cohomology-vs-topological-cohomology-in-the-case-of-non-trivial-action/15732#15732 Answer by Sammy Black for Group cohomology vs. topological cohomology in the case of non-trivial action Sammy Black 2010-02-18T17:26:54Z 2010-02-18T17:26:54Z <p>You have to introduce the language of sheaves. Then, the coefficients can be described succinctly as a locally constant sheaf.</p> <p>Alternatively, you can view the coefficient system as a bundle over your space $BG$ with $A$ fibers. Locally, the coefficients behave as if the $G$-action were trivial, but if one considers a loop that represents a nontrivial element of $\pi_1(BG) = G$, monodromy comes into play.</p> http://mathoverflow.net/questions/11305/constructing-the-hecke-algebra-from-the-burau-representation/12986#12986 Answer by Sammy Black for Constructing the Hecke-Algebra from the Burau representation Sammy Black 2010-01-26T00:02:31Z 2010-01-26T06:54:39Z <p><a href="http://sbseminar.wordpress.com/2009/04/13/how-to-get-an-algebra-from-a-knot-invariant/" rel="nofollow">Here</a>'s a rather nice description of how Hecke algebras come up whenever one has a skein relation explained by Ben Webster.</p> <p>Looking at representations of the braid group that factor through the Hecke algebra, one can construct the two-variable HOMFLY-PT polynomial.</p> <p>See <em>Hecke algebra representations of braid groups and link polynomials</em> by V.F.R. Jones. </p> <p>In this paper, Jones shows how the Burau representation arises by considering only certain irreps of $\mathcal{H}_n$ corresponding to hook-partitions, which can be identified (up to a sign) with exterior powers of the standard irrep.</p> <p>The HOMFLY-PT invariant specializes to the Alexander, Jones, etc. Each specialization corresponds to factoring through a further quotient of the Hecke algebra. For Jones, this is the Temperley-Lieb algebra.</p> <p>For Alexander, the quotient is more mysterious, but has to do with the Lie superalgebra $\mathfrak{gl}(1|1)$. I asked a question about this <a href="http://mathoverflow.net/questions/3299/does-the-super-temperley-lieb-algebra-have-a-z-form" rel="nofollow">here</a> and there was more discussed <a href="http://mathoverflow.net/questions/3366/is-there-a-version-of-temperley-lieb-using-sl3-rather-than-sl2" rel="nofollow">here</a>.</p> http://mathoverflow.net/questions/5144/is-murasugis-conjecture-still-open Is Murasugi's conjecture still open? Sammy Black 2009-11-12T03:24:44Z 2009-11-24T07:54:01Z <p>Normalize the Alexander polynomial (in $t$) so that the positive and negative exponents are balanced. For example in the Conway normalization, make the substitution $z = t^{1/2} - t^{-1/2}$. The trefoil gives $t^{-1} - 1 + t$.</p> <p>Suppose that $K$ is an alternating knot.</p> <p>CONJECTURE: The sequence of absolute values of the coefficients is unimodal. Specifically, </p> <blockquote> <p>if <img src="http://latex.mathoverflow.net/png?%24%5CDelta%20%5FK%28t%29%20%3D%20%5Csum%5Fi%20a%5Fi%20t%5Ei%24" alt="$\Delta \sb K(t) = \sum\sb i a\sb i t^i$" title="" />, then <img src="http://latex.mathoverflow.net/png?%24%7Ca%5F0%7C%20%5Cge%20%7Ca%5F1%7C%20%5Cge%20%7Ca%5F2%7C%20%5Cge%20%5Ccdots%24" alt="$|a\sb 0| \ge |a\sb 1| \ge |a\sb 2| \ge \cdots$" title="" /></p> </blockquote> <p>This is a conjecture due to Murasugi, I believe. Where is it written? Has this been proved (or disproved!)?</p> http://mathoverflow.net/questions/6394/lecture-notes-on-representations-of-finite-groups/6401#6401 Answer by Sammy Black for Lecture notes on representations of finite groups Sammy Black 2009-11-21T20:19:18Z 2009-11-21T20:19:18Z <p>The first section of <em>Representation Theory</em> by Fulton and Harris is a great introduction to representations of finite groups (about a quarter of the book, if I remember correctly). There are lots of examples and exercises. The rest of the book is devoted to Lie theory.</p> http://mathoverflow.net/questions/5372/dimension-leaps/5402#5402 Answer by Sammy Black for Dimension Leaps Sammy Black 2009-11-13T17:58:41Z 2009-11-13T17:58:41Z <p>The Euclidean ball takes up the most space in dimension 5.</p> <p>$V = \frac {8 \pi^2} {15} R^5 \approx 5.26\ldots R^5$</p> http://mathoverflow.net/questions/3366/is-there-a-version-of-temperley-lieb-using-sl3-rather-than-sl2/3413#3413 Answer by Sammy Black for Is there a version of Temperley-Lieb using sl(3) rather than sl(2)? Sammy Black 2009-10-30T06:26:54Z 2009-10-30T06:26:54Z <p>See the paper </p> <blockquote> <p>Schur–Weyl Reciprocity between the Quantum Superalgebra and the Iwahori–Hecke Algebra </p> </blockquote> <p>by Hideo Mitsuhashi.</p> http://mathoverflow.net/questions/3278/whats-a-reasonable-category-that-is-not-locally-small/3281#3281 Answer by Sammy Black for What's a reasonable category that is not locally small? Sammy Black 2009-10-29T16:20:58Z 2009-10-29T16:20:58Z <p>The category <strong>Cat</strong>, whose objects are categories and whose morphisms between two categories consist of functors.</p> <p>Whether this is "reasonable" is up to you to judge.</p> http://mathoverflow.net/questions/3271/variant-of-binomial-coefficients/3276#3276 Answer by Sammy Black for Variant of binomial coefficients Sammy Black 2009-10-29T15:52:44Z 2009-10-29T15:52:44Z <p>Let B<sub>k</sub>(z) denote the usual binomial coefficient, suitably generalized to allow z to be a formal variable. Assuming that you intend to have the same number of factors in the numerator and denominator in your definition of F<sub>(a,k)</sub>(z), then via the substitution z = aw,</p> <p>F<sub>(a,ab)</sub>(z) = B<sub>b</sub>(w).</p> <p>If 1/a is in your ring (still assuming that a divides k), then</p> <p>F<sub>(a,k)</sub>(z) = B<sub>k/a</sub>(z/a).</p> http://mathoverflow.net/questions/3134/whats-your-favorite-equation-formula-identity-or-inequality/3149#3149 Answer by Sammy Black for What's your favorite equation, formula, identity or inequality? Sammy Black 2009-10-28T21:43:38Z 2009-10-28T21:43:38Z <p>I think that Weyl's character formula is pretty awesome! It's a generating function for the dimensions of the weight spaces in a finite dimensional irreducible highest weight module of a semisimple Lie algebra.</p> <p><img src="http://upload.wikimedia.org/math/8/d/f/8df922b7028262e8ec0910790394127a.png" alt="alt text" /></p> http://mathoverflow.net/questions/2650/conjugacy-classes-in-finite-groups-that-remain-conjugacy-classes-when-restricted/2655#2655 Answer by Sammy Black for Conjugacy classes in finite groups that remain conjugacy classes when restricted to proper subgroups Sammy Black 2009-10-26T18:22:00Z 2009-10-26T18:22:00Z <p>Regarding <strong>Question 1</strong>, let G = S<sub>4</sub>, H = A<sub>4</sub>, and c = [(12)(34)]. This class does not split, and c &cong; C<sub>2</sub>&times;C<sub>2</sub>, which is not cyclic. I'm not sure if this is an example along the lines of your "semidirect product of N by Z/<sub>2<sup>k</sup>Z</sub>" since I forget which factor you expect to be the normal subgroup. In the example above, c is the normal subgroup, and C<sub>3</sub> acts by inner automorphisms of c to produce A<sub>4</sub>.</p> http://mathoverflow.net/questions/2065/computing-a-factor-group/2139#2139 Answer by Sammy Black for Computing a Factor Group Sammy Black 2009-10-23T18:07:04Z 2009-10-23T18:07:04Z <p>Since Z is a PID, it has projective dimension 1. Actually, a submodule of a free module is <em>free</em>! There is an obvious resolution of a quotient M of Z^n by the span of m vectors:</p> <blockquote> <p>0 --> Z^m --> Z^n --> M</p> </blockquote> <p>The (n&times;m)-matrix for the middle map has columns that are the given vectors. Reduction of this matrix to its Smith normal form (think row and column operations and some reductions using GCDs), one can read off the elementary divisors.</p> <p>Your first example has first column (3,6,9)<sup>T</sup> and zeroes in the other columns. Using row operations, we get a diagonal matrix with diagonal (3,0,0). These are the elementary divisors. Your group is Z/3&oplus;Z&oplus;Z.</p> http://mathoverflow.net/questions/127318/why-do-we-study-ideals-generated-by-subsets-and-not-rings-generated-by-subsets Comment by Sammy Black Sammy Black 2013-04-12T06:23:44Z 2013-04-12T06:23:44Z You can talk about a subring generated by a subset. (By the way there are several good answers to a similar question over at <a href="http://math.stackexchange.com/questions/42085/why-is-ideal-more-important-than-subring" rel="nofollow" title="why is ideal more important than subring">math.stackexchange.com/questions/42085/&hellip;</a>). http://mathoverflow.net/questions/125701/hey-guys-and-gals-new-guy-looking-for-a-book-or-two-any-helpz Comment by Sammy Black Sammy Black 2013-03-27T08:44:34Z 2013-03-27T08:44:34Z You probably ought to ask this question over at math.stackexchange.com. For what it's worth, you might enjoy <a href="http://books.google.com/books/about/The_Shape_of_Space.html?id=Lurp6nB4LtQC" rel="nofollow">books.google.com/books/about/&hellip;</a> http://mathoverflow.net/questions/65895/2-tqft-are-to-frobenius-algebras-as-are-to-hopf-algebras/65919#65919 Comment by Sammy Black Sammy Black 2011-05-25T19:24:56Z 2011-05-25T19:24:56Z But perhaps there are ways to add markings to the surfaces or otherwise limit the possible isotopies so that we don't expect those two maps to be equal? http://mathoverflow.net/questions/62197/representation-theory-of-s-n/62296#62296 Comment by Sammy Black Sammy Black 2011-04-19T21:38:13Z 2011-04-19T21:38:13Z I second this choice. http://mathoverflow.net/questions/55647/relationship-between-different-quantum-deformations/55653#55653 Comment by Sammy Black Sammy Black 2011-02-16T19:31:00Z 2011-02-16T19:31:00Z By the way, Peter, you ought to vote the answer up! http://mathoverflow.net/questions/42236/is-a-positive-link-the-closure-of-a-positive-braid/42253#42253 Comment by Sammy Black Sammy Black 2010-10-15T18:55:29Z 2010-10-15T18:55:29Z Thank you. Yamada's paper is indeed interesting. http://mathoverflow.net/questions/27490/geometric-interpretation-of-the-fundamental-groupoid Comment by Sammy Black Sammy Black 2010-06-08T17:12:55Z 2010-06-08T17:12:55Z The fundamental groupoid is &quot;how paths behave in the space,&quot; as in your description of the fundamental group. http://mathoverflow.net/questions/26719/unit-triangles-with-vertices-on-circles Comment by Sammy Black Sammy Black 2010-06-01T15:27:21Z 2010-06-01T15:27:21Z What is a &quot;unit regular triangle&quot;? http://mathoverflow.net/questions/26195/generators-for-hecke-algebra-quotients Comment by Sammy Black Sammy Black 2010-05-28T06:30:22Z 2010-05-28T06:30:22Z Can you define the $(k,r)$-quotient? Also, are you really asking for generators? What's wrong with considering the image of the generators of the Hecke algebra under the canonical map? http://mathoverflow.net/questions/26035/evaluation-of-the-following-series-s-1-2-times3-1-5-times6-1-7-tim Comment by Sammy Black Sammy Black 2010-05-26T17:14:21Z 2010-05-26T17:14:21Z Oh! Now I understand the question. Sorry. http://mathoverflow.net/questions/26035/evaluation-of-the-following-series-s-1-2-times3-1-5-times6-1-7-tim Comment by Sammy Black Sammy Black 2010-05-26T16:29:44Z 2010-05-26T16:29:44Z If you intend the sum as in Nate's comment above, then this is a standard calculus problem. The magic phrase is &quot;telescoping series.&quot; http://mathoverflow.net/questions/25747/is-there-a-non-trivial-knot-with-trivial-homfly-polynomial/25760#25760 Comment by Sammy Black Sammy Black 2010-05-24T14:48:58Z 2010-05-24T14:48:58Z You might be interested in these results of Matthew Hedden and Liam Watson about Khovanov homology and its colored variants detecting the unknot: <a href="http://front.math.ucdavis.edu/0805.4423" rel="nofollow">front.math.ucdavis.edu/0805.4423</a> and <a href="http://front.math.ucdavis.edu/0805.4418" rel="nofollow">front.math.ucdavis.edu/0805.4418</a> http://mathoverflow.net/questions/24453/how-does-singular-homology-h-n-capture-the-number-of-n-dimensional-holes-in-a-s/24468#24468 Comment by Sammy Black Sammy Black 2010-05-13T04:22:58Z 2010-05-13T04:22:58Z It seems that you are describing the cobordism functor $X \mapsto \Omega_n(X)$ that factors the Hurewicz map: $\pi_n \to \Omega_n \to H_n$. These maps are neither injective nor surjective in general. In particular, not all homology classes are represented by maps from manifolds. http://mathoverflow.net/questions/24122/subgroups-of-direct-product-of-groups Comment by Sammy Black Sammy Black 2010-05-10T16:42:47Z 2010-05-10T16:42:47Z I think that your second surjection in paragraph two should be $\pi_2:U_2 \to G$. http://mathoverflow.net/questions/21682/is-a-polynomial-group-law-on-mathbbrn-automatically-nilpotent Comment by Sammy Black Sammy Black 2010-04-17T19:13:23Z 2010-04-17T19:13:23Z Torsten, make it an answer (not just a comment).