User hauke reddmann - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T03:49:06Z http://mathoverflow.net/feeds/user/11504 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/133368/quantum-9j-symbols Quantum 9j symbols? Hauke Reddmann 2013-06-11T11:35:22Z 2013-06-12T13:10:15Z <p>A formula for (SU2) quantum 6j symbols exists. A formula expressing ordinary (q=1) 9j symbols in terms of 6j symbols is long known. Unfortunately, combining both (I tried it myself) got tricky - the associated graph K3,3 is nonplanar, at least one knot-type crossing is needed and first of all, this ruins the symmetry.</p> <p>Can I find the quantum analogon of the standard sum over the product of three 6j symbols in the literature (or can someone post it here)?</p> http://mathoverflow.net/questions/130830/a-funny-property-of-e8-family-clebsch-gordan-series A funny property of E8-family Clebsch-Gordan series Hauke Reddmann 2013-05-16T12:58:10Z 2013-05-16T12:58:10Z <p>Take any Lie group irrep $R$ with $R\otimes{R}=R_1+R_2+R_3+R_4+1$. (Of course you can take the defining dimension of any member of the $E_8$ family (in that case $R_1=R$), but, say, $1\lambda_8$ of $D_8$ works as well. We take this as actual example: $128^2=6435+8008+1820+120+1$.<br> And now for my magic trick. Abracadabra! $3*(128+2)*128*6435*8008*1820*120=749548800^2$.<br> I see doubt in the auditorium, let's repeat with $E_8$: $3*(248+2)*248*248*3785*27000*30380=12108600000^2$.<br> For the members of the $E_8$ family you can easily check the truth of the statement directly (that the RHS is a square), but it holds as well for all the others, like $1\lambda_4(A_7)$ etc.<br> Since I used magic to derive it, that's cheating - do you have a math proof instead (via the Vogel plane or suchlike)? I can't believe this neat property went unnoticed yet...</p> http://mathoverflow.net/questions/130267/invariant-tensors-that-are-neither-symmetric-nor-antisymmetric Invariant Tensors that are neither symmetric nor antisymmetric Hauke Reddmann 2013-05-10T14:25:37Z 2013-05-10T14:25:37Z <p>I recall only fully symmetric or antisymmetric tensors occuring in knot/graph theory (but then, what do I know besides "Birdtracks" ? :-) OK, so here is (one half of) a E-symmetrized tensor (where E is the two-dimensional irrep of $S_3$) derived from a "random" rank 3 tensor $T_{abc}$:<br> $E_{abc}=(T_{bca}-T_{cab})/2+(T_{acb}-T_{bac}/2-T_{cba}/2)/\sqrt3$<br> You can do useful stuff with these things (e.g. the defining equation of the $E_7$ family can be written as $E_{abcd}=0$ ...or at least that's what I computed) but is the existence of an <em>invariant</em> $E$ tensor compatible with Lie groups at all? (Handwave argument: cross legs, turn page over, cross legs again, is identity, character of any permutation must be $\pm1$, done. But I'm thinking in graphics again!) If no, how, pray tell, does one <em>compute</em> with an E pair, e.g. multiply two E tensors? Or if there is an E tensor pair, there also should be an R matrix pair of this symmetry (which sounds completely silly). </p> http://mathoverflow.net/questions/121017/diophantine-question Diophantine question Hauke Reddmann 2013-02-06T20:24:27Z 2013-02-08T22:36:08Z <p>This came up when I did a brand-new (or maybe it's just "birdtracks" in disguise :-) graph-based construction of the E8 family. x,y,z are dimensions and thus integer. x&lt;0 actually doesn't hurt. (For $y=x,z=(3 (-2 + o) o)/(10 + o)$ the standard E8 setup results, o must now divide 360 etc. pp.) So here is the equation:</p> <p>$3 x (2 + x) (-2 + x + x^2 - 2 y) y (-x + x^2 - 2 z) z=q^2$ </p> <p>With rational y or z I wouldn't pester MO - I'd solve it on the spot. But here some nasty division properties are involved, and I'm lousy in number theory. Is there a finite, easy describable solution list?</p> http://mathoverflow.net/questions/118634/symmetry-analysis-of-an-impossible-symmetry Symmetry analysis of an impossible symmetry Hauke Reddmann 2013-01-11T15:01:21Z 2013-01-11T15:01:21Z <p>I have a practical problem in chemistry. Consider a molecule MCp3 (Cp=C5H5) with the centroids of the Cp ring lying on an 60-60-60 triangle. The maximal symmetry you can get is C3h, because of the fivefold-axis Cp rings. (Orient one CH group into the centroid plane.) In praxis, though, the Cp rings rotate freely and the effective symmetry is D3h.</p> <p>Now there are the usual tools of vibrational analysis, giving for C3h that the normal vibrations are 16A'+14A''+16E'+12E'' (modulo my usual typos :-). I want to know how the 16A'+14A'' split into A1'+A2'+A1''+A2'', the irreps of the fictive D3h symmetry. This is impossible by normal means - e.g. the first step of a standard tool is to apply a symmetry operation on the atom set and count the trace of the operation matrix. But the symmetry operations aren't...</p> <p>I am quite sure that it is possible to work with half-integer characters and other abominations and in the end everything crosscancels and gives an answer that makes sense within the experimental data (which IS compatible with D3h).</p> <p>So my question: can you do representation analysis for objects with pseudo symmetry? (In this special case, it might be even doable this way: Compute the result for 4- and 6-rings, where D3h is a valid symmetry, and do the mean. :-)</p> http://mathoverflow.net/questions/115495/matrix-minimax-problem Matrix Minimax problem Hauke Reddmann 2012-12-05T12:11:50Z 2012-12-07T12:29:57Z <p>I have the equation $\Sigma_k(M_k{p_k})V=EV$, where the $M_k$ are n*n real Hermitian matrices, $V$ is a n*n eigenvector matrix, $E$ a dim-n energy eigenvector and the $p_k$ scalar parameters. The $M_k$ are known exactly, and I get m energies $e$ from experiment, where (usually) $k\lt{m}\lt{n}$.<br> I now want to find the set of $p_k$ such that $\Sigma_m(e_m-E_m)^2$ is minimized (where $E_m$ are the "theoretical" eigenvalues following from the above equation, and $e_m$ my experimental ones - knowing beforehand which $E_m$ and $e_m$ "belong" together is another can of worms).<br> Since 30 years I solve this problem with INVEIX, a FORTRAN kludge which starts from an initial set $p_k$ and converges to a miminum by a steepest gradient algorithm. Kludge or not, it works splendid since 30 years :-)<br> Still, I'm curious: Can this minimax problem be solved more analytically, maybe even as closed form? I'm very reluctant to differentiate this matrix equation after the $p_k$...</p> http://mathoverflow.net/questions/111893/cyclotomic-polynomial-question Cyclotomic polynomial question Hauke Reddmann 2012-11-09T12:14:58Z 2012-11-10T05:54:25Z <p>To avoid case distinction overload, I also call (say) $Z^3-Z^4$ cyclotomic. Just divide out the $Z=0$ solutions in the following if they offend.<br> In the following, all exponents are assumed to be positive integers.<br> Assume that $P=Z^a+Z^b+Z^c-Z^d-Z^e-Z^f$ is cyclotomic. The "standard" form (since $Z-1$ factors out immediately) would be something like $P'=Z^A(Z-1)(1\pm{Z}^B+Z^{2B})$. But note that, say, $P''=(Z-1)(1+Z^3+Z^4+Z^7+Z^8+Z^9+Z^{10})$ telescopes to the same form as $P$. Surely, $P''$ is not cyclotomic, but it was just a random example anyway :-) So: Is there a cyclotomic $P$ than can NOT be written in the form $P'$? OR even some $1\pm{Z}+Z^n, n>2$ that is cyclotomic (although to this my instinct says, no way - the proof is probably an one-liner in complex analysis...).</p> http://mathoverflow.net/questions/110646/links-with-same-jones-polynomial Links with same Jones polynomial Hauke Reddmann 2012-10-25T11:12:06Z 2012-10-27T04:00:50Z <p>Is there anything known about when two links have the same Jones polynomial (beyond a calculated list of small actual examples)? The first thing <em>I</em> would try is to compute the (formal - you would have something*split-horizontal+something*split-vertical) Jones polynomial of (4-)tangles and look <em>there</em> for two tangles with same Jones polynomial. Any instance would then generate an infinite example family. The snag would be, of course, that I'd first need a tangle table - my own goes to meagre 6 crossings. Aaaand the computation can't be automated that good. Still - any takers?</p> http://mathoverflow.net/questions/109807/asymptotic-length-of-clebsch-gordan-series Asymptotic length of Clebsch-Gordan series Hauke Reddmann 2012-10-16T12:09:32Z 2012-10-16T12:09:32Z <p>For $A_1$ (or should I say $SU_2$, since I use the familiar J) it's trivial: $J\bigotimes{J}$ has $2J+1$ terms in the CG expansion. It's still trivial with $J_1\bigotimes{J_2}$ but you immediately see increasing the larger, say $J_1$, of the two has no effect - the length stays $2J_2+1$.<br> I played a bit with $(m,m)\bigotimes(m,m)$ of $A_2$, and the CG length seems to increase with $m^2$. Since I never have a clue, but always a hypothesis :-), I conclude that the length of the CG series of $R_1\bigotimes{R_2}$, $R_1,R_2$ being irreps of the group $something_n$, scales like $m_1*...*m_n$ if the "smaller" irrep has weights $(m_1,...,m_n)$.<br> Is this correct (when properly formalized)?</p> http://mathoverflow.net/questions/109284/things-you-can-do-with-the-self-writhe Things you can do with the self-writhe Hauke Reddmann 2012-10-10T09:45:27Z 2012-10-10T10:13:35Z <p>I hope "self-writhe" is the established word. (0 for link-crossing, otherwise identical to writhe +1 or -1) I bet the following is known: Take some crossing of a link with self-writhe $w_a$. Flip it to get a link with $w_b$, call their arithmetic mean $w_{\times}$. Orient the crossing to overpass, split horizontally and vertically, respectively, to get links with self-writhe $w_-,w_|$. The three numbers are linear dependent: $w_1-w_2=w_2-w_3$ (where ${\times,|,-}={1,2,3}$ but which is which depends on the self-writhe of the crossing itself. It's nicely symmetric but I'm too idle to actually list the three subcases :-) Thus I defined $J=w_1-w_2$ (again with proper numbering, and a factor) to be the "angular momentum" of a crossing. E.g. Hopf link $J=-1/2$, positive trefoil $J=+1$. You can do neat things with it: R1 crossings have $J=0$, and R2 pairs $+J,-J$...too bad that one of the three crossings (the one that would make the R3 move pic alternating when flipped) of a R3 move changes J "randomly" or you could simply sum over all J of a link to get an invariant. Blast. (Still, I should go check now if there is a connection to the Thurston-Bennequin number!)<br> Is there some use of the self-writhe for knot polynomials (beyond the Kauffman bracket) ? (As usual, paper refs are welcome.)</p> http://mathoverflow.net/questions/108823/thurston-bennequin-number-vs-checkerboard-coloring-difference Thurston-Bennequin number vs. checkerboard coloring difference Hauke Reddmann 2012-10-04T14:40:26Z 2012-10-04T19:25:22Z <p>For an <em>alternating</em> knot K, checkerboard-color the knot (if this is a lousy ASCII crossing: %, white goes to the left/right and black to top/bottom). Assume no surplus Reidemeister 1 kinks exist (K has minimal crossing number =C). Call black-white areas D and the writhe W. Then for the Thurston-Bennequin numbers of K and mirror(K) (call them X and Y) the following equations hold (modulo sign error and typo :-):<br> $C=-X-Y-2;D+2*W=Y-X$<br> Surely this is known?!<br> The tricky part comes with nonalternating knots. I played around with minimal crossing number representants of knots, and with proper tweaking the second equation still seems to hold, while for the first, the difference somehow seems to be connected with Stasiaks "natural order of knots". I could conjecture a lot :-) but what is actually known about generalizing these equations to nonalternating knots? Oh, and does somebody have the T-B N for small links? E.g. for the link 4_2 I would assume them to be -5 and -1.</p> http://mathoverflow.net/questions/108253/schurs-di-lemma-finite-and-lie-groups-different Schur's Di-Lemma: finite and Lie groups different? Hauke Reddmann 2012-09-27T15:00:05Z 2012-09-27T15:00:05Z <p>For a finite group it's nothing special if two one-dimensional irreps pop up in a product, e.g. for $C_{3v}$ symmetry, $E\bigotimes{E}=A_1\bigoplus{A_2}\bigoplus{E}$ or in dimensions, $2*2=1+1+2$. Schur's Lemma merely forces that only one $A_1$ can be in the product.<br> Now I never saw a Lie group irrep with dimension $1$ which is <em>not</em> the trivial irrep. Have I just not looked around far enough? :-) (Maybe e.g. beyond semisimple ones?)<br> (Background: Playing around with tensors for 3-nodes and crossings, I found a six-dimensional irrep which gives an invariant for tangled graphs, the Clebsch-Gordan series being $6*6=1+1+6+8+8+12$. The second $1$ is kind of an "antisymmetric one", like $A_2$. In every respect (I could check), this invariant behaves like the invariants coming from Lie groups.) </p> http://mathoverflow.net/questions/107556/generalizing-the-reshitikhin-turaev-construction-possible Generalizing the Reshitikhin-Turaev construction possible? Hauke Reddmann 2012-09-19T12:30:21Z 2012-09-19T12:46:28Z <p>OK, I have to ask a dumb question again: Where do Lie groups enter in the construction of the Reshitikhin-Turaev invariant? The parts of the proof I understand are that 6j symbols take care of themselves. So why not define 6j symbols axiomatically (Biedenharn-Elliott plus symmetry plus function value at a 0 argument should suffice)? You also need an integer-valued triangle function f(a,b,c) giving the number of "irrep" c in the tensor product of a and b for a start (to construct an infinite version of a fusion category, or whatchamacallit), so {abc|def}=0 if f(a,b,c)=0. Not even talking of multiplicity hell. But these only look like "technical" difficulties to me, but not impossible. Of course, with a Lie group you get the 6j symbols "for free" (eh, semisimple? Otherwise I don't see why you can't use e.g. Vogels general Lie group).</p> <p>In fact, the last months I constructed general 6j symbols for the E7 series, just for fun, here is one: </p> <p>{JJA|VVV}=-I*Sqrt[Q20]*Sqrt[Q11]*Sqrt[Q10]*Sqrt[Q02]*Q11/Q32/Q30/Q43/Q22 </p> <p>(V defining, J adjoint, A antisymmetric, Qxy means QuantumInteger[x*m/2+y], where m is the parameter in Westburys "Magic" paper. Insert m=-2/3 to get the SU2 6j. If not, there is a typo...). Unfortunately, with higher irreps involved the 6j symbols lost such "pretty" form, I landed in phase choice hell and stopped.</p> <p>At the moment I construct 6j symbols without resorting to Lie groups at all, just with diagrams. (I'm not even past the Clebsch-Gordan series of V*V, since the technical difficulties are enormous.) So again, is there any reason why the Reshitikhin-Turaev construction with 6j symbols, but without Lie groups, shouldn't work? A non-constructive existence proof would completely suffice, and I could lay myself to rest finally after 20 years :-)</p> http://mathoverflow.net/questions/107098/can-reidemeister-3-be-weakened Can Reidemeister 3 be weakened? Hauke Reddmann 2012-09-13T14:41:10Z 2012-09-14T14:43:51Z <p>If you take the diagram of the Reidemeister 3 move and "shortcircuit" two ends, you get (click <a href="http://imgur.com/kRvZa" rel="nofollow">http://imgur.com/kRvZa</a> if Imgur hotlink doesn't work):<br> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<img src="http://i.imgur.com/kRvZa.jpg" alt="Weak R3"><br> I have circumstantial evidence that this weaker version is actually equivalent to R3. (Only in a computational sense! My hypothesis: If A and B are two diagrams of the same knot, while it might not be actually possible to go from A to B by applying weak R3 moves (+R2+R1, of course), the assumption that weak R3 holds forces invariant(A)=invariant(B) for any Lie group derived invariant. Likewise, in Kauffmans abstract tensor framework, just assume weak R3, solve and get the rest of the Yang-Baxter equation for free.)<br> Thus: Is there work on "alternative moves"? Can you construct a counterexample? (I.e. an pseudo-invariant which is constant under weak R3+R2+R1, but not under R3? The example must have a skein equation, though.)</p> http://mathoverflow.net/questions/106812/6j-symbols-trouble-when-irrep-multiplicity-1 6j symbols trouble when irrep multiplicity >1 Hauke Reddmann 2012-09-10T12:54:08Z 2012-09-10T12:54:08Z <p>With my uncanny guessing abilities :-) I finally derived the size of the vector space where a n-valent node of a graph, edge-colored with the irreps $R_k, 1\le{k}\le{n}$ "lives". (I.e. in how many linear independent diagrams you decompose it, see e.g. "Birdtracks".) It's (heavy notation abuse) $\sharp{(1)}\in\Pi^\bigotimes_n{R_k}$ or, since this likely causes a syntax crash, in words: Just tensor multiply all the $R_k$ and count how many copies of the trivial irrep $1$ you find in the product. (Can you verify my finding?) </p> <p>Especially, for n=3 this means (excuse my ASCII) a triangle =|>- reduces either to 6j* >- or zero if >- is inadmissible (the triple product doesn't contain $1$). But haaaalt! Assume all irreps here are $(1,1)$ from $A_2$. $(1,1)\bigotimes(1,1)$ contains TWO copies of $(1,1)$ and thus the triple product two $1$, i.e. there exist two linear independent graphs with three open ends. </p> <p>Which means =|>- should be something like 6j* >- + 6j* >- ...but there are no two >-, you can only build this one acyclic graph! On the other hand: 6j symbols being undefined due to multiplicity >1 - I would have heard of that. Where is the hole in my logic?</p> http://mathoverflow.net/questions/106046/phase-choice-for-6j-symbols Phase choice for 6j symbols Hauke Reddmann 2012-08-31T15:49:43Z 2012-09-02T11:54:29Z <p>If you define 6j symbols completely formally via trivalent graphs (take <a href="http://math.ucr.edu/home/baez/qg-fall2000/qg10.2.html" rel="nofollow">http://math.ucr.edu/home/baez/qg-fall2000/qg10.2.html</a> for a start, but be careful - looks like Racah coefficients to me...well, he already mentions "fudge factors" :-) I see no -1^whatever. Now in any standard work (e.g. M. Rotenberg et al.) you see a $(-1)^{S+k}$ in Biedenharn-Elliott, you see a $(-1)^{j1+j2+j3}$ when evaluating a 6j symbol containing a zero... which looks like a completely unnatural phase choice to me, especially given that outside $SU_2$, irreps are not simple integers and you can't take -1 to the power of that anyway.<br> What looks possible to me is that you could fix a gauge by setting the theta graph colored with $j1,j2,j3$ to $(-1)^{j1+j2+j3}$, and analogous that of $R_1,R_2,R_3$ to, eh, whatever is sensible, and thus have the 3-node 6j symbols concide with the standard ones (or so I think). Still, I find it much more natural to just set all thetas to 1.<br> Is there a "natural" phase choice? (AFAIK, my spectrocopist colleagues invented the 6j symbols and might had had other things in mind. :-) For $SU_2$? For another Lie group? (Please argue why it's more natural than the one I like best :-)</p> http://mathoverflow.net/questions/105318/fictive-irreps-of-the-enveloping-general-lie-algebra "Fictive" irreps of the enveloping general Lie algebra Hauke Reddmann 2012-08-23T11:07:09Z 2012-08-23T15:07:11Z <p>Notation abuse warning: I will use the E7 series irrep names. You'll soon see why. </p> <p>In the general Lie algebra, the irrep you "start" with is $J$, the adjoint. From the series for $J\bigotimes{J}=1+J+A...$ ($1$ is the 1-dimensional irrep) you get $A$, the antisymmetric one. (The quantum dimensions up to here are all given e.g. in Westbury's "R-matrices and the magic square" paper.) The next step would be $J\bigotimes{A}=J+A+S...$ where you can "fish out" the symmetric irrep $S$ from the Clebsch Gordan series. (Does anybody have the quantum dimension $Dim(S)=S(\alpha,\beta,\gamma)$ handy, just as for $J$ and $A$? In principle I could compute it myself, but this would speed up my work considerably.)<br> Now comes the silly part :-) In the E7 series, $V\bigotimes{V}=1+J+S+A$ and I can now backwards compute the dimension of the defining irrep $V$ and see what comes out for, say, E8. Well, nonsense is coming out, which is hardly surprising. Still, I would have expected integer nonsense or even better a zero ($324-273-52+1=0$ for F4 -the minus signs are only very mild cheating. :-)! Irrational nonsense is baaaad.<br> So: Is it possible to bring all irreps, even from different rows of the magic square, in an 1:1 correspondence (if neccessary by inventing fictive irreps which still have integer dimension) or am I completely barking up the wrong Lie? </p> http://mathoverflow.net/questions/105056/6j-symbols-for-u-q-mathfraksl-n-and-colored-homfly-polynomials/105316#105316 Answer by Hauke Reddmann for $6j$-symbols for $U_q({\mathfrak{sl}}_n)$ and colored HOMFLY polynomials Hauke Reddmann 2012-08-23T10:08:45Z 2012-08-23T10:08:45Z <p>Provided you have some "base data" (closed formulae for quadratic Casimirs and quantum dimensions, for instance), for "low" irreps the quantum 6j symbols <em>might</em> be calculated recursively using Biedenharn-Elliott and other equations (e.g. those following from Reidemeister for 3-nodes). At the moment I'm <em>exactly</em> doing this for the whole E7 series, which is even worse in complexity, and see no principal obstacles - Surgeons Warning: It's extremely time-consuming due to combinatoric explosion. Also note that as soon as a multiplicity>1 pops up (the irrep 11 of A2 coming to mind immediately) you are in even more trouble.<br> (I'm no expert and always rely on brute force calculations - for a more math oriented approach I'm no help at all.)</p> http://mathoverflow.net/questions/105151/mini-fusion-categories-via-6j-symbols "Mini" fusion categories via 6j symbols Hauke Reddmann 2012-08-21T11:31:27Z 2012-08-21T21:43:22Z <p>Just for fun, I set up the following scheme:<br> - A 6j symbol is everything that fulfils Biedenharn-Elliott. (Plus symmetry, orthogonality etc. if that doesn't follow from it anyway.)<br> - There are only a finite number $i$ of irreps (including $1$, the 1-dim "neutral"). Then I tried out which of the many possible Clebsch-Gordan expansions are mutally consistent with all the 6j equations via Biedenharn-Elliott. For two irreps I got only this: $1\bigotimes{X}=X (X=1,2), 2\bigotimes{2}=1+2.$<br> - Ha!<br> - Even I did see this before, it's the smallest Fibonacci fusion category. Now a quick research gave me a load of adjectives that come with fusion categories. Can you tell me which of them apply to my scheme?<br> - $i=3$ produces 3 "minis", with $i=4$ I get 6 (and I stopped here because of the dreaded combinatiorial explosion - are these already classified somewhere?).<br> - My main question, though, is whether you get a free parameter when you choose $i$ large enough. All my "minis" give only a set of fixed complex numbers for the values of the 6j symbols, the quantum dimension and the writhe normalizer (the minis all seem to be knot-theory compatible). The latter is a root of unity so I can speculate all the minis correspond to special values of the Jones polynomial. Or suchlike.</p> http://mathoverflow.net/questions/104678/are-there-unsociable-irreps-definition-inside Are there "unsociable" irreps? (Definition inside) Hauke Reddmann 2012-08-14T09:22:20Z 2012-08-14T18:04:02Z <p>If, with a bit abuse of notation, $U\notin R\bigotimes{R}\bigotimes{R}\bigotimes{...}$ (i.e., regardless how long you clebsch up $R$, $U$ won't appear in the expansion), I call $U$ unsociable with respect to $R$ (in the group $G$). Clearly everything else is unsociable with the $1$ irrep, since (duh) $1\bigotimes{1}\bigotimes...=1$. For finite groups there can be nontrivial (ahem) cases: For $G=C_{3v}$ (the symmetry group), $A_2\bigotimes{A_2}=A_1$ and thus $E$ is unsociable w.r.t. $A_2$.<br> But what is the deal with Lie groups? E.g. is the defining irrep of $E_7$ unsociable w.r.t. the adjoint one?</p> http://mathoverflow.net/questions/102880/simple-6j-symbol-question Simple 6j symbol question Hauke Reddmann 2012-07-22T17:07:39Z 2012-07-23T09:52:57Z <p>Consider a Clebsch-Gordan expansion $R_i\bigotimes{R_j}=\bigoplus_p{R_p}$. Assume the irrep $R_k$ does NOT appear in the sum on the right side. Does it now follow that the "triangle" ${R_i,R_j,R_k}$ is "inaccessible" and consequently the 6j symbol $\begin{Bmatrix} R_i &amp; R_j &amp; R_k\\ R_l &amp; R_m &amp; R_n \end{Bmatrix}$ vanishes for any entries ${R_l,R_m,R_n}$ ?</p> <p>(And what about a converse? After all, 6j symbols have accidental zeroes. But can ALL 6j symbols with some fixed upper row accidentally vanish even if the upper row forms an accessible triangle?)</p> http://mathoverflow.net/questions/102335/pictorial-explanation-of-dynkin-index-and-quadratic-casimir Pictorial explanation of Dynkin index and quadratic Casimir? Hauke Reddmann 2012-07-16T09:36:24Z 2012-07-16T10:03:31Z <p>Draw a colored loop. You explained quantum dimension. :-) OK, now you may use trivalent nodes and knot crossings too (with edges colored by irreps). You can now pictorialize 6j symbols, writhe normalizers, structure constants etc. My question is whether other things that also pop up all the time in Lie groups, as the (quadratic) Casimir or Dynkin index of an irrep can be visualized this way. (Since that's the only way I can do math :-) </p> <p>Background: a) In "Birdtracks" an additive formula for Dynkin indices (multiplied with dimensions somehow) of the irreps involved in a Clebsch-Gordan expansion is given, b) I deduced another formula involving writhe normalizers and dimensions that is also valid for any Clebsch-Gordan expansion. (In short: Let $R_i\bigotimes{R_j}=\bigoplus_k{R_k}$, and $W_k=w_i*w_j/w_k$, where the $w_k$ are writhe normalizers, let $o_k$ be the dimensions, then $\Sigma_k{(W_k-W_k^{-1})o_k}=0$. Using pictures the proof is an one-liner.) It's natural to speculate whether a) and b) are essentially the same (or maybe even more things are additive over a Clebsch-Gordan expansion).</p> http://mathoverflow.net/questions/98356/pseudo-dimensions-of-quantum-lie-groups Pseudo-dimensions of quantum Lie groups Hauke Reddmann 2012-05-30T10:28:31Z 2012-05-31T03:21:49Z <p>In my hunt for spurious "alternatives" to the $E_7$ family I always encounter "fake" solutions. They turn out to be mostly $E_7$ family solutions disguised by $q\rightarrow{i*q}$. The effect is that the dimension (at $q=1$) comes out totally wrong. Generalizing a bit:<br> Take a random quantum dimension, say $d(G_2)=1+q^2+1/q^2+q^8+1/q^8+q^{10}+1/q^{10}$ and plug in a random root-of-unity $q=(-1)^{m/n}$. (Since even I know that the "interesting" q are these.) For small $n$ almost all values of $d$ will be integer.<br> 1. This surely has to do with the fact that $d(q)$ is a ratio of quantum integers, i.e. cyclotomic?<br> 2. Do these pseudo-dimensions have some "intuitive" meaning? </p> http://mathoverflow.net/questions/94624/jonesunlinkphi Jones(unlink)=phi Hauke Reddmann 2012-04-20T11:15:12Z 2012-05-28T16:20:33Z <p>Somewhat nebulous question: there are many well known "special" values of the Jones polynomial, especially those at roots of unity. I always run into one that has unlink value $\phi$ (golden mean) and writhe factor $(-1)^{1/5}$. Is there something special about it (maybe it's "at the intersection" of the Lie groups A1 and G2 or whatnot)?</p> http://mathoverflow.net/questions/97952/when-is-xyz-1-x-1-y-1-z-cyclotomic When is x+y+z-1/x-1/y-1/z cyclotomic? Hauke Reddmann 2012-05-25T15:37:59Z 2012-05-25T15:37:59Z <p>Frame: I'm fighting with the "E7 family" again and look for spurious alternative solutions to the most general setup. One requirement seems to be that $uv+uw+vw-1/u/v-1/u/w-1/v/w$ is a cyclotomic polynomial (times t^whatever), where $u=t^i,v=-t^j,w=-t^k$, t free variable, i,j,k integer. The E7 solution corresponds to $w=-1/u^3$, BTW, but if you compute now, you notice a factor $v^2+vw+w^2$, so the cyclotomicity doesn't come out <em>totally</em> trivial. There are many other solutions, e.g. $w=-u$, and cyclotomicity isn't sufficient at all, but maybe <em>this</em> problem here can be solved to give a finite list of admissible equations in i,j,k and there are no others. (This would probably solve my original problem too, since integer dimensions force further divisibility rules.)<br> Obviously my knowledge of number theory isn't better than that of knot theory - is this a viable way to attack?</p> http://mathoverflow.net/questions/97563/again-size-of-lie-group-based-vector-spaces Again: size of Lie-group based vector-spaces Hauke Reddmann 2012-05-21T14:25:19Z 2012-05-21T14:25:19Z <p>When expressing tangles, with my colored 4-node scheme and $n$ colors I get $d_4=n+2$ independent elements of the $V^4$ vector space ($V$ a representation, usually the defining one, of the Lie group $L$), i.e. just the $n$ 4-nodes plus the $0$ and $\infty$ tangle.<br> For $V^6$ simple counting gives $X=5+6n+3n^2$ independent tangles within this approach. Let's see how $X$ compares to the actual $d_6=Dim(V^6)$ for small $n$:<br> $n=0$ Jones polynomial, $d_4=2, X=d_6=5$. This is just the usual Temperley-Lieb algebra.<br> $n=1$ Kauffman polynomial, $d_4=3, d_6=15$. Sensible as you can't simplify a triangular face of a knot by pass moves, so $15=X+1=5+6+3+1$ (zero, one, two, three crossings). Except (disregarding nontrivial cases) for $L=B_2$, where $X=d_6=14$.<br> $n=2$ A nuisance for me since 20 years :-) $X=29$. A brute force calculation gives $d_6=40$ (just from the skein relations, no Lie groups involved). Any try to reduce it promptly lets the $E_7$ family pop up ($d_6=35$), and from that only $L=B_3$ where $d_6=30$ comes close. ($A_1*A_1$, $d_6=25$ is uninteresting for me.)<br> $n=3$ And here comes the question: $X=50$, does any $L$ from the $E_8$ family (or another Lie group with $d_4=5$) have a $d_6$ value equal or below that? (I don't even dare to ask for higher values of $n$ :-)</p> http://mathoverflow.net/questions/95994/vector-product-diagonalization Vector "product" diagonalization Hauke Reddmann 2012-05-04T15:17:02Z 2012-05-07T00:42:02Z <p>Consider a vector space $V$ (with dimension $n+1$ and elements $v$) on which a (commutative and associative) "product" $\odot$ taking $V\odot V\rightarrow V$ is defined, and an $1$ element $v_0$ exists: $v\odot v_0=v$ for all $v\in V$. $\odot$ is now completely defined by choosing a basis $v_m$ $(0\le m\le n)$ ($v_0$ is predefined - hands off - you can't change it) and giving the "structure constants" $A^{ij}_k$ $(0\le i,j,k\le n)$ which are free parameters (except those with $i*j=0$ who are predefined by $v_0$ being the $1$ element): $v_i\odot v_j=\sum_k A^{ij}_k*v_k$.<br> After you defined the $A^{ij}_k$, you are free to chose a convenient other basis. First question: I have $n(n+1)$ free parameters to do so. Just from <em>counting</em>, I can use them to have $v_k\odot v_k=v_0$ for all $k$ and use up exactly all "gauge" freedom this way. My linear algebra is rusty - can I really? Or are there $A^{ij}_k$ sets where this fails?<br> Second question. Define an "outer product" $\otimes$ (with is distributive over vector sums, and follows $(v_i\otimes v_j)\odot v_k:=v_i\otimes (v_j\odot v_k)$ and $v_i\odot (v_j\otimes v_k):=(v_i\odot v_j)\otimes v_k$) and a quantity $S=\sum_i\sum_j a_{ij}*v_i \otimes v_j$. $S$ should be an "eigenvector": $S\odot v_k=v_k\odot S$ for all $k$. Express the allowed set of $a_{ij}$ in terms of the $A^{ij}_k$. (This is trivial for small n, where I do it with diagrams and by hand - in fact this is knot/graph theory in disguise as always when I ask :-) But a closed formula would be nice.)</p> http://mathoverflow.net/questions/93307/a-property-of-quantum-group-r-matrices A property of quantum group R matrices? Hauke Reddmann 2012-04-06T13:10:00Z 2012-04-06T18:51:05Z <p>Assume Q is a quantum Lie group which allows a R matrix (with the usual quantum Yang-Baxter equation). Take the Jordan normal form J of R. Does a Q exist with offdiagonal J elements (i.e. R has defective eigenvalues)? </p> http://mathoverflow.net/questions/93128/list-of-small-dimension-lie-group-irreps List of small dimension Lie group irreps Hauke Reddmann 2012-04-04T14:41:10Z 2012-04-04T14:58:06Z <p>For <em>semisimple</em> Lie groups it's just a lookup (I define "small" as dim(R)&lt;5 and put the dimensions of the Clebsch-Gordon series of RxR in parentheses): A1(1*1=1),A1(2*2=1+3),A1(3*3=1+3+5),A1(4*4=1+3+5+7),SO2(2*2=1+1+2),A2(3*3=3+6),B2(4*4=1+5+10), and the products SO2#SO2,SO2#A1 and A1#A1. Hope I forgot none.<br> But Lie groups can be non-this, non-that and non-bugsme. (In the "magic" series D(2,1,$\alpha$) and OSP(2,1) pop up, but these are super algebras and I have no idea what "dimension" means for a super algebra since they seem to have more than one.) Can you amend my list with, say, non-solvable Lie groups?<br> Especially interesting would be if RxR contains more than one copy of the 1 irrep...if that's possible at all! (SO2 doesn't count since the other 1 is no 1. Dim 1, but no unity irrep. Or whatever. Don't ask, I don't understand it :-) </p> http://mathoverflow.net/questions/90911/span-of-tangle-vector-space-for-different-lie-groups Span of tangle vector space for different Lie groups Hauke Reddmann 2012-03-11T16:19:21Z 2012-03-13T12:42:01Z <p>In his $G_2$ paper, Kuperberg gives the following numbers of acyclic freeways for n=0...6: 1, 0, 1, 1, 4, 10, 35. (Which is identical to $dim Inv(V^{⊗n}_{1,0})$, the spanning size of the tangle vector space. (??). But that is NOT identical with the number of crossingless trivalent tangle graphs: for n=6, the "hexagon" can't be resolved into the 34 acycling tangle graphs and must be added to the linear independent set.) The $G_2$ numbers (or so I think) are identical for the whole $E_7$ family (except $B_3(\Lambda_3)$, $A_1(3\Lambda_1)$ and $A_1*A_1(\Lambda_1*\Lambda_1)$ where the last number should be 30,34 and 25 if I computed correctly). (But since you now must use 2 irrep colors for any other than $G_2$, 5 of the 34 above graphs are forbidden.)<br> With some jump of faith, I suppose also for all members of the $E_8$ family the spanning space numbers are the same: 1, 0, 1, 1, 5, 15, 70 (???), except for some "special" groups of that family.</p> <p>Can you verify my $dim Inv(V^{⊗n}_{1,0})$ values for $E_8$ ? (Maybe with a list of exceptional members ? Directly I can only compute for $A_1(4\Lambda_1)$ and this surely gives less than 70.)</p> http://mathoverflow.net/questions/133368/quantum-9j-symbols Comment by Hauke Reddmann Hauke Reddmann 2013-06-12T10:57:48Z 2013-06-12T10:57:48Z @ Carlo: This would mean that the standard definition already is &quot;normalized&quot; with respect to the annoying crossing. (I &quot;pulled it through&quot; in my own computations; if it's not needed at all - the better! :-) My lib has the book, I can look up the details. THX!) http://mathoverflow.net/questions/118634/symmetry-analysis-of-an-impossible-symmetry Comment by Hauke Reddmann Hauke Reddmann 2013-05-10T13:46:54Z 2013-05-10T13:46:54Z (Mods, you can close the posting, I cracked the problem by unmathematical brute computing force. :-) For the record, the overall result is integer but as soon as you try to assign the found symmetries to specific atom movements, the half-integers rear their ugly head again.) http://mathoverflow.net/questions/121017/diophantine-question Comment by Hauke Reddmann Hauke Reddmann 2013-02-07T10:00:51Z 2013-02-07T10:00:51Z @Peter: No, not necessary. @Noam: Obviously (although: zero dimension...not sure if want)...the &quot;besides&quot; is the interesting part. (I hoped for a general applicable method.) Note that negative y or z is fishy - I'm still pondering if negative dimensions on the RHS of a Clebsch-Gordon expansion make any sense at all. http://mathoverflow.net/questions/115495/matrix-minimax-problem Comment by Hauke Reddmann Hauke Reddmann 2012-12-07T12:30:41Z 2012-12-07T12:30:41Z It is just multiplication. Edited. http://mathoverflow.net/questions/111893/cyclotomic-polynomial-question/111964#111964 Comment by Hauke Reddmann Hauke Reddmann 2012-11-10T12:03:22Z 2012-11-10T12:03:22Z Thanks. One counterexample suffices :-) (Although I would have preferred if there were none...) http://mathoverflow.net/questions/110646/links-with-same-jones-polynomial/110648#110648 Comment by Hauke Reddmann Hauke Reddmann 2012-10-26T14:41:01Z 2012-10-26T14:41:01Z @Paolo: THX for the paper link. (I should have added to my question that I know &quot;mutation&quot; and wanted to exclude this, but the paper answers my question.) @Bruce: AHA! You just answered a question I would have asked here eventually :-) @Czy: I wanted to exclude <i>that</i> either. Both are rather &quot;trivial&quot; constructions, I'm rather interested in the &quot;accidental&quot; cases. http://mathoverflow.net/questions/108823/thurston-bennequin-number-vs-checkerboard-coloring-difference/108850#108850 Comment by Hauke Reddmann Hauke Reddmann 2012-10-05T10:26:29Z 2012-10-05T10:26:29Z THX, that answer was spot-on. I check out the paper. http://mathoverflow.net/questions/108253/schurs-di-lemma-finite-and-lie-groups-different Comment by Hauke Reddmann Hauke Reddmann 2012-09-28T09:21:15Z 2012-09-28T09:21:15Z Thanks, I follow the leads. @Qiaochu: I hesitated to ask at all, but on Stackexchange noone answered a similar question. (If only the LiE software would be more versatile, I would pester MO less often...) http://mathoverflow.net/questions/107556/generalizing-the-reshitikhin-turaev-construction-possible/107558#107558 Comment by Hauke Reddmann Hauke Reddmann 2012-09-20T10:33:23Z 2012-09-20T10:33:23Z THX for explaining the &quot;technicalities&quot; - I probably read these names before, but I have already enough trouble with Lie algebras :-) Just for fun again, here are a few values for the q-E7 series I computed (just copypasta): e3=-1/e1^3; (<i>this specializes to E7 series</i>) u=e1*e2+e1*e3+e2*e3;v=1/e1/e2+1/e1/e3+1/e2/e3;w=e1*e2*e3; o=(-1+u*w^2-v*w^6+w^8)/((u-v)*w^4);(<i>unlink</i>) z=(-u+u^2*w^2-u*v*w^2+v^2*w^2-u^2*w^6+u*v*w^6-v^2*w^6+v*w^8)/((-u+v)*w^4);(<i>Hopf</i>) p3=(-1+2*u*v-v^2+u*w^2-u^3*w^2+u^2*v*w^2-u*v^2*w^2+u^3*w^6-v*w^6-u^2*v*w^6+u*v^2*w^6+w^8-2*u*v*w^8+v^2*w^8)/((u-v)*w^5);(<i>trefoil</i>) http://mathoverflow.net/questions/107098/can-reidemeister-3-be-weakened/107187#107187 Comment by Hauke Reddmann Hauke Reddmann 2012-09-15T14:37:46Z 2012-09-15T14:37:46Z (@Joseph - THX, I read the manual for image insertion twice and still got it wrong...) @Noah: The other skein relations are not THAT special - I'll try my best to write up a formal proof but I never excelled in that :-) My main observation was that I can write the conditions for Biedenharn-Elliott and R3 for 3-nodes with 4 boundary points instead of the usual 5. This must &quot;mean&quot; something :-) http://mathoverflow.net/questions/107098/can-reidemeister-3-be-weakened Comment by Hauke Reddmann Hauke Reddmann 2012-09-14T08:42:55Z 2012-09-14T08:42:55Z Blast, not even the diagram came through. I try a total workover. http://mathoverflow.net/questions/106812/6j-symbols-trouble-when-irrep-multiplicity-1 Comment by Hauke Reddmann Hauke Reddmann 2012-09-11T08:49:32Z 2012-09-11T08:49:32Z @Kevin (or anybody else) - do you know a good reference to start? (From my spectroscopic work, I'm accustomed to additional quantum numbers, like &quot;seniority&quot; or an additional counter index for equal $^{2S+1}L_J$ terms, but luckily, I'm exclusively working in SU(2) :-) http://mathoverflow.net/questions/105318/fictive-irreps-of-the-enveloping-general-lie-algebra Comment by Hauke Reddmann Hauke Reddmann 2012-08-24T09:49:57Z 2012-08-24T09:49:57Z Thus you can bring at least all irreps that pop up in some $J\bigotimes{n}$ of E8 in an 1:1 correspondence with those of E7. (No, I can't guarantee that you can REALLY say A is A and 8 is 8. Just from symmetry arguments A(E7) could be 8(E8) and vice versa. But from the lowest irrep CG expansions I checked, the correspondence seemed convincing to me.) ... And I wanted to do something for those irreps who feel lonesome now :-) http://mathoverflow.net/questions/105318/fictive-irreps-of-the-enveloping-general-lie-algebra Comment by Hauke Reddmann Hauke Reddmann 2012-08-24T09:38:05Z 2012-08-24T09:38:05Z I took the names from &quot;Birdtracks&quot;. OK, let's just take E7 and E8 (and hope weights are standardized): E7: V=0000001, J=1000000, A=0000010, S=0000002, 6=0010000, 7=0 (accidentally), 8=2000000. E8: V nope, J=00000001, A=10000000, S=0 (a.), 6=00000010, 7=0 (a.), 8=000000002. J*J=E+J+A+6+7+8. This works for E7 as well as E8, and you have the closed quantum dimension formula of Vogel for J,A,6,7,8, and counting terms in the Clebsch-Gordan you know that 6 corresponds to 6, 7 to 7 and 8 to 8 (duh :-). J*A=J+A+S+... Dito. You can say which irrep of E8 corresponds to S of E7. (It's a 0.) (cont.) http://mathoverflow.net/questions/105318/fictive-irreps-of-the-enveloping-general-lie-algebra/105326#105326 Comment by Hauke Reddmann Hauke Reddmann 2012-08-24T09:27:33Z 2012-08-24T09:27:33Z Nor do I :-) Yes, that's exactly the point: V is not defined for E8, but still you can compute the quantum dimension of this undefined dingbat (assuming V^2=1+J+S+A, of course). So I thought that maybe a whole series of fictive irreps could be defined such that all the dimension equations are fulfilled. (Hey, the square root of -1 doesn't exist either :-) But this is &quot;fun&quot; only with integer dimensions.