tag:blogger.com,1999:blog-288573692024-03-07T22:03:15.986+13:00Arcadian Functoroccasional meanderings in physics' brave new worldKeahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.comBlogger1018125tag:blogger.com,1999:blog-28857369.post-48509993668214208752010-04-10T13:01:00.000+12:002010-04-10T13:02:18.070+12:00New Format BlogNot sure if this will work out, but future posts should appear at <a href="http://pseudomonad.blogspot.com/">Arcadian Pseudofunctor</a>.Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com0tag:blogger.com,1999:blog-28857369.post-20215770506653116212010-04-10T09:35:00.002+12:002010-04-10T09:53:02.418+12:00Women at WorkUntil the 1200s women in Europe often worked for a living, even for themselves, and played prominent roles in public life. It was the priests and noble women who learned to read, because noble men had to focus on fighting. The slow demise of women was undoubtedly heavily influenced by Christian thought, but as an institution the Church was not solely responsible for curbing their rights. <br />An early culprit was that fine new European institution, the University. Historically, many women had practised empirical medicine, but from 1271 the <a href="http://en.wikipedia.org/wiki/University_of_Paris">University of Paris</a> insisted upon formal certification. By the early 1300s, women were being prosecuted for practicing medicine unlawfully. One example is <a href="http://books.google.co.nz/books?id=WmWsR_gP5zUC&pg=PA299&lpg=PA299&dq=%22Jacqueline+Felicie+de+Alemania%22&source=bl&ots=IFrNs84TFZ&sig=Fo6uzRpeYnv-Fo4CYiSSG6KPRRI&hl=en&ei=P52_S_zbDo7BngeG7ZCWCg&sa=X&oi=book_result&ct=result&resnum=1&ved=0CAkQ6AEwAA#v=onepage&q=%22Jacqueline%20Felicie%20de%20Alemania%22&f=false">Jacqueline Felicie de Alemania</a>. <br />In <a href="http://www.amazon.co.uk/Opera-Muliebra-Mediaeval-perspectives-European/dp/0075577445">Opera Muliebra</a>, David Herlihy argues that women continued to be employed in many professions up until the population growth of the 1400s, after the great plague. In this century the power of the professional <a href="http://en.wikipedia.org/wiki/Guild">guilds</a> was largely responsible for excluding women from apprenticeships. <br />The Church had tried to put women in their place for a millenium, without much success. The overriding factor in this story is that urban men wanted good work, once they had stopped fighting each other, and this was most easily achieved by excluding women, with excuses from Christian thought.Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com6tag:blogger.com,1999:blog-28857369.post-1486721646416422232010-04-08T11:46:00.003+12:002010-04-08T11:54:39.410+12:00Dwarf Mysteries<a href="http://www.universetoday.com/2010/04/06/mystery-object-found-orbiting-brown-dwarf/">Universe Today</a> reports on the <a href="http://hubblesite.org/newscenter/archive/releases/2010/03/image/d/">discovery</a> of a companion to the brown dwarf <a href="http://www.centauri-dreams.org/?p=11959">2M J044144</a>. <blockquote>Astronomers say it is the right size for a planet, but they believe the object formed in less than 1 million years — the approximate age of the brown dwarf — and much faster than the predicted time it takes to build planets according to conventional theories.</blockquote> <a href="http://riofriospacetime.blogspot.com/">Louise Riofrio</a> predicted years ago that such planets would be found, because planets can form around tiny black holes. Thanks to <a href="http://hubblesite.org/">Hubble</a> and <a href="http://www.gemini.edu/">Gemini</a> for these beautiful images. <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjEQ4gpoLUXX6MXgGBqS-Slt_JJDs51pyjrb9SzrQRpa9ItxISuWP7Q3OdDpeYCvmTkqrMdtDXsnYlJBv5DukIJ9cCtPZHCefJafyHG3NdfJRJ2RqcGNNXCGkZ2tstIcHtho1lN/s1600/brown_dwarf_companion1.jpg"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;width: 320px; height: 320px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjEQ4gpoLUXX6MXgGBqS-Slt_JJDs51pyjrb9SzrQRpa9ItxISuWP7Q3OdDpeYCvmTkqrMdtDXsnYlJBv5DukIJ9cCtPZHCefJafyHG3NdfJRJ2RqcGNNXCGkZ2tstIcHtho1lN/s320/brown_dwarf_companion1.jpg" border="0" alt=""id="BLOGGER_PHOTO_ID_5457547594105599138" /></a>Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com2tag:blogger.com,1999:blog-28857369.post-42969808308774030412010-04-06T17:11:00.005+12:002010-04-06T17:19:05.388+12:00M Theory Lesson 312Let us now take two quark matrices and multiply them together. <a href="http://kea-monad.blogspot.com/2010/04/m-theory-lesson-311.html">Using</a> the phase matrices (in terms of $12$-th roots) on the MUB side of the twisted Fourier transform, for anti-up and anti-down quarks, we find that the product is <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEibYj709gbC5mgw-_MDNwFnJ-ANmJpsPYYTMO0yaoi3ohFt0DxwBwwaRor3KdMA1EFY8pvtF7QjbiJ-2IifNHcel-WUKMbyrvMXrWAtUR73FQObJvTnQqWnOxbBLPHLUuXx3X68/s1600/quarkLepComp.png"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;width: 320px; height: 141px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEibYj709gbC5mgw-_MDNwFnJ-ANmJpsPYYTMO0yaoi3ohFt0DxwBwwaRor3KdMA1EFY8pvtF7QjbiJ-2IifNHcel-WUKMbyrvMXrWAtUR73FQObJvTnQqWnOxbBLPHLUuXx3X68/s320/quarkLepComp.png" border="0" alt=""id="BLOGGER_PHOTO_ID_5456887849610495490" /></a> a <a href="http://en.wikipedia.org/wiki/Positron">positron</a> in the dual <a href="http://kea-monad.blogspot.com/2010/03/ribbon-review.html">braid</a> space. This is a <a href="http://kea-monad.blogspot.com/2008/06/neutrinos-again-vi.html">complementarity</a> between quarks and leptons, taking full magic matrices to fun operators.Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com0tag:blogger.com,1999:blog-28857369.post-60672626532765788632010-04-05T18:36:00.001+12:002010-04-05T18:37:16.958+12:00Quote of the Week<a href="http://youngfemalescientist.blogspot.com/2010/04/scienciae-carnival-post-sustainability.html#comments">Disgruntled</a>, unemployed male postdoc: <blockquote>Science has withstood the Catholic Church and its death threats. Science is not gonna go down to a bunch of feminists. </blockquote>Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com3tag:blogger.com,1999:blog-28857369.post-61219111707702560842010-04-04T12:03:00.011+12:002010-04-04T12:55:24.463+12:00M Theory Lesson 311Under <a href="http://kea-monad.blogspot.com/2010/04/m-theory-lesson-310.html">the</a> correspondence between ribbon twists and roots of unity, as in <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjaHAC02XcgDnBduVCRR-qrquW1BGKxLHhJ2VwiO7V1FOLhJ3udHBUoaZT55f7Ro07_yfAuhUjj8Osiy-ZdEikBy-ybC-P6v544fY3hOSEaMt6NEy2zIgNwiORIMCosuT1-KFFi/s1600/twistDef.png"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;width: 274px; height: 152px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjaHAC02XcgDnBduVCRR-qrquW1BGKxLHhJ2VwiO7V1FOLhJ3udHBUoaZT55f7Ro07_yfAuhUjj8Osiy-ZdEikBy-ybC-P6v544fY3hOSEaMt6NEy2zIgNwiORIMCosuT1-KFFi/s320/twistDef.png" border="0" alt=""id="BLOGGER_PHOTO_ID_5456066496457491218" /></a> there is a matching of particle <a href="http://kea-monad.blogspot.com/2010/03/ribbon-review.html">ribbon</a> diagrams to matrices, giving in particular <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEipRCT95UZjqh43Hl4TrOkH9N-cGSGOZDYnM8TzFu5OA-vBi3bV3vO2sN1zF7nefpAL2RtZ3Vlj0mFUgrxFRrXJYUIHwj0A3S6Ih72xFHZanBOrMOfXigsh1Imntn7CWwewW2xR/s1600/particleMatr.png"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;width: 320px; height: 250px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEipRCT95UZjqh43Hl4TrOkH9N-cGSGOZDYnM8TzFu5OA-vBi3bV3vO2sN1zF7nefpAL2RtZ3Vlj0mFUgrxFRrXJYUIHwj0A3S6Ih72xFHZanBOrMOfXigsh1Imntn7CWwewW2xR/s320/particleMatr.png" border="0" alt=""id="BLOGGER_PHOTO_ID_5456066409340956818" /></a> Observe how two copies of each $1$-circulant set arise from the choice of crossings in the braid diagram. Complex conjugacy is a flip of crossings. Given these matrices, a <a href="http://kea-monad.blogspot.com/2010/03/m-theory-lesson-306.html">twisted</a> Fourier transform of the form $F C_{a} F^{\dagger} = DC_{b}$, for $D$ a phased diagonal, takes MUB type operators $C_{a}$ to particle circulants $C_{b}$. For example, <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgoTbjHxCCMTzDb6hqtJPkgFwpx3k6mbbAZZlZzZcAWxu5v0OUuFQZ1ON9o0mEbIRe5UVaR-S7iaCe6FgxtZNO6ONRvoRNepUDbXWwcuPVP4WHlNvbUr-Xa0bVmJqugh8P0-q9e/s1600/twistFouriNu.png"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;width: 219px; height: 188px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgoTbjHxCCMTzDb6hqtJPkgFwpx3k6mbbAZZlZzZcAWxu5v0OUuFQZ1ON9o0mEbIRe5UVaR-S7iaCe6FgxtZNO6ONRvoRNepUDbXWwcuPVP4WHlNvbUr-Xa0bVmJqugh8P0-q9e/s320/twistFouriNu.png" border="0" alt=""id="BLOGGER_PHOTO_ID_5456066302185388834" /></a> Let us not forget the quarks! Quarks do not correspond to basis circulants, so under the inverse twisted transform they come from more general matrices. For example, the $C_a$ phases of an up quark are given by <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiGCRYShk6vUHjfZINy1puSVQ4LltZg-f4XJeMmmgHi8weW71HzsRcAy4dhMSSaI77FuFGeciYAp4d6a9AWCmvIeBtM6D3Op947ahzhOnTflDasBzv5guFQuMe7xW1SBgL7NeXv/s1600/quarkThe.jpg"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;width: 133px; height: 144px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiGCRYShk6vUHjfZINy1puSVQ4LltZg-f4XJeMmmgHi8weW71HzsRcAy4dhMSSaI77FuFGeciYAp4d6a9AWCmvIeBtM6D3Op947ahzhOnTflDasBzv5guFQuMe7xW1SBgL7NeXv/s320/quarkThe.jpg" border="0" alt=""id="BLOGGER_PHOTO_ID_5456078706938548818" /></a> which is clearly not a MUB operator. Note, however, that this is a magic matrix, with all rows and columns summing to $- \pi / 6$.Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com2tag:blogger.com,1999:blog-28857369.post-38749947165498587952010-04-03T10:54:00.017+13:002010-04-03T11:25:49.892+13:00M Theory Lesson 310Permutations may be represented by symmetric <a href="http://en.wikipedia.org/wiki/Braid_group">braids</a>. A category theorist would usually draw a diagram like <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhWfSWhNHPJJOJTLcHqeAs18yyKzz_R6De-rnefdbRSMkGYbJCHwcublPUbq_cAZ8yzKPfcU7f8Agt1pl2Y2IxxzuR_8sWLYSR4w64_SYBIdJdZrTNsoFQKP0NlGmLuqjIye0tB/s1600/2strandBa.png"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;width: 134px; height: 112px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhWfSWhNHPJJOJTLcHqeAs18yyKzz_R6De-rnefdbRSMkGYbJCHwcublPUbq_cAZ8yzKPfcU7f8Agt1pl2Y2IxxzuR_8sWLYSR4w64_SYBIdJdZrTNsoFQKP0NlGmLuqjIye0tB/s320/2strandBa.png" border="0" alt=""id="BLOGGER_PHOTO_ID_5455662230470199858" /></a> indicating that the braid crossing goes neither over nor under, and that one can slide strands across each other. It would be nicer if we could draw a braid like <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiqFGeVHWZ-2p5HTNI-4IsEfA3HkSofAptCdGwdy7VD3ZjUHkNtXSbDLpOJc3V7joasDXdVIMdjg9_KmPAcqdZ6UgZEKL94FXKYudQCL_haFrLGGMymKgx-ikn7B1tWIHELkoiN/s1600/2strandCross.png"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;width: 138px; height: 126px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiqFGeVHWZ-2p5HTNI-4IsEfA3HkSofAptCdGwdy7VD3ZjUHkNtXSbDLpOJc3V7joasDXdVIMdjg9_KmPAcqdZ6UgZEKL94FXKYudQCL_haFrLGGMymKgx-ikn7B1tWIHELkoiN/s320/2strandCross.png" border="0" alt=""id="BLOGGER_PHOTO_ID_5455662138286815282" /></a> because this <a href="http://en.wikipedia.org/wiki/Reidemeister_move">obviously</a> untangles to the identity braid. However, the two crossings here are different, like in the law $(\sigma) \circ (- \sigma) = 1$, which is to say that $\sigma$ is a bit like the complex $i$, satisfying $i + i^{-1} = 0$. How about elements of $S_3$? The permutation $(231)$ cubes to the identity, as in the diagram <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgXxbS-Z9DJpOt8O_E62ARV3JCnr8jUlaOIIOQRrr5sojYtSpMJuxVP7cKb7zVPkv4WPbH_YyhP_lvCmdyqJylDVEn_yloFbJORRQ7wNYtKKCj7kRF3LumMGo5fj2AGljtYCtwF/s1600/3strandCross.png"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;width: 204px; height: 302px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgXxbS-Z9DJpOt8O_E62ARV3JCnr8jUlaOIIOQRrr5sojYtSpMJuxVP7cKb7zVPkv4WPbH_YyhP_lvCmdyqJylDVEn_yloFbJORRQ7wNYtKKCj7kRF3LumMGo5fj2AGljtYCtwF/s320/3strandCross.png" border="0" alt=""id="BLOGGER_PHOTO_ID_5455662040793533186" /></a> where the crossings have been chosen so that no strand actually links with any other. Note that each $(231)$ section is a <a href="http://kea-monad.blogspot.com/2010/03/ribbon-review.html">Bilson-Thompson</a> braid. What <a href="http://en.wikipedia.org/wiki/Field_with_one_element">fun</a>! By assigning $n$-th roots of unity to each strand one obtains a fun operator, for $S_d$. The choice $n = d+1$ (see <a href="http://kea-monad.blogspot.com/2010/04/m-theory-lesson-309.html">last time</a>) gives a <a href="http://en.wikipedia.org/wiki/Representation_theory_of_the_symmetric_group">representation</a> of $S_n$.Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com4tag:blogger.com,1999:blog-28857369.post-14352252338680089102010-04-02T11:38:00.008+13:002010-04-02T11:51:53.726+13:00M Theory Lesson 309Recall that much of the <a href="http://kea-monad.blogspot.com/2009/07/m-theory-lesson-286.html">tribimaximal</a> mixing literature discusses small finite groups, such as $S_3$ and $A_4$. For example, <a href="http://kea-monad.blogspot.com/2007/11/m-theory-lesson-120.html">Ernest Ma</a> considers a $2 \times 2$ representation of the <a href="http://en.wikipedia.org/wiki/Dihedral_group_of_order_6">six</a> element group given by the matrices <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhGrUxKE2lwiX-sB6vY3uXkNQp4R8INwQQG_ZRh9UslCrF1Q8a6fTTQZpsiXMNdL0k7epWLQQNuJIHeMVXs5870GAB6I7Vjlu9-zYN2HgwUmv-nXDmfJlXGzAT8Df9TCJnLKBmR/s1600/MaMatrixF1.png"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;width: 307px; height: 134px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhGrUxKE2lwiX-sB6vY3uXkNQp4R8INwQQG_ZRh9UslCrF1Q8a6fTTQZpsiXMNdL0k7epWLQQNuJIHeMVXs5870GAB6I7Vjlu9-zYN2HgwUmv-nXDmfJlXGzAT8Df9TCJnLKBmR/s320/MaMatrixF1.png" border="0" alt=""id="BLOGGER_PHOTO_ID_5455302320758835122" /></a> where $\omega$ is the usual root of unity. This is an example of a <a href="http://en.wikipedia.org/wiki/Field_with_one_element">fun</a> set, with $d = 2$ and $n = 3$. In fact, it is the full such set under the condition that the non zero entries multiply to $1$. So special representations need no longer live in boring categories of vector spaces!Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com2tag:blogger.com,1999:blog-28857369.post-75026034968462179662010-04-02T10:01:00.002+13:002010-04-02T10:05:12.449+13:00Meanwhile IISo I've found a suitable proof reader for my <a href="http://kea-monad.blogspot.com/2010/03/meanwhile.html">book</a>, am busy hunting down literary agents, and am making progress with exciting, although naturally temporary, alternative career options. Now if, and it is a big if, any of these avenues should land me in the UK again, where should I spend some time? Hmm. I wonder what Cambridge is like ...Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com4tag:blogger.com,1999:blog-28857369.post-78156250459313388632010-03-31T12:42:00.000+13:002010-03-31T12:43:41.021+13:00Quote of the Day<a href="http://wh40k.lexicanum.com/wiki/Thought_for_the_day">Courtesy</a> of the small grocery shop near my place, which places a thought for the day on the chalkboard outside, no matter the weather: <blockquote>The Emperor will not judge you by your medals and diplomas but by your scars.</blockquote>Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com10tag:blogger.com,1999:blog-28857369.post-62421111297251614632010-03-30T16:47:00.010+13:002010-03-30T17:42:40.491+13:00Quantum ComputationA well funded industry these days is research into the real world implementation of small sets of carefully chosen quantum operations, combinations of which are able to closely simulate <span style="font-weight:bold;">any</span> quantum computation. This is known as <span style="font-style:italic;">universal quantum computation</span>, or <a href="http://en.wikipedia.org/wiki/Quantum_gate#Universal_quantum_gates">UQC</a>. <br /><br />For example, in this <a href="http://arxiv.org/abs/quant-ph/0403025">paper</a> by <a href="http://www.cs.caltech.edu/cspeople/faculty/kitaev_a.html">Kitaev</a> and <a href="http://www.research.ibm.com/physicsofinfo/members/sergeyPub.htm">Bravyi</a>, it is shown that the Clifford operations (which stabilise Pauli operators) along with certain magic states are sufficient for UQC. They assume that Clifford operations may be implemented ideally, and that the preparation of magic states is faulty. However, they first consider ideal magic states, such as<br />$|H \rangle = \textrm{cos} \frac{\pi}{8} |0 \rangle + \textrm{sin} \frac{\pi}{8} |1 \rangle$<br />$|T \rangle = \textrm{cos} \beta |0 \rangle + \omega \textrm{sin} \beta |1 \rangle$<br />where $\omega$ is a primitive eighth root of unity and $\textrm{cos} 2 \beta = \sqrt{3}^{-1}$. In particular, $| T \rangle$ may be used to implement a one qubit <a href="http://en.wikipedia.org/wiki/Quantum_gate#Phase_shift_gates">phase gate</a> for the $12$-th root of unity. With the Clifford operations, this gate gives UQC. <br /><br />The only Clifford generator that is not obviously a <a href="http://kea-monad.blogspot.com/2010/03/magic-motives.html">fun</a> (field with one element) operation is the one qubit <a href="http://en.wikipedia.org/wiki/Quantum_gate#Hadamard_gate">Hadamard</a> gate (Fourier transform), but recall that (in MUB maths) this basis behaves like the zero of a finite field, or the standard choice of marked point for a fancy set! Now let's do UQC without complex numbers.Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com0tag:blogger.com,1999:blog-28857369.post-80779370204132405812010-03-29T12:20:00.003+13:002010-03-29T12:27:28.627+13:00MeanwhileWell, it seems that no matter how hard you try, you just can't keep a <a href="http://youngfemalescientist.blogspot.com/">good</a> woman down: a second peer reviewed <a href="http://www.aip.org/">AIP</a> paper was just accepted for publication; I have just finished a first draft of my scary book about life for women in Oxford; and I'm considering alternative careers (not really) that use my survival expertise.<br /><br />Meanwhile, I'll get back to wondering which streets to haunt when I become homeless once again. Have a nice day.Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com6tag:blogger.com,1999:blog-28857369.post-4347417267005503722010-03-27T16:12:00.014+13:002010-03-27T17:06:52.338+13:00M Theory Lesson 308Any elementary matrix, denoted $E_{ij}$, can be expressed as a Fourier transform of the <a href="http://kea-monad.blogspot.com/2009/02/matrix-power-i.html">democratic</a> matrix. <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgRZaxp0M3G_c8MqQw4hQ-NcFxBGY4VGOycRBxt7pau6SyleoEFYS6d8212eHhyBVPQZgILrGYJMgBTWfNGS02J5Ovv7_V2390VOKHvSI18pe2BmOvqHol56BwduNPCBwkF8AGV/s1600/fourierElem.jpg"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;width: 320px; height: 162px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgRZaxp0M3G_c8MqQw4hQ-NcFxBGY4VGOycRBxt7pau6SyleoEFYS6d8212eHhyBVPQZgILrGYJMgBTWfNGS02J5Ovv7_V2390VOKHvSI18pe2BmOvqHol56BwduNPCBwkF8AGV/s320/fourierElem.jpg" border="0" alt=""id="BLOGGER_PHOTO_ID_5453146434335015970" /></a> In the $3 \times 3$ case, the nine choices for $F$ give the $9$ elementary matrices. Thus any matrix at all may be expressed as a combination of such Fourier transforms. For example, the circulant permutation $(231)$ uses the three Fourier matrices that sum to <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgj4zoQNG4h8A1HeVAA4Cpg2CH6qtuoekfQwa2HoV4y66pNp81MSqjZ0M4YLtWG4siI7TPh_UNCxOFLpfYe5uVBUYNZ2pE_MLnbi5n5puLxQ_D9ReowsHfDfHxAimZxqBeWjeYm/s1600/threeCircF.png"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;width: 226px; height: 90px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgj4zoQNG4h8A1HeVAA4Cpg2CH6qtuoekfQwa2HoV4y66pNp81MSqjZ0M4YLtWG4siI7TPh_UNCxOFLpfYe5uVBUYNZ2pE_MLnbi5n5puLxQ_D9ReowsHfDfHxAimZxqBeWjeYm/s320/threeCircF.png" border="0" alt=""id="BLOGGER_PHOTO_ID_5453151660621448514" /></a> A general $1$-circulant is therefore a combination of three transforms, each of this form. An alternative choice of phases for $(231)$ would have given us <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi2_a_CKh3ndqYVaYR9L6Z5tLxIisEAWxeJGoVjHiPrthL4Btfo48xUpu8x0PYIvJxmFOvvNxmUrC4fC3ypNPq8nIE6MRSoM-63Dessg9iUlsblURPqvbdZ9xwjBGT4rJUTYcac/s1600/threeLikCKM.png"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;width: 228px; height: 89px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi2_a_CKh3ndqYVaYR9L6Z5tLxIisEAWxeJGoVjHiPrthL4Btfo48xUpu8x0PYIvJxmFOvvNxmUrC4fC3ypNPq8nIE6MRSoM-63Dessg9iUlsblURPqvbdZ9xwjBGT4rJUTYcac/s320/threeLikCKM.png" border="0" alt=""id="BLOGGER_PHOTO_ID_5453155500732467794" /></a> which is a bit more reminiscent of <a href="http://kea-monad.blogspot.com/2008/10/ckm-rules.html">CKM</a> symmetries. Since the <a href="http://kea-monad.blogspot.com/2009/07/m-theory-lesson-286.html">tribimaximal</a> mixing matrix may be expressed as a product $F_3 F_2$, one hopes that the CKM matrix is also easily written in terms of a natural transform.Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com1tag:blogger.com,1999:blog-28857369.post-17487982900891827952010-03-26T14:08:00.006+13:002010-03-26T14:21:37.945+13:00Magic Motives IIIf sets and vector spaces are secretly the <a href="http://kea-monad.blogspot.com/2010/03/magic-motives.html">same</a> thing, and spaces are built out of these new sets, then there should be one big category of spaces that has everything in it, including invariants!<br /><br />For instance, a vector space of dimension $n$ over a finite field may be represented by all $n$-tuplets of MUB operators representing the field. Now these tuplets are just collections of arrows in the category of fancy sets. A triplet of $3 \times 3$ matrices with cubed roots of unity would act on a $27$ element set, via Cartesian product. The collection of all finite dimensional vector spaces over $F_3$ would live in the set subcategory made up of Cartesian powers of the three element set. <br /><br />Similarly, spaces for the two element field $F_2$ may occupy powers of the two element set, usually denoted by $\Omega$ in the ordinary <a href="http://en.wikipedia.org/wiki/Topos">topos</a>. This suggests that a power set <a href="http://kea-monad.blogspot.com/2006/12/m-theory-lesson-8.html">monad</a> for fancy sets might have something to do with invariant functors for $F_2$ (or even $2$-adic numbers). How cool would that be? The world of motives would then return to <a href="http://www.jmilne.org/math/xnotes/MOT.pdf">Grothendieck's</a> dream.Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com1tag:blogger.com,1999:blog-28857369.post-65876829044575144842010-03-25T10:12:00.003+13:002010-03-25T10:24:49.107+13:00So FewThanks to <a href="http://science-professor.blogspot.com/2010/03/lounging-students.html">FSP</a> for mentioning the <a href="http://www.aauw.org/research/whysofew.cfm">Breaking Through Barriers</a> website. Their new report on the problems faced by women in science contains basic suggestions on what science departments can do to improve the situation, including:<br /><br />Teach professors about stereotype threat and the benefits of a growth mindset.<br />Take proactive steps to support women.<br />Ensure mentoring for all faculty.<br />Learn about your own implicit bias.<br />Take steps to correct for your biases.<br />Create clear criteria for success and transparency.<br />Raise awareness about bias against women.Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com11tag:blogger.com,1999:blog-28857369.post-35662947020983601742010-03-25T09:29:00.011+13:002010-03-25T09:55:39.716+13:00Job HuntingThere is <a href="http://www.frst.govt.nz/">one</a> Kiwi government source of science funding for which I am still eligible to apply. Other countries have similar funding processes. However, in order to apply for such grants I require a host institution. The Kiwi universities don't even reply to my emails, and neither do most of the foreign ones. Occasionally I receive a polite rejection. Fellowship applications are ongoing, although now clearly pointless. Harvard sent me an amusing rejection email, but most places don't bother. With so few jobs, one doubts that anybody bothers to read the Job <a href="http://academicjobsonline.org/ajo">Wanted</a> ads. <br /><br />Winter is fast approaching. Should I take up a cleaning job or a mountain job? They each have their advantages and disadvantages. Both would require me to move yet again. Neither would give me much opportunity to do research. The standard of living here is not high, for the working class. The average person spends 40% of their income on rent, and food prices are now international. Even with a full time job, my expert budgeting skills would come in handy. <br /><br />What are the alternatives? In the local news today, health law <a href="http://nz.news.yahoo.com/a/-/top-stories/6978690/starving-woman-refuses-offers-of-comfort/">expert</a> Jonathan Coates said the <a href="http://www.legislation.govt.nz/act/public/1990/0109/latest/DLM224792.html">Bill of Rights</a> Act and <a href="http://www.teara.govt.nz/en/law-and-the-economy/1">common law</a> were clear about a person's right to starve themselves to death, even if other people didn't like it. I guess effective hunger strikes are a real possibility here. <br /><br />In the past, disgruntled antipodeans would simply leave, never to return. This is becoming more difficult these days, with emigration to Europe or the States requiring either a job offer or tonnes of dosh. Well, at least the sky here is still blue.Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com2tag:blogger.com,1999:blog-28857369.post-69489750417662414162010-03-23T13:11:00.022+13:002010-03-23T14:39:46.835+13:00Magic MotivesThe most fashionable of all phrases amongst real mathematicians studying QFT, occasionally mentioned here, are the words <a href="http://en.wikipedia.org/wiki/Motivic_cohomology">motivic cohomology</a>. This is supposed to be the <a href="http://www.ams.org/ams/mathnews/motivic.html">mother</a> of all cohomology theories. But what is cohomology anyway?<br /><br />As a basic concept, cohomology relies on the logic of ordinary <a href="http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle">sets</a>, particularly the difference between unions and <a href="http://en.wikipedia.org/wiki/Disjoint_union">disjoint</a> unions. The latter type of union keeps track of which set each element actually came from, so it tends to double count intersections. In the category <span style="font-weight:bold;">Set</span>, disjoint union is a <a href="http://en.wikipedia.org/wiki/Coproduct">coproduct</a>. Ordinary unions are a little less natural, but intersections are <a href="http://en.wikipedia.org/wiki/Pullback_(category_theory)">pullbacks</a> of subset monics. <br /><br />Universality is a basic categorical concept. A cohomology theory usually assumes a standard set of <a href="http://en.wikipedia.org/wiki/Eilenberg-Steenrod_axioms">axioms</a> for a cohomology functor $H^{*}$ from some category of (commutative) spaces (which we want invariants for) to a category of groups (where invariants live). The existence of group inverses is related to the arbitrariness of path directions in a commutative space. A universal cohomology theory is supposed be about an elusive category of <a href="http://en.wikipedia.org/wiki/Motive_(algebraic_geometry)">motives</a>. <br /><br />But thinking of <a href="http://kea-monad.blogspot.com/2007/12/motive-madness-ii.html">the field</a> with one element, what happens if we have sets with funny actions, instead of just sets? Are our spaces necessarily commutative? Perhaps the underlying logic of intersection and union should be modified to properly account for the geometry of the absolute point. This might require drawing higher dimensional limit diagrams, or looking at higher dimensional target categories, not necessarily groupoids. Fortunately, some really smart people are now thinking about <a href="http://arxiv.org/abs/1001.0228">noncommutative</a> motives, but it may take a while for us poor physicists to see what is going on.Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com1tag:blogger.com,1999:blog-28857369.post-77887380119310331032010-03-22T10:46:00.005+13:002010-03-22T11:24:42.314+13:00M Theory Lesson 307Such <a href="http://kea-monad.blogspot.com/2010/03/m-theory-lesson-306.html">Hermitian</a> matrices provide a basis for all $3 \times 3$ Hermitian $1$-circulants in the form <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhq3QEWxcb4CH9YyYIXt_mWQ8Cts0sxuuxFkKUS5Oo-H7rvSz0LQpy9cI0saEkhckdX8dJ22BkuVbyjIbQxkbkDMRJJNabhTaZLSAFj96-64R8T4k6UBE-1Rv_vscJ5cJv2EGTm/s1600-h/basisOneTew.png"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;width: 320px; height: 144px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhq3QEWxcb4CH9YyYIXt_mWQ8Cts0sxuuxFkKUS5Oo-H7rvSz0LQpy9cI0saEkhckdX8dJ22BkuVbyjIbQxkbkDMRJJNabhTaZLSAFj96-64R8T4k6UBE-1Rv_vscJ5cJv2EGTm/s320/basisOneTew.png" border="0" alt=""id="BLOGGER_PHOTO_ID_5451206890001682290" /></a> Since the eigenvalues of each basis matrix take the values $(1,1,-1)$, the eigenvalues of the sum are of the form<br />$\lambda_1 = \alpha + \beta - \gamma$<br />$\lambda_2 = \alpha - \beta + \gamma$<br />$\lambda_3 = - \alpha + \beta + \gamma$<br />providing an alternative characterisation of <a href="http://carlbrannen.wordpress.com/2009/01/13/new-paper-on-hadrons-and-koides-mass-formula/">mass triplets</a>, replacing <a href="http://www.brannenworks.com/koidehadrons.pdf">Brannen's</a> $v$, $s$ and $\theta = 2/9$ with<br />$v = \alpha + \beta + \gamma$<br />$s = \frac{2 \gamma}{1.10265678}$<br />Note that when the Hermitian matrix is corrected by the appropriate phased diagonals, it arises as the Fourier transform of a circulant with only three non zero entries, representing the mass triplet. This Hermitian basis is only slightly different from the one derived by Brannen in his <a href="http://www.brannenworks.com/Gravity/EmergSpin.pdf">recent paper</a> on generations.Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com3tag:blogger.com,1999:blog-28857369.post-33803514301766759032010-03-21T14:22:00.019+13:002010-03-21T17:05:40.029+13:00M Theory Lesson 306The (<a href="http://en.wikipedia.org/wiki/Wreath_product">wreath</a>) product used in the theory of the field with one element is between the <a href="http://en.wikipedia.org/wiki/Symmetric_group">symmetric</a> group $S_d$ and $d$ copies of the $n$-th roots of unity. For $\mu$ the primitive $n$-th root, an element of the product group is usually written like <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhaj7unXpfbn6GjiXyclJaIq3U22WtI0wCOz9_j2uUgX-DXLMq4QpQyIJI10k8wz_Nicg8iiw1Yo_9h6vRwylC7OUwYAB_KDvdBaYD-YdL3eGT8hg_SMhqzn-z1oOsW7D4Ml-I6/s1600-h/wreProdDef.png"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;width: 320px; height: 73px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhaj7unXpfbn6GjiXyclJaIq3U22WtI0wCOz9_j2uUgX-DXLMq4QpQyIJI10k8wz_Nicg8iiw1Yo_9h6vRwylC7OUwYAB_KDvdBaYD-YdL3eGT8hg_SMhqzn-z1oOsW7D4Ml-I6/s320/wreProdDef.png" border="0" alt=""id="BLOGGER_PHOTO_ID_5450891986730623682" /></a> and the multiplication is given by <a href="http://en.wikipedia.org/wiki/Semidirect_product">semidirect</a> product<br />$(g_1 , \pi_1) \circ (g_2 , \pi_2) = (g_1 \cdot (\pi_1 g_2) , \pi_1 \pi_2)$ otherwise known as a <a href="http://www.tac.mta.ca/tac/volumes/12/14/12-14abs.html">2-group</a> composition. For this particular group, the semidirect product is given by ordinary matrix multiplication <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEinu7nTZUjHnQnTpoQAmIbC7BS68cUVAbvAXy7LlsnT29j4gChnRQggqEvPxs4IkCcyK-_nsgMKtnpzHJ6FE30EdVHLwNPqZ2_qGqrXD1qpAO_8oyemUyU42JDOcERTJBlLpMDk/s1600-h/wreathMateg.png"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;width: 320px; height: 58px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEinu7nTZUjHnQnTpoQAmIbC7BS68cUVAbvAXy7LlsnT29j4gChnRQggqEvPxs4IkCcyK-_nsgMKtnpzHJ6FE30EdVHLwNPqZ2_qGqrXD1qpAO_8oyemUyU42JDOcERTJBlLpMDk/s320/wreathMateg.png" border="0" alt=""id="BLOGGER_PHOTO_ID_5450891908516507330" /></a> The examples assume that $d = 3$, but $n$ is fairly arbitrary. How does the three dimensional Fourier transform act on such phased circulants? We may decompose the result into a diagonal and a $1$-circulant, as in <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgxlz8zDI3FpkfNMwgDMTF2txrF2PAgjx04a1W2CPI_ZloGe8OHJ7tTpwM1TsnbIMo86q9PLptYjTa0kKIDhXj3coYKg0VtvA4Hl7nCub-3UQyd6Tpc29J8ft_jd2DUFWByZFI5/s1600-h/fourier3phased.png"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;width: 320px; height: 88px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgxlz8zDI3FpkfNMwgDMTF2txrF2PAgjx04a1W2CPI_ZloGe8OHJ7tTpwM1TsnbIMo86q9PLptYjTa0kKIDhXj3coYKg0VtvA4Hl7nCub-3UQyd6Tpc29J8ft_jd2DUFWByZFI5/s320/fourier3phased.png" border="0" alt=""id="BLOGGER_PHOTO_ID_5450891803979784738" /></a> Consider the case where $\mu^a = \mu^c$, reducing the circulant to <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjNFZM1sIeQHf3saL3zlo1TkfNX1eujiOR9K7xImzHJn3nwvXXSOZ4RQwiZ_BW-SdmnuuP60z9TT8gqWUS-Nhkrjoi6g96dZgBx8RuVGih0SdW1EnzhyphenhyphenbfgjSHR_boh8eqgtuFG/s1600-h/wreathMatd.png"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;width: 320px; height: 78px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjNFZM1sIeQHf3saL3zlo1TkfNX1eujiOR9K7xImzHJn3nwvXXSOZ4RQwiZ_BW-SdmnuuP60z9TT8gqWUS-Nhkrjoi6g96dZgBx8RuVGih0SdW1EnzhyphenhyphenbfgjSHR_boh8eqgtuFG/s320/wreathMatd.png" border="0" alt=""id="BLOGGER_PHOTO_ID_5450899268292156242" /></a> where the diagonalisation of the ordinary $1$-circulant permutation is clear, if $\mu^a = \mu^b$. In order to be Hermitian, it is clear that the phases must satisfy $\mu^a = \pm 1$ and $\mu^b = \pm 1$. For example, when $(\mu^a , \mu^b)=(1,-1)$ then the two real eigenvalues for this operator are given by<br />$\lambda = \frac{1}{3} (1 - 4 \textrm{cos}(\frac{2 \pi k}{3})) \in -1, 1$ <br />which are the values of $\mu^i$ as square roots of unity. Note that the Hermitian condition can only be satisfied here if at least two of the roots are the same, so this is the general situation for such phased operators. The natural generalisation is to replace $S_d$ with the <a href="http://en.wikipedia.org/wiki/Braid_group">braid</a> group on $d$ strands, labelling strands with roots of unity.Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com0tag:blogger.com,1999:blog-28857369.post-23415986581864498222010-03-20T17:31:00.001+13:002010-03-20T17:32:30.808+13:00Quote of the MonthFrom <a href="http://scienceblogs.com/thusspakezuska/2010/03/you_femsplainers_just_see_sexi.php">Zuska</a>: <blockquote>You are not supposed to notice sexist behavior, and everything in our society is carefully designed to help you understand and accept it as natural and just the way things are and evolution and the battle of the sexes and God's will and girls like pink and boys love trucks and men are better at spatial hoo hah and women are so verbal and boys will be boys and act like a lady and don't be a slut and men can't help themselves and blah blah blah. If you should, by some amazing effort of will and education, manage to pull the curtain aside, there will be a great show of smoke and fire and booming voices declaiming from a huge green glaring disembodied head "pay no attention to that patriarchy behind the curtain!" Because, well, you might notice that it's a humbug.</blockquote>Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com3tag:blogger.com,1999:blog-28857369.post-31841557230640800092010-03-19T19:34:00.010+13:002010-03-19T19:50:25.089+13:00Quirky Quasars IIThe <a href="http://en.wikipedia.org/wiki/Halton_Arp">Arp</a> heresy is the hypothesis of intrinsic redshift for quasars, which are believed to be ejected from galaxies. An alternative, using a simple redshift-distance <a href="http://riofriospacetime.blogspot.com/">relation</a>, is to drop the assumption that classical geometry holds for the triangle shown. <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhNOJPJ8vq1f6zVSODavXzzjgDR7jQq_PUzZ_tUpsczuOSJFrxNhYZqGF15cIKYnIlWUkowsdqWEoYjABFitf2jlv5npZf3deYXhyphenhyphenM1_IdN6_OueEzmtFbC1q6WkJVsvF1yyi0r/s1600-h/QuasarTri.png"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;width: 225px; height: 204px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhNOJPJ8vq1f6zVSODavXzzjgDR7jQq_PUzZ_tUpsczuOSJFrxNhYZqGF15cIKYnIlWUkowsdqWEoYjABFitf2jlv5npZf3deYXhyphenhyphenM1_IdN6_OueEzmtFbC1q6WkJVsvF1yyi0r/s320/QuasarTri.png" border="0" alt=""id="BLOGGER_PHOTO_ID_5450230687411582530" /></a> That is, we assume that the galaxy (at low $z$) is fairly close to us and that the quasar (at high $z$) is very far away. But we do <span style="font-weight:bold;">not</span> assume that the distance between the galaxy and the quasar is determined by the pink curve. In fact, the observed filaments connecting quasars and galaxies suggest that the two objects are close together. From the perspective of Earth, it looks like the filament matter travels through cosmic time. If quasars are busy <a href="http://kea-monad.blogspot.com/2010/03/quirky-quasars.html">creating</a> galaxies in the early universe, perhaps they are also creating their futures!Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com1tag:blogger.com,1999:blog-28857369.post-90085779499888445032010-03-19T17:18:00.013+13:002010-03-19T17:38:46.465+13:00Cycles of Time<a href="http://www.maths.ox.ac.uk/contact/details/rouse">Penrose's</a> new book, <a href="http://www.amazon.co.uk/Cycles-Time-Extraordinary-View-Universe/dp/0224080369">Cycles of Time</a>, is due out soon. This discusses his own thermodynamic <a href="http://pirsa.org/06090005/">cosmology</a>, in which there are slowly <a href="http://kea-monad.blogspot.com/2007/07/penroses-landscape.html">decaying masses</a> and information loss. Wrong <a href="http://kea-monad.blogspot.com/2009/08/changing-light-speed.html">for sure</a>, but a step in the right direction. <br /><br />A recent <a href="http://www.newscientist.com/article/mg20527511.300-roger-penrose-nonstop-cosmos-nonstop-career.html">interview</a> also mentions another upcoming book by Penrose, entitled <span style="font-style:italic;">Fashion, Faith and Fantasy in the New Physics</span>. <span style="font-weight:bold;">Faith</span> questions the idea that quantum mechanics is the final word on the universe, <span style="font-weight:bold;">Fantasy</span> is the standard cosmology and <span style="font-weight:bold;">Fashion</span> is being a stringer. Some, such as <a href="http://www.math.columbia.edu/~woit/wordpress/">Woit</a>, would argue that being a stringer is no longer the fashion, but they still seem to have most of the jobs. <br /><br />Meanwhile, <a href="http://uduality.blogspot.com/2010/03/strings-2010-texas-m.html">kneemo</a> provides the link to <a href="http://mitchell.physics.tamu.edu/Conference/string2010/">Strings 2010</a> in Texas. My broadband limitations will not allow me to watch seminars online, so please comment if there is anything interesting here, or elsewhere.Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com4tag:blogger.com,1999:blog-28857369.post-88811863831664227292010-03-17T13:51:00.020+13:002010-03-17T14:30:56.842+13:00Quirky QuasarsThe 2009 <a href="http://www.aanda.org/index.php?option=article&access=doi&doi=10.1051/0004-6361/200912848&view=pdf">paper</a> by <a href="http://david.elbaz3.free.fr/">David Elbaz</a>, <span style="font-style:italic;">Quasar induced galaxy formation: a new paradigm?</span>, discusses the theory that supermassive black holes are responsible for galaxy formation. <br /><br />The team looked at the quasar <a href="http://en.wikipedia.org/wiki/HE0450-2958">HE0450-2958</a> from Chile and <a href="http://www.eso.org/public/teles-instr/vlt/index.html">noticed</a> that a jet of matter from the black hole pointed exactly in the direction of a star forming region in a nearby galaxy. As <a href="http://riofriospacetime.blogspot.com/">Louise</a> would ask: if galaxies are formed this way, even early in cosmic time, then where did all the black holes come from? <a href="http://www.newscientist.com/article/mg20527421.000-supermassive-black-holes--the-fathers-of-galaxies.html?full=true">New Scientist</a> calls this the 64000 dollar question. Goodbye LCDM. <br /><br />Note that this is a neat opposite of the <a href="http://en.wikipedia.org/wiki/Halton_Arp">Arp</a> theory, where galaxies eject quasars. It would explain why the <a href="http://kea-monad.blogspot.com/2010/03/agns-and-quasars.html">local</a> galaxy M33 is missing its central black hole: the black hole has not yet arrived in the centre. That is, the black holes can create galaxies from nearby gas clouds and only later migrate into the galactic centres. The quasar jets can contain so much matter that they might create a whole galaxy out of nowhere!Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com4tag:blogger.com,1999:blog-28857369.post-63317474068678633122010-03-16T09:18:00.012+13:002010-03-16T09:35:27.402+13:00Ribbon Review<span style="font-weight:bold;">Why you should love the <a href="http://matrix.cmi.ua.ac.be/fun/">Field</a> with One element</span><br /><br />In the theory of $F$, matrices are no longer interpreted as linear operators for complex number Hilbert spaces. If we can do quantum mechanics without ordinary vector spaces, using special sets instead, then we can consider the $n \times n$ quantum operators as a reduction of the larger set representing the <a href="http://en.wikipedia.org/wiki/Braid_group">braid group</a> $B_n$. <br /><br />For example, the two strand group $B_2$ consists of the unknot, the unlink, a series of Hopf links and the series of $(2,q)$ <a href="http://www.math.toronto.edu/~drorbn/KAtlas/TorusKnots/index.html">torus</a> knots. These knot diagrams look like twisted ribbons. Recall that a tripling of such ribbons, using six strands, was used in the <a href="http://kea-monad.blogspot.com/2007/03/ribbon-review.html">Bilson-Thompson</a> particle scheme, as shown. <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEif5I66BYnZHw7CJHqTA1UpHebQwX6kg68rJ1brwElymCnXYaamIJNTnq0XN218F-u8jqvA6jr2E1TCS2Kgk97XCfAwZJrSb4dBtYaM9qJ4cNZBoaOsvJazGSTWkyxYEo6d2Dy3/s1600-h/bilsonpics.JPG"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;width: 400px; height: 367px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEif5I66BYnZHw7CJHqTA1UpHebQwX6kg68rJ1brwElymCnXYaamIJNTnq0XN218F-u8jqvA6jr2E1TCS2Kgk97XCfAwZJrSb4dBtYaM9qJ4cNZBoaOsvJazGSTWkyxYEo6d2Dy3/s400/bilsonpics.JPG" border="0" alt=""id="BLOGGER_PHOTO_ID_5448958129402274066" /></a>Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com0tag:blogger.com,1999:blog-28857369.post-15673677309991205482010-03-15T12:27:00.029+13:002010-03-15T13:27:04.245+13:00Prime Evil CatsCategory theorists are fond of replacing the counting numbers by sets, in the category <span style="font-weight:bold;">Set</span>. The trouble is that prime numbers don't make a lot of sense for ordinary sets. If I buy 3 oranges and 5 apples, I can <span style="font-weight:bold;">add</span> them by putting them all in the same fruit bowl. On the other hand, what is special about <span style="font-weight:bold;">three</span> oranges, as opposed to four or six? The category <span style="font-weight:bold;">Set</span> does have <a href="http://en.wikipedia.org/wiki/Cartesian_product">Cartesian</a> product, but there is nothing special about sets with a prime number of elements.<br /><br />The simplest object for which the primes define a <span style="font-style:italic;">space</span> of some kind are the <a href="http://en.wikipedia.org/wiki/Integer">integers</a>, as a special object in the category of rings. But as <a href="http://matrix.cmi.ua.ac.be/fun/index.php/kapranov-smirnov-on-f_un.html">Kapranov and Smirnov</a> point out in their paper, this space has dimension greater than zero, with respect to any reasonable topology. Where are the points? <br /><br />This question leads to a study of the <a href="http://kea-monad.blogspot.com/2007/12/motive-madness-ii.html">field</a> with one element. Recall that ordinary sets are like vector spaces for this field $F$, but sets can also come with extra structure. In particular (for an extension $F(n)$ of the magical field $F$) one has <a href="http://en.wikipedia.org/wiki/Pointed_set">pointed</a> sets and actions by roots of unity. Now sets are usually acted on by <a href="http://en.wikipedia.org/wiki/Permutation">permutations</a>, so we want mixtures of permutations and roots of unity, which is what <a href="http://link.aip.org/link/JMAPAQ/v51/i2/p023507/s1">MUBs</a> were about! <br /><br />The correct actions for $F(n)$ are given by the <a href="http://en.wikipedia.org/wiki/Wreath_product">wreath product</a> of permutations $S_{d}$ and roots of unity. That is, $d \times d$ matrices with only one non zero element in each row and column, with every element an $n$-th root of unity. When $n$ is prime, and when $n$ corresponds with $d$, we get nice MUB type operators. <br /><br />Unfortunately, although quantum mechanics is suddenly a whole lot closer to arithmetic, it is still not clear why quantum $3$-oranges are so much better than quantum $4$-oranges, except that $n$-oranges had better be built from prime ones.Keahttp://www.blogger.com/profile/05652514294703722285noreply@blogger.com1