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0:00 uh thank you Uh so my abstract has been on the board for a while So hopefully
0:07 you looked at it The short version is I'm going to talk about uh not
0:12 invariance coming from um uh perturbed gausian integration find zero
0:18 dimensional sorry uh zero dimensional QFD so finite dimensional perturbed
0:24 gausian in integrals uh there is a short question and answer uh session which basically says are they
0:31 good for anything and the answer is yes Okay Now uh everything is joint work
0:36 with Roland Vanderbane and there is a URL omega epsilon beta over there and
0:43 there are various links to it throughout the talk You're welcome to check it You're also you welcome to use the QR
0:49 code Uh the talk follows a handout and I hope you all have a copy
0:56 though it seems that several people went up to the second floor without getting one There are more at the hands of
1:02 Matias over there and also here Uh and when I was preparing the handout I
1:09 thought oh wait this conference is at least partially in honor of Stavos's uh
1:15 birthday So there should be a happy birthday Savos uh banner at the top
1:21 right of the handout And uh I thought hey wait maybe I'll try experiment and
1:27 try and ask Chad GPT to make it Uh but first I had to know if Chad GPT knows
1:34 anything about Stavos So I asked Chad GPT who is
1:39 Stavos and it turns out that Chhat GPT knows quite a lot So here's a picture
1:46 over here and Star Wars Gophalides is a Greek American blah blah blah at
1:54 Shenzhen Shenzhen and he's also related to the Max Planck Institute of
1:59 Mathematics at Bomb That's a nice summary Academic background and career again it's all right Georgia Tech Greece
2:07 Chicago impressive and research interests Impressive list of research
2:13 interests We all know it Uh selective publications Again an impressive list of
2:18 publications I want to highlight on the Melvin Morton Rosanski conjecture because it's well it's joint with me but
2:26 it's also uh the basis in some sense of the talk I'm going to give today Uh
2:33 awards and honors lots of awards lots of honors Uh personal interests Well Chucky
2:40 PT even knows about his interest in beekeeping so it really knows uh
2:47 everything Uh so I said "Make a banner for his birthday." And here's what I
2:55 got That uh leaves something to be desired So I decided let me ask Chip to
3:05 try to improve it So first of all can you make the banner related to his professional interests and that's
3:13 certainly a slightly better one You know this does look like mathematics in the
3:18 background Uh but it's still Gophalites with two Fs So uh but Gophalites is with
3:25 one F not with two And I have no idea what went on in Chad GPT's brain when I
3:35 asked this question I really have no idea But it came out with this
3:40 [Music] answer Beat me Beat
3:48 me But at least the mask became a bit more relevant So I said "Stopos please
3:55 No draw." And here's what uh he gave And it's getting there It's actually getting
4:02 there except for the picture So uh can you replace the
4:07 portrait there with a portrait of Star Wars Geropalitis chap Chip says "Yes I can do that Please upload the photo." I
4:14 said "But are you already have a a portrait of Geropolitis because I sent it because you showed it before." Uh but
4:21 apparently they cannot use specific images because I don't know copyright
4:26 issues some issues whatever So they made me upload it made me upload a picture I
4:33 uploaded this one and got that and now that is acceptable Here is tavos correctly
4:41 spelled the right picture formulas that are vaguely relevant pictures of notes
4:47 So it's at the top of the handout Happy birthday dear Savos
4:54 Okay Uh you know I now have my own acknowledgement and uh I have to mention
5:00 prior art What I'm talking about is not in in in air it uh follows work of
5:06 several other people uh in particular that same paper that I mentioned before
5:11 which in some sense uh started this uh and let's start maybe but though I'll
5:18 still start with the dreams the dreams are usually at the end of the talk and therefore hardly noticed but let me tell
5:25 you what my dream is and then the talk will somewhat justify it so the dream is
5:30 that given a note K with a ciphered surface sigma Here is an example I found
5:37 it on the street in Mexico City So um here is a ciphered surface of the
5:44 boundary of this So it's a surface It's embedded in R3 It's orientable Its
5:50 boundary is a note and in fact it's a cipher surface of the TFO node And the
5:56 dream says is that there is a lrangian basically just a function which
6:03 look like a Gaussian plus a perturbation uh L sigma defined on six copies So the
6:11 direct sum of six copies of H1 of the first homology of the surface with
6:19 coefficients in the reals and with values in the reals and whose coefficients are low degree finite type
6:26 inving variance Uh so for example the quadratic part of the lambdian will should somehow
6:33 be related to the linking form of uh uh
6:38 the cipher surface So if you have a ciphered surface and you have two homology classes on it So uh here is
6:46 their picture then their union forms
6:51 a embedded graph because they intersect and then
6:58 you can look at finite typing varants of that embedded graph for example the linking number of the two components
7:04 when you split them apart a little bit and that's a quadratic form on H1 and
7:10 that would be the quadratic part of of of that lrangian and then the hope is
7:16 anyway that the perturbed Gausian integral that the formal Gaussian
7:23 integral of the exponential of that lrangian colute Z and this will be
7:28 explained in a few moments uh will be a naughty variant Furthermore
7:36 as I will show you it's a very very strong and very nicely behaved naughty
7:41 variance and we dream that the formulas are simple and natural and that they
7:47 have a universal generalization So there are many general I we expect many generalizations and the
7:55 dream is that there is some universal way to get them If this doesn't make sense yet that's fine This is my dream
8:02 Okay it's only a dream Uh that would be sweet And even sweeter is the following
8:09 So you see um special classes of form of knots For example ribbon have special
8:15 classes of ciphered surfaces If you have a ribon knot then it has a ciph So here
8:21 is a ribbon It's a knot in red which bounds a disk that is allowed to have
8:28 self intersection but only self intersection of one particular type the type shown here And if you have a ribon
8:36 knot then it turns out that it has a cipher surface So you can remove each of
8:42 the singularities by replacing this with that And then it has a cipher surface
8:48 that has a um um generally the homology will be 2G
8:55 dimensional It will have G uh uh classes of homology homology classes that are
9:02 represented by the unlink So all the formulas that will appear
9:09 here should be much simpler and therefore you should be be able to say something about ribbon The easiest
9:17 example of this is the Alexander polomial For exactly that reason you can
9:23 show that the Alexander polomial of a ribon knot is always satisfies the so-called fox minor condition whatever
9:30 that is Uh and the hope is to do better than that Anyway that was all dreams and
9:38 let's start with an example So the example is in some sense related to the
9:44 le algebra SL2 uh though we are not going to see it and uh so let me start
9:51 describing it So let t be an indeterminate Let epsilon be a formal
9:57 parameter whose square is equal to zero Later I will make the cube equal to zero
10:03 and the fourth power equal to zero For now the square is equal to zero and
10:09 start from a knot Draw the knot as a long knot So
10:15 kind of cut it open and draw it in such a way that at the bottom it's pointing
10:21 up at the top it's pointing up and at each of the crossings the two strands
10:28 are pointing up So you know the two strands at each crossing must be pointing up And of course you can
10:34 arrange it because if they're not pointing up you just turn them around Okay Now once you do that
10:42 uh you can label the edges 1 2 3 4 And
10:48 then each edge has uh begins starting up
10:55 and end sorry begins pointing up and ends pointing up So uh it has an overall
11:04 rotation number right if an edge begins pointing up and adds pointing up the tangent uh makes an integer number of uh
11:13 rotations So mark this rotation number as well Uh in fact it's zero here zero
11:19 here zero here and uh minus one for edge number
11:25 four Uh so the only rotation number that I've indicated is lowerase 54 is 54
11:33 which is minus one All the other rotation numbers are zero And let me say
11:38 that if an edge has a rotation number I cut it artificially one more time just
11:45 before the uh the next crossing So here is edge number five 6 7 and
11:54 8 Given a not presented in this way let
11:59 me tell you how to write a formal Gausian integral Even the word the word
12:07 formal gausian integral will be explained in a moment Formal formal procured gausian integral So first of
12:13 all it's an integral over uh r to the
12:19 14 14 being 1 through 7 You ignore the
12:24 last edge uh times two Uh and the variables will be called p i and x i So
12:33 you have P1 to P7 and X I X1 up to X7
12:38 And the measure of the real numbers is just for normalization It's this it's not quite the standard deb measure It's
12:45 2 pi to the minus one/2 times the standard deb measure And then the
12:52 integrant will be a product of uh four terms red green blue and purple where
13:01 the red term comes from the first crossing The the green term comes from
13:07 the second the blue from the third and the purple comes from the rotation numbers
13:13 number Each of these terms is essentially an exponential of a grand So
13:19 it's a product of exponential So it's really an exponential of the sum So it's
13:24 really the integral of the exponential of sum of terms each one having to do
13:31 with an individual feature of the knot And uh and and here are the formulas So
13:39 if you're looking at a crossing between strums I and J So my convention is that
13:49 a crossing is labeled by the incoming
13:55 upper strand and the incoming lower strand This one is called I This one is
14:01 called J So this one is a crossing between one and five Likewise this one
14:07 is a crossing between six and two and so on uh and then each crossing also has a
14:14 sign s In this example all the crossings are positive So s is always + one And
14:21 then the lagrangian corresponding the local lrangian corresponding to a
14:26 crossing is x i * p i + 1 - p i + xj *
14:32 pj + 1 - pj plus some other mess plus
14:38 epsilon * the sign time some further
14:44 mess I have nothing good to say about this mess except that it works You don't have to memorize it because it's on your
14:53 handout Likewise there is a lrangian corresponding to having a
14:58 rotation minus one on edge four So having rotation five on edge i adds the
15:07 following contribution to the lrang And again you don't have to memorize it That's not the point I don't remember it
15:13 Not remotely Okay Uh so therefore z
15:21 uh is the integral of e to the lrangian
15:26 of the tfoil dp1 to p7 dx1 to dx7 Well
15:32 where the lranian of the tfoil is the sum of all of the terms written before So a red term for the red crossing a
15:39 green term for the green crossing and so on uh and then you can compute this
15:45 integral I will show you how in a few minutes And the result is the Alexander
15:53 polomial times to to the power minus1 times the exponential of epsilon time 1
16:01 / the lrangian polomial squared uh times something else which is
16:07 actually new Okay And uh in general the
16:13 result will be delta minus1 * the exponential of epsilon * something
16:18 divided by delta squared where delta is the Alexander polomial times the
16:23 something new let us call it row one row after Rosansski and overbe who uh wrote
16:31 about it before and uh that's all now this is you can do it for any
16:40 not and you get a row one for any knot and
16:47 um you know what maybe I'll skip explaining the terms of the lrangian and
16:53 just tell you uh the next theorem so the theorem is that well this
17:00 row one or this Z is a not invariant namely it is invariant under the rider
17:09 moves except Except I have to modify the Rydermeister moves a little bit because
17:14 I also care about rotation numbers So for example there are two versions of Rydermeister 2 And in fact this is
17:21 slightly wrong because there is actually a third version of Rydermaster 2 Doesn't matter there There's a little bit more
17:29 moves to check because I also care about rotation numbers And then there is also the swirl move that says that if you
17:36 have a crossing and you swirl around it you add rotations all around it nothing
17:42 changes What's the proof so the proof is essentially to use Fubini Uh you you
17:50 need to show that the invariant of the left hand side of a rider 3 is equal to
17:56 the invariant of the right hand side you integrate over all the variables that
18:01 appear within the picture of the move and uh namely you decompose the
18:08 integration Instead of writing the overall integral as one big integral you
18:13 use fubini and you first integrate over the variables that appears on the appear
18:19 on the left hand side and the variables that appear on the right hand side If those integrals are equal then whatever
18:26 you're going to integrate later is not going to change that So uh basically that's the proof
18:35 Yes But for rather master one and two this is that mean there is more terms on
18:40 the right hand side than on the left hand side Yes So here you integrate over zero
18:46 variables Here you integrate over whatever it is four variables And the integral over four
18:53 variables is going to give the same answer as not integrating anything
18:59 Okay Okay Uh but but but now I have to
19:05 explain what do I mean by integration and how do you
19:11 integrate Okay easy said integrate but you know it's not so easy
19:17 So the goal for this uh little subsection is to understand how to
19:24 integrate over Rn So this is zero dimensional quantum field theory Uh the
19:31 exponential dx x runs over rn the exponential of a quadratic form So min -
19:39 one/2 a i jx iixj summation understood plus a term a perturbation
19:47 term that should in some sense be smaller So for example it could be preceded by epsilon where epsilon squar
19:55 is set to zero or epsilon to the 5 is equal to zero Okay So how do you
20:02 integrate this and the answer is uh physicists
20:07 know how to do it Mathematicians fewer mathematicians know how to do it But
20:12 it's something that everybody should do it And let me tell you how to do it Though my way of doing it is slightly
20:18 different than the usual way Okay So what I do is I define calligraphic Z
20:25 lambda of X So I slightly general or not general but massage this by dividing the
20:34 quadratic form by lambda integrating over changing the
20:39 variable to y but within the perturbation I write v
20:45 of x + y and then my goal is to find z
20:50 lambda of x and uh if I restrict my attention ion
20:56 to z1 of x is equal to zero If I set x equal to zero then x disappears and this
21:03 is the integral dy of the same thing as as is written here with dx So really my
21:09 goal is just to compute z1 of zero Okay
21:15 uh in the limit So if lambda is set to zero or rather in the limit as lambda
21:22 goes to zero um you have exponential of negative of a
21:28 quadratic form and you know that goes to a bump function which converges to a
21:34 delta function So uh Z0 of X can be computed and it's simply well it comes
21:41 out the determinant of the quadratic form uh times the exponential of V of X
21:47 and then uh uh with GI J the inverse
21:53 matrix matrix of AI it turns out that you can write a differential
22:00 equation for uh Z lambda of X and it's roughly a heat equation I'm going to
22:06 spare you the derivation It's here on the handout You differentiate under the integral You integrate under the
22:13 differential I don't know You do the usual tricks And at the end you find
22:18 that the derivative with respect to lambda of z lambda is kind of a llashian
22:26 operator applied to z lambda of x But that's a linear OD And the
22:33 solution of a linear OD So okay this is written here And the solution of a
22:39 nonlinear OD is at z of x is equal to the initial condition to which you apply
22:47 the exponential of the of the llian to to which you apply the exponential of
22:53 the operator Okay And this in fact is the birth of fineman diagrams So I'm not
23:02 going to show it because it's a little bit irrelevant But uh if you write the
23:08 if you think of the exponential as a sum of vertices then this is a sum of ways of
23:16 contracting these these ver vertices using uh the inverse of the quadratic
23:24 form And uh that ends up being uh so-called final diagrams
23:33 Anyway so again I'm not expecting you to understand the details That's not the point
23:39 Okay the point is to get the picture So in order to integrate
23:46 uh to make to do a Gaussian integral I compute something more general Z lambda
23:51 of X and I substitute lambda is one and X is zero and z lambda of x is some uh
23:59 differential operator applied to some function
24:04 Okay And then so so the problem of integration gets reduced to the problem
24:10 of differentiation which in principle is easier except it's not really what I
24:17 want because you see I want to apply fubini I want to integrate time after
24:23 time I want to integrate with respect to the first variable and then with respect to the second So I want the answer for
24:31 integration to be written in the same language as the input The input was an exponential was
24:40 an exponentiated thing So I want the answer not to be written straight I want
24:47 to write it as the exponential of something In other words I want oop sorry I went the wrong way In other
24:54 words I want to compute the logarithm of the answer So I let Z sub lambda not
25:02 calligraphic Z just Z sub lambda be the logarithm of uh calligraphic Z sub
25:09 lambda And then again you play game with differential equations If you have a differential equation for calligraphic z
25:16 sublambda you substitute this uh thing into it and you get a differential
25:21 equation for z lowerase z sub lambda And
25:27 uh uh again it's a differential equation now it's nonlinear and the the equation
25:33 is that the derivative with respect to lambda of z sub lambda is equal to a sum
25:38 of gig G and J is the inverse of the quadratic form times second derivatives
25:44 of Z lambda plus a term that has u this is the nonlinear terms dxi of z lambda*
25:52 dxj of z lambda and I'll short this for a functional f applied to z lambda So in
25:59 short the differential equation is d lambda of z lambda d lambda of z is
26:05 equal to f of z Okay where f is some complicated differential equation
26:10 differential operator nonlinear Uh I like to call this
26:16 differential equation the synthesis equation because uh if you look at it
26:22 very carefully you can explain it you can understand the terms in it and you
26:28 can see how it synthesizes connected fineman diagrams
26:33 uh but that will take me too far aside to explain so I will not explain it but it's a lovely uh
26:41 story anyway uh so how do you next come
26:46 how do you solve an OD uh but you know we all taught it when we
26:51 taught ODEs and you use picard iteration so if you want to solve the differential
26:59 equation d lambda of f lambda is equal to uh functional f applied to f lambda with
27:08 a given initial condition Then you iterate
27:14 uh f goes to f0 plus the integral of the functional applied to f lambda and you
27:21 seek a fixed point That's picard iteration That's how you solve differential equations That's
27:28 how you solve it in uh first year second year calculus Okay Uh and in our
27:37 cases the solution is always reached reached after finitely many iterations
27:44 So there is no limiting process here Okay You just do finitely many
27:49 iterations until the solution stops changing and then until the candidate solution stops changing and then you got
27:56 your uh uh solution of the differential equation which means the value of the
28:03 integral and then definition an integral with a g in it is the result of this
28:10 process ignoring convergence So I just define the
28:17 integral of the exponential of a perturbed quadratic to be the result of this process and I really don't care
28:23 about convergence Okay all the story about Lebe integration was only to motivate me
28:29 I really don't care about it anyway Um you know I kind of uh believe in
28:38 implementation I I I I never do mathematics unless I not never but I hardly ever do mathematics unless I can
28:45 write the code uh that shows that it works Otherwise I never believe myself
28:52 So here is an implementation Well I first load some libraries that have
28:57 nothing to do with the content of today's talk Then I define how I want to
29:02 the canonical form of of an expression how I want to display my expressions And
29:09 then I uh and then come the core and the core is in yellow and uh and then there
29:16 are hacks in very very light pink in the background So here is the core Well
29:24 first of all I define blackboard E to mean the exponential So E to the A * E
29:30 to the B is is get simplified to E to the A plus B That's how exponentials
29:36 behave Then I defined that the integral of e to
29:41 a langian d a list of variables v stands
29:46 for variables is a module in which first of all I make a table of the second
29:53 derivatives of the lrangian that's the quadratic form then
29:59 I if the if the quadratic form is uh has
30:04 determinant zero then I cannot invert it and the whole process from before
30:09 doesn't make sense So then you return the generate quadratic form in red and
30:15 stop the computation Otherwise you invert the quadratic form and then you
30:22 find the fixed point of this iterative solution So you define dz to be a table
30:29 of the first derivatives of z ddz to be a a table of the first derivatives of
30:34 the fir of the first derivatives So it's a table of second derivatives and then
30:39 you write the basically the formulas that I wrote before and you uh and and
30:45 you look for a fixed point at the and at the end you substitute lambda to one and
30:51 all the variables to zero because that's what you're supposed to do at the end according to my scheme And so this is a
30:59 messy program Oh and it has some hacks in the background So for example I
31:05 sometimes project using a projection that I call delta dollar pi And pi
31:12 starts as the identity projection So it doesn't do anything But later it will be
31:17 I will call it the wisdom projection Like sometimes I know that the integral
31:22 is equal of something is going to be zero and then I project it out to save
31:27 time and make convergence more reasonable faster Okay But uh so
31:32 basically this is the uh the program Let's check if it works
31:39 So first of all let's do Gausian integration Sorry just standard Gausian
31:46 integral So let's compute the integral of e to the minus ux mux^ 2 Oh sorry
31:54 it's not even gausian integral It's the for transform of gausian times plus e
32:01 plus iix relative to the variable x And as we
32:07 know the integral the the fet transform of a gausian is the gausian with the
32:14 inverted quadratic And that's exactly what you get here You get e to the x i^
32:20 2 / by u Then um let's make it a bit
32:25 more challenging So uh let's make it a a shifted gausian Compute the for
32:33 transform of a shifted gausian So it's the integral of the exponential of - mu
32:39 * x - a 2 This is the shift a plus iix
32:45 This is the fa transform bit and the result which I will call fa of g of
32:52 gausian is this ugly formula and just for fun now let's integrate let's do the
33:00 inverse for transform so let's integrate fa transform of g times the exponential
33:06 of minus iix x and the result is exactly what we started from this is essentially
33:14 uh the proof of the for inversion formula I mean if it works for Gausian
33:20 for shifted G gausian every other function in the world is a linear combination of shifted gausians So it
33:28 works for all functions in the world Okay roughly speaking Um let's uh do some more So now
33:37 let the lrangian the thing we inte the the quadratic form that we integrate be
33:43 a general quadratic form and let's compute its integral uh
33:50 sorry of dimension 2 with matrix a b c and let's compute its integral relative
33:57 to x1 and x2 and this is the ugly result that you
34:03 get On the other hand let's compute first of all the let's set Z1 to be the
34:11 integral of the same thing dx1 and let's check if z12 is equal to
34:20 the integral of z1 with respect to to to x2 Essentially I'm checking Fubini's
34:26 theorem and the computer runs the result uh runs the computation and outputs true
34:34 So we've checked Fubini and I have no idea why Fubini looks so miserable in the picture He should be
34:40 happy Okay Anyway uh one last example just to check that the program is
34:48 sound So uh now my wisdom projection is that if I see a power of epsilon greater
34:56 than 13 so I add O of epsilon to the 13 So if I see a power of epsilon greater
35:01 than 12 uh I I I drop everything So this
35:06 is a computation only up to epsilon to the^ 12 And then I compute the integral
35:12 of the exponential of minus 5^ 2 + epsilon 5 cubed
35:18 d Physicists would recognize it as phi cubed theory And I hit enter The computer
35:27 outputs that this is the answer I don't know if how to test it So I uh enter the
35:34 denominators 55 11105 No sorry that's the numerator into uh
35:42 Google a web search and the web search uh sends me to uh the online
35:48 encyclopedia of uh integer sequences and uh the online encyclopedia of integer
35:55 sequences tell me that the numerators of mass formula for connected vacuum graphs
36:01 on two end nodes for 5 cubed field theory is uh 1551 11 105 565 exactly my
36:10 series So we're on track Okay And uh now
36:16 let's finally compute the TF foil So uh
36:21 the features of the TF foil are that the number of variables the number of pairs
36:26 of variables is seven Uh the features are a rotation number a crossing a
36:31 crossing and a crossing And then the lrangian of a crossing is this is the
36:37 formula from a few pages ago The lranian of a rotation number is this is again
36:42 the formula for a few pages ago The variables is a table of pis and xis And
36:49 then I ask the computer what are the variables for the tfoil and what is the
36:54 lrangian for the tfoil and I get this answer And this was again displayed a
36:59 few pages ago except now this is with the higher authority with a stamp of approval by the computer Okay So draw
37:08 shouldn't where did you tell him that I that i + what is i + 1 and j + one
37:15 Sorry I mean at a crossing you would have four indices Yeah And so I was guessing that for 1 + one I mean one if
37:23 the if the numbers are 1 2 5 6 and you know 1 + 1 would actually be five No
37:29 like no no why no the crossings are um
37:34 uh so uh so does the crossing really have only two indices and not four
37:40 because the indexes in indexing is by consecutive integers So I the first step
37:46 was to label the edges of the note 1 2 3 4 5 So if the incoming uh index the
37:56 incoming over strand is indexed I the outgoing over strand will be indexed I +
38:04 1 And if the incoming under strand is index J the outgoing one will be index J
38:11 + one So there are four indices around the crossing But if I know I and J I
38:17 also know I + 1 and J + 1 I know the other the other two But because I saw a
38:23 P1 minus P5 P1 minus P5 comes from P I minus PJ
38:32 Okay Okay Thank you Okay Anyway uh so now I ask the computer to
38:41 integrate but again I set epsilon squared to be zero uh and this is the
38:48 answer that I get and this is the answer that I uh advertised before and it's
38:54 kind of a lousy way to compute row one but there is a faster program and more
39:00 stories at this reference uh which I will not show but let me just tell you I
39:06 mean all of this wasn't an idle exercise so uh the pair delta and row one it
39:14 makes sense to think of them as a pair because in In order to compute row one you have to compute delta
39:20 anyway So the pair delta in row one attains
39:27 270,000 distinct values on the 313,000 prime not with up to 15 crossing
39:34 and this is a deficit of 43,000 So it misses 43,000 things in some sense On
39:42 the other hand and compare it with the pair home polomial and hine of homology
39:49 uh both of those are exponential time to compute uh or at least non-polomial time
39:57 and they attain only 243,000 uh values on the same dot of same dot
40:05 not a deficit of 70,000 In a moment I will tell you about an
40:10 even better pair Delta and Teta And that one has a deficit of
40:17 6,700 which is really about 10 times better than Homefle and Hovanov taken
40:26 together Uh yet better than Homef and Hovanov
40:32 uh uh delta in row one and row two and theta can be computed in polomial time
40:39 and even for very very large knot and I'll show examples in a few minutes Uh
40:45 so in some sense these invariants are possibly the best we have Hate me hate the formulas I don't
40:54 care this is the best we have Okay Yes
41:00 How are these distinguished in the in the total set that you're considering what invarians do they use to distinguish these knots sorry what do
41:07 you Oh um like you you have you know so there are there are actually complete
41:14 invarants if you really want coming from hyperbolic geometry but they're much harder to compute
41:20 Yes I think these don't distinguish mutation Row one doesn't Teta does I'll
41:27 show you in a minute Okay Uh I think I want to skip the slide
41:34 that says that this is actually a polomial time computation Uh but
41:40 basically the point is that uh
41:45 um you have to invert this quadratic form That is a matrix inversion that is
41:53 in insanely complicated for humans but computers actually can do it rather quickly in fact in polomial time And
42:01 after that uh you have to uh uh sum over
42:07 finement diagrams But the fineman diagrams end up being like if if seeing
42:13 that epsilon squar is equal to zero you only get a like a sum of contribution
42:19 one from each crossing So you get a a sum of uh terms each one having to do
42:26 just a single summation over the not for theta it will become a double summation
42:32 uh but it's still just a summation over the note which is linear time and the
42:37 double summation well it's quadratic time So basically these are for a human this is god awful but for a computer
42:45 it's not so hard Okay Anyway uh let me
42:51 show you just to convince uh a proof of invariance under
42:56 ite So here is the left hand side and uh the edges are called I J and K And then
43:04 the next age edge along this I strand is I + 1 and the next one after is I + 2
43:11 and likewise J J plus And then J++ and likewise for K on the right hand side
43:17 there is a similar labeling and uh let's uh try to so you see um the variables I
43:28 plus J plus and K plus so the X I and so X I plus and PI plus and XJ plus and PJ
43:37 plus so the variables corresponding to the middle variables corresponding to
43:42 the single plus thing are never going to occur elsewhere in the lrangian So I can
43:49 just integrate them right away So I can write the lrangian of the left hand side
43:55 So a cross a positive crossing between strand i and j a positive crossing
44:01 between strand r + one and k that's the second crossing and so on Uh and I can
44:06 integrate it with respect to the middle variables Likewise I can write the lagrangian of
44:13 the right hand side and integrate it with respect to the middle variables and
44:20 ask the computer if the left hand side is equal to the right hand side If they are
44:25 equal we've proven invariance under meister two uh three sorry under meister
44:31 two Unfortunately the computer thinks that it's not they're not equal Bad means let's try let's make a
44:38 second attempt Now you have to look a little bit more carefully at the lrangeian And if you look a little bit
44:45 more carefully at the lrangeian you realize that
44:51 the xi variables so the bottom variables
44:57 but only the x bottom variables are also not going to appear
45:04 anywhere else So I can integrate them out also
45:09 Basically you look at the lrangian you see that it never has x i + one and and and and xj + one in it So therefore uh
45:19 this i and this j is never going to be the i + one of some other
45:24 crossing So uh or well is going to be the i + one of the other crossing but
45:30 that's not going to contribute to the lrandian So I can integrate the x i xj
45:36 xk variables also So my second attempt will be to oops write exactly the same
45:44 lranian but and integrate it with respect to the middle variables but also
45:51 with respect to the bottom x variables and I do it and I ask is the
45:57 left hand side equals to the right hand side and the answer is true So I go home
46:02 happily and then I check my notes and I find I print the left hand side and the
46:08 two sides were equal because both were illegal integrals Both had a degenerate
46:13 quadratic form So that's a bad proof So now I need to do something
46:21 cleverer So I end up needing to do something something cleverer So I need
46:26 to know that uh exactly this integral
46:32 these two integrals are equal except my program doesn't know my gausian integr
46:38 integration doesn't need know how to compute them So I need to show that these two things are equal for every
46:46 value of pi pj and pk because pi and pj and pk are not yet
46:53 set they they they might be coming their value might be disturbs disturbed by by
46:59 other processes So I need to know that this is true for every value of pip and
47:04 pk So what I do is I compute the fur transform with respect to pip and
47:12 pk So I take the same lrangian and I add
47:17 to it um I add to it uh I * pi I pi pi j pj
47:27 and pi k pk that's the for transform and now I integrate also with respect to the
47:34 ps and now the integral is more balanced because I have an integral of both the
47:40 the bottom p and the bottom x variable variable and also all the middle variables and that turns out to be a
47:47 good a computable gausian integral I compute ask if the left hand side is
47:53 equal to the right hand side The answer is true And now I'm just to be safe I
47:58 print the left hand side It's a big mess So it was an equality of non-trivial
48:03 things not of error messages And that actually is a valid proof of
48:09 Rydermeister 3 Uh likewise all other uh rider master
48:16 moves moves are proven in a similar way and if you really want you can check it on uh this
48:24 URL Uh I have to say that uh oh uh you
48:29 must be asking yourself where is this coming from i'll answer that in the last
48:35 minute Uh but before that there is more So there is a row two and this is the
48:41 lrangian for row two Uh there is a a an invariant peta Uh here we have two
48:49 formal variables t1 and t2 and t3 is just the product of the two and we have
48:54 to integrate over six variables for each edge So p1 i p2 i p3 i x1 i x2 i and x3
49:03 i Uh and the lagranian is a horrible mess Again I have nothing good to say
49:09 about it except that it works Uh and here is the verification of Rymeister 3
49:18 And uh again the answer is that it works You can compute it to the for the TFO Uh
49:25 and this is the invariant Teta that I mentioned before And I want to say a few
49:32 more words about it Okay So uh you see the main point of TEATA So uh is it's a
49:40 two variable polomial Laurand polomial and uh then uh there is a a you
49:49 know what I'll skip the faster program I'll just tell you that there is a nice way of visualizing two variable
49:56 polomials So basically okay let's start with a one variable polomial If you have
50:02 a one variable polomial it's a list of coefficients and the list of
50:07 coefficients you can turn it into a list of colors Make the positive coefficients red and the negative coefficients blue
50:15 and you get a sequence of reds and blues and let's call it the barcode Okay So in fact here is the
50:24 barcode for the tfoil knot or the Alexander polomial of the tfoil not it's
50:30 red blue red because the coefficients are positive negative positive okay uh
50:38 but if you have a two variable polomial so if you have a two variable polomial
50:43 then you get a two-dimensional array of coefficients and each one can be red or
50:50 blue So you get a two-dimensional array of colors of pixels and that's a QR
50:57 code instead of a barcode except it's kind of better to uh take the
51:05 coordinates to be uh not quite perpendicular So rather than using a
51:12 perpendicular coordinate system uh for the two powers of the two
51:18 variables you use uh one is in the direction of positive x-axis the other
51:25 one is in the direction of the cubic root of one the the a non-trivial cubic
51:31 root of one So uh uh so instead of kind of a standard QR code you get an
51:38 hexagonal QR code And here is the hexagonal QR code of the TFO knot So uh
51:45 here is the coefficient of minus T So so this would be the coefficient of one and it's zero The coefficient of T1 is also
51:54 zero The coefficient of T1 squared is negative So this is blue Uh this is T1^
52:01 2 * T2 So I went in the direction of the cubic root of one and that one is
52:06 positive so it's red So this is the tf knot Uh just for fun I also computed the
52:15 conway knot and the kinoshipa terasaka knots mutant
52:20 pairs and their con their lexander polomials are both one So these are not
52:26 very interesting uh barcodes but their QR codes are completely different and uh
52:35 Teta indeed sees uh mutation Um
52:42 um here are a few further example So these are the barcodes and the QR codes
52:49 for various Taurus knots Uh this is the barcode and the QR codes for the totals
52:56 not 22 over 7 approximately t subpi Uh this is the barcode and uh QR
53:06 codes of 300 crossing knots So let me emphasize you know for most other noting
53:12 variants just saying 300 crossings is science fiction fiction computing
53:17 science uh computing invariance of a 300 crossing note is science fiction this isn't science fiction here is the
53:24 results here is the result the result seems to have some patterns in it I have
53:29 no clue where they're coming from okay uh
53:35 uh an unproven fact so experimentally ally but also for various theoretical
53:41 reasons We know that the degree with respect to T1 of Teta is less than twice
53:47 the genus of the knot Strictly speaking it's unproven but there is no doubt that it's true The the evidence is just
53:55 overwhelming Um uh and here is an example So this is the 48 crossing
54:04 uh uh G charm not uh which might be a
54:09 counter example to the slice ribbon conjecture And uh here is the
54:15 invariant of the the barcode and the QR code of that knob It was computed by the
54:23 way in 17 seconds uh and if I compute the exponent of t in
54:30 the in the in the Alexander polomial it's eight If I compute the expo the the
54:36 the exponent the the the power of t the
54:41 degree in t2 of the uh uh teta part of the invariant and divide by two uh the
54:49 value is 10 and so uh um uh teta knows
54:55 things about uh this node that the Alexander polomial doesn't it gives you a better
55:01 bound Okay Uh and here is only the QR codes of
55:08 the notes of Indolson table Again just for fun There are patterns in it I have no idea uh what they mean And now it's
55:18 time to say where it came from
55:23 Uh yeah I went way too fast I wasn't supposed to have time to say where it came
55:31 Everything Okay So far it probably isn't because I went way
55:36 too fast Anyway I can tell you where it came from
55:45 I can also tell you that I don't really know And uh where it came from is
55:51 completely irrelevant It came from some roundabout thinking which went very very
55:58 complicated in crazy ways and that's not really where it is The result is more
56:06 the the these things are more fundamental than where they came from
56:11 And so the answer to where is it it came from is it's up to you to figure out or
56:19 maybe up to us We're working on it And I think the answer has to do with the
56:24 dream that I stated in the beginning So it has to do with some or one answer One
56:31 possible answer probably there isn't just one but one possible answer has to
56:36 do with uh lrangian defound defined by
56:43 uh low order finite type invariance of homology classes on a ciphered surface
56:50 for the note uh I think that's the correct kind of
56:57 future rewrite of where is it coming from but the
57:03 historical way in which it came from by which it came from is completely
57:09 different So really it came from um
57:14 um silly things in quantum algebra and
57:21 uh uh let me tell you in in in in in in very few words what these silly things
57:28 are So um in quantum algebra you end up looking
57:34 at universal quantized universal enveloping algebra So basically
57:39 universal enveloping algebra with an additional parameter Universal enveloping algebbras
57:47 are uh isomorphic by the pbw theorem as
57:52 vector spaces to polomial rings So basically you have polomial rings
58:00 with funny multiplications on them right because the uh uh universal enveloping
58:08 algebraas are isomeorphic to polomial rings but but with the wrong
58:13 multiplication So if you can import the multiplication from the universal enveloping algebra to the polomial ring
58:19 and um uh and then you get a funny
58:24 multiplication on the polomial ring Likewise the comm multiplication
58:30 will be some funny co multiplication on a polomial ring
58:35 So you need to learn how to write
58:40 functions between polomial rings or tens or powers of polomial rings which are again just polomial rings with more
58:48 variables For example the multiplication is a fun the multiplication of a of a
58:54 quantum uh universal enveloping algebra is some map from the polomial ring
58:59 tensor squared to the polomial ring So it's just some funny map from a polomial
59:05 ring to a polomial ring and then uh the trick is that uh uh
59:13 there is a an isomorphism So if you want a homorphism from one polomial ring to
59:20 another polomial ring and these are linear homorphisms not multiplicative
59:26 right the whole point is the multiplication gets changed So if you have a linear homomorphism between one
59:33 polomial ring to another you can write it as a a generating function This is
59:40 isomorphic to uh power series in variables dual to the variables here
59:46 with coefficients uh uh polomials in the variables that appear here And I think of this as a
59:55 generating function It's a function of two variables or two sets of variables
1:00:01 that encodes a homorphism between uh two polomial rings
1:00:08 Now uh you need to know how to compose basically you need to know how to
1:00:14 compose maps between polomial rings because like if multiplication is a map
1:00:20 between two polomial rings if you want to check associativity or if you want to do m iterated multiplication you need to
1:00:27 know how to compose multiplications So you need to know how to compose gener the generating
1:00:35 functions corresponding to homorphisms And then comes a funny thing
1:00:40 You can write a formal formula which is kind of true up to the laws of formal
1:00:47 Gausian integration that the composition the generating function of a composition
1:00:53 is equal to you multiply the two functions and you uh integrate it against a quadratic the exponential of a
1:01:00 quadratic form Uh then uh the last thing is that uh you
1:01:09 need to be careful which quantized universal enveloping algebbras to work
1:01:16 with and uh oh my god by now I'm over time uh but there is a little bit of a
1:01:24 story that tells you that uh like the
1:01:31 youq of G can be not quite deformed but
1:01:38 uh degenerated to be shown to be a degeneration of a family uq of
1:01:46 gepsilon So uh a and and then when epsilon is equal to zero everything come
1:01:53 turns out to be easy So you write things as powers of epsilon You expand
1:01:58 everything in powers of epsilon And then it turns out that all the
1:02:04 structure homorphisms which mean all the structure generating functions are
1:02:12 gaussian The integration sorry the composition is again the integration of
1:02:18 gausian things So that's exactly the techniques that I've introduced before and it boils down to the formulas that
1:02:26 I've intro that that I wrote Uh this is wrong This is
1:02:32 absolutely wrong This will this will not stay the right way to see these things
1:02:38 But it is at the moment the correct history of how we got to these things
1:02:44 Thank you for bearing with me