 Yeah So let me just say two words just as like a so good morning and welcome to the Medgar Evers College math Department Colloquium series. I wanted to say a few words just as a kind of background Zack who I'm meeting today for the first time in person Zack and I worked on this paper for almost what two years for about two years and So we've had this long ongoing relationship. We've done talks together at the joint meetings via zoom We've done sort of talk for the folks in Portugal via zoom and so I just wanted to say Thank you to Zack and and to say, you know, welcome to Medgar Evers College and we're quite happy to have you here and then and then to You know our department You know, we're we're in a kind of rebuilding mode and I think that most important is is is mathematics where mathematics department and so while you know, we're challenged with Managing the mission of the college to to integrate Sort of communities that have been underrepresented in mathematics in mathematics the the operative word here is Integrating into the word of mathematics. That's that's the tool that we have some expertise in and so You know, I just want us to always always always keep that in mind that You know part of what we're trying to do is to grow a culture of mathematics here at Medgar Evers College and in that respect I Want to thank you all for the work that you've done this semester and I just to say that I have two bottles of wine in my office So after this we will have a little bit of a little will infuse the math talk with a little bit of the Christmas spirit So having said that I'll I'll turn things over to Josh and he'll introduce you Thanks everybody Thanks for your introduction Thank you Josh for having me here and it's an honor to speak to you all. So my name is that I'm going to be speaking about topic and Please do stop me at any time and ask questions So Here's the bright outline of what we talked about And first I'm just gonna start with some pictures so here is the general problem We're taking a circle. Hopefully it comes out clearly Maybe in retrospect it was a mistake to make it like yellow on But we've got this circle. I'm gonna say it has length one instead of two by just The one will be a little clearer once we sort of move to algebra and symbols in the way from pictures But we're going to take this circle of length one We're gonna start with angle zero. We're going to go out by some angles And we're going to mark off that one And then what we're going to do is we're going to mark off and that's multiple So what I think about this is you went all the way around by this amount and then the same amount Around here and I do have chosen a pretty large angle So we get kind of chaos looking behavior. So the next multiple like this this this this this and I've done my rudimentary best to simulate a video or a gif inside this presentation. So hopefully this works. So Make the mail start from the beginning So hopefully that had the right effect. So What is happening is we're taking all these multiples and we're filling out so And these numbers are just to give me some runways that I run the video or video if I can know this so And Maybe it's a coincidence. Maybe it's not that I've chosen an angle that actually looks like these are pretty well spaced and This whole box is really going to be about quantifying what that means and Seeing what we can prove about this. So if we do This exact same procedure, but instead we mark off all the different lengths that we get Then we're going to get some very two-way. So to start we're going to have two different lengths We've got sort of the long way and the short one And then we'll do this again. We've got two of these long ones and one short one And it continues and now the color scheme changed So I was using orange and black to mean that we only have two different lengths and then Red blue green to mean that we have three different lengths and so always And these images red is going to be long, blue is going to be medium, green is going to be short And we keep going like this and like this and like this and like this and like this and Maybe you'll notice that I've never actually moved up to four, which I was kind of curious and also if we sort of go out kind of far Maybe it's a very very end of this sequence Here I'd say that these are actually pretty Even as far as our red, our long looks to be about twice as long as our green but It's not too bad I'd say and so We're going to see what we can improve about this So as I said before that sort of circle of length of total angle one and about this interaction line So we're going to specify irrational because a rational a space rational angle is not so interesting Right, you just take multiples and most we've taken denominator many multiples or back Because everything's modeling so So we're the only interesting thing that happens is irrational because Our rational one would not actually be that's right We're not going to fill out that's one because we're going to have all these gaps We're going to end up in the same number of slots, same denominator many slots And there's a statement No, actually never really seen it preferring down, but they kind of I think amazing fact is this You know we're observing empirically for these first 50 multiples was that we're never going to get more than three different lengths I Think that's our surprises, but I think it's something that is Possibly convince yourself up if you just kind of imagine well we've got these multiples going out and then when it wraps around sort of it has to sort of be a shifted version and just happen and By the time you end up with a fourth you've kind of deleted you've already interrupted You don't have one copy of your smallest length of your least common length and then that gets interrupted So that's kind of the behavior that happens. So this is an interesting fact and Then what we're really going to be after is no data and generate our circle to work and put any quotes that is what does that mean and Remind me in a sense. So I'm going to say these data is the longest of open market. So it was the red generically in the case before for the Orange when we only had two different When you say that the way you're doing is you're taking one point and the next iteration and that's what you're measuring, right? And that's red green and so three possibilities So I'm doing is so even though from say This one so our last point was up here. We're going from here all the way around to down there say We're then measuring instead of the order in which you wrote them out. We're sorting the message. So we're taking our multiples and So this is great as iteration. It's just like the next one. Yes The next one when you go around the circle not the previous iteration Okay, and there's only three of those distances exactly and this is Because if we were just tracking one multiple to the next it would just be the same length every time But there's this word behavior because of the record because everything is this model So that's what d theta is It's one we mark things up on the circle and allow for the wrapper What's the longest resulting work? And so this is always going to be bigger than one right because we've got all these arcs we've got exactly m arcs and And The longest one if you sort of replicated out them times is at least as big as every other arc but the total sum of those arcs has to be one because We started they just our partition of the circle and so There's all I speak one. So this has to be at least one And if we had some magical number that always marked off exactly equal to those This is being called We can't get a quality who actually quantified the extent of the failure So we're going to fix data They have not going to change through this and we're just going to look at for a very very long time What does this climb over? It's going to be bigger than uncertainty and it's always going to be at least this big So we'd hope okay for a magical number we can hit one and we're saying no The best we can do is 1.9 or so And we can also say exactly when this quality happens and all the final but This is this is the theorem of Gram and Van Rijn from the 60s and They are you'll do this with Some very nice continued fractions work and that's sort of what the both of us are talking about if you turn this problem into a problem continued fractions then You can't answer these questions, which I think is quite nice It's a very beautiful theory of continued fractions and things are surprising that they're able to answer these kinds of Circle geometry problems if you will so That's I think this really cool theorem of Gram and Van so I'm sure this is clear that sort of for very very large You can always find a larger and so some very large down such that this quantity is About And so sort of your failure to equally space is at least by a factor of 1.9 All right, so in order to talk about the proofs of the theorem that we're going to have here I'm using a set up in two fractions So every rational number has this infinity to the fraction So you can also write up the two fractions for a rational number It'll be finite and there's a correlation between if you're funny to the fraction then you have a rational number and Sort of the inverse converse is true. So these are these are equally fine and Rational versus infinite and irrational And we find this is you take your data, maybe it's number, but maybe it's not but you take the integer part and Then you take the fractional part, which is going to be half by the one over the else Sorry, so just beside all these a's are integers. These are all positive or This one can be zero and then you know And then we got the fractional part And a fractional part we just invert it and then you continue the fraction process So we look at here to break and then here. So it's very just sort of a symmetry to it And it's kind of an element procedure. So we'll see an example of it But it was sort of truncated and say, okay, what are the rational approximations is as you go further down the the end Valuing your continued fraction the coefficient in your continued fraction won't impact it that much It doesn't really matter if the hundred that number of the fraction is a thousand versus a thousand one That's not going to change value very material. So you think that these things is actually convergent to theta and so This is what we call it the convergence, right? They actually do convert in the proper sense And so we chop it up. We'll say that H over K to the number you're done, and this aspect is very nice to agree So it's actually this is the top line and the bottom line are equal because of some nice copranaral results, which is Actually kind of beautiful that We don't get any cancellation, but these are actual This is our current this is our current and there's no Constance very being omitted here for a canceled out So as an example if we just have all one so one's all the way out We're gonna call this one dot just to sort of save us some space and so one dot if any time we see it at the end of the communication we just do all of those and We can I have this be one plus three five all over two And there's a very nice Purpose actually quite a lot exactly you can call this this is only fee is the final size. We'll call this fee so maybe all So we have the Because we're able to replace the one one one one with the itself And it's all for fee We've got our formula for the and you come up with And it's only 16 and We can check through this version that these conversions are actually just good nothing And in fact if we see well our a n's are always just one so h n is the sum of previous to h n's k n is the sum of previous to k n's It has every meaning from the fifth not to the first so That's real nice and we're then able to recover the Sort of classical for the not to result that the limit of the quotients of successive if not to use is exactly the Which also goes under many guys's it's come from the next one but this is kind of conceptualization of Fn over fn minus one There's lots of other very cool things you can say about these convergence We won't need this fact, but they were to show one really cool fact that is that The even ones approach data from above and the odds from below But sort of an alternate being above below below below below which So That's the first set up of impressions and now I'm just going to choose a little more So from data we can introduce xn, which is just to take the first n plus one of these a we're not throw away a zero Just that a zero sort of we care about things model So we take the data mod one a zero doesn't matter all that matters is your main And so we're going to reverse it And it's a kind of cool fact actually that it's equal to the quotient of these successive denominators The proof I think is just And then say to end is we're just going to take data and chop off the first n circles are with a n And so there's also some extra versions there like We can write we can write the data in terms of data ends But these are just more for this concept. So we're just notation. That's going to show up later on This other thing this is a sort of equivalent Notation that I mentioned a couple slides ago. That was one of five. So all we're saying is Mu in sigma if we can chop off some prefixes of them possibly different ones And we were covered the same thing in the fraction then All one and then one two all ones we would just chop off that's pretty good This is an important class of Real numbers you don't write it Maybe a So this is this gives our coven classes and Our home classes show up enough there Because So Because all that matters is you've got prefix a zero and this sort of our number of choices is And we need a screech Perfect before you define that I assume that be in its conjure would be in the same Not No So Okay, actually I guess I said that this is on zero. We can also make that be Really what's going to be important here is for like this first one is It gives us an annual and we can think of going Rocking around negative times And then the last thing we're just on this is the number one over I showed up in that I It's nothing really special about the name. I just wanted to have a name for it and Now we're going to able to set up some of these facts that are going to useful for both proving the result of your commitment and then the So if you combine Some results of Slater and social and also So harding rate is just a Did not a beautifully written chapter of the fractions in their book in the same future of the theory of numbers and It's a wonderful chapter that's an incredible difference. I Think there's nothing too fancy from that chapter goes in here is just sort of nice computational simplifications. That way we'll do But what's important is that we can say okay pick a Constant alpha it's going to depend on and and then we pick a little and and our alpha depending on the land and then little M a whole number to be parameterized by these convergent denominators by these case Then we can actually break out exactly with D theta and you'll see that there's no m's We've got two different forms of this expression, but there's no So this number is constant across all and inside the And that's kind of the core mechanism that that's all these crystal that Through this Slater and social scenery Where you will just say D theta is precisely this and depends only on and doesn't depend on And so using this Years ago was able to show the problem They will show that on such an interval. So here we're going to pick alpha To be zero, which of course we can always do because and plus two is always at least one. It's always a non-negative We're going to start with theta, which is in this equivalence class And we're going to say that This starts to be true after n Well, then we can actually say exactly what this on this whole interval what md theta that is And of course md theta is actually constant on this interval, but The maximum children And so this is This is Very nice rates a quotient of things we have these x's but that's any x's eventually are going to look like You know all all ones But the point is it's just a nice book. We only have six terms Which is very convenient and so using the first of those results. So using This lemma the the top thing where I would prove Darren Van Lester. I'm going to skip over the algebra to see how they're all It's not the most It's really the idea of how they use the result. So using the case alpha zero And we're saying that this little soup of md theta then Is that these covenants? So just sticking in all the inside the argument of the little soup and then we transform here to here using And then we just do a little bit So it's that top on the bottom line Try to double it right Or try to just try double bottom to top. So if theta m plus 2 is theta term It is always bigger than 2 or at least 2 Then this whole quotient is at least 2 And this is happening a lot then the little soup will also be at least 2 could be bigger problem But certainly it will be at least 2 And we'll call that our sort of target value Our row is 1.89 right small Sort of saying if this thing happens Then this is going to be at least 2 But theta m plus 2 being at least 2 just means that for call that the theta m plus 2 Begins with a n or a n plus 2 Right, so this being at least 2 means that a n plus 2 is at least 2 So in order to have this little soup drop below Drop below 2 it's going to need to have theta m plus 2 be smaller than 2 always So this being very in 2 can only happen finally off and we want this to be smaller than 2 eventually And so these n plus 2s have better be 1 But our n plus 2s are always 1 eventually We're in the case of That's not the proof that This is a bit is this the proof. Yeah, this is their proof. So so this doesn't use This only uses so hard to write in a folder and they slight they cite state or social sensible So yeah, this is this is a grand business proof and I think you kind of do this in India of How to work with sort of how to transform into convergence and work with these convergence and these ways So now I'm going to talk about a Propulsion about what are good generators and what they mean So suppose we're starting with Again something in the equivalence class because this is already what we want. This is a set of fractions If you're practically good Well, what does it reach? Row from below so it's marked off in green the line And I'm plotting here MD theta of M For each amp. So see I'm axis. This is the MD theta the max is and this is just you know, totally out of control I Don't look at this and see any structure But these that I did choose status which are In the equivalence mass so the question is Can you characterize them that person below? One of the words is there this lower bound M theta For which after that point the M and the M's Stable and they never do this wild behavior Hopping above or maybe popping way above as you see for the second and the answer is yes, so We're going to be able to show that for all these days there does exist this M theta that makes this happen And if you really want me to compute our program It's going to be large, but we could say It'll be kind of indirect from what's written here. This is your error. Yes. Yes. This is this is a result in So what we're going to do is we're going to say go beyond P sorry go beyond and by and As this is guessed by this Hardy-Bricks later, so we're going to pick M between these two numbers to be incidental And so we're going to have of course that This is in this being from Jeff and so Well, anything of M is almost the max row If the whole algebraic condition just going to come down to this In sort of I know I really black box something to go from this to this but all that's happening is We use a sort of lower bound on KM on K little And compare these against sort of the last convergence that we have and I just sort of omitted some out the algebra Especially in my These are the observations and make Use an algebra to come up with this and then the crucial part is that n is fixed D is not So we just need to pick a little D So it's that if you have an empty number the D plus first one is at least that's not There's a function one way enough because this this number is fixed this depends on that data We get to choose this we get to choose however far out of it And so that's how we can compute on it. All we need is to ensure that F of D plus one is at least this and because we're given theta we compute on the data depending on one second And that's all And so sort of maybe one one of our observations is that This quotient here again about this two versus one of this X n term and the rest is just comparing sort of The extent of this portion So we see already to sort of first order It's going to be about two and then how can we enforce that actually drops below 1.9 Which is about what relative? So that is that is this whole thing And this is I've omitted some details, but I think none of the core ideas Actually that is That is full characterization of these good ones where good means it eventually dips below this row So now We're even going to quantify what the best ones are So the best ones Are going to completely bypass We want to start from below So starting I think in texture the goner issue of this so no since I did it and got goner issue this in texture signups and Jeff ms. Oh you proved in 2019 The question is what are the other ones are there other ones first of all and Can we Write them down and the answer is actually so this is this is the other main term in my paper about this that These are all So this one is going right here. So notice that This up to negative sign and up to shifting by a whole number is a goal because this number is too much So There are eight and I say distinct because I want to sort of draw the distinction between these generators and sort of the Further equivalence class of the same up to shifting by a whole number and up to a sign Right because we get the exact same behavior if we go back data or minus data And we just have the whole picture and so we flip But then we're just looking the whole picture. We're not the behavior itself is not any more or less interesting So that's really why I will make the zero problem I thought the a zero promises a one two three four five and those that five. It's all ones all And This is why I call you instead of 16 so each of these has sort of companion, which is it says you have like an extension of So Yeah, because Here's a specific number and then ask where the and so this provides an answer What is the note what is the immigration on the left? If I what applies a to one I believe they are Increasing in numerical value of the smallest representative in the year And then so that's what this subscript is and then these are sorted Let's say the graph so first in one column Then we'll admit that you come that I Sorry in the sense of the sense of that you have more of these things than Yes, I Don't really know Exactly what actually Yeah, I think there's a sign that I think wouldn't address that there So Just a one eyeball it You know these are a to one a to two and It appears that they are all below This row number obviously, it's not I just made my computer do this and it's sped out of this picture That's not true, but certainly it's very convincing That it's clearly approaching from below a good job, you know tip above, but I'm going to claim that I'm going to prove that And we've also eyeball it in this way so we take more of these circle numbers, right? so I think this is out to 75 interest and I have these eight it is and then pie pie I chose randomly as an anonymous pie has a non-repeating very chaotic thing to you fraction and We can look and look at how nice and evenly space this one in that one is an athlete And look at pie with all these very long intervals and then all these brunched-up green ones these long blues long reds and I think this is kind of telling of You know, maybe if we spend too much time staring at it a six and how nice the only space it is We can forget that the generic behavior of the real numbers is much worse much much worse And pie kind of wins it less and you get the same behavior if you choose E or Almost any other real numbers I mean, is there going to be like, I mean, I mean obviously so fee is algebraic Is there going to be if you choose other algebraic mountains? I know fee is really good, but is there going to be difference between trans events all over algebraic choices? I think the difference is going to be with respect to whether or not it's a quadratic serve Sir, so whether it's continued fraction or if it's not so Actually, this is one of the culprits in how you write that The ones with continued fractions with repeated interactions or about your pure I think traction is made with some precepts are exactly the ones that are roots of quadratic equations I guess not sure obviously But beyond that The sort of after these eight or after these Equivalence not to be the nicer forms are probably going to be the one for test repeating So now for this there I mean Yeah, I'm going to sort of You lie over the messy algebra And sort of showcase the nice clean expression so What we're going to do is Say a n is the last one Bigger than one the last continued fraction coefficient. It's very long We won't show that we never have to go further than six and actually if we look at look back to our table We never need to go beyond, you know and equals three suffices But we're going to be able to show algebraically that is actually six or that we can show And then a computer search reveals that it's really three But know what you really know this coming into the group. No, we know this Approaching it without that form and so using our and using Depending on can we're able to show that present exactly right down and the day of them using again our Slayer social and The condition that this expression small and raw after sort of you know, we're in because this a n plus one is just called ratio and all all sorts of mess implications We come out with This means more than all this equivalent to this algebraically Recovered this from that but Now look at this this Number these are all positive The two minus further positive those like one one or something. This is at least two So this we can go around this we can go around So when can they came I just one are a certain amount large This ceases to hold But we have lower bound them how fast these cans They grow as true natural. He said the biggest of them not use So in Anzalee six This has to be violated the cans have to have grown fast enough to violate this sum being small So we know right now that For a number to have for theta to be in the S the class S of Our our friends in three fractions to always say below wrote This n number that's not what I'm asked to be Sorry to be strict responses So now we're going to find some necessary conditions To hopefully reduce to a reasonable computer So we're going to do is we're gonna say for each of these coefficients a one two three four five We're going to look at em Like this interval that I was coming back to and write down every day of them And this means Just a n is done by this first time you're going to have smaller than bro Is it only really? Or again, I'm going to use these can grow super pretty much And this is just more sort of bounding pushing the other go down But if we just plug in n equals one two three four five these numbers Becoming And none of them are actually all that big but in particular We know that a one two three four five are all integers positive integers start at one and So the total number of numbers that we need to check is about 600,000 Not a very large number like Computers these days can handle that and heartbeat I think I ran into it Not a very long time So for each of these so your social intervals We just do the same scheme of little n plus d And then to that one check The reason you need to do that is the worst case Convergent group Or you adopt a non-native or for converting is from this number of this candidate and it turns out that for For this number case 6 plus k 5 over 5 plus 2 to 5 is about 25 25 There's going to be some Vietnamese where it turns out 24 is the one worse just the way the numerics came out and so By reducing down to the state or source argument or the sales of reduction will call Where it's like, okay, we have a finite amount of work to do for each candidate for each of the 60,000 candidates And we just check all 25 or 29 on these intervals And we know this there's this after this What's that? This is a weird question. So nothing is slipped here yet No Nope, and it's a service that allows us to conclude. Okay, we don't need to go up to 29 and we haven't missed anything by That and that's it. That's true So we just ask the computer check all these were most 18 and most 18 14 12 11 and then from there we say well Which ones passed the test And it's they or those or the 16 is it doesn't know about the whole symmetry thing, but that's That's what fits out. And so we can now ask all sorts of sort of heuristics numerics about this So here are limited chubby chubby little wood. So these are two different Sort of lines of inquiry. So chubby chub is a reference to chubby show the bias Is this very strange phenomenon where? I Don't know I don't know the elementary groups are like Chebacca destiny for example tells us that promise or one month for the three month for asymptomically, there's about the same number But if you plot you just count for really large and like number of times are one of four three mark four up to n There's more there three mark for one it seems and This is a result of Sarnac and Rubens the end of the 90s that this always happens that there's always more primes are three month more than one month for even though asymptomically The other six on the other end little wood has this striking result that If you compare the prime counting function, so how many primes are there to end I can see the logarithmic integral of The integral of one over log x from x equals 110 It's been known since like the 19th century that these are asymptomically the same and People had done sort of the numerics of how do these compare absolutely and they found that I think the prime counting function was always slightly larger And it was think actually it was always Little wood and I think the 19 teens or so proved actually they alternate and I Don't know if he established that they all came but there's it's not like minor and cottage industry I was establishing one of the first alternations one and the first alternation point is believed to be something like this So another time that most this number There's a first time that drops below the logarithmic integral It's actually known that there are into the number of alternations that they keep one up each other eventually but No, when in 1910 I was going to know Couldn't check this far out. We still can't check this far out and yet little was able to say no it happens And it happens at 10 to the 300,000 or whatever absurd number. I'm making a story put something Does the alternation happen in a regular way or I don't know, but I suspect not or the very least I suspect that we don't yet know enough Because you know the I think the true the true number actually made me like 10 to the 365 and Then improvements are like agreement the 10 to the 362 So these are, you know, my new confidence in the experiment but the quotient That I don't know if there's any nice regularity So we can ask kind of similar questions of well Does one of these eight numbers do much much better than the others asymptotically long-term term or what we say so we can call W and the data Which does best has the smallest maximum across one third the input capital M because each data rises really often and The answer is yes, each one appears at least as the best one with probability 2.7 percent Some of them are much more popular But the least popular one the least asymptotically good one is good with probability 2.7 percent Which is it's constant time. That's not bad that's like It has a real piece of the pot so I'm not going to talk about proof of this just because I'm not Conceptually as proof is it's mostly pushing around convergence and unaligning algebra algebra that's And it's not very fun to think about But I think it's a kind of punchy result kind of this way to tie this together and if Try eyeball it We'll see that eventually look at red But like this one Gets really high and then drops off and so stops being really bad high up is really bad So this one becomes the worst one and then becomes really good. So by the time we're over here red is pretty good and Right now we're taking my blue. So blue is much worse than red right here and this sort of Little would type behavior of each of the data that's being really good until it shoots up And then back again, and then slowly Was the inverse of you know slowly creeping up and then shooting down Just because we're sort of looking at different angles of it I think even though this I know this is in my store and I know this is kind of like distance-serving to look But that's kind of the way I read this graph So I think it's kind of neat in that way And so I don't idealize application and so maybe you all have heard of this pop-map fact of like the fifth 19 numbers show up in nature all the time and How many leaves do your sunflowers have then How do their leaves grow and how many petals of that kind of thing? I think even this You know cartoon drama I think it's clear that what's happening is that these are This is golden ratio But these are these are the vulnerable And so the pop-map back Is that the leaves of a sunflower to maximize sunlight to maximize the space around that they get around them But they can take the biggest share of sunlight grow not golden ratio and And now we kind of have a proof of this Right if we look at what are the Which are the ones that you value them so which are The best ones I color coded them and I think The golden ratio is a black one. So these are the leaves So if you believe that your sunflower plant is going to grow about 10 leaves or about 30 leaves Then in fact Golden ratio truly is the best one across all numbers and we can quantify that we now have a Rigorous way to say the golden ratio is the best angle for you leaves to grow Here's how now if you think you're sunflower, you know, maybe your sunflower doesn't know how many leaves it's going to grow But if it did if it typically grew between 23rd, then maybe the golden ratio isn't exactly the best maybe is whichever one corresponds to yellow or like But the point is that they're like There's a reasonable interpretation to what What each of these eight numbers means and hopefully it kind of addresses your question from earlier about One way to interpret obviously this is super duper idealized and this is not how nature works, but It's not too terrible in approximation So now I want to just close out by talking about briefly what happens in higher numbers And this is going to be where the quantum computing enter So this is not going directly related to quantum computing, but we have talked about the circle so far This is the first new interior group We can say what about other continuous groups, you know the new interior group as a circle. We have a circle metric Um, but in new other continuous groups often league groups And so we're going to talk about a group we're going to quickly generate a set and It's going to be finite generating set so it can't density generate, but we're going to look at words of length up to n And we're going to hope that they're dense and we're going to say assert that they're dense And we're going to say what Is the largest epsilon ball You know we've got this metric. We've got centered around g this epsilon ball That doesn't mean A set that doesn't mean our words of length up to n And that's what we've been studying this entire time in different languages. We've been studying this for g equals u1 And s a single number a single irrational generator That's exactly what we've been thinking about the largest empty space so largest open interval largest Ball because an interval in one dimension is a is the ball We also answer for s2. So s2 we use different machinery um But we can actually relate to quantum computers and motivation And say okay, well, we've got n qubits And we turn to a hardware design problem And we want to say for a target the gate is what we're going to be able to ask how Um, how far out do we need to go? And for n equals 1 so for 2 at the end um And this is a solid a cthaya theorem which is a sort of class result computing and it's actually the result of parents in mind that This can be improved almost everywhere to there should be a long face 59 59 something arcane and opaque where it's coming from um, but No point is that we get this lower in the dependence and we can fine tune the coefficient in front And now it's kind of open how to improve beyond seven So that means you're um, is your Innovation that is the seven part of the the base of the logarithm is important Our innovation is the application of getting the seven to the continuum seven So the seven number was known in the discrete setting in the context of um half time in romanian graphs Or in lpx graphs to be fair and Demonstration is that well, you can do approximately the same fabrication that they use continues This requires sort of its own um, but That's our main observation and then that's where seven comes from and in the 59 comes from the structure of the j-radar sets of the And so that's everything I say Um, so this is a part of the bibliography. Um, so you look at like this paper this paper they should have The full set of references that are relevant to And Yes, so this is relatively open. Um, ideally Ideally we'll understand I don't think anyone has any ideas at the moment Or or if they do they're you know, very much in their inconceivable um p3 has some progress but The difficulty I think is in setting up the correct Sort of corresponding Committorial objects So as I briefly observed the mention this whole the seven Regime comes from the observation that you can do very nice things with ramanusian graphs and lps graphs And for p3 they set up something called ramanusian complexes Which I don't know much about but they're the you know, there are some social conflicts that generalizes in ramanusian ground And then you actually understand these for the appropriate dimension for dimension that you choose And I think this is Um, quite difficult and Not as well understood. I think there's people but that's kind of That's eb on su tube setting I was trying to think about there is this describe a bit the the optimality like The notion of optimality in the context of the circle and then how to translate out that um, so this is The optimality I believe is in the sense of Covering So the covering exponent is Sort of not exactly or actually not. I think it is it's Maybe I'll try draw a So if you've got this is maybe our SU2 or pu2 or something that's a manifold with three real dimensions And we've got well it's covered by these By this s Up to n is what these dots are drawn and we're interested in What's the biggest ball we can fit that doesn't touch any of these? It's going to have radius I think the covering exponent is something like epsilon is going to be eventually up to constants one over n to the K s or K s is So what we know about only s is that this s is between four thirds and two And I think it's believed that four can never be four years and I don't I don't Know actually even if this is true and to me this is quite difficult. So this is This is done out and star next letter. So this is a letter to answer in politics and This is I think the way which one thing about optimality is What in terms of n is the exponent of the biggest radius we can fit inside Or it sort of fit around the words of like that. Yeah, I think this is the way to think about Not only because I know that we're in a circle I'll pull a point in the picture this is I want to draw a bad circle Um Again, we ask we've marked off those intervals because this is literally You know s upper n where s is a set Hi, right? um We're asking What is the biggest interval that fits inside this manifold here a manifold is one dimension but the base interval Is this red? and so It is actually also of size, you know best case we can still find one size one over hand And that is exactly what the main difference saying because sort of Yeah, something like this is saying well for large n we're actually going to have m c state of m There's no row and we just sort of push it onto the other side this is going to be about Over and so we get into the first That's our covering exponent is exactly So it's actually kind of amazing that With the golden gates in su2 you can get a covering exponent They're going to want right that We can actually do better than linear or one over one. We can do Maybe not quite a drag, but we know that we have sort of Worst case it's only into the fourth third. That's still kind of memorable That's better than the linear that we get on circles Oh, is that is that clarify? I'm sure So I'm just going to speak on the version pizza Yeah, so now the question is do you want to have it here or do you want to have it downstairs? Yeah, yeah, we never know so I'll give her some pizza You know what I'm sort of realizing now too. I did this is actually I think it's in a sense You mean this uh I haven't actually I don't know if I looked at It's possible