Thank you. It's great to be back. I'm really grateful to the organizers for the invitation. I think the number theory web seminar has developed into something really special. It's great to see people from all over the world. I see some old friends and hopefully some new ones in the participant list. I also like this venue because it's a very broad venue. I know we have brings together number theorists with many different backgrounds. And I want to make sure that I include everyone. So I'm going to take it slow. And I'd encourage you, if you have questions, please feel free to either interrupt or if you prefer put them in the chat. I have a chat window open on another computer, which I'll try to keep an eye on as I go. I should also thank my collaborators, Yung Huiliki, Kaikuan Lee, Thomas Oliver, and Alexi Posnikov. And I should also thank some of the people who helped contribute to this presentation, including some theorems I'm going to present. So I'm grateful to Nina and Zubelina and Peter Sarnak, also to Aaron Assoff, who helped with some of the computations I'm going to show you. And to John Boeber, Andrew Bookerman Lee, and David LaRiduda, who also have a very recent theorem that I'm excited to share with you. So before we jump into the math, I wanted to at least explain the first word in my title, which I've discovered is not necessarily familiar to everyone, but a murmuration refers to a movement of a group of birds. Originally it was starlings, and you can see an example of a murmuration of starlings in the picture, my front slide. But it was later extended to birds more generally and more recently to elliptic curves, as we shall see. Okay, let's get started. So like all good number 30 stories, the story begins with an elliptic curve over Q. So just to make sure we're all on the same page as far as notation goes. For the first part of this talk, I'll be talking about elliptic curves over Q, which we know can be defined for the virus stress equation, say Y squared equals X cubed plus AX plus B, where A and B are integers. We can reduce this equation modulo, our favorite prime P, as long as it's not a prime that divides the discriminant of a cubic, and maybe we need to stay away from two. And for each such prime, we get an elliptic curve over a finite field, and we can then compute the trace of Frobenius, A sub P at that prime, which we could define as taking P plus one minus the number of FP rational points on our curve. So this is some integer, which we know has absolute value at most to square root of P. And once we have these APs, or enough of them, we can use them to construct the L function of the elliptic curve. One also can define the APs at the bad primes, which I'm ignoring. But the good news is, in fact, the L function is completely determined by all the good primes, not just the L functions, many other prop, not just the L function itself, but many other invariance of the L function, including the conductor and the root number are determined by the good traces of Frobenius, the ACPs. We know, thanks to the modularity theorem, that the L function, the conjectured properties of an L function of an elliptic curve over Q, we now know are satisfied. In particular, it has a functional equation. And there's two key numerical invariance that appear in the functional equation, at least the way I've written it here, which are the root number, which is a plus or minus one, and the conductor, which is a positive integer, which divides the discriminant of the curve and is divisible by all the primes of that reduction. Most of this talk is really going to be about L functions. The L function of an elliptic curve determines, is in one-to-one correspondence with the isogenic class of the elliptic curve. And so for most of this talk, elliptic curves are simply going to be representatives of their isogenic class or a source of an L function that we're interested in. And because the Frobenius traces completely determine all the arithmetic properties of the L function, there has long been wide interest in understanding how these traces, Frobenius traces, behave and how they're distributed. So there are three well-known conjectures that all date from the 1960s and 70s. One of them is no longer a conjecture. It's now a theorem. The other two are still open. The first I'll mention is the Sater-Tate conjecture, which is about the distribution of the Frobenius traces sort of in the large. We know that they have absolute value at most two square root of p. So if we divide each Frobenius trace by the square root of p, we get a real number between minus two and two, and one could ask the question of how those real numbers are distributed. Are they completely random? Are they uniformly distributed or they have a particular distribution? And the Sater-Tate conjecture says that they do have a particular distribution, and in fact it's precisely the distribution you would get if you took the trace of a random matrix in SU2, sampled according to the Har measure, at least as long as our elliptic curve doesn't have any extra endomorphisms, but that applies, that holds generically. So if you were to write down an elliptic curve at random, it would almost certainly satisfy the Sater-Tate conjecture, which is now a theorem thanks to work of Richard Taylor and many collaborators. Second very well-known conjecture, and this one is still open, although there has been a lot of partial progress, is the Bertrand-Swintersen-Dyer conjecture, which was originally motivated by looking for information in the sequence of Frobenius traces and the sequence of APs. In particular, Bertrand-Swintersen-Dyer had the intuition that if the APs were more negative than one might expect, meaning there are an excess of rational points on our elliptic curve modulo small primes, then one might expect that that could come from extra rational points over Q, and in particular it might be an indication that the rank of the elliptic curve is larger than you might otherwise think. And they made a specific conjecture, which relates a limiting sum of, appropriately weighted sum of ASAP values to the rank of the elliptic curve in a very precise way, and I'll say more about this in a bit. This conjecture is still open, but it is at least known to hold in the case where the analytic rank is zero or one, which actually, at least conjecturally, applies to 100% of all elliptic curves, and we know that it applies to a lot of them, so we make good progress. The third conjecture I'll mention is the Langtrotter conjecture, which won't play a big role in this talk, but I mentioned it because it's another conjecture about how Frobenius traces are distributed, and it's an indication of how much work is still to be done, because this is a conjecture that is at this point completely open and appears, as far as I know, to be completely out of reach of current technology. Although having said that, there are results that are known about the Langtrotter conjecture on average, which maybe suggests that even if you can't say exactly the thing you might want to say about the sequence of Frobenius traces attached to an elliptic curve, maybe it makes sense to look at their behavior on average. Now, all of these conjectures are really just about the ASAPs, the sequence of integers indexed by primes, and so they only depend on the L function of the elliptic curve, and in fact, all of the L functions we're going to consider in this talk also have ASAPs, that are integers, and all of these conjectures can be generalized to those L functions as well. Now, I just wanted to sort of as a warm-up show one example of how one might try to gain intuition about an evidence for the Sauter-Tate conjecture and also see why Bertrand Swinton, Swinerton-Dyer, perhaps were led to their conjecture that an excess, a large rank, should lead to an excess of rational points. So this particular elliptic curve was found by No-Nelkes, and it has rank at least 28, and it's the elliptic curve of largest rank that we know. We don't know the rank is exactly 28, but under the generalized Riemann hypothesis, one can prove that. And the plot you're looking at here is a histogram of normalized Frobenius traces. So I've taken the ASAPs for each, every prime p up to 2 to the 10, I've divided by square root of p, I get a sequence of 172 real numbers between minus 2 and 2. I've divided the x-axis between minus 2 and 2 up into 13 subintervals. That's roughly the square root of the number of data points I have. And then I put a blue bar showing how many normalized Frobenius traces fall into that bucket. And at this point, the distribution looks very skewed. And that's because indeed this curve, as you might expect, has a lot of extra points module of small primes. Now the Sauter-Tate conjecture says that no matter what your rank is, asymptotically, the distribution of normalized Frobenius traces should converge to a semicircle. This is what you would get from applying the Har measure on SU2. So let's go ahead and take a look and see what happens as we increase the bound on our prime. So in each frame of this animation, we're roughly doubling the number of data points. We're looking at more and more ASAPs. We're dividing our line between minus 2 and 2 up into more and more subintervals. And our histogram appears to be converging to a nice smooth semicircle, which we now know asymptotically it must, because this elliptic curve does not have complex multiplication, so it obeys the Sauter-Tate conjecture. Now as I, it's standard, there is a much more precise version of the BSD conjecture that involves many other invariants of the elliptic curve. But I'm interested in this talk. I just want to talk about the distribution of properties of the Frobenius trace. And one interesting question you might ask is whether even if you don't know whether the BSD conjecture is true or not, you might look at this sum and try to use it as a way to predict ranks. And in fact, that is the way most large rank elliptic curves are found. They're found by starting with a family that you know has some excess rank to begin with. I think in the case of Elke's curve, that family already was guaranteed to have rank 19. And then you look for special examples in that family that have even more rank than they, than was already baked in. And how would you detect them? There's infinitely many curves to look at. Well, one way to detect them is to compute truncated versions of this sum, these are known as Mesternigals sums, and approximate the value. And when you get a value that's looking more negative than you might otherwise expect, that's maybe an indication that there's excess rank. Now we expect the minimalist conjecture and other conjectures predict that on average, half of the elliptic curves over Q should have rank zero, the other half should have rank one, and the rest, a negligible proportion have rank greater than one. So most elliptic curves aren't in the category of Elke's curve. And in the context of this equation, that means that this Mesternigals sum, whether it's positive or negative, is going to tell you whether you have rank zero or rank one. Now it's still an open question whether the VST conjecture is still open, but recently So-Yung Kim and Ron Murdy proved a result that establishes that if this limit on the left-hand side converges, it has to converge to the right-hand side. So that's a good sign that perhaps it should be true. Although they also prove another consequence of that conversion, this is that in that case, the L function of the elliptic curve also has to satisfy the Riemann hypothesis in all the zeroes around the critical line. That's something we don't expect to be able to prove anytime soon. So it's perhaps a suggestion that proving this equality may be difficult, but it's interesting nevertheless.  I think I took the bound on primes up to about 10 to the 12th, and I did it for an elliptic curve of each rank of where we know that we know occurs from zero to 28. And you can see for the small rank elliptic curves, the convergence looks really good. It hits the half integer it's supposed to almost right on the nose, and it hits it fairly quickly. But for the larger rank elliptic curves, it seems to be taking a long time to get there. For Elkie's rank 28 curve, it's still very far from minus 27 and a half. And the reason for that has to do with the conductor of the elliptic curves. In order to have large rank, you're necessarily going to have large conductor. You're going to be forced to have bad reduction to a lot of small primes, and that's going to make the conductor big when he can even prove explicit bounds. And having large conductor tends to make these sums converge more slowly and makes it difficult to predict the rank on the nose from the Mesternigals sums, but they still provide good evidence when there's access of rank. The chart on the right is a plot for elliptic curves with complex multiplication. These are all, I think, Mordell curves, but the BSD conjecture doesn't really care about whether the elliptic curve has complex multiplication or not. That sum should still converge to minus r plus a half, and you can see it seems to be doing a pretty good job of that. Now, I mentioned these Mesternigals sums in the BSD conjecture and efforts to predict rank because, in some sense, that's where the story started. There were a number of people, including the colleagues I mentioned here, who were conducting machine learning experiments on various sources of number of theoretic data, elliptic curves in particular. And so you might have seen a number of different papers that have been written about trying to learn how to say predict class numbers from information about a number field, or in the case of elliptic curves, trying to predict the rank from other information about the elliptic curve. In particular, trying to predict the rank from looking at the Frobenius traces. And a common source of data for these experiments is the L-functions and modular forms database, which has every elliptic curve of conductor up to 500,000 and as well as others, and information about a lot of detailed information about each elliptic curve, and also about many other types of L-functions. And so in the course of conducting experiments where they were trying to see if they could use a neural network to learn the rank by observing the APs for, say, the first so many small primes P, they observed an interesting phenomenon that wasn't what they were looking for. So the first step when you're going to run a machine learning experiment is typically to do a principal component analysis. So you make a matrix where each row in your matrix is an elliptic curve. Each column is a prime P, some invariant of the elliptic curve, say an AP value. And also the rank would be in one of the columns. And you're trying to see if you can learn from the information in the AP columns, you can come up with a good prediction for the rank column. And so the first step is to normalize your matrix by making each, you know, making every column have mean zero and the same standard deviation. So you're going to compute the average of each column, subtract that from each entry, and then divide by the standard deviation so that you have mean zero standard deviation one. And then there are many steps that take place after that, but even in that first step, I mentioned that first step because it's related to the picture you see here. So what you're seeing in the picture, in this picture is a plot where the x-axis is, the x-coordinate is primes. So two, three, five, seven, et cetera. Each blue dot represents the average value of AP for a particular prime P among all elliptic curves with rank zero and conductor in a certain range. In this case, the range is 7500 to 10,000. Again, it's sort of natural to organize your elliptic curves by conductor. This is how they come to you from the LMFDB. And as I mentioned before, the conductor plays a role in the rate of convergence. So not an unreasonable thing to do. And if you're trying to learn the rank, well, you want to separate the rank zero and the rank one, and most elliptic curves have ranks zero or one. So that's a natural case to look at. And when you do the first step in the principal component analysis and compute the average, then involves in computing the average value of each column, one of these researchers, Alexei Poltsnikov, happened to notice that the signs of these average values seem to oscillate. The first so many among the rank zero curves, say they were in the top half of the matrix, they were all positive for a while, while the rank one guys were negative, which is sort of what you'd expect from the BSD conjecture, right? You're expecting this, the Mesternigalsum to converge to a positive number for the blue team and to a negative number for the red team. So you're not at all surprised when you see that the averages are positive for the blue team and red negative for the red team at the beginning. But what is surprising is that they swap places. If you go as the prime increases, in this case, around the primes around 150, the averages are close to zero. And when you keep going, all of a sudden the blue team has decided that on average, they should have some access points, mod p, at least for primes in this particular range. And the red team has decided they should have fewer points, mod p. And then they switch again, and they keep switching. So this is quite unexpected. And the first thing I did when I saw their prepent, which appeared on Archive last year, I think around May, was to say, wait a second, how can this be?  And so the first thing I noticed is that in this experiment, they're only using about four or 5,000 elliptic curves on the blue team and four or 5,000 on the red team. And they're using a very narrow conductor range and the conductor is small. So I wanted to know, is this a robust phenomenon? How does it persist as you make the sample size larger and you increase the range of the conductor you're using? So I did something very similar to what I showed you in the cytotape plot, except rather than showing a histogram of normalized Frobenius traces, I'm showing you AP averages where in the top plot, I have a blue team and a red team again. The blue team is the rank zero team and the red team is the rank one team. Now, rather than looking at the rank, I'd rather, because I know that the root number is really the key invariant in the L function, and the root number is also something that's a lot easier to compute than the rank, I'm just going to separate them by root number. So the parity conjecture states that the algebraic rank as the same parity is even if the root number is positive and odd if the root number is negative. And we know that the analytic rank has to be lined up. So you can think of the root number as a proxy for the parity of the rank. And since almost all elliptic curves have rank zero or one plus one root number, you can identify with rank zero minus one root number, you could identify with rank one. Now, another thing you could do is since the plots look to be mirror images of each other, you might as well define a new parameter called, I've denoted m sub p, m for murmuration, which is a sign where I'm weighing the Frobenius trace by the root number. So I'm just taking the product of a root number in the Frobenius trace, and then I'm averaging over all the elliptic curves in our sample. And what are those elliptic curves? They're precisely the elliptic curves that have similar conductors. They're elliptic curves with conductors, in this case, in a dyadic interval, say between m and 2m. And so in the plot you see here on the screen, I've taken all the elliptic curves that I can find in that you can find in the element db with conductor between two to the 11 and two to the 12. I'm a computational member there, so I like powers of two, and so I tend to organize things that way. So just as in the, I did in the cytotate histogram, I'm now going to play the same game and let my powers of two grow. So in each frame, again, as in the cytotate plot, the range, the conductor range is doubling. The number of elliptic curves isn't quite doubling. We expect the number of elliptic curves to be growing something like x to the 5, 6, but it's still increasing at a steady rate. And what's remarkable about this, and I find it so remarkable, and remarkable, I'm going to play this animation again because it went kind of fast, each frame in this animation is a completely different set of elliptic curves. So this isn't about, you know, a Mester and a Galsum converging or anything. These are completely different datasets. The only thing they have in common is that I'm plotting them on the same scale. I'm always plotting the primes p relative to the conductor. So the x-axis, the range of the primes, which I've denoted from zero to one, I've done that intentionally. One here means the prime is going up all the way to the end of the conductor range, which is 2M. So in each frame, the scale on the x-axis is changing, but the ratio of p to the conductor is effectively being held constant. So I'm going to play this animation again, and you'll see that the crossover points are the peaks and valleys of this purple murmulation curve don't appear to change, even though it's a completely separate set of elliptic curves with completely different conductors, completely different sequences of APs. But somehow when you average them on, you separate them by root number, and you average the APs, and you organize, you plot your graph with p's on the x-axis ranging up to, in this case, you know, some multiple of the conductor. It doesn't have to be one, but I've chosen to normalize everything in this presentation. I'm measuring the prime relative to the top end of the conductor range. Okay, so this, having done this, I'm now convinced that this is something real. This is a real phenomenon. There's something interesting happening here. My first question is, surely someone has noticed this before. How could you not notice this? Once you sort of, once you see it, it's very hard to unsee, and you see it almost everywhere you look. So I immediately emailed everybody I could think of, in particular, I emailed Peter Sarnak and Mike Rubinstein, Andy Booker, John Boeber, Andrew Granville, Jordan Ellenberg, Bjorn Poon, and everybody I could think of who might have already seen this phenomenon or have a good explanation for why it's occurring. Now, one short answer is nobody had an immediate answer. You know, one possible connection might be to something like a Chevy Chavis bias that you see in prime races. This is one of the reasons why Mike Rubinstein and Peter Sarnak were the first people I emailed. And there may well be sort of a high level connection there, but it's not the same thing. Elliptic curve L functions are different than Dirichlet L functions, and the murmuration pattern we're seeing here is not the same as a Chevy Chavis bias in prime races. Now, the other thing that strikes you looking at this plot, so I'll write it one more time, is that when I separate it into the red and blue team, there's a lot of snow, there's a lot of drift downward drift. But when I take the signed murmur, you know, when I multiply by the root number to get the murmuration parameter m sub p, the drift goes away. And so that tells you something, well, first it tells you that multiplying by the root number is a good idea because it cancels out a negative bias that would be there otherwise. But it also tells you that this negative bias is really independent of the root number. And so every time you see a low hanging blue dot, there's a low hanging red dot that when you add them together to get the purple curve, the bias more or less cancels out. And so one way to see that would be to run the same experiment, but don't run by, you know, rather than separating by root number, throw all the elliptic curves in the conductor range together, but in one case include the root number. And in the other case, don't multiply by the root number. And so this is what you see when you do that experiment. And so you can see that just on average, there is a slight negative bias. Okay, this is the thing that happens when you look at elliptic curves up to a conductor bound. But the bias is really parity independent. It mostly depends on the shape of the prime, on the visibility of P minus one. And it cancels out when you include the root number because the negative bias is really coming from P and not the root number. And so when you insert the sign, it goes away. Now I don't think this negative bias has been completely explained either. But it's clear to me that it's separate from the murmuration phenomenon. So I don't want to distract us. I want to stay focused on the murmuration phenomenon. I'm not going to say anything more about the negative bias. But I think there's an interesting question for someone to look at here. And if there's questions at the end, I'd be happy to at least mention a few ideas for where I think this might be coming from. Some modular analog of Benford's law may be at work here. Okay, let's stay focused on the murmurations. So the next experiment that might occur to you to try would be to say, well, if I'm seeing some behavior on average, maybe the reason I'm seeing it is because it's something that occurs in every elliptic curve. Maybe I'm just averaging the same, the same bias occurs individually for each elliptic curve. And then we see it in aggregate when we average. And so, and this was something that Bjorn Poonam suggested. And so I did an experiment where I'm computing moving average, I'm averaging over a range of APs for each of each of eight, I chose eight different elliptic curves with conductors in a dyadic interval. And I've colored with a blue shade, the curves that have root number minus plus one, and then with a red shade, the curves that have root number plus one. And you can see that, well, there's oscillations, but they're not coordinated in any way. They're not murmurating. Each elliptic curve is doing its own thing. The murmuration really only occurs when you average a bunch of elliptic curves and look, and it's really the APs, the average APs that are murmurating in a pattern. The green line here is a line fit to the murmuration pattern, the purple curve that you saw in the previous slides. It looks much less pronounced here just simply because I had to enlarge the vertical scale so that you could even see what was happening with each of the elliptic curves, individual elliptic curves. So the green line here is an average over hundreds of thousands of elliptic curves. Each of the thin red and blue lines is one individual elliptic curve. So in some sense, the center of mass is following a very predictable pattern, but the individual birds are all over the place. They don't know what they're doing. Okay, so the next experiment that I tried, and this was suggested to me by several people who work in arithmetic statistics, the standard thing, while the conductor is a very natural invariant, if you're thinking about L functions in arithmetic properties of the elliptic curve, in arithmetic statistics, it's much more common to work with something that's easier to control, for example, the naive height. So if you wanted to compute a whole bunch of elliptic curves and compute some average over them, one way you could do would be rather than organizing elliptic curves by conductor. The conductor is not hard to compute, but it's actually quite hard to find all the elliptic curves of a given conductor or all elliptic curves within a conductor range. In fact, one of the organizers of this number three web seminar has spent a lot of time trying to compute all the elliptic curves with conductors in a given range or conductors in a given prime conductor. And I'm sure Mike Bennett could tell you it's a hard problem. We only know have complete lists of elliptic curves of all conductor up to 500,000, but it's easy to write down equations y squared equals x cubed plus a x plus b with a and b bounded by some box. So our naive height, a standard bound to use is the naive height, which is the maximum of four times the absolute value of a cubed and 27 b squared. And most of the results we know about arithmetic statistics of elliptic curves of over q, in particular, the average rank results of Bargavan Shankar are proved using the naive height. Now, the naive height and the conductor are not unrelated. I mean, the naive height is definitely going to impose is very closely related to the discriminant of the elliptic curve. And it's an upper bound on the discriminant. And so it's also an upper bound on the conductor. And one can even prove that asymptotically, the conductor most of the time can't be too much smaller than the naive height. But it can be a lot smaller. It can easily be as small as the square root of the naive height, for example, which is already that happens quite often. And that's already enough to completely destroy the murmuration pattern. So let's I'll run the plot here while I talk. So remember, when we saw this nice smooth curve that was correlated when we were moving up in dyadic intervals, remember those peaks and mallies really depended on the ratio of the prime to the conductor. But if you're mixing together curves with conductors that vary quite widely, much more than a dyadic interval, maybe they vary between m and m squared. And in fact, they might even be a larger interval than that. All of those oscillations are going to get muddled together. And you're going to get what looks like purple noise. No pattern to see. So unlike many other questions about elliptic curves over Q that arise in arithmetic statistics, where the naive height serves does serve as a good proxy, I mean, it works most of the time. And it's the general philosophy is if you can prove something on average using the naive height, it should also hold for discriminant conductor. In this case, the conductor really is critical. Another way you might organize elliptic curves elliptic curves over the complex numbers have a very natural and invariant, the J invariant. And over any field, the geometric isomorphism class of the elliptic curve is uniquely determined by a single parameter, it's J invariant, which can be computed explicitly in terms of the coefficients of the curve. And it's easy to write down elliptic curves with a given J invariant. So it's easy to enumerate all elliptic curves over Q with J invariant that are say minimal twists with J invariant in a particular range. And so that's what I did in this plot here. I took elliptic curve by minimal twist, I mean, say elliptic curve with the smallest naive height that occurs for that J invariant. And then I'm just putting a naive bound on the J invariant, the maximum of the absolute value of the numerator and the denominator. Now the height of the J invariant is much less closely correlated with the conductor because you're now you're taking a ratio and there's opportunity for lots and lots of cancellation. So you don't see any oscillations, you do see a different pattern than you see for the naive height, you get a much more like a tighter burst of random noise, but no murmurations. Now sort of coming in between the naive height in the conductor, you could try the minimal discriminant. Now we don't really know how to enumerate, we don't quite know how to enumerate all elliptic curves with a given minimal discriminant, but we're actually pretty close to having a good algorithm for doing that. The Stein-Walkins database was an attempt to enumerate all elliptic curves up to a much higher conductor bound with a bound on the discriminant. And there's an algorithm due to Elkies that actually makes it feasible, you know, modulo exceptions to Hall's conjecture to find all the elliptic curves with a given bound on the discriminant. And so I've done an experiment here where I'm ordering curves by discriminant rather than conductor, minimal discriminant. And you'll see here it looks pretty noisy, but not completely noisy. There's a little bit of a pattern, especially at the beginning. And if you sort of let your eyes blur, go out of focus a little bit and stare at the purple blob at the bottom, you might even be able to see the hint of a murmuration there. Now to make it easier to see, I've drawn a line fit. So what I'm doing in this is I'm dividing the horizontal axis up into subintervals, and then I'm just computing the best line fit to each, the scatter plot in each subinterval. And you can see that there is actually the average line fit does show a murmuration pattern hiding behind the purple noise. Now you look at this and you might think, well, maybe there was a pattern, the naive height plot that you just couldn't see. It's maybe it was too subtle for the human eye to detect. And so we should go back and try the same game. Well, we'll run a line, an average line plot. So this is building a sort of a best fit piecewise linear function. And on the one hand, there's not a murmuration, on the other hand, there is a pattern. This is not just straight down the horizontal axis, which you might have expected. And in fact, you might look at this and think that kind of maybe looks like the murmuration curve, but really spread out. And that's not an unreasonable thing to think because many of the curves with height in this box do have conductors that are quite close to the height. But some of them that there's going to be off by some non-trivial constant factor. And that's going to tend to spread out the murmuration curve, but there might still be enough data there to be able to see it on average. Okay, okay. And then the last plot I want to show before I leave the world of elliptic curves, one of the questions that was asked, this was asked by Akshay Venkatesh when I gave my talk about this in the spring. So the first question he asked was, you know, is this a small number's phenomenon? It's easy to get fooled, especially when you're ordering looking at elliptic curves ordered by conductor. There's a lot of things that tend to show up in excess for elliptic curves with small conductor. You get much more torsion when you expect asymptotically. You get a root number bias that you don't expect to necessarily see asymptotically and that you don't see when you order by naive height. And in particular, quite notably, when you compute average ranks, whether you order by naive height or by conductor, you tend to see an excess of rank for a long time. There was for a long time, there was a lot of tension between the prediction that the average rank overall elliptic curves over Q should be a half because when people computed the average up as high as they could go for quite some time, the average just kept growing and it was getting quite close to one. Now it was eventually found people went out far enough and naive height up into something like 10 to the 12th or 10 to the 13th, where that average rank started to turn around at around, I think a naive height around 10 to the 12th and head back towards a half where we believe it should be going. But there was a long period of time where people were perhaps mislead by the data at small conductors. So in this plot, I'm taking a much larger database of elliptic curves. So the LMDB has about 3 million. The Stein-Walkins database has over 100 million elliptic curves with conductors ranging up to 10 to the 8th. And I'm actually in the process of building an extension of the Stein-Walkins database that will increase the number of elliptic curves by another factor of 100. But for the purpose of this talk, the main purpose of this slide is I hope to convince you that this isn't a small numbers phenomenon. It seems to persist. Not only does it seem to persist the murmuration curve that I'm plotting both, you know, the purple scatter plot, but then I'm plotting the best fit piecewise linear curve on the bottom is very stable as the conductor grows. And I'd like to extend this plot another few frames where I would expect this curve to stay rock stable. And I'm willing to conjecture that there is an asymptotic murmuration curve here that elliptic curves over Q will converge to as the conductor tends to infinity. And it doesn't really depend on the interval being a dyadic interval. You could take a narrow interval or a wider interval, but something multiplicative is probably the right shape to take. Okay. All right. So the title of my talk was Mermorations of Arithmetic L Functions. And I have yet to tell you what an arithmetic L function is. Well, people have different definitions. But for me, an arithmetic L function first and foremost is a well-behaved L function by which I mean it has every property a good L function should have. But as an analytic continuation, functional equation, Euler product, temperateness, central character, all the things that we expect all L functions should have. And then among those good L functions, the nice L functions, we call them the ones that are arithmetic are the ones that have Dirichlet coefficients in the ring of integers of a number field when appropriately normalized. There's some power of the integer n in the index, some half integer power that you need to multiply to get an algebraic integer. And the exact power that you need to take for an arithmetic L function is called the motivic weight. And in the case of elliptic curve, that power is one half. The motivic weight is one and you need to scale by one half to get when you have the analytic normalization and you want to get an algebraic integer. So I'm going to restrict attention in this talk to arithmetic L functions where the coefficients are in the ring of integers of a number field. This includes all L functions of a billion varieties, all of which have motivic weight one. And it also includes a polymorphic cuss forms of weight k but note that the weight of the cuss form, the k and the weight of the cuss form is actually one greater than the motivic weight in this notation. So don't get confused by that. So weight two cuss forms have motivic weight one. I'm also going to, because I'm averaging over L functions, I'm going to always average over a gavel orbit. So if I have an L function whose dear side coefficients are algebraic integers, but not actual integers, I'm going to average APs over all of its gavel conjugates. So my averages are always going to be integers. And so I'm going to restrict my attention further to self-dual arithmetic the normalized L functions, which means that that also guarantees that the root number will always be plus or minus one, which is convenient so that I can separate L functions into blue teams and red teams. And then the final step, and this is really a key step that I think caused murmurations perhaps to be missed in other settings, because this normalization step wasn't done. But if you think about elliptic curves, it's kind of surprising you have all these coefficients that you're averaging together the coefficients are roughly on the order of square root. But when you average them together, you get fluctuations that are like O of one. From a statistical or a random perspective, that's maybe not so surprising. I think there are good reasons to think that that's the right scale. And so my instinct, whenever somebody hands me a sequence of APs of an L function that doesn't have motific weight one, is to normalize them so that they look like APs coming from something of motific weight one. I want them to have values that are bounded by something on the order of square root of P. I don't want to normalize them all the way down to have O of P, O of one, as you would naturally do in the analytic normalization. Because if you do that, then the averages are going to tend to zero. So square root P is sort of the sweet spot where you can see an interesting O of one fluctuation. Okay, so starting from coming from elliptic curves, the first thing you might try would be to look at weight two cuss forms for gamma naught with rational coefficients. Those are exactly the L functions of elliptic curves. We know this from the modularity theorem. But while we're at it, we may as well also consider cuss forms, holomorphic cuss forms of weights, other weights, that also have rational coefficients, just to see if they exhibit murmurations and if the murmuration pattern is any different. So this plot shows not a huge amount of data because it's much harder to compute L functions of modular forms at this level of granularity, at least when you want to isolate the ones that have rational coefficients. But hopefully you can see at least some suggestion that there is a murmuration pattern and that the murmuration pattern is slightly different depending on the weight. Now the great thing about moving into the world of modular forms is there's no reason to constrain ourselves to rational coefficients. Why not average over all modular forms? And since, again, since we're averaging over Galov orbits, we're eventually going to get things that have integer, that have integer averages. And so here's what happens when we, okay, so here we go. So these are averages over the entire space of weight two, four, six, eight cuss forms. So the top graph is weight two, then weight four, then weight six, weight eight. And I'm looking at in this last frame, we're looking at cuss forms with level between two to the 14 and two to the 15. And here I'm letting P go out further than the top end. I'm letting P go out to twice the top end of the conductor interval. So that's why there's a 2.0 at the end of the horizontal axis. And there's a lot more modular forms than there are elliptic curves, something on the order of n squared rather than x to the five, six. And one nice thing is that means that we're averaging over a lot more data. So we get much better conversions. Another nice thing about averaging over all modular forms is that it's much easier to compute these plots. These plots go out to much higher level than the plots I showed you on the previous screen, because they can be computed using trace formulas. And these plots in particular were computed using a very efficient implementation of the trace formula for the Ack and Lainer and Heck operators, implemented by Aaron Assoff. And the code is available on his GitHub repo if you want to try it out for yourself. But it makes it feasible to compute these averages over millions and millions of modular forms. I don't know if you can see it on your screen, but in the weight eight case, there's something like 66 million modular forms of root number plus one and 65 and a half million of root number minus one that are being included in this purple plot. So you see murmurations in each case, the murmurations are scale invariant. I just want to repeat that. So in this sequence of animations, I'm doubling the level range, but the location of the murmurations, because I'm scaling P over conductor, holding that constant stays fixed. Okay. All right. And I wanted to show you sort of a bigger picture of a few of the patterns so that you could really see them in all their beautiful detail. My computers have been very busy over the past few months computing these, but the top plot is showing the blue and the red team separated. The bottom plot is the signed murmuration variable, which is just basically folding the red and the red plot on top of the blue plot. And you can see that they line up perfectly. And here's the same thing for weight four, weight six and weight eight. Oh, and I even have weight 10 and weight 12. And the pictures get pretty funky. I don't know how well you can see them, but if you look closely, you'll see some, that there are some points in this curve where it's not looking anything like a smooth oscillation. There are points where the curve clearly looks to be only piecewise differentiable, even though it is continuous. You can see another point appear right at one. And if you look very closely, you'll notice that those apparent points of non-differentiability appear to be occurring roughly at squares of half integers. Again, we'll come back to that. Okay. So another nice thing about the trace formula is it gives you an analytic tool that you can actually apply to not just compute pretty plots, but to try to prove theorems. Now, anybody who's worked with these trace formulas know that's not such an easy task, but Nina has accomplished it. She announced this result for weight two earlier in the spring, and she very recently communicated to me and gave me permission to present her theorem, which covers arbitrary weights. And now I'm not going to have time to go into the details of this theorem, but I just want to highlight a few things. I mean, so the hard part about computing these averages over the trace formula is you're averaging a bunch of class numbers. And that's not an easy thing to do. But what comes out of her computations is actually quite beautiful. You get a leading term, which is some constant A is an explicit constant. It's something like 1.45 blah, blah, blah, times the square root of Y. That's telling you what the sort of the initial upward curve where we saw the blue team going up and the red team going down. But then as Y gets bigger in particular, as soon as Y is bigger than a fourth, there's additional terms that come in that include values of chubby chat polynomials. And so this is where you get something that is only piecewise differentiable, but nevertheless continuous, all the pieces line up together. And I'll also point out in this plot, and this will be important in a few slides that you could really think of this function on the right as written as a function of Y, which is the prime over the conductor. But it's really a function of the square root of Y. I could write everything in here. In weight two, there's one actual Y that occurs. But in the higher weights, everything is only square roots of Y that occur. And of course, Y is the square root of Y. Okay. So now that we have a theorem, let's compare it to the data. Now, Zuberlin's theorem is in general assuming that Y is going to be smaller than X. So I was using dyadic intervals where X and Y are the same. And so you can apply Zuberlin's theorem to compute by averaging over it to get an approximation of what you should expect to see in a dyadic interval. But I did something something different. I made the intervals narrower, or more precisely, I rather than just adding all of the APs together, I'm actually normalizing, I'm treating P over N as a variable in and of itself and computing average sums that way. And then fit the curve. So the blue curve is plotted directly from Zuberlin's theorem. And the purple plot is the plot that you've seen in all of the animations we've seen before. It looks a little different now because I'm constrained to a much narrower conductor interval. I want to think of X going to infinity, Y is also going to infinity, but it's going to infinity slower than X. And you can see that while there's maybe still a little tweaking, I'm not sure where exactly the tweaking needs to happen in the match. You know, at the beginning, the vertical axis to line up perfectly, I fit these so that I made them agree at the point one. The horizontal axis, the locations of the non-differential points line up perfectly right on the nose. And you can see there's one at a quarter and another at one. Square is a half integers. Okay. Now I said that it's really a function of the square root. And the memorization plots look a little prettier when you instead, rather than plotting them as a function of P over N, plot them as a function of the square root of P over N. So now the right hand, the X axis instead of going up to two, it goes up to the square root of two. And you can see that there's a little more regularity in where the peaks and values occur. And here's what you get if you go back to the very first memorization plot I showed you for elliptic curves, but instead use the square root normalization. And you can see a nice more evenly spaced peaks and valleys coming in. And in general, for now a function of degree D, the right, I believe the right normalization is really to be taking the Dth root of P over N. Now I'm running tight on time and I want to get to one other family of L functions before I finish, but I want to quickly share with you another very recent result which was shared with me just yesterday, which is to play the same game, but instead of looking at, but in the weight aspect rather than the level aspect. So one can define what's known as the analytic conductor of an L function. And in the case of modular forms, it's something that grows roughly on the scale of N times K squared. The precise formula is given here. And with the analytic conductor, you could ask, what would happen if we just fixed N? Keep the level one and let K grow. And so very recently, over Booker Lee and Larry Duda were able to prove a theorem showing murmurations in the weight aspect when they look at this family of cost forms.  The plot looks quite different from the ones we've seen before. You're actually going to get sort of point masses coming out of their murmuration function. So the step function is actually their prediction, and then the not quite stepping function is the averages that they're computing. And I'll just mention briefly that there were hints of this going back into the literature. This kind of question of studying AP averages when P is on the scale of the conductor, some power of the analytic conductor was actually looked back even back in 2000, and this paper by Yvonnex, Lou and Sarnak. But they mostly focused on the case where the power of the conductor is less than one or greater than one. And they noticed that there was really a phase transition when the power of the conductor is exactly one. And that's right where the murmurations are happening. When P is right on the scale of the conductor. And since I'm running out of time, I just want to finish up with the message that we don't have to confine ourselves to degree two L functions. One could look at L functions of genus two curves. These are degree four L functions, but they also have APs. One can also average them.  And you again get a negative bias drift that cancels out when you take the root number into account. The oscillations are more rapid. But that's because in this slide, I haven't normalized by taking the D through it. But here's what it looks like with the oscillations look like when you normalize by taking the D through the P of the prime over the conductor P over N. So this is the fourth root of P over N. And these are murmurations for genus two curves. And I'm going to skip ahead. One could also ask what happens when you look at other L functions. The message here is that it's important that the L functions you look at are primitive. If you have impermitted L functions, the merations are going to cancel each other out. And then the last slide I'll end on is what happens in genus three. So we don't have as much data, but we're starting to assemble a database of genus three curves and their L functions, which we want to add to the L and FDB. And here it's much harder to enumerate all the genus three curves of conductor up to a given bound, even harder than it is in genus two. And so there's a lot of noise. And that's just because we have a small data set, we're missing most of the of the genus three curves would conduct her in this range. But even within the small data set we have, you can see murmurations. Okay. And I think I had better stop there.


And now for your strong Europe spotlight. Army Europe has crowned its top warriors. 22 competitors endured exhaustion and put their warrior skills to the test with grueling obstacles and challenges during the five-day event. For the first time ever, a top warrant officer was selected. Chief warrant officer 2, Christopher Gillespie. It's very humbling to be named the first best warrior warrant officer. It's been my honor to be able to be out here and compete with all these great soldiers. First Lieutenant Christian Reeves was crowned top officer. Sergeant Jonathan Renteria and specialist Jacob Henriksen maintained motivation and came out on top as well. Renteria and Henriksen will move up to compete at the Department of the Army level event at Fort A.P. Hill, Virginia in early October.


Good morning. Welcome to church of the chair Where we celebrate all things we do while seated. I am your host E and today we're gonna be doing writing sprints First thing this morning Let people come in and chat for a little bit Do it about 15 minutes Those of you that don't know I use a gravity timer. This is because I'll have my My phone off and I suggest you do the same. This is $10 on Amazon Or you get an egg timer for five bucks at Walmart.  I'll try something else Good luck Do tomorrow and draw some moon kind of he's now instead. All right. Have fun Looking forward to seeing that one Give you all early birds a sneak peek. Well, not a sneak peek. I doubt you guys will ever read it, but It's uh, or if you do read it, it won't be by this name. So He dreams of starshine of collapsing gases coalescing a cacophony of creation destroyed cyclical and compounding A dark so bright it fills them with emptiness nothingness expanding within Eating away from the inside driving a hunger for starvation deprivation Cognitive dissonance hollowed out by an ever-present blindness with the same prefects The dichotomy of a man who's knowingly built his home on a fault line He won't budge What won't budge? He doesn't know Boom I can't wait for all the e is this horror Everything's horror. Horror is an emotion Horror is an emotion. It's not a genre. All genres are horror All horror is this conflict drama The next line is pretty good too. Hang on. Hang on. I'm I'm feeling to myself this morning. I'm feeling myself The earth is chuckberry on mdma rocking and rolling Some say the ground shakes during an earthquake. It doesn't it undulates like tide He dashes from one corner of his apartment to the other Saving what he can and missing more than he could ever replace A picture of mom Mid twirl in an open field circa 1980 something Crucified on sunshine plummets and explodes Glass and wood skitters skate across hardwood Struts and joists creak and wine Dust puffs powders his hair California structures are built to move. He is not Down he goes five nine two fifty lands hard Shards pierces back his nape Sunday bloody sunday Woop woop Good ish good ish I'm gonna be someone out there that saves this and tries to do research. He's like, I'm gonna find this book I'm gonna find it.  But there was no moon out is all thunder and lightning in the distance Maybe howlin at the stone. Yeah sure Howling at the slightly confused bat that got lost Well, I was on it when I when I wrote this one hold up now He dreams of falling a meaningless life ending with a boring death splat jars awake Still dark outside Not even a hint of purple over the interstate Not cave dark either. No never in San Bernardino The black above the mountains is speckled with stars Holes poked into the fabric of space glinting with the possibility of a brighter world beyond this one Whoo What I be doing over here though, what I be doing Shit existential crisis. Yeah, that's that's exactly what this person's going for So thank you for validating me because I love writing stuff that does not Basically, I don't like holding readers hands, you know, um, I want people to understand what I'm getting at without saying Without spelling it out for them. So Perfection. Thank you. I just deleted 1500 words that I didn't like because I didn't like them And before y'all like, how do you know it's bad because all the other stuff I know is good So I just went ahead and got rid of it because I am not happy with that Bye-bye darlings. Bye-bye Uh, yeah, rather than saying John was having an existential crisis. You showed it To use in Yeah, I use a metaphor. Yeah all metaphors and similes. That's uh, that's how I write. That's how I've always written And it's funny because I've had a numerous I've had numerous editors go Say What you mean But I'd rather mean what I say Um, and I know that's a little bit of word play um, but You know, don't don't use don't overuse metaphors and similes and all that stuff when I don't enjoy writing any other way. So why would I change to suit, you know, what what they're looking for it's another thing I tell people all the time is like if If you have an editor or an agent Who is constantly telling you to change what you wrote first off? I don't know why they accepted the book for publication to begin with um You either like what you're going what you acquired or you or you don't Um, so I I tell people all the time if they're asking you to change multiple things about the project Uh, you're in trouble. Um, and if you've already signed a contract There's not much you can do Especially early on You want to fight for creative control? and don't settle for nonsense like, uh The contracts that say that the the publisher or the agent or the editor has final say so of the product um And if they get to a point most most publishers if they get to a point where you're not making them happy They will just you know mutually they will say they need to mutually agree to get me to End the contract, uh, and a lot of them have that in there some of them Are truly shady and put stuff in their contracts Um, they word things so that they have control over your words over the uh over what comes out So just just be careful Uh, when you're when you're signing contracts Make sure you read your contracts and if you don't understand it get a lawyer or uh, even just read it to someone else Uh, because sometimes a lot of the time, uh, we have author blinders where uh, we're just so excited about the Scenario or the book or whatever we want to get it out there That we overlook, uh, distressing clauses in contracts And more more so now than ever we're seeing a lot of it with all of the uh publishers closing You know, he had silver shamrock dark views. Yeah, just they're just dropping like flies at this point What is the point in saying in just saying things outright if you can shape your writing style to capture the essence of a feeling right? Yeah Um, and I I I really do think that's what people mean when they say show.  Um, and unfortunately nowadays With attention spans being what they are Most publishers want you to just get to the point and most readers want you to get to the point But I think there's something to be said about beautiful language and Not over writing.  You know that that that kind of thing, you know um and most most readers like Building and then that's why we read so that we can build up the images in our head So give them just enough information and move the hell along Good morning, brian. How you doing? If all books just said what was happening outright with no frills, it just becomes monotonous. Yeah John was having a breakdown is not as effective as what you thank you. Thank you. I appreciate that there's There's getting to the point and just john felt sad, right? uh, another thing is there's a the Chapter title for this one is called cavitation And it's not a word that you hear all the time But if you google it or look up the meaning of it it fits perfectly So even the chapter title plays into the feel of what the character is going through Not bad. Thank you. You I'm doing fantastic brian. Um, it's been it's been a really really good morning And yeah, I think you guys saw I know hailey saw it over on the discord, but I woke up Uh, well first off yesterday was a really bad pain day because I spent so long in a chair playing resiny before and we'll get back to that tomorrow, but tonight's game night. Um, anyways We I played the game so long I was in the chair for like 12 13 hours because I'm also playing my own playthrough Trying to get a hundred percent over there as I go along with uh with the playthrough that I'm doing on youtube um with the with the streams and everything but Uh, so I spent so long in a chair.  You know the kind Um, I took some gummies and I passed out Um, and then I woke up at I think 230.  It's over on the discord. I I shot a little shot a quick video Trying to capture this but all you can really hear are crickets and all you can really see is You know some lightning off in the distance But it was there was thunder and lightning and coyotes Screaming out in the distance howling you could you hear him? It was amazing. Uh, it was a very magical Uh set piece. I guess you'd call it um But uh, yeah, I was just sitting there. I was like I I got to share this because this is cool So I tried to record it several times couldn't get it couldn't get Um, and the one that I did share I don't even think you can hear crickets in that one I didn't put my earbuds in Um, but uh, there was lightning off in the distance is very very pretty If other people like books that are literal and to the point they can go And to the point oh literal and to the point they can go read those exactly Um, I like your style. You know, I'm glad you weren't wrongly persuaded to change it. Yeah, um It is what makes your writing special. Thank you. Um the I think the the whole Old point I'm getting at is you got to have fun and do what you want It's just like if you're looking for a significant other if you're looking for another human being to spend your life with Um, do you Be yourself and the right people will want to be around you. It's the same thing with your writing um And it goes to the same thing with you know Going with a significant trying to find, you know, someone to date if it's not If your writing isn't going well if your relationship isn't going well Then maybe look for something new not everybody needs to be a writer not everyone can be a writer um, but it's it's also it's just You do you I guess that is the uh the the the point I'm trying to make is to be yourself and do What you want as long as you're not hurting anybody, but I beat that drum to death Uh, I have to find a video. Yeah, it's a it's over on discord.  It's a little deeper You know that kind of thing you can hear crickets in it. E Just you were surrounded by crickets so you couldn't hear the crickets over the crickets Yeah, I remember you saying that on discord too. That's funny Let's see here Yeah, somebody says writer get to the point me let the writer cook Yeah See the the thing the thing is with any good editor reads ahead. Um And I I think We're plagued Especially now with attention spans being what they are. I think we're plagued by people taking surface level Taking everything at surface level instead of getting into, you know Digging deeper drilling down and finding out what the point is of what the person is writing I've gotten so many. Um, I had an editor at one point in time. I no longer work with them And this is one of the reasons why um, where I have something there and they would say this doesn't make any sense Uh, please edit and then like three pages down Literally it would say disregard Previous comment about this not making sense. It makes perfect sense Why comment that to begin with? If it's if you are making the notes for yourself and get rid of them before you send you send them to to to the writer It's so hard to find decent editing these days because of that because of attention spans because of people not wanting to read past the surface level If I draw a concept that suits my own style I enjoy it and it goes well if I try To create some photorealistic art piece that isn't fitting with my own cartoony cutesy style It is a mess and a chore exactly and again Well other than artwork that you do for you know compensation for commission, whatever If you're not enjoying it, why do it, you know, mind the business that pays you Other than that do what you want to do just do it I understand You ain't the man you ain't the man Also, I'm trying a new setup So I'll be able to see comments as they pop up as I'm writing So We'll be on and off here. Um, probably take a break every 30 minutes or so. Also. I need to set my timer All right 30 minutes on the clock for the first sprint. Uh, you guys can chat as you want to I'm in no rush here Oh, it's raining Just just a little so it's like pratt pratt pratt Pratt pratt pratt It's uh good vibes this morning Yeah, like how you doing? I'm doing fine Doing real good He's all little I don't know if they're mosquitoes or gnats or what they are but I'm drawing a cody howling on a moon this night sharing my screen as I draw drawing between it howling at a lamppost Or a confused looking bat confused looking bats sounds funny. Ooh, that hits different That hits different confused moth flying by lamppost. That's awesome. All right.  Oh, welcome. Welcome in and lurk Phone's dead. My phone is dead I'm sorry. I shot you with subscriptions. Oh, we I have been subscribed to you. I subscriptuated How you doing this morning? You still here? You know, it's funny, uh, Brian There's uh, we we figured out because I'm one of those people, you know that likes to go into detail when they're, you know, playing games so Every single resident evil that has one of those opening sections be it the original version of four Be it the resident evil five um, I don't believe six has it, uh, but the opening and uh, resident evil eight has it seven doesn't But every single one of those sections at the beginning you have to survive for four and a half minutes We timed it this time and you actually have to survive for five minutes this time From the time that you were spotted till the time that the the bell tower or a clock or whatever it is starts ringing Um is five minutes and I'm wondering why they tacked on an extra 30 minutes I'm also wondering. I haven't played the the harder difficulties yet But I'm wondering if the harder difficulties last even longer. All right, Hailey All right. Have fun looking forward to the to the art awesome Hailey Oh, wow Pretty damn awesome. Nice. Thanks for sharing Hailey. I'm sorry Hailey The rub it out just had to rub it out Y'all know how it is So I'm so sorry Let me go back to work Sorry Yes, I saw I saw it Hailey, but I didn't want to comment on it I Yes, it needs some work sorry I'm just I'm trying to live my best life and you know be kind And if you don't have something nice to say don't say it at all, you know that kind of thing and uh, just I'm let I'm letting you say it this time, you know, normally I'm like, oh stop being so hard on yourself But if I'm quiet I love you Hailey Love you. It needs to be set on fire I've the I'm not going to argue with you Not gonna argue it. I'm gonna make something else and give him that because it's funny. Yeah It anyways, I here's the thing. I know you can do better. That's that's all I'm gonna say That's all I'll add. I know you can do better. So do better Now that just goes to show, you know, my honesty stands no matter what If if you don't but but this should make you realize that when I say I love something that you do I mean it because If I'm not saying something Anyways, I'm I'm gonna move on now. I'm gonna move on. I love you.  So it is what it is. I love you Do better Just amazing. Oh, I forgot to set my timer. How long have I been sitting here? Anybody know? Thanks. Yeah, I love you too. And uh, that's the best Do better be better. Am I right? Let's change the subject.  You can you can do it. I believe in you Here's the thing if I didn't if I if I didn't If I didn't care I would just be like Well, see that's the thing. Here's the problem with what I just said make kind of a bit of hypocrisy because I was like if Just keeping my mouth shut like if I didn't care I wouldn't say anything but I mean at this time I agreed with you, but I didn't say anything on a discord. Um, I do care.  That's what I find funny about it Like had you just done what you normally do? I think it would have came out fantastic But you tried to use the stencil and the stencil Made it look like someone just like through paint at I don't know. I don't know how it came out looking bad, but somehow It managed it. I don't it's it But here's the thing if you I know you shared it just to laugh But here's the thing for anybody else watching if you don't want an honest opinion from me Don't ask me because it's it's going to come out I have no filter when it comes to my friends. Like if someone's like, hey, look at this. Isn't this amazing? I'll be like Amazing's a word Not for this But amazing's a word Stay blessed. I don't Yeah, definitely that's that's one of those things and that's why I like uh people Who make things for other people like if you make me something I'm gonna treasure it far more than if you just go out and buy me something um, and uh, it's it's one of those it's I find it I I just think it means so much more when someone puts effort into something Even if it's terrible uh Then I would you know, it's like uh Some of the I still have a bunch of these beanies that my mom made for me that are steven king themed Like, uh, one of them is a cat. I don't want to dig it out. Um, because I don't want to go through those memories But the uh, there's one with like cat ears I use it for my pet cemetery review because she made me uh a beanie that looked like a cat had yellow cat eyes It was gray had cat ears And the thing looks god awful But it is one of my favorite Favorite things that my mother has ever made for me.  It looks terrible Um, but it means the world to me because she made it Um, and she put her heart into it Um, but yeah, I mean it does not look good And I loved it so much that I actually uh wore it when I did my pet cemetery Thursday theorist and I think I did it for the review also But uh, anyways, yeah, it's uh It was a hot mess But I love it I love it more than it had she went out and bought me like one of those actual cat hats I thought you I'm not trying to rush you off Hailey, but I thought you were leaving Are you hanging out? What is this line going down my forehead?  The light I need to change the lighting in here. Um, I got to order leds Uh ring lights. I have one I have one and I keep over here Um Either side, but uh, I need both of them so I can get some decent lighting Uh, I have a handmade card by rob that's hilarious and has a poem inside I love you so much with all of my heart even when you roll over in bed and far Yes, yes It's uh, yes Yeah I will most definitely be making a tough love positive animal sticker. I love you do better awesome Awesome. That's going on a t-shirt. That's definitely going on a t-shirt Hey tim. Yeah, it's an early one. I've been up since 230 I don't t-spring in red bubble now t-spring is more expensive, but the quality of t-shirts is a lot better Yeah, if you thought about six dollar tees, I'm thinking about that for our shop It's where I get all my shirts is where I got this one.  Um, they they do have some that are more expensive and because I'm a bigger dude um, I had to get uh Um, I had to pay like three dollars more because the shirts are bigger But uh, yeah, that's that's the only place I buy from anymore because you get this quality. I mean look how clean this is It's amazing I got more on the way too Thank you, hailey I've worn it several times on the channel. So All right. Yeah, I've been I've been working for more than 30 minutes. I'm gonna go take a break. Um, I'll come back We'll probably go for three hours today. I don't know. Um, if you guys want to sit and chat once I get done writing I'm going really really slow on this project. Uh, But it's it's coming along very nicely uh not doing anywhere near the uh the amount of writing that I used to do, uh, just because Slow and steady wins the race at the moment Uh, maybe in the future. I just don't have much time for editing anymore So I'm trying to get everything right and I know I've told people not to do this But I've also been doing this for almost 30 years now. So anyways Uh, just shy. No, not just shy. Let's see here when I was 19 I'm 42 now you guys can do the math Anyways, I'll be back in a little bit Break time Hello zoe. How are you doing? Are you doing good? I hope so If you missed it I told you earlier that I was uh that I subscribed to you You have a very interesting channel and I dig it very much Uh on this trailer Okay, let's get back to work I don't know. So funny glasses. I don't know what you're talking about Uh, let's see here. All right. I'm gonna go ahead and take my break when I get back. Uh We can chat for there's only two people in here But um, if there's anybody here when I get back we can chat If not, then I will probably write for another 30 minutes and head on out Anyway, y'all sit tight or go about your business. Whatever. I don't care. Bye Anyways, we are under a tornado watch So I'm gonna go ahead and go because I'm not about to be out in this office slash shed While uh, yeah Maybe possible in incoming tornadoes. But anyways, um, I am fine right now We have electricity obviously and all that stuff. I'll update you guys on discord. I'll take care talk to you next Whatever. Bye


UFC martial artist BJ Penn visited the Hickam Enlisted Club to meet service members and their families. He signed some autographs and talked with some of his military fans, a group he sees as having some things in common with his own profession.  Penn is originally from Hawaii, he's a world champion in Brazilian jiu-jitsu and a two-time champion in mixed martial arts. Tech Sergeant Michael Jackson, Joint Base Pearl Harbor Hickam, Hawaii.


In this video, I'm going to share the best practices I follow as a contest moderator when conducting a Toastmaster speech contest. This practices can be applied to any Zoom event where you want to control the audience and you want to ensure there are no distractions or disturbances when the event is in progress. If you follow all the steps, I can assure you that you will be the tech moderator and everyone will want for their event. So let's get started. The first thing I recommend is to ensure you are running the latest version of Zoom. The Zoom client is constantly updated with security patches and you want to make sure that you are up to date. To verify, log into your Zoom client, click on this profile button and click check for updates. If you see the message you are up to date, you are good to go. If not, install the recommended update and proceed. The next part is not really about Zoom or technical setup. To conduct successful events, preparation is everything. Don't just waltz into the meeting on the day of the event and expect to do a great job. You have to meet the event organizers and the participants before the contest and do a dress rehearsal. During the dress rehearsal, do an audio video test with all the participants to make sure that they are audible and visible. Here I have set up a meeting with four participants. Two of the participants cannot speak, but I think a participant Priscilla can. So let's ask her to unmute herself. Hi Priscilla, can you unmute yourself and say a few words? Hello, this is Priscilla, contestant at the Table Topic contest. And I'm looking forward to the contest. Great. I can hear you clearly and I can see that your video also is very good. So you're all set to go. The next thing we must make sure is that the contestant will be contesting using the same device. Let's verify with Priscilla. Yes, I will be using the same device. The next thing you must make sure is that all the participants are running the latest version of the Zoom client. Priscilla, can you verify that you're running the latest version? How do I do that? Open your Zoom client, go to your profile section and there is an option they're called check for updates. Okay, give me a second. Okay, it says here you are up to date. You are on the latest version 5.4.9. Excellent. And the last thing you need to validate with the participants is that they are comfortable to go in and out of the breakout rooms. The way I set up breakout rooms is as follows. I click on the breakout rooms and choose the option manually created breakout rooms. Click on the create button and we can give a name to our rooms. Let's say for the contestants, I want to create an isolation room. Next click on the options and the way I set up my breakout room is as follows. First of all, disable the option for the participants to choose their breakout rooms. Also, disable the option to allow participants to return to the main session at any time and this prevents them accidentally coming into the main session while a speech is going on. And the third option you must enable is to automatically move all the assigned participants to the breakout room. When we open the breakout rooms, all the participants will be automatically transferred into the breakout room without them having to press any buttons on their device. Now that the breakout room is set, let's assign Priscilla to the isolation room. I cannot assign Sam and Yo-Yo because I know they won't be able to press any buttons on their device. Let's select Priscilla and click the button open all rooms. As we can see, Priscilla is now in the breakout room. When it is time to bring the participants into the main session, all I have to do is click on the name of the participant, click move to and select main session. The participant will get an alert on their device indicating that the host wants them in the main session and within a few seconds, the participant should appear in the main session. And we have Priscilla with us in the main session. Additional things that you can verify during the dress rehearsal is that all the role players are ready with their scripts. If screen sharing is required, identify the person who will do the screen sharing as he, she will need co-host rights. We will be disabling screen sharing during the event itself. If you do all of this before the event and everyone feels comfortable, then and only then are you assured of a successful event. That's my take home message for all moderators, preparation. In summary, here's the checklist before the event. Update your Zoom client to the latest version. Have a dress rehearsal before the event, which includes an audio video test with all the participants. Ensure participants will use the same device that is tested. Ensure participants update to the latest version of Zoom. Test breakout rooms with the participants. Role players are ready with their script and finally identify role players who will do screen sharing to assign the co-host rights. Let's fast forward to the day of the event itself. Now it is all technical because everything is ready and everyone is prepared. Once you start the Zoom session, the first thing you want to ensure is the waiting room is enabled. You do that by clicking on this option security and ensure that the waiting room is enabled. This will ensure people do not enter the main session without being validated. In a contest environment, the next thing I like to do is to hide the profile pictures and what this does is any person whose video is not on will just show their name and we can see that by clicking on this button. Since Priscilla's video is not on, all I can see is her name and not her profile picture. You want to make sure that the share screen is off. This will prevent any Zoom bombing or any participant showing inappropriate content when a contest is in progress. I recommend keeping the chat on, but the way I configure chat is I click on the chat button. Next, click on these three buttons and select host only. This will ensure if there are any problems during the contest, people can reach out to you and highlight to you that there is some problem. If you completely disable the chat, then you would have no idea if something is going wrong. The rename themselves option I like to leave it on, especially the participants they like to name themselves once they enter the main session based on the contest that they are participating in. The unmute themselves option definitely needs to be disabled. We don't want people unmuting themselves in a middle of a speech and disturbing the speaker. For video, I recommend leaving it on because participants will want to switch on their video and you don't want that extra headache of asking participants to switch on their video. A few additional settings I like to do when it comes to participants is to click on the participants and at the bottom when we have these three buttons, I make sure that this is enable mute participants upon entry. I will get an alert which says all new participants will be muted. I will not allow them to unmute themselves. Click continue. And the final thing you want to make sure is that when someone enters or leaves the meeting room, Zoom doesn't play that annoying noise which you can disable by making sure this is off. Now we are ready to start the contest. Once the event starts, you have to pay full attention to everything that is happening. You can delegate responsibilities to other role players by assigning them co-host rights. With the new version of Zoom, co-hosts have exactly the same rights as a host. So there you have it. Follow the steps of preparing for this event, following the checklist of the event and setting up the security for the event itself. And you are guaranteed to have a smooth event. I wish you all the best.