 So thanks to the organizers for inviting me to speak. So in this talk, all of the representations will be complex representations. OK, and to remind everyone, the character of a representation row, like let's say of a finite group, is the function from the group to complex numbers defined by taking the trace of a row of G. And since we're taking the trace, characters are class functions. So they are constant on congenital classes. And another basic fact is that the number of irreducible representations of a finite group equals the number of congenital classes of that group. So you can organize all of the data of the values of the individual characters of your group in a square table called the character table. So this will be the character table of our group G. And to fix conventions, we'll label the columns by the congenital class. And we'll label the rows by irreducible representations. And where a row corresponding to an irreducible representation meets a column corresponding to the congenital class will put the value of the character corresponding to the irreducible representation evaluated at the congenital class C. And in the particular case we care about in this talk, G is the symmetric group on n letters. And then the congenital class is the symmetric group on one-to-one correspondence with the cycle types of permutations. And cycle types are then in one-to-one correspondence by just taking the sizes of the cycles with partitions of n. So that is decreasing sequences of positive integers that sum to n. And then we write lambda and then this symbol n to mean that lambda is a partition of n. And there is also a natural and beautiful one-to-one correspondence between irreducible representations of Sn and partitions of n, spec modules. And in this talk, we will denote the character value for the irreducible representation corresponding to a partition lambda of n evaluated at the congenital class corresponding to the partition mu of n at chi sub lambda sub mu. Yeah, so these are like the notations for the talk. So next I have an example of a character table symmetric group. So here is the character table of S5. So here I've labeled the irreducible representations in the congenital classes with the corresponding partitions of five. So there are five partitions of five. Here they all are. Sorry, there are seven partitions of five. Here they all are. So this is like a seven by seven table. And okay, I think one thing you might notice is that all of the entries of the character tables are integers. And this is not like in general true for character tables of like any finite group, but it turns out to always be the case for character tables of symmetric groups. So it is a general fact that the character values chi lambda mu of the symmetric group. This is always an integer. And this is the case for SN for all n. It wasn't just something for S5. And if you're a number theorist like a kind of natural question to ask if you are presented with these large arrays of integers, it's to ask like what are the statistical properties of these integers as n tends to infinity? Okay, so what are the statistical properties of the character table entries for SN as n tends to infinity? So Alex Miller in 2017 computed the whole character table of SN for n up to 38 and also computed the parity of all of the character values for n up to 76. Okay, and then he looked at kind of various, I guess arithmetic questions about the distribution of the entries in the character table. And I guess the most basic question you can ask is like as n tends to infinity, what proportion of entries are even and what proportion are odd? So here I have a graph taken from Alex's paper that plots the proportion of entries of the character table of SN that are even for all n between one and 76. Okay, I guess this is pretty suggestive. And next I have another table taken from Alex's paper in which he looks at how many or what proportion of entries of the character table of SN for n up to 38 are divisible by small prime powers. So he looks mod powers of two, three and five like up to the third power. And yeah, it's really difficult computationally to go much beyond 38. Like first of all, the size of the character table grows so quickly as n increases like pretty soon. I mean, probably at like 40, you can't even like fit the entire thing in your computer. And even kind of the, like even if you wanted to like compute a random character value, this also takes a pretty long time, typically it seems. But at least based on this small amount of data, Miller made the beautiful conjecture that as n tends to infinity almost all of the entries in the character table of SN are even. Okay, so he conjectured that n tends to infinity, the proportion of odd values in the character table tends to zero. And so this was like also a reasonable thing to conjecture because we've known since the 70s that it's true in the first column of the character table. So for the degrees of the irreplaceable representations. And this was shown by Mackay. This was 1972, but the character values, Thai, Lambda and then the partition of all ones, which just gives you the degree of the irreplaceable representation is almost always even. And he in fact found a formula for the number of odd degree irreplaceable representations. And then following Miller's conjecture, some partial progress was made and Morodi. And then in 2020, I was able to confirm that this conjecture is true. So almost every entry in the character table of SN is even. So the same in the same paper, Miller made the more general conjecture that the same holds for any prime. Okay, so now fixed P and we have the same thing as n tends to infinity, almost every entry in the character table of SN is a multiple of P. And this was proved by sound and I. And also with a degree of uniformity quantitatively in the prime P. So then in a later paper, Miller made the even more general conjecture that the same thing holds for all prime powers. Okay, so we'll fix now Q is a prime power. And again, as n tends to infinity, almost every entry of the character table of SN is a multiple of Q. And the topic of my talk today is about a proof of sound in me of this most general of Miller's conjectures. So precisely what we prove is like, first let's let n be large. So just larger than some absolute constant so that all the logs and log logs that are in our proof are actually defined. These will let n be large. And so we also can get a degree of like quantitative uniformity in this argument as well. And so then we let Q be a prime power. That's almost like 10 to the minus three log n over I think like log log n squared. So Q be a prime power. So then the proportion of entries in the character table of SN not divisible by Q is like big O of E to the minus log log n squared. So then by the union bound as a corollary, we get that if you fix any natural number, like say K, then as n tends to infinity, almost all entries of the character table of SN are divisible by that K. So before we had this for prime powers, we can only do it for like K square free. So maybe I'll pause here for a second to see if there's any questions about like the statement of the theorem. Then I assume there are no questions. So I'll start talking about the proof. And the proof relies on combining like four key intermediate results, which I'll tell you. So the first is a lemma. Should we call the combining parts lemma? If you've seen me talk about the prime stuff, this is just a more general first question. More general version of the combining parts lemma for my previous talks. So it says if we let P to the R be a prime power, okay, then we assume that mu is a partition of n and that new is another partition of n that is obtained from mu by replacing P to the R parts of the same size. Let's say M in mu by P to the R minus one parts of size P times M. So same total number of numbers were like same total size of the parts, though P to the R times M, but now I've combined them into P to the R minus one parts. So then the character values for mu and new are the same mod P to the R. So chi lambda mu is then congruent to chi lambda mu. Mod P to the R. Okay, so then I guess this is the case for all partitions lambda event. Okay, so for an example of what this combining parts operation is. So here we'll take the prime power to be four. So then the character values chi lambda 3, 1, 1, 1, 1. So four ones, these are the same as when we combine the four ones and the partition 3, 1, 1, 1, 1 into two parts of size too. So same mod for all character values. So for the people who know a bit about the representation theory, the best end, the way that this lemma is proven, it's really not hard. You just reduce the polynomials appearing in Frobenius's formula for character values, mod P to the R. So this is not very difficult. But to tell you about the next three lemmas, I'm gonna have to introduce a bit of, I guess definitions, notations, having to do with partitions. So for any partition, you can form the associated young diagram. Which is just like a left justified collection of boxes with each row containing number of boxes of the corresponding part of the partition. So for four to one, you get the young diagram with four boxes in the first row. So there's like, because the first part for the largest part of size four, then two boxes in the second row and then one box in the third. So this is the young diagram of the partition four to one. And each box in a young diagram has a corresponding hook. Again, this will be a definition by example. So I'm gonna just redraw young diagram above. I guess I could have copied it because this is an iPad. And then the hook for a box. So say this box in the like upper left-hand corner is the collection of all boxes to the right of the box in the same row. So all these boxes along with the boxes and the same column below. So all of these boxes I've drawn blue through form the hook corresponding to this box. And the length of a hook is just the number of boxes in it. So this is a hook length, there are six boxes. So this hook has length six. And for example, I guess, so this hook has length four. And you can do that with any box in the young diagram. So if you have a partition of N, the young diagram has N different hooks. And we say that a partition is a T-core. If none of its hooks have length, a multiple of T. Okay, so you can check that this partition four to one is a five-core. Okay, so longest hook has length six and you won't find a hook of length five in the diagram. So yeah, that's the definition of T-core. And the next key intermediate result, lemma that says that almost every partition lambda event is a T-core when T is large. So because we do things quantitatively, we'll let L be some parameter, which if you just care about a fixed prime power, like you can just take L to be some sufficiently large constant. So if you wanted to prove the result mod four, you could take L to be 25. So then for any natural number T, of size at least one plus one over L times square root of six over two pi square root of N log N. All but, oh, I don't know what happened there. All but a big O of log N over and the one over two L proportion of partitions of N are T-pores. Yeah, so maybe I'll give you some context for like this lower bound. So the expected largest part of a random partition of N is asymptotically square root of six over two pi square root of N log N. Okay, so yeah, that's exactly what appears here. So this is saying if T is like substantially larger, then the typical largest part of a random partition, then almost all partitions of N are T-pores. Okay, so that's what's meant by if T is large. So this is the second kind of key intermediate result. And the third is the like, like the kind of key new input into the argument that let us do prime powers instead of just primes. And it's a new divisibility condition for a character value and a prime power. So this is the divisibility condition. Okay, so we'll let M1 to MR just be distinct national numbers. And we'll assume that mu is a partition of N and it has parts of sizes M1 up to MR. Hi, Sarah, sorry to interrupt, but there's a question in the chat, a question of whether you could just provide an idea of how to prove the lemma for T-core partitions. Oh, yeah, so for this, I guess there's a couple of different ways you can do it. I guess you can use like, if you wanted even more information about the distribution of like the number of T-cores, you could use like On's like, like generating function identity for T-core partitions. Or I mean, there are just some, I guess basic inequalities, like for the number of T-core partitions that you can just plug in the asymptotic for the number of partitions of N into, it's the number of partitions of N and then the number of partitions of N minus T and like mess around with basic facts about T-cores to get this. Yeah, so this is not a like difficult lemma. Thank you. Yeah, I hope this is helpful. Okay, so for the divisibility condition, which is like much, much harder to prove, this is where like most of the work of the proof goes into. So we have distinct positive integers and M1 up to MR and then mu is a partition of N with parts of these sizes, all appearing with multiplicity at least P to the R, sorry, P to the R minus one. So each appearing at least P to the R minus one times in mu. So then lambda is a partition of N and it is a simultaneous sum K I M I core partition for all our tuples of Ks such that the Ks are non-negative integers between zero and R minus one R minus one, that sum to at least P to the R minus one. So if lambda is a simultaneous, this quantity core for all of this many our tuples, then the character value K lambda mu is a multiple of P to the R. Okay, and yeah, I will postpone saying anything about this proof until the end and then only if I have time. So I told you like three, the three intermediate results, three out of the four, the first one says we can do this like combining parts operation and not change the congruence class, my P to the R of our character values. So you might as well combine all the parts that we can until we can't combine any more parts. So nothing appears more than, or nothing appears at least P to the R times. And then it's, I think then reasonable to guess that if you combine parts in this way that you'll end up with many kind of largest parts M one up to MR that appear with large multiplicity and that the M one up to MR are large enough so that this quantity is sufficiently large that almost all partitions lambda are this quantity core partitions. So then the next key intermediate result is that we show that this is the case. So we show that this combining parts operation leads to many large parts as many largeish parts with large multiplicity, largeish multiplicity to be more consistent with my vague notions of large. So what we show is that like starting with the typical partition of N if you combine parts in the way that I told you from the first lemma until you can't any more then you're gonna have enough largest parts to satisfy the divisibility condition or so that typical lambda satisfy the divisibility condition from the previous lemma. Okay, so starting with the partition mu event you repeatedly replace every occurrence of P to the R parts of the same size M by P to the R minus one parts of size P times M. We do this until we arrive at a partition mu tilde also of N in which no part appears P to the R times. So you can't combine parts anymore. Okay, so then except for a big O of E to the minus N to the one over some constant depending on P to the R proportion of initial partitions mu, mu tilde has the parts we want. So it has at least R distinct parts. Let's say M sub J each appearing at least P to the R minus one times and satisfying P to the R minus one times M and J is at least a good bit larger than the typical largest part of a random partition event. Okay, and this is good enough to make the whole argument work. Yeah, so the proof of this is we can create a kind of convenient generating function whose terms are an upper bound for the number of like bad partitions mu. So those such that mu tilde like doesn't have this nice property. And yeah, we've picked this convenient generating function in such a way that we can kind of more easily evaluate more easily like upper bound its coefficients which then would give us an upper bound in the number of bad partitions mu. So now that I've told you the is four key intermediate results I'll let's give like now or let's write down an outline of the proof which I guess I've said verbally. So first, by repeatedly doing the combining parts operation the first lemma tells us that the accounts of the character table corresponding to mu and mu tilde are a congruent mod P to the R. Okay, so then it suffices to show that almost every partition mu character values chi lambda mu tilde are almost always multiples of P to the R. Okay, so then by combining the divisibility condition with the fact that almost every lambda is a T four or T large events suffices to show that mu tilde has many large parts with a high multiplicity, okay? But this is exactly what the last time I showed you said, okay, this is the lemma label that combining parts leads to many largest parts with a high multiplicity. Yeah, so these are how the different parts of the proof fit together. And it seems like I have enough time to say a bit about the divisibility condition and it's proof, which is like, yeah, this was the really fun part. So first I'll tell you about the Myrnihan-Nakiyama rule, which is a recursive formula for computing character values of the symmetric group. And first I need a bit more, it's notation. So here's the young diagram of the partition 333 or it will be. So it'll be like a three by three collection of boxes. And so here we have a hook of length four. And so any hook touches like two boxes or at least two boxes on the boundary of the young diagram. So we have the box all the way to the right on the horizontal arm and then we have the box leg all the way at the bottom on the vertical arm. And we can also connect these two boxes along the boundary of the young diagram. Okay, so right here we have a hook in blue and then we have the corresponding border strip, Bs of H where we connect the ending parts of the hook along the border strip or along the border of the young diagram. And for any hook, there's another quantity associated to it called the height of the hook, which is just the number of box or the number of rows the hook goes through minus one. So this hook has height equal to two because it goes through one, two, three rows. So then we subtract one to get a height of two. And the Murn hand Nakayama rule says, so if you let sigma be a permutation that factors as product of two permutations, one of which is a T cycle and then tau can be whatever else but the supports have to be disjoint. Then my partition lambda then the character value chi lambda sigma is a sum overall hooks in the young diagram of lambda of length T, negative one to the height of the hook and then the character value chi, the character corresponding to the partition whose young diagram you get by removing the corresponding border strip from the young diagram of lambda evaluated at tau. So both of these, so this is now a partition of N minus T and this has a, or corresponds to a, also a partition of N minus T. Okay, so it gives you character values of SN in terms of character values of like SN minus T. This is why it's like recursive. Okay, so it can give you like a very, I guess very vague sketch of the proof of the divisibility condition and it's basically just repeated applications of the following lemma does a lot of the work in the proof. So now more generally, you let sigma be a permutation in SN of the form of tau times now a product of, maybe you can't see this, P to the R minus one T cycles and all of the supports of these permutations are disjoint. Okay, so now we let for any lambda partition of N we'll let L denote the set of partitions lambda prime of N minus P to the R minus one times T that you can obtain by removing P to the R minus one border strips from lambda of length T. Okay, so then lambda is a partition of N. Oh no, I already said that. So then if lambda is a P to the R minus one times T core, then character value chi lambda sigma is the prime P times some sum of character values over L. It ends up being an integer linear combination of these character values. So then, okay, we've now gotten that some character value is multiple of P. And then the idea is to like repeatedly apply this lemma. I guess you apply it R many times to get a divisibility by P to the R. And the existence of this complicated looking like hook condition in the divisibility criterion basically comes from having to ensure that this always holds. Okay, so another part of the proof of the divisibility condition is checking that if you make this complicated core assumption on your initial lambda, then even when you remove like many different parts you still know that it's a, I guess, T prime core for enough T primes that you can apply this lemma. And the proof of this lemma was very fun. Basically the way it works is you apply because you have P to the R minus one T cycles. You apply the Myrna-Hend-Nakiyama rule, P to the R minus one many times. So then you definitely end up with a linear combination of character values of this form, right? Because either just the character values correspond to lambda that you get by removing P to the R minus one border strips of length T from lambda. And this is what the Myrna-Nakiyama rule like has in its recursion. And then, yeah, the thing you have to show is that each one of these appears with both the same sign and a multiple of P many times. And to do this, it's send this very combinatorial argument using the abacus representation of partitions. And yeah, that's all I'll say about the proof. I wanna end with there's still kind of many interesting opening questions about the character values of the symmetric group. So I'm gonna highlight two of these suggested by Miller's data. So the first is question of what proportion of the character table of SN is positive or negative. So based on his data, Miller conjectured that the portion of positive and negative values should be roughly equal. And then related to this is the question which is the most interesting question to me of how many zeros are in the character table of SN? So what proportion of the character table of SN is just zero. And for the second question, I'll show you some data in a second, but it's like, I think, yeah, it's hard to guess like what the answer should be based on the data. I'll say that like obtaining for this one, getting at least a constant times one over log N proportion is pretty easy. You can just use, there's this like old result of Erdition-Laner about the distribution of the largest part of a random partition event. So you can combine this with the Marinette-Nakayama rule and the fact that or the quantitative version of almost all partitions of N or T cores when T is large to get this one over log N proportion of zeros. And I think this is the best known. I don't think this has been improved at all. Okay, so yeah, I like these questions and here's some more data I stole from Alice's papers. So this one, this is the graph of the, or it plots both the density of positive character values in this set of non-zero character values and the same for negative in the character table of SN for all N between one and 38. And yeah, you can see they both appear to be approaching one half. And so this is actually from a different paper of Millers that's on the archive. Like maybe this was like the original version of the one that ended up published. And here is also the data for the portion of zeros in the character table of SN for N up to 38. Yeah, I think here it's really hard to tell like, yeah, what's going on? Like is the density of zeros approaching some positive constant or is it decaying like one over log N to zero? Yeah, who knows? So that's all I have.