I'm Corbin. I'm Rick. And he follows on Instagram and Twitter for more juice. A con tent. It's so juicy. Think we're supposed to send Patreon and follow on official Twitter account? Look at me. I'm a rat boy. Would you desist? Today, we are going to react to a Oostaji Zakira Hussain video. If you don't know who he is, prepare to have your mind blown.  Yeah, we're not anymore. A year ago, we were, but we're not because we're best friends with him now. We know him, but this one's called Oostaji Zakira Hussain playing set of tablets. Looks like he has a bunch of different instruments in front of him. And I'm guessing he's going to play them. And he, among other things that he's done in the short timeframe we've been exposed to him, he changed my perspective totally. I don't even use this term anymore that I've used my whole life when we were talking to him. And I said something, I think I said it about mastering an instrument and his respectfully taking strong exception to the fact that you don't master the instrument. The instrument allows you to participate with it. And he completely, I thought that his definition of that because yes, he would be considered a master of his instrument. Absolutely. But I love, it's not only a great healthy perspective but it's a really humble perspective which you can see when he performs. Absolutely.  Absolutely. Which is what makes him so freaking incredibly great. Here we go. Lean, lean, put your elbows on the drums and we're just gonna lean over. I'm gonna be your tight shot at you. Okay. And this is. This is actually, who's saying? Him, obviously playing, but you can tell his sense of humor. Yeah. From him playing. Yeah, you can. I didn't, his face was even showing and you could tell his playfulness. It's a very playful style that he does and it never ceases to amaze me, his talent. We've seen him live. We saw him in concert and obviously we got to talk to him but we saw his concert after we talked to him and it was as incredible as anything you'd ever, so if you ever get a chance to go to one of his concerts don't even hesitate and go. Go. Because it is a treat. It's an experience. It's especially a treat in America when nobody knows who this man is and they, I think when we went, they was handling two other people and I think most of the people in there knew the other. They knew the other two guys. Yeah, they knew the other guitarists who were incredible in jazz. And obviously we didn't know who this man was. Yeah. And then he just blew their minds with everybody when he does what he does in every video it's just like I'm having an eargasm. Yeah. Do what he can do with his fingers. Yeah. And people are like, holy. Yeah, and there's some moments where he's, cause in that trio there were moments where each of them got to do some soloing and when he does it, it's like the oxygen leaves the room. It's just this, everybody's holding a breath.  I know. In all, it was cool actually seeing a pose. Mm-hmm. The. Uh-huh. How, like, I don't know. And he's just a never ending fount of creativity. He to me. None of that was written. Yeah, he to me is comparable to the innocent creative joy of like when you watch little kids just in the backyard playing, they're just in a non-stop state of focused creativity just going and joyful creativity that it's just non-stop energy. That's what it's like watching him play.  Like, I can see when he gets some expressions in the space of joy, it's really his communicating with the instruments. It's almost like he's, oh, thank you for that. You know? And he'll probably say that's what it is. He's probably saying what it is. He's like, he did something and he was hoping to get the response he got and he got it and his heart's like saying, oh, thank you. That's what I was hoping you would do. Yeah. Yeah, it's beautiful. That's exactly what it is. Yeah, he's just, I can't get enough of him. Me too. I would love if somebody, because he's done a couple tea commercials that he told us about. Yeah. He did tea commercials. But they're on the internet and I don't think they have subtitles. But if somebody wants to create his compilation of his tea commercial, I really want to see them. And he acted in something, didn't he? Yeah, yeah. Give me a Zuckir who's saying compilation of his acting and his commercial. Yes, please. Because I want to see that. Yeah. I want to see that side of him. Also, he's acted in films and stuff, so I want to see all that. And any performance of his, I will watch and just be mesmerized. And I'd love to see some of the earliest recordings of him. I'd love to see some of his earliest playing. And I'm sure he was from the get-go. He was brilliant from his starting years.  My suspicion is he probably doesn't have anything prior to 17 or 18. But dad, I'd love to see that. Oh, that's insane. Love you, Ustadji Zuckir who's saying. Indeed.


What's going on everybody? This is optic blaster 10 box break number 19 to random teams each sold out on jaspy's case breaks Dot com so we're gonna set up the randomizers here, and we're gonna roll the dice To see how many times are going I think I do have the list ready, right? I might have not gone through orders here I'd be premature You might be actually I know maybe Joe did thanks Joe So here is a list of names. We're gonna copy it one more time and All NBA teams here Thanks everybody. Good luck We're gonna roll the dice and go six times four and a two six times There's one two three four Five and six and final six. We got Matthew down to Brett And now for the list of NBA teams gonna go six times one this might turn into a golf channel soon two three four five and six Washington Wizards down to the Denver Nuggets Matthews Wizards Joe Crowell pistons and Thunder Jennifer Raptors Brett Warriors Christopher Clippers Matthew Cavs Brett Sons Matthew King's Brett last Bob Mojo Jazz Joe Crowell Hornets Sixers Pacers Lakers Magic Nets Christopher with the Pelicans Matthew with the eat one with the Bucks Joe Crowell Celtics Juan Nicks Joe Crowell Rockets Hawks and Bulls Brett Spurs Joe Crowel Mavs and T-Wolves Jennifer Blazers Brett Grizzlies and Brett Nuggets Pause the video see if there's any trades. If not, we will be back to to rip these 10 blasters All right, there were no trades. So good luck everybody Thanks again for getting in Here's a list of teams Ten blasters here Good luck Yeah, but the thing is whence has a zoom ball ring Dolphins windows the last time they won 1972 and Then the the Chargers out of enough What is it like? What do they call it back then championships? Couldn't be couldn't be my quarterback Couldn't be my quarterback stick over whence I don't know if any you would have to have a hundred twenty three million dollars in in your bank account to buy it There's a Zion insert. So the only person who probably could do is who a handful of people Bezos Gates The Walmart sisters or whatever That was box one coming up after a specter basketball and then after that leafless football And then it's nighty night for Jaspie's Watching a little Oakland A's Boston Red Sox Don't know when but kind of entertaining game 2018 LeBron team line is three two one Guess Zuckerberg could afford it right culvert to 149 for the T-wolves Kendrick not Luca space Xbox I'm actually gonna open two at once. Yeah, no pressure Mac. I mean You hit the lottery you Give me a call first call It'll be we can rename the website Mac breaks and it's just personal breaks for you You don't sell that anybody else That'd be the easiest thing in the world Ship to one person Deal with one person That'd be fantastic Yeah, if they bought it though, they might never see me again. That's the problem. There's Devon Booker hollow Kari hollow Kari purple. We got a Cody Martin purple ray of rookie and that's that for two blasters. Oh Yeah, we would ship veteran base If we had just one on buyer I'd be ideally we would have we would be on 24-7 even if even if you're sleeping just in case you wake up I'm on a break. I'd have someone here whether it be me Jason Teddy Joe my dad Anybody so that's that's that deal stands for anybody watching too, you know, if you want your Your personal personal breaks done And you want and you have uh, what is it nine figures? No Ten fingers, I don't know ten fingers. Let's say and there has to be a one in there somewhere Not all zeros But for those people who don't have, you know, billion dollars, you can do jackpies.com You can buy any box that we have on our website you can Put in the order notes ship it sealed you can buy top loaders penny sleeves You can buy curbside pickup. You can come pick it up at our shop most of beach We've got hip packs on there all the time we got Whole ton of stuff teddy will be live saturday on instagram doing all your personal breaks starting around three o'clock pacific six eastern for all those that want to break And uh, have their own personal box We'll do that for you kevin durant Your dream is a reality at jazz piece Kobe white hollowed my house James harden sleep that Durant will ship Uh, the brown will ship and of course this john marant will ship to the grizzlies Brett f last by mojo No, I mean, we've had, you know famous people buy but Probably not charlie sheen level. I think he sold this whole collection When my dad worked at tops, he did the tops of all They would have uh, you know big-time buyers You know keith oberman former formerly of sports centers a big-time buyer of sports cards Kind of other guys. I'm sure All right four blasters to go Coming up next is spectra half caser and then the best of football That's crazy. Sean. Did you see the the kid who ran up his uh, his parents? credit card like a hundred thousand dollars on fortnight charges Yeah, I went to the I think my first or second ever national um I saw the dimitri young collection I think he was there too Maybe not. I don't know But yeah, he's got a cool collection. Yeah charlie sheen has sold his collection before Like auctioning it off. But yeah, my dad Told me a lot of stories one time he told me I think he needed extra five hundred dollars to pick up A t 206 honus waggoner and this is probably in like the 60s 70s I think it was extra five hundred dollars. He has his dad His dad said no and that card is probably a million dollars now the keel alexander walker But the pelicans I think it was one of the nicer graded ones Pelicans that's going on to christopher nice out of there. There's john wall There's seiku base There's kelly olynik. There's a ryu hachimora. I don't know what what that is on there Sleep it up. I don't know. I'm sure you can take that off Zion williamson Kendrick none and yeah, he said it was probably one of the better conditioned one and and it wasn't the one that was uh trimmed up either Marcus smart and space two more to go. I've never seen that. No went for four hundred and a half million a lot I did see all those kind of old moves that the uh The lebron logo man with who was at jordan nine hundred thousand dollars It's the highest modern sports card ever But yeah, it's crazy And yeah, I think there's still room to grow on a lot of cards. I don't think it's too late It's just like bitcoin, you know Oh out of 149 calyre and azion base for the pelicans christopher You know pit people said bitcoin was too much at what a thousand and then What did it go to december 2017 like 20 grand or something? I think they do William he he didn't really work in that kind of department. He worked in um, the tops vault and like tops.com So a lot of their like exclusive products and stuff But I know like for like artist sketch cards. They give uh artist proofs. They call them for like wacky packages and stuff So I would imagine players are the same thing, right? But maybe um, I'm wrong semi ojale auto for the celtics.  That was the break. Thanks everybody blaster number 20 in the books. Thanks everybody


So I wanna talk about two torsion in class groups of number fields. So everything I'm gonna say today is joint work with Dante Bonolis. Okay, so I'll start with an introduction. So what's the setting? So let's say we have a number field K. So I'm gonna let DK be its discriminant. I'm gonna let CLK be its class group. And for any problem number P, I'm gonna let CLK brackets P be the P torsion subgroup of CLK. All right, and here is a classical problem. Even two integers, M at least two, and P a prime number. People are interested in bounding the size of the subgroup CLK brackets P. So the P torsion subgroup of CLK. For instance, in terms of the discriminant of K, but could be something else, depending on K, as K runs over number fields of degree N. So the degree is fixed, the prime P is fixed, and we view these two integers as being fixed. So it means that when we will have bounds for the P torsion subgroup of the class group of K, we will have constants, which will depend on these two integers. And we will not care about these concepts, okay? All right, so here is a classical theorem. So it's due to land-out. So for any N at least two, and any epsilon positive, for any number field K of fixed degree N, then the past group of K is bounded above by roughly the square root of the discriminant, okay? Of the absolute value of the discriminant. And there is a constant hidden in this notation. So of course it depends on N. It may depend on N and on epsilon, but as I already said, we don't care about the dependence on either N or epsilon of this constant, okay? So that's a very classical result. And here is the conjecture. So it says that if now instead of looking at the full class group of K, you look at the P-torsion subgroup for fixed prime number P, then this time, this quantity should be, the cognitive of this group should be much, much smaller. And in particular, it should grow more slowly than any power of the absolute value of the discriminant, okay? So again, the constant here may depend on N, on P, on epsilon, but we don't care about these dependencies. Okay, so this conjecture was first raised as a question rather than a conjecture by Broomer and Superman in 96. And then it appeared in the literature in different places in the 90s. So for instance, in the work of Duke, and of the people, and then it became a classical, but there were for some time, absolutely no results close to this. So it became like a full-clone problem, which is like really hard, okay? Still there is of course one case for which the conjecture is known. So by Gauss-Genes theory, it's not hard to prove that in the case NP equals two, two. So you're looking at two-torsion of quadratic fields. Then you can bound the two-torsion subgroups of quadratic fields by roughly the divisor function of the discriminant. So in particular, it's smaller than any power of the discriminant. Okay, so that's the only case when the conjecture is known. And indeed, even proving that there is actually some saving for some fixed pair N and P, there is some delta and P positive, such that for any number 50 we have such a bound. So the P-torsion subgroup, the causality of the P-torsion subgroup is less than some power of the discriminant, which is less than one-half, then already this is very hard, okay? But still, if we are ready to assume some strong conjectures, then things can be done. So this is what Inenberg and Banquetes did in 2007. So for any N, at least two and P, any prime number, they prove that if you assume GRH, then you get some savings. So in this case, delta and P would be one over two P times N minus one. Okay, so why do they need GRH? Because they notice that if you have many small primes, which split completely in K, then the quotient of the class group by its delta ocean, it has to be somewhat big. So it means that the delta ocean subgroup, it has to be smaller than just the trivial bound. Okay, so that's what they do. And so to prove this in general, the existence of such primes and many small primes which split completely in K, they need an effective version of the Chebotaric density theorem. So that's why they need the GRH. So apart from these conditional results, most results which are available are statistical in some sense by which I mean, people have been looking at maybe the average of the two torsion of cubic fields, say with bounded discriminance, or maybe they've been looking at higher moments and they deduce results like for some pair N and P, most number fields K, which are bounded discriminant, must satisfy some bound for which there is a saving. And sometimes the saving is as good as what we obtain with GRH. Sometimes it's better, sometimes it's worse, it depends. But of course, even proving statistical results is hard because for number fields of large degree, we don't even know how many number fields have discriminant less than X, right? We don't know how to estimate this quantity as long as N is at least six, I believe, okay? So even statistical results are hard to obtain. So I had neither the courage nor the time to display all the results that have been proved over the years. So I just displayed many names. I hope people won't be too mad at me for doing this. So of course, it started with the work of that important high born in the 70s. And then there were works by Bargava at the beginning of the years, 2000s. And then many of the people, so Hit Brown and Pierce, Elenberg, Pierce, Machete Wood, Frey and Wiedmer, Pierce, Ternich, Bargava, Machete Wood and more recently, Briglino. So if people are interested in having like a complete or fairly complete list of works in the area, I recommend to go to the article by Pierce, Ternich, Bargava and Machete Wood. They have a very, I mean, it's like a survey of everything which has been done in the area. You go to section seven of the paper, okay? So I'll now move on to unconditional point wise results. Okay, so that's what I want to focus on today. So the first breakthrough in the area was done by Pierce in the years 20 or 500 years six. So she has two papers. So at the beginning, the saving was a bit less than the one I displayed there. And in the second paper, she obtained that savings. So she's looking at the three torsion in class groups of quadratic fields. Okay, so n is equal to two, p is equal to three and she proved that we can get saving like what I displayed, so one over 56 in the exponent. And to do this, she looked at some day of integration and she counted the integral points. I mean, the integral solutions of this day of integration which is what we are going to do later today, okay? Okay, so this was improved. First, I have got Venkatesh by counting the solutions of the same day of integration but differently using the results in their jumps paper about integral points on the curves. And finally, this was improved to one over six by Enandberg and Venkatesh not using a day of integration but using the strategy I told you about before. So with this time without assuming gyrates because that's some specific case for which this can be done. You can prove that there are primes which do what they need. Okay, so next, also using the same method, Enandberg and Venkatesh, look at the three torsion of cubic fields and they obtain the same savings, so one over six. And finally, they could also handle quartic fields. Again, P is equal to three but this time they didn't complete the saving but at least for many number fields K, the saving delta that I displayed there is something like one over 168 I think but in general, it's probably a bit smaller than this but I mean, still there is some delta positive such that for any quartic field K, we have this bound. Okay, so if we sum up, we know a non-trivial bound towards the conjecture in four cases. So NP is two, two, so two torsion of a quartic fields and then P is always three and we know it for N equals two, three and four. So three torsion for quartic fields, three torsion for cubic fields and three torsion for quartic fields. So there's four cases of pairs, NP such that there is some non-trivial step towards the conjecture. Okay, so this was the case until the work by Vagabha Shankar, Taniguchi, Thorn, Seymour Mann and Zhao. So 2020 is the year of publication but it was proved probably two or three years before and for the first time, they could handle infinitely many cases. So from now on, please gonna be two always, okay? And, but N can be just anything, okay? It's fixed, but it can be as large as we want and they obtain the saving one over two N in this case and also, yes, so when N is equal to three or four, they have a refinement of their approach which gives them a much better savings. So it's the saving in this case is something like 0.2215. So it's much better than one over eight say when N is equal to four, okay? So today I'm going to focus on this case. So P is going to be equal to two all the time but N is going to be at least five because their result is just too good when N is equal to three or four and we just don't know how to get close to what they achieved. Okay, all right, so it's probably a good sign to stop and ask if there are questions from people in the audience. Okay, so if that's, I'll just continue. So I would like to go through their proof. So I'm going to refer to their result by BSTTTZ, I hope they won't mind, but I mean there's six authors, so. Okay, so I'm going to need a lot of notation. So again, K is going to be a number field. As usual, okay, it's going to be its ring of integers.  All right, so then bold sigma is going to send K to R up to the N. Okay, so people call it the inventory embedding of K. So that's this map. So the first R coordinates are given by the sigma I's and then the next two S coordinates are given by the real and imaginary parts of sigma plus one up to sigma R plus S. Okay, so we have R plus two times S coordinates, so that's equal to N. Okay, so this map sends K to R to the N. And if we look at the image of the ring of integers of K, we get a full rank lattice in R to N. Okay, so I'm going to, sometimes people, sometimes people just call this lattice in R and just okay.  I find it somewhat confusing in some situations. So I prefer to call it LKs. Okay, so next I need to introduce the successive minima of LK, so any notation for that. So I guess it's pretty standard to call them lambda I of LK. So lambda one, what is lambda one? Lambda one is the shortest non-zero vector in LK. Then lambda two is the shortest vector in LK, which is not on the line generated by lambda one. I mean, some vector realizing lambda one. And then lambda three is the norm of some vector, which is in LK, but which is not in the plane generated by vectors realizing lambda one and number two. And we continue like that. So we end up with N numbers. Okay, so yeah, I should make this clear already. So lambda one up to lambda N, they are numbers. Maybe I said vectors, I meant norm of the vectors. Okay, so they are positive numbers. Okay, yes, so this equation is just one way to define them. They're exactly what I said. So in this notation, so I'm gonna call BN, the Euclidean bone, but I'm not sure I will use it again, but I will use the Euclidean norm quite often. Yes, so next, so in okay, there is a one. Okay, so it means that when you send one to our N, you get some vector, which is whose length depends only on N, okay, not on K. So it means that lambda one of LK is some number, which is bigger than one, okay? And another thing, two important thing to note is that the determinants of this lattice, if you just apply the definition of the determinant, you get that it's almost the absolute value of the discriminant of K with the square root, except that there is some factor, we need to divide by two to the S, where S is the number of complex embeddings of K. But the determinant of a K, it's essentially the square root of the discriminant of K. Okay, so it means that if you look at Minkowski's second theorem, so it says that the product of all successive minima is bounded by some constant, which depends on N times the determinant of the lattice. So in our case, lambda one is one, so it just goes away. And the determinant of the lattice is essentially the square root of the discriminant, okay? So that constant two to the S goes away as well. So in this, so we have two equalities there, the left equality is just the Adamao inequality, and the right inequality is really Minkowski's second theorem. Okay, okay, so we'll start now to go through the proof. So here is the first key lemma that they prove. So they prove that the cardinality of the two torsion of the class group of K, it's bounded by the pairs, the number of pairs Y and beta, so Y is gonna be some integer, beta is gonna be in the ring of integers of K, such that we have two conditions. So the Euclidean norm of the embedding of beta has to be bounded by dK to the one over N, and the norm of beta, so in K is the norm of K of Q, the norm of beta has to be equal to Y squared. So the right hand side is really the number of beta whose norms are squares and which are small, which have small Euclidean norms. Okay, so how do they prove this? I'm not going to go through the proof today, but I'll just say a few words. So if you have an ideal class whose square is trivial, that means that there is some ideal I, such that I squared equals its principle, so it's maybe equal to alpha times K, but then whenever you multiply alpha by the square of an element of K, of a non-zero element of K, then you're gonna remain in the same ideal class. Okay, and what they do is that they choose this element, so we have something like, so we have alpha and we multiply it by the square of some elements in K, maybe kappa, and they choose this kappa, sorry, kappa, we multiply alpha by kappa squared, and they choose this kappa so that the norm of beta is smaller like it's smaller than what I wrote, okay? And to do that, they use Minkowski's first theorem. So they build up some convex symmetric body and they say, well, if the volume of this body is large, which it is, then there must be some integral point, some, yes, some point in some lattice, which is going to be kappa, and so that's how they can locate beta. I mean, that's how they can make sure that the Euclidean norms of beta, the Euclidean norm of sigma beta as well, okay? And of course, the norm of beta is a square because beta times okay is gonna be J squared for some ideal J, okay? So that's how the proof goes. So it's fairly simple, but it's very clever. How are we going to use this? So let's take a basis omega one, omega two, omega n of okay, such that omega one is one, and we have this growth condition on the norms of sigma omega two, sigma omega n, and such that we have this condition, okay? So such a basis exists, some people call these basis reduced, some of the people call them minimal or almost orthogonal, they're all the same, they exist, and they satisfied these kind of properties.  So that's the basis for which you can do that, okay? Then we get the following. So what have I done there?  Because omega one, omega two, omega n is basis of okay. So I have replaced beta by this in the norm, so that's fine. And then I had norm of sigma beta that's most the absolute value of the discriminant to the one over n, but now by using the second property, this becomes for any i in one n, L i is at most dk to the one over n divided by the sigma of omega i, but by construction, these sigma of omega i, they have to be at least lambda i of lk, one i. So that's the bound that we get for the L i's, okay? So maybe I should say that BSTTZ in their proof, they don't need all these notation, but that's gonna be useful for what we want to do after, so that's why I'm introducing this notation there. But maybe that's just our interpretation of their proof. And they don't need the successive NEMA and norm of this mess, okay? All right, so next, so for any beta in okay, so what's the norm of beta? So it's gonna be the product of the embeddings, right? So it's gonna be bounded by the Euclidean norm by definition of sigma of beta to the N, okay? So I guess that's just the, how do you call it in English at AGM? The arithmetic geometric means the inequality. So we get this inequality for Y, right? If I go back, so I had Y squared equals the norm, blah, blah. So if the norm is at most sigma of beta, Euclidean norm to the N, then I get that Y squared is at most decay. So Y is at most the square root of decay, okay? So then they use the Bombieri-Pilla bound for discounting function. So we are counting the number of Y and L1 such that Y squared equals the norm of L1 omega one plus blah, blah, plus LN omega N and Y and L1 are in some box, okay? So it's typically the kind of situations where you want to use the Bombieri-Pilla bound. So I have hidden some a few things there because sometimes it can happen that for some choice of L2 or L3 and N, the polynomial NK of L1 omega one blah, blah, blah, the polynomial in L1 is going to be a square, okay? So if you're in this situation, then the variety, the curve you are looking at, it's not going to be irreducible. So you're going to be in trouble, but there are very few choices of L2 or L3 and N for which this can happen. So in the end, it's fine, okay? But I have oversimplified a bit there. So what do we get for the bound? So the polynomial NK of blah, blah, as a polynomial in L1, it has degree N. So Y is bounded by DK to the one half. So we get DK to the one over two N in the exponent in the right-hand side, okay? This is the Bombieri-Pilla bound, okay? Key limit two. So it says that our lattices, LK, they cannot be just any lattices, okay? So in particular, the largest successive minimum of LK, it has to be smaller than DK to the one over N. So of course, I mean, in general, if you take just a very, very lattice with, I don't know, N minus one, very small vectors, non-independent small vectors and one very large, the next one very large, then sometimes it can happen that lambda N is about the same size as the discriminant of the lattice. But for lattices which come from number fields, this cannot happen. Okay, so why? Because so when we send okay to our N, we get a lattice, okay? But so that's a lattice because okay is the Z-module, good? But once we're there in our N, we have forgotten about the fact that okay is actually a ring, right? In okay, you can multiply elements and you still remain in okay. Okay, so by using this, you can prove that they prove that lambda N has to be as small as this. And also it's a consequence of Minkowski's second theorem. Okay, so key lemma one was a consequence of Minkowski's first theorem and key lemma two is a consequence of Minkowski's second theorem. Okay, good. So what does it tell us? So what have we done? So far we have counted the number of Y and N1 satisfying our dive-on-t equation, but now we need to sum over all the remaining LIs. Okay, so we have L2, L3 up to LN in some box and what the lemma says is that for each coordinate, the size of the box is at least one. Okay, so it means that there is no coordinate for which LN has to be exactly zero and nothing else. Okay, because if that were the case, then we would be in trouble because we want to bound the number of possible LN say by a decay to the one over N divided by lambda N, but if these were very small, if these were very small, very close to zero, then the number of LN would be only one and that would be much larger than this ratio. Okay, but since we have the key lemma two, we have this bound, okay? Which is good because by Adamard's inequality that I displayed two slides ago, then this is exactly bounded about by the square root of this increment minus one over N in the exponent, okay? Yes, and of course there is something that I didn't see and that I should have said, when you use the boundary pillar bound, you get absolutely no dependence in the constant on L2, L3 or LN and same, no dependence on omega one, omega two, omega N. So it means that the constant that you get in the boundary pillar bound, the only dependence on K is in this power of the discriminant, nothing else, okay? So now you can just multiply the two bounds that we got. So the number of Y and L1 and the number of L2, L3, LN and you get half minus one over N plus one over two N so that's a half minus one over two N which is what they approve, okay? So that's a proof, okay? Any questions? Maybe about their proof. So let me just continue. So I'll continue with, there is a question. So does this work for the motion? Well, no, no, why? Because so you see, let me just go back. Yes, there. Yes, there, so there if you drop the condition that the norm of beta has to be a square, if you just forget about it, then the number of beta in this set, it's exactly the square root of the discriminant, okay? So it means that even there, you have already your recover Lendau's with epsilon equals zero, okay? However, if you look at the three torsion say, then in this set of beta, I think you're gonna get that if you drop the condition that the norm of beta has to be a cube this time. So I'm looking at P equals three now. So the norm of beta would be a cube. If I drop this condition, then the number of all possible betas it's gonna be, I think just decay and not square root of decay. So it means that to get some saving over decay to the one half, you're gonna have to do, you're gonna have to do something huge because you have approximately decay possible betas, which satisfies some diluentine equation. And by using this diluentine equation, you want to save it which is more than a half in the exponent, okay? So that's just impossible. And there we got one over two N, a save, okay? So hoping for a half in the saving is just completely crazy. So no, it doesn't work for higher primes, all right? So now we'll just continue. You're welcome. Yes, so I'll talk about Salbergeal's work now. Okay, so what did he do? So I learned about his work through Tim Browning because last year I met Tim at some point in the summer in Zurich. And I told him that I had the idea to maybe improve upon what BSTTZ did by not looking at interval points on a curve, but maybe by looking at interval points on surfaces and using some more general versions of the determinate, the Bombay area. And Tim stopped me and he told me per Salbergeal did exactly what you just told me. So for some time I posed and so I didn't know exactly what Salbergeal's results were. Now I do. So I'm going to explain exactly, I'm going to state his result. Okay, so for this I need notation. So again, K is a number field and we're going to let mu K be defined by this. Okay, so mu K is log of lambda two divided by log of the discriminant. Okay, so that's new notation. And if you remember Minkowski's second theorem, so we had lambda two times lambda three times lambda times lambda N is bounded above by the square root of the discriminant, but lambda two is the smallest of all the elements in this product. Okay, so there's N minus one of them. So it means that lambda two to the N minus one, it's at most the square root of the discriminant. So it means that lambda two, I mean the exponent of lambda two in the K, it has to be at most a half of N minus one. I mean one over two times N minus one. Okay, all right, so we are going to need to keep that in mind all the time. Okay, so here is Salvego's results. So I wasn't so sure how to state it, but yeah, that's what it is. So it gets this exponent when mu K is less than one over two N. Okay, so that improves upon the BSTTTZ results in the whole range, mu K between zero and one over two N, okay? So there's two ways to view this improvement. So the optimistic way is to say, well, when N is large, one over two N is almost one over two times N minus one, so that's not so bad. And the pessimistic way is to say, well, for generate number fields, actually all the successive minima, they should be of the same size, right? So lambda two, it should be for most number fields, it should already be very close to one over two times N minus one. So in this case, this result is not gonna apply. Okay, so I'll give a very rough sketch of Salvego's first, so what does he do? Instead of looking at integral points on affine curves, he looks at integral points on affine surfaces, instead of using the boundary pillar method, he uses global determinant method, okay? So I'm not giving any details, and again, I'm hiding a lot of stuff there because there are many surfaces for which, and there are some surfaces for which the bound that I displayed there will not be true or will not be provable, but then in this case, one needs to prove that there are few surfaces for which this happens, okay? But I mean, I guess the outcome is really this bound, okay? And so then you need to sum over all the remaining variables, and when you do that, you get the exponent that I displayed in the result, okay? So and again, here, I mean, in the first bound there, it's crucial that nothing depends, I mean, in the constant, nothing depends on a two and three and N or we get one and we get two and we get N, okay? All right, so new bounds. So here is a DRM, so when N is at least five, we have this bound, okay? So yes, so it's not so clear that it's not only does it improve upon BSTTTZ, but it also improves upon some bagels result in the whole range, new K in zero, one over 20, okay? So I drew some plots, so that's in the case N equals five, and the range of new K is zero to one over 10. So I'm only displaying the range for which our result is better than both results, but in the remaining range, one over 10, one over eight, the BSTTTZ result is still the best, okay? So in these plots, the pink line is corresponds to our result, the green, the mint curve corresponds to San Berger's result and the horizontal line black, 0.4, is just a half minus one over 10, so that's BSTTZ, okay? All right, so how can we use this result? Here is the corollary. Let's say you fix F, a number field, which is not Q, then for any number field K of degree N containing F, you get this new savings. So now you get two over three N as saving, which is much better than the one over two N from BSTTTZ, but in this case, of course, the constant may depend on F, okay? That's the idea. Let's say you're looking only at 60 fields, which contains Q adjoints square root two, then you're gonna get saving one over nine instead of the one over 12 coming from BSTTZ, and the constant is going to depend on square root two, but that's fine, you don't care about that, okay? All right, so why is that? Yeah, because then that two is gonna be bounded in terms of F only, okay? Another corollary. So for any prime number Q, the two torsion, the class group of the field, a Q adjoint Nth root of Q is gonna be bounded by this. So here you get a saving, which is two over three N, okay, which is much better than the one over two N we had before, minus something, but something which looks like one over N squared. So when N is large, this is gonna be small, okay? So that's another example where you can apply our results. Okay, so why is that? Well, in K, there is Q to the one over N, so if you send it to RN using sigma, and if you compute lambda two, you get this bound, but the discriminant, is it computed? It's Q to the N minus one, so it means that lambda two, it's really the discriminant to the one over N, N minus one, okay, so that's why we get, I would say, all right. So I'd like to say a few words about our proof, but time is flying. So we use the same strategy as our regular, but instead of using the determinant method, we use some C, we use the square C, okay? So for this, we use, so we consider the same counting function as we had in Selberg's proof, then we smooth it because it's just easier, and so what does the square C do? It's a tool to bound sums like the sum in the last line. So sum of omega of y squared, when omega is such a function, like this one, okay? So when you apply this square C, you find that our counting function, it's bounded by two terms, the first one is what it is, okay?  But the second term, you need to bound it, and for this, I mean, depends on exactly what it is, but you need to, this is where the work actually is, right? Okay, but if you look at this one, so I didn't say it, of course, M over PQ is the Jacobi symbol, okay? So if you look at this sum, and if you just use the definition of our smooth function, our counting function, W there, so this is exactly what you get, okay? So you get a smooth function times the Jacobi symbol, where the what's upstairs in the Jacobi symbol is the norm of N1 omega 1 plus N2 omega 2 plus value, okay? And P and Q are fixed in this calculation, okay? And then you apply the Poisson summation formula, and you get something like this, where subside hat, of course, is the Fourier transform of psi, and T is a complicated exponential sum, okay?  So then the game is to bound the exponential sum, and for this, you need to find an appropriate set of primes, P for which you can bound the exponential sum, okay? So, I mean, I don't really have time to say anything about that, but we can do it, and we can prove that there is full square reconciliation in this exponential sum. So these bounds is as good as we can expect, okay? And the outcome is as follows. So we get that the first term in the question was just this, and then the second term, we get P squared, okay? So then we choose the best possible P, which is this. So we get this bound for the counting function, and then as in Salvega's proof, we sum over all the remaining variables, and there, of course, I didn't say, but when you bound the exponential sum, there is no dependence, you remember we had NK of L1 omega 1 plus blah blah in the Jacobi symbol, but there is no dependence on, in the constant there, there is no dependence on omega 1, omega 2, or L3, L4, okay? So we can just multiply these two bounds, and we get, we get arbitrary, okay? All right, so that was a bit fast, but I didn't really have time, so sorry for that. So further expectations. So what's gonna be in this slide, and the next one are not there yet, but we expect the statements in these two slides to be there and very simple, okay? So here is another, not there, but statement. So we expect this to be true, okay? So it's, you get a different exponent for DK and a different exponent for M.2. And I'm claiming that this improves on all the other results in the range one over five N, one over two, and minus one. So it means that it improves upon everything, even for the maximum possible value of mu K, okay? So situation is like that now.  Again, what you see on the right of these three curves is that even for the maximum possible value, so this time I've plotted the three exponents in the range zero, one over eight. So before I only considered the range zero, one over 10, because between one over 10 and one over eight, we couldn't be BSTTZ, but now we can beat what they had even there, okay? So in particular, we should be able to prove the following. So that's a new bound. In this statement, there is no lambda two anymore, which I just replaced lambda two by its maximum possible value given by Minkowski's second theorem, and we get that. So remember, BSTTZ, they had a saving which was one over two N, and there, if you look at the first term, we get minus 15 over 28N, so that's minus one over two N, minus one over 28N, so that's always smaller than what they have, okay? So for instance, I mean, for N large, it's clear that it's gonna be better because the second term is in one over N squared, so it's gonna be small, but even when N is equal to five, we save something. We save one over 500 something, okay? Okay, so do I have maybe one minute to tell you? I'm running out of time, so if you want to know about how the proof of this result should go, then feel free to ask at the end. Yeah, I guess that's it. Thanks for your attention.


and welcome to Inside Hawaii Real Estate, a community real estate show dedicated to providing up-to-date information used to Hawaii homebuyers, sellers, and investors. I'm Wotanaka with my co-host, business partner, and wife, Leonie Lam, a realtor with over 20 years of experience in various leadership roles in the Hawaii real estate industry. Thanks, Will. Will is a full-time realtor, a lawyer, a law school professor, and the former head of Hawaii Title and Astral Company. Together, as full-time realtor professionals, Will and I work as a team to bring you the latest in Hawaii real estate. So how well do you understand taxes? I don't know about you, but for me, I find taxes pretty complicated, and I'm always in awe of CPAs who are able to articulate tax matters, so they make sense to me. On this show, we're talking about taxes in Hawaii real estate investments, and our very special guest is going to help us to understand some of the basics. We're going to start with tax basis, gains and losses, depreciation, and the tax table, and probably so much more. Our local state expert, Hawaii CPA, Brad Konishi, was licensed in 2001 as a certified public accountant. He's an educator. He's a very, very trusted CPA in Hawaii, and he specializes his tax expertise as the president and owner of HARPTA Help LLC. They assist home sellers with all the many issues related to HARPTA and FERPTA. Will and I are so honored to have you as our guest. Welcome, Brad. Thank you, Will and the only. Thanks for having me back. Thanks for being back. We're so excited. Yeah, let's talk about taxes. Well, Brad, welcome again, and we often support sellers with the sale of their home, and tax basis always comes up, and it's a whole different world, this tax world. So let's get started with the basics of foundation. Would you help us better understand what tax basis is and why it's so important? Okay, sure. Well, there are actually a couple of numbers that are really important when calculating gain or loss for the sale of real estate, and one of those two numbers is your tax basis, in particular how it starts and how it's adjusted. So your tax basis is, it depends on how you acquire the property. For many people who purchase property, usually their original tax basis is whatever they pay in so far as their cost. But of course, people can come into property different ways. They can receive it as a gift, or they might inherit property, which also happens as well. So tax basis is one of those really important numbers that is really important to understand in order to understand your gain on the sale of your property. Okay, so you kind of talked about tax basis being established. It depends how you acquire the property. So how is it determined? What are the different ways? Okay, yeah, great. Thanks. For most people who purchase the property, like I said, their original basis is what they paid for the property. You pay 500,000 for rental property that you're buying. Your original basis is going to be 500,000. But sometimes people acquire property through different ways. If you inherit property, and this is something that some people understand about, what happens is when you receive the property, you receive it at something called the stepped up basis. So in the case of, say, a parent who's owned the property for many, many years, their basis might be really low because maybe they purchased the property back in the 1970s when prices were a lot, lot lower. But then as time went on and then finally maybe the parent passes away, what happens is the fair market value as of the date that the parent passes away is the new basis for the person who inherits the property, the beneficiary. And so that concept is known as the stepped up basis. And you usually determine that by getting an appraisal from a third party appraiser. Right. And that's called a date of death appraisal. That's correct. Okay. Okay. And there's parameters like you have to get it done within a certain time frame and things like that.  We've had, you know, clients who actually didn't get a date of death appraisal when their parents may have originally passed away. But then, you know, now that they're faced with this tax quandary that you have to, you know, justify their stepped up basis, there's actually an appraiser in town that we go to sometimes that actually helps us get appraisals that are retroactive. I see. So they look back at the records of sales maybe during that time period and they're able to establish. Yes. Yes. And I know it's a little bit, a little bit more expensive because, you know, you can't actually go to the property and see what the condition is, right? Because it's been many years past. But nevertheless, it can still be done. Right. Okay. Is there anything else that they would need, you know, to be aware of if it was an investment property as it comes, as it pertains to this? Like, do they have to make sure they're current with taxes? Oh, yeah, yeah. That's another thing is sometimes, you know, when somebody sells rental property that they've inherited or maybe that they own themselves and they're not a resident of Hawaii, sometimes there's a withholding. The withholding is known as Harpta, which just stands for the Hawaii Real Property Tax Act. It's kind of a harsh withholding, a 7.25% of the gross sales price. But what happens is during the escrow process, when somebody's selling a home, the escrow officer will actually set aside that money. After closing, they'll send that money over to the state and then the state will hold on to it until the seller can prove that their liability is less. And if they want a refund of some of their money, they have to make sure that they're current with, you know, things like general excise tax, transient accommodations tax, which is a killer, right? 10.25%. And also Hawaii income tax as well. Hey, Brad. Yeah, just going back to the tax basis, how about you often have situations where they receive a gift from a family member, their parents, their grandparents. How is the tax basis determined in those type of situations? That happens quite a bit. You know, if somebody's still alive, I guess this is a definition of a gift as opposed to an inheritance. With an inheritance, somebody actually has to pass away first and then the beneficiary acquires the property. However, in the case of a gift, you know, both the giver and the receiver of the gift are still alive when the gift actually happens. So, you know, in the case of a gift, what happens is the beneficiary receives the real estate at the giver's basis. So if the giver, say, paid $500,000 for the property and their basis at the time of the gift was still $500,000, what would happen is the receiver of that gift would have that same basis. Their basis would also be $500,000. So the basis would just be carried over from the giver to the receiver of the gift. I see. So regardless of when it was gifted, it doesn't matter. It's when the gift or purchase the property. It's the giver's tax basis on the date that they give the property to the receiver. Got it, got it. Okay. So we started off with the tax basis and, you know, when we're determining taxes, there's something called the net selling price. So can you get into that a little bit? Sure, sure, sure. The net selling price is the other number that's really important for determining your gain or loss when selling real estate. The net selling price essentially is really simple. It's basically your gross selling price, whatever it is you're selling the property for, less certain expenses. Certain expenses like commission expenses, escrow fees, title insurance, transfer taxes, those types of expenses can actually be used to offset your gain and to lower your net selling price and get you to the point where your gain is reduced hopefully as much as possible. But yeah, those are some of the expenses that are allowed to be introduced into the transaction to actually lower your gain. Okay, so how does the calculation of the gain or loss work? This is tricky. First, it's important to understand that the gain is taxed at a marginal rate. And, you know, the rates I believe for Hawaii purposes are anywhere from 1.4 up to, I believe it's 1.4 up to seven and a quarter percent. Whereas for federal purposes, the capital gains tax rate, long-term capital gains tax rate is actually zero percent in some instances. It can be 15 percent or it can be 20 percent. So it depends, you know, for state and federal purposes, it's, you know, the capital gains tax rates are much higher for federal purposes than state. And long-term capital gains tax rates are usually generally lower than the ordinary income tax rates.  Whereas capital gains tax rates are taxed at a little bit lower rate, which is preferable. Okay, so in terms of just talking about numbers, so we have the adjusted tax basis, net selling price. And how would you calculate gains or losses, you know, using like real-world numbers, like a million-dollar price, for example? Okay, that's, you know, million dollars is always easy to do the math in my head. That's great. Let's just say somebody is selling a house for a million dollars. Let's just say their net selling price is a million dollars. It's going to make it easy. Okay. But let's just say that they purchased, you know, the property, say 15 years ago, you know, when prices were much lower, let's say their basis in the property was 500,000. What happens is when you own the property, if you use it as a rental, which, you know, a lot of people do, if they're not actually living in a house and it's not a second home, which is, you know, fairly infrequent phenomena, they're renting it, right? Because, you know, if you got a property, might as well make money off of it. So they're renting the property, but when they're renting it, they have to depreciate this property. And the depreciation becomes actually integral into their basis. All the depreciation that accumulates during their rental period lowers their basis. And because it lowers their basis when they finally sell it, they're going to have more gain as a result of that depreciation.  So a million dollars in that selling price, adjusted basis, 500,000, you come up with a gain of about 500,000. So that's going to be taxed for federal and for state at different marginal rates. Okay. Kind of like that chart. So that's the tax basis, about 500,000, net selling price, a million dollars. So the gain is this half a million dollars. Exactly. And I should state that this applies to rental real estate, right? Because personal real estate, real estate that you actually live in, you know, there's a something called the gain exclusion, section 121, also known as the main home gain exclusion, where people who are living in the home as their main home may be given a really large tax break, but we're talking about that. We are talking about investment properties today. So, but I just thought I'd mentioned that because, you know, some people might be confused because a lot of people know about that main home gain exclusion. And so they might be thinking, well, you know, I shouldn't be taxed on all of this, you know, if it's a home that you live in, you may not be taxed on it at all. But if it's a rental property, you get taxed from the very first dollar of gain that you make. And Brad, you know, you mentioned depreciation. I remember prior to buying my first investment property, I always heard about depreciation, and I could never really understand the concept until I went through it myself. So can we get into the basics of depreciation and how that works for investment properties and tax benefits? Okay, sure, sure, sure. You know, that's a great example. And I think of all the areas of real estate investment, when it comes to taxes, this is the area that causes the most confusion, because I think the concept of depreciation works very differently from the way we think in the real world. But let me describe what depreciation is. When you own real estate, and you're using it as an investment, you're renting it out, what you need to do is the very first thing is you need to split your cost or whatever your basis is. You need to split it between land and building, because land does not get depreciated whereas building does. But as a concept, depreciation essentially is the recognition of an expense over a period of time that you're actually using the rental property, right? Because let's just say you purchase a real estate for a million dollars. It's going to be a rental real estate that you're going to rent out. You're not going to recognize all of that expense in the very first year, right? Or else it's going to be lopsided. You know, the benefit that you get from purchasing the real estate and then renting it out over time, those benefits are seen over time. So what happens is the IRS allows you to, well, forces you to recognize that expense over time. In the case of residential rentals, it's 27.5 years. So let's just say you bought a property, a million dollar property, but the actual building value, let's just say most of the value is in land, but the building value is $275,000. If you depreciate that over 27.5 years, that's going to be what, $10,000 per year of depreciation. So that $10,000 of depreciation is recognized. You put it on your tax returns, you get a benefit from that, and that's what depreciation would be. So that's the upside of depreciation is the expense that you get to lower your taxable income.  And when you lower the basis, what happens is when you finally sell, lowering the basis, if you remember the way that example looked, that creates a larger and larger gap that increases your gain. So yeah, depreciation will lower your basis, as you can see from that chart there. It'll lower your basis, but it increases your gain. Here's the kicker though. The IRS says, with depreciation, your basis is lowered by the depreciation that has been allowable or allowed. So even if you had depreciation that you could have taken but didn't take, you have to lower your basis when you sell by depreciation that you could have taken but you didn't take. So that's one thing to keep in mind. Make sure you depreciate because if you don't, you could find yourself really at the short end of the stick when you sell. And that's where you had kind of shared about certain clients, maybe like military clients and stuff, if they're not doing taxes with a professional that's advising them about depreciation, maybe they're doing it on their own that sometimes they miss out on that. Exactly. We've seen it a lot because you know at Harpta, that's one of the things we do is a lot of our clientele, we see people who've actually lived in the house first and then converted to a rental, particularly military. And it's hard because their circumstance a lot of times is they're getting PCS, so they're actually moving off Island while they're vacating their home and then trying to establish a new home and then trying to be assigned to new work while their children go to a new school and maybe their spouse has a new job themselves. So there's so many things going on that for them, they unfortunately don't have the time to research what's the proper way to convert a home for your main home into a rental. And so as a result, they might skip steps or they might not know exactly all the requirements that they have owning a rental property. So yeah, that's one of the challenging parts for a lot of our clients. So the bottom line is in terms of depreciation, you know, for investment property, I mean work with the CPA because I think you had mentioned like when you're doing your own taxes, TurboTax or whatnot and people are just trying to save money, there's a section on depreciation and they just kind of bypass that. Yes. So make sure that I mean, because from the higher standpoint, they're going to depreciate anyway, right? I mean, the value of your property. Exactly, exactly. So yeah, your basis is going to be lowered by the amount of depreciation that you took or that you could have taken.  And then when you go to sell, you got to lower your basis by the amount of depreciation that you didn't actually take but could have taken. I mean, talk about getting kicked in the face. The worst thing that happened and it just it breaks my heart. There are ways of making up that depreciation that you couldn't have you that you could have taken but you didn't take. That's just, you know what? That's a separate lecture. There's something called a Form <IP_ADDRESS>. It's an IRS form. And I used to do this form, crazy, crazy long form. I think just the form itself is like somewhere around eight or nine pages. And so it's just... That's enough. Exactly. And that was just, you know, the separate form that gets attached to the Form 1040. But, you know, so I won't go into too much into that. But yeah, if you have rental property, depreciate or at least, you know, work with the tax professional the first year or two, get as much information from them. If you're a real cheapskate like I am, I'll admit it, right? You want to try to get the best value. And so a lot of times, you know, having a tax, you know, a CPA or an enrolled agent, somebody with a lot of credentials, it can be expensive to do it every single year. So maybe what you might want to do is maybe the first two years of your rental, go and see that person, the tax professional, ask as many questions as you possibly can. Once you understand the process yourself and how the process is reported on your tax returns, then maybe you can, you know, if you feel comfortable, you can do it yourself after that, you know?  That's something that I, you know, clients used to do that with me when I did income taxes and that was just fine with me. I understood, you know, that's a good way to save money. You know, to pay $700 to $1,000 plus, you know, every year to a CPA to get your taxes done can be expensive. It adds up. How about if the investor owner does, you know, renovations on the property? Does that, that helps the tax basis to increase? Yes, it does. It does. You know, here's the thing is when they do improvements on the property, sometimes that, those improvements may get folded into the actual, you know, entire basis of the the home itself, or you might be able to depreciate those improvements over 15 years. You know, there's a separate category of improvement for 15 years. And if you can depreciate it over 15 years, great. You know, that's what some people do. Oh, that sounds good. And then you got to make sure to keep your records and everything because basically this form is, I mean, this, that amount of renovation work is kind of reported, right? On your tax form, but then it never got audited. Exactly.  If they find evidence of fraud, they can go back as far as six years. There's no statute of limitations for, you know, forms unfiled. But a lot of people hear the three to six years and they think, oh, I'll be, you know, I hold my records for six years and then toss it out. I'm okay. But that only applies for, you know, things that have already been reported on your tax returns.  Those types of documents, you need to hold them for three to six years after you actually sell that asset, you know, at the very minimum. So, you know, when it comes to record retention, we hear this a lot because, you know, sometimes when we're desperate to try to get somebody to, you know, change their game into a loss because we want to try to get them a heart to waiver, you know, we start asking questions about improvements. And one of the things we sometimes hear is, I don't keep those records, you know, that happened like, you know, 15 years ago, 20 years ago, I don't have those records. The unfortunate thing is, if you don't have records to justify that expense, you know, if you get audited, they're allowed to disallow it. So, yeah, that happens. Super good point. So, as we're in 2024, just kind of pivoting, are you as a CPA, as a tax professional expert, are you watching out for anything this year? Yeah, I think, you know, every year I watch for 1031s, what they're going to do at 1031s. They always talk about that. 1031 transactions are, you know, this kind of ties into what we're talking about because this applies for commercial or real estate investors, people who have rental property. 1031 exchange is a way that you can exchange rental property from one to another and not have to pay tax. It's a very, very powerful vehicle for investors. But every single time a new administration takes over or anytime there's a tax proposal that comes out, usually one of the first things that they put at the very top of it is that, yeah, we're going to do away with 1031s. But 1031s are still here for real estate and they haven't been done away with it. The reason why is when they take a look at this 1031s, and they look at not only the 1031 exchanges itself, but all the side businesses that have popped up because of 1031s and all the benefits that investors have because of the 1031s, eventually they decide to do away with it. And that's probably what they'll do again. But I always, always watch because something they always mention. Because you just never know. You just never know, right? Exactly. Just never know. And it depends on who the new administration is. Exactly. Exactly. But that's, you know, for new administrations, it's always tempting because if they did away with 1031, the tax savings, or the tax revenue, additional tax revenue to the government would be tremendous, you know. So because of that, that's a, I guess you would say on the proposals that they put forward, it's kind of a low hanging fruit because it's a number that's really large and really helps their case for more tax collections. But when they look at it, I think you try to take that away, the benefit goes away. It's actually probably less beneficial than actually having 1031 exchange. Okay. So we'll be keeping a close eye on 1031s and the new administration. And is there anything else that you're kind of tracking on for this year? Of course, I always look at the, listen to the proposals locally. One of the ones that has just been put forward by, I think it was Brenton Awa, who is, you know, he's really, really, I appreciate what he's trying to do for the real estate industry, make it for the locals, but he's trying to do some things with some of the taxation and make it difficult for foreigners or non-residents to purchase property in Hawaii. So that's one of the things that I was watching for. Got it. Got it. In terms of like the IRS budget staffing, that kind of stuff, why would that be important? That's another thing. For anyone who's worked in the tax area, it's no secret that the IRS has been starved for many, many years. And a budget cuts over the last decade have really, really cut back the amount of services that the IRS can provide. Many times as tax professionals, we find when we try to contact them, they don't pick up their phone, so we can't get answers a lot of times. So we've been seeing an uptick over the last year or so. And you know what? I've actually been seeing results, not too long ago. I had a problem that we had to iron out with the IRS for one of our FERPA clients. And I actually got a remote meeting where we scheduled something like this over camera where I spoke to an agent in Las Vegas. And by the time we were done with our meeting, our client's issue was all taken care of and they got their check within two weeks from that day. So it's getting better, but it's not to the point where it used to be where you could always call and speak to somebody. It's really difficult to get in contact with them. Well, it's good to hear that it's getting better. And, you know, Will and I always turn to Brad for any and all guidance or questions regarding Harpta and FERPA for our clients. Brad and his team, you know, as I mentioned earlier, they specialize in helping with any issues regarding Harpta and FERPA. Thank you. Brad is awesome. And, you know, thank you so much, Brad, for all your time and your energy, information. You're so valuable. So we really appreciate you. Well, thank you so much. So this is Will and Leonie on Think Tech, Hawaii Inside Hawaii Real Estate. We've been talking about taxes and investment properties with Brad Konishi, CPA, owner of Harpta LLC. So thank you so much, Brad and Aloha. Aloha. Thank you.