Some students struggle with certain conventions when they write. This is especially true when it comes to grammatical elements, like subject-verb agreement, pronoun antecedent agreement, verb forms, and parallelism. This lesson focuses on subject-verb agreement. The first thing we'll do is define these terms, subject, verb, and agreement. The subject is the word in a sentence that is the focus of the action, or who or what the sentence speaks about. For example, macaela is the subject in macaela went to the store. They is the subject in they lived a long time ago. The verb is the part of the sentence that is the action word, such as a word that describes what the subject is doing, will do, or has done. The verb and the subject always relate to one another. For instance, went is the verb in macaela went to the store. And lived is the verb in they lived a long time ago. Agreement refers to parts of a sentence that are similar in gender, number, and person. This is critical when dealing with nouns and verbs or pronouns and their antecedents. For noun verb agreement in the sentence, the man drove his car. The noun man and the verb drove would be in agreement. Well, now we have three new terms, gender, number, and person. So we will define these terms. Gender refers to classification groups such as masculine, feminine, and neuter. Number refers to one singular or two or more plural. Nouns, pronouns, adjectives, and verbs are all affected by number. The man was 30 years old is an example of a singular noun and verb. They were all 15 years old is an example of a plural pronoun and verb. Person, this grammatical category applies to pronouns and verbs and is used to distinguish between the speaker of a sentence, who or what is being talked to and who or what is being talked about. There are three categories of person, first person, second person, and third person. We're going to put gender aside for this lesson and focus on number and to a limited extent person. As you have probably figured out by now, we like to use mnemonics to help us remember steps to take when doing activities. Here's another one for you. Wee-wee three. The I to the third means that I appears three times. Wee-wee three. What is the subject? Is the subject singular or plural? What is the verb? Is the verb singular or plural? If the subject is singular, is the verb singular? If the subject is plural, is the verb plural? Let's look at what each line means. Wee-wee three. What is the subject? It is important to be able to recognize the subject in a sentence. We look at a sentence and ask, what is being written about in this sentence? Usually, this is pretty easy to do, as in the sentence, Tom ran in the marathon. Tom is what is being written about, so Tom is the subject. Find the subject in this sentence. The girls walked their dogs. Girls is the subject. Another point deserves mention. Sometimes several words are found between the subject and the verb, and this can cause confusion as to which word is the subject. For example, consider this sentence. Mark, who is always among the fastest runners, lost the race by 20 feet. Here, the singular subject is Mark, but one could get confused by the clause, who is always among the fastest runners, because runners is plural. Again, the key is to remember to identify the person, place, and thing the sentence is about. In this case, Mark. Now let's look at the next step. Is the subject singular or plural? You probably remember that singular means one, and plural means more than one. So a singular subject in a sentence involves one person, place, or thing in a sentence. And a plural subject involves two or more people, places, or things. Remember our two sentences? Tom ran in the marathon, and the girls walked their dogs. Tom is singular because he is one person. Girls is plural, because it refers to more than one girl. Some sentences can be tricky. Take, for instance, this sentence. The box of oranges is in the fruit department. Here, box is the subject. Is it singular or plural? It's singular, but oranges is plural, so we could be fooled into thinking the subject is plural. Don't be fooled when you see sentences like this. Pronouns can be particularly important Pronouns can be particularly troublesome. Indefinite pronouns like someone, somebody, anybody, everybody, each, either one, everyone, or anyone are all singular pronouns that can serve as subjects. And some pronouns like any or all can be either singular or plural, depending on whether they mean a group of or separate individuals. While we're talking about pronouns, this is a good place to set aside a few minutes to deal with the first person, second person, and third person singular and plural pronouns. I is a first person singular pronoun, and when alone, always goes with a singular verb. You is a second person singular pronoun, but it always goes with a plural verb.  He, she, and it are third person singular pronouns, and when alone, go with singular verbs. We is a plural first person pronoun, you is a second person plural pronoun, and they is a third person plural pronoun. I'll go with plural verbs. Let's pick up where we left off. Have you ever heard of compound subjects? When a subject has more than one person, place, or thing, and the words are connected by and, it is considered a compound subject and is plural. As an example, consider the sentences below. Tim and Jessica were friends. The cat, dog, and parrot are three pets. To hit, to run, and to field are the goals of baseball practice. Now let's go to our next step. What is the verb? By definition, a verb is an action word. What word describes the action in this sentence? The girls walked their dogs. Walked is the verb. Is the verb singular or plural? Verbs can be either singular or plural. Consider, for example, singular. Is, was, am, plural. Are, were, are. Many verbs can be both singular and plural, and the decision as to whether they are singular or plural depends on the subject. Look at these two sentences. I travel to Spain every year. Mary and Jane travel to Spain every year. In the first sentence, travel is singular, but in the second sentence, travel is plural. You just have to decide within the context of the sentence whether the verb is singular or plural, depending on the subject. The last two steps go together. If the subject is singular is the verb singular, and if the subject is plural is the verb plural. These steps ask us to examine the sentence for subject verb agreement. We use the information already presented to answer these questions, but there are some other points to be made. Do you remember our earlier discussion of compound subjects? There are specific instances that occur in sentences that need attention. Sometimes a compound subject in a sentence is made up of two singular subjects that are connected by the words either, or, or, neither, nor. In this case, the verb will be singular. Let's look at the following examples. Neither mom nor dad is home, and either Tim or Mary is going to be promoted to a higher position. Notice that the verb is singular. That is because of the words neither or either in the sentence. This is about the only time when you will have a compound subject with a singular verb. Finally, here's something else you will see that might cause confusion. Every once in a while, you will find a sentence where the subject comes after the verb instead of before it. For instance, under the bridge swims a lone duck. There were several large dogs on the field. In the first sentence, the subject is duck, even though it comes at the end of the sentence. The verb is swims. The subject and verb are singular. In the second sentence, the subject is dogs, and the verb is where. The verb where is plural because the subject, dogs, is plural. Keep an eye out for these types of sentences. They can be tricky. Congratulations! You have completed this activity, Subject Verb Agreement.


Green, green, green, green, green! For as long as I can remember, my favourite colour has always been green. I must have popped out of the womb with some knife-vision goggles or something because this has been a lifetime obsession of mine, literally from as early as I can remember. Don't. At the tablet giveaway! The amazing people over at GEMON have once again said to be an art tablet to test and review, which I then want to give away to one of you guys!  This is the PD1320, a 13.3 inch drawing tablet with a fully laminated 1080p HD screen. Perfect for animation and illustration! The device has a generous working area, but is still lightweight and portable enough to take around with you. Slice! In the box, you get a leather case bracket, the pen along with a pen holder and replacement nibs, some cables and instructions, a smudge glove and cleaning cloth, and of course the tablet itself! Tada! It's very comfortable to draw with the tablet, and the pen creates super accurate lines. Not to mention, it's also completely battery-less, meaning you'll never have to charge it before use! With 8192 levels of pressure sensitivity and 60 degrees of tilt functionality, it really is a joy to draw with! Damn! That's some beautiful artwork right there sir! The tablet is on sale right now, so check out the links in the description if you want to purchase one for yourself while they're discounted! If you want to take your chances on the giveaway, however, here's how to enter! Head over to any or all of the following! My Instagram post, my tweet, and or my TikTok about the giveaway, they all look like this! Give their accounts a follow and drop a comment on the Instagram and TikTok post to submit two entries! On Twitter, give it a retweet and that's a third entry! The link to my socials up below and good luck! Big thanks to Gammon for giving me the opportunity to give back to you guys like this, please also check out their links below as well! Damn! Green is known to be a soothing, relaxing and youthful colour.  Hmm, then why do my plants keep dying, huh? Explain that one punk! Shhh. Shush. While doing some research for this video, I was typing in a question. And this was the top question. I don't get why people tell their kids that things will just be a phase. I knew full well that it's not a phase, mom, you don't understand me! Took my job mixing yellow and blue paint very seriously in kindergarten. It was my duty. I also remember my teacher reading the class a book where each page would talk about a different colour. So you bet I was like, oh can we see green next? Green is my favourite colour! Please, please, please, please! Green. I don't think my teacher liked me very much. Even before that, my mom happened to give my brothers and I a $5 note to spend on whenever we next plan to go to a toy shop. Now a $5 note in Australian currency is a purpley pink, aka my second favourite colour. Let me know if you want a video on that too.  No, the green note is $100. It was a nice try. Now don't even get me started on the range. I would enjoy if I wasn't the green piece saying a board game of some sort.  What are you doing? I want to be green this time. Give it to me! My piece! The green is known to be a soothing, relaxing and useful colour. My piece! I'm always green! Back off! This is mine! Stop!   Stop hitting me! Ahhhh! Only 2% of the world have green eyes! And guess what? That doesn't include me. One thing that made me so devastated as a child is when my eldest brother first began school. He was the one who got to pick the sporting house colour for the rest of my family! And guess what he picked, huh? You're never going to guess what he picked. The choices were out of green, blue, yellow and red. He picked yellow! Excuse me, ah. I didn't agree to wearing this urine on my shirt every sports day. Okay, so it's to the people who actually love the colour yellow. My bad. But the yellow house team also sucked because they never won any sports carnivals. Then again, green wasn't that great either. Anyway, green fun fact number two. Disneyland used a certain shade of green which they refer to as, go away green. Oh man. This shade is used throughout the park to camouflage things like construction buildings, fences, bins, etc.  Ah, whoa. Jess, what about all the different types of shades of green? Do you like all of them? All of them are acceptable. Except for this, it reminds me of, without a doubt, I'm guaranteed to be that type of old lady that just lives amongst her favourite colour. Like come on, they are living the dream. Green fun fact number three. In the 1900s, green wallpaper was actually very toxic. Ha ha, fun fact. Right? No. It contained things like arsenic and lead, and it is to believe that it supposedly caused Napoleon's death. And no, not to get confused with Neapolitan ice cream. That would be far more tragic. Oh. Now it's time for some honourable mentions to green characters that I loved as a child. And not going to lie? Still do. Let's go. Let's start off with Shrek. Look, is there much I need to say here? He is a god. Next. The green telly-tubby Dipsy. Dipsy. He was definitely the first person to jump on that cow print trend 20 years ago. Creeper. Oh man. Minty from G3, My Little Pony. She's iconic. I remember that was the first pony I ever got, and out of all places, I got it at Woolworth? A grocery store? That's all so green. Thank you. Thank you very much. Out of Care Bears, you know it had to be Good Luck Bear and Wish Bear. It had toys of these two. Wish Bear would glow in the dark, and Good Luck Bear would talk to you. Kinda creepy. Let's speed it up a little. The green M&M, Pascal, Mike Wazowski, Yoshi, Franklin, Buttercup, Kiff, Tinkerbell. There are so many. But do you want to know what one is the worst? That got me thinking. The thing that's on Good Luck Bears, um... What do you call it? Stomach marking? Oh. I guess I called belly badges. Well, on his, it's a four-leaf clover, and it reminded me that I actually found one of those in my front yard once, so that's pretty... ...rad. I think my mum slammed it in a dictionary over 10 years ago, so who knows if it exists now? I'll run and check real quick. B-R-B! Ah, excuse me. Does that mean you have heaps of green clothes? You better do! For both my year 10 and year 12 formals, a.k.a. proms, etc. I actually wore green dresses, so that was pretty cool. We don't want to talk about my identity crisis in U6. I see no green! Year 10 was a pale green, while year 12 was a foresty, bodily green shade. Green fun fact number four! Green is the second most common favourite colour after blue. Again, not a fun fact. Green deserves better! Also, I know no one asked, but Mylokite is the best rock out there. They're minerals. So, um, yeah. Glad that's off my chest. Um, heh. Now let's quickly go through some green foods! Let me know if you agree. Lettuce. Yes. Broccoli. Yes. Spinach leaves. Yes. Peas. Ha! Grapes. Yes. Capsicum. Ha! Beans. Ha! Spring onions. Yes. Celery. Ha! Asparagus. It's a superior colour that gives you the power to do this. I know red and blue are also commonly used for this, but we say green screen for a reason. It's the best. Okay, anyway, how have I made it this far in the video and not mention nature properly? No, we aren't talking about this. Nature is amazing. It comes in all shapes and sizes, and guess what? The majority of it is green! Green! I love how people can shape shrubs into weird things. Like, what? How is that possible? Anyway, enjoy this quick vlog of some nature I filmed eight years ago. So, I'm in the forest. I'm running. I'm running. And I'm going to put some stairs. Stairs. Stairs. I was so cringe as a kid. And now for the final green fun fact. Number five. The green light on traffic lights allow you to go. Thanks guys! That fact was crazy! I'll catch you all next time. Stay safe! For real though, I highly recommend adding some more green to your everyday life, and if you want to see me do a video on my second favorite color, purple, leave a comment below and I'll see what I can do!


Hi everyone. Happy New Year. I'm Joe for JaspiesCaseBrakes.com coming at you with 2021 Leaf Autographed Mini Football Helmets. Eight box, random team break number three. One spot gets you two teams in this eight mini helmet break. Big thanks to this group for getting in on it. Let's double your names up. And all 32 teams are in. Let's roll it and randomize names and teams. One out of three, four times. One, two, three, and four. Got Oliver down to lead by. One out of three, four times for the teams. One, two, three, and fourth and final. Jordan, you can hang out with us while you're delayed at LAX. We've got Dallas down to the Cleveland Browns who are down 10 to the Steelers. Steelers 10, Browns nothing in the third quarter. Oliver, you got the Dallas Cowboys. Justin with the Titans. Charles with the Saints. Steve with the Colt and the Chargers. Charles with the Eagles. Tim with the Rams. Steve Birch with the Giants and Jags. Levi with the Ravens. Karen with the Panthers. Charles with the Niners. Justin with the Falcons. Tyler with the Packers. Karen with the Chiefs. Levi with the Bears. Tyler with the Buccaneers. Tim with the Jets. Justin with the Texans. Levi with the Steelers. Steve with the Cardinals. Steve Wright with my Raiders. Jack with the Dolphins. Charles with the Lions. Steve Wright with the Bengals. Jack with the Seahawks. Levi with the Washington Football Team. Oliver with the Patriots. Steve with the Vikings. Levi with the Bills. Justin with the Broncos and Levi with the Cleveland Browns. All right, let's alphabetize by team here. Tyler's happy. He's got his Packers. Let's get all this on the one screen right here. Okay, I'm going to pause the video. Actually, before I pause the video, remember, there's no checklist for these or college rules apply or no checklist rules apply, which you've seen in the Frequently Asked Questions section of our website, Jaspy'scasebreaks.com. So obviously that college helmet, that player is active, so it'll go to the team he's currently on. Trevor Lawrence is on the Jags. If it's a retired player, it'll go to the team they played for the longest. All right, so if it's a college helmet of a retired player, it'll go to the team they played for the longest, according to proflipallreference.com. And we'll keep going from left to right on tiebreakers if necessary. We'll cross that bridge if and when we get there. All right, now with all that being said, now I'm going to pause the video. When we come back, we'll see if there's any trades. Then we'll have the break, stick around, BRB. All right, welcome back, everybody. They're a little bit of trade chatter, but in the end, no deals were done here in this, I know it's upside down, but the eight box break mini helmets. Let's pop this case open here. A little bit of awkward size. We're doing all eight boxes, right? Yeah, we're doing all eight. All right, well, since we have a little bit of time, nothing sold out after this. So we've got a little extra time to play some Who Am I? America's favorite game where I give you the clues, new ID, tell me who the player is. And don't forget our no checklist or non-pro uniform rules. If they're active, they'll go to the team they're currently on. If they're inactive, they'll go to the team they played for the longest point of pro football reference.com. All right, first helmet is this person for that team. Okay, I am currently 89 years old and I am a executive. Okay, I'm an American former football executive who was the vice president of player personnel in the NFL for this team from 1960 to 1988. Wow. He was with the Rams from 1950, 1955 to 57, with the Niners from 1958 to 1959, and then was with this team from 1960 to 1988, two-time Super Bowl champion and in the Hall of Fame. Hey, happy New Year to you too, Mark. I think this is going to be kind of tough to get. He's a Dallas Cowboys executive from 1960 to 1988, so this is going to go to Oliver. Anyone remember the director of president of player personnel from 1960 to 1988 for the Dallas Cowboys? Oh, Jack got it. Nice. Gil Brant. You're absolutely right. There it is. Hall of Fame executive Gil Brant. Kind of hard, the little stickers in the way, but there's his autograph right in there. Beckett card inside. Beckett sticker on the back. Gil Brant. 89 years old. Born in, it's for Oliver, by the way, at the Dallas Cowboys. Born in Milwaukee, Wisconsin, went to Wisconsin. I don't know what he's, what is he doing these days? I'm just part of the Cowboys ring of honor. According to Wikipedia, I don't know how updated this is, but according to Wikipedia, it says that he's a regular on Sirius XM NFL radio as a commentator on late hits, the end zone and the Sirius XM NFL tailgate show. I don't know if he's still doing that, but yeah, some commentating. Another retired, this time not an executive, natural player. I'm currently 51 years old, born in Pine Bluff, Arkansas, 65320. He was an offensive tackle, played for two teams entire career, and he was an eighth overall pick in the early 90s. Who am I? Nickname Nasty. And went to Louisiana Tech. Who am I? Offensive tackle, active between 1993 and 2005. Two teams, six time all pro, three time second team all pro, 11 time pro bowler, pro football hall of fame, college football hall of famer, 1990s all decades team, NFL 2000s all decade team, part of nice Jack, Jack B again. New Orleans Saints Hall of Fame and Ring of Honor. It's Willie Rhofe. It's a cool looking helmet right there, Willie Rhofe. Played with the Chiefs at the end towards the end of the last few years of his career with this, was with Kansas City, but I think the Saints is where most of you will recognize him. And that is for Charles and the Saints. Nice Jack is two for two. The big dude. What did I say? 65320. Next mini helmet. A active wide receiver, drafted in the late first round in 2018, was a two time college football national champion. First team all SEC, part of the PFWA all rookie team. Currently have 248 receptions, 3,342 receiving yards and 28 receiving touchdowns. He's pretty good, but for 2021, he stepped away from the NFL citing mental health issues. So he took some time off for himself. But when he's on, dude's really good. 61, 190. Went to Alabama. Yep, Justin, you got it. Justin and Jack, Calvin Ridley. And that's for you, Justin. Justin and the ATL. Tried to trade, couldn't trade mojo. Love these mini helmets. I think they're great. Hopefully, hopefully he, hopefully he's back next season. Stronger, healthier, better than ever. There you are. All right, here is a, there is a retired offensive guard born in Detroit, went to Michigan State, drafted in 1973, first round, late first round, 26 pick overall, played for one, two, three, four teams. We'll check that played for two teams my entire career. The team that drafted me, the team that this, that is this helmet is my second team. And then went back to the team that drafted me. And then a number of years later played for the Charlotte Rage, six time pro bowler, eight time all pro part of the 1970s, all decade team. And part of both teams, ring of honor and wall of fame. Offensive guard between 1973 and 1985 with a, he's got the same first name as me and a difficult to pronounce last name. Who am I? There you go, Jack. Yep. It's, it's Joseph or Joe, uh, Delemi Lour. Delemi Lour for the Cleveland Browns. That's for Levi. Classic Brown. Tell me, that's a big, big bold autograph right there, Joe. Nice. I like it. Kind of hard to get a good angle on this right there, but there it is. There's your Beckett sticker right there. There should be a corresponding card inside. You can see it right over there. Cleveland. This is for you. Delemi, Delemi Lour, Delemi Lour, Delemi Lour. That's what it says on the Wikipedia pronunciation. All right. Halfway through this full case break. Correct. He did. He did play for Buffalo. He was drafted by the Buffalo bills from 73 to 79. He played with the bills and then with the Browns. That was his helmet there from 80 to 84. Back to the bills in 85. And then a number of years later in 1992, he went to, he played for the Charlotte Rage, which was an arena football league. All right. I'm a retired quarterback. Played, I'm trying. He's pretty good. I don't know. I guess I'll just give you, I'll just give you the highlights. You'll be able to figure it out. Played for one team, my entire 17-year career. One team, my entire 17-year career, and is currently a special advisor to this team as well. I was an MVP, an offensive player of the year, a combat player of the year, a nine-time pro bowler, a five-time passing yards leader, a three-time passing touchdowns leader, passer rating leader, 100th anniversary all-time team. I have 420 touchdowns to 250 to 200 receptions, 61,361 passing yards, and an 86.4 lifetime passer rating, college football hall of famer, pro football hall of famer, Walter Payton, man of the year, blah, blah, blah, blah, blah, blah. Yeah. There you go. Jack B has got it. Oliver has got it too. Dan the man, Marino. It's a pretty nice helmet. And that goes to Jack B. There you go, Jack. Got randomized the Dolphins. Gets the Dan Marino. The Beckett sticker, the card. Classic logo right here. Very nice. Yeah, Jack. Got it. So Joe P, no. Manning played for, Payton played for two teams his entire career. I don't know if Eli played 17-season city. All right, next one. All right, we've got an active player here, active player, active wide receiver, and active wide receiver in just two seasons I've been in the league, 185 receptions, 2851 receiving yards, and 16 receiving touchdowns. Just just two seasons. That's pretty good. Late first round pick, two-time pro bowler, PFWL rookie team. Should be pretty easy. Yep, Justin's got it. Justin Jefferson is correct. Man, I thought year one, he was pretty good. There's the card right there. And you know, a lot of times with these football players, especially if you have a breakout season like that, now you got tape on Justin Jefferson, and defenses can adjust, right? No, year two has been pretty awesome as well. Minnesota viking Steve Birch, two more to go. And ladies and gentlemen, if you want to play more, who am I? A little trivia is always fun to play. We've got another eight box case of this in the store right now, jaspyscasebreaks.com. One spot gets you two teams in this eight box break. Back to a retired player. Currently 47 years old, stood 66-244 in my playing career, was a defensive end slash outside linebacker, played for one, two, three teams my entire career, and played for the team that drafted me three different times in my career, was a third round pick in the late 90s. An NFL defensive player of the year, a Walter Payton man of the year, two-time AFC defensive player of the year, six-time pro bowler, an NFL forced fumbles leader one season, in a different season, the NFL sacks leader, part of the NFL 2000s all decade team, ended up with 775 tackles, 193.5 quarterback sacks, eight interceptions, 46 forced fumbles, 29 fumbles, recovered nine touchdowns. Hall of Famer, he's pretty good. He's pretty good. Born in Pittsburgh and went to Akron College, not Cornelius Bennett. His sister is calling Coward's co-host. That's right. It's Jason Taylor. You think it's Miami edition? He did play for Miami. Then the Redskins for a season, back to Miami. Jets for a season, then back to Miami. Now it's obviously Jason Taylor. Dolphins edition. That's another one for Jack B. Nice one, Jack. Another Miami dolphin. Another dolphin legend, really. Final box. All right. Who am I? Retired running back. Currently 61 years old. Once stood at 6-2-2-10, which at the time, I think, the time of his playing days, I was considered really tall. If I remember the stories correctly, I really wasn't. It really was not their first-hand information, but 6-10-2-10. Born in San Diego, went to Lincoln High School in San Diego, went to USC. My playing career is from 1982 to 1997. Two teams my entire career. Super Bowl Champion, a Super Bowl MVP, NFL Most Valuable Player, NFL Offensive Player of the Year, NFL Offensive Rookie of the Year, NFL Comeback Player of the Year, 6-time Pro Bowler, 2-time Rushing Touchdowns Leader, NFL Rushing Yards Leader, was a national champion at USC, Heisman Trophy winner. Ended up with 12,243 rushing yards. It's 4.1 yards a carry. 123 rushing touchdowns. Could catch the ball a little bit. 587 reception for 5,411 yards and 21 receiving touchdowns. College Football Hall of Famer, Pro Football Hall of Famer. And he never should have left this original team. Yeah, Kevin's got it. There you go, Marcus Allen. Raiders edition, the silver and black. Big fan of the Marcus Allen. Never should have left the Raiders, had some issues with the owner, which has been well documented. Nice. Who's got my Raiders? That'll be Steve Wright with my Raiders. And there you go, ladies and gentlemen, another eight boxes in the books. I'm Joe for jaskiescasebreaks.com. We'll see you next time for another mini-helmet break. Bye-bye.


So I'd like to thank the organizers for bringing this together. It seems like a really nice really nice thing So, let's see Like to start with a couple kind of basic questions a little bit big First question is what kinds of structural theorems are possible for finitely generated semi groups of finite endomorphisms of quasi-projective Rides say a little bit more about what I mean by structural theorem later on Sort of want to pin this down a little bit by saying what might these structural theorems say about pre-periodic. So in other words One information about pre-periodic points of elements of a semi-group of finite endomorphisms What can they say about the structure of the semi-group? I'll start with a really really simple example. Let's say that I have two maps pn to pn Okay, each with a degree greater than one and let's say they commute, okay Then they have the same pre-periodic moments, okay? That's that's pretty easy to see you can get it from from Northcott for example So, you know if G has finitely many pre-periodic points to find over K and they commute the net is going to shuffle those around So they're going to have to be pre-periodic for F2 and vice versa So basically what happens here is a very simple finiteness result means that if you commute each morphism Shuffles around the pre-periodic points of the other one in sense. There's finally many Over over various field extensions, right? You will get it'll have those left to be pre-periodic as well So in particular right and a billion semi-group let's say of rational functions of degree greater than one all the elements will have the same exact set Okay, so how much so commuting implies same set of pre-periodic points proof as I said Is quite simple when your degree is greater than one It's true and more a lot more generality which I'll get to later, but it's not true in complete general Very silly example, but it illustrates something that I'll get to later if I take say F of X equals X cubed and G of X equals minus X then obviously F and G commute, but Clearly right everything's pre-periodic for G G is coercion Whereas not everything was pre-periodic for F so clearly the pre-periodic focus is not this Okay, you need a little bit more so and a lot of what you could there's an easy fix for this that I'll get to later But it's not quite exactly true if you use this sort of naive notion of which of what you mean by pre-periodic points So there's something actually a lot more general than that than that commuting statement, which is the following If I take an endomorphism of PN of degree greater than one To state my theorem. I want to do a little bit of notation that might look funny to you We're gonna be right, so we're gonna be composing maps of degree greater than one Right, so we're not in a group. These things don't have inverses. We're just in a semi group So I'm gonna use the notation Generating, you know left angle blah blah right angle. You mean the semi group generated by Not the group generated by A and B because there is no group generated Okay So I'm gonna use this to state my theorem. This is due to Jason Bell, Kepping, Hwang, Lane, Peng and me Let's say I just take two maps PN to PN Let's say they each have degree greater than one if the pre-periodic focus is not the same Then there is an L such that F to the L, comma G to the L is the free semi group on two genres So in other words, if I raise F and G up to a high enough power, there's no connection at all between those powers This means there's no non-equivalent words W1 and W2 NFL and G to the L such that that are equal to each other as Morphisms, okay So there's literally no relations at all any kind between F to the L and G to the L Of course, if they if F and G commuted right there'd be millions of relations because the word F to the L G to the L It's not equivalent right to the word G to the L F to the L, but through the same Okay, so this is quite a lot more general I'll say it maybe a tiny bit more about this later But one thing we don't actually know yet if we need this L at all Maybe it's true that FG itself is just free. We don't have any counter examples to that at all And in fact for polynomials, you can often show this is true with it So the proof of this is actually pretty simple Once you once you set up your height functions correctly So what you do is this? We have I'm gonna be a little bit vague about what I mean by height functions for now Because it's gonna turn out you actually have a lot of flexibility in choosing your height function They don't have to be Bay Heights in particular So we have good canonical heights attached to my F and G and they're just given by the usual tape limiting procedure Okay Where D is the degree of F and I can do the same thing for G So like if this is the Bay height you can do this But you can also do this for more walkie heights as well And I don't really know for all I know there's other Functions out there with some of the same properties as I functions You know, maybe not all the properties, but where you can where you can do this So what do I really need for this canonical height? There's a really crucial thing that I need Which is this right here? If the pre periodic points are not the same then the heights are not the same. Okay now Where does this come from exactly right this comes from the fact that? That the pre periodic if I choose my height function wisely enough, okay It will vanish at exactly the pre periodic points and nowhere else Okay, so The pre periodic look this will be exactly the place for each of them where the canonical height this canonical height is zero So in other words a zero cannot like basically picks out the pre periodic points So if the pre periodic points are not the same then the height functions are not equal to each other Now we can use this trick. It's this is very simple What I'll do is this I'll start with some word w1 and w2 in FG. I sorry I think I should yeah take any words like this, okay? If the word if my first word okay my w1 F to the L Okay, if I take a word that starts with F to the L It could be anything after that not anything just an F to the L and G to the L But even just an F and G take a word like this I can attach sort of what you might call like a normalized height attached to it right where I take a point Z I hit it with the word and then I divide by the degree of the word turns out that if L is large That will be almost exactly the same thing as the canonical light It's very easy to prove this again using the T limiting procedure Then I can do the same thing for any word that begins with G to the L So so this is any word in F So it's a little bit in the end what I get is freeness of F to the L comma G to the L But what I actually have here is a little bit stronger in a way Because it's actually telling you something where the words the w1 the w2. They're just words in F and G And I do the same thing here, okay? And This thing here for any Z will be about the canonical height of Z and this thing here will be So this thing here will be about the height of the F canonical height of Z This thing here will be about the G But those aren't equal to each other So if these two things aren't equal to each other these words cannot be equal to each other as maps Right because if they were the same map, they'd give the same height. They were the same map They'd have the same degree. They do the same thing to the points. The height would obviously have to come out the same Okay, that means that this w1 F to the L not equal to w2 G to the L And so then using some sort of there's a little bit of so morphisms in general the right they're not Really cancel a tip but you can cancel stuff off from the right-hand side because they're surjective So once you have this you can play a little game with induction and you'll get that F to the L come in G to the L It's free in other words. There's no word and F There's no to there's no non-equivalent words and F to the L and G to the L that are equal to each other as maps So let me say a little bit more here This works it doesn't really have to be P to the N P to the N It would work exactly the same for any polarizable math as long as I have good canonical heights It can work. It might work even more generally. There might even be some situations where you have Something that captures enough good properties of a canonical height that you can use the same argument Okay, so I don't really use really for example We don't really use anywhere enough proof that this canonical height It's is similar to a they hide or a more a walkie-hide or anything like that We really just use this formal you sort of formal properties about what happens when you iterate and that's really good The height could be a they hide or more a walkie height those are the two that we use one really crucial thing I should mention here is that Because we have only two maps F and G right we're automatically defined over a finitely generated field So we do have some height we do have these height functions we can attach for that reason This wouldn't work quite the same if I tried to do so you could try to do this for three maps four maps five You can do various things where I added more maps in Mostly things work the same Crucially though it has to be a finite set of maps because I do have to have a finally generated field that I'm working on to get these to get these Maybe it's how yeah one thing we're curious about is can you always just do this without a leak with one? So either F and G generate a free semi group on two elements or the or the pre periodic locals is equal to So we can prove a nice converse for rational functions in dimension one in Characteristic zero It's some case we can treat some non characteristic zero cases But in general to get something really nice we have to be rational functions in dimension one of Overfueled the characteristics zero we have a converse for this if the pre periodic locals is the same then they don't contain a free We can say a little bit more than that I will say there's no nice converse in general though. At least if I consider all polarizable maps It's possible the converse is true for maps PN to PN But if I look at more general polarizable maps There's lots of counter examples probably the easiest one is you take an a billion variety with paternity and multiplication. Okay So then all the endomorphisms, right? We'll have the same all the group endomorphisms We'll have the same pre periodic locals that ie the torsion points, right? But if you just look at the quaternions as a ring Under multiplication, it does contain three. It does contain three semi groups That seems a little bit strange, but it's not that hard to contain a free semi group Generally they contain it to can to take two elements that don't have any relation between them when I just take positive Powers of them. It's generally not that difficult to do So I think this works with a one plus i and one plus j or something The group that one plus i one plus j generates. It's not a free group on two elements The semi group that it generates really is just the free semi group So you really cannot expect a converse of this in general although there is a nice one I'll say a little bit more about heights So with very heights over function fields, right? There's a problem. There could it could be Right, uh, it may not be true that only the pre periodic holds up Now in characteristic p if you choose your initial day if you choose your bay height to have field of constants It's finite. It is true So you can handle any characteristic p case this way Characteristic zero try to use bay heights because it's a lot more complicated. There's a lot of nice papers of robinadetto and um Matt Baker handle it in a dimension in dimension one this work of chassadakas rushaski in higher dimension This work of bogey's emi in higher dimension are any probably others It's really there's a lot of really interesting questions about these about canonical heights attached to day heights in higher dimension Turns out we don't need them because we can just use more walkie heights So they have the same for they have a complicated definition, but they have the same formal properties That bay heights do and trim those formal properties we can direct Like as I said, we're using very very little about heights Just a few formal properties drives everything basically you need to know That uh, they vanish that you end up with the tape limiting procedure works. Okay Uh, and that what you end up with is a function that only vanishes on pre period. That's it Okay, so here's a more general question Is it true that for any finite morphisms x to x okay x here, let's say quasi-projective variety I don't really know exactly what conditions we need to put on Let's say quasi-projective variety for now Uh that the semi-group f and g generates contains a free semi-group on two generators as long as the pre periodic locus is not going to say So the answer is no We can't that's not true in general. Okay Um, we actually already saw an example like when we did f of the x equals x cubed Okay, and in g of x equals minus x Not every pre periodic point in g is pre periodic Now it's pretty easy to modify the question to avoid this kind of counter example I'm going to use what I call isolated pre periodic points, right? So let's notice that with g of x equals minus x. What's really the problem? The problem is that everything is pre periodic For for my x equals to minus x So you're not going to be able to say anything that great about the pre periodic points when it's just all of p one Um, so what we do is we use isolated ones and it's the sort of the natural definition here Well, we start with a map x to x You define What I'll call prep and n of f to be all the x such that f to the n plus n of x is equal to p That'll be the risk Now notice, I'm not saying this is so this is a this is maybe a slightly funny definition It's tempting to look at this and say That the period is m or something like that, but that doesn't have to be the case the period could divide m The pre period doesn't really have to be exactly n either. It could be less than n Okay So this m and n they're not for an x. They're not the exact period and pre period at all. Okay, they're just They're related to it. They're not exactly there um So this is a risky closed, right? So since it's a risky closed I I I can look at its components, right? I'm working in some quality projective variety. So I come up with the components So we'll define the isolated pre periodic points to be the elements of the prep of f that are not in a positive dimensional component prep and n of f So isolated is a Just to be a little careful about the definition It's isolated relative to other points With the property that f f to the n plus out of x equal to the n So it's not isolated with respect to all pre periodic points because very typically, right? The all pre periodic points will be risky dense So if I took the it's a risky closure of all the pre periodic points, I'd get everything in it So so the isolation is relative to other pre periodic points That meet this condition. It's it's not it's not they're not isolated relative to all pre periodic points Um a little bit about the notation. Why do I call it star star? so originally when um jayson and kepping and I were working on this we actually had another notion of isolated that's um a little bit different Where I take m n I I want to just not be in a positive dimensional locus of prep n m prep m n Of f or m and n are minimal for my x, but it turns out that doesn't work quite right. So you need this sort of limiting notion But you need a somewhat limiting notion of isolated to make it work. We had this other notion that I think is actually more natural Um harder to say but I think more natural than was prep star But most of the things we wanted to prove we needed to be in this prep double star Um, let's just think back on our earlier example When f of x equals minus x perhaps star star bath is actually empty There's no it doesn't have any isolated. It doesn't have any isolated pre periodic points Even though everything is a point of period too Um, and I can do the same thing with periodic points as well. So the theorems I state will actually be for periodic points for for automorphisms We can't prove that much Along with what I'd like for a general map. So we'll talk about period as well. It'll be the same thing same definitions So here's a much better way of phrasing my general our general question I have two finite morphisms x to x probably we want x to be quasi-projective variety But maybe it's true more generally for schemes with some property You do need f and g to be finite. I'll say a little bit more about that later um Everything fails. It's essentially the most spectacular possible way If f and g are not finite It literally everything fails um Is it true that if I take these two finite maps x to x that the semi-group they generate contains a free semi-group on two generators Whenever x has an isolated prepuri point that is not a prepuri So it is so so if we went back to minus x and x cubed, right? This is true in fact Because one of the maps has no prepuriotic points, right? They're isolated the other one all its prepuriotic points are prepurated for the minus x because minus x says everything is a The sketch around the example is quite natural I want to say a little bit about I want to say a tiny bit about the What the free semi-group on two generators is going to look like so back when I did pn to pn The semi-group had the form f to the l generated by f to the l g to the l Can you always do something like that? Can so can you could you not only say you've got a free semi-group? Two generators, but here's the generators The answer is actually no kind of a cool example very simple Let's say I take s to be generated by a 2t and t squared So obviously these do satisfy some relation right t squared 2t This and that'll be true no matter what I raise 2t to No matter what I raise 2t to and what I raise t squared to I'll always have a relation kind of like this So right so what is this relation? Let's call these f and g right this is f g Equals f f. Sorry. This is going to be g f equals f f g Okay, and no matter what I rate no matter what I Take power I take to t2 and what power I take t squared to I'll get some kind of relation like this, but oops, sorry s does contain 2t squared comma t squared which actually This doesn't look like it should be free, but it is because Uniqueness of two attic expense. So, you know, you can look at your word Basically, it'll give you a two attic expansion of your Of your of your of your coefficient Okay, um, so we can give a bit of an answer to question six in the case of linear groups Turn that so the following we can prove in a pretty pretty bare hands manner This is bell Cut thing and me Let's say I take a finally Generated group of automorphisms of variety x no condition on field of death of field or anything like that If g is virtually nil potent and for any g1 and g2 the Isolated of g1 The isolated periodics of g1 Are periodic for g2 I should say what virtually nil potent means It means it contains a nil potent start group of finite events So Uh, so if I combine this with work of teats rose and black or nitsky and saw what we're going to get the following Let g1 and g2 be almost at the same linear group If there is x and per star g1 Is not periodic for g2 Then the semi group they generate contains a free semi group bunch of Sort of basically just just what we want Um, so how do we prove this? I'm a most of it isn't this theorem a and then we use some sort of powerful stuff um That comes from combining the teats alternative with some work of rose and black or nitsky and saw a on free semi groups and saw the Okay, so let me say a little bit of what this what this is when you piece it all together This is stated in a paper about nitsky and saw a If I take a finally generated semi group in a linear group One of the following holds either s contains a non cyclic free semi group or the group s generates Okay, so either the group they generate virtually nil potent or you contain a free non-ability semi group Um, so the really heavy power theorem under this the really deep deep theorem the drivers This is something called the teats alternative the teats alternative Says that any finally generated linear group either contains a non-cyclic free group or is virtually solved with And this teats alternative was sort of the inspiration for all the questions that we're asking here I want to say a little bit here and i'll come back to this again is that Having a non-cyclic free group is very very different from having a non-cyclic free semi group It's very very easy to have a non-cyclic free semi group. It's much harder to have non-cyclic free groups. Okay, so um The way this really works then is teats breaks it down if you if you have a free group you obviously have a free semi group Otherwise, you're solvable and then there's a whole line of theorems unsolvable groups Okay, I will say this to one one slightly funny thing was that I had hoped that when I looked at a nitsky saw what what they actually proved is that this is Did if I take any solvable group? Okay, one of these two things holds But they don't actually prove that it's only this really is specific the solvable Um, yeah, so here's some other groups for which the teats alternative is known Teats alternative is either you're virtually solvable or you contain a non-cyclic free group The chroma group some really really beautiful work of kanta Barbara and lamby provingness The group of automorphisms of any projective variety and characteristics here. That was another example So it's a chem a we saw in shang Yeah, so I want to really emphasize one thing here It's much much much easier to contain a non-cyclic free semi group to contain a non-cyclic free group so, uh Any virtually solvable group size I said this well any virtually solvable linear group That is not virtually no potent contains enough so So lots of things contained Oh, sorry the way I said this is no sorry the way I said this is correct here So any virtually solvable group that is not virtually no potent contains enough Uh, what I actually want for the what we actually wanted on the last page was a tiny tiny bit stronger because we wanted it We wanted our free semi group to be in this particular semi group um Now here's something else quite interesting. There is this everything fails pretty much completely In the worst possible manner If you stop having your semi groups consist only of invertible If I allow myself to take matrices you can do this with three by three matrices with integer coefficients If I allow some of my matrices to be non invertible Everything fails completely Um, there's non-cancelative winners semi groups meaning that not everything's invertible Of what's called intermediate growth? Uh, so you this uh, so turns out say a little bit about growth at the end There's an ocean of growth in a group and the sort of the two Obvious cases of what can happen one of them is called is a virtually no potent case in that case You have what's called polynomial growth the number of words grows very slowly On the other hand on the other side you have something called exponential growth You automatically have exponential growth anytime you contain a free Not a free non-abiliate semi group But you could also have it without having Um, so just with very simple three by three matrices integer coefficients Um, they actually have examples of things that are basic completely, you know A complete failure of any kind of principle that we're suggesting here So at the beginning when I started to ask these questions for finite Morphisms finite is really really crucial things like this just won't work if you're not obviously for The linear map right, uh, you're either finite if you're finite, you're not a morphism right if you're not Finite then you have an infinite kernel Okay, so I want to so I want to talk about one other interesting structure Called the borrel fixed point theorem So the borrel fixed point theorem, um, it's really quite nice and easy to state Take a connected solvable affine algebraic group acting on a projective variety over an algebraically closed field Then uh, if you're solvable, right? There's an x that's a fixed point for everything in G So this generalizes the result basically saying The same thing for linear groups saying, but they have a common eigenvalue basically if you're solvable common eigenvector This is sort of generalizing every comment Uh, really really beautiful proof if you haven't seen it before it's it's very easy to describe You use induction on the length of the derived series for g right you take commutators over and over again You eventually reach the identity Um, so right when I take when I take the commutator, right that'll have a I don't have a shorter derived series, right? Um, so you just take uh, you pass the gg and you use induction So, uh, there'll be something that where gg fixes everything, right? You'll take that y Where ggx trivially It's easy to see that that'll be closed So y will be a projective variety in its own right And ggx trivially Okay, um, now the isotropy groups g y g acting on y are normal in g basically because g my gg is appealing It follows quite easily from that Now There's a y in g y that's closed in y that follows pretty uh, pretty simple application of shovel lace there The g y will have if I look at g y It'll have to be uh, it'll have to be constructable You can get that there's some so it doesn't mean not all g y But there'll be some some that it's like and you can it's really just shovel like Plus induction on the dimension Now it turns out this isn't too hard to show either that if I take a Um, I take an affine group g Okay, affine algebraic group and I am mod out by a normal subgroup g y What I get will still be affine Um, so g y which is a similar effect to g my g y And it's closed Because g y is closed Uh, so if you're affine enclosed in a projective variety, you're a point for g y That's a really really so it's a really really elegant So, uh, what we wanted is how much can you generalize this like, uh, for example Trying to do this for any finitely generated Um, group is is more general right because um any Any, uh, affine algebraic group will contain some finitely generated group that's that's the risky depth in it Okay, and obviously, uh Being a fixed point when we made right when you passed it with a risky closure So you might say, uh, can you do something like this for finitely generated groups? Um, secondly, you can ask, um What if the variety's not projective? Well, the variety's not projective. It's quite easy to see. There's no guarantee that you'll get a You can just delete all the public experts and what you'll have will be quasi-productive Um, yeah, so what do we get here? So here's something that we can prove Now here it's only characteristic zero So here we take any finitely generated solvable group of automorphisms of the variety x to find over a new characteristic zero Suppose that every element of g has an has at least one uh isolated periodic point Then there's a subrub H of my index and gene That fixes it that there's a subrub H. Sorry. There's an x and a subrub H that That fixes x. Well, so it's so H. I have a I have a subrub of finite index for the common fix Um, the proof of this is quite similar to the Borel fix multiple here Um The key fact is really this we do the same trick we pass over. Okay to the uh, to the commutator group And then we use induction Um, what we do is this we show that if I have an isolated Periodic point just for any element of it just so this so I have an x it's isolated for some h Not for everything, but for some h And and is periodic for everything Then x will actually be periodic for everything. So it's a subtle lift property here It has to be isolated for something in the commutator and periodic for everything But not necessarily isolated for other things Okay, um, then when I go up to g it'll stay periodic for every Um, so the way that this side is really similar to sort of the inductive argument in the Borel fix point theorem There uses something about closed orbits. Um, what we do here instead is um, we basically Whoops, we basically look at the orbit of this x and show that it's and show that it's a finite for every element Gee So how do we do that? Well, we end up using as a sort of a burnside type here for group actions along with a A really general version of norc So, uh, the following so that you know, there's the the famous burnside problem for groups, right? Is that it asks is a finitely generated torsion group necessarily fine The answer is no, uh, there were examples about maybe 50 years after burnside posed the question We have sort of a similar You can ask a sort of similar question for group actions. So Let's say I take a quasi-projective variety over a field of characters d zero Let g be if finally generated a group of automorphisms of x Okay, if x is periodic for all g and g then the orbit is finite so it's sort of like, um So it's sort of like this, right with the burnside problem for groups You have the group acting on it. You have the group acting on itself, right? And uh, all the toward everything being torsion and sort of analogous to this this periodicity result, okay So yes, this is saying if you're finally generated for a group action And something is periodic for everything in g the most important So in other words, its orbit is finite under any every Every single element of g means that it's actually finite for all of g Um, this ends up being a pretty simple application of the belpun and arcland You start you take your x. Okay, take any x that you want actually periodic or not There's a p and a subgroup h of g a finite index such that for every h and h You have a periodic analytic map theta h From z p to q p such that theta h of n is just h of n That's the arc lemma um If sorry, I think yes, if and if x is periodic under h and it's got to be fixed And so we have this finite index thing where anything that's periodic is fixed Okay, since that's finite index it acts trivially on the periodic points And so that gives you a finite limit The other thing we use is a version of north cod for integral points um So with the usual the usual north cod theorem i find this uh, Sorry finding us a pre periodic points. It's not no over finally generated fields, but for integral points We really do have it. Okay. Um, so I start with the n characteristic zero So let's take a variety over a finally generated field k of characteristic zero Let's let x be a model for x over a finally generated ring We'll field the fractions k Let's let f x to x be the finite morphism that extends to a morphism on Then there are finally many isolated pre periodic points that extend to integral Um, this turns out not to be really very difficult to prove. Um There's a really really nice paper of fachrudence where he bounds the possible periods of our integral points Uh, and then you can get about in the pre periods using a nice argument of scan Um, so while north cod's theorem for pre periodic points is not known really at all over finally generated fields It's a relatively easy and it's fairly precise too. You can you can write down a relatively precise bound on the possible periods Um Crucially here the the argument about the pre periods the characteristics doesn't matter at all um, but the fachrudent argument on bounding the uh, the fachrudent argument on bounding the Um periods it's very crucial Okay, so I think what I might do is end up, uh, I'll pose some just kind of general questions if I generalize these so, uh Would be nice to have a truly general north cod theorem So let's say I take a finite morphism for x of the projected variety over k And let's say that k is finally generated Is it true That the isolated points isolated pre periodic points, there's finally many of them enough Uh I haven't yeah, I don't I don't really know any results along these lines at all But there's a there's a few there's a few cases right? So, I mean obviously this is true if you're polarizable the question is it's still true If you're not polarizable You do need to be fun. You do need to be finite probably and you also do need To do this for isolated right if you take the non isolated there's tons and tons and tons of What do we get this for We only get this for integral pre periodic points and even then only in characteristics So it's not known in characteristic p or for rational points Um, I asked a few people I sort of thought well, is there maybe some obvious counter example in characteristic p I talked to drago schioka and a few other people about it and There's not any obvious. Well, I mean if there is a really obvious counter example in characteristic p Let no one seem to know what it is So, uh, I think that's yeah, I think it's a very very natural question. I don't know the answer to Here's another thing and this uh, let s be a finally generated semi group of finite morphos in terms of x to the south Suppose that x is pre periodic under every element of s Is it true that the orbit of x under s must be fine? So when did so when can what can we prove this for? We can prove it when you're in a group in characteristic zero. This is our karma Can't prove it at all in characteristic p because there's no arc lima in characteristic zero there is a slightly weaker version, okay of the arc lima that applies that's sort of an almost arc lima that jason bell came up with and It seems that it may actually be enough Uh, you may be able to use that to prove that this is true if x is actually periodic under every element It seems to possibly work for that But handling pre periodic point seems much much harder So again, this one I really this one I really suspect is true I'd be surprised if there was a covered. Okay, so finally the last one. I'll need a little bit of terminology to really Do this one here um So, uh, I talked a little bit earlier about uh rates of growth Which probably many of you seen before so the rate of growth of a finely generated semi group is measured by the number of words of length at most then In some finite set of generator graphs, right? This is you might have only seen this before for groups It's actually I think it's a slightly more general natural definition In a way over semi groups because you don't have to consider inverses. You really just taking Taking words um So the growth is said to be poly it's said to be polynomially bounded if this number is bounded by a polynomial in n so Yeah, let's see. That's before sort of intuitively What's something that would clearly give you polynomially bounded growth of the alien for example, right? It's very it's very easy to count the number of words In n generators if they can meet with each other Right, you can you can check that it's you can easily check that it's um polynomial in n plus one It's said to be exponential if it grows at least as quickly as even here So if you were free, it's very very easy to see that it grows that it grows like that it grows like even here And at least as fast as even here So there was an open question for a long time Are there any groups that have and are there any groups or semi groups that have Growth it's faster than polynomial, but slower than exponential Okay, uh gorgorchak found groups of growth that are in between It's a if you haven't seen it before it's an extremely extremely simple example It can be described in a couple pages. It comes from A group action on a binary rooted tree. It's not it's not really a very complicated example proving that the group that you get Has this weird intermediate group is growth is part but describing the group is really surprising and surprising Um, I might say one more thing doesn't really necessarily come up in what we've been doing But really really beautiful proof of Gromov is that a group Has polynomially bounded growth if and only if it's virtually no If and only if it has an importance of group of finite Really really big uh result of Gromov. So it doesn't really come into If you think of this to some of the things we're talking about earlier with virtually with virtually, uh, no potent It's virtually no points are very important All right, so, uh Yeah, so here's the question that I hope I'll end with here Are are there any examples of finitely generated semi groups of finite self maps of pn having intermediate growth? Uh, or are there any semi groups of finite self maps of anything? Maybe scheme widely projected variety, whatever having intermediate growth We'd be very interested to know the answers to this So I think I will end a little early if I didn't get any questions during the talk