 All right. Well, thank you to the organizers you guys are the invitation to speak and I do see some people in the audience that I know so I also are the people I don't know So I'm gonna do something. I've never done before which is give a slide talk on my iPad with this Stylus so that I can circle stuff Probably not as pointless stuff is doing that, but we'll try to keep it relevant So throughout this talk X is going to be a projective variety sometimes I'll use quasi-projective But that's okay for the purposes of it. We can always assume X is Projective and F is going to be a dominant rational self-map. So I just mean that it will have dense image and Given that it's a dominant self-map. We get a dominant a dynamical system With the variety X and the map F Where we can take a point in X and we can just repeatedly iterate F And as long as we stay away from the indeterminacy locus, we're fine and So what one wants to do is one wants to understand how this The behavior of this dynamical system and one of the more important most important Invariance for these sorts of self-maps are the dynamical degrees and in particular the first dynamical degrees Which we'll denote by lambda F So what's the definition? I'll give a more intuitive definition, but I'll give the precise definition here We take an ample divisor H of our variety X and this Notation isn't so good that I've done but F to the end is the nth Iterate the under composition of F and then what we do is we take the pullback along H And then we take the dot with H to the dim X minus one so perhaps the easiest case to think of is with you when you take a surface and What basically you're just pulling along pulling an ample pullback of an ample divisor and then taking an intersection with another Ample divisor and just counting the number of points a Generic one so the thing is that if we take this limit it actually does exist and It doesn't depend on what ample divisor you choose and The other important fact is that it's it doesn't it doesn't change under birational conjugacy So I've written it here that if we conjugate with a birational Map from X to X prime then we get a new Self-map on a variety X prime and it doesn't change the dynamical degree of the map. Okay, so I'll give some examples and try to make it more intuitive So What we want to say is that these dynamical degrees and I've only given the first one sort of gives you a very coarse understanding of the dynamical system and I'll try to say a little bit more about that and I do say that in general one has higher dynamical degrees one in each co-homology class No, I wanted for each for each eye And they were introduced by Friedland and They were originally introduced used a limb soup rather than a limit But they have this sub multiplicative property that implies that you bakes this with a limit Okay, I just wanted just okay, that's good So the dynamical degree is mostly studied in complex dynamics But I want to say that recently in arithmetic dynamics it's gotten a lot of attention do this work of cow Gucci and Silverman and What the idea is here now if you have a global field and you take some points Some k point and global field k you take some k point in your variety and you iterate You can try to understand the orbit by understanding the heights of the points in the orbit and how they grow the asymptotics and cow Gucci and Silverman show That it's related in a strong way to the dynamical degree It's it gives an upper bound in terms you can get an upper bound for the heights for how the heights grow in terms of this dynamical degree And there's a question of when equality holds and it's Suspected that equality should hold in most general cases as long as the orbit of the starting point is the risky dense so that's the cow Gucci Silverman conjecture and it's gotten a lot of attention in the last few years and It's a very interesting Conjecture relating heights and this dynamical degree Okay, so one thing we pointed out was that this sequence Is sub multiplicative and I'll try to explain why that's the case So the limit always exists So that's why we have this limit and I'd like to just to a more intuitive way of viewing the dynamical degree just so you can See it so I'm going to look at the special case when X is Projective D space and here it's much more intuitively clear So if here we have a rational self map from projective D space to itself then on a dense open set we can give it by this D plus one tuple of polynomials and Of course, we want to assume they have no common factors So we have polynomials P zero through PD. They're homogeneous of the same degree M And they have no common factors. Otherwise, we would just cancel those out and that M is going to be the degree of our F and So what we can do is now we can just iterate this repeatedly and we can compute the degree of F to the N and take this one over N So now we can see why we should expect submultiplicativity if F to the M is given by this D plus one projective tuple of polynomials and F to the N is given by the these coordinates Then if I look at F to the M plus N, I'm just composing these two. It doesn't matter which way I do it I've chosen it to do it this way Then we get P zero and we plug in Q zero through QD and so on and so forth So if F to the M has degree A and F to the N has degree B then these polynomials are all going to have degree A times P and So what we get is that the degree of F to the N plus M is less than or equal to the product of the degree And I've put a less than or equal to Because there could be some cancellation, right? So it's not inequality in general although you know We usually expect it to be an equality if we if we did something like took something in a generic element in the Modulite space of degree A maps or something like this So let me look at an example straightforward example where we can compute the dynamical degree and I've started off with the simplest example F is just going to be the map that takes X a colon Y to X square colon Y squared and This is a degree 2 map and if I iterate this n times I get a map of degree 2 to the N and so I can see that the dynamical degree is just 2 in this case So not so hard to compute in this case. Although in general other can be cancellation So we don't get this Equality in general because oh, that's ugly. Let me get rid of that because there can be cancellation And so I've given the easy example which I saw from Jeff Diller Which he used to motivate the cancellation. This is a I don't know what this map is called I call it I call it the nature map, but maybe it's got another name But basically the group of birational automorphisms of P2 is called the Cromona group and it's generated by PGL to C and this map F and it's almost a the free product a free product of those two Groups, this is it going to be a cyclic group, but it's got some one relation So this this is this Nurture map or maybe it's called the Cromona map. I don't know what people call it But it's it's basically a birational involution of order two. So it takes x y z to y z x z x y and F has degree 2 obviously, but if I apply f to itself if I apply f twice Then if I just do the computation I get x squared y z x y squared z x y z squared And I see I have a common factor of x y z and so I actually get the identity map And so if I look at f to the end Every time I apply f an even number of times I'm just going to get the identity and every time I applied an odd number of times I'm just going to get f so it's going to have degree 2 if n is odd and it's going to have degree 1 if n is even So when I take the nth root I get a dynamical degree is 1 Okay So in general the dynamical degree doesn't have to be an integer and so I'm going to work with a to which and that Contradicts what I said about pig taking a projective variety, but like I said, we can just we can lift this to a a Rational self map of p2. So I'm not really losing anything The only thing that I might be hiding is the degree calculation, but this is just homogenization and you can check what I say is true Um So we're going to take this map f u v is u v comma u and so this is like a torus torus Toric map and Notice that if we apply f squared of u v So what does f do it takes for the first coordinate just multiplies the two coordinates together to get the new first coordinate And it takes the first coordinate over to the second coordinate And so if you do it twice you get u squared v u v if you do it three times you get u cubed v squared u squared v And in general if you do it n times you get these Fibonacci numbers showing up in the degree and this is really because it's U v u v going to u v comma u so that's a u to the one v to the u v comma u so it's really like U is going to one one and v is going to one zero. Hopefully I did that, right? But anyway, this should be the matrix that generates the Fibonacci numbers when you take powers so F to the n has degree fn plus two and We know of course that when we take the nth root of that that a plot that tends to the golden ratio One plus root five over two. So the dynamical degree of f is now this irrational number, but still algebraic Let me look at another example. Just to this example is kind of similar to the other example But it just shows what the dynamical degree tells us about the two dynamical systems So if I take another one, which is f u v equals u v v Well, this is now u v goes to u v comma v So now if I were to write down the matrix I'm getting a unipotent matrix and if I apply f to the n I can see that I get u v to the nv and the degree is just n plus one and if I take the limit I just get a one and So this has dynamical degree one and having dynamical degree one is the smallest dynamical degree you can have it's very special and It says in some sense that this dynamical system with this map this tame map corresponding to a unipotent matrix is Somehow the dynamics are somehow tamer than the one coming from this that has eigenvalues Row and one over row if I've yeah, okay Okay, so I want to just give a few cases where the dynamical degree has been worked out so Ciboni has That if f is algebraically stable and what that means here is that if you take your f So I should explain what this notation means. I guess I've got this ns sub rx So what I'm doing is I'm taking the Neuron-Severet group, which is the Picard group of x Modulo pic zero the connected component of the identity And this ends up being a finitely generated abelian group, but it might have some torsion And what I'm going to do is I'm going to tensor it up with r which kills the torsion and just gives me a finite dimensional real vector space and My map f Well, if it's algebraically stable, so imagine you have something like an endomorphism. It'll actually induce Self map on this vector space But in general there's a class of Rational maps called algebraically stable where you get the same sort of thing and where they behave nicely So that if you take f star to the n that's the same as f to the n star And in that case lambda f is just the spectral radius of this z linear operator f star on the specter space And in that case you get that lambda f is an algebraic integer Because the spectral radius at the f and integer matrix Diller and fava Showed that by rational maps of p2 you can always achieve algebraic stability. So you can Show that the dynamical degree of any birational self map of p2 Is an algebraic integer And fava and johnson did a similar thing, but their thing is actually very difficult They looked at dominant endomorphisms of a2 the dynamical degrees are always algebraic integers In fact, they can even say more. They're either integers. So things like, you know, this map That we saw x y goes to x squared y squared where the dynamical degree is just two because the iterates are all just powers of two um Or there are things like row. They're they're basically, um, you know, uh quadratic Irrational's And that's all you can get and it's a very difficult uh paper And then I should point out that uh bonifant bonifant for nascent urich Showed there are only countably many different dynamical degrees. So it's a countable set and We've seen that under a lot of conditions we get algebraic numbers So it's a very natural question to ask Is the dynamical degree always an algebraic integer or at least an algebraic number? And the purpose of the talk well the title gives it away But uh, the answer is no And this is work with um Jeff diller and matthias yaunsen And at the end of the talk, I'll say we're doing some work us three along with Holly kreger I'll talk about that. Hopefully at the end But our main theorem is that there exists a dominant rational map f from p2 to p2 Whose dynamical degree is a transcendental number? And I should just make a remark That if you look here, we have this diller fava result that says birational maps of p2 The dynamical degrees are algebraic integers So our map is clearly not going to be a birational self map. It's Uh, it's not birational Um, did I want to say anything else about this? I should point out that I was fortunate enough to go to a simons workshop Sometime last year and that's where I met jeff. I already knew matthias and We just happened to They they just happened to talk to me about this and that's how the collaboration started But they have been working on this before so I was sort of in the right place in the right time In some sense. So I'll try to give an overview of how this works And say what the construction is So the construction Is they're all maps of the form where we take a tau composed worse with a sigma Where this sigma is a fixed birational involution of p2 and I've it's easier for me just to To use an affine piece rather than than p2. So if that's okay And then we have this this map which is really an endomorphism of GM squared Which is tau of y1 y2 is this map and this is where you can see where we're not birational That's clearly not birational So this is a monomial map and the only property we need for the transcendence argument to go through Is that a plus bi to the n should never be real? Okay, so Basically saying that when we write this in polar form Maybe I should what do I want to say I want to say theta is not A rational multiple of pi. I guess I want to say something like that. Oops. What's going on? Okay, so if we have these properties then every map of this form is going to have a transcendental dynamical degree Okay So um fava in 2003 He showed that under these conditions the map tau that we give this monomial map Can't be conjugated to an algebraically stable map Although still we do have that. Sorry. This should be So lambda of tau is still an algebraic integer though And lambda of sigma is equal to one. It's sort of like that Neuter map I gave earlier on So the key fact and this is really all matias and jeff Or jeff and matias, I guess depending on the order is that the degrees of these f to the n's Are really closely related to the tau to the n's and this is really the key fact That if we look at the the degree sequence of tau to the j this this nice map and we call it dj Then our dynamical degree is the unique positive solution on zero infinity That satisfies this equation Okay, so the question then becomes we've got this power series And we just say, okay, we want to prove that the value we plug in that gives us one has to be transcendent It becomes that question. So it becomes a completely A completely nice problem in diafantine approximation And there are other problems of this type that hopefully I'll get to talk about So This part of the is difficult and I can say it's difficult because it is all jeff and matias And it really relies on a careful analysis of how these sigma and tau's act on space evaluations of Apollo of the function field in two variables um And again, it's a very it's very delicate because it relies carefully on how this Byrational map sigma is chosen so that it behaves well with the tau But what they are able to show recursively Is that if you let dfn d sub n be the degree of tau to the n and e of n be the degree of f n f to the n, which is the thing we're after Then we have this nice um relationship That en is equal to this expression Where we have to set d zero equal to one and e zero equal to two And what that means is that if we recast this in terms of generating functions Then what we can say is that we've got this Uh this relationship e of z is one minus d of z equals two and that's why we picked our our starting values as we did So now what we get is that the dynamical degree of f Well, we know it's just uh one over the radius of convergence of Uh e of z because these are the Right the dynamical degree is just this minute By definition en is the degree of f to the n. So the dynamical degree of f is just going to be one over the radius of convergence of this And similarly the dynamical degree of tau is one over the radius of convergence of d um So one thing that's well known is that if you take a map if you take a generating series like this Associated to a map and you approach it As you approach the radius of convergence from the left it has to go to infinity And that's just an easy exercise using sub multiplicativity So what that means is because we go off to infinity there's going to be some point In the interval from zero to to the radius of convergence Where d of z zero is equal to one And then that means that this has a pole there or a singularity there And that has to be the radius of convergence because Um two over one minus d z is analytic inside that region so That tells us that the dynamical degree which we said was One over the radius of convergence satisfies this it's the expression. It's where the d of z is one Okay, so hopefully that made sense but now We can forget all the all the complex dynamics and we have a purely Difantine question of proving that a lambda that is the solution to this equation is transcendental. So it's A totally different world now we leave I had a quick question Yes So in the previous slide, what happened with the sigma? In the previous slide What happened with sigma Uh f was was f tau sigma or sigma tau f was It it might actually be easier to say that's f is tau sigma um Yeah, so the sigma. I mean it it comes into play with this computation but um I see so the so the sigma comes into play when you're proving that recursive relation between the e n's and the d n's Yes, got it. Yeah, thank you So now the question is how do we show that this solution is transcendental? Uh, okay, so let's do this in terms of difantine approximation And let's look at something that um, well people who who do difantine approximation know this much better than I do But there's a general philosophy that if you have a power series It's not algebraic and let's say it has some additional structure that allows us to produce rational approximations to it A good rational approximations Then what one expects is that when you specialize At q bar points inside the radius of convergence, you should get something that's transcendental Unless there's a compelling reason why why not And well, that seems like it's saying either it's transcendental or it's not which I am but I'm saying there's some there's some explanation For what's going on for for when transcendence doesn't hold And let me give you some examples So the first is a seagull shidlowski Theorem and I'll mention the result of boykers boykers as well So as an example if you take the function e to the x so this is what's called an e function And it's obviously transcendental for e function And what we know is that when we plug in E to the alpha the radius convergence here is everything so if we plug in any e to the alpha for any Algebraic value other than zero we get a transcendental number And this is a general phenomenon that if you have an e function You can make some system Describing the the differential equation satisfied by it And you can just you just want to throw out poles and zeros of Rational functions in the system and throw it zero as well And those are bad values and it says if you take any other algebraic Value and you evaluate there you should get it one always has you should get a transcendental value specialization and then With difference equations you can look at say Mahler's method and i'll point out work of philippin and adam chesky and colin favreau That says if you take an irrational Mahler series over a number field Then when you specialize You get a transcendental number Except outside of a computable set of bad Values and bad values again come from a system from you know, it pulls inside of a system for this series Okay, so let me see where I am. Okay. I better hurry up a bit So now we see a general approach to how we show that lambda f is transcendental We're going to use the structure of d to show that d of alpha is transcendental For algebraic alpha inside the radius convergence outside of bad values And then we're going to show somehow that one over lambda can't be bad But now we know that d of alpha equals one so that means alpha can't be algebraic Because it can't it's not bad So if it were algebraic then d of alpha would have to be transcendental So that's our that's the strategy Sorry just a question from uh just a question from uh an asking for a clarification. Just what is an irrational Mahler series Ah, okay. So when I just say all I mean is that f of z is a power series and I just mean that it's not a rational power series so it's not Given by the expansion of a rational function. It's at equals zero and in fact if it's irrational It's it's uh, I think it's due to a besavan that it has to be transcendental Uh, so hopefully that's okay. So that's all I mean by irrational. I mean it's confusing. I mean not a rational power series So it's it's it's coefficients. Don't satisfy a linear recurrence Where was I um here? So I wanted to say how we do this in practice So the idea is you take your series d of z And you show you have good approximations by rational functions Uh in the sense that if phi n has degree a n Then I want I want the difference to be big o of z to the a times a n where a is some big number bigger than two And this will actually apply that my d of z is transcendental Unless it just by some fluke happens to be equal to all my phi n's for n large um And then one wants to argue that phi n alpha is a good approximation of d alpha which sort of makes sense if you look at this And then um, you want to just use the you know, the typical methods in transcendence theory to show that d of alpha has to be transcendental unless There were some compelling reasons not which in this case would be that the phi n of alphas Happen to agree with d of alpha. So that's our compelling reason why it might not be transcendental okay, so We want to apply this strategy. So we need a closed form for our series d of z And what i'm going to do is i'm going to let d j be the degree of tau to the j So remember this was what our our series was And this is now the other thing I mentioned before where it was this You know this recursive for expression involving the e n's and d n's that was very difficult, but this is more of an elementary computation This is some somehow corresponding to the matrix a b Minus b a and taking powers of it And just doing a degree computation And so what you can show is that the degree of tau to the j is given by this expression So here zeta is a plus bi and gamma j is one of these elements And you choose whichever one maximizes this right hand side And maybe it's easier if I draw a picture to it's probably easier if I draw a picture We're going to let zeta We've got this a plus i b And what we do is we compute a to the a we compute z to the n and we're going to end up in one of these regions And if we end up here Then we're just going to take c Minus 2d and that's going to be positive because the imaginary part is Negative so depending on what region we end up on and when we apply zeta to the n wherever zeta to the n ends up That's the expression we use to compute d of n We take the real part of zeta to the n plus two times the imaginary part if if we're in this region okay, so And now what we're going to basically use is the fact that our are the argument of our zeta is irrational. So You know, we're going to There's not going to be any real periodicity in terms of which region we lie in So that's the idea So we're interested in the case when the angle theta is irrational. That was the hypothesis before so we're going to make that assumption from now on And hassle hot and prop hassle flatten prop show it. Sorry showed that the sequence dj does not satisfy a linear recurrence And so in particular it's an irrational power series It's not rational And in fact our argument somehow actually shows it's not even algebraic although there should be a direct argument for showing It's not algebraic Typically to show something's not an algebraic series You either show it doesn't satisfy a homogeneous linear differential equation or You show by reduction my p that it's not algebraic or use asymptotic methods or there's there's But anyway, we didn't try this out, but it's it would be an interesting question to show this is not algebraic Um, so we're going to use the fact that theta has reasonably good rational approximations And I when I say recently good, I mean I don't mean anything really just that it's an irrational number and irrational numbers You know have a continued fraction Expansion and that you these give good approximates to the rational approximates to theta And so we're then going to do obtain transcendence using pietic subspace theorem And I should point out that it was really corvaya and zanyi Who used? pietic subspace Theorem in this way first for proving transcendence as far as I know they were the first ones to do it Um To prove transcendence of constants this way And then adam chesky and bougio Showed automatic real numbers were either rational or transcendental using this Corvaya zanyi modification of the technique So we're going to use this we're going to need a lot more linear forms because we don't really have an understanding of the angle theta But I'll hopefully be able to explain how this works So I'll talk about really there's a few parts. So I'll talk first about the easiest part Which is to construct the rational approximations and this is sort of intuitively clear The idea is that if we take the a good rational approximation to theta Then m over n then zeta to the n Is really close to being a positive real number. It's either here a little above or a little below And if we go back to our picture That means that if we look at where zeta to the j lies It's going to lie in one of this contrast then zeta to the j plus n Is probably also going to lie in the same region because zeta to the n just Is very close to a real number. So it's it's only it's not going to change the argument very much Now what could happen is we could have some bad luck where zeta to the j is very close to the boundary And when we multiply it by zeta to the n it pushes us to the other side So but what we can say is that we expect zeta gamma of j plus n to be gamma of j So we expect to be in the same region Unless zeta j is very close to one of these boundaries So what's that saying is saying that gamma j is going to be nearly n periodic So we expect d of z to be approximated by this function Well, this would be a good approximation if the gamma j's were actually n periodic, but they're not So we need some correction And what we can show is that the number of bad indices up to a constant cn is uniformly bounded So we don't need that many corrections And also we can show that the bad indices repel in some sense and that will ultimately be important So what we're going to do is we're going to adjust dn of z this approximation we produced By just adding on k other at most k other terms that correct for those bad indices And then we get a new function That we're going to call phi n of z and that's going to be our good approximation Okay So where am I so showing that this phi n alpha is a good approximation So i'm going to maybe i'll regret doing this, but I thought I should do it just in case people Uh, I know this is a broad number theory seminar. So maybe people don't know the piatic subspace theorem But let me just give an idea an overview So we're going to have k as a number fields of degree d And I I decided to go all in and say we're going to let m of k denote the set of places of k so really what I mean is that Each place We have two types of places finite places and infinite places and I'll say what these are But these basically give us an absolute value really up to some equivalence And we're going to pick a pick one for each v in a in a sort of canonical way Um, I should say that if you haven't seen this you probably have seen it at least in the case when k is q And then you know that the absolute values are precisely the the piatic absolute values where p is a prime And then the ordinary euclidean Absolute value and those are my places of q, but in general I have other places So let me just explain how this works. I'll try to go over this quickly We have finite places. So here Um, we're going to use the notation m sub fin of k to be the set of finite places And what we do is we take the ring of integers and we take a prime ideal p In that ring of integers. So it's a maximal ideal and so if we take The quotient okay mod p we get a finite fields of cardinality and And what I have is I have a rank one discrete valuation on this Okay, uh coming By looking at what power of p I lie in right the biggest power of p I lie in so I get an order So if x is in p to the m but not in p to the m plus one I can say that its order is m And I can then put a my An order on all non zero elements of my field By just saying okay, if it's a fraction a over b then I can say what's the biggest power of p that The biggest power of the prime ideal p in which a lies minus the biggest power of prime p that b lies And I can now define my valuation this way and this is exactly like the paedic valuation Where it was p to the minus Right, it's the exact same thing Um, I don't even know if this is actually all that important But I I thought I should say it just because I I'm going to say Sit have a theorem that has all this notation in there. So I might as well state it We have the infinite places which people probably know a bit better. We only have finitely many infinite places that correspond to our embeddings into c And basically we have two types of embeddings. We have real embeddings and complex embeddings And if we have a Real embedding, we're just going to take our place just defined like this And then to a pair of complex embeddings We're going to define our place like this And the way the Like I said a place is really in a equivalence class But the the nice thing about doing it in this Way is that we have this nice product formula That if you take the product over all places of a non-zero number you get one And I should also add that This makes sense because the the absolute value Of c is one for all but finitely many places. So this is actually collapses to a finite product okay So now Throughout now we're going to be working with a finite set of places and we're always going to assume that it includes all infinite places And if we do that we can make a ring o sub k of s Which is the ring of s integers. So it's just the set of elements where the absolute value Is less than or equal to one for every place not in s And notice that if we just take the infinite places Then we just recover. Okay. So this is some generalization of this. So for instance, if I took For q if I took the two attic place And the euclidean place Then the ring of integers Here would be the two attic rationals right because One half has absolute value one for all places other than two And the last bit of notation we want is we want to have some notion of height of an m tuple And this is what our height sub s is so we just take the max over the The x i's at each of my places And take the product. Okay. So if you've seen heights before that sort of makes sense Okay, so that was a lot of stuff Just for one Thing which I didn't say so instead of using a pietic subspace theorem. We're going to use a theorem of averica, which is more or less A version of pietic subspace, but it's it works better for us And here's the theorem. So we've got a finite set of places And it has to contain all the infinite places. So that's exactly like pietic subspace theorem And we've got an integer m bigger than or equal to two And a positive constant epsilon that can be whatever we want And there's a constant c. It's not really it's not an effective constant if you're wondering, but it's Well, it's something it's a constant Such that if you take some your vector x1 through xm in the ring of s integers And we suppose that every subsum is non-zero Every non-trivial subsum is non-zero Then we've got this inequality Which might be a little hard to look at But how you can think of it is saying that the sum of the x i's in some sense can't be too small Right, so maybe the easiest case to look at would be is if you look at things like s units So things that are units in this ring and then in that case These these products all just become one And it's basically saying that a a non vanishing sum of s units Can't be too small compared to the biggest element in that non vanishing sum And in fact you can make it In fact, you can take any power of epsilon and it just relates to the height of the epsilon so it can't be too small So that's the theorem But notice it's more general than beyond s units. So that's the power of it And why this theorem is useful to us rather than subspace theorem is because We have these correction factors Which when we specialize will actually end up being s units So they behave well with respect to avert this theorem, but not everything is an s unit. So We need to work with that So I see I've got about six minutes left. So let me try to wrap this up in almost six minutes So how do we use a veritz's theorem? So you might remember we made this approximation phi n is that which looked like this Uh truncated series But we had to add on some bounded uniformly bounded number of monomials Which depend on n Which took account for those bad j approximation that where the periodicity failed And so really our Approximation looks like this And the significance of this this has no real pietic significance to us, but if we take an infinite place It does have some significance because it's saying that at an infinite place When we plug in we should get something Something small if we plug in inside the radius of convergence So at infinite places this this expression is quite strong So now remember Our goal is to show that if alpha is algebraic and inside the radius convergence then d of alpha is transcendental Unless there's some reason otherwise, which is that the phi n of alpha that is an exact approximation So we're going to use the corvaya zonye original strategy, which is really great The original strategy was you you want to say okay, suppose this is algebraic Okay, so now we can make consider The galois closure of the field where we throw in beta and alpha and some other constants And we're going to use some sort of pietic subspace variant And then we're going to show that this can't happen that we that we're going to get a contradiction So let's return to our Our expression and maybe I'll have to go over this a little fast Um, but we're going to take s to be the set of all places where we're going to take all the infinite places of this field k that we formed Along with the places at which the non-zero elements from our element beta alpha these gamma j's these ci's which are all in a finite Set that doesn't depend on n Where these where the places where they're the absolute values are not one Which I guess I said the places at which um then like we said if we Specialized we get this and this approximation is really only useful at the infinite places So now what we're going to do is we're going to use technical estimates where we apply the aversa theorem To show that d of alpha is algebraic, then it has to be equal to phi n of alpha Well, what's the idea? We've got an infinite place Where this is very small And if we expand this out And we expand these out These when we expand these out. These are just now s units According to our construction of s and then we've got this awkward thing, which is just an s integer But what we can do is we can apply A aversa's theorem and we show that no subsum can vanish And so we get that something has to be kind of big at a fixed place a fixed infinite place but We know at a fixed infinite place we can make this Smaller than that and we get a contradiction that way I went over this a little fast, but I don't even think if I went over it slowly it would make More sense, but that's just the intuitive idea. It's you know these transcendence proofs are You sort of have to sit down and do it yourself sometimes So now finally the last part is we use an ad hoc argument to show that if alpha is positive and real And inside the radius of convergence, then you always get this inequality It doesn't matter if alpha is transcendental algebraic, whatever So that means that these phi and alpha these approximations can't be exact And so what we get is then that alpha Which is one over the dynamical degree Can't be algebraic Because When we specialize we don't get a transcendental value Okay, so that's how the proof works. I don't know if I Made it so much sense, but let me just say a few things because I mentioned at the beginning of the talk there's joint work with matias and jeff, but also with holly kreger So um One question one can ask is our maps are not by rational self maps, right? We said that the diller Favre result shows that by rational self maps of p2 have algebraic dynamical degree So you can ask what happens for by rational maps, maybe a higher dimensional projective varieties And there's also a kind of an interesting question You can look at the field generated by all Dynamical degrees and now once we know it's not algebraic Well, we should expect that maybe it has infinite transcendence degree Um And I'll say a little bit about these things. So this is work in progress With the three authors, but also with holly kreger um So like we said the we we sort of did this In a very special setting where we had this Very special map d of z but it obviously applies more broadly And so we want to give a there should be a sort of general Uh transcendence criterion that applies to lots of maps coming from dynamics And so with this we think I don't want to claim anything because it's not Uh been nailed down But we think we should be able to find by rational self maps of p3 that have transcendental dynamical degrees So that once to the other question And this will also just Maybe of interest for its own sake it will it gives rise to a sort of general transcendence problem Um, which I decided not to write down because it's a little tricky, but it's basically a transcendence problem Where you look at special values of series that are defined on these piecewise constant functions in this way and with matrices with certain eigenvalues And you ask whether you get a transcendental number number when you when you look at the value that gives you A one or something like that The difficulty now lies in the fact that maybe You don't have a dominant eigenvalue or or or something like this or or you know, you have multiple dominant eigenvalues I guess I should say Now the infinite transcendence degree question I think is an interesting one the answer is probably yes And probably the collection of maps that we produce we produce an infinite family of them Probably you could prove infinite transcendence degree of these maps If you really sat down and were motivated to do so, but it's more involved and I don't know if it's Something I plan to revisit anytime soon. So I think my time is up So I'll say thanks and I will end the talk Right now. So thank you