 Tako da. Zato je toga kompetija. Zato pravno tudi, da sem očeli, da sem ne ga poznala. Na tituljiéc smo razlučili 1. Všeč smo všli 2. Poživaj, da bomo, zelo, ovo. Vsih moramo zvrkni v 2. in zato v 4. in zato v tom na wordsi moramo jonka semja umreči z splashov na 5. Dončanje k dvej točnih vse je. Tudi pričaj sem se umožela, da sem posledal je tutorial na 4.30h, tudi katarina Kukić. Zelo je zelo vzvečenje v ovom delu, in vzvečenje vzvečenje na fri dnev, z Anani in Edwardem, je zelo vzvečenje v ovom delu. Tudi 3-5 svojov svojov, nekaj ne so vzvečenje v zelo, ali zelo se prišlično prišlično. Tudi je kubiko in električno funkcije. Zelo smo prišlično izgledati, da se vse, ki se vse izgleda, vse je, da se vse izgleda, kaj je zelo, da imamo dve vrste, kaj je vrste, kaj je vrste, kaj je vrste, kaj je vrste, kaj je vrste, In tudi je TORUS, da se tako pravimo, kaj smo obtežili, da je vse fundamentalne paralogrami za dv periodične funkcije, kaj je omega-1 in omega-2. To je bilo različ, da je omega-2, da je omega-1. Here we have two structures. Here on a cubic curve, this structure has been presented in the following way. So this is a circle b, and then this involution gives us a plus b. You remember that for a given a, there was also involution this tau, and then we map a to minus a in this structure. And here we have just a fundamental addition of vectors in a, if you want in a R2 or complex numbers in C, calculated modulo the lattice generated by omega 1 and omega 2. And then we constructed here very important elliptic function of order 2. So this is a Weistras p function, and we show that if you use the Weistras p function, satisfies a very important differential equation, which allow us to construct this map from this picture to that. Denoting this as a polynomial of degree 3 in p function, we have a map. Map is, so map phi from torus 2, let's say, gamma mapping z2 p of z p prime of z1. I mean, this is okay that this formula goes across this slide, because this is a map from this window to that one. And then the final statement was that this map is not only a bijection, it is an isomorphism of groups, isomorphism of groups. So in a sense, for example, if we have three points here in this picture, and we have three points, let's say, and w, such that u plus v plus w is zero, then we know that this map would map these three points here into, so you see, this is a plus b, that means that a circle b is tau of a plus b, that means this is minus of a plus b. So a b and minus a plus b are collinear, so three points here are collinear, if and only if, it's sum is zero, that means that three points here are mapped here in collinear points. And we are now going to use this very simple thing, but just before that let me make a few very simple remarks, but important. So, of course, here we are getting a polynomial of a specific form. Here we are getting polynomial of a specific form. It doesn't contain terms with quadratic terms, but of course with the appropriate change of coordinates we can always reduce general polynomial of degree three to that form. Maybe I can leave using this good example of Sergei, some exercises. So, talking about this, immediately we have an exercise to prove, so what was exercised last time was that we have these midpoints here, omega one, omega two, and omega one plus omega two, and here if you know these points e1, e2, e3, then if you remember exercise from the last time, we said that p prime is zero in these three points. So, from the last time we have that p prime of, for example, omega i is equal to p prime of omega one plus omega two is zero, so that means that at this point this is going to be zero, so that means that p of w1, p of w2, and p of w1 plus w2 are going to give us e1, e2, e3. So, e1 is, for example, p of w1, e2 is p of w2, and e3 is, for example, p of w1 plus w2. Exercise is to prove that these three points obtained in this way are distinct, so prove that e1 is not equal, e2 is not equal, e3 is not equal, e1. So, this is one exercise. Another remark, of course here there is some sort of choice, because we can denote p of omega one to be e2, p of omega two to be e3, and, for example, to use some permutation of this, and, of course, we cannot expect from an elliptic curve to know what was our choice. So, the question is when two elliptic curves are isomorphic, so when can we, in a sense, identify two elliptic curves, and the answer is given in terms, so we have this, you remember the picture of last time, so we have e1, e2, e3, and p infinity as four points where we made these cuts, where we glued these two planes or two Cp ones together, ramified over these four points. So, we have four significant points, which correspond to those points, which satisfy that two times this point is zero, and for four points, let's say e1, e2, e3, and p infinity in our last notation, what we can do as follows from the instructions obtained in the Sergei's lectures is that having four points we can calculate their cross ratio. So, we have a cross ratio of these four points, and as Sergei observed, cross ratio of four points is not uniquely determined, it depends on this choice of order, so we can get numbers like, and make mistake, but you will, lambda 1 minus lambda, what else, 1 over lambda, 1 over 1 minus lambda, and lambda over lambda minus 1, and lambda minus 1 over lambda. I think this 6 typically you can get, and then we can say that, I mean this is a theorem, two elliptic curves are isomorphic if this set of six values coincide for these two elliptic curves. Of course, there are some specific elliptic curves where you don't have six different values, but this is in a sense an additional story. So, what I want to say is that we should in a sense start accepting a principle that whenever you have something which is double cover over Cp1 with four branching points, it's elliptic curve. So, this is in a sense guiding informal principle. This principle, of course, can be formalized by Riemann-Hurwitz formula, which says if you have a holomorphic map between two compact Riemann surfaces, then the Euler characteristic of origin is equal degree of F times Euler characteristic of image minus something, which we call degree of ramification. So, this is a general story. So, what it is about, so we have something, let's say gamma somehow mapped to gamma prime. Here is gamma, here is gamma prime. A general fiber, if you take here some general point and consider inverse image, general this fiber has some fixed number of n points. This n is called degree of a map. This is a degree of a map. In our case, it's two. But it could happen that there are some points where you don't have so many points in preimage, like here, and you have some gluing, and these are ramification divisor, in a sense, counts. This number of points where you have this sort of effect and degree of this R is a total count of this sort of collapses of numbers of preimages above some points. So, in our case, since g prime is cp1 is of genus 0, this is elliptic curve of genus 1. This degree is 2 and this degree of R, since we have four points and we have this ramification, this is four. So, in our case, so we have 2 minus 2 times 1 equal 2 times 2 times 0 minus 4. So, as you can see, this identity is correct. So, this is a verification of this principle that whenever you have something, double ramified, some double cover over cp1, ramified at four points, this is an elliptic curve. Okay, so, now, let's, as a sort of example, use this in a sense, if you want to fully appreciate what we derived, we need somehow to apply this. So, one way how we can apply, we can use this fact that having three points in this picture with the sum 0 means that, in this image, they are going to collinear points. This information is enough to derive a very nice classical formula, addition formula for VHSP function. So, let me just, as a sort of example, this is addition formula for VHSP function. So, if I may allow a small remark, I see that all of us are a little bit today, in a sense, out of shape. I don't know if you had some interesting nights watching football or so. So, please try to follow, because I can easily make now mistake in the calculations on the board. I need your full attention in order to prevent some creation of new unknown formula. So, please, let's try to do it together. It's a simple exercise, but still it requires some sort of concentration. So, we have three points, u, v, we have three points, u, v, and w is nothing but minus u plus v. So, that means that take u and v and map here, you will get some, these two points that remind some line, and then that means that a point obtained as a map of minus u plus v belong to that line. So, suppose I will now, that the formula for that line is something like so, let me, something like this. So, p prime of z minus some a p of z minus b equal zero is satisfied for z equal u is satisfied for z equal for z equal u, v, and minus u plus v. So, when we plug these three concrete values then these three points belong to the same line here determined by coefficients one minus a minus b. On the other hand we know that these three points are also situated on the curve. They belong to this cubic curve so that means that let me just write this p prime of z square is equal this polynomial there. So, this is a four times p cubed z minus g2 p z minus g3. And again if you plug these three values here we are getting this is also satisfied. So, so, this system is satisfied in these three points. So, from this relation I can exclude p prime and I can write here a p plus b square equal this four p cubed minus g2 p minus g3. So, now we need to calculate something so this is a so let's now try to rewrite everything in a just as a polynomial. So, u v and minus to b argument of p function we satisfy the following polynomial polynomial equation. So, the equation is four z cubed minus a square z square minus g2 plus 2ab I hope it's correct z minus g3 minus b square equal 0. So, this is a polynomial in z for which we know that it has three zeros we know three zeros. z1 equal u z2 equal p of v and z3 equal p of minus u minus v but we know that p is even function we can use it not use this in my mind. So, it's a p of u plus v. So, we know three zeros of this polynomial what would you suggest me to do? We have a polynomial and we know it's I mean we know we know these three zeros what we really want to do is we want to express this zero in terms of these two this is our ultimate goal. To get addition formula means to express p of u plus v in terms of values in u and v and this is our all the time the same dilemma so tell me what do you suggest me to do? If we have a polynomial and we know three zeros we can use what is the proper English term for this? Vieta formula Vieta this is in English Vieta in Italian I don't know why it should be in English Vieta formula saying that what? The sum of zeros is equal what? a squared over four sum of zeros so p of u plus p of v plus p of u plus v is equal a squared over four we are almost done we are almost done but not yet we need to calculate somehow a from here knowing that this is satisfied for u and v so we plug now here u and v and we are getting two equations from where we can calculate a in terms of p of u p of v p prime of v so what is a? A is how can we get? a is from here a is something like p prime of v minus p prime of u divided by p of v minus p of u this is a and finally we are getting the formula that we want to express to express p of u plus v minus p of u minus p of v plus two times p of v minus p of u p prime of v minus p prime of u square so please remember this formula if time permits in section 5 we will use it once so that's why we need it because we are going to if not use at least mention so this is additional formula for vajštas function we are getting it as a simple consequence of this of this situation okay and let me say just some remarks since we are here regarding situation of multi-dimensional billiards so for billiards and ponselet theorem we will see billiards in the within ellipse and for ponselet theorem in the plane we will see today or tomorrow that it is related to this addition structure on a cubic for higher dimensional situation for billiards within quadrics in dimension d it appears that relevant relevant objects are hyper elliptic curves so here in dimension d equal 2 so what we had here in dimension d equal 2 the relevant was genus g equal 1 this is elliptic curve in d equal d it appears that the relevant object is going to be hyper elliptic curve of genus equal d minus 1 so hyper elliptic curve in similar fashion as elliptic curve can be defined by its equation of the form y square equal p 2g plus 1 of x where p is a polynomial of degree 2g plus 1 or this is also true for for elliptic curves you can go to even dimensional and then here put p of 2g plus 2 we could have here also tell the story with the case of polynomials of degree 4 it just depends on this so here we have degree 3 because one of these 4 important points is chosen to be at infinity of course in a sense there is a subtle difference talking about elliptic curves of presentation in this way because in presentation in this way somehow this point at infinity is in a sense significant this significance is lost here because here in a sense all these 4 points are equally equally good but let me come back to this case of hyper elliptic curves of genus of any genus so the point that the genus is of course immediately bigger than one so in that case what I want to say is that still there are in a sense there exist there are these 2 screens one screen here is a curve itself of genus g so curve of genus g just schematically we can something like this so here is a genus 3 and here in this case object is here it was a torus and in the case of curves of high genus is torus of higher dimension so what is at the same time nice significant and in a sense transparent in the case of elliptic curves this is at the same time misleading in the case of elliptic curves so in the case of elliptic curves we have 2 pictures of objects which are isomorphic so we have here a curve here we have torus and double theorem says torus is isomorphic to the curve in higher general situation here we have curves which are not torus anymore not tori anymore but there exist some very important objects some high dimensional tori which play the role this torus here played here it's just a sort of a framework to understand that there is a sort of higher generalization which is going to be relevant for higher dimensional ok having said that let me also mention few few general important the theorems and I will formulate them for the case of elliptic curves and we are going to this way or another use so when we are talking about complex algebraic curves and elliptic and high elliptic curves are examples then there are two main classical questions so suppose you have some two collections of points let's say p1 pn and this is one and another qn these points belong to the curve you can also assume that some of these points have a multiplicity and that's why they use a sort of fancy name for this called divisor, just a formal sum where you can put some coefficients here if you want and here q1 but here there is no actual addition this is just a way of saying points with maybe some multiplicity and then the question could be given for example these two collections of points, is there a metamorphic function on gamma having these points as zeros and these points as poles this is one classical question and another classical question is so the existence of a function such that this is the notation divisor of function means zeros minus poles is p1 plus pn minus, this is just notation so the existence of such a function and the second question is, so this is the first and the second question is if you provide a sort of a set of points of some possible multiplications the question is what is the dimension of the space of functions having poles not worse than those prescribed here, so if I denote this by d, then we can say that l of d is a space of functions metamorphic functions such that these functions don't have poles worse than those described by d, this notation here is for the following so if we have some d1 which is some combination of n1p1 plus nkpk where pI are points on gamma and nI are integers, then we say that d1 is effective d1 is this is the notation and the name is effective if all nI are no negative so the answer to the first question is given by so called Abel's Theorem again Abel's Theorem and I will formulate this for elliptic functions so Abel's Theorem for elliptic curves so the answer is as simple as this so you have this n points one set of n points and another set of n points this plus here is just formal notation but the Abel's Theorem says since this function in the case of elliptic curve which is group, belong to a group then this is a real summation do summation in the terms of the group so calculate so having given p1pn and q1qn then do real summation in the group p1 plus let me do like this plus pn where here elliptic curve so gamma plus I want to indicate that here this is a real addition as described here and here so sum this n points as n elements in the group sum this n points as elements in the group and if this is equal in that group which means in this picture means equal as a sum of complex number up to the lattice and in this picture means here if and only if this is satisfied then there exists a function such that is this pp1 plus pn minus q1 plus qn so here this plus is just a formal notation saying we are taking these points and here the plus is real summation in this group structure so this is a Abel's theorem and this theorem measuring the dimension of this space is known as Riemann-Roch theorem it says that for the case of elliptic curve it says that dimension of this space is equal I will explain now degree of d so degree of degree of so I will explain here degree of d1 is sum of these numbers on I so sum of these coefficients this is the degree so this is for g equal 1 for g greater than 1 is little bit more delicate so for any g I can easily write the following the dimension of L of d is greater or equal degree of d plus 1 this I can easily say and I can say that for example equality is for sure so divisors d for which here is equality are called non-special and then for example if if degree of d is greater or equal d2g minus 1 then d is for sure non-special ok I am just mentioning this because it can have some relevance in higher dimension billiards but for us actually this is what is important so the dimension of this space dimension of this space in g equal 1 depends only on pure arithmetic how many how big this divisor is how many this is the dimension of the space it doesn't depend on geometry of these points can be can be scattered arbitrary there is no geometric component in this formula however in higher genera it can be important where what is the geometric position of these points because sometimes divisor of the same degree it can be special or non-special so here can be some some difference in this formulation that for arbitrary g dimension is greater or equal this expression here let me just mention this for reference this is a remaining equality but in a sense we are safe here because we know from here we know many gradients how to construct elliptic functions so maybe I can give again some exercise this is a suppose you have an exercise suppose you have you need to have at least devices of degree 2 so suppose you have 4 points on an elliptic curve and suppose that a1 plus a2 is equal I can maybe say b1 b2 b1 b2 is equal b1 plus b2 so construct a function such that divisor of f is a1 plus a2 minus b1 plus b2 so you need to play with vajstras function so vajstras function provides you LEGO cubes and you need somehow with these LEGO cubes to construct a function satisfying this ok now let us answer one more question and then we can go to direct and then we can go directly to Poncelli theorem so this is the question how to the so given cubic given elliptic curve this way or that way the goal is to describe points which satisfy the condition that so the points which satisfy the condition that now I need to think I don't want to erase now what I need later so let me think so definitely exercises I don't need so this no exercise no exercise no talk general anymore plan we don't need ok now we will see so the question now we are interested in is question so in that group structure we want to describe points such that nP zero so you sum several times point with itself and you end up at zero these points are some times called division points so can we here see these division points for example if n is equal to if n is equal to where are these division points for n equal to let me erase this so where are these points for n equal to do you see them the answer is they are there they are presented on the board this is omega one small omega one, small omega two and small omega one plus omega two they are already there so this is the solution of 2P equals zero 3P equals solutions of this we will call them flexes so the solutions are flex can we see these here those where 3P equals zero so for 2P equals zero we need to have here half half and then to sum for 3P what we need to do we need to take I think what one seventh or what one third two thirds one third, two thirds and then to produce these points this is how we can catch these points here but what we want to do is we want to somehow determine some analytic condition in terms of this equation here y squared equals polynomial of degree three in x so this is what we need to derive so suppose that nP equals zero what it means is that we can say nP is equal n zero n zero is in that presentation there is this point at infinity this is this point at infinity so we see that nP should be equal n times P at infinity and this equality means sum this in a group and this should be equal so if this is satisfied then by Abel's theorem that means there exist a metamorphic function having pole of order n at P infinity and having zero at P order n so by Abel's theorem that means there exist in infinity such that P is zero of order n this is translation of our problem using Abel's theorem so we see that this space is this one nP infinity so this is the space of functions having pole at zero of order not more than n this is space of functions this is the space of functions having so this is the space of functions metamorphic functions having pole only at zero of order not bigger than n what is the dimension of this space what is the dimension of this space so by Roch formula this is degree of this divisor so this is degree of nP infinity and this is n so take a basis take a basis f1 fn a basis in l infinity and then we can even derive a little bit more than what we need so suppose that instead of looking for one point P such that nP equal zero we can consider even more general to check if for given n different or some equal points is P1 plus Pn equal zero so take this points and so what we want to so if P1 plus plus Pn equal zero that means that there exist a function f in P infinity such that f of Pi is zero for i going from one to n but if f is in this space that means that there exist c1 cn constants such that f is equal c1f1 plus plus cnfn because if this is a element of the space it is expressed by the elements of the basis and we know so these are constants when we evaluate this at P1 to Pn we are getting zero so we are getting a system of equations we are getting a system of equations this is known basis and we evaluate in a known point so we are getting a system on unknown coefficient so what we are getting we are getting a system c1f1 of P1 plus plus cnfn of Pn of P1 is zero c1f1 of P2 plus cnfn of P2 is zero so on c1 of f1 of Pn cnfn of Pn is zero this is what we are getting so in order of such a system to have a non-trivial solution in c1 to cn we need to the condition is that this determinant calculated at P1 Pn of f1 P1 let me si f1 of P1 so c1 so what should go here what should I put here what what come on what is here what f2 of P1 very good let's see fn of P1 now is easy fn of P1 now it goes f1 is easy f1 of P2 f2 of P2 fn of P2 f1 Pn f2 Pn fn Pn should be equal to zero this is the condition so now these points P1 to Pn are going to coincide we are getting a limit case and for limit case we are getting Vronskian just to make it a little bit quicker is f1 f1 prime fn prime here is f1 n-1 fn n-1 calculated at this point P should be equal to zero so the condition that n times P is equal to zero P is division point if and only if this determinant is zero this is the condition now we can still use this picture we have had here because we can provide an explicit basis using this morphism we had here so how can we provide this basis so there are two now we need to we need to I'm just thinking if I can erase now this Abel's theorem suppose I can I need some space just for some small technical reasons we should now distinguish two cases when n is even and n is odd so suppose for first case first case if n is odd if n is odd I want to give you a specific basis or I want to ask you to give me a specific basis so what we know here is this map still here so you see p function goes to x so this is our x on the cubic side this is our y of a cubic side so what was the pole of p function 2 so x is of order 2 and y is of order 3 we don't have anything of order 1 we have x of order 2 and we have y of order 3 and we know that y square is equal polynomial of degree 3 x so if you know that x is of order 2 and y is of order 3 how can we provide the basis of this space which should be so we need to have 2m plus 1 function satisfying the condition that order at 0 is not bigger than this so first what any ideas x is of order 2 then we can go with x 2 this is order 2 so if we put here m we are still good because this is of order 2m and this is 2m plus 1 but we cannot go further with x now we can put y this is of order 3 we can put yx this is of order 5 and we can go on until here we need to be precise y times x of something what? this is where I need your help so what I can put here so y is 3 2m plus 1 minus 3 is 2m minus 2 so we can put here m minus 1 and now first of all we already have 2m plus 1 we already have 2m plus 1 we already have so we don't need to go further so this is the basis we could have for example go y square but y square is already a function of x so you cannot give us anything new ok so that means that we need to produce this sort of calculation we need to produce this sort of calculation for these particular functions and this is differentiation with respect to x so this is the differentiation with respect to x and the good thing with this differentiation with respect to x is that differentiating x on n with respect to x k times is easy so lemma d over dx j times of x of k and we are doing this at 0 even better is almost always 0 and it is we evaluated here now at 0 only at 0 it can be equal k so it is 0 if j is not equal k and if it is j equal k it is what? k factorial or j factorial if j equal k so it's good by the way and what is the now you can tell me for n equal 2m just for reference what is the basis for n equal 2m what is the basis for n equal 2m it's 1x x m y yx on what should be there m minus 1 is it m minus 2 m minus 2 ok so this is the basis in this case ok and now we need to calculate this there is no we need to do some calculation there is some tedious calculation thanks to this lemma thanks to this lemma I am now just referring to the first case this can be so here we have a here we have b 0 and here we have some c where a due to this lemma a is 1 1 2 factorial m factorial so this is m plus 1 times m plus 1 so it's a diagonal matrix it is a diagonal matrix which means that what is the consequence of the fact that this matrix is diagonal the consequence we are looking for the condition determinant equals 0 and this is a diagonal matrix with this positive coefficient so that means you see now we have to be very happy because that means we don't care about b we don't care about b we only care about c and of course c includes this now derivatives of y so if we suppose that so y square equal polynomial of degree 3 of x and now y is square root of polynomial of degree of 3 of x if we expand this in Taylor expansion plus and this Sergei say this is implicit exercise that c is of this form ok, there are some really coefficients which are not so important but let me write correct formula so and since we are talking about the condition so m plus 1 factorial is coefficient of the whole column m plus 2 factorial and so on so since we are looking for the condition when this determinant equals 0 we can forget about these coefficients and we can rewrite finally our answer in the form we can rewrite our answer in the form ok for n equal 2n plus 1 the answer is and so forget about these coefficients forget about these coefficients then we are getting we see that this formula this metric has a very specific form this is a very specific type if you just forget about these coefficients so this is so I'll write transpose of this matrix here so c prime so determinant is c2 hat c3 hat cn plus 1 hat then this one goes here c3 hat c4 hat then this one goes 1 higher m plus 2 until cm plus 1 hat then here one more cm plus 2 hat so the question is what is this one this is to check if you are still present what is this coefficient here so this is 2n the condition is this is equal to 0 so you see by the way this determinant has a very specific form what is this to be honest I am always confused because there are two types of matrices so one is Henkel matrix and another is teplet so one of these is Henkel teplet if you forget these coefficients then this one is one type and this one is another type this is Henkel that's correct so this one is Henkel and this one is teplet so how can we memorize this no way I think so this one is teplet I think that you should say the teplet is more important and then if you remember the teplet is more important then on the main diagonal is constant this is somehow I don't know other way to memorize this okay so what so this is so this one is teplet and this one is Henkel for this particular problem I think it's so this is in a sense a challenge is there a way somehow to explore in a more elaborate way why is this form appearing here and so on okay there was a question a few days ago about some polynomial entering into this game there was a question okay probably the answer was not satisfactory so personally so you see now we are looking for a specific c1 to cm that this function f what we get at some point has zero for the n so plug instead of f1 and fn the functions we have here these are some polynomials in x y times polynomial in x so when we do this here we will get polynomial in x plus y times polynomial in x so y is square root of this polynomial and what we want is in a sense we are looking for a solution which has a good approximation in certain point so this is the connection here is how these polynomials enter into a game so here we really get very specific polynomials okay now let me so this is for n equal to n plus 1 and now I can also just for the reference I should write this also for n equal to m so in particular we see that for n let me see so n equal example n equal 3 m is equal 1 what we are getting so m equal 1 we should put here m equal 1 we are getting c2 so all we have to get this is c2 equal 0 for n equal 4 c2 we should put here m equal 2 we get 3 so we are getting c3 equal 0 and for n equal 5 this is m equal 2 we are getting c2 c4 minus c3 square equal 0 so this is formulate an exercise is to you remember we had this curve yesterday and the point p 38 so can you somehow apply this machinery for this particular case so this is an exercise this is an exercise with this we somehow derived everything we needed in this section 2 and still we have some time to I think go to section 4 ok are there any questions so far ah this is no no this is going to be exercise next time so we still didn't come to connect this with we derived machinery but we still didn't so this is this is exercise yes not to derive fuchs but at least to compare ok so now we need to answering to Sergei's question now we need to see why do we need all this why did we use all this ok have you copied this formulas and everything can erase that is the point I plan to reach on the previous lesson so this is sort of this compatibility of our plans and our capability to realize the plans so now we are going to introduce one configuration this presentation now follows closely very nice paper of Griffiths and Harris of seven years ago or something I think it's 77 or 78 ah so so we are now looking this sort of configuration so given two conics B and C2 we are looking so let me at least say once again so we are thinking about now so we are thinking about two conics B and C2 and we are we want to code the situation where we have this tangent to C2 which intersect at B so tangents to C2 belong to dual of C2 so Griffiths and Harris suggest to consider this configuration configuration of lines and points such that line belongs to C2 star point belongs to B and there is one more condition that q belongs to L so this is a sort of configuration and as not exactly but similarly to what was mentioned today by Sergey here we have two natural here we have two natural involušen here in this set we have two natural involušen one involušen is so given L if this is L there is q1 and q2 there are two points and with one q1 there are two L L1 and L2 which pass to this q1 so there are two involušen associated with this set given line there are two points given point there are two tangents so the first involušen I denoted it is C2 star of L and q is L prime q meaning that L prime is another tangent C2 through and there is IB of L q this is L q prime meaning that q prime is another intersection and how can we look at this we can say this this configuration arises as alternative usage of these so what are we doing we are starting from a point and a tangent and switch the point so this is a new point with the first tangent then we keep the point but switch the tangent then we again keep the tangent and switch the point then we keep the point and switch the tangent and so on so this is how it goes so in this row of p or row of C2 star B is composition of these two involutions this is what it is now for example let me think so we can so what are the fixed points so let's think what are the fixed points of these involutions what are the fixed points of all of the first one when do we have a situation that in a given point we have only one tangent to that we have only one tangent to C2 so the fixed points of the first involution are those correspond to those q where we have only one tangent so where are those where we have only one tangent we don't see them in this picture this picture is misleading because it's too real too realistic we need to have something more complex actually it looks like this it always looks like this so now if this is B and this is C2 so we have this intersection points I just wanted to tell a joke but still we cannot ok so there are these four tangents and these four tangents are let's say L1 L1, L2, L3, L4 these are fixed points of the involution so actually we don't know what this P exactly is but we see that it can be seen as a double cover for example over B with four ramification points or it can be seen as a double cover over C2 with four ramification points depending which of these two involutions you choose on the other hand we know that conic so this is a Cp1 so we have double cover over Cp1 with four ramification points and using the principle I tried to convey to you at the beginning so this is like how it goes if it sounds like a duck if it moves like a duck and I don't know what else then it's a duck so since that is duck this is elliptic curve so using this principle we actually proved or scientifically speaking applying Riemann-Hurwitz formula we immediately see that Pc2 star B is an elliptic so let's look like this applying Riemann-Hurwitz formula we get that P of C 2 star B is elliptic curve and we know of course that elliptic curves have different appearances if they appear as cubics then involutions which have fixed points can be described by the following actually I mentioned this last time we get two points let's say T1 and T2 and we can define evolution from last time tau T1 as tau T1 of x is T1 circle x and we can define evolution similarly tau T2 of x is T2 x so this picture of having dynamics induced by two involutions in this screen can be seen in the following way we are applying two involutions all the time we are starting from some point x so tau T1 of x is this point here then we apply tau of T2 we are getting some point somewhere here applying again the first one we are going here then we from here applying this one and we are going somewhere come we are going somewhere here and so on so this is the way how we can see dynamics generated by two involutions so I will just finish with the following of course if we are looking at an elliptic curve as we did in this sense of a torus in this fundamental parallelogram that I will leave this as an exercise that so we are considering elliptic curve as a factor of a complex line by a lattice then there are two types of involution so if tau is an involution with a fixed point then tau of z is minus z plus b this is the form and if tau is an involution without fixed points so I will leave this as an exercise so if tau is an involution with fixed points then this is necessarily of the form in this presentation tau of z equals minus z plus b however there are involutions without fixed points and they have a little bit different presentation so here we have involutions here we have involutions with fixed points so in this presentation we can say that the first one Ib of z is minus z plus b and Ic2 star of z is minus z plus c2 then rho we are looking for of poncele which is composition of these two involutions is nothing but z plus c2 minus b if I'm not mistaken in the calculation so that means that these biller dynamics in this picture is going to be nothing but a shift for this element c2 minus b denoted as p and this shift doesn't depend on the choice of the point so that means if we want to have this poncele configuration to be periodic starting from some point close after n steps that means that that means that at that point we conclude that np is equal to 0 and then since np is equal to 0 if we apply this to any other point we conclude that rho n p is going to be identically equal not 0, it should be the identity identity so we are coming to this condition np equal to 0 on an elliptic curve and just 15 minutes ago we derived a machinery how to investigate if a given point is satisfying this condition or not to do that we need one more step this is to pass from given two matrices to a cubic curve to curve of the form y2 equal a polynomial of third degree in x and this curve is y2 equal polynomial t where this polynomial of third degree of t is determinant of tB plus c2 so this would close the loop once we show that this elliptic curve is isomorphic to that elliptic curve so in order to show that as I told you how we compare to elliptic curves we compare them as ramified coverings and we have these four points over which they are ramified if this cross ratio coincide then these two curves are isomorphic next time we will show even simpler thing we will show that actually we can realize this curve as branch over exactly the same B was