Tropical Jacobian Conjecture
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Grigoryev, Dimitri
We prove that, for a tropical rational map if for any point the convex hull of Jacobian matrices at smooth points in a neighborhood of the point does not contain singular matrices then the map is an isomorphism. We also show that a tropical polynomial map on the plane is an isomorphism if all the Jacobians have the same sign (positive or negative). In addition, for a tropical rational map we prove that if the Jacobians have the same sign and if its preimage is a singleton at least at one regular point then the map is an isomorphism.
 Okay, so I will talk about what happened with the tropical version of the Jacobian conjecture. First, I will briefly remind the classical Jacobian conjecture. I would not dwell much on it because there are very good overviews on it including the very rich and dramatic history. Then I will say a few words about actually giving notations for the tropical settings, actually repeating some parts of what Stefan just did. And then I will tell about results on the tropical version of Jacobian conjecture. So, first of all, let me briefly remind classical Jacobian conjecture. So we have a polynomial map for the characteristic field zero and consider the Jacobian of this polynomial map. So, consisting the matrix consisting of the partial derivatives and this is the classical Jacobian conjecture due to Keller very old source of 39 that if the Jacobian equals to one then f so the polynomial is nine. And it's inverse is also a polynomial map. Well, there were a lot of results around it. I will just mention a few of them which have a flavor common with what I will talk about tropical stuff. Okay, so for an algebraically closed field f if f is injective, then f is by objective as well. So, actually, the proof due to ox was model theoretic and it's very easy in few lines, and it's a reduction to finite fields using no stellen that and if if you formulate this statement for finding fields it's trivial, because in a finite set if there is an objective map it's also a budget. So, it's a very easy, nice result. And one can actually beat a Jacobian conjecture is a local isomorphism. So due to the implicit function theorem applies the global isomorphism. And another, maybe the second and the last result in this area which I will mention is a counter example in fact, which shows that the whole issue of Jacobian conjecture is very delicate. So if you can see that the real field is a ground field and the counter example shows that assumption that just them Jacobian is positive is not enough. So it is the other counter example that we have an isomorphism with a positive with a positive Jacobian. So that that it's necessary to have the job and equals actually equals to constant so you can consider this constant to be one. Okay, that's all what I wanted to say about the classical just just to remind the classical conjecture and Now let me briefly introduce the notations about the tropical world. Okay, so the basic object of them tropical algebra is a tropical submarine, and it's in doubt with the operations, which is denoted in this way, and say, the main source of the tropical submarines is the following we take an order to semi group, and then it says it's a submarine with inhibited operations as the sum for the minimum and the product as the group semi group operation. So if say you have not just a semi group, but rather a group. It could be a billion. So, always it is ordered, then. So if it is a group, then we are talking about that tropical semi skew field. And then we can introduce also the division in the tropical semi field, which is, in fact, the subtraction in the subtraction in the group. Okay, the typical examples are say the non negative, non negative integers. All the non negative integers with the added infinity. The examples of community tropical semi rings and infinity plays a role of the neutral element. And then it's turned zero place the role of the unity. And other examples are just so they were these were the examples of the semi rings. And if we can see them, not just the non negative but the integers or the integers with the added infinity these are semi fields because they allow allow subtraction so the division division in the tropical division and an example of the non commutative tropical semi ring is the semi ring of them and by and matrices say over over any of the semi fields say over Z with infinity with the usual operation for the matrix multiplication. Another important object is the tropical polynomial. So to define it with the polynomial, which is just the product of a variable. Then a monomial. He actually it is a isn't the classical case in the coefficient tropical times the tropical monomials. So consider its tropical degree is the sum. Again, as a classical situation, the sum of the degree is the sum of the degrees and some of the powers and classically we can look at this as a linear adjust as a linear function. Well, this was a monomial and again similar to that classical case we define tropical polynomial is the tropical sub the monomial and this could be viewed as a classical as the minimum of linear functions so a convex convex function. Convex piecewise linear function and well what is so this was before was very typical and similar to classical case but what it differs that tropical geometry and what is maybe psychological difficulty usually is understanding is the concept of the tropical zero. So, as I say, and it has a lot of justifications the definitions that we say that x is a tropical zero, if say of the polynomial of the polynomial F, if the minimum is attained at for at least two different values of of J so for a two different for two different tropical So that means that in other words if we can see the tropical polynomial as a piecewise linear function, then the tropical zero is the points at which the polynomial is is not smooth. Okay, so, well, we can continue these definitions and extend them to tropical algebraic rational functions. So, the tropical fractions, and we have seen that the fraction is a subtraction in the classical sense so this minimum is a tropical polynomial and this is a, we can treat as a numerator and minimum of queues as a denominator and their subtraction. So, the second difference is the tropical tropical fraction. And this we can view as a tropical algebraic rational function and where the peas and queues are linear function with rational coefficients. So, electrically, it is a piecewise linear function. So that means that one can partition the whole space into a finite number of and dimension polyhedra on each of which this function is linear. Actually, convisally, conversely, any piecewise linear function can be represented in this form. So, as the difference of two tropical polynomials. Moreover, if you can see the any continuous function, it is known that it's a difference of two convex functions. So this is a particular case of this theorem for piecewise linear functions. So, generally, and for our considerations would be also valid. One can assume that the coefficients are real not not just rational or integers so more coefficients as it is in the tropical world, but we can consider the coefficients to be real so just consider any piecewise piecewise linear function and the results would be true. So, the question arises how to how to replace the Jacobin for non smooth tropical algebraic say rational maps. So we have now this will be our main object at tropical algebraic rational map so each coordinate is a tropical is a tropical algebraic rational function. So, how to how to replace the Jacobin in the same Jacobin conjecture. Well, actually, unlike unlike the classical situation. Our situation is a little bit well not a little bit essentially easier, because we need to prove only that the map, the map is invertible. So the inverse does exist, because if it does exist, then it is also a tropical map. So it is a piecewise piecewise linear function. And so it can be represented in the same way. So we need to test. So, actually one can view the Jacobin conjecture is the criterion for a map to be in isomorphism. Okay. What will do. First, I would be no unique version of the tropical Jacobin conjecture but there would be a one week and one strong version. So we start with the week version. And first, we need the following definition. Okay, consider it tropical map and the point. And the old. Demand and dimensional polyhedral containing P on which if it is linear so we know that we it is a tropical map it is piecewise linear so we take these pieces on which F is F is linear. And for each for each of these for each of these linear maps with which are which are now Jacobin matrices simultaneously. We denote them a one a K so in the neighborhood in the so that in the neighborhood of the point P. So we assume that P say is in the boundary of say K, polyhedral on which the map is linear. And we can see this. And then we take the codes. So the determinants of these matrices. And we consider the convex hull of these matrices and denote them by DP of F. So the first proposition which is which what I call them a week, the week version of the tropical Jacobin conjecture that if for each for each point P. If it doesn't contain a singular matrix. Then F is an isomorphism. So this is a, this DP replaces the role of the Jacobin. So we assume that that it doesn't contain a singular matrix that's the, that's the assumption. And then we state that it's in isomorphism, actually, the, I can give the proof because it is, it is in the goals in few lines. It relies essentially on them. Clarks a theorem that F actually Clarks theorem holds for any leapshitz map is under this under this condition. So then each DP doesn't contain a singular matrix, then it's a local homeomorphism so it's a local theorem better to say that if DP for a given P doesn't contain a singular matrix then it's a local homeomorphism. And this is true more over for not only for piecewise linear but also for leapshitz maps. We use the easy observation that tropical map is a proper map so the pre-image of every compact is again compact. And this implies that So we know already that it's a local homeomorphism and that then because it's proper then it's a global global homeomorphism so that's the whole proof. Okay, unfortunately, so we have a sufficient condition for a map to be nice amorphism, but unfortunately it is not necessary and I will give a counter example for that which is also quite instructive. Okay, so we can see the tropical map on the plane. It would be. It would be isomorphism, and it's a composition of a lower triangle and upper triangle isomorphism. Okay, so this is the lower triangle and this is the triangle. If we consider their composition, then it is a, well, it's a, it's a linear in piecewise linear in four pieces. And on so on four sectors on four sectors of the origin. Like on the picture, and if we can see the D at the origin, then it's a convex hull of the four following Jacobian matrices. Well, just for from the, from the formula for the lower triangle and upper triangle isomorphisms. And this, so if we take the sum of the second and the third matrices, we see that it is, it is singular when, so when the following condition holds. So when one of the either beta equals to alpha or B equals to a. Okay. Oh, sorry. When so when the product when the product is equals to four. So for example, one can take alpha and a equals to zero and beta, beta equals to two. And so this is the counter example which shows that this sufficient condition is not necessary. And it. Okay, still it would be nice to have a necessary sufficient condition. So we start with the following is the remark. A tropical map is a nice amorphism then all the Jacobians have the same sign. So either all of them are positive or negative. This is due to due to orientation and the degree on the degree of the map equals equals one. The question arises when this condition is sufficient so when the condition of the constancy of the science of of the Jacobians is sufficient to be to be in isomorphism. And it's true on the plane when we can see the tropical polynomial map so that both F1 and the twat tropical polynomials. That means so they're convex. And then indeed that if all the Jacobians say are positive, then F is an isomorphism, but beyond beyond this conditions on the plane and for the tropical polynomials. This can sufficient, sufficiency condition is is not necessary condition is not sufficient, unfortunately. And the following example shows that okay consider the following now we consider noted polynomial map tropical polynomial but but rather tropical rational map. So we write here module function and clearly module function is a is a tropical tropical rational function you can easily write it with the with using the subtraction. And we consider the following tropical rational map. It has one can easily verify that it has positive Jacobians in all its linear pieces in all the pieces where the function the map is linear. But because this function is central symmetric. It is not an isomorphism. So we can modify slightly modify this example to get a now a tropical polynomial map rather than rational map but in the three dimensional space with all positive with all positive Jacobians. And again being again being not an isomorphism. So we see that we have a now necessary condition of positivity say of all the Jacobians but it's not. It's not sorry necessary condition but it's not sufficient. So this one can formulate a one can formulate a nested and efficient condition but it's now it looks not not so natural but it's on the other hand it's good to have an algorithm to verify whether tropical map is an isomorphism. So make sure you remind that actually this is this definition is holds for any for any for any map. So we say that the point is regular if for any. Point in the in the target. If for any pre image from from for from this point is Jacobian is not zero. Okay, so this then this point is regular and then by the set of regular points. So the value is dense. Okay, and then we can simulate now at simultaneously a necessary sufficient condition for a tropical map to be an isomorphism. And namely that means that the condition from the previous slide that all the Jacobians have the same sign. And now we require that at least for one regular value. The pre image is unique. Okay, well, the necessity is trivial but the statement of the theorem and this this condition is sufficient also. If such a point does exist, at least one point then the whole tropical map is an isomorphism and relying on this theorem, we can now design an algorithm to verify whether a tropical, whether a tropical map is an isomorphism. Namely, an algorithm yields a partition into polyhedral such that F is linear on each PI. This can be done. This can be done by means of linear programming. Okay, then if we take any point in the target, which is out of the union of the boundaries, boundaries of this polyhedral. So we take the boundaries of this polyhedral which are polyhedral of one less dimension and take the images and subtract them, then any point out of this out of this union is regular. So we have the criterion from the previous theorem. Test that the pre image is unique of this point. And if it is the case, then we know from the previous theorem then we dealing with isomorphism. All, all this can be performed have an algorithm which tests whether a tropical map is in fact an isomorphism. Okay, and another issue which is related to the Jacobian conjecture is the timeness of the of the automorphisms. And it is a classical Dixmere problem and what do we have in the tropical world in the tropical world. Similar again to the classical, but setting but actually we cannot cannot we cannot use the statement of the classical in the setting but we can we can prove this independently that on the plane indeed, indeed any automorphism is staying. So we define two classes, two classes of automorphism. The first class is triangle. Actually, we had already an example before. So triangle triangle that means with the change we change the triangle we change one only one coordinate at that time. And also, we consider an analog of linear tropical rational automorphisms. So these are automorphism, all with linear and with a determinant with a determinant equals plus or minus one. And, okay, so the proposition states that the group, the group of tropical rational homogeneous automorphisms. So on the on the plane is generated generated by triangle and linear linear automorphism so it is staying. In classic in the classical world we know that in the free three dimensional case. The group of automorphism is not tame so it is not generated by triangle and linear automorphisms. Here. This is an here this is an open question. And my conjecture is that this is indeed the case the group is is also not the. Okay, thank you. Thank you very much for for attention. Thank you. So, I have a question for the map. I have a question with respect to last proposition. So is it related that if we have convex function that we can approximate it by the by the sum of this triangular function. No, I don't think so because this is not approximation this is an exact statement you you have a automorphism is and you need to represent it as a product of triangle and linear linear automorphisms. Yeah, but but in some sense of the same ingredients you have triangular and linear. Yeah, in one approximate as a sum but here you have exact. And then in multivariate case you have to replace triangular by simplex's. Yes, yes, yes, well you can approximate but here it is it is it is an exact it is an exact it is an exact equality of automorphisms. It's different. So if you have a function at piecewise linear. Yeah, so therefore, for piecewise linear it would be exact. So if you have, so if you have just convex function. Yeah. Then for convex function you have approximation but if you have linear then it will be exact approximation by the sum of linear and angles. But people have this lemma by it's called hardy little wood and someone so it is funny funny so great people with such funny lemma. And so here they have piecewise linear then it will be exact. Just obtain this piecewise linear function as a sum. I don't know. Well, you mean the sum but the sum. Well, look, the sum. No, I don't I, I'm not sure. Well, I agree that it's related but I'm not sure that it will give the exact result. Well, it's a good point I will think but I don't think that I think that is I think it is just the opposite the statement is. I mean, it is not tame. Okay, but it's a good point I'll think about the connection. Because I just see the same ingredients. So therefore, yeah, and gradients are the same but it's a different statement. Yeah, and then it was supposed to do with sepulxis and then it was supposed to estimate but but I think there is a problem so no not only such triangle, you have maybe allow some some rotation. But the exact triangles it's not true in the for the for general. And then you have, I have got the exact statement but but I can found the reference. Oh yeah, okay, let's send me then the link. Yeah, that would be interesting. Thank you for the nice talk. Oh, there are no patients so we stop here we have a price until 1150. Thank you very much. Thank you. Thank you for the interaction and so.
 Okay, so I will talk about what happened with the tropical version of the Jacobian conjecture.  First, I will briefly remind the classical Jacobian conjecture.  I would not dwell much on it because there are very good overviews on it including the very rich and dramatic history.  Then I will say a few words about actually giving notations for the tropical settings, actually repeating some parts of what Stefan just did.  And then I will tell about results on the tropical version of Jacobian conjecture.  So, first of all, let me briefly remind classical Jacobian conjecture.  So we have a polynomial map for the characteristic field zero and consider the Jacobian of this polynomial map.  So, consisting the matrix consisting of the partial derivatives and this is the classical Jacobian conjecture due to Keller very old source of 39 that if the Jacobian equals to one then f so the polynomial is nine.  And it's inverse is also a polynomial map.  Well, there were a lot of results around it. I will just mention a few of them which have a flavor common with what I will talk about tropical stuff.  Okay, so for an algebraically closed field f if f is injective, then f is by objective as well.  So, actually, the proof due to ox was model theoretic and it's very easy in few lines, and it's a reduction to finite fields using no stellen that and if if you formulate this statement for finding fields it's trivial, because in a finite set if there is an objective map it's also a budget.  So, it's a very easy, nice result.  And one can actually beat a Jacobian conjecture is a local isomorphism. So due to the implicit function theorem applies the global isomorphism.  And another, maybe the second and the last result in this area which I will mention is a counter example in fact, which shows that the whole issue of Jacobian conjecture is very delicate.  So if you can see that the real field is a ground field and the counter example shows that assumption that just them Jacobian is positive is not enough.  So it is the other counter example that we have an isomorphism with a positive with a positive Jacobian.  So that that it's necessary to have the job and equals actually equals to constant so you can consider this constant to be one. Okay, that's all what I wanted to say about the classical just just to remind the classical conjecture and  Now let me briefly introduce the notations about the tropical world. Okay, so the basic object of them tropical algebra is a tropical submarine, and it's in doubt with the operations, which is denoted in this way, and say,  the main source of the tropical submarines is the following we take an order to semi group, and then it says it's a submarine with inhibited operations as the sum for the minimum and the product as the group semi group operation.  So if say you have not just a semi group, but rather a group. It could be a billion. So, always it is ordered, then. So if it is a group, then we are talking about that tropical semi skew field.  And then we can introduce also the division in the tropical semi field, which is, in fact, the subtraction in the subtraction in the group.  Okay, the typical examples are say the non negative, non negative integers.  All the non negative integers with the added infinity.  The examples of community tropical semi rings and infinity plays a role of the neutral element. And then it's turned zero place the role of the unity.  And other examples are just so they were these were the examples of the semi rings. And if we can see them, not just the non negative but the integers or the integers with the added infinity these are semi fields because they allow allow subtraction so the division division in the  tropical division and an example of the non commutative tropical semi ring is the semi ring of them and by and matrices say over over any of the semi fields say over Z with infinity with the usual operation for the matrix multiplication.  Another important object is the tropical polynomial. So to define it with the polynomial, which is just the product of a variable.  Then a monomial. He actually it is a isn't the classical case in the coefficient tropical times the tropical monomials.  So consider its tropical degree is the sum. Again, as a classical situation, the sum of the degree is the sum of the degrees and some of the powers and classically we can look at this as a linear adjust as a linear function.  Well, this was a monomial and again similar to that classical case we define tropical polynomial is the tropical sub the monomial and this could be viewed as a classical as the minimum of linear functions so a convex convex function.  Convex piecewise linear function and well what is so this was before was very typical and similar to classical case but what it differs that tropical geometry and what is maybe psychological difficulty usually is understanding is the concept of the tropical zero.  So, as I say, and it has a lot of justifications the definitions that we say that x is a tropical zero, if say of the polynomial of the polynomial F, if the minimum is attained at for at least two different values of of J so for a two different for two different tropical  So that means that in other words if we can see the tropical polynomial as a piecewise linear function, then the tropical zero is the points at which the polynomial is is not smooth.  Okay, so, well, we can continue these definitions and extend them to tropical algebraic rational functions.  So, the tropical fractions, and we have seen that the fraction is a subtraction in the classical sense so this minimum is a tropical polynomial and this is a, we can treat as a numerator and minimum of queues as a denominator and their subtraction.  So, the second difference is the tropical tropical fraction. And this we can view as a tropical algebraic rational function and where the peas and queues are linear function with rational coefficients.  So, electrically, it is a piecewise linear function. So that means that one can partition the whole space into a finite number of and dimension polyhedra on each of which this function is linear.  Actually, convisally, conversely, any piecewise linear function can be represented in this form.  So, as the difference of two tropical polynomials.  Moreover, if you can see the any continuous function, it is known that it's a difference of two convex functions.  So this is a particular case of this theorem for piecewise linear functions.  So, generally, and for our considerations would be also valid.  One can assume that the coefficients are real not not just rational or integers so more coefficients as it is in the tropical world, but we can consider the coefficients to be real so just consider any piecewise piecewise linear function and the results would be true.  So, the question arises how to how to replace the Jacobin for non smooth tropical algebraic say rational maps.  So we have now this will be our main object at tropical algebraic rational map so each coordinate is a tropical is a tropical algebraic rational function.  So, how to how to replace the Jacobin in the same Jacobin conjecture.  Well, actually, unlike unlike the classical situation.  Our situation is a little bit well not a little bit essentially easier, because we need to prove only that the map, the map is invertible.  So the inverse does exist, because if it does exist, then it is also a tropical map.  So it is a piecewise piecewise linear function.  And so it can be represented in the same way.  So we need to test.  So, actually one can view the Jacobin conjecture is the criterion for a map to be in isomorphism.  Okay.  What will do.  First, I would be no unique version of the tropical Jacobin conjecture but there would be a one week and one strong version.  So we start with the week version.  And first, we need the following definition.  Okay, consider it tropical map and the point.  And the old.  Demand and dimensional polyhedral containing P on which if it is linear so we know that we it is a tropical map it is piecewise linear so we take these pieces on which F is F is linear.  And for each for each of these for each of these linear maps with which are which are now Jacobin matrices simultaneously.  We denote them a one a K so in the neighborhood in the so that in the neighborhood of the point P.  So we assume that P say is in the boundary of say K, polyhedral on which the map is linear.  And we can see this.  And then we take the codes.  So the determinants of these matrices.  And we consider the convex hull of these matrices and denote them by DP of F.  So the first proposition which is which what I call them a week, the week version of the tropical Jacobin conjecture that if for each for each point P.  If it doesn't contain a singular matrix.  Then F is an isomorphism.  So this is a, this DP replaces the role of the Jacobin.  So we assume that that it doesn't contain a singular matrix that's the, that's the assumption.  And then we state that it's in isomorphism, actually, the, I can give the proof because it is, it is in the goals in few lines.  It relies essentially on them.  Clarks a theorem that F actually Clarks theorem holds for any leapshitz map is under this under this condition.  So then each DP doesn't contain a singular matrix, then it's a local homeomorphism so it's a local theorem better to say that if DP for a given P doesn't contain a singular matrix then it's a local homeomorphism.  And this is true more over for not only for piecewise linear but also for leapshitz maps.  We use the easy observation that tropical map is a proper map so the pre-image of every compact is again compact.  And this implies that  So we know already that it's a local homeomorphism and that then because it's proper then it's a global global homeomorphism so that's the whole proof.  Okay, unfortunately, so we have a sufficient condition for a map to be nice amorphism, but unfortunately it is not necessary and I will give a counter example for that which is also quite instructive.  Okay, so we can see the tropical map on the plane.  It would be.  It would be isomorphism, and it's a composition of a lower triangle and upper triangle isomorphism.  Okay, so this is the lower triangle and this is the triangle.  If we consider their composition, then it is a, well, it's a, it's a linear in piecewise linear in four pieces.  And on so on four sectors on four sectors of the origin.  Like on the picture, and if we can see the D at the origin, then it's a convex hull of the four following Jacobian matrices.  Well, just for from the, from the formula for the lower triangle and upper triangle isomorphisms.  And this, so if we take the sum of the second and the third matrices, we see that it is, it is singular when, so when the following condition holds.  So when one of the either beta equals to alpha or B equals to a.  Okay.  Oh, sorry.  When so when the product when the product is equals to four.  So for example, one can take alpha and a equals to zero and beta, beta equals to two.  And so this is the counter example which shows that this sufficient condition is not necessary.  And it.  Okay, still it would be nice to have a necessary sufficient condition.  So we start with the following is the remark.  A tropical map is a nice amorphism then all the Jacobians have the same sign.  So either all of them are positive or negative.  This is due to due to orientation and the degree on the degree of the map equals equals one.  The question arises when this condition is sufficient so when the condition of the constancy of the science of of the Jacobians is sufficient to be to be in isomorphism.  And it's true on the plane when we can see the tropical polynomial map so that both F1 and the twat tropical polynomials. That means so they're convex.  And then indeed that if all the Jacobians say are positive, then F is an isomorphism, but beyond beyond this conditions on the plane and for the tropical polynomials.  This can sufficient, sufficiency condition is is not necessary condition is not sufficient, unfortunately.  And the following example shows that okay consider the following now we consider noted polynomial map tropical polynomial but but rather tropical rational map.  So we write here module function and clearly module function is a is a tropical tropical rational function you can easily write it with the with using the subtraction.  And we consider the following tropical rational map.  It has one can easily verify that it has positive Jacobians in all its linear pieces in all the pieces where the function the map is linear.  But because this function is central symmetric.  It is not an isomorphism.  So we can modify slightly modify this example to get a now a tropical polynomial map rather than rational map but in the three dimensional space with all positive with all positive Jacobians.  And again being again being not an isomorphism. So we see that we have a now necessary condition of positivity say of all the Jacobians but it's not.  It's not sorry necessary condition but it's not sufficient.  So this one can formulate a one can formulate a nested and efficient condition but it's now it looks not not so natural but it's on the other hand it's good to have an algorithm to verify whether tropical map is an isomorphism.  So make sure you remind that actually this is this definition is holds for any for any for any map. So we say that the point is regular if for any.  Point in the in the target. If for any pre image from from for from this point is Jacobian is not zero. Okay, so this then this point is regular and then by the set of regular points.  So the value is dense. Okay, and then we can simulate now at simultaneously a necessary sufficient condition for a tropical map to be an isomorphism.  And namely that means that the condition from the previous slide that all the Jacobians have the same sign. And now we require that at least for one regular value.  The pre image is unique. Okay, well, the necessity is trivial but the statement of the theorem and this this condition is sufficient also.  If such a point does exist, at least one point then the whole tropical map is an isomorphism and relying on this theorem, we can now design an algorithm to verify whether a tropical, whether a tropical map is an isomorphism.  Namely, an algorithm yields a partition into polyhedral such that F is linear on each PI.  This can be done. This can be done by means of linear programming.  Okay, then if we take any point in the target, which is out of the union of the boundaries, boundaries of this polyhedral.  So we take the boundaries of this polyhedral which are polyhedral of one less dimension and take the images and subtract them, then any point out of this out of this union is regular.  So we have the criterion from the previous theorem.  Test that the pre image is unique of this point.  And if it is the case, then we know from the previous theorem then we dealing with isomorphism.  All, all this can be performed have an algorithm which tests whether a tropical map is in fact an isomorphism.  Okay, and another issue which is related to the Jacobian conjecture is the timeness of the of the automorphisms.  And it is a classical Dixmere problem and what do we have in the tropical world in the tropical world.  Similar again to the classical, but setting but actually we cannot cannot we cannot use the statement of the classical in the setting but we can we can prove this independently that on the plane indeed, indeed any automorphism is staying.  So we define two classes, two classes of automorphism. The first class is triangle. Actually, we had already an example before.  So triangle triangle that means with the change we change the triangle we change one only one coordinate at that time.  And also, we consider an analog of linear tropical rational automorphisms.  So these are automorphism, all with linear and with a determinant with a determinant equals plus or minus one.  And, okay, so the proposition states that the group, the group of tropical rational homogeneous automorphisms. So on the on the plane is generated generated by triangle and linear linear automorphism so it is staying.  In classic in the classical world we know that in the free three dimensional case.  The group of automorphism is not tame so it is not generated by triangle and linear automorphisms.  Here.  This is an here this is an open question. And my conjecture is that this is indeed the case the group is is also not the.  Okay, thank you. Thank you very much for for attention.  Thank you.  So, I have a question for the map.  I have a question with respect to last proposition. So is it related that if we have convex function that we can approximate it by the by the sum of this triangular function.  No, I don't think so because this is not approximation this is an exact statement you you have a automorphism is and you need to represent it as a product of triangle and linear linear automorphisms.  Yeah, but but in some sense of the same ingredients you have triangular and linear.  Yeah, in one approximate as a sum but here you have exact.  And then in multivariate case you have to replace triangular by simplex's.  Yes, yes, yes, well you can approximate but here it is it is it is an exact it is an exact it is an exact equality of automorphisms.  It's different.  So if you have a function at piecewise linear. Yeah, so therefore, for piecewise linear it would be exact.  So if you have, so if you have just convex function. Yeah.  Then for convex function you have approximation but if you have linear then it will be exact approximation by the sum of linear and angles.  But people have this lemma by it's called hardy little wood and someone so it is funny funny so great people with such funny lemma.  And so here they have piecewise linear then it will be exact.  Just obtain this piecewise linear function as a sum.  I don't know.  Well, you mean the sum but the sum. Well, look, the sum. No, I don't I, I'm not sure. Well, I agree that it's related but I'm not sure that it will give the exact result.  Well, it's a good point I will think but I don't think that I think that is I think it is just the opposite the statement is.  I mean, it is not tame. Okay, but it's a good point I'll think about the connection.  Because I just see the same ingredients. So therefore, yeah, and gradients are the same but it's a different statement.  Yeah, and then it was supposed to do with sepulxis and then it was supposed to estimate but but I think there is a problem so no not only such triangle, you have maybe allow some some rotation.  But the exact triangles it's not true in the for the for general.  And then you have, I have got the exact statement but but I can found the reference.  Oh yeah, okay, let's send me then the link. Yeah, that would be interesting.  Thank you for the nice talk.  Oh, there are no patients so we stop here we have a price until 1150.  Thank you very much. Thank you.  Thank you for the interaction and so.
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