Definition of Tropical Hypersurface












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Given the tropical semiring $(mathbb{T},oplus,otimes)$, a tropical hypersurface associated to a tropical polynomial is the set of points where it is non-differentiable.



I'm wondering how incidental the non-differentiable part is, and how this part is derived. For instance, had we given a different semiring $S$, would there be a canonical way of associating to $S$-polynomials $S$-hypersurfaces? I imagine this might have something to do with Universal Algebra, of which I am ignorant, so it would be good to get a light answer on this as much as possible, and if not possible a light reference would be greatly appreciated!










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    0












    $begingroup$


    Given the tropical semiring $(mathbb{T},oplus,otimes)$, a tropical hypersurface associated to a tropical polynomial is the set of points where it is non-differentiable.



    I'm wondering how incidental the non-differentiable part is, and how this part is derived. For instance, had we given a different semiring $S$, would there be a canonical way of associating to $S$-polynomials $S$-hypersurfaces? I imagine this might have something to do with Universal Algebra, of which I am ignorant, so it would be good to get a light answer on this as much as possible, and if not possible a light reference would be greatly appreciated!










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Given the tropical semiring $(mathbb{T},oplus,otimes)$, a tropical hypersurface associated to a tropical polynomial is the set of points where it is non-differentiable.



      I'm wondering how incidental the non-differentiable part is, and how this part is derived. For instance, had we given a different semiring $S$, would there be a canonical way of associating to $S$-polynomials $S$-hypersurfaces? I imagine this might have something to do with Universal Algebra, of which I am ignorant, so it would be good to get a light answer on this as much as possible, and if not possible a light reference would be greatly appreciated!










      share|cite|improve this question









      $endgroup$




      Given the tropical semiring $(mathbb{T},oplus,otimes)$, a tropical hypersurface associated to a tropical polynomial is the set of points where it is non-differentiable.



      I'm wondering how incidental the non-differentiable part is, and how this part is derived. For instance, had we given a different semiring $S$, would there be a canonical way of associating to $S$-polynomials $S$-hypersurfaces? I imagine this might have something to do with Universal Algebra, of which I am ignorant, so it would be good to get a light answer on this as much as possible, and if not possible a light reference would be greatly appreciated!







      algebraic-geometry universal-algebra tropical-geometry






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      asked Dec 2 '18 at 16:12









      Andrew WhelanAndrew Whelan

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          $begingroup$

          On not initially receiving an answer, I did some further reading on this and found a lovely light exposition: https://www.mathenjeans.fr/sites/default/files/documents/Sujets2012/tropical_geometry_-_casagrande.pdf



          The idea is to construct the tropical semiring $(mathbb{T},oplus,otimes)$ in analogy to the reals via closure of operations from the more primitive $(mathbb{N},oplus,otimes)$. What a root is turns out to be a pivotal concept. So the root of an $mathbb{R}$-monomial $P(x) = ax+b$ is a number $x_0$ for which $P(x_0)=0$, the additive identity. If we try transferring this over naively, we get the following:
          $$ begin{align} a otimes x oplus b &= -infty \
          iff max(a+x,b) &= - infty, end{align} $$

          which is not analogously informative. In the above paper, they provide an analogue of the fundamental theorem of algebra, and give a justification for the 'non-differentiability' aspect as a consequence of wanting our $mathbb{T}$-roots to give us the information on the factorisation implied by the fundamental theorem.



          So the associated hypersurface of a tropical polynomial is its set of tropical roots as defined above. Very happy to accept answers that might go into more detail or on anything I missed here!






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            $begingroup$

            On not initially receiving an answer, I did some further reading on this and found a lovely light exposition: https://www.mathenjeans.fr/sites/default/files/documents/Sujets2012/tropical_geometry_-_casagrande.pdf



            The idea is to construct the tropical semiring $(mathbb{T},oplus,otimes)$ in analogy to the reals via closure of operations from the more primitive $(mathbb{N},oplus,otimes)$. What a root is turns out to be a pivotal concept. So the root of an $mathbb{R}$-monomial $P(x) = ax+b$ is a number $x_0$ for which $P(x_0)=0$, the additive identity. If we try transferring this over naively, we get the following:
            $$ begin{align} a otimes x oplus b &= -infty \
            iff max(a+x,b) &= - infty, end{align} $$

            which is not analogously informative. In the above paper, they provide an analogue of the fundamental theorem of algebra, and give a justification for the 'non-differentiability' aspect as a consequence of wanting our $mathbb{T}$-roots to give us the information on the factorisation implied by the fundamental theorem.



            So the associated hypersurface of a tropical polynomial is its set of tropical roots as defined above. Very happy to accept answers that might go into more detail or on anything I missed here!






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              On not initially receiving an answer, I did some further reading on this and found a lovely light exposition: https://www.mathenjeans.fr/sites/default/files/documents/Sujets2012/tropical_geometry_-_casagrande.pdf



              The idea is to construct the tropical semiring $(mathbb{T},oplus,otimes)$ in analogy to the reals via closure of operations from the more primitive $(mathbb{N},oplus,otimes)$. What a root is turns out to be a pivotal concept. So the root of an $mathbb{R}$-monomial $P(x) = ax+b$ is a number $x_0$ for which $P(x_0)=0$, the additive identity. If we try transferring this over naively, we get the following:
              $$ begin{align} a otimes x oplus b &= -infty \
              iff max(a+x,b) &= - infty, end{align} $$

              which is not analogously informative. In the above paper, they provide an analogue of the fundamental theorem of algebra, and give a justification for the 'non-differentiability' aspect as a consequence of wanting our $mathbb{T}$-roots to give us the information on the factorisation implied by the fundamental theorem.



              So the associated hypersurface of a tropical polynomial is its set of tropical roots as defined above. Very happy to accept answers that might go into more detail or on anything I missed here!






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                On not initially receiving an answer, I did some further reading on this and found a lovely light exposition: https://www.mathenjeans.fr/sites/default/files/documents/Sujets2012/tropical_geometry_-_casagrande.pdf



                The idea is to construct the tropical semiring $(mathbb{T},oplus,otimes)$ in analogy to the reals via closure of operations from the more primitive $(mathbb{N},oplus,otimes)$. What a root is turns out to be a pivotal concept. So the root of an $mathbb{R}$-monomial $P(x) = ax+b$ is a number $x_0$ for which $P(x_0)=0$, the additive identity. If we try transferring this over naively, we get the following:
                $$ begin{align} a otimes x oplus b &= -infty \
                iff max(a+x,b) &= - infty, end{align} $$

                which is not analogously informative. In the above paper, they provide an analogue of the fundamental theorem of algebra, and give a justification for the 'non-differentiability' aspect as a consequence of wanting our $mathbb{T}$-roots to give us the information on the factorisation implied by the fundamental theorem.



                So the associated hypersurface of a tropical polynomial is its set of tropical roots as defined above. Very happy to accept answers that might go into more detail or on anything I missed here!






                share|cite|improve this answer











                $endgroup$



                On not initially receiving an answer, I did some further reading on this and found a lovely light exposition: https://www.mathenjeans.fr/sites/default/files/documents/Sujets2012/tropical_geometry_-_casagrande.pdf



                The idea is to construct the tropical semiring $(mathbb{T},oplus,otimes)$ in analogy to the reals via closure of operations from the more primitive $(mathbb{N},oplus,otimes)$. What a root is turns out to be a pivotal concept. So the root of an $mathbb{R}$-monomial $P(x) = ax+b$ is a number $x_0$ for which $P(x_0)=0$, the additive identity. If we try transferring this over naively, we get the following:
                $$ begin{align} a otimes x oplus b &= -infty \
                iff max(a+x,b) &= - infty, end{align} $$

                which is not analogously informative. In the above paper, they provide an analogue of the fundamental theorem of algebra, and give a justification for the 'non-differentiability' aspect as a consequence of wanting our $mathbb{T}$-roots to give us the information on the factorisation implied by the fundamental theorem.



                So the associated hypersurface of a tropical polynomial is its set of tropical roots as defined above. Very happy to accept answers that might go into more detail or on anything I missed here!







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 4 '18 at 0:17

























                answered Dec 4 '18 at 0:03









                Andrew WhelanAndrew Whelan

                1,363415




                1,363415






























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