Decomposition of product of two Plucker coordinates












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Let $Gr(k,n)$ be the set of all $k$-dimensional subspaces of an $n$-dimensional vector space. Then $Gr(k,n)$ is a projective variety and it has Plucker coordinates $P_{i_1, ldots, i_k}$ ($i_1<ldots<i_k$) which is the determinant of the matrix $(x_{ij})_{i in [k], j in {i_1, ldots, i_k}}$. Certain Plucker coordinates satisfy the Plucker relation. For example, for $Gr(2,n)$, $P_{12}P_{34} + P_{23}P_{14}-P_{13}P_{24}=0$. Therefore $P_{13}P_{24} = P_{12}P_{34} + P_{23}P_{14}$ can be viewed as a decomposition of the product of $P_{13}$ and $P_{24}$. For $P_{12}$, $P_{34}$, we say that their product is irreducible. That is $P_{12}P_{34}$ cannot be written as a sum (with two or more terms in the summation, each summand has positive coefficient) of products of Plucker coordinates.



Given two Plucker coordinates $P_{i_1, ldots, i_k}$, $P_{j_1, ldots, j_k}$, is there some formula for the decomposition of $P_{i_1, ldots, i_k} P_{j_1, ldots, j_k} = sum_T c_T P_T$ (P_T is a product of certain Plucker coordinates, $c_T>0$) in the literature? Thank you very much.










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    Let $Gr(k,n)$ be the set of all $k$-dimensional subspaces of an $n$-dimensional vector space. Then $Gr(k,n)$ is a projective variety and it has Plucker coordinates $P_{i_1, ldots, i_k}$ ($i_1<ldots<i_k$) which is the determinant of the matrix $(x_{ij})_{i in [k], j in {i_1, ldots, i_k}}$. Certain Plucker coordinates satisfy the Plucker relation. For example, for $Gr(2,n)$, $P_{12}P_{34} + P_{23}P_{14}-P_{13}P_{24}=0$. Therefore $P_{13}P_{24} = P_{12}P_{34} + P_{23}P_{14}$ can be viewed as a decomposition of the product of $P_{13}$ and $P_{24}$. For $P_{12}$, $P_{34}$, we say that their product is irreducible. That is $P_{12}P_{34}$ cannot be written as a sum (with two or more terms in the summation, each summand has positive coefficient) of products of Plucker coordinates.



    Given two Plucker coordinates $P_{i_1, ldots, i_k}$, $P_{j_1, ldots, j_k}$, is there some formula for the decomposition of $P_{i_1, ldots, i_k} P_{j_1, ldots, j_k} = sum_T c_T P_T$ (P_T is a product of certain Plucker coordinates, $c_T>0$) in the literature? Thank you very much.










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      $begingroup$


      Let $Gr(k,n)$ be the set of all $k$-dimensional subspaces of an $n$-dimensional vector space. Then $Gr(k,n)$ is a projective variety and it has Plucker coordinates $P_{i_1, ldots, i_k}$ ($i_1<ldots<i_k$) which is the determinant of the matrix $(x_{ij})_{i in [k], j in {i_1, ldots, i_k}}$. Certain Plucker coordinates satisfy the Plucker relation. For example, for $Gr(2,n)$, $P_{12}P_{34} + P_{23}P_{14}-P_{13}P_{24}=0$. Therefore $P_{13}P_{24} = P_{12}P_{34} + P_{23}P_{14}$ can be viewed as a decomposition of the product of $P_{13}$ and $P_{24}$. For $P_{12}$, $P_{34}$, we say that their product is irreducible. That is $P_{12}P_{34}$ cannot be written as a sum (with two or more terms in the summation, each summand has positive coefficient) of products of Plucker coordinates.



      Given two Plucker coordinates $P_{i_1, ldots, i_k}$, $P_{j_1, ldots, j_k}$, is there some formula for the decomposition of $P_{i_1, ldots, i_k} P_{j_1, ldots, j_k} = sum_T c_T P_T$ (P_T is a product of certain Plucker coordinates, $c_T>0$) in the literature? Thank you very much.










      share|cite|improve this question









      $endgroup$




      Let $Gr(k,n)$ be the set of all $k$-dimensional subspaces of an $n$-dimensional vector space. Then $Gr(k,n)$ is a projective variety and it has Plucker coordinates $P_{i_1, ldots, i_k}$ ($i_1<ldots<i_k$) which is the determinant of the matrix $(x_{ij})_{i in [k], j in {i_1, ldots, i_k}}$. Certain Plucker coordinates satisfy the Plucker relation. For example, for $Gr(2,n)$, $P_{12}P_{34} + P_{23}P_{14}-P_{13}P_{24}=0$. Therefore $P_{13}P_{24} = P_{12}P_{34} + P_{23}P_{14}$ can be viewed as a decomposition of the product of $P_{13}$ and $P_{24}$. For $P_{12}$, $P_{34}$, we say that their product is irreducible. That is $P_{12}P_{34}$ cannot be written as a sum (with two or more terms in the summation, each summand has positive coefficient) of products of Plucker coordinates.



      Given two Plucker coordinates $P_{i_1, ldots, i_k}$, $P_{j_1, ldots, j_k}$, is there some formula for the decomposition of $P_{i_1, ldots, i_k} P_{j_1, ldots, j_k} = sum_T c_T P_T$ (P_T is a product of certain Plucker coordinates, $c_T>0$) in the literature? Thank you very much.







      ag.algebraic-geometry co.combinatorics






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      asked Mar 30 at 12:57









      Jianrong LiJianrong Li

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          Your exact question seems a little strange to me because, beyond the 3-term relation, more general Plücker relations will have many terms with both positive and negative signs. So for $k>2$ it is not clear that we can ever do decompositions of the type you're describing. But the question of which subsets of Plücker coordinates are algebraically independent and generate the coordinate ring of the Grassmannian, and how do we write arbitrary elements of the coordinate ring in the corresponding basis, is the beginning of the study of cluster algebras. See e.g. https://arxiv.org/abs/math/0311148.






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            $begingroup$

            Your exact question seems a little strange to me because, beyond the 3-term relation, more general Plücker relations will have many terms with both positive and negative signs. So for $k>2$ it is not clear that we can ever do decompositions of the type you're describing. But the question of which subsets of Plücker coordinates are algebraically independent and generate the coordinate ring of the Grassmannian, and how do we write arbitrary elements of the coordinate ring in the corresponding basis, is the beginning of the study of cluster algebras. See e.g. https://arxiv.org/abs/math/0311148.






            share|cite|improve this answer









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              6












              $begingroup$

              Your exact question seems a little strange to me because, beyond the 3-term relation, more general Plücker relations will have many terms with both positive and negative signs. So for $k>2$ it is not clear that we can ever do decompositions of the type you're describing. But the question of which subsets of Plücker coordinates are algebraically independent and generate the coordinate ring of the Grassmannian, and how do we write arbitrary elements of the coordinate ring in the corresponding basis, is the beginning of the study of cluster algebras. See e.g. https://arxiv.org/abs/math/0311148.






              share|cite|improve this answer









              $endgroup$
















                6












                6








                6





                $begingroup$

                Your exact question seems a little strange to me because, beyond the 3-term relation, more general Plücker relations will have many terms with both positive and negative signs. So for $k>2$ it is not clear that we can ever do decompositions of the type you're describing. But the question of which subsets of Plücker coordinates are algebraically independent and generate the coordinate ring of the Grassmannian, and how do we write arbitrary elements of the coordinate ring in the corresponding basis, is the beginning of the study of cluster algebras. See e.g. https://arxiv.org/abs/math/0311148.






                share|cite|improve this answer









                $endgroup$



                Your exact question seems a little strange to me because, beyond the 3-term relation, more general Plücker relations will have many terms with both positive and negative signs. So for $k>2$ it is not clear that we can ever do decompositions of the type you're describing. But the question of which subsets of Plücker coordinates are algebraically independent and generate the coordinate ring of the Grassmannian, and how do we write arbitrary elements of the coordinate ring in the corresponding basis, is the beginning of the study of cluster algebras. See e.g. https://arxiv.org/abs/math/0311148.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 30 at 15:23









                Sam HopkinsSam Hopkins

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                5,24212560






























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