Understanding homomorphism from coalgebra to algebra












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Could someone please explain what exactly a homomorphism from coalgebra to algebra (from this paper: 1, page 10, definition 5.1). I understand a homomorphism as a map between two structures which preserves operations and their neutral elements, but which operations would it preserve between coalgebra and algebra? Thank you.






hopf-algebras







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  • $begingroup$
    Usually this is done if the domain and range are bialgebras, so they have both a product and a coproduct.
    $endgroup$
    – Matt Samuel
    Jan 6 at 15:46










  • $begingroup$
    Maybe linear map from coalgebra to algebra makes more sense, but I still don't understand how it can be defined. Found it there: maths.qmul.ac.uk/~whitty/LSBU/MathsStudyGroup/SeligHopf.pdf , on page 5, section 4 (in the beginning).
    $endgroup$
    – Ordev Agens
    Jan 7 at 18:31








  • 1




    $begingroup$
    Well it defines it explicitly with a formula. And it's exactly as I said: it's a map between bialgebras (specifically here a bialgebra with itself). In terms of linear maps, you can define a nontrivial linear map between any two nontrivial vector spaces over the same field. It's not a homomorphism, it's just a composition of particular important functions here, including the product and the coproduct.
    $endgroup$
    – Matt Samuel
    Jan 7 at 19:22










  • $begingroup$
    Thank you for explaining. So a linear map from coalgebra ($A$, $mu$, $nu$) to algebra ($A$, $Delta$, $epsilon$) is just a linear map $Ato A$?
    $endgroup$
    – Ordev Agens
    Jan 7 at 19:47






  • 1




    $begingroup$
    Yes it is. Comment too short.
    $endgroup$
    – Matt Samuel
    Jan 7 at 20:36
















0












$begingroup$


Could someone please explain what exactly a homomorphism from coalgebra to algebra (from this paper: 1, page 10, definition 5.1). I understand a homomorphism as a map between two structures which preserves operations and their neutral elements, but which operations would it preserve between coalgebra and algebra? Thank you.






hopf-algebras







share|cite|improve this question









$endgroup$












  • $begingroup$
    Usually this is done if the domain and range are bialgebras, so they have both a product and a coproduct.
    $endgroup$
    – Matt Samuel
    Jan 6 at 15:46










  • $begingroup$
    Maybe linear map from coalgebra to algebra makes more sense, but I still don't understand how it can be defined. Found it there: maths.qmul.ac.uk/~whitty/LSBU/MathsStudyGroup/SeligHopf.pdf , on page 5, section 4 (in the beginning).
    $endgroup$
    – Ordev Agens
    Jan 7 at 18:31








  • 1




    $begingroup$
    Well it defines it explicitly with a formula. And it's exactly as I said: it's a map between bialgebras (specifically here a bialgebra with itself). In terms of linear maps, you can define a nontrivial linear map between any two nontrivial vector spaces over the same field. It's not a homomorphism, it's just a composition of particular important functions here, including the product and the coproduct.
    $endgroup$
    – Matt Samuel
    Jan 7 at 19:22










  • $begingroup$
    Thank you for explaining. So a linear map from coalgebra ($A$, $mu$, $nu$) to algebra ($A$, $Delta$, $epsilon$) is just a linear map $Ato A$?
    $endgroup$
    – Ordev Agens
    Jan 7 at 19:47






  • 1




    $begingroup$
    Yes it is. Comment too short.
    $endgroup$
    – Matt Samuel
    Jan 7 at 20:36














0












0








0





$begingroup$


Could someone please explain what exactly a homomorphism from coalgebra to algebra (from this paper: 1, page 10, definition 5.1). I understand a homomorphism as a map between two structures which preserves operations and their neutral elements, but which operations would it preserve between coalgebra and algebra? Thank you.






hopf-algebras







share|cite|improve this question









$endgroup$




Could someone please explain what exactly a homomorphism from coalgebra to algebra (from this paper: 1, page 10, definition 5.1). I understand a homomorphism as a map between two structures which preserves operations and their neutral elements, but which operations would it preserve between coalgebra and algebra? Thank you.






hopf-algebras




hopf-algebras






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 6 at 13:02









Ordev AgensOrdev Agens

205




205












  • $begingroup$
    Usually this is done if the domain and range are bialgebras, so they have both a product and a coproduct.
    $endgroup$
    – Matt Samuel
    Jan 6 at 15:46










  • $begingroup$
    Maybe linear map from coalgebra to algebra makes more sense, but I still don't understand how it can be defined. Found it there: maths.qmul.ac.uk/~whitty/LSBU/MathsStudyGroup/SeligHopf.pdf , on page 5, section 4 (in the beginning).
    $endgroup$
    – Ordev Agens
    Jan 7 at 18:31








  • 1




    $begingroup$
    Well it defines it explicitly with a formula. And it's exactly as I said: it's a map between bialgebras (specifically here a bialgebra with itself). In terms of linear maps, you can define a nontrivial linear map between any two nontrivial vector spaces over the same field. It's not a homomorphism, it's just a composition of particular important functions here, including the product and the coproduct.
    $endgroup$
    – Matt Samuel
    Jan 7 at 19:22










  • $begingroup$
    Thank you for explaining. So a linear map from coalgebra ($A$, $mu$, $nu$) to algebra ($A$, $Delta$, $epsilon$) is just a linear map $Ato A$?
    $endgroup$
    – Ordev Agens
    Jan 7 at 19:47






  • 1




    $begingroup$
    Yes it is. Comment too short.
    $endgroup$
    – Matt Samuel
    Jan 7 at 20:36


















  • $begingroup$
    Usually this is done if the domain and range are bialgebras, so they have both a product and a coproduct.
    $endgroup$
    – Matt Samuel
    Jan 6 at 15:46










  • $begingroup$
    Maybe linear map from coalgebra to algebra makes more sense, but I still don't understand how it can be defined. Found it there: maths.qmul.ac.uk/~whitty/LSBU/MathsStudyGroup/SeligHopf.pdf , on page 5, section 4 (in the beginning).
    $endgroup$
    – Ordev Agens
    Jan 7 at 18:31








  • 1




    $begingroup$
    Well it defines it explicitly with a formula. And it's exactly as I said: it's a map between bialgebras (specifically here a bialgebra with itself). In terms of linear maps, you can define a nontrivial linear map between any two nontrivial vector spaces over the same field. It's not a homomorphism, it's just a composition of particular important functions here, including the product and the coproduct.
    $endgroup$
    – Matt Samuel
    Jan 7 at 19:22










  • $begingroup$
    Thank you for explaining. So a linear map from coalgebra ($A$, $mu$, $nu$) to algebra ($A$, $Delta$, $epsilon$) is just a linear map $Ato A$?
    $endgroup$
    – Ordev Agens
    Jan 7 at 19:47






  • 1




    $begingroup$
    Yes it is. Comment too short.
    $endgroup$
    – Matt Samuel
    Jan 7 at 20:36
















$begingroup$
Usually this is done if the domain and range are bialgebras, so they have both a product and a coproduct.
$endgroup$
– Matt Samuel
Jan 6 at 15:46




$begingroup$
Usually this is done if the domain and range are bialgebras, so they have both a product and a coproduct.
$endgroup$
– Matt Samuel
Jan 6 at 15:46












$begingroup$
Maybe linear map from coalgebra to algebra makes more sense, but I still don't understand how it can be defined. Found it there: maths.qmul.ac.uk/~whitty/LSBU/MathsStudyGroup/SeligHopf.pdf , on page 5, section 4 (in the beginning).
$endgroup$
– Ordev Agens
Jan 7 at 18:31






$begingroup$
Maybe linear map from coalgebra to algebra makes more sense, but I still don't understand how it can be defined. Found it there: maths.qmul.ac.uk/~whitty/LSBU/MathsStudyGroup/SeligHopf.pdf , on page 5, section 4 (in the beginning).
$endgroup$
– Ordev Agens
Jan 7 at 18:31






1




1




$begingroup$
Well it defines it explicitly with a formula. And it's exactly as I said: it's a map between bialgebras (specifically here a bialgebra with itself). In terms of linear maps, you can define a nontrivial linear map between any two nontrivial vector spaces over the same field. It's not a homomorphism, it's just a composition of particular important functions here, including the product and the coproduct.
$endgroup$
– Matt Samuel
Jan 7 at 19:22




$begingroup$
Well it defines it explicitly with a formula. And it's exactly as I said: it's a map between bialgebras (specifically here a bialgebra with itself). In terms of linear maps, you can define a nontrivial linear map between any two nontrivial vector spaces over the same field. It's not a homomorphism, it's just a composition of particular important functions here, including the product and the coproduct.
$endgroup$
– Matt Samuel
Jan 7 at 19:22












$begingroup$
Thank you for explaining. So a linear map from coalgebra ($A$, $mu$, $nu$) to algebra ($A$, $Delta$, $epsilon$) is just a linear map $Ato A$?
$endgroup$
– Ordev Agens
Jan 7 at 19:47




$begingroup$
Thank you for explaining. So a linear map from coalgebra ($A$, $mu$, $nu$) to algebra ($A$, $Delta$, $epsilon$) is just a linear map $Ato A$?
$endgroup$
– Ordev Agens
Jan 7 at 19:47




1




1




$begingroup$
Yes it is. Comment too short.
$endgroup$
– Matt Samuel
Jan 7 at 20:36




$begingroup$
Yes it is. Comment too short.
$endgroup$
– Matt Samuel
Jan 7 at 20:36










1 Answer
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Maybe an answer is not needed anymore, because the OP question has already been solved, but I think a clarification is needed for those who are not familiar with the subject.



If we have two vector spaces $V,W$ over a field $Bbbk$ then we can consider the set $mathsf{Hom}_{Bbbk}(V,W)$ of all $Bbbk$-linear maps from $V$ to $W$.



In the particular case in which $V$ has additionally a $Bbbk$-coalgebra structure $(V,Delta,varepsilon)$ and $W$ has an algebra structure $(W,m,u)$, then $mathsf{Hom}_{Bbbk}(V,W)$ can be endowed with a product $*$ (called the convolution product) and a unit element $1$ that makes of it a monoid, namely
$$f*g:=mcirc (fotimes g)circ Delta qquad text{and} qquad 1:=ucircvarepsilon,$$
which are still $Bbbk$-linear maps from $V$ to $W$.



If $V=W=(B,m,u,Delta,varepsilon)$ bialgebra, then $mathsf{Hom}_{Bbbk}(B,B)=mathsf{End}_{Bbbk}(B)$ is a monoid as above and an inverse of the identity $mathsf{Id}_B$ with respect to the convolution product (when it exists) is called and antipode. A bialgebra with an antipode is nowadays called a Hopf algebra.






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    $begingroup$

    Maybe an answer is not needed anymore, because the OP question has already been solved, but I think a clarification is needed for those who are not familiar with the subject.



    If we have two vector spaces $V,W$ over a field $Bbbk$ then we can consider the set $mathsf{Hom}_{Bbbk}(V,W)$ of all $Bbbk$-linear maps from $V$ to $W$.



    In the particular case in which $V$ has additionally a $Bbbk$-coalgebra structure $(V,Delta,varepsilon)$ and $W$ has an algebra structure $(W,m,u)$, then $mathsf{Hom}_{Bbbk}(V,W)$ can be endowed with a product $*$ (called the convolution product) and a unit element $1$ that makes of it a monoid, namely
    $$f*g:=mcirc (fotimes g)circ Delta qquad text{and} qquad 1:=ucircvarepsilon,$$
    which are still $Bbbk$-linear maps from $V$ to $W$.



    If $V=W=(B,m,u,Delta,varepsilon)$ bialgebra, then $mathsf{Hom}_{Bbbk}(B,B)=mathsf{End}_{Bbbk}(B)$ is a monoid as above and an inverse of the identity $mathsf{Id}_B$ with respect to the convolution product (when it exists) is called and antipode. A bialgebra with an antipode is nowadays called a Hopf algebra.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Maybe an answer is not needed anymore, because the OP question has already been solved, but I think a clarification is needed for those who are not familiar with the subject.



      If we have two vector spaces $V,W$ over a field $Bbbk$ then we can consider the set $mathsf{Hom}_{Bbbk}(V,W)$ of all $Bbbk$-linear maps from $V$ to $W$.



      In the particular case in which $V$ has additionally a $Bbbk$-coalgebra structure $(V,Delta,varepsilon)$ and $W$ has an algebra structure $(W,m,u)$, then $mathsf{Hom}_{Bbbk}(V,W)$ can be endowed with a product $*$ (called the convolution product) and a unit element $1$ that makes of it a monoid, namely
      $$f*g:=mcirc (fotimes g)circ Delta qquad text{and} qquad 1:=ucircvarepsilon,$$
      which are still $Bbbk$-linear maps from $V$ to $W$.



      If $V=W=(B,m,u,Delta,varepsilon)$ bialgebra, then $mathsf{Hom}_{Bbbk}(B,B)=mathsf{End}_{Bbbk}(B)$ is a monoid as above and an inverse of the identity $mathsf{Id}_B$ with respect to the convolution product (when it exists) is called and antipode. A bialgebra with an antipode is nowadays called a Hopf algebra.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Maybe an answer is not needed anymore, because the OP question has already been solved, but I think a clarification is needed for those who are not familiar with the subject.



        If we have two vector spaces $V,W$ over a field $Bbbk$ then we can consider the set $mathsf{Hom}_{Bbbk}(V,W)$ of all $Bbbk$-linear maps from $V$ to $W$.



        In the particular case in which $V$ has additionally a $Bbbk$-coalgebra structure $(V,Delta,varepsilon)$ and $W$ has an algebra structure $(W,m,u)$, then $mathsf{Hom}_{Bbbk}(V,W)$ can be endowed with a product $*$ (called the convolution product) and a unit element $1$ that makes of it a monoid, namely
        $$f*g:=mcirc (fotimes g)circ Delta qquad text{and} qquad 1:=ucircvarepsilon,$$
        which are still $Bbbk$-linear maps from $V$ to $W$.



        If $V=W=(B,m,u,Delta,varepsilon)$ bialgebra, then $mathsf{Hom}_{Bbbk}(B,B)=mathsf{End}_{Bbbk}(B)$ is a monoid as above and an inverse of the identity $mathsf{Id}_B$ with respect to the convolution product (when it exists) is called and antipode. A bialgebra with an antipode is nowadays called a Hopf algebra.






        share|cite|improve this answer









        $endgroup$



        Maybe an answer is not needed anymore, because the OP question has already been solved, but I think a clarification is needed for those who are not familiar with the subject.



        If we have two vector spaces $V,W$ over a field $Bbbk$ then we can consider the set $mathsf{Hom}_{Bbbk}(V,W)$ of all $Bbbk$-linear maps from $V$ to $W$.



        In the particular case in which $V$ has additionally a $Bbbk$-coalgebra structure $(V,Delta,varepsilon)$ and $W$ has an algebra structure $(W,m,u)$, then $mathsf{Hom}_{Bbbk}(V,W)$ can be endowed with a product $*$ (called the convolution product) and a unit element $1$ that makes of it a monoid, namely
        $$f*g:=mcirc (fotimes g)circ Delta qquad text{and} qquad 1:=ucircvarepsilon,$$
        which are still $Bbbk$-linear maps from $V$ to $W$.



        If $V=W=(B,m,u,Delta,varepsilon)$ bialgebra, then $mathsf{Hom}_{Bbbk}(B,B)=mathsf{End}_{Bbbk}(B)$ is a monoid as above and an inverse of the identity $mathsf{Id}_B$ with respect to the convolution product (when it exists) is called and antipode. A bialgebra with an antipode is nowadays called a Hopf algebra.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 30 at 10:54









        Ender WigginsEnder Wiggins

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