Subalgebra of a group algebra Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Characterization of Complex Group AlgebrasClassification of Hopf algebra with exactly two 1-dimensional modulesCriterion for nilradical of a maximal parabolic subalgebra to be abelian?Iwahori-Hecke algebras as endomorphism (or convolution) algebra?Jacobson radical and group rings/subalgebrasSoluble group algebras and centralizersMotivational ideas for the Gelfand-Graev character of a finite group of Lie typeWhen is the exterior algebra a Hopf algebra?Symplectic group over finite field and quaternionsCartan subalgebra and group measure space construction

Subalgebra of a group algebra



Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Characterization of Complex Group AlgebrasClassification of Hopf algebra with exactly two 1-dimensional modulesCriterion for nilradical of a maximal parabolic subalgebra to be abelian?Iwahori-Hecke algebras as endomorphism (or convolution) algebra?Jacobson radical and group rings/subalgebrasSoluble group algebras and centralizersMotivational ideas for the Gelfand-Graev character of a finite group of Lie typeWhen is the exterior algebra a Hopf algebra?Symplectic group over finite field and quaternionsCartan subalgebra and group measure space construction










6












$begingroup$


Let $k$ be a field, $G$ a finite group, and $k[G]$ the group algebra.
Let $A$ be a subalgebra of $k[G]$. In general, $A$ is not the group algebra of some subgroup $H$ of $G$.



Question: Is there any criterion for when $A = k[H]$ for some subgroup $H$? Also, in that case, how do we read of the generating subgroup $H$? Will the situation become better/easier if I assume $A$ to be a sub-Hopf-algebra?










share|cite|improve this question









$endgroup$
















    6












    $begingroup$


    Let $k$ be a field, $G$ a finite group, and $k[G]$ the group algebra.
    Let $A$ be a subalgebra of $k[G]$. In general, $A$ is not the group algebra of some subgroup $H$ of $G$.



    Question: Is there any criterion for when $A = k[H]$ for some subgroup $H$? Also, in that case, how do we read of the generating subgroup $H$? Will the situation become better/easier if I assume $A$ to be a sub-Hopf-algebra?










    share|cite|improve this question









    $endgroup$














      6












      6








      6





      $begingroup$


      Let $k$ be a field, $G$ a finite group, and $k[G]$ the group algebra.
      Let $A$ be a subalgebra of $k[G]$. In general, $A$ is not the group algebra of some subgroup $H$ of $G$.



      Question: Is there any criterion for when $A = k[H]$ for some subgroup $H$? Also, in that case, how do we read of the generating subgroup $H$? Will the situation become better/easier if I assume $A$ to be a sub-Hopf-algebra?










      share|cite|improve this question









      $endgroup$




      Let $k$ be a field, $G$ a finite group, and $k[G]$ the group algebra.
      Let $A$ be a subalgebra of $k[G]$. In general, $A$ is not the group algebra of some subgroup $H$ of $G$.



      Question: Is there any criterion for when $A = k[H]$ for some subgroup $H$? Also, in that case, how do we read of the generating subgroup $H$? Will the situation become better/easier if I assume $A$ to be a sub-Hopf-algebra?







      gr.group-theory rt.representation-theory noncommutative-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Apr 18 at 20:19









      StudentStudent

      1394




      1394




















          2 Answers
          2






          active

          oldest

          votes


















          8












          $begingroup$

          The characteristic of the field is important here, when considering Hopf sub-algebras. The Cartier-Kostant-Milnor-Moore theorem says that a cocommutative Hopf algebra $H$ over an algebraically closed field $k$ of characteristic 0, is a semidirect product of a group algebra and an enveloping algebra of a Lie algebra. In particular, if $H$ is finite dimensional cocommutative Hopf algebra over $k$, then $H$ is isomorphic to a group algebra.



          This theorem is not true over algebraically closed fields in positive characteristics, as there are restricted enveloping $p$-Lie algebras that are finite dimensional cocommutative Hopf algebras but not isomorphic to group algebras.



          See the introduction of https://cel.archives-ouvertes.fr/cel-00374383/document (by Nicolas Andruskiewitsch) for further information.






          share|cite|improve this answer









          $endgroup$








          • 1




            $begingroup$
            The CKMM theorem is very powerful! Is there a way to read off the group algebra part?
            $endgroup$
            – Student
            Apr 19 at 12:24










          • $begingroup$
            I have only used this when I start with something finite dimensional, and then, as the result states, the Lie algebra part is not there. I don't know how to you find the group algebra part when you start with something which is infinite dimensional.
            $endgroup$
            – Oeyvind Solberg
            Apr 20 at 7:35


















          6












          $begingroup$

          If $A$ is the group algebra of a subgroup, then $k[G]$ will be free as a module over $A$, and that may help to rule out some cases. This is not at all an "if and only if" statement, though.



          Proposition 3.2.1 in Sweedler's book Hopf Algebras says that a cocommutative Hopf algebra is a group algebra if and only if it has a basis of group-like elements (those elements $x$ which satisfy $Delta(x) = x otimes x$). So if you have a sub-Hopf algebra, you can try to see if it has such a basis.



          Ravenel says in Theorem 6.2.3 in Complex Cobordism and Stable Homotopy Groups of Spheres: "this is equivalent to the existence of a dual basis of idempotent elements $y$ satisfying $y_i^2=y_i$ and $y_iy_j=0$ for $i neq j$."






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Is there a characteristic assumption on the last part? Is there division by 2 in formula for y?
            $endgroup$
            – AHusain
            Apr 18 at 22:29










          • $begingroup$
            Great! Also, I think Sweedler provided a fairly nice answer already, since if A comes from some subgroup, then A is automatically a sub-Hopf-algebra!
            $endgroup$
            – Student
            Apr 19 at 0:03










          • $begingroup$
            For Ravenel's statement, I do not have access to the book for now. Would you mind pointing it out that in his content, is $A$ assumed to be a Hopf algebra?
            $endgroup$
            – Student
            Apr 19 at 0:04






          • 2




            $begingroup$
            Ravenel's book is available from his web page: web.math.rochester.edu/people/faculty/doug/mybooks/ravenel.pdf. He is certainly working in positive characteristic, but I don't know if this is necessary for his condition. He is certainly working with Hopf algebras, since otherwise there wouldn't be a multiplication on the dual.
            $endgroup$
            – John Palmieri
            Apr 19 at 5:28






          • 2




            $begingroup$
            @JoshuaGrochow: Ravenel uses his version to show that a particular Hopf algebra is a group algebra, so he found it useful.
            $endgroup$
            – John Palmieri
            Apr 19 at 5:29











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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          8












          $begingroup$

          The characteristic of the field is important here, when considering Hopf sub-algebras. The Cartier-Kostant-Milnor-Moore theorem says that a cocommutative Hopf algebra $H$ over an algebraically closed field $k$ of characteristic 0, is a semidirect product of a group algebra and an enveloping algebra of a Lie algebra. In particular, if $H$ is finite dimensional cocommutative Hopf algebra over $k$, then $H$ is isomorphic to a group algebra.



          This theorem is not true over algebraically closed fields in positive characteristics, as there are restricted enveloping $p$-Lie algebras that are finite dimensional cocommutative Hopf algebras but not isomorphic to group algebras.



          See the introduction of https://cel.archives-ouvertes.fr/cel-00374383/document (by Nicolas Andruskiewitsch) for further information.






          share|cite|improve this answer









          $endgroup$








          • 1




            $begingroup$
            The CKMM theorem is very powerful! Is there a way to read off the group algebra part?
            $endgroup$
            – Student
            Apr 19 at 12:24










          • $begingroup$
            I have only used this when I start with something finite dimensional, and then, as the result states, the Lie algebra part is not there. I don't know how to you find the group algebra part when you start with something which is infinite dimensional.
            $endgroup$
            – Oeyvind Solberg
            Apr 20 at 7:35















          8












          $begingroup$

          The characteristic of the field is important here, when considering Hopf sub-algebras. The Cartier-Kostant-Milnor-Moore theorem says that a cocommutative Hopf algebra $H$ over an algebraically closed field $k$ of characteristic 0, is a semidirect product of a group algebra and an enveloping algebra of a Lie algebra. In particular, if $H$ is finite dimensional cocommutative Hopf algebra over $k$, then $H$ is isomorphic to a group algebra.



          This theorem is not true over algebraically closed fields in positive characteristics, as there are restricted enveloping $p$-Lie algebras that are finite dimensional cocommutative Hopf algebras but not isomorphic to group algebras.



          See the introduction of https://cel.archives-ouvertes.fr/cel-00374383/document (by Nicolas Andruskiewitsch) for further information.






          share|cite|improve this answer









          $endgroup$








          • 1




            $begingroup$
            The CKMM theorem is very powerful! Is there a way to read off the group algebra part?
            $endgroup$
            – Student
            Apr 19 at 12:24










          • $begingroup$
            I have only used this when I start with something finite dimensional, and then, as the result states, the Lie algebra part is not there. I don't know how to you find the group algebra part when you start with something which is infinite dimensional.
            $endgroup$
            – Oeyvind Solberg
            Apr 20 at 7:35













          8












          8








          8





          $begingroup$

          The characteristic of the field is important here, when considering Hopf sub-algebras. The Cartier-Kostant-Milnor-Moore theorem says that a cocommutative Hopf algebra $H$ over an algebraically closed field $k$ of characteristic 0, is a semidirect product of a group algebra and an enveloping algebra of a Lie algebra. In particular, if $H$ is finite dimensional cocommutative Hopf algebra over $k$, then $H$ is isomorphic to a group algebra.



          This theorem is not true over algebraically closed fields in positive characteristics, as there are restricted enveloping $p$-Lie algebras that are finite dimensional cocommutative Hopf algebras but not isomorphic to group algebras.



          See the introduction of https://cel.archives-ouvertes.fr/cel-00374383/document (by Nicolas Andruskiewitsch) for further information.






          share|cite|improve this answer









          $endgroup$



          The characteristic of the field is important here, when considering Hopf sub-algebras. The Cartier-Kostant-Milnor-Moore theorem says that a cocommutative Hopf algebra $H$ over an algebraically closed field $k$ of characteristic 0, is a semidirect product of a group algebra and an enveloping algebra of a Lie algebra. In particular, if $H$ is finite dimensional cocommutative Hopf algebra over $k$, then $H$ is isomorphic to a group algebra.



          This theorem is not true over algebraically closed fields in positive characteristics, as there are restricted enveloping $p$-Lie algebras that are finite dimensional cocommutative Hopf algebras but not isomorphic to group algebras.



          See the introduction of https://cel.archives-ouvertes.fr/cel-00374383/document (by Nicolas Andruskiewitsch) for further information.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Apr 19 at 8:18









          Oeyvind SolbergOeyvind Solberg

          4864




          4864







          • 1




            $begingroup$
            The CKMM theorem is very powerful! Is there a way to read off the group algebra part?
            $endgroup$
            – Student
            Apr 19 at 12:24










          • $begingroup$
            I have only used this when I start with something finite dimensional, and then, as the result states, the Lie algebra part is not there. I don't know how to you find the group algebra part when you start with something which is infinite dimensional.
            $endgroup$
            – Oeyvind Solberg
            Apr 20 at 7:35












          • 1




            $begingroup$
            The CKMM theorem is very powerful! Is there a way to read off the group algebra part?
            $endgroup$
            – Student
            Apr 19 at 12:24










          • $begingroup$
            I have only used this when I start with something finite dimensional, and then, as the result states, the Lie algebra part is not there. I don't know how to you find the group algebra part when you start with something which is infinite dimensional.
            $endgroup$
            – Oeyvind Solberg
            Apr 20 at 7:35







          1




          1




          $begingroup$
          The CKMM theorem is very powerful! Is there a way to read off the group algebra part?
          $endgroup$
          – Student
          Apr 19 at 12:24




          $begingroup$
          The CKMM theorem is very powerful! Is there a way to read off the group algebra part?
          $endgroup$
          – Student
          Apr 19 at 12:24












          $begingroup$
          I have only used this when I start with something finite dimensional, and then, as the result states, the Lie algebra part is not there. I don't know how to you find the group algebra part when you start with something which is infinite dimensional.
          $endgroup$
          – Oeyvind Solberg
          Apr 20 at 7:35




          $begingroup$
          I have only used this when I start with something finite dimensional, and then, as the result states, the Lie algebra part is not there. I don't know how to you find the group algebra part when you start with something which is infinite dimensional.
          $endgroup$
          – Oeyvind Solberg
          Apr 20 at 7:35











          6












          $begingroup$

          If $A$ is the group algebra of a subgroup, then $k[G]$ will be free as a module over $A$, and that may help to rule out some cases. This is not at all an "if and only if" statement, though.



          Proposition 3.2.1 in Sweedler's book Hopf Algebras says that a cocommutative Hopf algebra is a group algebra if and only if it has a basis of group-like elements (those elements $x$ which satisfy $Delta(x) = x otimes x$). So if you have a sub-Hopf algebra, you can try to see if it has such a basis.



          Ravenel says in Theorem 6.2.3 in Complex Cobordism and Stable Homotopy Groups of Spheres: "this is equivalent to the existence of a dual basis of idempotent elements $y$ satisfying $y_i^2=y_i$ and $y_iy_j=0$ for $i neq j$."






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Is there a characteristic assumption on the last part? Is there division by 2 in formula for y?
            $endgroup$
            – AHusain
            Apr 18 at 22:29










          • $begingroup$
            Great! Also, I think Sweedler provided a fairly nice answer already, since if A comes from some subgroup, then A is automatically a sub-Hopf-algebra!
            $endgroup$
            – Student
            Apr 19 at 0:03










          • $begingroup$
            For Ravenel's statement, I do not have access to the book for now. Would you mind pointing it out that in his content, is $A$ assumed to be a Hopf algebra?
            $endgroup$
            – Student
            Apr 19 at 0:04






          • 2




            $begingroup$
            Ravenel's book is available from his web page: web.math.rochester.edu/people/faculty/doug/mybooks/ravenel.pdf. He is certainly working in positive characteristic, but I don't know if this is necessary for his condition. He is certainly working with Hopf algebras, since otherwise there wouldn't be a multiplication on the dual.
            $endgroup$
            – John Palmieri
            Apr 19 at 5:28






          • 2




            $begingroup$
            @JoshuaGrochow: Ravenel uses his version to show that a particular Hopf algebra is a group algebra, so he found it useful.
            $endgroup$
            – John Palmieri
            Apr 19 at 5:29















          6












          $begingroup$

          If $A$ is the group algebra of a subgroup, then $k[G]$ will be free as a module over $A$, and that may help to rule out some cases. This is not at all an "if and only if" statement, though.



          Proposition 3.2.1 in Sweedler's book Hopf Algebras says that a cocommutative Hopf algebra is a group algebra if and only if it has a basis of group-like elements (those elements $x$ which satisfy $Delta(x) = x otimes x$). So if you have a sub-Hopf algebra, you can try to see if it has such a basis.



          Ravenel says in Theorem 6.2.3 in Complex Cobordism and Stable Homotopy Groups of Spheres: "this is equivalent to the existence of a dual basis of idempotent elements $y$ satisfying $y_i^2=y_i$ and $y_iy_j=0$ for $i neq j$."






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Is there a characteristic assumption on the last part? Is there division by 2 in formula for y?
            $endgroup$
            – AHusain
            Apr 18 at 22:29










          • $begingroup$
            Great! Also, I think Sweedler provided a fairly nice answer already, since if A comes from some subgroup, then A is automatically a sub-Hopf-algebra!
            $endgroup$
            – Student
            Apr 19 at 0:03










          • $begingroup$
            For Ravenel's statement, I do not have access to the book for now. Would you mind pointing it out that in his content, is $A$ assumed to be a Hopf algebra?
            $endgroup$
            – Student
            Apr 19 at 0:04






          • 2




            $begingroup$
            Ravenel's book is available from his web page: web.math.rochester.edu/people/faculty/doug/mybooks/ravenel.pdf. He is certainly working in positive characteristic, but I don't know if this is necessary for his condition. He is certainly working with Hopf algebras, since otherwise there wouldn't be a multiplication on the dual.
            $endgroup$
            – John Palmieri
            Apr 19 at 5:28






          • 2




            $begingroup$
            @JoshuaGrochow: Ravenel uses his version to show that a particular Hopf algebra is a group algebra, so he found it useful.
            $endgroup$
            – John Palmieri
            Apr 19 at 5:29













          6












          6








          6





          $begingroup$

          If $A$ is the group algebra of a subgroup, then $k[G]$ will be free as a module over $A$, and that may help to rule out some cases. This is not at all an "if and only if" statement, though.



          Proposition 3.2.1 in Sweedler's book Hopf Algebras says that a cocommutative Hopf algebra is a group algebra if and only if it has a basis of group-like elements (those elements $x$ which satisfy $Delta(x) = x otimes x$). So if you have a sub-Hopf algebra, you can try to see if it has such a basis.



          Ravenel says in Theorem 6.2.3 in Complex Cobordism and Stable Homotopy Groups of Spheres: "this is equivalent to the existence of a dual basis of idempotent elements $y$ satisfying $y_i^2=y_i$ and $y_iy_j=0$ for $i neq j$."






          share|cite|improve this answer









          $endgroup$



          If $A$ is the group algebra of a subgroup, then $k[G]$ will be free as a module over $A$, and that may help to rule out some cases. This is not at all an "if and only if" statement, though.



          Proposition 3.2.1 in Sweedler's book Hopf Algebras says that a cocommutative Hopf algebra is a group algebra if and only if it has a basis of group-like elements (those elements $x$ which satisfy $Delta(x) = x otimes x$). So if you have a sub-Hopf algebra, you can try to see if it has such a basis.



          Ravenel says in Theorem 6.2.3 in Complex Cobordism and Stable Homotopy Groups of Spheres: "this is equivalent to the existence of a dual basis of idempotent elements $y$ satisfying $y_i^2=y_i$ and $y_iy_j=0$ for $i neq j$."







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Apr 18 at 22:23









          John PalmieriJohn Palmieri

          2,35011726




          2,35011726











          • $begingroup$
            Is there a characteristic assumption on the last part? Is there division by 2 in formula for y?
            $endgroup$
            – AHusain
            Apr 18 at 22:29










          • $begingroup$
            Great! Also, I think Sweedler provided a fairly nice answer already, since if A comes from some subgroup, then A is automatically a sub-Hopf-algebra!
            $endgroup$
            – Student
            Apr 19 at 0:03










          • $begingroup$
            For Ravenel's statement, I do not have access to the book for now. Would you mind pointing it out that in his content, is $A$ assumed to be a Hopf algebra?
            $endgroup$
            – Student
            Apr 19 at 0:04






          • 2




            $begingroup$
            Ravenel's book is available from his web page: web.math.rochester.edu/people/faculty/doug/mybooks/ravenel.pdf. He is certainly working in positive characteristic, but I don't know if this is necessary for his condition. He is certainly working with Hopf algebras, since otherwise there wouldn't be a multiplication on the dual.
            $endgroup$
            – John Palmieri
            Apr 19 at 5:28






          • 2




            $begingroup$
            @JoshuaGrochow: Ravenel uses his version to show that a particular Hopf algebra is a group algebra, so he found it useful.
            $endgroup$
            – John Palmieri
            Apr 19 at 5:29
















          • $begingroup$
            Is there a characteristic assumption on the last part? Is there division by 2 in formula for y?
            $endgroup$
            – AHusain
            Apr 18 at 22:29










          • $begingroup$
            Great! Also, I think Sweedler provided a fairly nice answer already, since if A comes from some subgroup, then A is automatically a sub-Hopf-algebra!
            $endgroup$
            – Student
            Apr 19 at 0:03










          • $begingroup$
            For Ravenel's statement, I do not have access to the book for now. Would you mind pointing it out that in his content, is $A$ assumed to be a Hopf algebra?
            $endgroup$
            – Student
            Apr 19 at 0:04






          • 2




            $begingroup$
            Ravenel's book is available from his web page: web.math.rochester.edu/people/faculty/doug/mybooks/ravenel.pdf. He is certainly working in positive characteristic, but I don't know if this is necessary for his condition. He is certainly working with Hopf algebras, since otherwise there wouldn't be a multiplication on the dual.
            $endgroup$
            – John Palmieri
            Apr 19 at 5:28






          • 2




            $begingroup$
            @JoshuaGrochow: Ravenel uses his version to show that a particular Hopf algebra is a group algebra, so he found it useful.
            $endgroup$
            – John Palmieri
            Apr 19 at 5:29















          $begingroup$
          Is there a characteristic assumption on the last part? Is there division by 2 in formula for y?
          $endgroup$
          – AHusain
          Apr 18 at 22:29




          $begingroup$
          Is there a characteristic assumption on the last part? Is there division by 2 in formula for y?
          $endgroup$
          – AHusain
          Apr 18 at 22:29












          $begingroup$
          Great! Also, I think Sweedler provided a fairly nice answer already, since if A comes from some subgroup, then A is automatically a sub-Hopf-algebra!
          $endgroup$
          – Student
          Apr 19 at 0:03




          $begingroup$
          Great! Also, I think Sweedler provided a fairly nice answer already, since if A comes from some subgroup, then A is automatically a sub-Hopf-algebra!
          $endgroup$
          – Student
          Apr 19 at 0:03












          $begingroup$
          For Ravenel's statement, I do not have access to the book for now. Would you mind pointing it out that in his content, is $A$ assumed to be a Hopf algebra?
          $endgroup$
          – Student
          Apr 19 at 0:04




          $begingroup$
          For Ravenel's statement, I do not have access to the book for now. Would you mind pointing it out that in his content, is $A$ assumed to be a Hopf algebra?
          $endgroup$
          – Student
          Apr 19 at 0:04




          2




          2




          $begingroup$
          Ravenel's book is available from his web page: web.math.rochester.edu/people/faculty/doug/mybooks/ravenel.pdf. He is certainly working in positive characteristic, but I don't know if this is necessary for his condition. He is certainly working with Hopf algebras, since otherwise there wouldn't be a multiplication on the dual.
          $endgroup$
          – John Palmieri
          Apr 19 at 5:28




          $begingroup$
          Ravenel's book is available from his web page: web.math.rochester.edu/people/faculty/doug/mybooks/ravenel.pdf. He is certainly working in positive characteristic, but I don't know if this is necessary for his condition. He is certainly working with Hopf algebras, since otherwise there wouldn't be a multiplication on the dual.
          $endgroup$
          – John Palmieri
          Apr 19 at 5:28




          2




          2




          $begingroup$
          @JoshuaGrochow: Ravenel uses his version to show that a particular Hopf algebra is a group algebra, so he found it useful.
          $endgroup$
          – John Palmieri
          Apr 19 at 5:29




          $begingroup$
          @JoshuaGrochow: Ravenel uses his version to show that a particular Hopf algebra is a group algebra, so he found it useful.
          $endgroup$
          – John Palmieri
          Apr 19 at 5:29

















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