Subalgebra of a group algebra












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?






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






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






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




      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 yesterday









      StudentStudent

      1373




      1373






















          2 Answers
          2






          active

          oldest

          votes


















          7












          $begingroup$

          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
            15 hours ago



















          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
            yesterday










          • $begingroup$
            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
            yesterday










          • $begingroup$
            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
            yesterday






          • 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
            22 hours ago






          • 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
            22 hours ago












          Your Answer








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          2 Answers
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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          7












          $begingroup$

          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
            15 hours ago
















          7












          $begingroup$

          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
            15 hours ago














          7












          7








          7





          $begingroup$

          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 19 hours ago









          Oeyvind SolbergOeyvind Solberg

          4614




          4614








          • 1




            $begingroup$
            The CKMM theorem is very powerful! Is there a way to read off the group algebra part?
            $endgroup$
            – Student
            15 hours ago














          • 1




            $begingroup$
            The CKMM theorem is very powerful! Is there a way to read off the group algebra part?
            $endgroup$
            – Student
            15 hours ago








          1




          1




          $begingroup$
          The CKMM theorem is very powerful! Is there a way to read off the group algebra part?
          $endgroup$
          – Student
          15 hours ago




          $begingroup$
          The CKMM theorem is very powerful! Is there a way to read off the group algebra part?
          $endgroup$
          – Student
          15 hours ago











          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
            yesterday










          • $begingroup$
            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
            yesterday










          • $begingroup$
            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
            yesterday






          • 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
            22 hours ago






          • 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
            22 hours ago
















          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
            yesterday










          • $begingroup$
            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
            yesterday










          • $begingroup$
            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
            yesterday






          • 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
            22 hours ago






          • 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
            22 hours ago














          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 yesterday









          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
            yesterday










          • $begingroup$
            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
            yesterday










          • $begingroup$
            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
            yesterday






          • 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
            22 hours ago






          • 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
            22 hours ago


















          • $begingroup$
            Is there a characteristic assumption on the last part? Is there division by 2 in formula for y?
            $endgroup$
            – AHusain
            yesterday










          • $begingroup$
            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
            yesterday










          • $begingroup$
            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
            yesterday






          • 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
            22 hours ago






          • 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
            22 hours ago
















          $begingroup$
          Is there a characteristic assumption on the last part? Is there division by 2 in formula for y?
          $endgroup$
          – AHusain
          yesterday




          $begingroup$
          Is there a characteristic assumption on the last part? Is there division by 2 in formula for y?
          $endgroup$
          – AHusain
          yesterday












          $begingroup$
          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
          yesterday




          $begingroup$
          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
          yesterday












          $begingroup$
          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
          yesterday




          $begingroup$
          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
          yesterday




          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
          22 hours ago




          $begingroup$
          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
          22 hours ago




          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
          22 hours ago




          $begingroup$
          @JoshuaGrochow: Ravenel uses his version to show that a particular Hopf algebra is a group algebra, so he found it useful.
          $endgroup$
          – John Palmieri
          22 hours ago


















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          迪纳利峰 位于迪纳利国家公园和保留区的迪纳利峰（旧名麦金利峰） 最高点 海拔 20,237英尺（6,168米）   NGVD 29 . [1] [2] 相對高度 20,073英尺（6,118米） 列表 各洲最高峰列表 特独立山峰 各国最高点列表 美国最高点  排名第一 [3] 坐标 63°04′10″N 151°00′27″W  /  63.06944°N 151.00750°W  / 63.06944; -151.00750  ( Mount McKinley ) 坐标： 63°04′10″N 151°00′27″W  /  63.06944°N 151.00750°W  / 63.06944; -151.00750  ( Mount McKinley )   地理 位置   美國阿拉斯加州迪纳利自治市镇 迪纳利国家公园和保留区 山脈 阿拉斯加山脉 攀山 首次登頂 （ 英语 ： First ascent ） 1913年6月7日 哈德逊·史塔克 哈里·卡斯坦斯 沃尔特·哈珀 罗伯特·塔特姆 最簡路線 西坡路线（攀登冰川/雪） 迪纳利 （北阿萨巴斯卡语支、 英语： Denali ， / d ᵻ ˈ n ɑː l i / ，直译“高山”），1917年至2015年官方名称为 麦金利山 （另译作 麦金利峰 ， 英语： Mount McKinley ），位于阿拉斯加州东南部、阿拉斯加山脉中段，海拔6194米，是北美洲最高峰，也是美国的最高峰。它的俄语名称是 Большая Гора ， 罗马化：Bolshaya Gora ，意思是“大山”。 迪纳利山地区拥有变幻莫测的高山风、典型的北极植被以及野生动植物。这里大部分地区终年积雪，山间经常浓雾不断，雾气在皑皑白雪中缭绕弥漫时，几百米之外的景物便不可见。夏季，迪纳利山的青青山坡上鲜花盛开，紫色的杜鹃和精巧的铃状石南花随处可见。迪纳利山还是世界登山爱好者的汇集之地。每年5月到7月有数百人登山，只有一半不到的人能登顶成功。一次登山约花3个星期。 2015年8月31日，美國政府宣佈將山名由“麦金利山”改回原名“迪纳利山”。美國內政部發表聲
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          南乌拉尔铁路局

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          南乌拉尔铁路局 Южно-Уральская железная дорога 成立 1934年1月4日 国家   蘇聯（1934年—1991年）   俄羅斯（1991年—） 总部 车里雅宾斯克 隶属 俄罗斯铁路股份公司 路局代码 80 铁路里程 4,934.9公里 奖项 网址 http://yuzd.rzd.ru/ 查 论 编 南乌拉尔铁路局 （俄语： Южно-Уральская железная дорога ， 羅馬化： Yuzhno-Uralskaya zheleznaya doroga ，简称 ЮУЖД ）是俄罗斯铁路股份公司属下的铁路管理局之一，总部位于车里雅宾斯克，负责管辖俄罗斯联邦乌拉尔南部地区的铁路系统，范围包括奥伦堡州、车里雅宾斯克州、库尔干州的大部分铁路，以及萨马拉州、斯维尔德洛夫斯克州、巴什科尔托斯坦共和国的少部分铁路，并有部分铁路延伸至或穿过哈萨克斯坦境内。南乌拉尔铁路局的营业里程为4,934.9公里，与伏尔加铁路局、古比雪夫铁路局、斯维尔德洛夫斯克铁路局、西西伯利亚铁路局相邻，并与南方的哈萨克斯坦铁路连接。 外部链接 （俄文） Южно-Уральская железная дорога — филиал ОАО РЖД （俄文） 南乌拉尔铁路局管辖路线图（一） （俄文） 南乌拉尔铁路局管辖路线图（二） 查 论 编 俄罗斯铁路（集团）股份公司 十月铁路局 · 莫斯科铁路局 · 北高加索铁路局 · 高尔基铁路局 · 古比雪夫铁路局 · 北方铁路局 · 伏尔加铁路局 · 东南铁路局 · 斯维尔德洛夫斯克铁路局 · 南乌拉尔铁路局 · 西西伯利亚铁路局 · 克拉斯诺亚尔斯克铁路局 · 东西伯利亚铁路局 · 后贝加尔铁路局 · 远东铁路局 · 加里宁格勒铁路局 This page is only for reference, If you need detailed information, please check here
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