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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          數位音樂下載 為非以「實體」方式來販賣音樂的相關產品，而使用數位格式如mp3、AAC等文件格式來進行銷售販賣的一种音乐产业形式。著名的數位音樂下載服務商有iTunes Store、RecoChoku、Google Play音乐和微软Zune等。数字音乐下载在二十一世纪初一度风靡全球， [1] [2] 使得传统唱片产业走向衰落，但其自身也在2010年代流媒体音乐服务如Apple Music、Spotify等的挤压下迅速失去市场份额。 [3] [4] 数字音乐下载有便于携带、传播迅速、制作简单成本低等优点，与流媒体音乐相比也能更好保障音乐人权益； [5] 但它同时很可能带来侵犯版权和滥用的情形， [6] 也给音乐产业带来了诸多不利影响。数字音乐下载为音乐人和乐迷提供了更加便捷直接的交流渠道，也一定程度上推动了网络音乐人的出现和独立音乐的迅速发展。 [7] 目录 1 历史 1.1 技术积淀（1987-2000） 1.2 出现与兴起（2000-2004） 1.3 唱片市场的颠覆者（2004-2007） 1.4 巅峰和衰落（2005-2013） 1.5 摇摇欲坠的现状与不可预知的明天（2014至今） 2 主要文件格式 2.1 PCM编码格式 2.1.1 WAV 2.1.2 APE 2.1.3 FLAC 2.2 苹果推出的无损文件格式 2.2.1 AIFF 2.2.2 ALAC 2.3 有损压缩格式 2.3.1 MP3 2.3.2 AAC 3 主要市场 3.1 iTunes Store 3.2 Google Play音乐商店 3.3 亚马逊音乐商店 3.4 部分其他音乐商店 4 影响与评价 4.1 对音乐产业的影响 4.2 对科技行业的影响 4.3 正面评价 4.4 负面评价 4.4.1 侵权问题 4.4.2 经营模式不明晰 4.4.3 不适应移动时代 4.5 文化现象 5 延伸阅读 6 参考资料 历史 苹果iTunes+iPod的软硬件搭配让数字音...
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          When can things happen in Etherscan, such as the picture below?

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          2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  ...
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          格利澤436b

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          格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP...
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