Are there finite dimensional $mathbb{R}$-algebras which are not Banach algebras?












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It is known that not every algebra (over a ground field $mathbb{R}$ or $mathbb{C}$) is a Banach algebra. It might be a silly question, but are there examples of finite dimensional ($mathbb{R}$- or $mathbb{C}$-)algebras which are not Banach algebras (i.e. for which no submultiplicative norm exists)?






abstract-algebra norm banach-algebras







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    3












    $begingroup$


    It is known that not every algebra (over a ground field $mathbb{R}$ or $mathbb{C}$) is a Banach algebra. It might be a silly question, but are there examples of finite dimensional ($mathbb{R}$- or $mathbb{C}$-)algebras which are not Banach algebras (i.e. for which no submultiplicative norm exists)?






    abstract-algebra norm banach-algebras







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      It is known that not every algebra (over a ground field $mathbb{R}$ or $mathbb{C}$) is a Banach algebra. It might be a silly question, but are there examples of finite dimensional ($mathbb{R}$- or $mathbb{C}$-)algebras which are not Banach algebras (i.e. for which no submultiplicative norm exists)?






      abstract-algebra norm banach-algebras







      share|cite|improve this question









      $endgroup$




      It is known that not every algebra (over a ground field $mathbb{R}$ or $mathbb{C}$) is a Banach algebra. It might be a silly question, but are there examples of finite dimensional ($mathbb{R}$- or $mathbb{C}$-)algebras which are not Banach algebras (i.e. for which no submultiplicative norm exists)?






      abstract-algebra norm banach-algebras




      abstract-algebra norm banach-algebras






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Apr 3 '16 at 18:02









      Bib-lostBib-lost

      1,998627




      1,998627






















          2 Answers
          2






          active

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          3












          $begingroup$

          Every finite dimensional unitary algebra $A$ is isomorphic to a subalgebra of a matrix algebra $M_n(mathbb R)$. Retricting the norm of the latter gives a norm on $A$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            This is what I was starting to think. Can you give a reference or short proof as to why such an isomorphism always exists?
            $endgroup$
            – Bib-lost
            Apr 3 '16 at 18:47






          • 1




            $begingroup$
            @Bib-lost: You can embed $A$ in $End(A)$ since $A$ acts on itself by multiplication.
            $endgroup$
            – Eric Wofsey
            Apr 3 '16 at 18:55










          • $begingroup$
            Ok, got it to work, thanks!
            $endgroup$
            – Bib-lost
            Apr 3 '16 at 20:25



















          3












          $begingroup$

          I think it's clear that in the finite-dimensional case multiplication is going to be continuous with respect to any norm, just as any linear transformation is bounded. If so then any norm is going to satisfy $||xy||le c||x||,||y||$, and then $|||x|||=c||x||$ is submultiplicative.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @user43208 I always get that backwards - thanks.
            $endgroup$
            – David C. Ullrich
            Dec 3 '18 at 14:54











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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









          3












          $begingroup$

          Every finite dimensional unitary algebra $A$ is isomorphic to a subalgebra of a matrix algebra $M_n(mathbb R)$. Retricting the norm of the latter gives a norm on $A$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            This is what I was starting to think. Can you give a reference or short proof as to why such an isomorphism always exists?
            $endgroup$
            – Bib-lost
            Apr 3 '16 at 18:47






          • 1




            $begingroup$
            @Bib-lost: You can embed $A$ in $End(A)$ since $A$ acts on itself by multiplication.
            $endgroup$
            – Eric Wofsey
            Apr 3 '16 at 18:55










          • $begingroup$
            Ok, got it to work, thanks!
            $endgroup$
            – Bib-lost
            Apr 3 '16 at 20:25
















          3












          $begingroup$

          Every finite dimensional unitary algebra $A$ is isomorphic to a subalgebra of a matrix algebra $M_n(mathbb R)$. Retricting the norm of the latter gives a norm on $A$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            This is what I was starting to think. Can you give a reference or short proof as to why such an isomorphism always exists?
            $endgroup$
            – Bib-lost
            Apr 3 '16 at 18:47






          • 1




            $begingroup$
            @Bib-lost: You can embed $A$ in $End(A)$ since $A$ acts on itself by multiplication.
            $endgroup$
            – Eric Wofsey
            Apr 3 '16 at 18:55










          • $begingroup$
            Ok, got it to work, thanks!
            $endgroup$
            – Bib-lost
            Apr 3 '16 at 20:25














          3












          3








          3





          $begingroup$

          Every finite dimensional unitary algebra $A$ is isomorphic to a subalgebra of a matrix algebra $M_n(mathbb R)$. Retricting the norm of the latter gives a norm on $A$.






          share|cite|improve this answer









          $endgroup$



          Every finite dimensional unitary algebra $A$ is isomorphic to a subalgebra of a matrix algebra $M_n(mathbb R)$. Retricting the norm of the latter gives a norm on $A$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Apr 3 '16 at 18:33









          Mariano Suárez-ÁlvarezMariano Suárez-Álvarez

          111k7155282




          111k7155282








          • 1




            $begingroup$
            This is what I was starting to think. Can you give a reference or short proof as to why such an isomorphism always exists?
            $endgroup$
            – Bib-lost
            Apr 3 '16 at 18:47






          • 1




            $begingroup$
            @Bib-lost: You can embed $A$ in $End(A)$ since $A$ acts on itself by multiplication.
            $endgroup$
            – Eric Wofsey
            Apr 3 '16 at 18:55










          • $begingroup$
            Ok, got it to work, thanks!
            $endgroup$
            – Bib-lost
            Apr 3 '16 at 20:25














          • 1




            $begingroup$
            This is what I was starting to think. Can you give a reference or short proof as to why such an isomorphism always exists?
            $endgroup$
            – Bib-lost
            Apr 3 '16 at 18:47






          • 1




            $begingroup$
            @Bib-lost: You can embed $A$ in $End(A)$ since $A$ acts on itself by multiplication.
            $endgroup$
            – Eric Wofsey
            Apr 3 '16 at 18:55










          • $begingroup$
            Ok, got it to work, thanks!
            $endgroup$
            – Bib-lost
            Apr 3 '16 at 20:25








          1




          1




          $begingroup$
          This is what I was starting to think. Can you give a reference or short proof as to why such an isomorphism always exists?
          $endgroup$
          – Bib-lost
          Apr 3 '16 at 18:47




          $begingroup$
          This is what I was starting to think. Can you give a reference or short proof as to why such an isomorphism always exists?
          $endgroup$
          – Bib-lost
          Apr 3 '16 at 18:47




          1




          1




          $begingroup$
          @Bib-lost: You can embed $A$ in $End(A)$ since $A$ acts on itself by multiplication.
          $endgroup$
          – Eric Wofsey
          Apr 3 '16 at 18:55




          $begingroup$
          @Bib-lost: You can embed $A$ in $End(A)$ since $A$ acts on itself by multiplication.
          $endgroup$
          – Eric Wofsey
          Apr 3 '16 at 18:55












          $begingroup$
          Ok, got it to work, thanks!
          $endgroup$
          – Bib-lost
          Apr 3 '16 at 20:25




          $begingroup$
          Ok, got it to work, thanks!
          $endgroup$
          – Bib-lost
          Apr 3 '16 at 20:25











          3












          $begingroup$

          I think it's clear that in the finite-dimensional case multiplication is going to be continuous with respect to any norm, just as any linear transformation is bounded. If so then any norm is going to satisfy $||xy||le c||x||,||y||$, and then $|||x|||=c||x||$ is submultiplicative.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @user43208 I always get that backwards - thanks.
            $endgroup$
            – David C. Ullrich
            Dec 3 '18 at 14:54
















          3












          $begingroup$

          I think it's clear that in the finite-dimensional case multiplication is going to be continuous with respect to any norm, just as any linear transformation is bounded. If so then any norm is going to satisfy $||xy||le c||x||,||y||$, and then $|||x|||=c||x||$ is submultiplicative.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @user43208 I always get that backwards - thanks.
            $endgroup$
            – David C. Ullrich
            Dec 3 '18 at 14:54














          3












          3








          3





          $begingroup$

          I think it's clear that in the finite-dimensional case multiplication is going to be continuous with respect to any norm, just as any linear transformation is bounded. If so then any norm is going to satisfy $||xy||le c||x||,||y||$, and then $|||x|||=c||x||$ is submultiplicative.






          share|cite|improve this answer











          $endgroup$



          I think it's clear that in the finite-dimensional case multiplication is going to be continuous with respect to any norm, just as any linear transformation is bounded. If so then any norm is going to satisfy $||xy||le c||x||,||y||$, and then $|||x|||=c||x||$ is submultiplicative.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 3 '18 at 14:54

























          answered Apr 3 '16 at 18:26









          David C. UllrichDavid C. Ullrich

          59.6k43893




          59.6k43893












          • $begingroup$
            @user43208 I always get that backwards - thanks.
            $endgroup$
            – David C. Ullrich
            Dec 3 '18 at 14:54


















          • $begingroup$
            @user43208 I always get that backwards - thanks.
            $endgroup$
            – David C. Ullrich
            Dec 3 '18 at 14:54
















          $begingroup$
          @user43208 I always get that backwards - thanks.
          $endgroup$
          – David C. Ullrich
          Dec 3 '18 at 14:54




          $begingroup$
          @user43208 I always get that backwards - thanks.
          $endgroup$
          – David C. Ullrich
          Dec 3 '18 at 14:54


















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