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
asked Apr 3 '16 at 18:02
Bib-lostBib-lost
1,998627
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add a comment |
$begingroup$
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
asked Apr 3 '16 at 18:02
Bib-lostBib-lost
1,998627
$endgroup$
add a comment |
$begingroup$
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
asked Apr 3 '16 at 18:02
Bib-lostBib-lost
1,998627
$endgroup$
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
asked Apr 3 '16 at 18:02
Bib-lostBib-lost
1,998627
asked Apr 3 '16 at 18:02
Bib-lostBib-lost
1,998627
asked Apr 3 '16 at 18:02
Bib-lostBib-lost
1,998627
asked Apr 3 '16 at 18:02
Bib-lostBib-lost
1,998627
asked Apr 3 '16 at 18:02
Bib-lostBib-lost
1,998627
1,998627
add a comment |
add a comment |
2 Answers
2
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oldest
votes
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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 $A$.
answered Apr 3 '16 at 18:33
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7155282
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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?
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– Bib-lost
Apr 3 '16 at 18:47
1
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@Bib-lost: You can embed $A$ in $End(A)$ since $A$ acts on itself by multiplication.
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– Eric Wofsey
Apr 3 '16 at 18:55
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Ok, got it to work, thanks!
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– Bib-lost
Apr 3 '16 at 20:25
add a comment |
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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.
answered Apr 3 '16 at 18:26
David C. UllrichDavid C. Ullrich
59.6k43893
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$begingroup$
@user43208 I always get that backwards - thanks.
$endgroup$
– David C. Ullrich
Dec 3 '18 at 14:54
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
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 $A$.
answered Apr 3 '16 at 18:33
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7155282
$endgroup$
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
add a comment |
$begingroup$
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 $A$.
answered Apr 3 '16 at 18:33
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7155282
$endgroup$
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
add a comment |
$begingroup$
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 $A$.
answered Apr 3 '16 at 18:33
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7155282
$endgroup$
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 $A$.
answered Apr 3 '16 at 18:33
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7155282
answered Apr 3 '16 at 18:33
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7155282
answered Apr 3 '16 at 18:33
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
111k7155282
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
add a comment |
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
add a comment |
$begingroup$
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.
answered Apr 3 '16 at 18:26
David C. UllrichDavid C. Ullrich
59.6k43893
$endgroup$
$begingroup$
@user43208 I always get that backwards - thanks.
$endgroup$
– David C. Ullrich
Dec 3 '18 at 14:54
add a comment |
$begingroup$
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.
answered Apr 3 '16 at 18:26
David C. UllrichDavid C. Ullrich
59.6k43893
$endgroup$
$begingroup$
@user43208 I always get that backwards - thanks.
$endgroup$
– David C. Ullrich
Dec 3 '18 at 14:54
add a comment |
$begingroup$
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.
answered Apr 3 '16 at 18:26
David C. UllrichDavid C. Ullrich
59.6k43893
$endgroup$
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.
answered Apr 3 '16 at 18:26
David C. UllrichDavid C. Ullrich
59.6k43893
edited Dec 3 '18 at 14:54
answered Apr 3 '16 at 18:26
David C. UllrichDavid C. Ullrich
59.6k43893
answered Apr 3 '16 at 18:26
David C. UllrichDavid C. Ullrich
59.6k43893
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
add a comment |
$begingroup$
@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
add a comment |
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