Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was wondering wether one can always find/construct a norm which turns an involutive algebra into a C*-algebra. For sure, if it exists it is unique, but does it always exist. If not can you provide a simple example of an involutive algebra such that there's no norm which turns it into a C*-algebra. Thanks alot! Kind regards, Alex

share|improve this question
3  
You don't just have to find a norm. You also have to have $x \mapsto x^*$. –  kahen May 27 '13 at 22:31
2  
See also Why is $\ell^1(\mathbb{Z})$ not a C*-algebra? –  Martin May 27 '13 at 22:35
add comment

2 Answers

up vote 10 down vote accepted

(Below by "algebra" I mean "complex algebra.")

You can't even always find a norm that turns an algebra into a Banach algebra. For example, the Weyl algebra $\mathbb{C} \langle x, y \rangle/(xy - yx - 1)$ can't embed into a Banach algebra (see this math.SE question).

Another constraint on being a Banach algebra is that the spectral radius (a purely algebraic concept) of every element of a Banach algebra is finite (in fact bounded by its norm), and this fails in general. For example, the spectral radius of every nonzero element of $\mathbb{C}[x]$ is infinite.

Yet another constraint is that, by the Gelfand-Mazur theorem, the only Banach algebra which is also a division algebra is $\mathbb{C}$. This means that, for example, $\mathbb{C}(x)$ cannot be a Banach algebra, but it means more generally that any commutative algebra $A$ with a maximal ideal $m$ such that $A/m$ is not isomorphic to $\mathbb{C}$ cannot be a Banach algebra (since if $A$ were a Banach algebra then $m$ would be closed, so $A/m$ would also be a Banach algebra).


In the case of C*-algebras there are even heavier restrictions. For example, the finite-dimensional C*-algebras are products of matrix algebras, and these are precisely the semisimple algebras. Most finite-dimensional algebras are very far from semisimple; the smallest example is $\mathbb{C}[x]/x^2$ (which can be made into a Banach algebra).

More generally, the Jacobson radical of a C*-algebra is always zero (so a C*-algebra is always semiprimitive). This is a strong and purely algebraic constraint, and it is easy to write down many examples of algebras which are not semiprimitive, e.g. a commutative algebra with nilpotent elements can't be semiprimitive.

share|improve this answer
    
Ok that shows that it cannot even be equipped with a norm turning it into a normed rather than a banach algebra since any norm gives rise to a completion ...the construction is outlined in math.ksu.edu/~nagy/real-an/2-05-b-alg.pdf –  Freeze_S Apr 10 at 0:24
    
Btw thx for the awesome answer ;) –  Freeze_S Apr 10 at 0:28
add comment

Im not sure about $M_2(\mathbb{C})$. Because if it had an involution, then the restriction of the norm of $M_2(\mathbb{C})$ would make it a $C^*$-algebra. But we know it is not a $C^*$-algebra.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.