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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

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You don't just have to find a norm. You also have to have $x \mapsto x^*$. – kahen May 27 '13 at 22:31
See also Why is $\ell^1(\mathbb{Z})$ not a C*-algebra? – Martin May 27 '13 at 22:35
up vote 11 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.

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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 – Alexander Frei Apr 10 '14 at 0:24
Btw thx for the awesome answer ;) – Alexander Frei Apr 10 '14 at 0:28

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.

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Restriction of the norm of $M_2(\mathbb{C})$ to $M_2(\mathbb{C})$??? I am sorry, but I have some difficulty to follow your argument, can you please explain? Many thanks in advance! – UwF Jul 15 '14 at 19:34

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