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Let $A$ be a C*-algebra. Consider its cartesian square $A^2$ and define a multiplication on $A^2$ by the identity $$ (x_0,x_1)\cdot (y_0,y_1)=(x_0y_0,x_0y_1+x_1y_0),\qquad x_0,x_1,y_0,y_1\in A $$ This turns $A^2$ into the algebra of "polynomials of one variable of degree 1 with coefficients in $A$": if we denote by $t$ this variable, then each element of $A^2$ can be represented as a polynomial $$ p(t)=x_0+x_1\cdot t,\qquad x_0,x_1\in A, $$ and the multiplication in $A^2$ will be the usual multiplication of polynomials (under the agreement that $t^2=0$): $$ p(t)=x_0+x_1\cdot t,\quad q(t)=y_0+y_1\cdot t $$ $$ \Downarrow $$ $$ p(t)\cdot q(t)=(x_0+x_1\cdot t)\cdot (y_0+y_1\cdot t)=x_0\cdot y_0+(x_0y_1+x_1y_0)\cdot t $$

Is it possibe to define a natural C*-norm on $A^2$?

I mean a C*-norm $\|\cdot\|_{A^2}$ such that $$ \|(x_0,0)\|_{A^2}=\|x_0\|_A,\qquad x_0\in A. $$

Similarly one can define the algebra of polynomials of $m$ variables of degree $n$ with coefficients in $A$, and I am curious in the same question for this case.

Michael Hardy
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Sergei Akbarov
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1 Answers1

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"Polynomials" seems like a misleading name; what you're actually doing is trying to adjoin nilpotents.

Anyway, the answer is no, this is already impossible when $A$ is commutative (with no naturality assumption: it is not possible to define any C*-norm on the underlying algebra). The reason is that adjoining nilpotents in the way you want to a commutative C*-algebra gives another commutative C*-algebra, but commutative C*-algebras have no nonzero nilpotent elements by the commutative Gelfand-Naimark theorem.

Qiaochu Yuan
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  • Thank you! I need a name for such algebras, what do people call them? – Sergei Akbarov Dec 21 '14 at 23:42
  • No, I mean a name for the class of such algebras. I thought they can be called "algebras of polynomials of a given degree with coefficients in a C*-algebras". Are you saying that this is inapt? – Sergei Akbarov Dec 21 '14 at 23:49
  • I'm saying it's misleading because polynomials of a given degree form a vector space, not an algebra; you've chosen a multiplication so that the variables you're adjoining are nilpotent, but you could choose others. – Qiaochu Yuan Dec 21 '14 at 23:57
  • If we define a multiplication like I describe, polynomials of given degree form an algebra. – Sergei Akbarov Dec 22 '14 at 00:00
  • Yes, but that's a choice you've made that you're hiding with the use of the term "polynomial." I'm not saying your definition isn't meaningful, just that using the term "polynomial" for it is misleading. – Qiaochu Yuan Dec 22 '14 at 00:33
  • This is strange, I can't find the definition for adjoined elements in an algebra. Is it assumed automatically that the adjoined elements must commute with each other and with the elements of the initial algebra $A$? – Sergei Akbarov Dec 22 '14 at 08:17
  • @Sergei: no, of course you can adjoin whatever you want. When I write $A[t]$ and so forth I have in mind primarily the commutative case. If $A$ is noncommutative I don't know what the standard notation is for adjoining a new element with no commutativity requirements. – Qiaochu Yuan Dec 22 '14 at 08:42
  • One of the most unpleasant things in our work is seeking a proper term... My algebra $A$ is not commutative, but those adjoined elements must commute with each other and with elements of $A$. – Sergei Akbarov Dec 22 '14 at 09:12
  • @QiaochuYuan: Any hint on why $t$ in $A[t]$ would have spectrum ${0}$? To me, if $A[t]$ denotes the polynomial algebra in $t$ with coefficients in $A$, the spectrum of $t$ should be $\mathbb{C}$. Perhaps $A[t]$ denotes the algebra of entire functions in $t$ with coefficients in $A$. This makes sense too. In this case, which norm is this space equipped with? – Jeyrome Sapin Jun 05 '21 at 09:22
  • @Jeyrome: whoops, you're right. – Qiaochu Yuan Jun 25 '21 at 01:39