# existence of the extension of “derivation”

First, we can define a "derivation" (a linear operator) $\delta$ on $\mathbb{C}[x_{1},\cdots\\,\\,]$ by defining $$\delta(x_{k})=x_{k+1},\forall k\geq 1,\delta(x_{m}x_{n})=\delta(x_{m})x_{n}+x_{m}\delta(x_{n})=x_{m+1}x_{n}+x_{m}x_{n+1},$$ for example, $\delta(x_{2}^2x_{3}+x_{4}+2)=2x_{2}x_{3}^2+x_{2}^2x_{4}+x_{5}$.

Second, define $P_m$ to be the set of all polynomials in $m$ indeterminates, and $C(A),C(B)$ to be the $C^*$ algebras, where $A, B$ denote bounded closed sets in $\mathbb{C}^N,\mathbb{C}^{N+1}$ respectively. Define $P\_N|\_A$ and $P\_{N+1}|\_B$ to be the set of $P_N$, $P_{N+1}$ defined on $A,B$ respectively. Then the previous derivation $\delta$ can be considered as a linear operator from $P\_N|\_A$ to $P\_{N+1}|\_B$, and the set $P\_N|\_A$ to $P\_{N+1}|\_B$ are respectively dense in $C(A)$,$C(B)$ with respect to its norm. I want to ask the following question:

Can we extend the derivation $\delta$ to a "derivation" from $C(A)$ to $C(B)$, i.e., does there exist a derivation $\tilde{\delta}$ from $C(A)$ to $C(B)$, such that the restriction of $\tilde{\delta}$ to $P_N|_A$ is the previous one $\delta$ ? Here, the latter "derivation"$\tilde{\delta}$ means the common one defined for $C^*$ algebras, i.e, a linear operation which satisfies the Lebniz relation: $\tilde{\delta}(fg)=\tilde{\delta}(f)g+f\tilde{\delta}(g),\forall f,g\in C(A)$, and note that all the derivations here are not assumed to be bounded.

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I've fixed the LaTeX in your post, and made a few very minor changes I think will help the question be more readable. If you don't like any of the changes I made you are welcome to undo them. –  Zev Chonoles Apr 7 '11 at 15:49
Thanks Zev, you have helped a lot! –  Jiang Apr 7 '11 at 15:51
To talk about a derivation $\delta : A\to V$ you need to specify an $A$-bimodule structure on $V$, for the Leibniz rule to make sense. Can you explain how $C(A)$ acts on $C(B)$ (I don't exactly follow this since $A$ and $B$ seem to be unrelated sets, unless I am misreading the question). It may be a useful fact that any derivation from a $C^*$-algebra $A$ into a Banach $A$-bimodule is automatically continuous and bounded (J. Ringrose,J. London Math. Soc. (2), 5 (1972), 432-438). Thus if the bimodule structure you want is a Banach bimodule one, my guess would be that the answer is "no". –  Dima Shlyakhtenko Apr 8 '11 at 1:43
Oh,maybe I have to add the condition that $B=A*L$,here $L$ means a bounded interval in $\mathbb{C}$, and regard $C(A)$ as a natural subspace of $C(B)$.But considering the fact that you have mentioned, the answer is indeed "no". However, I really want to know the question:Could the "derivation" $\delta$ in the first paragraph be really considered as a restriction of a derivation from a $C^*$ algebra to another one? Of course, these $C^*$ algebras should be noncommutative. –  Jiang Apr 8 '11 at 4:11
In this generality, yes, at least for $N=1$ (I suppose for other $N$ it's similar), since you can always take an inner derivation from $A_1 \to A_1\otimes A_1$ given by $\delta(x)=1\otimes [x,T]$ for $T\in A_1$ and arrange for $A_2=A_1\otimes A_1$ to contain a non-self-adjoint element $x_1$, and $A_2$ to contain two commuting elements called $x_1\otimes 1$ and $1\otimes [x,T]$. You can even arrange that both $x_1$ and $x_2$ generate the polynomial algebra on $2$ variables. I think this does what you want? –  Dima Shlyakhtenko Apr 8 '11 at 5:05