# Need help determining whether a certain map is a $C^\ast$ homomorphism

Hello, I need help determining whether the map I defined between two algebras is a well-defined homomorphism of $C^\ast$-algebras. I ran into this problem while trying to define a "rotation map" between algebras.

Here are the notations:

• $S=C_0(\mathbb{R})$, viewed as a $\mathbb{Z}_2$-graded algebra (odd and even functions corresponding exactly to odd and even grading degree)
• $V, W$ are finite dimensional real Euclidean spaces.
• $Cliff(V)$ is the complexified Clifford algebra over V. $C_\tau(V)=C_0(V, Clifford(V))$
• All the $\otimes$ you see here actually mean "graded tensor product" (can somebody tell me how to type that in LaTeX?), with which we have the well-known identities $Cliff(V\oplus W)=Cliff(V)\otimes Cliff(W)$ and $C_\tau(V\oplus W)=C_\tau(V)\otimes C_\tau(W)$
• $A(V)$ is the subalgebra of $S\otimes C_\tau(V)$ generated by elements of the form $f(s\otimes1+1\otimes \Psi(v-a))$ for some vector $a \in V$ (in technical terms, this is a Bott-type element with center at a). $\Psi \colon V \to Cliff(V)$ and maps a vector to the corresponding degree 1 element in Clifford algebra.
• For a given $f(s)\in S$. Let $f(s) = f_0(s^2) + sf_1(s^2)$ (odd and even part). Then $f(s\otimes1+1\otimes \Psi(v-a))=f_0(s^2+||v-a||^2)+sf_1(s^2+||v-a||^2)+\Psi(v-a)f_1(s^2+||v-a||^2)$

My question: Is the following map

$\beta_t \colon A(V) \to A(V \oplus W)$, $f(s \otimes 1 + 1 \otimes \Psi(x-a)) \mapsto f(s \otimes 1 \otimes 1 + 1 \otimes \Psi(x_1 - \cos t \cdot a) \otimes 1 + 1 \otimes 1 \otimes \Psi(x_2 - \sin t \cdot a))$

a homomorphism of $C^\ast$-algebras?

My first concern is: I define only the map on generators, how do I know if it actually extends to a well-defined map on the whole algebra? (or at least the dense subalgebra generated by the generators?) Thank you!

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It seems like neither $s\otimes 1$, nor $1\otimes \Psi(v-a)$ belong to $C_0(\mathbb{R})\otimes C_\tau(V)$, so what you mean by $f(s\otimes 1+1\otimes\Psi(v-a))$ is unclear –  Ollie Margetts May 2 '13 at 11:47
Thanks for your comment! You are right that neither of $s \otimes 1$ nor $1 \otimes \Psi(v-a)$ is in the algebra, but after applying "functional calculus" (see my point 6), the resultant function will be in the algebra. You may try using $f(s)=e^{-s^2}$ or $f(s)=se^{-s^2}$ and see for yourself :) –  Clark Chong May 5 '13 at 5:39