I am having trouble with this proof: Let $X$ be a Hilbert $A$-module and let $A$ be a $C^*$-algebra then the direct sum $A\oplus X$ is also a Hilbert $A$-module.

Useful information:

$X$ be a Hilbert $A$-module if $X$ is an inner-product $A$-module which is complete with respect to the norm $\| x \| = \| \langle x, x \rangle \| ^{1/2}$.

  • $\begingroup$ I'm confused: so the obvious candidate "inner product" $\langle (a,x),(b,y) \rangle_{A \oplus X} = a^\ast b + \langle x,y \rangle_X$ where $\langle \cdot,\cdot \rangle_X \colon X \times X \to A$ is the "inner product" on $X$ doesn't work? Why? I don't see what goes wrong, as all the axioms here seem to be satisfied $\endgroup$ – t.b. Jul 3 '12 at 2:42
  • 1
    $\begingroup$ @t.b.: Nothing goes wrong, and that is the usual definition. Peter: Where are you having trouble? $\endgroup$ – Jonas Meyer Jul 3 '12 at 2:52
  • $\begingroup$ @Jonas: thanks! $\endgroup$ – t.b. Jul 3 '12 at 3:01

$A$ is made a right module over $A$ in the usual way. It is a Hilbert $A$-module with the inner product $\langle a,b\rangle = a^*b$.

If $X$ and $Y$ are Hilbert $A$-modules, then $X\oplus Y$ is a Hilbert $A$-module with componentwise operations and $\langle (x_1,y_1),(x_2,y_2)\rangle=\langle x_1,x_2\rangle_X + \langle y_1,y_2\rangle_Y$. (This generalizes immediately to finite direct sums, and infinite direct sums also exist.)

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