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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 $\| x \| = \| \langle x, x \rangle \| ^{1/2}$.

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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 –  t.b. Jul 3 '12 at 2:42
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@t.b.: Nothing goes wrong, and that is the usual definition. Peter: Where are you having trouble? –  Jonas Meyer Jul 3 '12 at 2:52
    
@Jonas: thanks! –  t.b. Jul 3 '12 at 3:01
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1 Answer 1

$A$ 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.)

If you are having trouble proving one of these assertions, please let us know by making your question more specific.

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