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Is there a compact Reiemannian manifold $M$ for which the following complex $C^{*}$ algebra does not have a nontrivial idempotent:

$A=Hom(E,E)$ where $E$ is the complexification of $TM$.

Of course any such $M$ can not have an almost complex structure.

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No: if $M^n$ is $n$-dimensional then $E = TM \otimes_{\mathbb{R}} \mathbb{C}$ is $n$-complex dimensional so classified by a map $M \to BU(n)$. By obstruction theory this lifts to $BU(n-1)$ (the obstructions are in the groups $H^{i+1}(M^n; \pi_i(S^{2n-1}))$, which all vanish).

Hence there is an isomorphism of vector bundles $E \cong \mathbb{C} \oplus E'$, and projection to $E'$ gives a nontrivial idempotent.

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  • $\begingroup$ To nitpick, in the case $n=1$ this argument doesn't work (as $E'$ will be $0$), and indeed $S^1$ is a counterexample. To nitpick further, a point and the empty manifold are also counterexamples (though arguably for the right definition of "no nontrivial idempotents" the empty manifold is not a counterexample). $\endgroup$
    – Eric Wofsey
    Commented Jun 23, 2015 at 10:38
  • $\begingroup$ @Oscar I would appreciate if you more explain about the first part of your answer. $\endgroup$
    – Ali Taghavi
    Commented Jun 23, 2015 at 15:36
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    $\begingroup$ In other words: a generic section of $E \to M$ will be nowhere zero (as its zeroes will have real codimension $2n$ in an $n$-manifold). Such a section gives a trivial 1-dimensional sub-bundle. $\endgroup$
    – Oscar Randal-Williams
    Commented Jun 23, 2015 at 15:54

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