It is shown that if $\alpha ,\zeta $ are ordinals such that $1\leq \zeta <\alpha <\zeta \omega ,$ then there is an operator from $C(\omega ^{ \omega ^\alpha })$ onto itself such that if $Y$ is a subspace of $C(\omega ^{ \omega ^\alpha })$ which is isomorphic to $C(\omega ^{ \omega ^\alpha })$ $,$ then the operator is not an isomorphism on $Y.$ This contrasts with a result of J. Bourgain that implies that there are uncountably many ordinals $\alpha $ for which any operator from $C(\omega ^{ \omega ^\alpha })$ onto itself there is a subspace of $C(\omega ^{ \omega ^\alpha })$ which is isomorphic to $% C(\omega ^{ \omega ^\alpha })$ on which the operator is an isomorphism.
Categories: math.FA