Proof. [03I2]

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Proof.

Since $\omega$ is harmonic, by Bochner’s formula,

(4.169)

\frac{1}{2}\Delta_{g}|\omega|^{2}=|\nabla\omega|^{2}+\Ric_{g}(\omega,\omega)\geq 0,

then $|\omega|^{2}$ is subharmonic. Given the asymptotic property (4.168), applying the maximum principle to the above subharmonic function $|\omega|^{2}$ , we have $\omega\equiv 0$ on $M^{n}$ .

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