$ \langle\Delta_\partial\omega,\omega\rangle = || \partial \omega||^2 + ||\partial^*\omega||^2 $ on compact Kahler manifold

Question

Why do we need compactness to have $ \langle\Delta_\partial\omega,\omega\rangle = || \partial \omega||^2 + ||\partial^*\omega||^2 $?

I think $ \langle\Delta_\partial\omega,\omega\rangle =<\partial\partial^*\omega,\omega>+<\partial^*\partial\omega,\omega>=<\partial^*\omega,\partial^*\omega>+<\partial\omega,\partial\omega>= || \partial \omega||^2 + ||\partial^*\omega||^2 $. Why do we need compactness then?

Answer 1

The problem is that the adjoint operator $\partial^*$ in general only exists on compact manifolds.

We want an operator $\partial^*$, such that $\left< \partial \alpha,\beta\right>=\left< \alpha, \partial^*\beta\right>$ for all $\alpha, \beta$.

Let $\alpha$ a $(p-1,q)$-form and $\beta$ a $(p,q)$-form on $M$. We calculate: (see Huybrechts p.125) $$\left<\partial \alpha, \beta\right>= \int_M \partial \alpha\wedge *\overline {\beta} \stackrel{Leibniz~Rule}= \int_M \partial(\alpha \wedge * \overline \beta)-(-1)^{p+q-1}\int_X\alpha\wedge \partial(*\overline \beta)$$ The first summand vanishes if $M$ is compact due to Stokes' theorem as $\alpha \wedge *\overline \beta$ is of type $(n-1,n)$ and hence $\partial(\alpha \wedge *\overline \beta)=d(\alpha \wedge *\overline \beta)$.

For the second summand, note that $\partial(*\overline \beta)$ is of type $(n-p+1,n-q)$. Hence $$\partial(*\overline \beta)= (-1)^{(2n-p-q+1)(p+q-1)}**\partial(*\overline \beta)= (-1)^{-(p+q-1)^2}**\partial(*\overline \beta)=(-1)^{-(p+q-1)}**\partial(*\overline \beta).$$ Inserting this in the second summand we get: $$-(-1)^{p+q-1}\int_X\alpha\wedge \partial(*\overline \beta)= -(-1)^{p+q-1}\int_X\alpha\wedge (-1)^{-(p+q-1)}**\partial(*\overline \beta)\\= -(-1)^{p+q-1-(p+q-1)} \left< \alpha, \overline{*\partial*\overline\beta}\right>=\left< \alpha,- *\overline \partial*\beta\right> $$

We conclude $$\left< \partial \alpha, \beta\right>= \left< \alpha,- *\overline \partial*\beta\right> + \int_M \partial(\alpha \wedge * \overline \beta).$$ If $M$ is compact, then the second summand vanishes and $\partial^*=-*\overline\partial *$.
If $M$ is not compact, then the second summand does in general not vanish and could take any value, so there is no adjoint operator in this case.