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

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?

• @peterag I am just confused, it seems that it’s true even without compactness – Danny Feb 21 at 6:56
• how did you define $\partial ^*$? – klirk Feb 21 at 17:15

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.

• you have a typo on "We want an operator $\partial^*$, such that $\left< \partial \alpha,- \partial^*\beta\right>$ for all $\alpha, \beta$" – Danny Feb 21 at 23:57
• Also, O'Wells' book, p.169 says $\partial^*=-\bar{*}\partial\bar{*}$ – Danny Feb 22 at 0:00
• @Danny: oh, fixed it – klirk Feb 22 at 11:06
• @Danny: $\bar *\neq *$, but $\bar * \partial \bar*$ should still be equal to $*\bar \partial *$. – klirk Feb 22 at 11:25