Dual Lefschetz Operator and Contraction with the Fundamental Form Let $M$ be a Kahler manifold, with metric $g$, Kahler form $\omega$, Lefschetz operator $L$, and dual Lefschetz operator $\Lambda$.
$\Lambda$ and contraction with $\omega$ both map $k$-forms to ($k-2$)-forms (for $k \geq 2$): are they equal?

Is it correct to argue this way: for any $k$-form $\alpha$,
$$\Lambda\alpha = g(\Lambda\alpha,1)=g(\alpha,L1)=g(\alpha,\omega)\ ?$$
No. What is $1$? The constant $0$-form $1$ is by definition orthogonal to any $k$-form with $k\neq 0$.

The correct answer is $L^* = \mathrm{sgn} *L*$, where $\mathrm{sgn}(\varphi)=(-1)^k$ on $k$-forms. To see that this is a contraction with $\omega^\sharp$, write $L$ and $\omega$ in holomorphic normal coordinates so that at a point $p$ we have $\omega(p) = \sum \mathrm{d}z^i \wedge \mathrm{d}\bar{z}^i$. Then use that $L$ is an algebraic operator.

Finally, to see that $L^*=\mathrm{sgn} *L*$, we compute for $\varphi\in\wedge^{k-2}$ and $\psi\in\wedge^{k}$
\begin{align*}
 \langle L \varphi , \psi\rangle \mathrm{vol}
&amp;= L\varphi \wedge \bar * \psi \\
&amp;= \omega \wedge \varphi \wedge *\bar \psi \\
&amp;= \varphi \wedge \omega \wedge * \bar \psi \\
&amp;= \varphi \wedge L*\bar\psi \\
&amp;= \varphi \wedge (*\mathrm{sgn}*)(L*\bar\psi) \\
&amp;= \varphi \wedge * (\mathrm{sgn}*L*\bar\psi) \\
&amp;= \varphi \wedge * \overline{\mathrm{sgn} *L* \psi} \\
&amp;= \langle \varphi , \mathrm{sgn} *L*\psi\rangle \mathrm{vol}.
\end{align*}
To see that $*\mathrm{sgn}* = \mathrm{id}$ on $k$-forms, just use that $*^2=(-1)^k$ on $k$-forms. A good reference is Wells Differential Analysis on Complex Manifolds.
No. What is $1$? The constant $0$-form $1$ is by definition orthogonal to any $k$-form with $k\neq 0$.

The correct answer is $L^* = \mathrm{sgn} *L*$, where $\mathrm{sgn}(\varphi)=(-1)^k$ on $k$-forms. To see that this is a contraction with $\omega^\sharp$, write $L$ and $\omega$ in holomorphic normal coordinates so that at a point $p$ we have $\omega(p) = \sum \mathrm{d}z^i \wedge \mathrm{d}\bar{z}^i$. Then use that $L$ is an algebraic operator.

Finally, to see that $L^*=\mathrm{sgn} *L*$, we compute for $\varphi\in\wedge^{k-2}$ and $\psi\in\wedge^{k}$
\begin{align*}
 \langle L \varphi , \psi\rangle \mathrm{vol}
&amp;= L\varphi \wedge \bar * \psi \\
&amp;= \omega \wedge \varphi \wedge *\bar \psi \\
&amp;= \varphi \wedge \omega \wedge * \bar \psi \\
&amp;= \varphi \wedge L*\bar\psi \\
&amp;= \varphi \wedge (*\mathrm{sgn}*)(L*\bar\psi) \\
&amp;= \varphi \wedge * (\mathrm{sgn}*L*\bar\psi) \\
&amp;= \varphi \wedge * \overline{\mathrm{sgn} *L* \psi} \\
&amp;= \langle \varphi , \mathrm{sgn} *L*\psi\rangle \mathrm{vol}.
\end{align*}
To see that $*\mathrm{sgn}* = \mathrm{id}$ on $k$-forms, just use that $*^2=(-1)^k$ on $k$-forms. A good reference is Wells Differential Analysis on Complex Manifolds.