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I refer to the text Complex Geometry by David Huybrechts. In remark 3.2.7 iii) he stated that the surjectivity of the map $L^{n-k}:\mathcal{H}^{p,q}(X,g)\rightarrow \mathcal{H}^{n-q,n-p}(X,g)$ can be deduced from the fact that the dual Lefschetz operator maps harmonic forms to harmonic maps.

However I have trouble seeing how this leads to showing surjectivity. Can anyone kindly enlighten me?

Bernard
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Soby
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1 Answers1

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$L^{n-k}$ is an isomorphism from $\mathcal{A}^{k}_{\mathbb{C}}(X)$ to $\mathcal{A}^{2n-k}_{\mathbb{C}}(X).$ Thus$\Delta L^{n-k} \alpha =0 \iff L^{n-k}\Delta\alpha=0 \iff \Delta\alpha=0.$

For any $\beta \in \mathcal{H}^{n-q,n-p}(X,g),\,$ we can find$\,\alpha \in \mathcal{A}^{k}_{\mathbb{C}}(X),\,L^{n-k}(\alpha)=\beta.\,$By the argument above,$\,\alpha$ must be harmonic, that is, $\alpha \in \mathcal{H}^{k}(X,g).$ Since $L^{n-k}$ is of bidegree $(n-k,n-k),\,\alpha \in \mathcal{H}^{p,q}(X,g).$

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