## First eigenvalue of $\Delta$ on Kaehler manifold with $Ricci\ge k$.

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

Let $M$ be a Kaehler manifold of complex dimension $n$. Let $\Delta$ be the real Laplacian of the underline Riemannian manifold. Let's assume the Ricci curvature of $M$ satisfies $\text {Ric}\ge k>0$. Denoted the first eigenvalue of $\Delta$ by $\lambda_1$. Then we have $$\lambda_1\ge 2k.$$ Notice that this estimate is better than the non-Kaehler case namely Lichnerowicz's estimates for Riemannian manifolds would yield the lower bound: $$\lambda_1\ge \frac{2n}{2n-1}k.$$

Some authors attributes this results to Lichnerowicz: G´eom´etrie des groupes de transformations, Travaux et Recherches Math´ematiques, III. Dunod, Paris, 1958.

Some author attributes it to Udagawa: Compact Kaehler manifolds and the eigenvalues of the Laplacian Colloq. Math. 56 (1988), no. 2, 341–349.

Also Urakaw also give a proof of this lower bound in "Stability of harmonic maps and eigenvalues of the Laplacian" Trans of AMS 1987. However the proof given in this paper is not direct and depends on the more sophisticated Harmonic map theory, which is not the way I prefer.

However I can't find the first two references. So my question is: who is the first to prove this theorem and is there any book or reference that contains the proof.(Chavel's book on eigenvalues does not contain Kaehler case.)

I suppose the proof is not hard based on the length of paper of Udagawa.

-

## 1 Answer

I am not sure about the history of this result. You can find in Thierry Aubin's book "Some nonlinear problems in Riemannian geometry", Theorem 4.20. He attributes it to his own 1978 paper.

The proof is really simple, though. If you write $\Delta f=g^{i\overline{j}}f_{i\overline{j}}$ for the complex Laplacian, and assume that $\Delta f=-\lambda f$ with $\lambda$ the first eigenvalue of $\Delta$ (which is half of the first eigenvalue of the real Laplacian), then commuting covariant derivatives you can easily see that $$\int_M (\Delta f)^2=\int_M g^{i\overline{j}}g^{p\overline{q}}f_{i\overline{j}} f_{\overline{q}p}=-\int_M g^{i\overline{j}}g^{p\overline{q}}f_{i} f_{\overline{q}p\overline{j}}=-\int_M g^{i\overline{j}}g^{p\overline{q}}f_{i} f_{\overline{q}\overline{j}p} +\int_M R^{i\overline{j}} f_i f_{\overline{j}}$$ $$=\int_M |f_{ij}|^2+\int_M R^{i\overline{j}} f_i f_{\overline{j}}.$$ If you assume that $\mathrm{Ric}\geq k>0$, then $R_{i\overline{j}}\geq kg_{i\overline{j}}$ and so $$\lambda^2\int_M f^2 =\int_M(\Delta f)^2\geq k\int_M |\partial f|^2=k\lambda\int_M f^2,$$ and so $\lambda\geq k$.

-

@YangMills, I was looking at your answer to a question here: mathoverflow.net/questions/120286/…. It seems there is factor $2$ in front of the Real part. But why there is no $2$ in the first equality before $\int_M (\Delta f)^2$. I think this term comes from integration by part of the Real part of your answer. Did I miss something? – Ralph Feb 1 at 20:16
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 19, "mathjax_display_tex": 5, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8992531299591064, "perplexity_flag": "head"}