What is the exact value of the first eigenvalue of the laplacian for complex projective space viewed as $SU(n+1)/S(U(1)\times U(n))$?

1$\begingroup$ Hmmm, I guess answer is  zero. Constant function is zero eigen for Laplace is not it ? Otherwise you might want to fix some normalization of metric ? Like volume = 1 ? $\endgroup$ – Alexander Chervov Jan 12 '12 at 13:57

6$\begingroup$ I guess the OP means the first NONZERO eigenvalue. $\endgroup$ – Alain Valette Jan 12 '12 at 16:35
The spectrum of the Laplacian of $\mathbb C P^n$ with the FubiniStudy metric is
$$Spec(\Delta_{\mathbb C P^n})=\{4k(n+k):k\in\mathbb N\} \quad\quad(*)$$
So, the first nonzero eigenvalue of $\mathbb C P^n$ is $\lambda_1=4n+4$.
Note this matches with the fact that $\mathbb C P^1$, with the FS metric, is isometric to the $2$sphere of radius $1/2$, whose first nonzero eigenvalue is $\lambda_1=8$.
Let me quote a brief justification of (*) that I had written here:
Using the classic spherical harmonics theory, one obtains the $k$th eigenvalue of the $n$dimensional round sphere $S^n$ to be $k(k+n1)$, and its multiplicity is $\binom{n+k}{k}\binom{n+k2}{k2}$. By looking at eigenfunctions of the Laplacian on $S^n$,$S^{2n+1}$ and $S^{4n+3}$ (note they are the unit spheres of $\mathbb R^{n+1}$, $\mathbb C^{n+1}$ and $\mathbb H^{n+1}$) that are respectively invariant under the natural actions of $\mathbb Z_2$, $S^1$ and $S^3$, one can obtain the eigenfunctions hence the $k$th eigenvalue of the projective spaces $\mathbb R P^n$, $\mathbb C P^n$ and $\mathbb H P^n$, respectively. These are, respectively, $2k(n+2k1)$, $4k(n+k)$ and $4k(k+2n+1)$.
If you can understand some French, you will find a thorough explanation of the above in the book by Berger, Gauduchon, Mazet, "Le spectre d'une variete Riemannienne", Lecture Notes in Math, Springer, vol 194; but beware of a small typo [pointed out to me by G. Wei] on the multiplicity of the $k$th eigenvalue of $S^n$ (cf. above).
More generally, it is possible to compute the Laplace spectrum of a compact homogeneous space using some representation theory, as explained e.g. in Nolan Wallach's "Minimal immersions of symmetric spaces into spheres" or in more detail on his book "Harmonic analysis on homogeneous spaces" (Pure and applied mathematics, 19).

$\begingroup$ What is decomposition of $L^2(CP^n)$ in irreps of su(n). From the "general theory(??)" there should be only tautological representations in $C^n$ and its symmetric powers. Is it true? what is their multiplicity ? (I guess should be 1  by analogy with BorelWeil). I mean by "General theory" the following argument which is not formal  one can look on "functional dimension" of irrep  which corresponds to 1/2 of dimension of corresponding orbit  the point ALL other irreps correspond to higherdimensional orbits  so they cannot be realized by differential operators on $CP^n$. $\endgroup$ – Alexander Chervov Jan 20 '12 at 6:59

$\begingroup$ @Alexander: The way I indicate how the spectrum of $CP^n$ can be computed does not use (directly) the rep theory arguments you mention. Assuming one knows the eigenfunctions of the Laplacian on spheres (by using spherical harmonics), the invariant ones will be eigenfunctions of the Laplacian on projective spaces. Now, following your suggestions, one can also compute the spectrum using irreducible representations of $SU(n)$. From the PeterWeyl Theorem one gets a decomposition of $L^2(SU(n))$ and then of $L^2(CP^n)$ by picking only invariant rep's ... $\endgroup$ – Renato G. Bettiol Jan 20 '12 at 17:39

1$\begingroup$ @Alexander: ... The Casimir element acts on each representation that appears in this decomposition of $L^2(CP^n)$ as multiplication by a certain scalar, that depends on half the sum of the positive roots of the representation and its highest weight. The formula can be found e.g. in N. Wallach's book. Thus, since the irred rep's of $SU(n)$ are wellknown, one can look at the ones that appear in the decomposition of $L^2(CP^n)$ and compute its highest weight and half sum of positive roots to obtain the corresponding eigenvalue. Doing this for all such representations gives the entire spectrum. $\endgroup$ – Renato G. Bettiol Jan 20 '12 at 17:42

$\begingroup$ @Renato thank you for your comments ! But let me formulate in the other words my remark. What is (or is there) the easiest (1sentence) way to get spectrum for $L^2(CP^n)$ ??? What I advocated that it should be $S^k(C^n)$. May be argument is unclear (I can try to explain if so)  but it is quite short. $\endgroup$ – Alexander Chervov Jan 21 '12 at 14:59

$\begingroup$ @Alexander: I'm afraid there's no "1sentence" way to get that... If I understood correctly what you suggest, the point is to use the PeterWeyl theorem to decompose $L^2(G/K)=\bigoplus o_\rho(G/K)$ where we sum over equivalence classes of representations $\rho$ of $G$ for which $K$ has some nontrivial fixed vector. [for details, see Thm 1.3, p. 17 of Takeuchi's book "Modern spherical functions", AMS] Then, applying this to $G/K=SU(n+1)/S(U(1)\times U(n)$ we get a decomposition of $L^2(CP^n)$ and we know how the Casimir element acts on each factor (by mult. by the correspondent eigenvalue). $\endgroup$ – Renato G. Bettiol Jan 21 '12 at 17:32
See SPECTRA AND EIGENFORMS OF THE LAPLACIAN ON $S^n$ AND $P^(C)$ (osaka j math 1977, you can skip to page 529 or if really lazy look at Theorem 5.2
Not an answer but advise.
Laplace probably(?) comes(should I explain this "comes"???) from quadratic Casimir in U(g) up to scalar factor which depends on volume.
Casimir can be written as \sum_i e^ie_i, where e^i and e_i are dual basises in g, with respect to CartanKilling form.
So the question is what is the minimal eigen of quadratic Casimirs in finitedim representations which enter decomposition of L^2(CP^n).
I guess(?) standard vector representation of su(n) in C^n enters this decomposition.
I guess(?) minimal eigen of quadratic Casimir in ALL irreps corresponds to this C^n.
If all guesses are correct you need just to calculate e^ie_i value in C^n and also care about the scalar which is related to volume normalization.