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

edited Mar 16 2012 at 22:32

Renato G Bettiol
2,070●5●12

asked Jan 12 2012 at 11:02

Soma
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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 ? – Alexander Chervov Jan 12 2012 at 13:57

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I guess the OP means the first NONZERO eigenvalue. – Alain Valette Jan 12 2012 at 16:35

Renato G Bettiol · Answer 1 · 2012-03-16 15:33:16Z UTC

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The spectrum of the Laplacian of $\mathbb C P^n$ with the Fubini-Study metric is

$$Spec(\Delta_{\mathbb C P^n})=\{4k(n+k):k\in\mathbb N\} \quad\quad(*)$$

So, the first non-zero 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 non-zero 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 $k(k+n-1)$, and its multiplicity is $\binom{n+k}{k}-\binom{n+k-1}{k-1}$, see e.g. [Berger, Gauduchon,Mazet, "Le spectre d'une variété riemannienne", Lecture Notes in Mathematics, Vol. 194 Springer-Verlag]. 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, $2k(n+2k-1)$, $4k(n+k)$ 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.

edited Mar 16 2012 at 15:33

answered Jan 13 2012 at 18:05

Renato G Bettiol
2,070●5●12

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 Borel-Weil). 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 higher-dimensional orbits - so they cannot be realized by differential operators on $CP^n$. – Alexander Chervov Jan 20 2012 at 6:59

@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 Peter-Weyl Theorem one gets a decomposition of $L^2(SU(n))$ and then of $L^2(CP^n)$ by picking only invariant rep's ... – Renato G Bettiol Jan 20 2012 at 17:39

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@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 well-known, 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. – Renato G Bettiol Jan 20 2012 at 17:42

@Renato thank you for your comments ! But let me formulate in the other words my remark. What is (or is there) the easiest (1-sentence) 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. – Alexander Chervov Jan 21 2012 at 14:59

@Alexander: I'm afraid there's no "1-sentence" way to get that... If I understood correctly what you suggest, the point is to use the Peter-Weyl 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 non-trivial 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). – Renato G Bettiol Jan 21 2012 at 17:32

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Igor Rivin · Answer 2 · 2012-01-12 11:18:46Z UTC

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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

answered Jan 12 2012 at 11:18

Igor Rivin
31.1k●3●37●71

Alexander Chervov · Answer 3 · 2012-01-12 18:28:06Z UTC

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 Cartan-Killing form.

So the question is what is the minimal eigen of quadratic Casimirs in finite-dim 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.

The first eigenvalue of the laplacian for complex projective space

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