Timeline for The first eigenvalue of the laplacian for complex projective space
Current License: CC BY-SA 3.0
16 events
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| Jun 6, 2017 at 19:51 | history | edited | Renato G. Bettiol | CC BY-SA 3.0 |
Rectified a typo, included further references
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Mar 16, 2012 at 15:33 | history | edited | Renato G. Bettiol | CC BY-SA 3.0 |
Fixed typo; improved exposition
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| Jan 23, 2012 at 17:13 | comment | added | Alexander Chervov | @Renato what are the irreps which enter decomposition $L^2(CP^n)$ ? Is it true that they are only $S^k(C^n)$ ? i.e. those which have highest weight (k,0,0,0,...0) ? | |
| Jan 22, 2012 at 17:08 | comment | added | Renato G. Bettiol | @Alexander: I'm sorry, but I'm afraid I don't understand your remark. Anyways, independently of what method is use, the final result is the one posted :) Just one comment regarding the multiplicities: the multiplicity of the $k^{th}$ eigenvalue $4k(n+k)$ of the Laplacian on $CP^n$ is $\binom{n+k}{k}^2-\binom{n+k-1}{k-1}^2$. As in any other compact symmetric space of rank one, the multiplicities form a strictly increasing sequence (the higher the eigenvalue, the higher the multiplicity). The formula above can be computed with the same ideas of actions on spheres I used to describe spectrum. | |
| Jan 22, 2012 at 11:22 | comment | added | Alexander Chervov | By orbit I mean the coadjoint orbit which corresponds to this irrep. So we can conclude that $L^2(CP^n)$ contains only $S^k(C^n)$. However this argument does shed light what are the multiplicities. I guess all of them equal to 1. By analogy with Borel-Weil, but I am not sure this is true... | |
| Jan 22, 2012 at 11:19 | comment | added | Alexander Chervov | @Renato your remark is very interesting. However my argument was different. Let me try to express it again. For any M irreps in $L^2(M)$ are realized by differential operators, agree of course ? Now I want to say that $CP^n$ is $n$-dimensional - so we get map from $U(su(n))$ to differential operators in $n$ variables, agree of course ? The non-trivial point - only $S^k(C^n)$ - symmetric power of tautological representation can be realized by by differential operators in $n$ variables - the other irreps you need MORE VARIABLES ! Why? Because number of variables is half of dim(orbit). | |
| Jan 21, 2012 at 17:35 | comment | added | Renato G. Bettiol | @Alexander: ... anyways, if we're aiming at "1-sentence" ways, I feel like Berger-Gauduchon-Mazet's approach that I mention is much more straight-forward, and also requires less technical tools. Nevertheless, those representation theory techniques can certainly be applied to a much larger set of problems. | |
| Jan 21, 2012 at 17:32 | comment | added | Renato G. Bettiol | @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). | |
| Jan 21, 2012 at 14:59 | comment | added | Alexander Chervov | @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. | |
| Jan 20, 2012 at 17:42 | comment | added | Renato G. Bettiol | @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. | |
| Jan 20, 2012 at 17:39 | comment | added | Renato G. Bettiol | @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 ... | |
| Jan 20, 2012 at 6:59 | comment | added | Alexander Chervov | 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 $CP^n$. | |
| Jan 15, 2012 at 17:04 | history | edited | Renato G. Bettiol | CC BY-SA 3.0 |
Fixed typo.
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| Jan 14, 2012 at 17:22 | history | edited | Renato G. Bettiol | CC BY-SA 3.0 |
added 24 characters in body
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| Jan 13, 2012 at 18:05 | history | answered | Renato G. Bettiol | CC BY-SA 3.0 |