MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know the round $n$-sphere has $f_i=\cos(dist(e_i, x))$ as the set of first eigenfunctions for $e_i=(0, \cdots, 1, \cdots, 0)\in \mathbb R^{n+1}$. i.e. $\Delta f_i=\lambda_1 f$, where $\lambda_1$ is the first eigenvalue of $\Delta$.

So I am wondering whether there is a similar description for the 'round' complex projective spaces?

share|cite|improve this question
up vote 13 down vote accepted

Of course, Igor's answer points the way to working out the answer the OP wanted, but it may not be clear, even after you have got the eigenvalues, what the corresponding eigenfunctions are, or that they have a simple geometric interpretation analogous to the one for the sphere, as the OP asks.

The nice way to describe the first eigenfunctions on the $n$-sphere with its standard metric is to start with the standard isometric embedding $X:S^n\to\mathbb{R}^{n+1}$ and then, for each vector $v\in\mathbb{R}^{n+1}$, one defines the function $f_v:S^n\to\mathbb{R}$ by $$ f_v(p) = v\cdot X(p). $$ Every eigenfunction for the first nonzero eigenvalue of the Laplacian is of this form for a unique $v$.

Similarly, for $\mathbb{CP}^n$, there is a natural embedding $X:\mathbb{CP}^n\to\mathsf{H}_{n+1}$, where $\mathsf{H}_{n+1}$ is the (real) vector space of $(n{+}1)$-by-$(n{+}1)$ Hermitian symmetric matrices, given by $$ X\bigl([v]\bigr) = \frac{v\ v^\ast}{|v|^2} = \frac{v\ v^\ast}{(v^\ast v)}, $$ where $v$ is any nonzero vector in $\mathbb{C}^{n+1}$ (thought of as column vectors of height $(n{+}1)$ and $v^\ast$ is its conjugate transpose, and $[v]\in\mathbb{CP}^n$ is the line in $\mathbb{C}^{n+1}$ that is spanned by $v$. (With the right scaling and inner product on $\mathsf{H}_{n+1}$, this $X$ is an isometric embedding of $\mathbb{CP}^n$ endowed with the Fubini-Study metric. In some sense, it is the simplest such. Note that it is equivariant with respect to the natural actions of $\mathrm{SU}(n{+}1)$ on the domain and range.)

Now let $w\in\mathsf{H}_{n+1}$ be traceless and define $$ f_w\bigl([v])\bigr) = \mathrm{tr}\bigl(\ w X\bigl([v]\bigr)\ \bigr). $$ Then every eigenfunction of lowest nontrivial eigenvalue of the Laplacian is of the form $f_w$ for some traceless $w\in\mathsf{H}_{n+1}$. Thus, as in the case of the $n$-sphere, the lowest eigenfunctions are essentially the components of the 'natural' isometric embedding of $\mathbb{CP}^n$. (The reason for the 'essentially' qualifier is that, as defined, $X$ actually maps $\mathbb{CP}^n$ into the hyperplane in $\mathsf{H}_{n+1}$ that consists of the matrices of trace $1$, and one really ought to subtract the term $\tfrac1{n+1}\ I_{n+1}$ from $X$ so that it maps into the hyperplane of traceless matrices. Then the statement is literally true. Those who are familiar with symplectic geometry will recognize that if one then multiplies by $i$, so that the map goes into the traceless skew-Hermitian matrices, then the result is the canonical embedding of $\mathbb{CP}^n$ as a (co)-adjoint orbit of $\mathrm{SU}(n{+}1)$.)

To get something like the formula in terms of distance that held for the $n$-sphere, you just need to observe that as $[v]$ varies, the traceless elements $$ W\bigl([v]\bigr) = X\bigl([v]\bigr) - \tfrac{1}{n+1} I_{n+1} $$ span the space of traceless Hermitian $(n{+}1)$-by-$(n{+}1)$ matrices, so this means that this eigenspace is spanned by the functions of the form $$ h_{[w]}\bigl([v]\bigr) = \frac{\bigl|\langle w,v\rangle\bigr|^2}{|w|^2|v|^2} - \frac{1}{n+1} $$ as $[w]$ varies over $\mathbb{CP}^n$. Obviously, the function $h_{[w]}$ can be expressed as a function of the distance from $[w]$ in the Fubini-Study metric, even though the generic linear combinations of such eigenfunctions aren't usually expressible in terms of distance from a point.

share|cite|improve this answer
@Robert: Thanks for such a nice answer! I've learnt a lot from it that I didn't know about! :) – Renato G. Bettiol Nov 22 '12 at 17:23
@Robert, Thank you so much for the detailed answer. I am just wondering about two things. 1) Is this point of view written in the literature? 2) If not, then as a students, what is a good way to get these interesting and geometric insights of thinking analysis properties? – J. GE Dec 1 '12 at 21:09
@GB: I'm sure that it's written in the literature in various places, usually on the way to working out some other property. By the way, the above story generalizes in a straightforward way to all of the compact rank one symmetric spaces, so you probably will find some version of it in Besse's book "Manifolds all of whose geodesics are closed". More generally, the lowest nontrivial eigenspace on any compact irreducible Riemannian symmetric space $M=G/K$ provides a $G$-equivariant isometric embedding of $M$ into an irreducible representation space of $G$ that often has interesting properties. – Robert Bryant Dec 1 '12 at 23:20
@GB: Your second question is more philosophical in nature, and I'm not sure that I have a helpful answer. For me (but, apparently, not for everyone), working out a lot of explicit examples when I'm learning the basics of a subject seems to necessary for developing intuition. It also makes it easier to present those examples concretely without having to rely on a lot of theoretical machinery. Hiding the theoretical machinery is a little misleading, though, because it is a critical part of organizing the plethora of examples one examines into a working understanding. – Robert Bryant Dec 2 '12 at 1:12
@Robert, your two answers are really helpful. Thanks again. – J. GE Dec 2 '12 at 14:11

See this question, and the answers to it.

share|cite|improve this answer
Thanks for the link. The origional paper M. Takeuchi: Gendai no kyukansu, Iwanami Shoten 1974 is in Japanese, do you know any English survey? – J. GE Nov 21 '12 at 22:35
The book is translated by the AMS:… – YangMills Nov 22 '12 at 1:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.