For a Riemannian manifold $\mathbb M$, let $0=\lambda_0<\lambda_1<\cdots$ be the eigenvalues of (negative of) its Laplace-Beltrami $-\Delta_{\mathbb M}$, with corresponding eigenspaces $\mathcal E_0,\mathcal E_1,\cdots$. Suppose $\mathbb M$ is a compact connected semi simple Lie group, with maximal torus $\mathbb T$. By Borel-Weil, all the irreducible representations are of the form $H^0(\mathbb M/\mathbb T,\mathcal L_{-\lambda})^\star$ for dominant integral weights $\lambda$ (and the weights giving rise to line-bundles $\mathcal L_{-\lambda}$), and by Peter-Weyl, the representations \begin{equation*}H^0(\mathbb M/\mathbb T,\mathcal L_{-\lambda})^\star\otimes H^0(\mathbb M/\mathbb T,\mathcal L_{-\lambda})\end{equation*} are the Laplacian eigenspaces $\mathcal E_j$. I would like to know if the following is true: if $f_{j_1}\in \mathcal E_{\lambda_{\ell_1}}$ and $f_{j_2}\in \mathcal E_{\lambda_{\ell_2}}$ for some $\ell_1,\ell_2>0$, then \begin{equation*}f_{j_1}f_{j_2}\in \bigoplus_{0<\lambda_i\leq \lambda_{\ell_1}+\lambda_{\ell_2}}\mathcal E_{\lambda_i}\,.\end{equation*} This is certainly true for spheres (even though these are not Lie groups in general) and hence for $SU(2)$. But is it true for all compact Lie groups, or at least classical compact Lie groups?
1 Answer
Too long for a comment, but this is not a full answer.
I'm not sure about the sums of LB eigenvalues, but the analogue with highest weights instead of the corresponding eigenvalues should hold.
Recall that the LB eigenspaces for compact Lie groups correspond to irreducible representations, and the latter are numbered by their highest weights, and for the HW $\lambda$ the corresponding eigenvalue equals $\|\rho\|^2-\|\lambda+\rho\|^2$ (it sometimes happens that distinct representations give the same eigenvalue, hence the correspondence is not bijective). Now the eigenspace corresponding to a representation can be realized inside $L^2(G)$ as the linear span of its matrix coefficients (fix a basis of a representation space, present group elements as matrices, take the matrix elements). Now if you take two representations, with the highest weights $\lambda$ and $\mu$, then their tensor product, while usually not irreducible, has highest weight $\lambda+\mu$. On the other hand, in the tensor basis the matrix elements of the tensor product are given by pairwise products of matrix elements of the tensor product constituents. Combining, we see that the product of matrix coefficients of irreducible representations (i.e. eigenfunctions) is a matrix coefficient of their tensor product, hence decomposes into a sum of matrix coefficients of representations with highest weights at most the sum of original highest weights.
