On eigenfunctions of the Laplace Beltrami operator How can we generate the eigenspace for the Laplace Beltrami operator on SU(2)?
 A: For $\mathrm{SU}(2)$, with the scale for the biïnvariant metric so that it becomes isometric to the unit $3$-sphere $S^3$ in Euclidean $4$-space, it is well-known what the eigenvalues of the Laplace-Beltrami operator are and, indeed, what the eigenfunctions are as well.  The eigenvalues are $\lambda_k = k(k{+}2)$ for $k=0,1,2,\ldots$, and the $k$-th eigenspace consists of the restrictions to $S^3$ of the harmonic polynomials homogenous of degree $k$ on $\mathbb{R}^4$. Hence, the multiplicity of $\lambda_k$ is $(k{+}1)^2$.  For example, see:

*

*the classic reference: Berger, M., Gauduchon, P., Mazet, E.: Le spectre d'une variété riemannienne, Springer Lecture Notes No.194, 1971.

*these lecture notes from MIT (theorem 2.9): https://math.mit.edu/~dav/spheres.pdf
More generally, for the biïnvariant metric on a compact, simply-connected simple Lie group $G$ corresponding to the (negative of the) Cartan-Killing form on the Lie algebra, the computation of the eigenvalues of its Laplace-Beltrami operator and their multiplicities can be reduced to a combinatorial problem by means of the Peter-Weyl Theorem, Cartan's classification of the irreducible representations of $G$, the Casamir operator, and various representation-theoretic formulas such as Weyl's character formula, Steinberg's formula, and formulae of Kostant.  It is not a trivial combinatorial problem, though.  I would suggest getting a reference such as A. Knapp's Lie Groups: Beyond an Introduction and then searching the internet for phrases such as the spectrum of compact simple groups to get started.
In principle, one could work out the spectrum for any left-invariant metric on a compact Lie group, but the combinatorial problems that one runs into in carrying this out explicitly can be thorny for all but the simplest cases.
Addendum: Because of IGT's question/comment below from June 24, 2022, I decided to include the example of a left-invariant metric on $\mathrm{SU}(2)\simeq S^3$ in the first few cases to illustrate what's going on.  If we let $g_0$ be the biïnvariant metric on $\mathrm{SU}(2)$ with sectional curvature $1$, and let $g$ be any other left-invariant metric on $\mathrm{SU}(2)$, then there are left-invariant $1$-forms $\omega_i$ and positive constants $c_i$ such that $g_0 = {\omega_1}^2+{\omega_2}^2+{\omega_3}^2$ while $g = {\omega_1}^2/c_1+{\omega_2}^2/c_2+{\omega_3}^2/c_3$.  If $X_i$ are the dual left-invariant vector fields, then the Laplacian with respect to the metric $g$ is easily seen to be the second-order operator $\Delta = -c_1\,{X_1}^2 -c_2\, {X_2}^2 -c_3\, {X_3}^2$.  Since differentiation with respect to $X_i$ (whose flow is an isometric rotation of the sphere) clearly preserves the $(k{+}1)^2$ dimensional subspace $H_k$ that consists of restrictions to $S^3$ of the harmonic polynomials homogeneous of degree $k$ on $\mathbb{R}^4$, it follows that $\Delta$ preserves this space as well, i.e., $\Delta(H_k)=H_k$ for $k\ge1$.
On $H_1$, $\Delta$ has a single eigenvalue: $c_1 + c_2 + c_3$, of multiplicity $4$.
on $H_2$, $\Delta$ has three eigenvalues:
$4(c_i + c_j)$  ($i\not=j$), each of multiplicity $3$.
on $H_3$, $\Delta$ has two eigenvalues: $5(c_1+c_2+c_3)\pm 4\delta$, where
$$
\delta = \sqrt{{c_1}^{2}+{c_2}^{2}+{c_3}^{2}-c_1c_2-c_2c_3-c_3c_1},
$$
and each eigenvalue has multiplicity $8$.  (Note that $\delta=0$ if and only if $c_1=c_2=c_3$.)
Finally, on $H_4$, $\Delta$ has five eigenvalues:
Three of them are $4(c_1+c_2+c_3)+12\,c_i$ for $i=1,2,3$, and
the other two are $8(c_1+c_2+c_3\pm\delta)$.
Each of these $5$ eigenvalues has multiplicity $5$.
As one can see, the eigenvalues of $g$ need not all be integer multiples of a single value, or even all rational if the $c_i$ are rational.  Also, note that if one of the $c_i$ is much larger than the other two (for example, when $c_1 > 3(c_2{+}c_3)$), the eigenfunction belonging to the lowest eigenvalue of $\Delta$ may not be the restriction of a linear function on $\mathbb{R}^4$ to $S^3$.
