Heat kernel of left-invariant metric on 3-sphere

Question

This post concerns the heat equation on $S^3 (\simeq \mathrm{SU}(2))$ endowed with an arbitrary left-invariant metric. We think of $S^3$ as the space of unit quaternions, and its Lie algebra $\mathfrak{g} = T_e S^3$ as the set of pure quaternions (i.e., zero real part) endowed with bracket given by the cross product on $\mathbb{R}^3$. It is known that any left-invariant metric on $S^3$ is given by left-translates of the Euclidean inner product $\langle x , Q^{-1} y \rangle$ on $\mathfrak{g}$, where $Q$ is a positive-definite, symmetric matrix. Without loss of generality, we take $Q = \mathrm{diag}(a^2 , b^2 , c^2)$. The metric is bi-invariant if and only if $a=b=c$ (because $\mathfrak{g}$ is simple and $\langle x, Q^{-1} y \rangle \propto \mathrm{Tr}(\mathrm{ad}_x \mathrm{ad}_y)$ if and only if $Q = a^2 I_3$ for some $a \in \mathbb{R}$).

My question concerns equation 3.7.32 of [1], which we restate here:

Let $\Delta_g$ be the Laplace–Beltrami operator on a Riemannian space $(M,g)$ of dimension $n$. Let $R$ be the Ricci scalar curvature of the space, which will be assumed constant. Let $d(x,y)$ denote the Riemannian distance between the points $x,y \in M$. The heat kernel for the elliptic operator $\Delta_g$ is given by $$ e^{t \Delta_g} (x,y) = (4 \pi t)^{-n/2} \sqrt{D(x,y)} e^{Rt} e^{-\frac{d^2 (x,y)}{4t}} , \tag{1} $$ where $$ D(x,y) = |g(x)|^{-1/2} \det{ \left( - \frac{1}{2} \mathrm{Hess}(d^2 )(x,y)\right) } |g(y)|^{-1/2} . \tag{2} $$

Formula (1) originates from the work of Schulman [2] and Dowker [3,4]. Note, it should only apply for $y \notin \mathrm{Cut}(x)$. Schulman started with (1), claiming that it is an expression for the "short-time" propagator (a result apparently due to DeWitt). Schulman then showed the expression that results from (1) is actually "exact for finite-time", thus providing a closed-form for the heat kernel on $S^3$ with the standard round metric. I believe Dowker identified those spaces for which the "short-time" propagator is actually "exact for finite-time" - and these spaces are simple Lie groups with constant $R$. (I am being intentionally cautious with assertions here, because it is not clear to me if formula (1), as written, is correct.)

Now, for the case of $S^3$, we know from a famous paper of Milnor [5] that $S^3$ with left-invariant metric has constant Ricci scalar curvature$^{\ast}$ $R$. In our notation, this curvature equals (c.f. Proposition 2.3 of [6])

$$ R = 4 (a^2 + b^2 + c^2) - 2 \left( \frac{b^2 c^2}{a^2} + \frac{a^2 c^2}{b^2} + \frac{a^2 b^2}{c^2} \right) . $$

Now, it is clear from (1) that $\Delta_g \sqrt{D(x,y)} = R \sqrt{D(x,y)}$. Proposition 6.1 of [6] states that $R/2 \leq \lambda_1$ with equality if and only if $a=b=c$, where $\lambda_1$ is the smallest non-zero eigenvalue of $\Delta_g$. ($g$ is now our left-invariant metric on $S^3$.) So, it appears somewhat plausible that $R$ could be an eigenvalue.

With this background set, I have two main questions.

Q1: Is formula (1) correct and how could we prove it?

Q2: Is there a simple formula for $d (x,y)$ for $y \notin \mathrm{Cut} (x)$?

Regarding Q2, I believe I have a formula, but I am not completely convinced. Introduce exponential coordinates for $q \in S^3$ as $q = e^x$, where $x \in \mathbb{R}^3$ is thought of as a quaternion with zero real part and $e^x$ is the quaternion exponential. Then, $$ d^2 (e^x , e^y) = \langle \log{(e^{-x} e^y)} , Q^{-1} \log{(e^{-x} e^y)} \rangle, \tag{3} $$ where $\log$ is the quaternionic logarithm. Note that $\log{(e^{-x} e^y)}$ is given by the BCH expansion. It is easy to verify that this reduces to the angle between the quaternions $q^{-1}(x) = e^x$ and $q(y) = e^y$ when $Q = I_3$. So, Q2 could be rephrased as

Q2': Is (3) correct?

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$^{\ast}$ I believe the $R$ used in (1) and the papers [5,6] is the sum of sectional curvatures (not the average).

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[1] O. Calin, D.-C. Chang, K. Furutani, and C. Iwasaki, Heat kernels for elliptic and sub-elliptic operators. Springer, 2010.

[2] L. Schulman, “A path integral for spin,” Phys. Rev., vol. 176, pp. 1558–1569, Dec 1968.

[3] J. Dowker, “When is the sum over classical paths’ exact?,” Journal of Physics A: General Physics, vol. 3, no. 5, p. 451, 1970.

[4] J. Dowker, “Quantum mechanics on group space and huygens’ principle,” Annals of Physics, vol. 62, no. 2, pp. 361–382, 1971.

[5] J. Milnor, “Curvatures of left invariant metrics on lie groups,” Advances in Mathematics, vol. 21, no. 3, pp. 293–329, 1976.

[6] E. A. Lauret, “The smallest laplace eigenvalue of homogeneous 3-spheres,” Bulletin of the London Mathematical Society, vol. 51, no. 1, pp. 49–69, 2019.

Answer 1

I don't know a formula for $d(e^x,e^y)^2$, and I suspect that there is no simple formula, but the answer to Q2' is 'no'. The right hand side of (3) is linear in $Q^{-1}$, but it is not hard to see that $d(e^x,e^y)^2$ cannot be linear in $Q^{-1}$.

Here is an outline of the argument: Since the metric $g$ is left-invariant, the function $f(p,q) = d(p,q)^2$ satisfies $f(p,q) = f(I,p^{-1}q)=f(I,q^{-1}p)$ for all $p,q\in S^3 = \mathrm{SU}(2)$, so it suffices to compute a Taylor expansion for $f(I,q)$ for $q$ sufficiently near the identity $I\in \mathrm{SU}(2)$. Now, a convenient parametrization by $\mathbb{R}^3$ of the open ball $\mathrm{tr}(q)>0$ for $q\in\mathrm{SU}(2)$ is $$ q = \frac{1}{R}\begin{pmatrix}1+i x_1&-x_2+ix_3\\x_2+ix_3&1-ix_1\end{pmatrix}\qquad\text{where}\quad R = \sqrt{1+x_1^2+x_2^2+x_3^2}, $$ This gives $$ q^{-1}\mathrm{d}q = \begin{pmatrix}i\,\omega_1&-\omega_2+i\,\omega_3\\\omega_2+i\,\omega_3&-i\,\omega_1\end{pmatrix}, $$ where the $\omega_i$ are a basis for the left-invariant 1-forms expressed in terms of $x$. Computation yields $$ \omega_i = \frac{\mathrm{d}x_i-x_k\,\mathrm{d}x_j+x_j\,\mathrm{d}x_k}{1+x_1^2+x_2^2+x_3^2}. $$ for $(i,j,k)$ an even permutation of $(1,2,3)$. The dual basis of vector fields is $$ X_i = \partial_i + x_i\,(x_\ell\,\partial_\ell) + x_k\,\partial_j - x_j\,\partial_k\,. $$

Write $g=\lambda_1\,{\omega_1}^2+\lambda_2\,{\omega_2}^2+\lambda_3\,{\omega_3}^2$ where the $\lambda_i$ are positive constants. Note that the $\lambda_i$ are the diagonal coefficients of the OP's $Q^{-1}$.

Let $f(x)$ be the squared $g$-distance from $0\in\mathbb{R}^3$ to $x\in\mathbb{R}^3$. Note that $f$ is a smooth (in fact real-analytic) on a neighborhood of $0\in\mathbb{R}^3$ and has a nondegenerate minimum at $x=0$.

As is well-known, the function $\sqrt{f}$ satisfies the eikonal equation $\left\|\mathrm{d}\sqrt{f}\right\|_g = 1$ on a punctured neighborhood of $x=0$, which implies $\left\|\mathrm{d} f\right\|^2_g = 4f$, or, equivalently, $$ \lambda_1^{-1}(X_1f)^2+\lambda_2^{-1}(X_2f)^2+\lambda_3^{-1}(X_3f)^2 - 4f = 0, $$ which holds on an open neighborhood of $0\in\mathbb{R}^3$. Straightforward calculation (using the above formula and the fact that $f$ has a nondegenerate minimum at $x=0$) yields the Taylor expansion $$ \begin{aligned} f(x) &= \lambda_1\,{x_1}^2+\lambda_2\,{x_2}^2+\lambda_3\,{x_3}^2\\ &\quad -\frac23\bigl(\lambda_1\,{x_1}^2+\lambda_2\,{x_2}^2+\lambda_3\,{x_3}^2\bigr)\bigl({x_1}^2+{x_2}^2+{x_3}^2\bigr)\\ &\quad-\frac13\left( \frac{(\lambda_2{-}\lambda_3)^2}{\lambda_1}{x_2}^2{x_3}^2 +\frac{(\lambda_3{-}\lambda_1)^2}{\lambda_2}{x_3}^2{x_1}^2 +\frac{(\lambda_1{-}\lambda_2)^2}{\lambda_3}{x_1}^2{x_2}^2\right)\\ &\qquad +O\bigl(|x|^6\bigr) \end{aligned} $$ Note that $f$ is already not linear in the $\lambda_i$ in the $|x|^4$ terms, hence the OP's proposed formula $(3)$ cannot hold.