Indeed, the unique geodesic ray $\gamma$ that we are going to construct necessarily crosses $\underline{S}$ at a single point $\underline{\sigma}$ (otherwise $\gamma$ would fail to be a minimizing geodesic) and it is such that $\underline{\gamma}(t) = \underline{x}$ for a unique time $t > 0$. Conversely, a pair $(\underline{\sigma}, t)$ would uniquely determine a point $\underline{\gamma}(t) \in \underline{X}$.

Let us proceed now with the construction. Take $\underline{x} \in \underline{X} \setminus \underline{\{o\}}$. Let $\{x_i\}_{i \subset X}$ be such that $x_i \xrightarrow{\text{GH}} \underline{x}$. For any $i$, let $c_i$ be the minimizing $d_i$-geodesic joining $o$ to $x_i$. As done previously, up to extracting a subsequence we can assume that $\{c_i\}_{i}$ converges uniformly on compact subsets of $[0, +\infty)$ to a geodesic ray $\gamma : [0, +\infty) \to \underline{X}$. We know from Claim 4.3 and the previous step that

$$ (4.5) \qquad \beta(\underline{H}(\underline{x}), H(y)) = \underline{d}(\underline{o}, \underline{x})b_{\gamma}(y) $$

holds for any $y \in \underline{X}$, where $b_{\gamma}$ is the Busemann function associated with $\gamma$. Consider now a minimizing geodesic $c$ in $(\underline{X}, \underline{d})$ joining $\underline{o}$ to $\underline{x}$ and set

$$ \underline{D} := \underline{d}(\underline{o}, \underline{x}). $$

For any $s \in [0, \underline{D}]$, acting as we did to establish (4.5), we can prove that

$$ \beta(\underline{H}(c(s)), H(y)) = sb_{\gamma}(y) $$

holds for any $y \in \underline{X}$. Subtracting (4.5) to this latter equality yields to

$$ \beta\left(\underline{H}(c(s)) - \frac{s}{\underline{D}} \underline{H}(\underline{x}), H(y)\right) = 0 $$

for any $y \in \underline{X}$. By Claim 4.2, this implies

$$ \beta\left(\underline{H}(c(s)) - \frac{s}{\underline{D}} \underline{H}(\underline{x}), \xi\right) = 0 $$

for any $\xi \in \mathbb{R}^n$ and then

$$ (4.6) \qquad \underline{H}(c(s)) = \frac{s}{\underline{D}} \underline{H}(\underline{x}) $$

since $\beta$ is non-degenerate. Uniqueness of $c$ follows: if $c_1$ and $c_2$ are two minimizing geodesics joining $\underline{o}$ to $\underline{x}$, for any $s \in [0, \underline{D}]$ one has

$$ \underline{H}(c_1(s)) = \frac{s}{\underline{D}} \underline{H}(\underline{x}) = \underline{H}(c_2(s)) $$

and thus $c_1(s) = c_2(s)$ since $\underline{H}$ is injective.

Let us show now that $c$ extends in an unique way into a geodesic ray. Our argument is inspired by the analysis done by Cheeger about generalized linear functions [Che99, §8]. For any $i$, set $D_i := d(o, x_i)$ and write $\gamma_i : [0, +\infty) \to \underline{X}$ for the geodesic ray in $(\underline{X}, \underline{d}_i)$ defined by:

$$ \gamma_i(s) = \gamma(st_i + D_i) \quad \forall s > 0. $$

On the one hand, by Proposition 2.12, we know that up to extracting a subsequence we can assume that $\{\gamma_i\}_{i}$ converges uniformly on compact subsets of $[0, +\infty)$ to a geodesic ray $\tilde{\gamma} : [0, +\infty] \to \underline{X}$ whose associated Busemann function we denote by $b_{\tilde{\gamma}}$.
Indeed, the unique geodesic ray $\gamma$ that we are going to construct necessarily crosses $S$ at a single point $\sigma$ (otherwise $\gamma$ would fail to be a minimizing geodesic) and it is such that $\gamma(t) = \underline{x}$ for a unique time $t > 0$. Conversely, a pair $(\underline{\sigma}, t)$ would uniquely determine a point $\underline{\gamma}(t) \in \underline{X}$.

Let us proceed now with the construction. Take $\underline{x} \in \underline{X} \setminus \{\underline{o}\}$. Let $\{x_i\}_{i \subset X}$ be such that $x_i \xrightarrow{\text{GH}} \underline{x}$. For any $i$, let $c_i$ be the minimizing $d_i$-geodesic joining $\underline{o}$ to $x_i$. As done previously, up to extracting a subsequence we can assume that $\{c_i\}_{i}$ converges uniformly on compact subsets of $[0, +\infty)$ to a geodesic ray $\gamma : [0, +\infty) \to \underline{X}$. We know from Claim 4.3 and the previous step that

$$ (4.5) \qquad \beta(\underline{H}(\underline{x}), H(y)) = \underline{d}(\underline{o}, \underline{x})b_{\gamma}(y) $$

holds for any $y \in \underline{X}$, where $b_{\gamma}$ is the Busemann function associated with $\gamma$. Consider now a minimizing geodesic $c$ in $(\underline{X}, \underline{d})$ joining $\underline{o}$ to $\underline{x}$ and set

$$ \underline{D} := \underline{d}(\underline{o}, \underline{x}). $$

For any $s \in [0, \underline{D}]$, acting as we did to establish (4.5), we can prove that

$$ \beta(\underline{H}(c(s)), H(y)) = sb_{\gamma}(y) $$

holds for any $y \in \underline{X}$. Subtracting (4.5) to this latter equality yields to

$$ \beta\left(\underline{H}(c(s)) - \frac{s}{\underline{D}} \underline{H}(\underline{x}), H(y)\right) = 0 $$

for any $y \in \underline{X}$. By Claim 4.2, this implies

$$ \beta\left(\underline{H}(c(s)) - \frac{s}{\underline{D}} \underline{H}(\underline{x}), \xi\right) = 0 $$

for any $\xi \in \mathbb{R}^n$ and then

$$ (4.6) \qquad \underline{H}(c(s)) = \frac{s}{\underline{D}} \underline{H}(\underline{x}) $$

since $\beta$ is non-degenerate. Uniqueness of $c$ follows: if $c_1$ and $c_2$ are two minimizing geodesics joining $\underline{o}$ to $\underline{x}$, for any $s \in [0, \underline{D}]$ one has

$$ \underline{H}(c_1(s)) = \frac{s}{\underline{D}} \underline{H}(\underline{x}) = \underline{H}(c_2(s)) $$

and thus $c_1(s) = c_2(s)$ since $\underline{H}$ is injective.

Let us show now that $c$ extends in an unique way into a geodesic ray. Our argument is inspired by the analysis done by Cheeger about generalized linear functions [Che99, §8]. For any $i$, set $D_i := d(\underline{o}, x_i)$ and write $\gamma_i : [0, +\infty) \to \underline{X}$ for the geodesic ray in $(\underline{X}, \underline{d}_i)$ defined by:

$$ \gamma_i(s) = \gamma(st_i + D_i) \quad \forall s > 0. $$

On the one hand, by Proposition 2.12, we know that up to extracting a subsequence we can assume that $\{\gamma_i\}_{i}$ converges uniformly on compact subsets of $[0, +\infty)$ to a geodesic ray $\tilde{\gamma} : [0, +\infty] \to \underline{X}$ whose associated Busemann function we denote by $b_{\tilde{\gamma}}$.
On the other hand, if we write $b_{\gamma_i}$ for the Busemann function associated with $\gamma_i$, we can proceed as in Step 2 with $d_i, H_i, \gamma_i$ in place of $d, H, \gamma$ respectively to get

$$b_{\gamma_i}(y) = \frac{\beta(H(x), H_i(y))}{\underline{D}}$$

for any $y \in X$. Then the sequence $(b_{\gamma_i})$ converges pointwise to the function

$$F: \underline{X} \ni \underline{y} \mapsto \frac{\beta(H(\underline{x}), H(\underline{y}))}{\underline{D}}$$

and we have the following:

CLAIM 4.6

$$(4.7) \qquad F = \underline{b}_{\tilde{\gamma}} + \underline{D}.$$

*Proof.* — Observe first that $F$ is strongly harmonic since it is a linear combination of the strongly harmonic functions $\underline{h}_1, \dots, \underline{h}_\ell$. Let us show that $\underline{b}_{\tilde{\gamma}}$ is strongly harmonic too. For any $i$, set

$$p_i(x, y, t) := \frac{1}{(4\pi t)^{\alpha/2}} e^{-d_i^2(x,y)/4t} = r_i^\alpha p(x, y, r_i^2 t)$$

for any $x, y \in X$ and $t > 0$, and

$$\underline{p}(\underline{x}, \underline{y}, t) := \frac{1}{(4\pi t)^{\alpha/2}} e^{-d^2(\underline{x}, \underline{y})/4t}$$

for any $\underline{x}, \underline{y} \in X$ and $t > 0$. Then for any $x, y \in X$ and $s, t > 0$,

$$\begin{align*}
p_i(x, y, t+s) &= r_i^\alpha p(x, y, r_i^2 t + r_i^2 s) = r_i^\alpha \int_X p(x, z, r_i^2 t) p(z, y, r_i^2 s) \, d\mu(z) \\
&= \int_X p_i(x, z, t) p(z, y, r_i^2 s) r_i^\alpha \frac{d\mu(z)}{r_i^\alpha} = \int_X p_i(x, z, t) p_i(z, y, s) \, d\mu_i(z).
\end{align*}$$

For any $\underline{x}, \underline{y} \in \underline{X}$ and $\{x_i\}_i, \{y_i\}_i \in X$ such that $x_i \xrightarrow{\text{GH}} \underline{x}$ and $y_i \xrightarrow{\text{GH}} \underline{y}$, the convergence $d_i(x_i, y_i) \to d(\underline{x}, \underline{y})$ implies $p_i(x_i, y_i, t) \to \underline{p}(\underline{x}, \underline{y}, t)$ for any $t > 0$, hence:

$$\underline{p}(\underline{x}, \underline{y}, t+s) = \int_X \underline{p}(\underline{x}, \underline{z}, t) \underline{p}(\underline{z}, \underline{y}, s) \, d\underline{\mu}(\underline{z}) \quad \forall \underline{x}, \underline{y} \in \underline{X}, \forall t, s > 0.$$

By a standard procedure described for instance in [Gri10, §2], we can construct a Dirichlet form $\mathcal{E}$ on $(\underline{X}, d, \underline{\mu})$ admitting a heat kernel given by $\underline{p}$. In particular, $(\underline{X}, d, \underline{\mu}, \mathcal{E})$ has an $\alpha$-dimensional Euclidean heat kernel. Writing $\underline{L}$ for the associated self-adjoint operator, we deduce from Lemma 2.15 that $\underline{b}_{\tilde{\gamma}}$ is locally $\underline{L}$-harmonic, then Lemma 3.14 implies that $\underline{b}_{\tilde{\gamma}}$ is strongly harmonic.

Let us show now that $F - \underline{b}_{\tilde{\gamma}} \ge 0$. Take $i \in \mathbb{N}$, $s > 0$ and $y \in X$. Then $b_{\gamma}(y) \ge r_i s + D_i - d(\gamma(r_i s + D_i), y)$ by definition of a Busemann function, hence
On the other hand, if we write $b_{\gamma_i}$ for the Busemann function associated with $\gamma_i$, we can proceed as in Step 2 with $d_i, H_i, \gamma_i$ in place of $d, H, \gamma$ respectively to get

$$b_{\gamma_i}(y) = \frac{\beta(H(x), H_i(y))}{\underline{D}}$$

for any $y \in X$. Then the sequence $(b_{\gamma_i})$ converges pointwise to the function

$$F: \underline{X} \ni \underline{y} \mapsto \frac{\beta(H(x), H(y))}{\underline{D}}$$

and we have the following:

CLAIM 4.6

$$(4.7) \qquad F = \underline{b}_{\tilde{\gamma}} + \underline{D}.$$

*Proof.* — Observe first that $F$ is strongly harmonic since it is a linear combination of the strongly harmonic functions $\underline{h}_1, \dots, \underline{h}_\ell$. Let us show that $\underline{b}_{\tilde{\gamma}}$ is strongly harmonic too. For any $i$, set

$$p_i(x, y, t) := \frac{1}{(4\pi t)^{\alpha/2}} e^{-d_i^2(x,y)/4t} = r_i^\alpha p(x, y, r_i^2 t)$$

for any $x, y \in X$ and $t > 0$, and

$$\underline{p}(\underline{x}, \underline{y}, t) := \frac{1}{(4\pi t)^{\alpha/2}} e^{-d^2(\underline{x}, \underline{y})/4t}$$

for any $\underline{x}, \underline{y} \in X$ and $t > 0$. Then for any $x, y \in X$ and $s, t > 0$,

$$\begin{align*}
p_i(x, y, t+s) &= r_i^\alpha p(x, y, r_i^2 t + r_i^2 s) = r_i^\alpha \int_X p(x, z, r_i^2 t) p(z, y, r_i^2 s) \, d\mu(z) \\
&= \int_X p_i(x, z, t) p(z, y, r_i^2 s) r_i^\alpha \frac{d\mu(z)}{r_i^\alpha} = \int_X p_i(x, z, t) p_i(z, y, s) \, d\mu_i(z).
\end{align*}$$

For any $\underline{x}, \underline{y} \in \underline{X}$ and $\{x_i\}_i, \{y_i\}_i \in X$ such that $x_i \xrightarrow{\text{GH}} \underline{x}$ and $y_i \xrightarrow{\text{GH}} \underline{y}$, the convergence $d_i(x_i, y_i) \to d(\underline{x}, \underline{y})$ implies $p_i(x_i, y_i, t) \to \underline{p}(\underline{x}, \underline{y}, t)$ for any $t > 0$, hence:

$$\underline{p}(\underline{x}, \underline{y}, t+s) = \int_X \underline{p}(\underline{x}, \underline{z}, t) \underline{p}(\underline{z}, \underline{y}, s) \, d\underline{\mu}(\underline{z}) \quad \forall \underline{x}, \underline{y} \in \underline{X}, \forall t, s > 0.$$

By a standard procedure described for instance in [Gri10, §2], we can construct a Dirichlet form $\mathcal{E}$ on $(\underline{X}, d, \underline{\mu})$ admitting a heat kernel given by $\underline{p}$. In particular, $(\underline{X}, d, \underline{\mu}, \mathcal{E})$ has an $\alpha$-dimensional Euclidean heat kernel. Writing $\underline{L}$ for the associated self-adjoint operator, we deduce from Lemma 2.15 that $\underline{b}_{\tilde{\gamma}}$ is locally $\underline{L}$-harmonic, then Lemma 3.14 implies that $\underline{b}_{\tilde{\gamma}}$ is strongly harmonic.

Let us show now that $F - \underline{b}_{\tilde{\gamma}} \ge 0$. Take $i \in \mathbb{N}$, $s > 0$ and $y \in X$. Then $b_{\gamma}(y) \ge r_i s + D_i - d(\gamma(r_i s + D_i), y)$ by definition of a Busemann function, hence
$r_i^{-1}b_{\gamma}(y) \ge s + r_i^{-1}D_i - d_i(\gamma_i(s), y)$. Since

$$
\begin{align*}
r_i^{-1}b_{\gamma}(y) &= \lim_{s \to +\infty} r_i^{-1}s - r_i^{-1}d(\gamma(s), y) \\
&= D_i r_i^{-1} + \lim_{s' \to +\infty} s' - d_i(\gamma(D_i + r_i s'), y), \quad \text{where } s' = (s - D_i)r_i^{-1} \\
&= D_i r_i^{-1} + b_{\gamma_i}(y),
\end{align*}
$$

we get $b_{\gamma_i}(y) \ge s - d_i(\gamma_i(s), y)$. Letting $i$ tend to $+\infty$ provides $F(y) \ge s - d(\tilde{\gamma}(s), y)$, after what letting $s$ tend to $+\infty$ gives $F \ge b_{\tilde{\gamma}}$.

By Lemma 2.6, we get that $F - b_{\tilde{\gamma}}$ is a constant function. Since $F(\underline{x}) = \underline{D} = \underline{d}(\underline{o}, \underline{x})$ and $b_{\tilde{\gamma}}(\underline{x}) = 0$, the claim is proved. $\square$

Let $\gamma : [0, +\infty) \to \underline{X}$ be the concatenation of $\underline{c}$ and $\tilde{\gamma}$, i.e.,

$$
\underline{\gamma}(t) := \begin{cases} \underline{c}(t) & \text{if } 0 < t \le \underline{D}, \\ \tilde{\gamma}(t - \underline{D}) & \text{if } t \ge \underline{D}. \end{cases}
$$

By construction, $\underline{\gamma}$ is 1-Lipschitz: $d(\underline{\gamma}(t), \underline{\gamma}(s)) \le |t-s|$ for any $s,t>0$. Moreover, (4.6) implies $F(\underline{\gamma}(t)) = t$ when $0<t\le\underline{D}$ while (4.7) implies $F(\underline{\gamma}(t)) = t$ when $t\ge\underline{D}$. Since the function $F$ is 1-Lipschitz we get $|t-s|\le d(\underline{\gamma}(t), \underline{\gamma}(s))$ for any $s,t>0$, thus $\underline{\gamma}$ is a geodesic ray that extends $\underline{c}$.

Let us show that this extension $\underline{\gamma}$ is unique. By (4.7), we have

$$
(4.8) \qquad \beta(H(\underline{x}), H(\underline{y})) = \underline{D}(b_{\tilde{\gamma}}(\underline{y}) - \underline{D})
$$

for any $\underline{y} \in \underline{X}$ and we can obtain

$$
\beta(H(\tilde{\gamma}(t)), H(\underline{y})) = t(b_{\tilde{\gamma}}(\underline{y}) - \underline{D})
$$

for any $\underline{y} \in \underline{X}$ and $t > \underline{D}$ by a similar reasoning. Then if $\underline{\gamma}'$ is another extension of $\underline{c}$ obtained from a geodesic ray $\tilde{\gamma}'$ emanating from $\underline{x}$, we get

$$
\beta(H(\tilde{\gamma}(t)) - H(\tilde{\gamma}'(t)), H(\underline{y})) = 0
$$

for any $\underline{y} \in \underline{X}$ and $t > \underline{D}$, from which Claim 4.5 yields $\tilde{\gamma}(t) = \tilde{\gamma}'(t)$.

REMARK 4.7. — Note that (4.8) implies $\beta(H(\tilde{\gamma}(t)), H(\underline{y})) = t\beta(H(\tilde{\gamma}(1)), H(\underline{y}))$ for all $\underline{y} \in \underline{X}$, hence

$$
(4.9) \qquad \underline{H}(\tilde{\gamma}(t)) = t \underline{H}(\tilde{\gamma}(1)).
$$

*Step 4. Construction of the isometry.* — Let $\Phi$ be the inverse of the bijection constructed in the previous step, i.e.,

$$
\Phi: \begin{array}{ccc} (0, +\infty) \times \underline{S} & \to & \underline{X} \setminus \{\underline{o}\} \\ (t, \underline{\sigma}) & \mapsto & \underline{\gamma}_{\underline{\sigma}}(t), \end{array}
$$

where $\underline{\gamma}_{\underline{\sigma}}$ is the geodesic ray obtained by extending the minimizing geodesic joining $\underline{o}$ to $\underline{\sigma}$. Note that (4.9) implies

$$
(4.10) \qquad \underline{H}(\Phi(t, \underline{\sigma})) = t \underline{H}(\Phi(1, \underline{\sigma}))
$$
$r_i^{-1}b_{\gamma}(y) \ge s + r_i^{-1}D_i - d_i(\gamma_i(s), y)$. Since

$$
\begin{align*}
r_i^{-1}b_{\gamma}(y) &= \lim_{s \to +\infty} r_i^{-1}s - r_i^{-1}d(\gamma(s), y) \\
&= D_i r_i^{-1} + \lim_{s' \to +\infty} s' - d_i(\gamma(D_i + r_i s'), y), \quad \text{where } s' = (s - D_i)r_i^{-1} \\
&= D_i r_i^{-1} + b_{\gamma_i}(y),
\end{align*}
$$

we get $b_{\gamma_i}(y) \ge s - d_i(\gamma_i(s), y)$. Letting $i$ tend to $+\infty$ provides $F(y) \ge s - d(\tilde{\gamma}(s), y)$, after what letting $s$ tend to $+\infty$ gives $F \ge b_{\tilde{\gamma}}$.

By Lemma 2.6, we get that $F - b_{\tilde{\gamma}}$ is a constant function. Since $F(\underline{x}) = \underline{D} = \underline{d}(\underline{o}, \underline{x})$ and $b_{\tilde{\gamma}}(\underline{x}) = 0$, the claim is proved. $\square$

Let $\gamma : [0, +\infty) \to \underline{X}$ be the concatenation of $\underline{c}$ and $\tilde{\gamma}$, i.e.,

$$
\underline{\gamma}(t) := \begin{cases} \underline{c}(t) & \text{if } 0 < t \le \underline{D}, \\ \tilde{\gamma}(t - \underline{D}) & \text{if } t \ge \underline{D}. \end{cases}
$$

By construction, $\underline{\gamma}$ is 1-Lipschitz: $d(\underline{\gamma}(t), \underline{\gamma}(s)) \le |t-s|$ for any $s,t>0$. Moreover, (4.6) implies $F(\underline{\gamma}(t)) = t$ when $0<t\le\underline{D}$ while (4.7) implies $F(\underline{\gamma}(t)) = t$ when $t\ge\underline{D}$. Since the function $F$ is 1-Lipschitz we get $|t-s| \le d(\underline{\gamma}(t), \underline{\gamma}(s))$ for any $s,t>0$, thus $\underline{\gamma}$ is a geodesic ray that extends $\underline{c}$.

Let us show that this extension $\underline{\gamma}$ is unique. By (4.7), we have

$$
(4.8) \qquad \beta(H(\underline{x}), H(\underline{y})) = \underline{D}(b_{\tilde{\gamma}}(\underline{y}) - \underline{D})
$$

for any $\underline{y} \in \underline{X}$ and we can obtain

$$
\beta(H(\tilde{\gamma}(t)), H(\underline{y})) = t(b_{\tilde{\gamma}}(\underline{y}) - \underline{D})
$$

for any $\underline{y} \in \underline{X}$ and $t > \underline{D}$ by a similar reasoning. Then if $\underline{\gamma}'$ is another extension of $\underline{c}$ obtained from a geodesic ray $\tilde{\gamma}'$ emanating from $\underline{x}$, we get

$$
\beta(H(\tilde{\gamma}(t)) - H(\tilde{\gamma}'(t)), H(\underline{y})) = 0
$$

for any $\underline{y} \in \underline{X}$ and $t > \underline{D}$, from which Claim 4.5 yields $\tilde{\gamma}(t) = \tilde{\gamma}'(t)$.

REMARK 4.7. — Note that (4.8) implies $\beta(H(\tilde{\gamma}(t)), H(\underline{y})) = t\beta(H(\tilde{\gamma}(1)), H(\underline{y}))$ for all $\underline{y} \in \underline{X}$, hence

$$
(4.9) \qquad \underline{H}(\tilde{\gamma}(t)) = t \underline{H}(\tilde{\gamma}(1)).
$$

*Step 4. Construction of the isometry.* — Let $\Phi$ be the inverse of the bijection constructed in the previous step, i.e.,

$$
\Phi: \begin{array}{ccc} (0, +\infty) \times \underline{S} & \to & \underline{X} \setminus \{\underline{o}\} \\ (t, \underline{\sigma}) & \mapsto & \underline{\gamma}_{\underline{\sigma}}(t), \end{array}
$$

where $\underline{\gamma}_{\underline{\sigma}}$ is the geodesic ray obtained by extending the minimizing geodesic joining $\underline{o}$ to $\underline{\sigma}$. Note that (4.9) implies

$$
(4.10) \qquad \underline{H}(\Phi(t, \underline{\sigma})) = t \underline{H}(\Phi(1, \underline{\sigma}))
$$
for any $(t, \underline{\sigma}) \in (0, +\infty) \times \underline{S}$. Let $d_C$ be the cone distance on $(0, +\infty) \times \underline{S}$ defined by

$$d_C^2((t, \underline{\sigma}), (t', \underline{\sigma}')) := (t - t')^2 + 4tt' \sin^2(\underline{d}_S(\underline{\sigma}, \underline{\sigma'})/2)$$

for any $(t, \underline{\sigma}), (t', \underline{\sigma}') \in (0, +\infty) \times \underline{S}$, where $\underline{d}_S$ is the length distance associated with the distance on $\underline{S}$ obtained by restricting $\underline{d}$ to $\underline{S} \times \underline{S}$. We are going to establish

$$(4.11) \qquad \underline{d}(\underline{x}, \underline{x}') = \underline{d}_C((t, \underline{\sigma}), (t', \underline{\sigma}'))$$

for any $\underline{x} = \Phi(t, \underline{\sigma}), \underline{x}' = \Phi(t', \underline{\sigma}') \in \underline{X} \setminus \{\underline{o}\}$.

**CLAIM 4.8.** — There exists $\delta(\underline{\sigma}, \underline{\sigma}') \in [0, \pi]$ such that

$$(4.12) \qquad \underline{d}^2(\underline{x}, \underline{x}') = (t - t')^2 + 4tt' \sin^2(\delta(\underline{\sigma}, \underline{\sigma'})/2).$$

*Proof.* — Choose $\{x_i\}_i, \{x'_i\}_i \subset X$ such that $x_i \xrightarrow{\text{GH}} \underline{x}$ and $x'_i \xrightarrow{\text{GH}} \underline{x}'$. For any $i$, divide (4.4) by $r_i^2$ to get $\underline{d}_i^2(x_i, x'_i) = Q(H_i(x_i) - H_i(x'_i))$. Letting $i$ tend to $+\infty$ implies $\underline{d}^2(\underline{x}, \underline{x}') = Q(\underline{H}(\underline{x}) - \underline{H}(\underline{x}'))$, hence

$$\underline{d}^2(\underline{x}, \underline{x}') = Q(t\underline{H}(\underline{\Phi}(1, \underline{\sigma})) - t'\underline{H}(\underline{\Phi}(1, \underline{\sigma}'))),$$

thanks to (4.10). To compute $Q(t\underline{H}(\underline{\Phi}(1, \underline{\sigma})) - t'\underline{H}(\underline{\Phi}(1, \underline{\sigma}')))$, let us use $\underline{h}_i(\underline{\sigma})$ as a shorthand for $\underline{h}_i(\underline{\Phi}(1, \underline{\sigma}))$. Then:

$$\begin{align*}
\underline{d}^2(\underline{x}, \underline{x}') &= Q(\underline{h}_1(\underline{\sigma}) - t'\underline{h}_1(\underline{\sigma}'), \dots, \underline{h}_\ell(\underline{\sigma}) - t'\underline{h}_\ell(\underline{\sigma}')) \\
&= \sum_{i,j=1}^\ell B(x_i, x_j)(\underline{h}_i(\underline{\sigma}) - t'\underline{h}_j(\underline{\sigma}'))(\underline{h}_j(\underline{\sigma}) - t'\underline{h}_j(\underline{\sigma}')) \\
&= \left(\sum_{i,j=1}^\ell B(x_i, x_j)\underline{h}_i(\underline{\sigma}')\underline{h}_j(\underline{\sigma}')\right)t^2 + \left(\sum_{i,j=1}^\ell B(x_i, x_j)\underline{h}_i(\underline{\sigma}')\underline{h}_j(\underline{\sigma}')\right)(t')^2 \\
&\quad - 2tt'\left(\sum_{i,j=1}^\ell B(x_i, x_j)\underline{h}_i(\underline{\sigma}')\underline{h}_j(\underline{\sigma}')\right) \\
&= Q(\underline{H}(\underline{\sigma}))t^2 + Q(\underline{H}(\underline{\sigma}'))(t')^2 - 2tt'\beta(\underline{H}(\underline{\sigma}), \underline{H}(\underline{\sigma}')) \\
&= \underline{d}^2(\underline{\sigma}, \underline{o})t^2 + \underline{d}^2(\underline{\sigma}', \underline{o})(t')^2 - 2tt'\beta(\underline{H}(\underline{\sigma}), \underline{H}(\underline{\sigma}')) \\
&= t^2 + (t')^2 - 2tt'\beta(\underline{H}(\underline{\sigma}), \underline{H}(\underline{\sigma}')).
\end{align*}$$

Set $\underline{B}(\underline{x}, \underline{x}') := \frac{1}{2}(\underline{d}^2(\underline{o}, \underline{x}) + \underline{d}^2(\underline{o}, \underline{x}') - \underline{d}^2(\underline{x}, \underline{x}'))$. Write (4.3) with $x = x_i$, $x' = x'_i$, divide by $r_i^2$ and let $r_i$ tend to $+\infty$ to get $\underline{\beta}(\underline{H}(\underline{x}), \underline{H}(\underline{x}')) = \underline{B}(\underline{x}, \underline{x}')$ and then

$$\underline{d}^2(\underline{x}, \underline{x}') = t^2 + (t')^2 - 2tt'\underline{B}(\underline{\sigma}, \underline{\sigma}').$$

Assuming $t = t' = 1$ provides $\underline{B}(\underline{\sigma}, \underline{\sigma}') = 1 - \frac{1}{2}\underline{d}^2(\underline{\sigma}, \underline{\sigma}')$. The triangle inequality implies $\underline{d}^2(\underline{\sigma}, \underline{\sigma}') \le 4$, thus $\underline{B}(\underline{\sigma}, \underline{\sigma}') \in [-1, 1]$, so we can set

$$\cos(\delta(\underline{\sigma}, \underline{\sigma}')) := \underline{B}(\underline{\sigma}, \underline{\sigma}')$$

for some $\delta(\underline{\sigma}, \underline{\sigma}') \in [0, \pi]$. $\square$

<figure><img src="image_7.png" /></figure>

In particular, (4.12) implies $\underline{d}(\underline{\sigma}_0, \underline{\sigma}_1) = 2 \sin(\delta(\underline{\sigma}_0, \underline{\sigma}_1)/2)$ for any $\underline{\sigma}_0, \underline{\sigma}_1 \in \underline{S}$.

**CLAIM 4.9.** — The function $\delta$ defines a geodesic distance on $\underline{S}$.
for any $(t, \underline{\sigma}) \in (0, +\infty) \times \underline{S}$. Let $d_C$ be the cone distance on $(0, +\infty) \times \underline{S}$ defined by

$$d_C^2((t, \underline{\sigma}), (t', \underline{\sigma}')) := (t - t')^2 + 4tt' \sin^2(\underline{d_S}(\underline{\sigma}, \underline{\sigma}'))/2$$

for any $(t, \underline{\sigma}), (t', \underline{\sigma}') \in (0, +\infty) \times \underline{S}$, where $\underline{d_S}$ is the length distance associated with the distance on $\underline{S}$ obtained by restricting $\underline{d}$ to $\underline{S} \times \underline{S}$. We are going to establish

$$(4.11) \qquad \underline{d}(\underline{x}, \underline{x}') = \underline{d}_C((t, \underline{\sigma}), (t', \underline{\sigma}'))$$

for any $\underline{x} = \Phi(t, \underline{\sigma}), \underline{x}' = \Phi(t', \underline{\sigma}') \in \underline{X} \setminus \{\underline{o}\}$.

**CLAIM 4.8.** — There exists $\delta(\underline{\sigma}, \underline{\sigma}') \in [0, \pi]$ such that

$$(4.12) \qquad \underline{d}^2(\underline{x}, \underline{x}') = (t - t')^2 + 4tt' \sin^2(\delta(\underline{\sigma}, \underline{\sigma}'))/2.$$

*Proof.* — Choose $\{x_i\}_i, \{x'_i\}_i \subset X$ such that $x_i \xrightarrow{\text{GH}} \underline{x}$ and $x'_i \xrightarrow{\text{GH}} \underline{x}'$. For any $i$, divide (4.4) by $r_i^2$ to get $\underline{d}_i^2(x_i, x'_i) = Q(H_i(x_i) - H_i(x'_i))$. Letting $i$ tend to $+\infty$ implies $\underline{d}^2(\underline{x}, \underline{x}') = Q(\underline{H}(\underline{x}) - \underline{H}(\underline{x}'))$, hence

$$\underline{d}^2(\underline{x}, \underline{x}') = Q(t\underline{H}(\underline{\Phi}(1, \underline{\sigma})) - t'\underline{H}(\underline{\Phi}(1, \underline{\sigma}'))),$$

thanks to (4.10). To compute $Q(t\underline{H}(\underline{\Phi}(1, \underline{\sigma})) - t'\underline{H}(\underline{\Phi}(1, \underline{\sigma}')))$, let us use $\underline{h}_i(\underline{\sigma})$ as a shorthand for $\underline{h}_i(\underline{\Phi}(1, \underline{\sigma}))$. Then:

$$
\begin{align*}
\underline{d}^2(\underline{x}, \underline{x}') &= Q(\underline{h}_1(\underline{\sigma}) - t'\underline{h}_1(\underline{\sigma}'), \dots, \underline{h}_\ell(\underline{\sigma}) - t'\underline{h}_\ell(\underline{\sigma}')) \\
&= \sum_{i,j=1}^\ell B(x_i, x_j)(\underline{h}_i(\underline{\sigma}) - t'\underline{h}_j(\underline{\sigma}'))(\underline{h}_j(\underline{\sigma}) - t'\underline{h}_j(\underline{\sigma}')) \\
&= \left(\sum_{i,j=1}^\ell B(x_i, x_j)\underline{h}_i(\underline{\sigma}')\underline{h}_j(\underline{\sigma}')\right)t^2 + \left(\sum_{i,j=1}^\ell B(x_i, x_j)\underline{h}_i(\underline{\sigma}')\underline{h}_j(\underline{\sigma}')\right)(t')^2 \\
&\quad - 2tt'\left(\sum_{i,j=1}^\ell B(x_i, x_j)\underline{h}_i(\underline{\sigma}')\underline{h}_j(\underline{\sigma}')\right) \\
&= Q(\underline{H}(\underline{\sigma}))t^2 + Q(\underline{H}(\underline{\sigma}'))(t')^2 - 2tt'\beta(\underline{H}(\underline{\sigma}), \underline{H}(\underline{\sigma}')) \\
&= \underline{d}^2(\underline{\sigma}, \underline{o})t^2 + \underline{d}^2(\underline{\sigma}', \underline{o})(t')^2 - 2tt'\beta(\underline{H}(\underline{\sigma}), \underline{H}(\underline{\sigma}')) \\
&= t^2 + (t')^2 - 2tt'\beta(\underline{H}(\underline{\sigma}), \underline{H}(\underline{\sigma}')).
\end{align*}
$$

Set $\underline{B}(\underline{x}, \underline{x}') := \frac{1}{2}(\underline{d}^2(\underline{o}, \underline{x}) + \underline{d}^2(\underline{o}, \underline{x}') - \underline{d}^2(\underline{x}, \underline{x}'))$. Write (4.3) with $x = x_i$, $x' = x'_i$, divide by $r_i^2$ and let $r_i$ tend to $+\infty$ to get $\underline{\beta}(\underline{H}(\underline{x}), \underline{H}(\underline{x}')) = \underline{B}(\underline{x}, \underline{x}')$ and then

$$\underline{d}^2(\underline{x}, \underline{x}') = t^2 + (t')^2 - 2tt'\underline{B}(\underline{\sigma}, \underline{\sigma}').$$

Assuming $t = t' = 1$ provides $\underline{B}(\underline{\sigma}, \underline{\sigma}') = 1 - \frac{1}{2}\underline{d}^2(\underline{\sigma}, \underline{\sigma}')$. The triangle inequality implies $\underline{d}^2(\underline{\sigma}, \underline{\sigma}') \le 4$, thus $\underline{B}(\underline{\sigma}, \underline{\sigma}') \in [-1, 1]$, so we can set

$$\cos(\delta(\underline{\sigma}, \underline{\sigma}')) := \underline{B}(\underline{\sigma}, \underline{\sigma}')$$

for some $\delta(\underline{\sigma}, \underline{\sigma}') \in [0, \pi]$. $\square$

<figure><img src="image_7.png" /></figure>

In particular, (4.12) implies $\underline{d}(\underline{\sigma}_0, \underline{\sigma}_1) = 2 \sin(\delta(\underline{\sigma}_0, \underline{\sigma}_1)/2)$ for any $\underline{\sigma}_0, \underline{\sigma}_1 \in \underline{S}$.

**CLAIM 4.9.** — The function $\delta$ defines a geodesic distance on $\underline{S}$.
Proof. — Let us first show that $\delta$ defines a distance on $\mathcal{S}$. We only prove the triangle inequality since the two other properties are immediate. For given $\underline{\sigma}_0, \underline{\sigma}_1, \underline{\sigma}_2 \in \mathcal{S}$, let us set $\alpha := \delta(\underline{\sigma}_0, \underline{\sigma}_1)$, $\beta := \delta(\underline{\sigma}_1, \underline{\sigma}_2)$ and $\gamma := \delta(\underline{\sigma}_0, \underline{\sigma}_2)$. We can assume $\alpha + \beta \le \pi$ because otherwise we would have $\alpha + \beta > \pi \ge \gamma$, thus nothing to prove. For any $t, s, r > 0$, the triangle inequality for $\underline{d}$ written with (4.12) gives

$$ \sqrt{(t-s)^2 + 4ts \sin^2(\alpha/2)} + \sqrt{(s-r)^2 + 4sr \sin^2(\beta/2)} \ge \sqrt{(t-r)^2 + 4tr \sin^2(\gamma/2)}. $$

Considering the three complex numbers $z_0 = t$, $z_1 = se^{i\alpha}$ and $z_2 = re^{i(\alpha+\beta)}$, this can be rewritten as

$$ |z_0 - z_1| + |z_1 - z_2| \ge \sqrt{(t-r)^2 + 4tr \sin^2(\gamma/2)}. $$

Choosing $s$ so that $z_0, z_1, z_2$ are aligned implies $|z_0 - z_2| = |z_0 - z_1| + |z_1 - z_2|$ thus

$$ \sqrt{(t-r)^2 + 4tr \sin^2((\alpha + \beta)/2)} = |z_0 - z_2| \ge \sqrt{(t-r)^2 + 4tr \sin^2(\gamma/2)}, $$

which yields to $\alpha + \beta \ge \gamma$.

Let us show now that $\delta$ is geodesic. For given $\underline{\sigma}_0, \underline{\sigma}_1 \in \mathcal{S}$ with $\underline{\sigma}_0 \ne \underline{\sigma}_1$, we aim at finding $\underline{\sigma}_m \in \mathcal{S}$ such that

$$ \delta(\underline{\sigma}_0, \underline{\sigma}_m) = \delta(\underline{\sigma}_m, \underline{\sigma}_1) = \frac{1}{2}\delta(\underline{\sigma}_0, \underline{\sigma}_1). $$

Let $c: [0, \underline{d}(\underline{\sigma}_0, \underline{\sigma}_1)] \to X$ be the minimizing $\underline{d}$-geodesic between $\underline{\sigma}_0$ and $\underline{\sigma}_1$. Assume first $\delta(\underline{\sigma}_0, \underline{\sigma}_1) < \pi$ so that $c(\underline{d}(\underline{\sigma}_0, \underline{\sigma}_1)/2) \ne 0$. Then $c(\underline{d}(\underline{\sigma}_0, \underline{\sigma}_1)/2)$ writes as $\Phi(s, \underline{\sigma}_m)$ for some $(s, \sigma_m) \in (0, 1) \times \mathcal{S}$. We have

$$ \underline{d}(\underline{\sigma}_0, \underline{\sigma}_m) = \underline{d}(\underline{\sigma}_m, \underline{\sigma}_1) = \frac{1}{2} \underline{d}(\underline{\sigma}_0, \underline{\sigma}_1), $$

from which follows

$$ (4.13) \qquad (1-s)^2 + 4s \sin^2(\alpha_0/2) = (1-s)^2 + 4s \sin^2(\alpha_1/2) = \sin^2(\beta/2) $$

thanks to (4.12), where we have set $\alpha_0 := \delta(\underline{\sigma}_0, \underline{\sigma}_m)$, $\alpha_1 := \delta(\underline{\sigma}_m, \underline{\sigma}_1)$ and $\beta := \delta(\underline{\sigma}_0, \underline{\sigma}_1)$. Note first that (4.13) immediately implies $\alpha_0 = \alpha_1$. Moreover, for any $t > 0$,

$$ \underline{d}(\underline{\sigma}_0, \Phi(t, \sigma_m)) + \underline{d}(\underline{\sigma}_1, \Phi(t, \sigma_m)) \ge \underline{d}(\underline{\sigma}_0, \underline{\sigma}_1) $$

implies

$$ \sqrt{(1-t)^2 + 4t \sin^2(\alpha_0/2)} + \sqrt{(1-t)^2 + 4t \sin^2(\alpha_1/2)} \ge 2 \sin(\beta/2), $$

thus

$$ (1-t)^2 + 4t \sin^2(\alpha_0/2) \ge \sin^2(\beta/2). $$

Therefore, the polynomial function $F: t \mapsto (1-t)^2 + 4t \sin^2(\alpha_0/2) - \sin^2(\beta/2)$ is non-negative and vanishes only at $t = s$, so $F'(s) = 0$ hence

$$ 2(1-s) = 4 \sin^2(\alpha_0/2). $$

Plugging this in (4.13) leads to $\sin^2(\beta/2) = \sin^2(\alpha_0/2)$ hence $\alpha_0 = \beta_2$.

□

Claim 4.9 and Lemma 2.14 implies $\delta = \underline{d}_{\mathcal{S}}$ which yields (4.11) by Claim 4.8.
*Proof.* — Let us first show that $\delta$ defines a distance on $\mathcal{S}$. We only prove the triangle inequality since the two other properties are immediate. For given $\underline{\sigma}_0, \underline{\sigma}_1, \underline{\sigma}_2 \in \mathcal{S}$, let us set $\alpha := \delta(\underline{\sigma}_0, \underline{\sigma}_1)$, $\beta := \delta(\underline{\sigma}_1, \underline{\sigma}_2)$ and $\gamma := \delta(\underline{\sigma}_0, \underline{\sigma}_2)$. We can assume $\alpha + \beta \le \pi$ because otherwise we would have $\alpha + \beta > \pi \ge \gamma$, thus nothing to prove. For any $t, s, r > 0$, the triangle inequality for $\underline{d}$ written with (4.12) gives

$$ \sqrt{(t-s)^2 + 4ts \sin^2(\alpha/2)} + \sqrt{(s-r)^2 + 4sr \sin^2(\beta/2)} \ge \sqrt{(t-r)^2 + 4tr \sin^2(\gamma/2)}. $$

Considering the three complex numbers $z_0 = t$, $z_1 = se^{i\alpha}$ and $z_2 = re^{i(\alpha+\beta)}$, this can be rewritten as

$$ |z_0 - z_1| + |z_1 - z_2| \ge \sqrt{(t-r)^2 + 4tr \sin^2(\gamma/2)}. $$

Choosing $s$ so that $z_0, z_1, z_2$ are aligned implies $|z_0 - z_2| = |z_0 - z_1| + |z_1 - z_2|$ thus

$$ \sqrt{(t-r)^2 + 4tr \sin^2((\alpha + \beta)/2)} = |z_0 - z_2| \ge \sqrt{(t-r)^2 + 4tr \sin^2(\gamma/2)}, $$

which yields to $\alpha + \beta \ge \gamma$.

Let us show now that $\delta$ is geodesic. For given $\underline{\sigma}_0, \underline{\sigma}_1 \in \mathcal{S}$ with $\underline{\sigma}_0 \ne \underline{\sigma}_1$, we aim at finding $\underline{\sigma}_m \in \mathcal{S}$ such that

$$ \delta(\underline{\sigma}_0, \underline{\sigma}_m) = \delta(\underline{\sigma}_m, \underline{\sigma}_1) = \frac{1}{2}\delta(\underline{\sigma}_0, \underline{\sigma}_1). $$

Let $c: [0, \underline{d}(\underline{\sigma}_0, \underline{\sigma}_1)] \to X$ be the minimizing $\underline{d}$-geodesic between $\underline{\sigma}_0$ and $\underline{\sigma}_1$. Assume first $\delta(\underline{\sigma}_0, \underline{\sigma}_1) < \pi$ so that $c(\underline{d}(\underline{\sigma}_0, \underline{\sigma}_1)/2) \ne 0$. Then $c(\underline{d}(\underline{\sigma}_0, \underline{\sigma}_1)/2)$ writes as $\Phi(s, \underline{\sigma}_m)$ for some $(s, \sigma_m) \in (0, 1) \times \mathcal{S}$. We have

$$ \underline{d}(\underline{\sigma}_0, \underline{\sigma}_m) = \underline{d}(\underline{\sigma}_m, \underline{\sigma}_1) = \frac{1}{2} \underline{d}(\underline{\sigma}_0, \underline{\sigma}_1), $$

from which follows

$$ (4.13) \qquad (1-s)^2 + 4s \sin^2(\alpha_0/2) = (1-s)^2 + 4s \sin^2(\alpha_1/2) = \sin^2(\beta/2) $$

thanks to (4.12), where we have set $\alpha_0 := \delta(\underline{\sigma}_0, \underline{\sigma}_m)$, $\alpha_1 := \delta(\underline{\sigma}_m, \underline{\sigma}_1)$ and $\beta := \delta(\underline{\sigma}_0, \underline{\sigma}_1)$. Note first that (4.13) immediately implies $\alpha_0 = \alpha_1$. Moreover, for any $t > 0$,

$$ \underline{d}(\underline{\sigma}_0, \Phi(t, \sigma_m)) + \underline{d}(\underline{\sigma}_1, \Phi(t, \sigma_m)) \ge \underline{d}(\underline{\sigma}_0, \underline{\sigma}_1) $$

implies

$$ \sqrt{(1-t)^2 + 4t \sin^2(\alpha_0/2)} + \sqrt{(1-t)^2 + 4t \sin^2(\alpha_1/2)} \ge 2 \sin(\beta/2), $$

thus

$$ (1-t)^2 + 4t \sin^2(\alpha_0/2) \ge \sin^2(\beta/2). $$

Therefore, the polynomial function $F: t \mapsto (1-t)^2 + 4t \sin^2(\alpha_0/2) - \sin^2(\beta/2)$ is non-negative and vanishes only at $t = s$, so $F'(s) = 0$ hence

$$ 2(1-s) = 4 \sin^2(\alpha_0/2). $$

Plugging this in (4.13) leads to $\sin^2(\beta/2) = \sin^2(\alpha_0/2)$ hence $\alpha_0 = \beta_2$. $\square$

<figure><img src="image_8.png" /></figure>

Claim 4.9 and Lemma 2.14 implies $\delta = \underline{d}_{\mathcal{S}}$ which yields (4.11) by Claim 4.8.
4.3. EQUALITY $\ell = \alpha$ AND POSITIVE DEFINITENESS OF $Q$. — Since $\beta$ is non-degenerate, we can write

$$ (4.14) \qquad \mathbb{R}^{\ell} = E_{+} \oplus E_{-}, $$

where $E_+$ is a subspace of $\mathbb{R}^\ell$ with maximal dimension where $\beta$ is positive definite and $E_-$ is its $\beta$-orthogonal complement; $\beta$ is negative definite on $E_-$. We call $p_+$ the dimension of $E_+$ and $p_-$ the dimension of $E_-$. Note that $\ell = p_+ + p_- \text{ so in particular, } \ell \ge p_+$. Let us prove $p_+ \ge \alpha$, then $\ell = \alpha$, in order to reach our conclusion that is $\ell = p_+ = \alpha$.

Step 1. $p_+ \ge \alpha$. — Let us write $\underline{H} = (\underline{H}_+, \underline{H}_-)$, where $\underline{H}_+ := \text{proj}_{E_+} \circ \underline{H}$ and $\underline{H}_- := \text{proj}_{E_-} \circ \underline{H}$, and $\text{proj}_{E_+}, \text{proj}_{E_-}$ are the projections associated to the decomposition (4.14). Moreover, we set $q_+(v_+) := \beta(v_+, v_+)$ for any $v_+ \in E_+$ and $q_-(v_-) := \beta(v_-, v_-)$ for any $v_- \in E_-$. Then for any $\underline{x}, \underline{y} \in \underline{X}$,

$$ Q(\underline{H}(\underline{x}) - \underline{H}(\underline{y})) = q_+(\underline{H}_+(\underline{x}) - \underline{H}_+(\underline{y})) + q_-(\underline{H}_-(\underline{x}) - \underline{H}_-(\underline{y})), $$

thus

$$ (4.15) \qquad \underline{d}^2(\underline{x}, \underline{y}) - q_-(\underline{H}_-(\underline{x}) - \underline{H}_-(\underline{y})) = q_+(\underline{H}_+(\underline{x}) - \underline{H}_+(\underline{y})). $$

Since $q_- \le 0$, it follows from (4.15) that $q_+(\underline{H}_+(\underline{x}) - \underline{H}_+(\underline{y})) \ge \underline{d}^2(\underline{x}, \underline{y})$. Moreover, $-q_-(\underline{H}_-(\underline{x}) - \underline{H}_-(\underline{y}))$ is bounded from above by $\lambda \underline{d}^2(\underline{x}, \underline{y})$, where $\lambda$ is the largest modulus an eigenvalue of $Q$ can have, so $q_+(\underline{H}_+(\underline{x}) - \underline{H}_+(\underline{y})) \le (1 + \lambda) \underline{d}^2(\underline{x}, \underline{y})$. Finally, since $\underline{H}$ is injective, then $\underline{H}_+$ is injective too. Therefore, the map $\underline{H}_+$ is a bi-Lipschitz embedding of $(\underline{X}, \underline{d})$ into $(E_+, \underline{d}_{q_+})$ where $\underline{d}_{q_+}(v_+, v'_+) := \sqrt{q_+(v_+ - v'_+)}$ for any $v_+, v'_+ \in E_+$. This implies that $p_+$ is greater than or equal to the local Hausdorff dimension of $\underline{X}$ which is equal to $\alpha$.

Step 2. $\ell = \alpha$. — Set $\tilde{\mu} := (\Phi^{-1})_{\#}(\mu \llcorner \underline{X} \setminus \underline{\sigma})$. Then $\tilde{\mu}$ is a Borel measure on $(0, +\infty) \times \underline{S}$ equipped with $\underline{d}_C$. We complete $(0, +\infty) \times \underline{S}$ by adding the point $\underline{\sigma}$ corresponding to the tip of this metric cone. Let $\underline{d} \otimes \nu_t$ be the disintegration of $\tilde{\mu}$ with respect to the first variable $t$ (we refer to [AFP00, 2.5] for the definition of disintegration of a measure). Since for any $\lambda, r > 0$, we have $\underline{\mu}(h_\lambda(B_r(\underline{\sigma}))) = \lambda^\alpha \underline{\mu}(B_r(\underline{\sigma}))$, where $h_\lambda : (t, \underline{\sigma}) \mapsto (\lambda t, \underline{\sigma})$, then $\underline{d}\nu_t = t^{\alpha-1}\nu_1$ for any $t > 0$ and $\underline{\nu}_1(\underline{S}) = \alpha\omega_\alpha$. Let us write $\underline{\nu}$ instead of $\underline{\nu}_1$.

CLAIM 4.10. — For any $\underline{\sigma} \in \underline{S}$ and $\underline{h} \in \underline{V} := \text{Span}(\underline{h}_1, \dots, \underline{h}_\ell)$,

$$ (4.16) \qquad \underline{h}(\underline{\sigma}) = \frac{\alpha}{\underline{\nu}(\underline{S})} \int_{\underline{S}} \cos(\underline{d}_{\underline{S}}(\underline{\sigma}, \underline{\varphi})) \underline{h}(\underline{\varphi}) \underline{d}\underline{\nu}(\underline{\varphi}). $$

Proof. — Take $\underline{h} \in \underline{V}$ and $t > 0$. Since $\underline{h}_1, \dots, \underline{h}_\ell$ are locally $\underline{L}$-harmonic, then for any $\underline{x} \in \underline{X}$,

$$ (4.17) \qquad \underline{h}(\underline{x}) = \int_{\underline{X}} \underline{p}(\underline{x}, \underline{y}, t) \underline{h}(\underline{y}) \underline{d}\underline{\mu}(\underline{y}). $$
### 4.3. EQUALITY $\ell = \alpha$ AND POSITIVE DEFINITENESS OF $Q$. — Since $\beta$ is non-degenerate, we can write

$$ (4.14) \qquad \mathbb{R}^{\ell} = E_{+} \oplus E_{-}, $$

where $E_+$ is a subspace of $\mathbb{R}^\ell$ with maximal dimension where $\beta$ is positive definite and $E_-$ is its $\beta$-orthogonal complement; $\beta$ is negative definite on $E_-$. We call $p_+$ the dimension of $E_+$ and $p_-$ the dimension of $E_-$. Note that $\ell = p_+ + p_- \text{ so in particular, } \ell \ge p_+$. Let us prove $p_+ \ge \alpha$, then $\ell = \alpha$, in order to reach our conclusion that is $\ell = p_+ = \alpha$.

**Step 1.** $p_+ \ge \alpha$. — Let us write $\underline{H} = (\underline{H}_+, \underline{H}_-)$, where $\underline{H}_+ := \text{proj}_{E_+} \circ \underline{H}$ and $\underline{H}_- := \text{proj}_{E_-} \circ \underline{H}$, and $\text{proj}_{E_+}$, $\text{proj}_{E_-}$ are the projections associated to the decomposition (4.14). Moreover, we set $q_+(v_+) := \beta(v_+, v_+)$ for any $v_+ \in E_+$ and $q_-(v_-) := \beta(v_-, v_-)$ for any $v_- \in E_-$. Then for any $\underline{x}, \underline{y} \in \underline{X}$,

$$ Q(\underline{H}(\underline{x}) - \underline{H}(\underline{y})) = q_+(\underline{H}_+(\underline{x}) - \underline{H}_+(\underline{y})) + q_-(\underline{H}_-(\underline{x}) - \underline{H}_-(\underline{y})), $$

thus

$$ (4.15) \qquad \underline{d}^2(\underline{x}, \underline{y}) - q_-(\underline{H}_-(\underline{x}) - \underline{H}_-(\underline{y})) = q_+(\underline{H}_+(\underline{x}) - \underline{H}_+(\underline{y})). $$

Since $q_- \le 0$, it follows from (4.15) that $q_+(\underline{H}_+(\underline{x}) - \underline{H}_+(\underline{y})) \ge \underline{d}^2(\underline{x}, \underline{y})$. Moreover, $-q_-(\underline{H}_-(\underline{x}) - \underline{H}_-(\underline{y}))$ is bounded from above by $\lambda \underline{d}^2(\underline{x}, \underline{y})$, where $\lambda$ is the largest modulus an eigenvalue of $Q$ can have, so $q_+(\underline{H}_+(\underline{x}) - \underline{H}_+(\underline{y})) \le (1 + \lambda) \underline{d}^2(\underline{x}, \underline{y})$. Finally, since $\underline{H}$ is injective, then $\underline{H}_+$ is injective too. Therefore, the map $\underline{H}_+$ is a bi-Lipschitz embedding of $(\underline{X}, \underline{d})$ into $(E_+, \underline{d}_{q_+})$ where $\underline{d}_{q_+}(v_+, v'_+) := \sqrt{q_+(v_+ - v'_+)}$ for any $v_+, v'_+ \in E_+$. This implies that $p_+$ is greater than or equal to the local Hausdorff dimension of $\underline{X}$ which is equal to $\alpha$.

**Step 2.** $\ell = \alpha$. — Set $\tilde{\mu} := (\Phi^{-1})_{\#}(\mu \llcorner \underline{X} \setminus \{\underline{\sigma}\})$. Then $\tilde{\mu}$ is a Borel measure on $(0, +\infty) \times \underline{S}$ equipped with $\underline{d}_C$. We complete $(0, +\infty) \times \underline{S}$ by adding the point $\underline{\sigma}$ corresponding to the tip of this metric cone. Let $\underline{d} \otimes \nu_t$ be the disintegration of $\tilde{\mu}$ with respect to the first variable $t$ (we refer to [AFP00, 2.5] for the definition of disintegration of a measure). Since for any $\lambda, r > 0$, we have $\underline{\mu}(h_\lambda(B_r(\underline{\sigma}))) = \lambda^\alpha \underline{\mu}(B_r(\underline{\sigma}))$, where $h_\lambda : (t, \underline{\sigma}) \mapsto (\lambda t, \underline{\sigma})$, then $\underline{d}\nu_t = t^{\alpha-1}\nu_1$ for any $t > 0$ and $\underline{\nu}_1(\underline{S}) = \alpha\omega_\alpha$. Let us write $\underline{\nu}$ instead of $\underline{\nu}_1$.

**CLAIM 4.10.** — For any $\underline{\sigma} \in \underline{S}$ and $\underline{h} \in \underline{V} := \text{Span}(\underline{h}_1, \dots, \underline{h}_\ell)$,

$$ (4.16) \qquad \underline{h}(\underline{\sigma}) = \frac{\alpha}{\underline{\nu}(\underline{S})} \int_{\underline{S}} \cos(\underline{d}_{\underline{S}}(\underline{\sigma}, \underline{\varphi})) \underline{h}(\underline{\varphi}) \underline{d}\underline{\nu}(\underline{\varphi}). $$

**Proof.** — Take $\underline{h} \in \underline{V}$ and $t > 0$. Since $\underline{h}_1, \dots, \underline{h}_\ell$ are locally $\underline{L}$-harmonic, then for any $\underline{x} \in \underline{X}$,

$$ (4.17) \qquad \underline{h}(\underline{x}) = \int_{\underline{X}} \underline{p}(\underline{x}, \underline{y}, t) \underline{h}(\underline{y}) \underline{d}\underline{\mu}(\underline{y}). $$
Use the notation $\underline{x} = \Phi(r, \underline{\sigma})$ and $\underline{y} = \Phi(s, \underline{\varphi})$ and note that $\underline{h}(\Phi(t, \underline{\sigma})) = r\underline{h}(\underline{\sigma})$ and $\underline{h}(\Phi(s, \underline{\varphi})) = s\underline{h}(\underline{\varphi})$ thanks to (4.9). Then (4.17) writes

$$ (4.18) \qquad r\underline{h}(\underline{\sigma}) = \frac{1}{(4\pi t)^{\alpha/2}} \int_{0}^{+\infty} \int_{\underline{S}} e^{(-r^2 - s^2 + 2rs \cos(\underline{d}_{\underline{S}}(\underline{\sigma}, \underline{\varphi}))) / 4t} \underline{h}(\underline{\varphi}) s^{\alpha} \, ds \, d\underline{\nu}(\underline{\varphi}). $$

Since

$$ \frac{d}{dr} \left( e^{(-r^2 + 2rs \cos(\underline{d}_{\underline{S}}(\underline{\sigma}, \underline{\varphi}))) / 4t} \right) = \left( -\frac{r}{2t} + \frac{s \cos(\underline{d}_{\underline{S}}(\underline{\sigma}, \underline{\varphi}))}{2t} \right) e^{(-r^2 + 2rs \cos(\underline{d}_{\underline{S}}(\underline{\sigma}, \underline{\varphi}))) / 4t}, $$

differentiating (4.18) with respect to $r$ and evaluating at $r = 0$ gives

$$ \underline{h}(\underline{\sigma}) = \frac{1}{(4\pi t)^{\alpha/2}} \int_{0}^{+\infty} e^{-s^2/4t} \frac{s^{\alpha+1}}{2t} \, ds \int_{\underline{S}} \cos(\underline{d}_{\underline{S}}(\underline{\sigma}, \underline{\varphi})) \underline{h}(\underline{\varphi}) \, d\underline{\nu}(\underline{\varphi}). $$

A direct computation using the change of variable $\xi = s^2/4t$ shows that

$$ \frac{1}{(4\pi t)^{\alpha/2}} \int_{0}^{+\infty} e^{-s^2/4t} \frac{s^{\alpha+1}}{2t} \, ds = \frac{1}{\omega_{\alpha}} = \frac{\alpha}{\underline{\nu}(\underline{S})}. \quad \square $$

Set $\underline{W} := \{\underline{h}|_{\underline{S}} : \underline{h} \in \underline{V}\}$. Note that (4.9) implies that the restriction map $\underline{\nu} : \underline{W} \to \underline{W}$ is a bijection, hence $\dim \underline{W} = \ell$. We introduce the operator $\mathcal{K}: L^2(\underline{S}, d\underline{\nu}) \to L^2(\underline{S}, d\underline{\nu})$ defined by

$$ \mathcal{K}(f)(\underline{\sigma}) := \frac{\alpha}{\underline{\nu}(\underline{S})} \int_{\underline{S}} \cos(\underline{d}_{\underline{S}}(\underline{\sigma}, \underline{\varphi})) f(\underline{\varphi}) \, d\underline{\nu}(\underline{\varphi}) $$

for any $f \in L^2(\underline{S}, d\underline{\nu})$ and $\underline{\mu}$-a.e. $\underline{\sigma} \in \underline{S}$. Since for any $\underline{\sigma}, \underline{\sigma}' \in \underline{S}$,

$$ \cos(\underline{d}_{\underline{S}}(\underline{\sigma}, \underline{\sigma}')) = \underline{B}(\underline{\sigma}, \underline{\sigma}') = \beta(\underline{H}(\Phi(1, \underline{\sigma})), \underline{H}(\Phi(1, \underline{\sigma}'))) = \sum_{i,j} B(x_i, x_j) \underline{h}_i(\underline{\sigma}) \underline{h}_j(\underline{\sigma}'), $$

then the image of $\mathcal{K}$ is contained in $\underline{W}$ and according to (4.16), we have

$$ \mathcal{K}f = f \quad \text{for every } f \in \underline{W}. $$

Hence $\mathcal{K}$ is the orthogonal projection onto $\underline{W}$ and if $\underline{k}_1, \dots, \underline{k}_\ell$ form an orthonormal basis of $\underline{W}$ for the $L^2(\underline{S}, \underline{\nu})$ scalar product, then for any $f \in L^2(\underline{S}, d\underline{\nu})$:

$$ \mathcal{K}(f)(\underline{\sigma}) = \sum_{i=1}^{\ell} \underline{k}_i(\underline{\sigma}) \int_{\underline{S}} \underline{k}_i(\underline{\varphi}) f(\underline{\varphi}) \, d\underline{\nu}(\underline{\varphi}). $$

This implies

$$ (4.19) \qquad \frac{\alpha}{\underline{\nu}(\underline{S})} \cos(\underline{d}_{\underline{S}}(\underline{\sigma}, \underline{\varphi})) = \sum_{i=1}^{\ell} \underline{k}_i(\underline{\sigma}) \underline{k}_i(\underline{\varphi}) $$

for $\underline{\nu} \otimes \underline{\nu}$-a.e. $(\underline{\sigma}, \underline{\varphi}) \in \underline{S} \times \underline{S}$. Since for any $i$, the function $\underline{k}_i$ admits a continuous representative – still denoted by $\underline{k}_i$ – defined by

$$ \underline{k}_i(\underline{\sigma}) = \int_{\underline{S}} \cos(\underline{d}_{\underline{S}}(\underline{\sigma}, \underline{\varphi})) \underline{h}(\underline{\varphi}) \, d\underline{\nu}(\underline{\varphi}) $$
Use the notation $\underline{x} = \Phi(r, \underline{\sigma})$ and $\underline{y} = \Phi(s, \underline{\varphi})$ and note that $\underline{h}(\Phi(t, \underline{\sigma})) = r\underline{h}(\underline{\sigma})$ and $\underline{h}(\Phi(s, \underline{\varphi})) = s\underline{h}(\underline{\varphi})$ thanks to (4.9). Then (4.17) writes

$$ (4.18) \qquad r\underline{h}(\underline{\sigma}) = \frac{1}{(4\pi t)^{\alpha/2}} \int_{0}^{+\infty} \int_{\underline{S}} e^{(-r^2 - s^2 + 2rs \cos(\underline{d}_{\underline{S}}(\underline{\sigma}, \underline{\varphi}))) / 4t} \underline{h}(\underline{\varphi}) s^{\alpha} \, ds \, d\underline{\nu}(\underline{\varphi}). $$

Since

$$ \frac{d}{dr} \left( e^{(-r^2 + 2rs \cos(\underline{d}_{\underline{S}}(\underline{\sigma}, \underline{\varphi}))) / 4t} \right) = \left( -\frac{r}{2t} + \frac{s \cos(\underline{d}_{\underline{S}}(\underline{\sigma}, \underline{\varphi}))}{2t} \right) e^{(-r^2 + 2rs \cos(\underline{d}_{\underline{S}}(\underline{\sigma}, \underline{\varphi}))) / 4t}, $$

differentiating (4.18) with respect to $r$ and evaluating at $r=0$ gives

$$ \underline{h}(\underline{\sigma}) = \frac{1}{(4\pi t)^{\alpha/2}} \int_{0}^{+\infty} e^{-s^2/4t} \frac{s^{\alpha+1}}{2t} ds \int_{\underline{S}} \cos(\underline{d}_{\underline{S}}(\underline{\sigma}, \underline{\varphi})) \underline{h}(\underline{\varphi}) d\underline{\nu}(\underline{\varphi}). $$

A direct computation using the change of variable $\xi = s^2/4t$ shows that

$$ \frac{1}{(4\pi t)^{\alpha/2}} \int_{0}^{+\infty} e^{-s^2/4t} \frac{s^{\alpha+1}}{2t} ds = \frac{1}{\omega_{\alpha}} = \frac{\alpha}{\underline{\nu}(\underline{S})}. \quad \square $$

Set $\underline{W} := \{\underline{h}|_{\underline{S}} : \underline{h} \in \underline{V}\}$. Note that (4.9) implies that the restriction map $\underline{\nu} : \underline{W} \to \underline{W}$ is a bijection, hence $\dim \underline{W} = \ell$. We introduce the operator $\mathcal{K}: L^2(\underline{S}, d\underline{\nu}) \to L^2(\underline{S}, d\underline{\nu})$ defined by

$$ \mathcal{K}(f)(\underline{\sigma}) := \frac{\alpha}{\underline{\nu}(\underline{S})} \int_{\underline{S}} \cos(\underline{d}_{\underline{S}}(\underline{\sigma}, \underline{\varphi})) f(\underline{\varphi}) d\underline{\nu}(\underline{\varphi}) $$

for any $f \in L^2(\underline{S}, d\underline{\nu})$ and $\underline{\mu}$-a.e. $\underline{\sigma} \in \underline{S}$. Since for any $\underline{\sigma}, \underline{\sigma}' \in \underline{S}$,

$$ \cos(\underline{d}_{\underline{S}}(\underline{\sigma}, \underline{\sigma}')) = \underline{B}(\underline{\sigma}, \underline{\sigma}') = \beta(\underline{H}(\Phi(1, \underline{\sigma})), \underline{H}(\Phi(1, \underline{\sigma}'))) = \sum_{i,j} B(x_i, x_j) \underline{h}_i(\underline{\sigma}) \underline{h}_j(\underline{\sigma}'), $$

then the image of $\mathcal{K}$ is contained in $\underline{W}$ and according to (4.16), we have

$$ \mathcal{K}f = f \quad \text{for every } f \in \underline{W}. $$

Hence $\mathcal{K}$ is the orthogonal projection onto $\underline{W}$ and if $\underline{k}_1, \dots, \underline{k}_\ell$ form an orthonormal basis of $\underline{W}$ for the $L^2(\underline{S}, \underline{\nu})$ scalar product, then for any $f \in L^2(\underline{S}, d\underline{\nu})$:

$$ \mathcal{K}(f)(\underline{\sigma}) = \sum_{i=1}^{\ell} \underline{k}_i(\underline{\sigma}) \int_{\underline{S}} \underline{k}_i(\underline{\varphi}) f(\underline{\varphi}) d\underline{\nu}(\underline{\varphi}). $$

This implies

$$ (4.19) \qquad \frac{\alpha}{\underline{\nu}(\underline{S})} \cos(\underline{d}_{\underline{S}}(\underline{\sigma}, \underline{\varphi})) = \sum_{i=1}^{\ell} \underline{k}_i(\underline{\sigma}) \underline{k}_i(\underline{\varphi}) $$

for $\underline{\nu} \otimes \underline{\nu}$-a.e. $(\underline{\sigma}, \underline{\varphi}) \in \underline{S} \times \underline{S}$. Since for any $i$, the function $\underline{k}_i$ admits a continuous representative – still denoted by $\underline{k}_i$ – defined by

$$ \underline{k}_i(\underline{\sigma}) = \int_{\underline{S}} \cos(\underline{d}_{\underline{S}}(\underline{\sigma}, \underline{\varphi})) \underline{h}(\underline{\varphi}) d\underline{\nu}(\underline{\varphi}) $$
for any $\underline{\sigma} \in \underline{S}$, then (4.19) holds for all $(\underline{\sigma}, \varphi) \in \underline{S} \times \underline{S}$. In particular, we can take $\underline{\sigma} = \varphi$ in (4.19) to get

$$ \frac{\alpha}{\nu(\underline{S})} = \sum_{i=1}^{\ell} k_i(\underline{\sigma})^2. $$

Integrating over $\underline{S}$ with respect to $\nu$ gives $\alpha = \ell$.

4.4. CONCLUSION. — From the previous subsections, we get that $H$ is an isometric embedding of $(X, \mathbf{d})$ into $(\mathbb{R}^{\ell}, \mathbf{d}_Q)$ or, as explained at the beginning of this section, into $(\mathbb{R}^{\ell}, \mathbf{d}_e)$. Therefore, $H(X)$ equipped with the restriction of $\mathbf{d}_e$ is geodesic. Minimizing geodesics in $(\mathbb{R}^{\ell}, \mathbf{d}_e)$ being straight lines, this implies that $H(X)$ is convex. Being also closed, $H(X)$ is equal to its closed convex hull that is equal to $\mathbb{R}^{\ell}$ by the proof of Claim 4.2, hence Theorem 1.1 is proved.

5. ALMOST RIGIDITY RESULT FOR THE HEAT KERNEL

In this section, we show how our rigidity result (Theorem 1.1) provides an almost rigidity result (Theorem 1.2). We fix a positive constant $T > 0$, a positive integer $n$, and we recall that $\mathbb{B}_r^n$ stands for an Euclidean ball in $\mathbb{R}^n$ with radius $r > 0$ (where this ball is centered as no importance), and $\mathbf{d}_{GH}$ for the Gromov-Hausdorff distance.
We begin with the following lemma:

LEMMA 5.1. — If $(X, \mathbf{d}, \mu)$ is a complete metric measure space endowed with a symmetric Dirichlet form $\mathcal{E}$ admitting a heat kernel $p$ such that for some $\gamma > 1$,

$$ (5.1) \qquad \frac{\gamma^{-1}}{(4\pi t)^{n/2}} e^{-\gamma \mathbf{d}^2(x,y)/4t} \le p(x,y,t) \le \frac{\gamma}{(4\pi t)^{n/2}} e^{-\gamma \mathbf{d}^2(x,y)/4\gamma t} $$

for all $x, y \in X$ and $t \in (0, T]$, then there exists positive constants $c(n, \gamma)$, $C(n, \gamma)$ such that for any $x \in X$ and $r \le \sqrt{T}$,

$$ c(n, \gamma) r^n \le \mu(B_r(x)) \le C(n, \gamma) r^n. $$

REMARK 5.2. — The upper bound is quite classical, the novelty is the lower bound which was nonetheless known for stochastically complete spaces (see [Gri10, Th. 2.11]).

Proof. — For any $x \in X$ and $r > 0$, integrating the lower bound in (5.1) gives

$$ e^{-\gamma r^2 / 4t} \mu(B_r(x)) \le \int_{B_r(x)} e^{-\gamma \mathbf{d}^2(x,y)/4t} d\mu(y) \le \gamma (4\pi t)^{n/2}, $$

hence $\mu(B_r(x)) \le \gamma(4\pi t)^{n/2} e^{\gamma r^2 / 4t}$ for any $t \in (0, T]$. Consequently, when $r \le \sqrt{T}$, choosing $t = r^2$ provides

$$ (5.2) \qquad \mu(B_r(x)) \le e^{\gamma/4} \gamma (4\pi)^{n/2} r^n, $$

while when $r \ge \sqrt{T}$, choosing $t = T$ gives

$$ (5.3) \qquad \mu(B_r(x)) \le \gamma (4\pi T)^{n/2} e^{\gamma r^2 / 4T}. $$
for any $\underline{\sigma} \in \underline{S}$, then (4.19) holds for all $(\underline{\sigma}, \underline{\varphi}) \in \underline{S} \times \underline{S}$. In particular, we can take $\underline{\sigma} = \underline{\varphi}$ in (4.19) to get

$$ \frac{\alpha}{\nu(\underline{S})} = \sum_{i=1}^{\ell} k_i(\underline{\sigma})^2. $$

Integrating over $\underline{S}$ with respect to $\nu$ gives $\alpha = \ell$.

4.4. CONCLUSION. — From the previous subsections, we get that $H$ is an isometric embedding of $(X, \mathbf{d})$ into $(\mathbb{R}^\ell, \mathbf{d}_Q)$ or, as explained at the beginning of this section, into $(\mathbb{R}^\ell, \mathbf{d}_e)$. Therefore, $H(X)$ equipped with the restriction of $\mathbf{d}_e$ is geodesic. Minimizing geodesics in $(\mathbb{R}^\ell, \mathbf{d}_e)$ being straight lines, this implies that $H(X)$ is convex. Being also closed, $H(X)$ is equal to its closed convex hull that is equal to $\mathbb{R}^\ell$ by the proof of Claim 4.2, hence Theorem 1.1 is proved.

5. ALMOST RIGIDITY RESULT FOR THE HEAT KERNEL

In this section, we show how our rigidity result (Theorem 1.1) provides an almost rigidity result (Theorem 1.2). We fix a positive constant $T > 0$, a positive integer $n$, and we recall that $\mathbb{B}_r^n$ stands for an Euclidean ball in $\mathbb{R}^n$ with radius $r > 0$ (where this ball is centered as no importance), and $\mathbf{d}_{GH}$ for the Gromov-Hausdorff distance.
We begin with the following lemma:

LEMMA 5.1. — If $(X, \mathbf{d}, \mu)$ is a complete metric measure space endowed with a symmetric Dirichlet form $\mathcal{E}$ admitting a heat kernel $p$ such that for some $\gamma > 1$,

$$ (5.1) \qquad \frac{\gamma^{-1}}{(4\pi t)^{n/2}} e^{-\gamma \mathbf{d}^2(x,y)/4t} \le p(x,y,t) \le \frac{\gamma}{(4\pi t)^{n/2}} e^{-\gamma \mathbf{d}^2(x,y)/4\gamma t} $$

for all $x, y \in X$ and $t \in (0, T]$, then there exists positive constants $c(n, \gamma)$, $C(n, \gamma)$ such that for any $x \in X$ and $r \le \sqrt{T}$,

$$ c(n, \gamma) r^n \le \mu(B_r(x)) \le C(n, \gamma) r^n. $$

REMARK 5.2. — The upper bound is quite classical, the novelty is the lower bound which was nonetheless known for stochastically complete spaces (see [Gri10, Th. 2.11]).

Proof. — For any $x \in X$ and $r > 0$, integrating the lower bound in (5.1) gives

$$ e^{-\gamma r^2 / 4t} \mu(B_r(x)) \le \int_{B_r(x)} e^{-\gamma \mathbf{d}^2(x,y)/4t} d\mu(y) \le \gamma (4\pi t)^{n/2}, $$

hence $\mu(B_r(x)) \le \gamma(4\pi t)^{n/2} e^{\gamma r^2 / 4t}$ for any $t \in (0, T]$. Consequently, when $r \le \sqrt{T}$, choosing $t = r^2$ provides

$$ (5.2) \qquad \mu(B_r(x)) \le e^{\gamma/4} \gamma (4\pi)^{n/2} r^n, $$

while when $r \ge \sqrt{T}$, choosing $t = T$ gives

$$ (5.3) \qquad \mu(B_r(x)) \le \gamma (4\pi T)^{n/2} e^{\gamma r^2 / 4T}. $$
Note that (5.2) is the desired upper bound. Take $t \in (0, T/2]$. Combining (5.1) with the Chapman-Kolmogorov formula, we get

$$ \frac{\gamma^{-1}}{(8\pi t)^{n/2}} \le p(x, x, 2t) = \int_X p(x, y, t)^2 d\mu(y) \le \frac{\gamma^2}{(4\pi t)^n} \int_X e^{-d^2(x,y)/2\gamma t} d\mu(y), $$

hence

$$ (5.4) \quad \gamma^{-3} (2\pi)^{n/2} t^{n/2} \le \int_X e^{-d^2(x,y)/2\gamma t} d\mu(y) \le \mu(B_r(x)) + \int_{X \setminus B_r(x)} e^{-d^2(x,y)/2\gamma t} d\mu(y). $$

From now on, assume $r \le \sqrt{T}$. By Cavalieri's principle and the estimates (5.2) and (5.3), we get

$$ (5.5) \quad \int_{X \setminus B_r(x)} e^{-d^2(x,y)/2\gamma t} d\mu(y) = \int_r^{+\infty} e^{-\rho^2/2\gamma t} \frac{\rho}{\gamma t} \mu(B_\rho(x)) d\rho \\ \le (4\pi)^{n/2} e^{\gamma/4} \int_r^{\sqrt{T}} e^{-\rho^2/2\gamma t} \frac{\rho}{t} \rho^n d\rho + (4\pi T)^{n/2} \int_{\sqrt{T}}^{+\infty} e^{-\rho^2/2\gamma t} \frac{\rho}{t} e^{\gamma\rho^2/4T} d\rho. $$

A direct computation shows that for any $n \in \mathbb{N}$, there exists $C_0 > 0$ depending only on $n$ such that for any $A \ge 1$,

$$ \int_A^{+\infty} e^{-\xi^2/2} \xi^{n+1} d\xi \le C_0 A^n e^{-A^2/2}. $$

Therefore, using the change of variable $\xi = \rho/\sqrt{\gamma t}$ to get

$$ \int_r^{\sqrt{T}} e^{-\rho^2/2\gamma t} \frac{\rho}{t} \rho^n d\rho \le \int_r^{+\infty} e^{-\rho^2/2\gamma t} \frac{\rho}{t} \rho^n d\rho = \gamma^{(n/2)+1} t^{n/2} \int_{r/\sqrt{\gamma t}}^{+\infty} e^{-\xi^2/2} \xi^{n+1} d\xi, $$

we obtain that $r \ge \sqrt{\gamma t}$ implies

$$ (5.6) \quad \int_r^{\sqrt{T}} e^{-\rho^2/2\gamma t} \frac{\rho}{t} \rho^n d\rho \le C_0 \gamma r^n e^{-r^2/2\gamma t}. $$

To bound the second term in (5.5), assume $t \le T/\gamma^2$. Then a straightforward computation shows that $-\rho^2/2\gamma t + \gamma\rho^2/4T \le -\rho^2/4\gamma t$ holds, thus

$$ (5.7) \quad \int_{\sqrt{T}}^{+\infty} e^{-\rho^2/2\gamma t} \frac{\rho}{t} e^{\gamma\rho^2/4T} d\rho \le \int_{\sqrt{T}}^{+\infty} e^{-\rho^2/4\gamma t} \frac{\rho}{t} d\rho = 2\gamma e^{-T/4\gamma t}. $$

Combining (5.5), (5.6) and (5.7), we get existence of a constant $C > 0$ depending only on $n$ such that if $r^2 \ge \gamma^2 t$ (this implies both $r \ge \sqrt{\gamma t}$ and $t \le T/\gamma^2$), then

$$ \int_{X \setminus B_r(x)} e^{-d^2(x,y)/2\gamma t} d\mu(y) \le C(\gamma e^{\gamma/4} r^n e^{-r^2/2\gamma t} + \gamma T^{n/2} e^{-T/4\gamma t}). $$

Then (5.4) implies

$$ \gamma^{-3} (2\pi)^{n/2} t^{n/2} \le \mu(B_r(x)) + C(\gamma e^{\gamma/4} r^n e^{-r^2/2\gamma t} + \gamma T^{n/2} e^{-T/4\gamma t}) $$

for any $t \in (0, r^2/\gamma^2)$, what can be rewritten as

$$ c'(n, \gamma) t^{n/2} \le \mu(B_r(x)) + C t^{n/2} (\gamma e^{\gamma/4} F(r^2/t) + \gamma G(T/t)), $$
Note that (5.2) is the desired upper bound. Take $t \in (0, T/2]$. Combining (5.1) with the Chapman-Kolmogorov formula, we get

$$ \frac{\gamma^{-1}}{(8\pi t)^{n/2}} \le p(x, x, 2t) = \int_X p(x, y, t)^2 \, d\mu(y) \le \frac{\gamma^2}{(4\pi t)^n} \int_X e^{-d^2(x,y)/2\gamma t} \, d\mu(y), $$

hence

$$ (5.4) \quad \gamma^{-3} (2\pi)^{n/2} t^{n/2} \le \int_X e^{-d^2(x,y)/2\gamma t} \, d\mu(y) \le \mu(B_r(x)) + \int_{X \setminus B_r(x)} e^{-d^2(x,y)/2\gamma t} \, d\mu(y). $$

From now on, assume $r \le \sqrt{T}$. By Cavalieri's principle and the estimates (5.2) and (5.3), we get

$$ (5.5) \quad \int_{X \setminus B_r(x)} e^{-d^2(x,y)/2\gamma t} \, d\mu(y) = \int_r^{+\infty} e^{-\rho^2/2\gamma t} \frac{\rho}{\gamma t} \mu(B_\rho(x)) \, d\rho \\ \le (4\pi)^{n/2} e^{\gamma/4} \int_r^{\sqrt{T}} e^{-\rho^2/2\gamma t} \frac{\rho}{t} \rho^n \, d\rho + (4\pi T)^{n/2} \int_{\sqrt{T}}^{+\infty} e^{-\rho^2/2\gamma t} \frac{\rho}{t} e^{\gamma\rho^2/4T} \, d\rho. $$

A direct computation shows that for any $n \in \mathbb{N}$, there exists $C_0 > 0$ depending only on $n$ such that for any $A \ge 1$,

$$ \int_A^{+\infty} e^{-\xi^2/2} \xi^{n+1} \, d\xi \le C_0 A^n e^{-A^2/2}. $$

Therefore, using the change of variable $\xi = \rho/\sqrt{\gamma t}$ to get

$$ \int_r^{\sqrt{T}} e^{-\rho^2/2\gamma t} \frac{\rho}{t} \rho^n \, d\rho \le \int_r^{+\infty} e^{-\rho^2/2\gamma t} \frac{\rho}{t} \rho^n \, d\rho = \gamma^{(n/2)+1} t^{n/2} \int_{r/\sqrt{\gamma t}}^{+\infty} e^{-\xi^2/2} \xi^{n+1} \, d\xi, $$

we obtain that $r \ge \sqrt{\gamma t}$ implies

$$ (5.6) \quad \int_r^{\sqrt{T}} e^{-\rho^2/2\gamma t} \frac{\rho}{t} \rho^n \, d\rho \le C_0 \gamma r^n e^{-r^2/2\gamma t}. $$

To bound the second term in (5.5), assume $t \le T/\gamma^2$. Then a straightforward computation shows that $-\rho^2/2\gamma t + \gamma\rho^2/4T \le -\rho^2/4\gamma t$ holds, thus

$$ (5.7) \quad \int_{\sqrt{T}}^{+\infty} e^{-\rho^2/2\gamma t} \frac{\rho}{t} e^{\gamma\rho^2/4T} \, d\rho \le \int_{\sqrt{T}}^{+\infty} e^{-\rho^2/4\gamma t} \frac{\rho}{t} \, d\rho = 2\gamma e^{-T/4\gamma t}. $$

Combining (5.5), (5.6) and (5.7), we get existence of a constant $C > 0$ depending only on $n$ such that if $r^2 \ge \gamma^2 t$ (this implies both $r \ge \sqrt{\gamma t}$ and $t \le T/\gamma^2$), then

$$ \int_{X \setminus B_r(x)} e^{-d^2(x,y)/2\gamma t} \, d\mu(y) \le C \left( \gamma e^{\gamma/4} r^n e^{-r^2/2\gamma t} + \gamma T^{n/2} e^{-T/4\gamma t} \right). $$

Then (5.4) implies

$$ \gamma^{-3} (2\pi)^{n/2} t^{n/2} \le \mu(B_r(x)) + C \left( \gamma e^{\gamma/4} r^n e^{-r^2/2\gamma t} + \gamma T^{n/2} e^{-T/4\gamma t} \right) $$

for any $t \in (0, r^2/\gamma^2)$, what can be rewritten as

$$ c'(n, \gamma) t^{n/2} \le \mu(B_r(x)) + C t^{n/2} \left( \gamma e^{\gamma/4} F(r^2/t) + \gamma G(T/t) \right), $$
where $c'(n, \gamma) := \gamma^{-3}(2\pi)^{n/2}$ and $F(s) := s^{n/2}e^{-s/2\gamma}$, $G(s) := s^{n/2}e^{-s/4\gamma}$ for any $s \ge 0$. The function $G$ is decreasing on $(2n\gamma, +\infty)$ so if $r^2/t \ge 2n\gamma$, since $r^2 \le T$, we get $G(T/t) \le G(r^2/t)$. As $\lim_{s \to +\infty} F(s) = \lim_{s \to +\infty} G(s) = 0$, then there exists $s(n, \gamma) > 0$ such that if $s \ge s(n, \gamma)$,

$$C(\gamma e^{\gamma/4}F(s) + \gamma G(s)) \le \frac{c'(n, \gamma)}{2}.$$

Then for any $t > 0$ such that $r^2/t \ge \max(\gamma^2, 2n\gamma, s(n, \gamma)) =: \theta(n, \gamma)$, we get

$$\frac{c'(n, \gamma)}{2} t^{n/2} \le \mu(B_r(x)).$$

Choosing $t = t(r)$ such that $\theta(n, \gamma)t \le r^2 \le 2\theta(n, \gamma)t$, we get

$$\frac{c'(n, \gamma)}{2^{n/2+1}\theta(n, \gamma)^{n/2}} r^n \le \mu(B_r(x)). \quad \square$$

We shall also need the next proposition.

**Proposition 5.3.** — Let $(X, \mathbf{d}, \mu)$ be a measure metric space satisfying the local doubling condition, namely there exists $r_o > 0$ and $C_D > 0$ such that $\mu(B_{2r}) \le C_D\mu(B_r)$ for any $r \in (0, r_o)$, and such that for some $\alpha > 0$, we have

$$\int_X \frac{1}{(4\pi t)^{\alpha/2}} e^{-d^2(x,z)/4t} \frac{1}{(4\pi s)^{\alpha/2}} e^{-d^2(z,y)/4s} \mathrm{d}\mu(z) = \frac{1}{(4\pi(t+s))^{\alpha/2}} e^{-d^2(x,y)/4(t+s)}$$

for all $x, y \in X$ and $t, s \in (0, T)$. Then there exists a symmetric Dirichlet form $\mathcal{E}$ on $(X, \mathbf{d}, \mu)$ admitting an $\alpha$-Euclidean heat kernel.

*Proof.* — By [Car19, Lem. 3.9], the space $(X, \mathbf{d}, \mu)$ satisfies $\mu(B_R(x))/\mu(B_r(x)) \le c_o e^{c_1 R/r}$ for any $x \in X$, $r \in (0, r_o)$ and $R \ge r$, where $c_o$ and $c_1$ depend only on $C_D$. For any $C > 0$ and $z \in X$, applying (3.3) with $\varphi(\lambda) = \lambda^2 e^{-C\lambda^2}$ and $g(y) = \mathbf{d}(z, y)$ yields to

$$\begin{align*}
\int_X e^{-Cd^2(z,y)} \mathrm{d}\mu(y) &= \int_0^{+\infty} 2C\lambda e^{-C\lambda^2} \mu(B_\lambda(y)) \mathrm{d}\lambda \\
&\le \left( \int_0^r 2C\lambda e^{-C\lambda^2} \mathrm{d}\lambda + \int_r^{+\infty} 2C\lambda c_o e^{-C\lambda^2+c_1\lambda/r} \mathrm{d}\lambda \right) \mu(B_r(y)) \\
&\le c_2\mu(B_r(y))
\end{align*}$$

for any $r \in (0, r_o)$, where $c_2$ depends only on $r$, $C_D$ and $C$. For any $x, y \in X$ and $t \in \mathbb{C} \setminus \{0\}$, we set

$$\mathbb{P}_{\alpha}(x, y, t) := \frac{1}{(4\pi t)^{\alpha/2}} e^{-d^2(x, y)/4t}.$$

Take $x, y \in X$ and $t \in (0, T)$. By assumption, the identity

$$\int_X \mathbb{P}_\alpha(x, z, t) e^{-d^2(z, y)/4s} \mathrm{d}\mu(z) = \left(\frac{s}{t+s}\right)^{\alpha/2} e^{-d^2(z, y)/4(t+s)}$$

is valid for any $s \in (0, T - t)$. However both expressions are holomorphic in $s \in \mathbb{C}_+ := \{z \in \mathbb{C} : \mathrm{Re}z > 0\}$ (the left-hand side can be proved holomorphic by a suitable application of the dominated convergence theorem using (5.8)), thus the identity holds
where $c'(n, \gamma) := \gamma^{-3}(2\pi)^{n/2}$ and $F(s) := s^{n/2}e^{-s/2\gamma}$, $G(s) := s^{n/2}e^{-s/4\gamma}$ for any $s \ge 0$. The function $G$ is decreasing on $(2n\gamma, +\infty)$ so if $r^2/t \ge 2n\gamma$, since $r^2 \le T$, we get $G(T/t) \le G(r^2/t)$. As $\lim_{s \to +\infty} F(s) = \lim_{s \to +\infty} G(s) = 0$, then there exists $s(n, \gamma) > 0$ such that if $s \ge s(n, \gamma)$,

$$C(\gamma e^{\gamma/4}F(s) + \gamma G(s)) \le \frac{c'(n, \gamma)}{2}.$$

Then for any $t > 0$ such that $r^2/t \ge \max(\gamma^2, 2n\gamma, s(n, \gamma)) =: \theta(n, \gamma)$, we get

$$\frac{c'(n, \gamma)}{2} t^{n/2} \le \mu(B_r(x)).$$

Choosing $t = t(r)$ such that $\theta(n, \gamma)t \le r^2 \le 2\theta(n, \gamma)t$, we get

$$\frac{c'(n, \gamma)}{2^{n/2+1}\theta(n, \gamma)^{n/2}} r^n \le \mu(B_r(x)). \quad \square$$

We shall also need the next proposition.

**Proposition 5.3.** — Let $(X, \mathbf{d}, \mu)$ be a measure metric space satisfying the local doubling condition, namely there exists $r_o > 0$ and $C_D > 0$ such that $\mu(B_{2r}) \le C_D\mu(B_r)$ for any $r \in (0, r_o)$, and such that for some $\alpha > 0$, we have

$$\int_X \frac{1}{(4\pi t)^{\alpha/2}} e^{-d^2(x,z)/4t} \frac{1}{(4\pi s)^{\alpha/2}} e^{-d^2(z,y)/4s} \mathrm{d}\mu(z) = \frac{1}{(4\pi(t+s))^{\alpha/2}} e^{-d^2(x,y)/4(t+s)}$$

for all $x, y \in X$ and $t, s \in (0, T)$. Then there exists a symmetric Dirichlet form $\mathcal{E}$ on $(X, \mathbf{d}, \mu)$ admitting an $\alpha$-Euclidean heat kernel.

*Proof.* — By [Car19, Lem. 3.9], the space $(X, \mathbf{d}, \mu)$ satisfies $\mu(B_R(x))/\mu(B_r(x)) \le c_o e^{c_1 R/r}$ for any $x \in X$, $r \in (0, r_o)$ and $R \ge r$, where $c_o$ and $c_1$ depend only on $C_D$. For any $C > 0$ and $z \in X$, applying (3.3) with $\varphi(\lambda) = \lambda^2 e^{-C\lambda^2}$ and $g(y) = \mathbf{d}(z, y)$ yields to

$$\begin{align*}
\int_X e^{-Cd^2(z,y)} \mathrm{d}\mu(y) &= \int_0^{+\infty} 2C\lambda e^{-C\lambda^2} \mu(B_\lambda(y)) \mathrm{d}\lambda \\
(5.8) \qquad &\le \left( \int_0^r 2C\lambda e^{-C\lambda^2} \mathrm{d}\lambda + \int_r^{+\infty} 2C\lambda c_o e^{-C\lambda^2+c_1\lambda/r} \mathrm{d}\lambda \right) \mu(B_r(y)) \\
&\le c_2\mu(B_r(y))
\end{align*}$$

for any $r \in (0, r_o)$, where $c_2$ depends only on $r$, $C_D$ and $C$. For any $x, y \in X$ and $t \in \mathbb{C} \setminus \{0\}$, we set

$$\mathbb{P}_{\alpha}(x, y, t) := \frac{1}{(4\pi t)^{\alpha/2}} e^{-d^2(x, y)/4t}.$$

Take $x, y \in X$ and $t \in (0, T)$. By assumption, the identity

$$\int_X \mathbb{P}_\alpha(x, z, t) e^{-d^2(z, y)/4s} \mathrm{d}\mu(z) = \left(\frac{s}{t+s}\right)^{\alpha/2} e^{-d^2(z, y)/4(t+s)}$$

is valid for any $s \in (0, T - t)$. However both expressions are holomorphic in $s \in \mathbb{C}_+ := \{z \in \mathbb{C} : \mathrm{Re}z > 0\}$ (the left-hand side can be proved holomorphic by a suitable application of the dominated convergence theorem using (5.8)), thus the identity holds
for any $s \in \mathbb{C}_+$. Freezing $s$ and letting $t$ be variable, we can apply the same reasoning to get the identity valid for any $s, t \in \mathbb{C}_+$. In particular, we obtain:

$$ \int_X \mathbb{P}_\alpha(x, z, t) \mathbb{P}_\alpha(z, y, s) \mathrm{d}\mu(z) = \mathbb{P}_\alpha(x, y, t+s) \quad \forall s, t > 0. $$

Thus for any $x \in X$ and $t > 0$,

$$ \begin{aligned} \int_X \mathbb{P}_\alpha(x, z, t) \mathrm{d}\mu(z) &= (4\pi t)^{\alpha/2} \int_X \left( \frac{1}{(4\pi t)^{\alpha/2}} e^{-d^2(x,z)/8t} \right)^2 \mathrm{d}\mu(z) \\ &= (16\pi t)^{\alpha/2} \int_X \mathbb{P}_\alpha(x, z, 2t)^2 \mathrm{d}\mu(z) \\ &= (16\pi t)^{\alpha/2} \mathbb{P}_\alpha(x, x, 4t) = 1. \end{aligned} $$

This easily implies that for any $f \in L^2(X, \mu)$, if $f_t(x) = \int_X \mathbb{P}_\alpha(x, z, t)f(z) \mathrm{d}\mu(z)$, then

$$ \lim_{t \to 0+} \|f_t - f\|_{L^2} = 0. $$

Then by a standard procedure described for instance in [Gri10, §2], we can build a symmetric Dirichlet form whose heat kernel is $\mathbb{P}_\alpha$. $\square$

We can now prove Theorem 1.2.

*Proof.* — The metric spaces considered in this proof are all complete. Assume that the result is not true. Then there exists some $\epsilon > 0$ such that for any $\delta > 0$ we can find:

$$ -T_{\delta} > 0, $$

— a metric measure space $(X_\delta, \mathbf{d}_\delta, \mu_\delta)$ endowed with a symmetric Dirichlet form $\mathcal{E}_\delta$ admitting a heat kernel $p_\delta$ satisfying

$$ (1 - \delta) \frac{1}{(4\pi t)^{n/2}} e^{-d_{\delta}^2(x,y)/4(1-\delta)t} \le p_{\delta}(x, y, t) \le (1 + \delta) \frac{1}{(4\pi t)^{n/2}} e^{-d_{\delta}^2(x,y)/4(1+\delta)t} $$

for any $x, y \in X_\delta$ and $t \in (0, T_\delta]$,

$$ -x_{\delta} \in X_{\delta} \text{ and } r_{\delta} \in (0, \sqrt{T_{\delta}}] \text{ such that } \mathbf{d}_{\text{GH}}(B_{r_{\delta}}(x_{\delta}), \mathbb{B}_{r_{\delta}}^{n}) \ge \epsilon r_{\delta}. $$

By a rescaling of the distance and of the measure, we can assume that $r_\delta = 1$ and $T_\delta = 1$. It follows from Lemma 5.1 that the set of pointed metric measure space

$$ \{(X_\delta, \mathbf{d}_\delta, \mu_\delta, x_\delta)\}_{\delta \in (0, 1/2)} $$

satisfies a uniform local doubling condition, thus it is precompact for the pointed measure Gromov-Hausdorff topology. Therefore, we can consider an infinitesimal sequence $\{\delta_\ell\}_\ell \subset (0, 1/2)$ and a sequence of pointed metric measure spaces

$$ \{(X_\ell, \mathbf{d}_\ell, \mu_\ell, x_\ell)\}_\ell $$

converging to some pointed metric measure space $(X_\infty, \mathbf{d}_\infty, \mu_\infty, x_\infty)$ such that for any $\ell$:
for any $s \in \mathbb{C}_+$. Freezing $s$ and letting $t$ be variable, we can apply the same reasoning to get the identity valid for any $s, t \in \mathbb{C}_+$. In particular, we obtain:

$$ \int_X \mathbb{P}_\alpha(x, z, t) \mathbb{P}_\alpha(z, y, s) \, d\mu(z) = \mathbb{P}_\alpha(x, y, t+s) \quad \forall s, t > 0. $$

Thus for any $x \in X$ and $t > 0$,

$$ \begin{aligned} \int_X \mathbb{P}_\alpha(x, z, t) \, d\mu(z) &= (4\pi t)^{\alpha/2} \int_X \left( \frac{1}{(4\pi t)^{\alpha/2}} e^{-d^2(x,z)/8t} \right)^2 \, d\mu(z) \\ &= (16\pi t)^{\alpha/2} \int_X \mathbb{P}_\alpha(x, z, 2t)^2 \, d\mu(z) \\ &= (16\pi t)^{\alpha/2} \mathbb{P}_\alpha(x, x, 4t) = 1. \end{aligned} $$

This easily implies that for any $f \in L^2(X, \mu)$, if $f_t(x) = \int_X \mathbb{P}_\alpha(x, z, t)f(z) \, d\mu(z)$, then

$$ \lim_{t \to 0+} \|f_t - f\|_{L^2} = 0. $$

Then by a standard procedure described for instance in [Gri10, §2], we can build a symmetric Dirichlet form whose heat kernel is $\mathbb{P}_\alpha$. $\square$

We can now prove Theorem 1.2.

*Proof.* — The metric spaces considered in this proof are all complete. Assume that the result is not true. Then there exists some $\epsilon > 0$ such that for any $\delta > 0$ we can find:

$$ -T_{\delta} > 0, $$

— a metric measure space $(X_\delta, \mathbf{d}_\delta, \mu_\delta)$ endowed with a symmetric Dirichlet form $\mathcal{E}_\delta$ admitting a heat kernel $p_\delta$ satisfying

$$ (1 - \delta) \frac{1}{(4\pi t)^{n/2}} e^{-d_{\delta}^2(x,y)/4(1-\delta)t} \le p_{\delta}(x,y,t) \le (1 + \delta) \frac{1}{(4\pi t)^{n/2}} e^{-d_{\delta}^2(x,y)/4(1+\delta)t} $$

for any $x, y \in X_\delta$ and $t \in (0, T_\delta]$,

$$ -x_{\delta} \in X_{\delta} \text{ and } r_{\delta} \in (0, \sqrt{T_{\delta}}] \text{ such that } \mathbf{d}_{GH} (B_{r_{\delta}}(x_{\delta}), \mathbb{B}_{r_{\delta}}^{n}) \ge \epsilon r_{\delta}. $$

By a rescaling of the distance and of the measure, we can assume that $r_\delta = 1$ and $T_\delta = 1$. It follows from Lemma 5.1 that the set of pointed metric measure space

$$ \{(X_\delta, \mathbf{d}_\delta, \mu_\delta, x_\delta)\}_{\delta \in (0, 1/2)} $$

satisfies a uniform local doubling condition, thus it is precompact for the pointed measure Gromov-Hausdorff topology. Therefore, we can consider an infinitesimal sequence $\{\delta_\ell\}_\ell \subset (0, 1/2)$ and a sequence of pointed metric measure spaces

$$ \{(X_\ell, \mathbf{d}_\ell, \mu_\ell, x_\ell)\}_\ell $$

converging to some pointed metric measure space $(X_\infty, \mathbf{d}_\infty, \mu_\infty, x_\infty)$ such that for any $\ell$:
– the space $(X_\ell, d_\ell, \mu_\ell)$ is endowed with a symmetric Dirichlet form $\mathcal{E}_\ell$ admitting a heat kernel $p_\ell$ satisfying

$$ (5.9) \quad (1 - \delta_\ell) \frac{1}{(4\pi t)^{n/2}} e^{-\frac{d_\ell^2(x,y)}{4(1-\delta_\ell)t}} \le p_\ell(x, y, t) \\ \le (1 + \delta_\ell) \frac{1}{(4\pi t)^{n/2}} e^{-\frac{d_\ell^2(x,y)}{4(1+\delta_\ell)t}} $$

for any $x, y \in X_\ell$ and $t \in (0, 1]$,

$$ - d_{GH} (B_1(x_\ell), \mathbb{B}_1^n) \ge \epsilon. $$

In particular, letting $\ell$ tend to $+\infty$ gives:

$$ (5.10) \qquad d_{GH} (B_1(x_\infty), \mathbb{B}_1^n) \ge \epsilon. $$

Since for any $\ell$ we have

$$ \int_{X_\ell} p_\ell(x, z, t) p_\ell(z, y, s) d\mu_\ell(z) = p_\ell(t + s, x, y) $$

for all $x, y \in X_\ell$ and $t, s > 0$, we deduce from (5.9) that when $t + s < 1$,

$$ \frac{(1 - \delta_\ell)^{(n/2)+1}}{(1 + \delta_\ell)^{n+1}} \mathbb{P}_n \left( x, y, \frac{1 - \delta_\ell}{1 + \delta_\ell} (t + s) \right) \le \int_{X_\ell} \mathbb{P}_n(x, z, t) \mathbb{P}_n(z, y, s) d\mu_\ell(z) $$

and

$$ \int_{X_\ell} \mathbb{P}_n(x, z, t) \mathbb{P}_n(z, y, s) d\mu_\ell(z) \le \frac{(1 + \delta_\ell)^{(n/2)+1}}{(1 - \delta_\ell)^{n+1}} \mathbb{P}_n \left( x, y, \frac{1 + \delta_\ell}{1 - \delta_\ell} (t + s) \right). $$

From this, we obtain for any $x, y \in X_\infty$ and any $t, s > 0$ with $t + s < 1$,

$$ \int_{X_\infty} \mathbb{P}_n(x, z, t) \mathbb{P}_n(z, y, s) d\mu_\infty(z) = \mathbb{P}_n(x, y, t + s). $$

Then Proposition 5.3 and Theorem 1.1 imply that $(X_\infty, d_\infty)$ is isometric to $(\mathbb{R}^n, d_e)$. But this is in contradiction with (5.10). □

## 6. A NEW PROOF OF COLDING'S ALMOST RIGIDITY THEOREM

In this section, we show how our almost rigidity result (Theorem 1.2) can be used to give an alternative proof of the almost rigidity theorem for the volume of Riemannian manifolds with non-negative Ricci curvature (Theorem 1.4). Here again $n$ is a fixed positive integer and $\mathbb{B}_r^n$ is an Euclidean ball in $\mathbb{R}^n$ with radius $r > 0$.

We recall that whenever $(M^n, g)$ has non-negative Ricci curvature, the Bishop-Gromov comparison theorem states that the function $r \mapsto \omega_n^{-1} r^{-n} \text{vol}(B_r(x))$ is non-increasing for any $x \in M$ and the quantity

$$ (6.1) \qquad \theta = \lim_{r \to +\infty} \frac{\text{vol}(B_r(x))}{\omega_n r^n} $$

does not depend on $x$. When $\theta > 0$, we say that $(M^n, g)$ has Euclidean volume growth, in which case one has

$$ (6.2) \qquad \text{vol}(B_r(x)) \ge \theta \omega_n r^n $$
– the space $(X_\ell, d_\ell, \mu_\ell)$ is endowed with a symmetric Dirichlet form $\mathcal{E}_\ell$ admitting a heat kernel $p_\ell$ satisfying

$$ (5.9) \quad (1 - \delta_\ell) \frac{1}{(4\pi t)^{n/2}} e^{-\frac{d_\ell^2(x,y)}{4(1-\delta_\ell)t}} \le p_\ell(x, y, t) \\ \le (1 + \delta_\ell) \frac{1}{(4\pi t)^{n/2}} e^{-\frac{d_\ell^2(x,y)}{4(1+\delta_\ell)t}} $$

for any $x, y \in X_\ell$ and $t \in (0, 1]$,

$$ - d_{GH} (B_1(x_\ell), \mathbb{B}_1^n) \ge \epsilon. $$

In particular, letting $\ell$ tend to $+\infty$ gives:

$$ (5.10) \qquad d_{GH} (B_1(x_\infty), \mathbb{B}_1^n) \ge \epsilon. $$

Since for any $\ell$ we have

$$ \int_{X_\ell} p_\ell(x, z, t) p_\ell(z, y, s) d\mu_\ell(z) = p_\ell(t + s, x, y) $$

for all $x, y \in X_\ell$ and $t, s > 0$, we deduce from (5.9) that when $t + s < 1$,

$$ \frac{(1 - \delta_\ell)^{(n/2)+1}}{(1 + \delta_\ell)^{n+1}} \mathbb{P}_n \left( x, y, \frac{1 - \delta_\ell}{1 + \delta_\ell} (t + s) \right) \le \int_{X_\ell} \mathbb{P}_n(x, z, t) \mathbb{P}_n(z, y, s) d\mu_\ell(z) $$

and

$$ \int_{X_\ell} \mathbb{P}_n(x, z, t) \mathbb{P}_n(z, y, s) d\mu_\ell(z) \le \frac{(1 + \delta_\ell)^{(n/2)+1}}{(1 - \delta_\ell)^{n+1}} \mathbb{P}_n \left( x, y, \frac{1 + \delta_\ell}{1 - \delta_\ell} (t + s) \right). $$

From this, we obtain for any $x, y \in X_\infty$ and any $t, s > 0$ with $t + s < 1$,

$$ \int_{X_\infty} \mathbb{P}_n(x, z, t) \mathbb{P}_n(z, y, s) d\mu_\infty(z) = \mathbb{P}_n(x, y, t + s). $$

Then Proposition 5.3 and Theorem 1.1 imply that $(X_\infty, d_\infty)$ is isometric to $(\mathbb{R}^n, d_e)$. But this is in contradiction with (5.10). □

## 6. A NEW PROOF OF COLDING'S ALMOST RIGIDITY THEOREM

In this section, we show how our almost rigidity result (Theorem 1.2) can be used to give an alternative proof of the almost rigidity theorem for the volume of Riemannian manifolds with non-negative Ricci curvature (Theorem 1.4). Here again $n$ is a fixed positive integer and $\mathbb{B}_r^n$ is an Euclidean ball in $\mathbb{R}^n$ with radius $r > 0$.

We recall that whenever $(M^n, g)$ has non-negative Ricci curvature, the Bishop-Gromov comparison theorem states that the function $r \mapsto \omega_n^{-1} r^{-n} \text{vol}(B_r(x))$ is non-increasing for any $x \in M$ and the quantity

$$ (6.1) \qquad \theta = \lim_{r \to +\infty} \frac{\text{vol}(B_r(x))}{\omega_n r^n} $$

does not depend on $x$. When $\theta > 0$, we say that $(M^n, g)$ has Euclidean volume growth, in which case one has

$$ (6.2) \qquad \text{vol}(B_r(x)) \ge \theta \omega_n r^n $$
for any $x \in X$ and $r > 0$. Note that a manifold satisfying (1.2) has Euclidean volume growth with $\theta \ge 1 - \delta$. Our proof of Theorem 1.4 is a direct application of Theorem 1.2 together with the following heat kernel estimate.

**THEOREM 6.1.** — There exists a function $\gamma: [0, 1] \to [1, \infty)$ satisfying $\lim_{\theta \to 1^-} \gamma(\theta) = 1$ such that whenever $(M^n, g)$ is a complete Riemannian manifold with non-negative Ricci curvature and Euclidean volume growth, then the heat kernel $p$ of $(M^n, g)$ satisfies

$$ \frac{1}{(4\pi t)^{n/2}} e^{-d^2(x,y)/4t} \le p(x, y, t) \le \gamma(\theta) \frac{1}{(4\pi t)^{n/2}} e^{-d^2(x,y)/\gamma(\theta)4t} $$

for all $x, y \in M$ and $t > 0$, where $\theta$ is given by (6.1).

**REMARK 6.2.** — Our proof of the above heat kernel upper bound follows the arguments of P. Li, L-F. Tam and J. Wang [LTW97].

*Proof.* — The lower bound is the comparison theorem of J. Cheeger and S-T. Yau [CY81]: for any $t > 0$ and $x, y \in M$, we have

$$ (6.3) \qquad \mathbb{P}_n(x, y, t) \le p(x, y, t), $$

where $\mathbb{P}_n(x, y, t) = (4\pi t)^{-n/2} e^{-d^2(x,y)/4t}$. Consequently we only need to prove the upper bound.

Take $x, y \in X$ and $t > 0$. We shall need the following estimates from P. Li and S-T. Yau (see [LTW97, Form. (2.1)]): for any $r, \tau > 0$,

$$ \int_{B_r(x)} p(x, z, \tau) \, d\operatorname{vol}(z) \ge \int_{\mathbb{B}_r^n} \frac{1}{(4\pi\tau)^{n/2}} e^{-\|\xi\|^2/4\tau} \, d\xi $$

and

$$ (6.4) \qquad \int_{M \setminus B_r(x)} p(x, z, \tau) \, d\operatorname{vol}(z) \le \int_{\mathbb{R}^n \setminus \mathbb{B}_r^n} \frac{1}{(4\pi\tau)^{n/2}} e^{-\|\xi\|^2/4\tau} \, d\xi. $$

For $\delta > 0$ to be precisely chosen later, set $r := (1 + \delta)^{-1}d(x, y)$ and $\tau := (1 + \delta)t$. Note that $B_r(x) \cap B_{\delta r}(y) = \emptyset$. By the Harnack inequality of P. Li and S-T. Yau [LY86], we have

$$ p(x, y, t) \le \left(\frac{\tau}{t}\right)^{n/2} e^{d(z,y)^2/4(\tau-t)} p(x, z, \tau) $$

for every $z \in M$, so that averaging over the ball $B_{\delta r}(y)$ gives

$$ (6.5) \qquad \begin{aligned} p(x, y, t) &\le e^{\delta^2 r^2/4(\tau-t)} \left(\frac{\tau}{t}\right)^{n/2} \int_{B_{\delta r}(y)} p(x, z, \tau) \, d\operatorname{vol}(z) \\ &\le e^{\delta d^2(x,y)/4(1+\delta)^2t} \left(\frac{\tau}{t}\right)^{n/2} \int_{B_{\delta r}(y)} p(x, z, \tau) \, d\operatorname{vol}(z). \end{aligned} $$
for any $x \in X$ and $r > 0$. Note that a manifold satisfying (1.2) has Euclidean volume growth with $\theta \ge 1 - \delta$. Our proof of Theorem 1.4 is a direct application of Theorem 1.2 together with the following heat kernel estimate.

**THEOREM 6.1.** — There exists a function $\gamma: [0, 1] \to [1, \infty)$ satisfying $\lim_{\theta \to 1^-} \gamma(\theta) = 1$ such that whenever $(M^n, g)$ is a complete Riemannian manifold with non-negative Ricci curvature and Euclidean volume growth, then the heat kernel $p$ of $(M^n, g)$ satisfies

$$ \frac{1}{(4\pi t)^{n/2}} e^{-d^2(x,y)/4t} \le p(x, y, t) \le \gamma(\theta) \frac{1}{(4\pi t)^{n/2}} e^{-d^2(x,y)/\gamma(\theta)4t} $$

for all $x, y \in M$ and $t > 0$, where $\theta$ is given by (6.1).

**REMARK 6.2.** — Our proof of the above heat kernel upper bound follows the arguments of P. Li, L-F. Tam and J. Wang [LTW97].

*Proof.* — The lower bound is the comparison theorem of J. Cheeger and S-T. Yau [CY81]: for any $t > 0$ and $x, y \in M$, we have

$$ (6.3) \qquad \mathbb{P}_n(x, y, t) \le p(x, y, t), $$

where $\mathbb{P}_n(x, y, t) = (4\pi t)^{-n/2}e^{-d^2(x,y)/4t}$. Consequently we only need to prove the upper bound.

Take $x, y \in X$ and $t > 0$. We shall need the following estimates from P. Li and S-T. Yau (see [LTW97, Form. (2.1)]): for any $r, \tau > 0$,

$$ \int_{B_r(x)} p(x, z, \tau) \, d\operatorname{vol}(z) \ge \int_{\mathbb{B}_r^n} \frac{1}{(4\pi\tau)^{n/2}} e^{-\|\xi\|^2/4\tau} \, d\xi $$

and

$$ (6.4) \qquad \int_{M \setminus B_r(x)} p(x, z, \tau) \, d\operatorname{vol}(z) \le \int_{\mathbb{R}^n \setminus \mathbb{B}_r^n} \frac{1}{(4\pi\tau)^{n/2}} e^{-\|\xi\|^2/4\tau} \, d\xi. $$

For $\delta > 0$ to be precisely chosen later, set $r := (1 + \delta)^{-1}d(x, y)$ and $\tau := (1 + \delta)t$. Note that $B_r(x) \cap B_{\delta r}(y) = \emptyset$. By the Harnack inequality of P. Li and S-T. Yau [LY86], we have

$$ p(x, y, t) \le \left(\frac{\tau}{t}\right)^{n/2} e^{d(z, y)^2/4(\tau-t)} p(x, z, \tau) $$

for every $z \in M$, so that averaging over the ball $B_{\delta r}(y)$ gives

$$ (6.5) \qquad \begin{aligned} p(x, y, t) &\le e^{\delta^2 r^2/4(\tau-t)} \left(\frac{\tau}{t}\right)^{n/2} \int_{B_{\delta r}(y)} p(x, z, \tau) \, d\operatorname{vol}(z) \\ &\le e^{\delta d^2(x, y)/4(1+\delta)^2 t} \left(\frac{\tau}{t}\right)^{n/2} \int_{B_{\delta r}(y)} p(x, z, \tau) \, d\operatorname{vol}(z). \end{aligned} $$
Now

$$
\begin{align*}
\int_{B_{\delta r}(y)} p(x, z, \tau) \, \mathrm{d} \mathrm{vol}(z) & \\
&= \int_{M \setminus B_r(x)} p(x, z, \tau) \, \mathrm{d} \mathrm{vol}(z) - \int_{M \setminus (B_r(x) \cup B_{\delta r}(y))} p(x, z, \tau) \, \mathrm{d} \mathrm{vol}(z) \\
&\le \int_{\mathbb{R}^n \setminus \mathbb{B}_r^n} \frac{1}{(4\pi\tau)^{n/2}} e^{-\|\xi\|^2/4\tau} \, \mathrm{d}\xi - \int_{M \setminus (B_r(x) \cup B_{\delta r}(y))} \mathbb{P}_n(x, z, \tau) \, \mathrm{d} \mathrm{vol}(z)
\end{align*}
$$

thanks to (6.4) and (6.3). Continuing,

$$
\begin{align*}
\int_{B_{\delta r}(y)} p(x, z, \tau) \, \mathrm{d} \mathrm{vol}(z) &\le \int_{\mathbb{R}^n \setminus \mathbb{B}_r^n} \frac{1}{(4\pi\tau)^{n/2}} e^{-\|\xi\|^2/4\tau} \, \mathrm{d}\xi \\
&\quad - \int_{M \setminus B_r(x)} \mathbb{P}_n(x, z, \tau) \, \mathrm{d} \mathrm{vol}(z) + \int_{B_{\delta r}(y)} \mathbb{P}_n(x, z, \tau) \, \mathrm{d} \mathrm{vol}(z) \\
&\le \int_{\mathbb{R}^n \setminus \mathbb{B}_r^n} \frac{1}{(4\pi\tau)^{n/2}} e^{-\|\xi\|^2/4\tau} \, \mathrm{d}\xi \\
&\quad - \int_{M \setminus B_r(x)} \mathbb{P}_n(x, z, \tau) \, \mathrm{d} \mathrm{vol}(z) + \mathrm{vol}(B_{\delta r}(y)) \frac{1}{(4\pi\tau)^{n/2}} e^{-(d(x,y)-\delta r)^2/4\tau}.
\end{align*}
$$

By Cavalieri's principle and (6.2), we have

$$
\begin{align*}
\int_{M \setminus B_r(x)} \mathbb{P}_n(x, z, \tau) \, \mathrm{d} \mathrm{vol}(z) &= \int_r^{+\infty} \frac{1}{(4\pi\tau)^{n/2}} e^{-s^2/4\tau} \frac{s}{2\tau} \mathrm{vol}(B_s(x)) \, \mathrm{d}s \\
&\geq \int_r^{+\infty} \frac{1}{(4\pi\tau)^{n/2}} e^{-s^2/4\tau} \frac{s}{2\tau} \theta \omega_n s^n \, \mathrm{d}s = \theta \int_{\mathbb{R}^n \setminus \mathbb{B}_r^n} \frac{1}{(4\pi\tau)^{n/2}} e^{-\|\xi\|^2/4\tau} \, \mathrm{d}\xi,
\end{align*}
$$

hence

$$
\begin{align*}
& (6.6) \quad \int_{B_{\delta r}(y)} p(x, z, \tau) \, \mathrm{d} \mathrm{vol}(z) \\
&\le (1-\theta) \int_{\mathbb{R}^n \setminus \mathbb{B}_r^n} \frac{1}{(4\pi\tau)^{n/2}} e^{-\|\xi\|^2/4\tau} \, \mathrm{d}\xi + \mathrm{vol}(B_{\delta r}(y)) \frac{1}{(4\pi\tau)^{n/2}} e^{-(d(x,y)-\delta r)^2/4\tau}.
\end{align*}
$$

As pointed out in [LTW97, Form. (2.6)], direct computations show that there exists a
constant $C = C(n) > 0$ such that

$$
\int_{\mathbb{R}^n \setminus \mathbb{B}_r^n} \frac{1}{(4\pi\tau)^{n/2}} e^{-\|\xi\|^2/4\tau} \mathrm{d}\xi \le C \left(1 + \left(r/\sqrt{4\pi\tau}\right)^n\right) e^{-r^2/4\tau}.
$$

This together with (6.6) and (6.5) yields to

$$
p(x, y, t) \le (1 - \theta) C \left(1 + \left(r/\sqrt{4\pi\tau}\right)^n\right) e^{-r^2/4\tau} e^{\delta d^2(x,y)/4(1+\delta)^2 t} \left(\frac{\tau}{t}\right)^{n/2} \frac{1}{\mathrm{vol}(B_{\delta r}(y))} \\
+ \frac{1}{(4\pi t)^{n/2}} e^{-(d(x,y)-\delta r)^2/4\tau} e^{\delta d^2(x,y)/4(1+\delta)^2 t}.
$$
Now

$$
\begin{align*}
\int_{B_{\delta r}(y)} p(x, z, \tau) \, \mathrm{d} \mathrm{vol}(z) \\
&= \int_{M \setminus B_r(x)} p(x, z, \tau) \, \mathrm{d} \mathrm{vol}(z) - \int_{M \setminus (B_r(x) \cup B_{\delta r}(y))} p(x, z, \tau) \, \mathrm{d} \mathrm{vol}(z) \\
&\le \int_{\mathbb{R}^n \setminus \mathbb{B}_r^n} \frac{1}{(4\pi\tau)^{n/2}} e^{-\|\xi\|^2/4\tau} \, \mathrm{d}\xi - \int_{M \setminus (B_r(x) \cup B_{\delta r}(y))} \mathbb{P}_n(x, z, \tau) \, \mathrm{d} \mathrm{vol}(z)
\end{align*}
$$

thanks to (6.4) and (6.3). Continuing,

$$
\begin{align*}
\int_{B_{\delta r}(y)} p(x, z, \tau) \, \mathrm{d} \mathrm{vol}(z) &\le \int_{\mathbb{R}^n \setminus \mathbb{B}_r^n} \frac{1}{(4\pi\tau)^{n/2}} e^{-\|\xi\|^2/4\tau} \, \mathrm{d}\xi \\
&\quad - \int_{M \setminus B_r(x)} \mathbb{P}_n(x, z, \tau) \, \mathrm{d} \mathrm{vol}(z) + \int_{B_{\delta r}(y)} \mathbb{P}_n(x, z, \tau) \, \mathrm{d} \mathrm{vol}(z) \\
&\le \int_{\mathbb{R}^n \setminus \mathbb{B}_r^n} \frac{1}{(4\pi\tau)^{n/2}} e^{-\|\xi\|^2/4\tau} \, \mathrm{d}\xi \\
&\quad - \int_{M \setminus B_r(x)} \mathbb{P}_n(x, z, \tau) \, \mathrm{d} \mathrm{vol}(z) + \mathrm{vol}(B_{\delta r}(y)) \frac{1}{(4\pi\tau)^{n/2}} e^{-(d(x,y)-\delta r)^2/4\tau}.
\end{align*}
$$

By Cavalieri's principle and (6.2), we have

$$
\begin{align*}
\int_{M \setminus B_r(x)} \mathbb{P}_n(x, z, \tau) \, \mathrm{d} \mathrm{vol}(z) &= \int_r^{+\infty} \frac{1}{(4\pi\tau)^{n/2}} e^{-s^2/4\tau} \frac{s}{2\tau} \mathrm{vol}(B_s(x)) \, \mathrm{d}s \\
&\geq \int_r^{+\infty} \frac{1}{(4\pi\tau)^{n/2}} e^{-s^2/4\tau} \frac{s}{2\tau} \theta \omega_n s^n \, \mathrm{d}s = \theta \int_{\mathbb{R}^n \setminus \mathbb{B}_r^n} \frac{1}{(4\pi\tau)^{n/2}} e^{-\|\xi\|^2/4\tau} \, \mathrm{d}\xi,
\end{align*}
$$

hence

$$
\begin{align*}
& (6.6) \quad \int_{B_{\delta r}(y)} p(x, z, \tau) \, \mathrm{d} \mathrm{vol}(z) \\
&\le (1-\theta) \int_{\mathbb{R}^n \setminus \mathbb{B}_r^n} \frac{1}{(4\pi\tau)^{n/2}} e^{-\|\xi\|^2/4\tau} \, \mathrm{d}\xi + \mathrm{vol}(B_{\delta r}(y)) \frac{1}{(4\pi\tau)^{n/2}} e^{-(d(x,y)-\delta r)^2/4\tau}.
\end{align*}
$$

As pointed out in [LTW97, Form. (2.6)], direct computations show that there exists a
constant $C = C(n) > 0$ such that

$$
\int_{\mathbb{R}^n \setminus \mathbb{B}_r^n} \frac{1}{(4\pi\tau)^{n/2}} e^{-\|\xi\|^2/4\tau} \mathrm{d}\xi \le C \left(1 + \left(r/\sqrt{4\pi\tau}\right)^n\right) e^{-r^2/4\tau}.
$$

This together with (6.6) and (6.5) yields to

$$
p(x, y, t) \le (1 - \theta) C \left(1 + \left(r/\sqrt{4\pi\tau}\right)^n\right) e^{-r^2/4\tau} e^{\delta d^2(x,y)/4(1+\delta)^2 t} \left(\frac{\tau}{t}\right)^{n/2} \frac{1}{\mathrm{vol}(B_{\delta r}(y))} \\
+ \frac{1}{(4\pi t)^{n/2}} e^{-(d(x,y)-\delta r)^2/4\tau} e^{\delta d^2(x,y)/4(1+\delta)^2 t}.
$$
It is easily checked that $-r^2/4\tau = -(d(x, y) - \delta r)^2/4\tau = -d^2(x, y)/4(1+\delta)^3t$ hence
$p(x, y, t)$

$$
\begin{align*}
&\le \left[ (1-\theta)C \left(1 + \left(r/\sqrt{4\pi\tau}\right)^n\right) (1+\delta)^{n/2} \underbrace{\frac{1}{\operatorname{vol}(B_{\delta r}(y))}}_{\le 1/(\theta\omega_n\delta^n r^n)} + \frac{1}{(4\pi t)^{n/2}} \right] \\
&\qquad \cdot e^{-(1-\delta-\delta^2)d^2(x,y)/4(1+\delta)^3t} \\
&\le \left[ (\theta^{-1}-1)C \left(1 + \left(r/\sqrt{4\pi\tau}\right)^n\right) \frac{(4\pi\tau)^{n/2}}{(4\pi t)^{n/2}} \frac{1}{\omega_n\delta^n r^n} + \frac{1}{(4\pi t)^{n/2}} \right] \\
&\qquad \cdot e^{-(1-\delta-\delta^2)d^2(x,y)/4(1+\delta)^3t} \\
&= \left[ (\theta^{-1}-1) \left(1 + \left(r/\sqrt{4\pi\tau}\right)^n\right) \frac{C}{\omega_n} \frac{(4\pi\tau)^{n/2}}{\delta^n r^n} + 1 \right] \frac{1}{(4\pi t)^{n/2}} \\
&\qquad \cdot e^{-(1-\delta-\delta^2)d^2(x,y)/4(1+\delta)^3t}.
\end{align*}
$$

Now we distinguish two cases. According to [LTW97, Form. (2.4)], if $d(x, y) \le \delta\sqrt{t}$, then

$$
p(x, y, t) \le \frac{1}{\theta} \frac{1}{(4\pi t)^{n/2}} e^{-d^2(x,y)/4t} e^{\delta^2/4}.
$$

If $d(x, y) \ge \delta\sqrt{t}$, then $r/\sqrt{\tau} \ge \delta/(1+\delta)^2$, thus

$$
\left(1 + \left(r/\sqrt{4\pi\tau}\right)^n\right) \frac{(4\pi\tau)^{n/2}}{\delta^n r^n} = \left(\frac{(4\pi\tau)^{n/2}}{r^n} + 1\right) \delta^{-n} \le (4\pi)^{n/2} \left(\frac{\delta+1}{\delta}\right)^{2n} + \left(\frac{1}{\delta}\right)^n.
$$

Therefore, if $\delta < 1/2$, we get

$$
\left(1 + \left(r/\sqrt{4\pi\tau}\right)^n\right) \frac{(4\pi\tau)^{n/2}}{\delta^n r^n} \le C'\delta^{-2n},
$$

where $C'$ depends only on $n$, which yields to

$$
p(x, y, t) \le \left[ (\theta^{-1} - 1) \Lambda \delta^{-2n} + 1 \right] \frac{1}{(4\pi t)^{n/2}} e^{-(1-\delta-\delta^2)d^2(x,y)/4(1+\delta)^3t},
$$

where $\Lambda := CC'/\omega_n$ depends only on $n$. Now we choose

$$
\delta = \delta(\theta) := \min \left\{ \frac{1}{2}, \left( (\theta^{-1} - 1) \Lambda \right)^{1/(2n+1)} \right\},
$$

so that when $((\theta^{-1} - 1) \Lambda)^{1/(2n+1)} < 1/2$ then

$$
\delta(\theta) = (\theta^{-1} - 1) \Lambda \delta(\theta)^{-2n},
$$

hence

$$
p(x, y, t) \le \frac{\delta(\theta) + 1}{(4\pi t)^{n/2}} e^{-(1-\delta(\theta)-\delta(\theta)^2)d^2(x,y)/4(1+\delta(\theta))^3t},
$$

and when $((\theta^{-1} - 1) \Lambda)^{1/(2n+1)} \ge 1/2$ - which corresponds to the case $\theta \le 1 - \epsilon_n$
with $\epsilon_n := (1 + 2^{2n+1}\Lambda)^{-1}$ depending only on $n$ - then $\delta(\theta) = 1/2$ implies

$$
p(x, y, t) \le \frac{(\theta^{-1} - 1) \Lambda 2^{2n} + 1}{(4\pi t)^{n/2}} e^{-(1-\delta(\theta)-\delta(\theta)^2)d^2(x,y)/4(1+\delta(\theta))^3t}.
$$
It is easily checked that $-r^2/4\tau = -(d(x, y) - \delta r)^2/4\tau = -d^2(x, y)/4(1+\delta)^3t$ hence
$p(x, y, t)$

$$
\begin{align*}
&\le \left[ (1-\theta)C \left(1 + \left(r/\sqrt{4\pi\tau}\right)^n\right) (1+\delta)^{n/2} \underbrace{\frac{1}{\mathrm{vol}(B_{\delta r}(y))}}_{\le 1/(\theta\omega_n\delta^n r^n)} + \frac{1}{(4\pi t)^{n/2}} \right] \\
&\qquad \cdot e^{-(1-\delta-\delta^2)d^2(x,y)/4(1+\delta)^3t} \\
&\le \left[ (\theta^{-1}-1)C \left(1 + \left(r/\sqrt{4\pi\tau}\right)^n\right) \frac{(4\pi\tau)^{n/2}}{(4\pi t)^{n/2}} \frac{1}{\omega_n\delta^n r^n} + \frac{1}{(4\pi t)^{n/2}} \right] \\
&\qquad \cdot e^{-(1-\delta-\delta^2)d^2(x,y)/4(1+\delta)^3t} \\
&= \left[ (\theta^{-1}-1) \left(1 + \left(r/\sqrt{4\pi\tau}\right)^n\right) \frac{C}{\omega_n} \frac{(4\pi\tau)^{n/2}}{\delta^n r^n} + 1 \right] \frac{1}{(4\pi t)^{n/2}} \\
&\qquad \cdot e^{-(1-\delta-\delta^2)d^2(x,y)/4(1+\delta)^3t}.
\end{align*}
$$

Now we distinguish two cases. According to [LTW97, Form. (2.4)], if $d(x, y) \le \delta\sqrt{t}$, then

$$
p(x, y, t) \le \frac{1}{\theta} \frac{1}{(4\pi t)^{n/2}} e^{-d^2(x,y)/4t} e^{\delta^2/4}.
$$

If $d(x, y) \ge \delta\sqrt{t}$, then $r/\sqrt{\tau} \ge \delta/(1+\delta)^2$, thus

$$
\left(1 + \left(r/\sqrt{4\pi\tau}\right)^n\right) \frac{(4\pi\tau)^{n/2}}{\delta^n r^n} = \left(\frac{(4\pi\tau)^{n/2}}{r^n} + 1\right) \delta^{-n} \le (4\pi)^{n/2} \left(\frac{\delta+1}{\delta}\right)^{2n} + \left(\frac{1}{\delta}\right)^n.
$$

Therefore, if $\delta < 1/2$, we get

$$
\left(1 + \left(r/\sqrt{4\pi\tau}\right)^n\right) \frac{(4\pi\tau)^{n/2}}{\delta^n r^n} \le C'\delta^{-2n},
$$

where $C'$ depends only on $n$, which yields to

$$
p(x, y, t) \le \left[ (\theta^{-1} - 1) \Lambda \delta^{-2n} + 1 \right] \frac{1}{(4\pi t)^{n/2}} e^{-(1-\delta-\delta^2)d^2(x,y)/4(1+\delta)^3t},
$$

where $\Lambda := CC'/\omega_n$ depends only on $n$. Now we choose

$$
\delta = \delta(\theta) := \min \left\{ \frac{1}{2}, \left( (\theta^{-1} - 1) \Lambda \right)^{1/(2n+1)} \right\},
$$

so that when $((\theta^{-1} - 1) \Lambda)^{1/(2n+1)} < 1/2$ then

$$
\delta(\theta) = (\theta^{-1} - 1) \Lambda \delta(\theta)^{-2n},
$$

hence

$$
p(x, y, t) \le \frac{\delta(\theta) + 1}{(4\pi t)^{n/2}} e^{-(1-\delta(\theta)-\delta(\theta)^2)d^2(x,y)/4(1+\delta(\theta))^3t},
$$

and when $((\theta^{-1} - 1) \Lambda)^{1/(2n+1)} \ge 1/2$ - which corresponds to the case $\theta \le 1 - \epsilon_n$
with $\epsilon_n := (1 + 2^{2n+1}\Lambda)^{-1}$ depending only on $n$ - then $\delta(\theta) = 1/2$ implies

$$
p(x, y, t) \le \frac{(\theta^{-1} - 1) \Lambda 2^{2n} + 1}{(4\pi t)^{n/2}} e^{-(1-\delta(\theta)-\delta(\theta)^2)d^2(x,y)/4(1+\delta(\theta))^3t}.
$$
Note that $\delta(\theta) \to 0$ when $\theta \to 1$. Therefore, setting

$$F(\theta) := (1 + \delta(\theta))^3 / (1 - \delta(\theta) - \delta(\theta)^2)$$

$$\text{and} \quad \gamma(\theta) := \begin{cases} \max(1 + \delta(\theta), F(\theta)) & \text{if } 1 - \epsilon_n < \theta < 1, \\ \max(2^{2n}\Lambda(\theta^{-1} - 1) + 1, F(\theta)) & \text{if } 0 < \theta \le 1 - \epsilon_n, \end{cases}$$

we get the result.

<figure><img src="image_8.png" /></figure>

For completeness, let us provide a short proof of Theorem 1.4.

*Proof.* — Take $\epsilon > 0$. By Theorem 1.2, there exists $\delta' = \delta'(n, \epsilon) > 0$ such that if $(M^n, g)$ is complete and satisfying Ric $\ge 0$ and (1.1) with $\delta$ replaced by $\delta'$, then any ball with radius $r$ in $M$ is $(\epsilon r)$-GH close from a ball with same radius in $\mathbb{R}^n$. But Theorem 6.1 implies that there exists $\delta = \delta(n, \delta') = \delta(n, \epsilon) > 0$ such that if $1 - \delta \le \theta$ holds, then $\gamma(\theta) - 1 \le \delta$ and thus (1.1) is true. The result follows.

□

<figure><img src="image_9.png" /></figure>

## 7. CASE OF A SPHERICAL HEAT KERNEL

In this section, for any Riemannian manifold $(M^n, g)$, we define the operator $L$ acting on $L^2(M)$ as the Friedrich extension of the operator $\tilde{L}$ defined by the formula:

$$-\int_M (\tilde{L}u)v = \int_M \langle \nabla u, \nabla v \rangle \quad \forall u, v \in C_c^\infty(M).$$

The spectral theorem implies that $L$ generates a semi-group $(e^{tL})_{t>0}$ which admits a smooth heat kernel.

The heat kernel of $(e^{tL})_{t>0}$ on the sphere $\mathbb{S}^n$ equipped with the canonical spherical metric $g_{\mathbb{S}^n}$ admits a well-known expression, namely

$$K_t^{(n)}(\mathbf{d}_{\mathbb{S}^n}(x, y))$$

for any $x, y \in \mathbb{S}^n$ and $t > 0$, where $\mathbf{d}_{\mathbb{S}^n}$ is the Riemannian distance canonically associated with $g_{\mathbb{S}^n}$ and

$$(7.1) \qquad K_t^{(n)}(r) := \sum_{i=0}^{+\infty} e^{\lambda_i t} C_i^{(n)}(r)$$

for any $r > 0$, with

$$\lambda_i := -i(i + n - 1) \quad \text{and} \quad C_i^{(n)}(\cdot) := (2i + n - 1)(n - 1)^{-1} \sigma_n^{-1} G_i^{(n-1)/2}(\cos(\cdot))$$

for any $i \in \mathbb{N}$. Here the functions $G_i^\alpha$ are the Gegenbauer polynomials (see e.g. [AH12]). For our purposes, it is worth mentioning that

$$C_0^{(n)}(r) = \frac{1}{\sigma_n} \quad \text{and} \quad C_1^{(n)}(r) = \frac{n+1}{\sigma_n} \cos(r)$$

for any $r > 0$. Moreover, the sum in (7.1) converges uniformly in $C([0, +\infty))$.

**THEOREM 7.1.** — Let $(X, \mathbf{d}, \mu)$ be a complete metric measure space equipped with a Dirichlet form $\mathcal{E}$ admitting a spherical heat kernel $p$, that is

$$(7.2) \qquad p(x, y, t) = K_t^{(n)}(\mathbf{d}(x, y))$$

for any $x, y \in X$ and $t > 0$. Then $(X, \mathbf{d})$ is isometric to $(\mathbb{S}^n, \mathbf{d}_{\mathbb{S}^n})$.
Note that $\delta(\theta) \to 0$ when $\theta \to 1$. Therefore, setting

$$F(\theta) := (1 + \delta(\theta))^3 / (1 - \delta(\theta) - \delta(\theta)^2)$$

$$\text{and} \quad \gamma(\theta) := \begin{cases} \max(1 + \delta(\theta), F(\theta)) & \text{if } 1 - \epsilon_n < \theta < 1, \\ \max(2^{2n}\Lambda(\theta^{-1} - 1) + 1, F(\theta)) & \text{if } 0 < \theta \le 1 - \epsilon_n, \end{cases}$$

we get the result.

<figure><img src="image_9.png" /></figure>

For completeness, let us provide a short proof of Theorem 1.4.

*Proof.* — Take $\epsilon > 0$. By Theorem 1.2, there exists $\delta' = \delta'(n, \epsilon) > 0$ such that if $(M^n, g)$ is complete and satisfying Ric $\ge 0$ and (1.1) with $\delta$ replaced by $\delta'$, then any ball with radius $r$ in $M$ is $(\epsilon r)$-GH close from a ball with same radius in $\mathbb{R}^n$. But Theorem 6.1 implies that there exists $\delta = \delta(n, \delta') = \delta(n, \epsilon) > 0$ such that if $1 - \delta \le \theta$ holds, then $\gamma(\theta) - 1 \le \delta$ and thus (1.1) is true. The result follows.

□

<figure><img src="image_10.png" /></figure>

## 7. CASE OF A SPHERICAL HEAT KERNEL

In this section, for any Riemannian manifold $(M^n, g)$, we define the operator $L$ acting on $L^2(M)$ as the Friedrich extension of the operator $\tilde{L}$ defined by the formula:

$$-\int_M (\tilde{L}u)v = \int_M \langle \nabla u, \nabla v \rangle \quad \forall u, v \in C_c^\infty(M).$$

The spectral theorem implies that $L$ generates a semi-group $(e^{tL})_{t>0}$ which admits a smooth heat kernel.

The heat kernel of $(e^{tL})_{t>0}$ on the sphere $\mathbb{S}^n$ equipped with the canonical spherical metric $g_{\mathbb{S}^n}$ admits a well-known expression, namely

$$K_t^{(n)}(\mathbf{d}_{\mathbb{S}^n}(x, y))$$

for any $x, y \in \mathbb{S}^n$ and $t > 0$, where $\mathbf{d}_{\mathbb{S}^n}$ is the Riemannian distance canonically associated with $g_{\mathbb{S}^n}$ and

$$(7.1) \qquad K_t^{(n)}(r) := \sum_{i=0}^{+\infty} e^{\lambda_i t} C_i^{(n)}(r)$$

for any $r > 0$, with

$$\lambda_i := -i(i + n - 1) \quad \text{and} \quad C_i^{(n)}(\cdot) := (2i + n - 1)(n - 1)^{-1} \sigma_n^{-1} G_i^{(n-1)/2}(\cos(\cdot))$$

for any $i \in \mathbb{N}$. Here the functions $G_i^\alpha$ are the Gegenbauer polynomials (see e.g. [AH12]). For our purposes, it is worth mentioning that

$$C_0^{(n)}(r) = \frac{1}{\sigma_n} \quad \text{and} \quad C_1^{(n)}(r) = \frac{n+1}{\sigma_n} \cos(r)$$

for any $r > 0$. Moreover, the sum in (7.1) converges uniformly in $C([0, +\infty))$.

**THEOREM 7.1.** — Let $(X, \mathbf{d}, \mu)$ be a complete metric measure space equipped with a Dirichlet form $\mathcal{E}$ admitting a spherical heat kernel $p$, that is

$$(7.2) \qquad p(x, y, t) = K_t^{(n)}(\mathbf{d}(x, y))$$

for any $x, y \in X$ and $t > 0$. Then $(X, \mathbf{d})$ is isometric to $(\mathbb{S}^n, \mathbf{d}_{\mathbb{S}^n})$.
Proof. — Let $L$ be the self-adjoint operator canonically associated with $\mathcal{E}$ and $(P_t)_{t>0}$ the associated semi-group. Assumption (7.2) implies that for any $t > 0$ and $f \in L^2(X, \mu)$,

$$ \int_{-\infty}^{+\infty} e^{t\lambda} \mathrm{d}(f, E_\lambda f) = \sum_{i=0}^{+\infty} e^{\lambda_i t} \iint_{X \times X} C_i(\mathbf{d}(x, y)) f(x) f(y) \mathrm{d}\mu(x) \mathrm{d}\mu(y) $$

holds, where $(f, E_\lambda f)$ is the projection-valued measure of $L$ associated with $f$, see e.g. [RS80, p. 262–263]. Uniqueness of the map $f \mapsto (f, E_\lambda f)$ implies that the spectrum of $L$ is given by $\lambda_0, \lambda_1, \lambda_2, \dots$ and that the projection operators $P_i : L^2(X, \mu) \to E_i := \text{Ker}(L - \lambda_i \text{Id})$, for any $i \in \mathbb{N}$, have a kernel $p_i$ such that for any $x, y \in X$,

$$ p_i(x, y) = C_i(\mathbf{d}(x, y)). $$

Since $P_i$ commutes with $L$ for any $i$, we have $P_i L g = \lambda_i P_i g$ for any $g \in \mathcal{D}(L)$, thus $\langle p_i(x, \cdot), L g \rangle_{L^2} = \lambda_i \langle p_i(x, \cdot), g \rangle_{L^2}$ for any $x \in X$. This implies $p_i(x, \cdot) \in \mathcal{D}(L)$ with

$$ L p_i(x, \cdot) = \lambda_i p_i(x, \cdot) $$

for any $x \in X$. In case $i = 0$, as $\lambda_0 = 0$ and $p_0(x, y) = C_0(\mathbf{d}(x, y)) = 1/\sigma_n$ for any $x, y \in X$, we get $L \mathbf{1} = 0$ thus $P_0 \mathbf{1} = \mathbf{1}$. This implies $\int_X p_0(x, y) \mathrm{d}\mu(y) = 1$ for any $x \in X$, hence

$$ (7.3) \qquad \mu(X) = \sigma_n. $$

In case $i = 1$, we have $\lambda_1 = -n$ and

$$ p_1(x, y) = C_1(\mathbf{d}(x, y)) = \frac{n+1}{\sigma_n} \cos(\mathbf{d}(x, y)) $$

for any $x, y \in X$, hence

$$ (7.4) \qquad L_x \cos(\mathbf{d}(x, y)) = -n \cos(\mathbf{d}(x, y)). $$

Let $\varphi_1, \dots, \varphi_l$ be continuous functions forming a $L^2(X, \mu)$-orthogonal basis of $E_1$. Observe that

$$ (7.5) \qquad P_1 f(x) = \int_X p_1(x, y) f(y) \mathrm{d}\mu(y) = \int_X \frac{n+1}{\sigma_n} \cos(\mathbf{d}(x, y)) f(y) \mathrm{d}\mu(y) $$

and

$$ P_1 f(x) = \sum_{j=1}^{\ell} \left( \int_X \varphi_i(y) f(y) \mathrm{d}\mu(y) \right) \varphi_i(x) = \int_X \left[ \sum_{j=1}^{\ell} \varphi(y) \varphi(x) \right] f(y) \mathrm{d}\mu(y) $$

holds for any $f \in L^2(X, \mu)$ and $x \in X$. This implies

$$ \sum_{i=1}^{\ell} \varphi(x)^2 = \frac{n+1}{\sigma_n} $$

for any $x \in X$, hence integration over $X$ and (7.3) provides

$$ \ell = n + 1. $$

Setting

$$ \mathcal{V} := \operatorname{Span}\{\cos(\mathbf{d}(x, \cdot)) : x \in X\}, $$
Proof. — Let $L$ be the self-adjoint operator canonically associated with $\mathcal{E}$ and $(P_t)_{t>0}$ the associated semi-group. Assumption (7.2) implies that for any $t > 0$ and $f \in L^2(X, \mu)$,

$$ \int_{-\infty}^{+\infty} e^{t\lambda} \mathrm{d}(f, E_\lambda f) = \sum_{i=0}^{+\infty} e^{\lambda_i t} \iint_{X \times X} C_i(\mathbf{d}(x, y)) f(x) f(y) \mathrm{d}\mu(x) \mathrm{d}\mu(y) $$

holds, where $(f, E_\lambda f)$ is the projection-valued measure of $L$ associated with $f$, see e.g. [RS80, p. 262–263]. Uniqueness of the map $f \mapsto (f, E_\lambda f)$ implies that the spectrum of $L$ is given by $\lambda_0, \lambda_1, \lambda_2, \dots$ and that the projection operators $P_i : L^2(X, \mu) \to E_i := \mathrm{Ker}(L - \lambda_i \mathrm{Id})$, for any $i \in \mathbb{N}$, have a kernel $p_i$ such that for any $x, y \in X$,

$$ p_i(x, y) = C_i(\mathbf{d}(x, y)). $$

Since $P_i$ commutes with $L$ for any $i$, we have $P_i L g = \lambda_i P_i g$ for any $g \in \mathcal{D}(L)$, thus $\langle p_i(x, \cdot), L g \rangle_{L^2} = \lambda_i \langle p_i(x, \cdot), g \rangle_{L^2}$ for any $x \in X$. This implies $p_i(x, \cdot) \in \mathcal{D}(L)$ with

$$ L p_i(x, \cdot) = \lambda_i p_i(x, \cdot) $$

for any $x \in X$. In case $i = 0$, as $\lambda_0 = 0$ and $p_0(x, y) = C_0(\mathbf{d}(x, y)) = 1/\sigma_n$ for any $x, y \in X$, we get $L \mathbf{1} = 0$ thus $P_0 \mathbf{1} = \mathbf{1}$. This implies $\int_X p_0(x, y) \mathrm{d}\mu(y) = 1$ for any $x \in X$, hence

$$ (7.3) \qquad \mu(X) = \sigma_n. $$

In case $i = 1$, we have $\lambda_1 = -n$ and

$$ p_1(x, y) = C_1(\mathbf{d}(x, y)) = \frac{n+1}{\sigma_n} \cos(\mathbf{d}(x, y)) $$

for any $x, y \in X$, hence

$$ (7.4) \qquad L_x \cos(\mathbf{d}(x, y)) = -n \cos(\mathbf{d}(x, y)). $$

Let $\varphi_1, \dots, \varphi_l$ be continuous functions forming a $L^2(X, \mu)$-orthogonal basis of $E_1$. Observe that

$$ (7.5) \qquad P_1 f(x) = \int_X p_1(x, y) f(y) \mathrm{d}\mu(y) = \int_X \frac{n+1}{\sigma_n} \cos(\mathbf{d}(x, y)) f(y) \mathrm{d}\mu(y) $$

and

$$ P_1 f(x) = \sum_{j=1}^{\ell} \left( \int_X \varphi_i(y) f(y) \mathrm{d}\mu(y) \right) \varphi_i(x) = \int_X \left[ \sum_{j=1}^{\ell} \varphi(y) \varphi(x) \right] f(y) \mathrm{d}\mu(y) $$

holds for any $f \in L^2(X, \mu)$ and $x \in X$. This implies

$$ \sum_{i=1}^{\ell} \varphi(x)^2 = \frac{n+1}{\sigma_n} $$

for any $x \in X$, hence integration over $X$ and (7.3) provides

$$ \ell = n + 1. $$

Setting

$$ \mathcal{V} := \mathrm{Span}\{\cos(\mathbf{d}(x, \cdot)) : x \in X\}, $$
we get $\mathcal{V} \subset E_1$ thanks to (7.4). Since $E_1$ is the image of $L^2(X, \mu)$ by $P_1$, the reverse inclusion follows from (7.5), hence

$$ \mathcal{V} = E_1. $$

Acting as in Section 4.1, we can show that there exist $x_1, \dots, x_{n+1} \in X$ such that $\{\delta_{x_1}, \dots, \delta_{x_{n+1}}\}$ is a basis of $\mathcal{V}^*$ whose associated basis $\{h_1, \dots, h_{n+1}\}$ of $\mathcal{V}$ permits to write

$$ (7.6) \qquad \cos(\mathbf{d}(x, y)) = \sum_{i,j=1}^{n+1} c_{ij} h_i(x) h_j(y) $$

for any $x, y \in X$, where $c_{ij} := \cos(\mathbf{d}(x_i, x_j))$ for any $i, j$. Let $\beta$ be the bilinear form defined by

$$ \beta(\xi, \xi') = \sum_{i,j=1}^{n+1} c_{ij} \xi_i \xi_j' $$

for any $\xi = (\xi_1, \dots, \xi_{n+1})$, $\xi' = (\xi'_1, \dots, \xi'_{n+1}) \in \mathbb{R}^{n+1}$ and $Q$ the associated quadratic form. Set

$$ H: \begin{array}{rcl} X & \to & \mathbb{R}^{n+1} \\ x & \mapsto & (h_1(x), \dots, h_{n+1}(x)). \end{array} $$

Then (7.6) writes as

$$ \cos(\mathbf{d}(x, y)) = \beta(H(x), H(y)). $$

Choosing $y = x$ implies $H(x) \in \Sigma := \{\xi \in \mathbb{R}^{n+1} : \beta(\xi, \xi) = 1\}$, so $H(X)$ is a subset of $\Sigma$. A direct computation provides:

$$ Q(H(x) - H(y)) = 4 \sin^2 \left( \frac{\mathbf{d}(x, y)}{2} \right) \quad \forall x, y \in X, $$

from which follows that $H$ is an injective map. Writing $\mathbb{R}^{n+1} = E_+ \oplus E_- \oplus \text{Ker } \beta$ where $E_+$, $E_-$ are subspaces of $\mathbb{R}^{n+1}$ where $\beta$ is positive definite and negative definite respectively, we can proceed as in Section 4.3, Step 1 (using the same notations) to get that $H_+$ is a bi-Lipschitz embedding of $(X, \mathbf{d})$ onto its image in $(E_+, q_+)$. Therefore, $\dim(E_+)$ is greater than the Hausdorff dimension of $X$.

**CLAIM 7.2.** — The Hausdorff dimension of $X$ is $n$.

*Proof.* — The short-time expansion of the heat kernel on Riemannian manifolds [MP49] and the Cheeger-Yau estimate [CY81] implies that for some $C > 0$ and $t_o > 0$,

$$ \frac{1}{(4\pi t)^{n/2}} e^{-r^2/4t} \le K_t^{(n)}(r) \le \frac{C}{(4\pi t)^{n/2}} e^{-r^2/5t} $$

holds for any $r \in (0, \pi)$ and $t \in (0, t_0)$. Therefore, proceeding as in the proof of Lemma 5.1, we get existence of a positive constant $C$ such that for any $x \in X$ and any $r \in (0, \sqrt{t_0})$,

$$ C^{-1}r^n \le \mu(B_r(x)) \le Cr^n. $$

Hence the claim is proved.

□
we get $\mathcal{V} \subset E_1$ thanks to (7.4). Since $E_1$ is the image of $L^2(X, \mu)$ by $P_1$, the reverse inclusion follows from (7.5), hence

$$ \mathcal{V} = E_1. $$

Acting as in Section 4.1, we can show that there exist $x_1, \dots, x_{n+1} \in X$ such that $\{\delta_{x_1}, \dots, \delta_{x_{n+1}}\}$ is a basis of $\mathcal{V}^*$ whose associated basis $\{h_1, \dots, h_{n+1}\}$ of $\mathcal{V}$ permits to write

$$ (7.6) \qquad \cos(\mathbf{d}(x, y)) = \sum_{i,j=1}^{n+1} c_{ij} h_i(x) h_j(y) $$

for any $x, y \in X$, where $c_{ij} := \cos(\mathbf{d}(x_i, x_j))$ for any $i, j$. Let $\beta$ be the bilinear form defined by

$$ \beta(\xi, \xi') = \sum_{i,j=1}^{n+1} c_{ij} \xi_i \xi_j' $$

for any $\xi = (\xi_1, \dots, \xi_{n+1}), \xi' = (\xi'_1, \dots, \xi'_{n+1}) \in \mathbb{R}^{n+1}$ and $Q$ the associated quadratic form. Set

$$ H: \begin{array}{rcl} X & \to & \mathbb{R}^{n+1} \\ x & \mapsto & (h_1(x), \dots, h_{n+1}(x)). \end{array} $$

Then (7.6) writes as

$$ \cos(\mathbf{d}(x, y)) = \beta(H(x), H(y)). $$

Choosing $y = x$ implies $H(x) \in \Sigma := \{\xi \in \mathbb{R}^{n+1} : \beta(\xi, \xi) = 1\}$, so $H(X)$ is a subset of $\Sigma$. A direct computation provides:

$$ Q(H(x) - H(y)) = 4 \sin^2 \left( \frac{\mathbf{d}(x, y)}{2} \right) \quad \forall x, y \in X, $$

from which follows that $H$ is an injective map. Writing $\mathbb{R}^{n+1} = E_+ \oplus E_- \oplus \text{Ker } \beta$ where $E_+, E_-$ are subspaces of $\mathbb{R}^{n+1}$ where $\beta$ is positive definite and negative definite respectively, we can proceed as in Section 4.3, Step 1 (using the same notations) to get that $H_+$ is a bi-Lipschitz embedding of $(X, \mathbf{d})$ onto its image in $(E_+, q_+)$. Therefore, $\dim(E_+)$ is greater than the Hausdorff dimension of $X$.

**CLAIM 7.2.** — The Hausdorff dimension of $X$ is $n$.

*Proof.* — The short-time expansion of the heat kernel on Riemannian manifolds [MP49] and the Cheeger-Yau estimate [CY81] implies that for some $C > 0$ and $t_o > 0$,

$$ \frac{1}{(4\pi t)^{n/2}} e^{-r^2/4t} \le K_t^{(n)}(r) \le \frac{C}{(4\pi t)^{n/2}} e^{-r^2/5t} $$

holds for any $r \in (0, \pi)$ and $t \in (0, t_0)$. Therefore, proceeding as in the proof of Lemma 5.1, we get existence of a positive constant $C$ such that for any $x \in X$ and any $r \in (0, \sqrt{t_0})$,

$$ C^{-1}r^n \le \mu(B_r(x)) \le Cr^n. $$

Hence the claim is proved.

□
Thus $n + 1 \ge \dim(E_+) > n$, so $\dim(E_+) = n + 1$. This shows that $\beta$ is positive definite, thus the distance $d_Q$ is well-defined. The associated length distance $\delta$ on $\Sigma$ is then given by:

$$d_Q(\xi, \xi') = 2 \sin(\delta(\xi, \xi')/2) \quad \forall \xi, \xi' \in \Sigma,$$

so that one eventually has:

$$\delta(H(x), H(y)) = d(x, y) \quad \forall x, y \in X,$$

i.e., $H$ is an isometric embedding of $(X, d)$ into $\Sigma$ equipped with $\delta$. Since

$$\lim_{t \to 0^+} -4t \log K_t^{(n)}(r) = r^2,$$

we get from Remark 3.5 that $(X, d)$ is a geodesic space. Then $H(X)$ is a closed totally geodesic subset of $\Sigma$, meaning that minimizing geodesics joining two points in $H(X)$ are all contained in $H(X)$. We assume that there exists $p \in \Sigma \setminus H(X)$ and set $r := \delta(p, H(X))$.

**CLAIM 7.3.** — We have $r < \pi/2$.

*Proof.* — Assume $r \ge \pi/2$. Then $H(X)$ is contained in the hemisphere

$$\{\sigma \in \Sigma : \beta(\sigma, p) \le 0\}.$$

Set $\lambda(\xi) = \beta(\xi, p)$ for any $\xi \in \mathbb{R}^{n+1}$. Then $\lambda \circ H : X \to \mathbb{R}$ is non-positive, and $\lambda \circ H(x) = 0$ if and only if $H(x) = 0$, which is impossible, so $\lambda \circ H$ is actually negative. But $\lambda \circ H$ is a linear combination of $h_1, \dots, h_n$ thus it is an element of $\mathcal{V}$. Since functions in $\mathcal{V} = E_1$ are $L^2$-orthogonal to constant functions, we reach a contradiction, namely $\int_X \lambda \circ H d\mu = 0$. $\square$

In fact, the same reasoning can be used to prove that $H(X)$ is contained in no hemisphere of $\Sigma$.

We are now in a position to conclude. Since $H(X)$ is closed there exists $q \in H(X)$ such that $\delta(p, q) = r$. The convexity of $H(X)$ implies that any minimizing geodesic of length $< \pi$ starting at $q$ and passing through the open ball $B_r(p)$ cannot meet $H(\Sigma)$. But the union of these minimizing geodesics is an open hemisphere, so $H(X)$ is contained in the complementary hemisphere, hence a contradiction. $\square$

## REFERENCES

[AGG19] T. ADAMOWICZ, M. GACZKOWSKI & P. GÓRKA – “Harmonic functions on metric measure spaces”, *Rev. Mat. Univ. Complut. Madrid* **32** (2019), no. 1, p. 141–186.

[ACT21] C. L. ALDANA, G. CARRON & S. TAPIE – “$A_\infty$ weights and compactness of conformal metrics under $L^{n/2}$ curvature bounds”, *Anal. PDE* **14** (2021), no. 7, p. 2163–2205.

[ACDM15] L. AMBROSIO, M. COLOMBO & S. DI MARINO – “Sobolev spaces in metric measure spaces: convexity and lower semicontinuity of slope”, in *Variational methods for evolving objects*, Adv. Stud. Pure Math., vol. 67, Mathematical Society of Japan, Tokyo, 2015, p. 1–58.

[AFP00] L. AMBROSIO, N. FUSCO & D. PALLARA – *Functions of bounded variation and free discontinuity problems*, Oxford Math. Monographs, The Clarendon Press, Oxford University Press, New York, 2000.

[AT04] L. AMBROSIO & P. TILLI – *Topics on analysis in metric spaces*, Oxford Lecture Series in Math. and its Applications, vol. 25, Oxford University Press, Oxford, 2004.
Thus $n + 1 \ge \dim(E_+) > n$, so $\dim(E_+) = n + 1$. This shows that $\beta$ is positive definite, thus the distance $d_Q$ is well-defined. The associated length distance $\delta$ on $\Sigma$ is then given by:

$$d_Q(\xi, \xi') = 2 \sin(\delta(\xi, \xi')/2) \quad \forall \xi, \xi' \in \Sigma,$$

so that one eventually has:

$$\delta(H(x), H(y)) = d(x, y) \quad \forall x, y \in X,$$

i.e., $H$ is an isometric embedding of $(X, d)$ into $\Sigma$ equipped with $\delta$. Since

$$\lim_{t \to 0^+} -4t \log K_t^{(n)}(r) = r^2,$$

we get from Remark 3.5 that $(X, d)$ is a geodesic space. Then $H(X)$ is a closed totally geodesic subset of $\Sigma$, meaning that minimizing geodesics joining two points in $H(X)$ are all contained in $H(X)$. We assume that there exists $p \in \Sigma \setminus H(X)$ and set $r := \delta(p, H(X))$.

**CLAIM 7.3.** — We have $r < \pi/2$.

*Proof.* — Assume $r \ge \pi/2$. Then $H(X)$ is contained in the hemisphere

$$\{\sigma \in \Sigma : \beta(\sigma, p) \le 0\}.$$

Set $\lambda(\xi) = \beta(\xi, p)$ for any $\xi \in \mathbb{R}^{n+1}$. Then $\lambda \circ H : X \to \mathbb{R}$ is non-positive, and $\lambda \circ H(x) = 0$ if and only if $H(x) = 0$, which is impossible, so $\lambda \circ H$ is actually negative. But $\lambda \circ H$ is a linear combination of $h_1, \dots, h_n$ thus it is an element of $\mathcal{V}$. Since functions in $\mathcal{V} = E_1$ are $L^2$-orthogonal to constant functions, we reach a contradiction, namely $\int_X \lambda \circ H d\mu = 0$. $\square$

In fact, the same reasoning can be used to prove that $H(X)$ is contained in no hemisphere of $\Sigma$.

We are now in a position to conclude. Since $H(X)$ is closed there exists $q \in H(X)$ such that $\delta(p, q) = r$. The convexity of $H(X)$ implies that any minimizing geodesic of length $< \pi$ starting at $q$ and passing through the open ball $B_r(p)$ cannot meet $H(\Sigma)$. But the union of these minimizing geodesics is an open hemisphere, so $H(X)$ is contained in the complementary hemisphere, hence a contradiction. $\square$

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[AGG19] T. ADAMOWICZ, M. GACZKOWSKI & P. GÓRKA – “Harmonic functions on metric measure spaces”, *Rev. Mat. Univ. Complut. Madrid* **32** (2019), no. 1, p. 141–186.

[ACT21] C. L. ALDANA, G. CARRON & S. TAPIE – “$A_\infty$ weights and compactness of conformal metrics under $L^{n/2}$ curvature bounds”, *Anal. PDE* **14** (2021), no. 7, p. 2163–2205.

[ACDM15] L. AMBROSIO, M. COLOMBO & S. DI MARINO – “Sobolev spaces in metric measure spaces: convexity and lower semicontinuity of slope”, in *Variational methods for evolving objects*, Adv. Stud. Pure Math., vol. 67, Mathematical Society of Japan, Tokyo, 2015, p. 1–58.

[AFP00] L. AMBROSIO, N. FUSCO & D. PALLARA – *Functions of bounded variation and free discontinuity problems*, Oxford Math. Monographs, The Clarendon Press, Oxford University Press, New York, 2000.

[AT04] L. AMBROSIO & P. TILLI – *Topics on analysis in metric spaces*, Oxford Lecture Series in Math. and its Applications, vol. 25, Oxford University Press, Oxford, 2004.
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[SC92] L. SALOFF-COSTE – “A note on Poincaré, Sobolev, and Harnack inequalities”, *Internat. Math. Res. Notices* (1992), no. 2, p. 27–38.
[AH12] K. ATKINSON & W. HAN – *Spherical harmonics and approximations on the unit sphere: an introduction*, Lect. Notes in Math., vol. 2044, Springer, Heidelberg, 2012.

[BD59] A. BEURLING & J. DENY – “Dirichlet spaces”, *Proc. Nat. Acad. Sci. U.S.A.* **45** (1959), p. 208–215.

[BBI01] D. BURAGO, Y. BURAGO & S. IVANOV – *A course in metric geometry*, Graduate Studies in Math., vol. 33, American Mathematical Society, Providence, RI, 2001.

[Car19] G. CARRON – “Geometric inequalities for manifolds with Ricci curvature in the Kato class”, *Ann. Inst. Fourier (Grenoble)* **69** (2019), no. 7, p. 3095–3167.

[Che99] J. CHEEGER – “Differentiability of Lipschitz functions on metric measure spaces”, *Geom. Funct. Anal.* **9** (1999), no. 3, p. 428–517.

[CC97] J. CHEEGER & T. H. COLDING – “On the structure of spaces with Ricci curvature bounded below. I”, *J. Differential Geom.* **46** (1997), no. 3, p. 406–480.

[CY81] J. CHEEGER & S. T. YAU – “A lower bound for the heat kernel”, *Comm. Pure Appl. Math.* **34** (1981), no. 4, p. 465–480.

[Coh07] A. M. COHEN – *Numerical methods for Laplace transform inversion*, Numerical Methods and Algorithms, vol. 5, Springer, New York, 2007.

[Col97] T. H. COLDING – “Ricci curvature and volume convergence”, *Ann. of Math. (2)* **145** (1997), no. 3, p. 477–501.

[CM97] T. H. COLDING & W. P. MINICOZZI, II – “Harmonic functions on manifolds”, *Ann. of Math. (2)* **146** (1997), no. 3, p. 725–747.

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[Gri94] – –, “Integral maximum principle and its applications”, *Proc. Roy. Soc. Edinburgh Sect. A* **124** (1994), no. 2, p. 353–362.

[Gri10] – –, “Heat kernels on metric measure spaces with regular volume growth”, in *Handbook of geometric analysis, No. 2*, Adv. Lect. Math. (ALM), vol. 13, Int. Press, Somerville, MA, 2010, p. 1–60.

[Gro07] M. GROMOV – *Metric structures for Riemannian and non-Riemannian spaces*, Modern Birkhäuser Classics, Birkhäuser Boston, Inc., Boston, MA, 2007.

[Hei01] J. HEINONEN – *Lectures on analysis on metric spaces*, Universitext, Springer-Verlag, New York, 2001.

[Hon15] S. HONDA – “Ricci curvature and $L^p$-convergence”, *J. reine angew. Math.* **705** (2015), p. 85–154.

[Hua11] B. HUA – “Harmonic functions of polynomial growth on singular spaces with nonnegative Ricci curvature”, *Proc. Amer. Math. Soc.* **139** (2011), no. 6, p. 2191–2205.

[HKX16] B. HUA, M. KELL & C. XIA – “Harmonic functions on metric measure spaces”, 2016, *arXiv*: 1308.3607v2.

[KZ12] P. KOSKELA & Y. ZHOU – “Geometry and analysis of Dirichlet forms”, *Adv. Math.* **231** (2012), no. 5, p. 2755–2801.

[LTF97] P. LI, L.-F. TAM & J. WANG – “Sharp bounds for the Green’s function and the heat kernel”, *Math. Res. Lett.* **4** (1997), no. 4, p. 589–602.

[LY86] P. LI & S.-T. YAU – “On the parabolic kernel of the Schrödinger operator”, *Acta Math.* **156** (1986), no. 3-4, p. 153–201.

[MP49] S. MINAKSHISUNDARAM & A. PLEIJEL – “Some properties of the eigenfunctions of the Laplace-operator on Riemannian manifolds”, *Canad. J. Math.* **1** (1949), p. 242–256.

[RS80] M. REED & B. SIMON – *Methods of modern mathematical physics I. Functional analysis*, second ed., Academic Press, Inc., New York, 1980.

[SC92] L. SALOFF-COSTE – “A note on Poincaré, Sobolev, and Harnack inequalities”, *Internat. Math. Res. Notices* (1992), no. 2, p. 27–38.
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Manuscript received 11th March 2021
accepted 25th October 2021

GILLES CARRON, Université de Nantes, Département de Mathématiques
2 rue de la Houssinière, BP 92208, 44322 Nantes cedex 03, France

E-mail: Gilles.Carron@univ-nantes.fr

Url: https://www.math.sciences.univ-nantes.fr/~carron/

DAVID TEWODROSE, CY Cergy Paris University, AGM

2 av. Adolphe Chauvin, 95302 Cergy-Pontoise cedex, France

E-mail: David.Tewodrose@univ-nantes.fr

Url: http://www.davidtewodrose.com/
[Stu94] K.-T. STURM – “Analysis on local Dirichlet spaces. I. Recurrence, conservativeness and $L^p$-Liouville properties”, *J. reine angew. Math.* **456** (1994), p. 173–196.

[Stu95] ———, “Analysis on local Dirichlet spaces. II. Upper Gaussian estimates for the fundamental solutions of parabolic equations”, *Osaka J. Math.* **32** (1995), no. 2, p. 275–312.

[Stu96] ———, “Analysis on local Dirichlet spaces. III. The parabolic Harnack inequality”, *J. Math. Pures Appl. (9)* **75** (1996), no. 3, p. 273–297.

Manuscript received 11th March 2021
accepted 25th October 2021

GILLES CARRON, Université de Nantes, Département de Mathématiques
2 rue de la Houssinière, BP 92208, 44322 Nantes cedex 03, France

E-mail: Gilles.Carron@univ-nantes.fr

Url: https://www.math.sciences.univ-nantes.fr/~carron/

DAVID TEWODROSE, CY Cergy Paris University, AGM

2 av. Adolphe Chauvin, 95302 Cergy-Pontoise cedex, France

E-mail: David.Tewodrose@univ-nantes.fr

Url: http://www.davidtewodrose.com/