Volume growth in euclidean and hyperbolic space

The radial density of the nonpositively curved model. [F3, given] Let $v\in S_{p}M$ and let ${\gamma }_v(t)={\exp }_p(tv)$ be the radial geodesic, a unit-speed geodesic along which the sectional curvature is constantly $k\le 0$. By [F3] the radial Jacobi tensor is $A(t)={\mathrm{sn}}_k(t)P_t$ and the radial volume Jacobian is $\det a_v(t)=J_p(t,v)={\mathrm{sn}}_k(t{)}^{n-1},0<t<c_p(v)\mathrm{.}$

There are no cut points in nonpositive curvature. [F5, F6, given] The manifold is complete, connected, boundaryless and simply connected with sectional curvature $K=k\le 0$, so by [F6] the exponential map ${\exp }_p:T_{p}M\to M$ is a diffeomorphism, in particular injective. Let $v\in S_{p}M$, $t>0$ and $q:={\gamma }_v(t)={\exp }_p(tv)$. By [F6] (Hopf–Rinow) there is $w\in T_{p}M$ with ${\exp }_p(w)=q$ and $\mathrm{\mid }w{\mathrm{\mid }}_{g_p}=d_g(p,q)$; injectivity forces $w=tv$, so $d_g(p,q)=\mathrm{\mid }w{\mathrm{\mid }}_{g_p}=t$. As $t>0$ was arbitrary, the minimizing-time set $A_p(v)$ of [F5] contains every positive time, so its supremum is $c_p(v)=+\mathrm{\infty }$.

The ball volume is the model volume. [F2, F4, F5, F7, step 1.1, step 1.2, given] Fix $p\in M$ and $r>0$. The ball $B(p,r)$ is open, hence Borel, so by [F7] the polar formula [F4] applies to $f={\mathbf{1}}_{B(p,r)}$: ${\mathrm{vol}}_g(B(p,r))={\int }_{S_{p}M}{\int }_0^{c_p(v)}{\mathbf{1}}_{B(p,r)}({\gamma }_v(t))\det a_v(t)dtd{\sigma }_p(v)\mathrm{.}$ For $0<t<c_p(v)$ the minimizing property in [F5] gives $d_g(p,{\gamma }_v(t))=t$, so ${\mathbf{1}}_{B(p,r)}({\gamma }_v(t))={\mathbf{1}}_{\{t<r\}}$, while step 1.1 gives $\det a_v(t)={\mathrm{sn}}_k(t{)}^{n-1}$. By step 1.2 the cut time is infinite, so the inner integral equals ${\int }_0^{\min \{c_p(v),r\}}{\mathrm{sn}}_k(t{)}^{n-1}dt={\int }_0^r{\mathrm{sn}}_k(t{)}^{n-1}dt,$ a finite number independent of $v$. Therefore the outer integral is the constant inner value times ${\sigma }_p(S_{p}M)={\omega }_{n-1}$ by [F4, F2], and ${\mathrm{vol}}_g(B(p,r))={\omega }_{n-1}{\int }_0^r{\mathrm{sn}}_k(t{)}^{n-1}dt={\int }_0^r{A}_k(t)dt=V_k(r)=V_k^{\star }(r),$ using $A_k={\omega }_{n-1}{\mathrm{sn}}_k^{n-1}$ and $V_k^{\star }=V_k$ for $k\le 0$ from [F2].

The flat case. [F1, F8, F11, step 2.1] Let $k=0$. Then ${\mathrm{sn}}_0(t)=t$ by [F1], so step 2.1 gives ${\mathrm{vol}}_g(B(p,r))=V_0(r)={\omega }_{n-1}{\int }_0^r{t}^{n-1}dt\mathrm{.}$ The power integral [F8] with $m=n-1\ge 1$ and $s=0$ evaluates this as ${\omega }_{n-1}{r}^n\mathrm{/}n$. This is the polynomial flat volume $V_0$; Euclidean $n$-space, of identically zero curvature and metrically complete, realizes the case $k=0$ [F11].

The hyperbolic case, closed form. [F1, step 2.1] Let $k=-a^2$. Then $\sqrt{-k}=a$, so ${\mathrm{sn}}_{-a^2}(t)=\sinh (at)\mathrm{/}a$ by [F1], and step 2.1 gives ${\mathrm{vol}}_g(B(p,r))=V_{-a^2}(r)={\omega }_{n-1}{\int }_0^r{\left(\frac{\sinh (at)}{a}\right)}^{n-1}dt\mathrm{.}$

Exponential growth in negative curvature. [F9, F10, step 3.2, given] Write $f(t):=(\sinh (at)\mathrm{/}a{)}^{n-1}\ge 0$ for $t\ge 0$. If $t\ge 1\mathrm{/}a$ then $at\ge 1$, and [F9] gives $\sinh (at)\ge e^{at}\mathrm{/}4$, hence $f(t)\ge \frac{e^{a(n-1)t}}{(4a{)}^{n-1}}\mathrm{.}$ Now let $r\ge 2\mathrm{/}a$ and $t\in [r\mathrm{/}2,r]$. Then $t\ge 1\mathrm{/}a$ and $t\ge r\mathrm{/}2$, so $e^{a(n-1)t}\ge e^{a(n-1)r\mathrm{/}2}$ because $n-1\ge 1$ and the exponential is strictly increasing by [F9]; consequently $f\ge e^{a(n-1)r\mathrm{/}2}\mathrm{/}(4a{)}^{n-1}$ on the whole interval $[r\mathrm{/}2,r]$. Since $f$ is nonnegative, the additivity and monotonicity of the integral [F10] give ${\int }_0^{r}f\ge {\int }_{r\mathrm{/}2}^{r}f\ge \frac{r}{2}\cdot \frac{e^{a(n-1)r\mathrm{/}2}}{(4a{)}^{n-1}},$ and multiplying by ${\omega }_{n-1}>0$ and combining with step 3.2 yields $V_{-a^2}(r)\ge \frac{{\omega }_{n-1}r}{2(4a{)}^{n-1}}{e}^{a(n-1)r\mathrm{/}2},r\ge \frac{2}{a}\mathrm{.}$ The exponent rate $a(n-1)\mathrm{/}2$ is strictly positive because $n\ge 2$, so the hyperbolic volume grows at least exponentially, in contrast with the polynomial flat volume of step 3.1; the constants $n$, $a$ and ${\omega }_{n-1}$ are fixed by the model.

Realizations, Bishop–Gromov consistency and edge cases. [F11, F12, step 2.1, step 3.1, step 4.1, given] Euclidean $n$-space has zero curvature and the upper-half-space metric of parameter $a$ has curvature $-a^2$, by [F11], so the two computed cases carry the flat and hyperbolic curvature normalizations (the general statement is proved for every $(M,g)$ satisfying the hypotheses). By [F12] a space form of curvature $k$ has $\mathrm{Ric}=(n-1)kg$, so both cases satisfy the hypotheses of the Bishop–Gromov comparison [F12], whose conclusion ${\mathrm{vol}}_g(B(p,r))\le V_k^{\star }(r)$ is attained with equality by step 2.1. Thus the comparison is sharp on the model, and step 4.1 shows that on the negative-curvature side the volume is allowed to grow exponentially, whereas the flat volume $V_0(r)={\omega }_{n-1}{r}^n\mathrm{/}n$ of step 3.1 is polynomial: a Ricci lower bound $\mathrm{Ric}\ge -(n-1)a^{2}g$ does not force polynomial volume growth. The degenerate cases are excluded or explained: $a>0$ and $r>0$ are fixed positive numbers; the threshold $r=2\mathrm{/}a$ is finite; $n\ge 2$ makes $n-1\ge 1$, which is exactly what the exponential rate needs (for $n=1$ the integrand is constant and the growth is linear); and $k=0$, $k<0$ exhaust the stated nonpositive curvature. The lower bound is stated only for $r\ge 2\mathrm{/}a$, while the closed formula of step 3.2 holds for every $r>0$; no upper bound and no exact asymptotic is claimed. No choice beyond the inherited [A1] is used: the radial direction, the minimizing geodesic and the model functions are fixed or explicit, and the cut-time, polar-integration, Cartan–Hadamard and Hopf–Rinow interfaces carry exactly ${\mathrm{A}\mathrm{C}}_{\omega }$.