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Post Undeleted by Sam Nead
typesetting -- in the previous I edit completely rewrote the answer. :)
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Sam Nead
Sam Nead
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Not an answer, but an intuition for an answer (suggesting that balls do notshould have exponentially small expansionCheeger constant).

Consider the three-regular infinite treeSuppose that $T_3$$n \geq 2$. All balls The volume of $B_R$$B^n(R)$, the ball of radius $R$ are isometric; they are all finite almost-binary trees. The number of vertices in $B_R$$n$-dimensional hyperbolic space, is

$1 + 3(1 + 2 + 2^2 + \cdots + 2^{R-1}) = 3 \cdot 2^R - 2$ basically some constant (depending on $n$) times $e^{(n - 1)R}$. In dimension one ($n = 1$) we instead have some constant times $R$.

The number of leaves isSuppose now that $3 \cdot 2^{R-1}$. This$M$ is close enough to onea closed connected hyperbolic $n$-half of the vertices that we will ignore the errormanifold. The one-neighbourhood of the leaves, taken Pick a basepoint $x$ in $B_R$, of course has$M$ and pick a graph $\Gamma$ in $M$ that nicely carries the same numberfundamental group of edges$M$. So the expansion Lift all of this to the set of leavesuniversal cover $M'$ to get a graph $\Gamma'$ which is onequasi-isometric to $n$-dimensional hyperbolic space. So (If we instead take the other definition$\Gamma'$ will be expansive, and instead take the numerator towill be the number of verticesroughly as expansive as balls in the one-neighbourhood of the leaves, then the expansion is $3/2$ambient hyperbolic space.)

The intuition is that this isSo we can move interchangably between large balls in the extreme casegraph $\Gamma'$ and large balls in hyperbolic space. If Let's assume that $n$ is correctat least three. Now consider $B^n(R)$, then the expansionball of radius $B_R$$R$ in $n$-dimensional hyperbolic space. This is close to some constantcut exactly in half by its "equatorial disk", and thusa copy of not exponentially small in$B^{n-1}(R)$. The volume of half of $R$$B^n(R)$ is roughly $e^{(n-1)R}$. The volume of the equatorial disk is roughly $e^{(n-2)R}$. So the ratio is $e^{-R}$, as suggested by the original question.

To make these examples work, we needed a notion of dimension. So it is not exactly clear how this generalises.

Not an answer, but an intuition for an answer (suggesting that balls do not have exponentially small expansion).

Consider the three-regular infinite tree $T_3$. All balls $B_R$ of radius $R$ are isometric; they are all finite almost-binary trees. The number of vertices in $B_R$ is

$1 + 3(1 + 2 + 2^2 + \cdots + 2^{R-1}) = 3 \cdot 2^R - 2$

The number of leaves is $3 \cdot 2^{R-1}$. This is close enough to one-half of the vertices that we will ignore the error. The one-neighbourhood of the leaves, taken in $B_R$, of course has the same number of edges. So the expansion of the set of leaves is one. (If we instead take the other definition, and instead take the numerator to be the number of vertices in the one-neighbourhood of the leaves, then the expansion is $3/2$.)

The intuition is that this is the extreme case. If that is correct, then the expansion of $B_R$ is close to some constant, and thus not exponentially small in $R$.

Not an answer, but an intuition for an answer (suggesting that balls should have exponentially small Cheeger constant).

Suppose that $n \geq 2$. The volume of $B^n(R)$, the ball of radius $R$ in $n$-dimensional hyperbolic space, is basically some constant (depending on $n$) times $e^{(n - 1)R}$. In dimension one ($n = 1$) we instead have some constant times $R$.

Suppose now that $M$ is a closed connected hyperbolic $n$-manifold. Pick a basepoint $x$ in $M$ and pick a graph $\Gamma$ in $M$ that nicely carries the fundamental group of $M$. Lift all of this to the universal cover $M'$ to get a graph $\Gamma'$ which is quasi-isometric to $n$-dimensional hyperbolic space. So $\Gamma'$ will be expansive, and will be roughly as expansive as balls in the ambient hyperbolic space.

So we can move interchangably between large balls in the graph $\Gamma'$ and large balls in hyperbolic space. Let's assume that $n$ is at least three. Now consider $B^n(R)$, the ball of radius $R$ in $n$-dimensional hyperbolic space. This is cut exactly in half by its "equatorial disk", a copy of $B^{n-1}(R)$. The volume of half of $B^n(R)$ is roughly $e^{(n-1)R}$. The volume of the equatorial disk is roughly $e^{(n-2)R}$. So the ratio is $e^{-R}$, as suggested by the original question.

To make these examples work, we needed a notion of dimension. So it is not exactly clear how this generalises.

Post Deleted by Sam Nead
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Sam Nead
Sam Nead
  • 28.2k
  • 5
  • 72
  • 131

Not an answer, but an intuition for an answer (suggesting that balls do not have exponentially small expansion).

Consider the three-regular infinite tree $T_3$. All balls $B_R$ of radius $R$ are isometric; they are all finite almost-binary trees. The number of vertices in $B_R$ is

$1 + 3(1 + 2 + 2^2 + \cdots + 2^{R-1}) = 3 \cdot 2^R - 2$

The number of leaves is $3 \cdot 2^{R-1}$. This is close enough to one-half of the vertices that we will ignore the error. The one-neighbourhood of the leaves, taken in $B_R$, of course has the same number of edges. So the expansion of the set of leaves is one. (If we instead take the other definition, and instead take the numerator to be the number of vertices in the one-neighbourhood of the leaves, then the expansion is $3/2$.)

The intuition is that this is the extreme case. If that is correct, then the expansion of $B_R$ is close to some constant, and thus not exponentially small in $R$.