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Theo
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The concentration of measure phenomena on the sphere:

If $A\subset\mathcal{S}^{n-1}$ is a measurable set on the sphere with $\lambda(A)=1/2$ and $A_\epsilon$ is an epsilon neighborhood of $A$ on $\mathcal{S}^{n-1}$, then

$$\lambda(A_\epsilon)\geq 1-\frac{2}{e^{n\epsilon^2/2}}$$

ItSo say you take $A$ to be a cap on the sphere and fix a small $\epsilon$. As the dimension of the sphere increases, eventually the $\epsilon$ enlargement of $A$ will have almost the entire area of the sphere! Playing with the upper and lower cap and the corresponding enlargements, one sees that area is concentrated around the equator.

Imagine you have a lawnmower and you cut the grass moving along the equator. What percentage of the sphere do you mow? Well, in 3 dimensions, not that much. But as you cut the grass on higher and higher dimensional spheres, moving centered along the equator, the surface area covered becomes almost 100% of the entire area of the sphere!

This result felt pretty counter-intuitive to me the first time I saw it.

The concentration of measure phenomena on the sphere:

If $A\subset\mathcal{S}^{n-1}$ is a measurable set on the sphere with $\lambda(A)=1/2$ and $A_\epsilon$ is an epsilon neighborhood of $A$ on $\mathcal{S}^{n-1}$, then

$$\lambda(A_\epsilon)\geq 1-\frac{2}{e^{n\epsilon^2/2}}$$

It felt pretty counter-intuitive to me the first time I saw it.

The concentration of measure phenomena on the sphere:

If $A\subset\mathcal{S}^{n-1}$ is a measurable set on the sphere with $\lambda(A)=1/2$ and $A_\epsilon$ is an epsilon neighborhood of $A$ on $\mathcal{S}^{n-1}$, then

$$\lambda(A_\epsilon)\geq 1-\frac{2}{e^{n\epsilon^2/2}}$$

So say you take $A$ to be a cap on the sphere and fix a small $\epsilon$. As the dimension of the sphere increases, eventually the $\epsilon$ enlargement of $A$ will have almost the entire area of the sphere! Playing with the upper and lower cap and the corresponding enlargements, one sees that area is concentrated around the equator.

Imagine you have a lawnmower and you cut the grass moving along the equator. What percentage of the sphere do you mow? Well, in 3 dimensions, not that much. But as you cut the grass on higher and higher dimensional spheres, moving centered along the equator, the surface area covered becomes almost 100% of the entire area of the sphere!

This result felt pretty counter-intuitive to me the first time I saw it.

Source Link
Theo
  • 2.4k
  • 15
  • 23

The concentration of measure phenomena on the sphere:

If $A\subset\mathcal{S}^{n-1}$ is a measurable set on the sphere with $\lambda(A)=1/2$ and $A_\epsilon$ is an epsilon neighborhood of $A$ on $\mathcal{S}^{n-1}$, then

$$\lambda(A_\epsilon)\geq 1-\frac{2}{e^{n\epsilon^2/2}}$$

It felt pretty counter-intuitive to me the first time I saw it.

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