{"version":1,"lectureId":"01M14TZ4DMWK0M8EYWYQWJ0SZ2","attempt":0,"publication":{"slug":"concentration-on-the-sphere","title":"Concentration of Measure on the Sphere","subject":"mathematics","summary":"On a high-dimensional sphere, a band you would barely notice on a globe holds essentially all of the area. This lecture shows that phenomenon before explaining it, then builds the explanation in full: the epsilon-extension of a set, Levy's spherical isoperimetric inequality argued by comparing one cap against two, and the concentration inequality with its universal constants, cashed in at measure one half. The spheres are then named a normal Levy family, Lipschitz functions are forced to their medians in three lines, and the payoff is Milman's proof of Dvoretzky's theorem: every n-dimensional normed space has an almost Euclidean subspace of dimension on the order of log n, with the intuition carried by the straight and diagonal slices of a cube.","metaDescription":"Why nearly all of a high-dimensional sphere lies within a whisker of any equator: Levy's isoperimetric inequality, concentration, and Dvoretzky's theorem.","transcript":"Take a sphere. Not the one in front of you, but its cousin in a thousand dimensions. Almost everything you know about the round one quietly fails up there, and it fails in a useful direction: by the end of this lecture, one fact about where a sphere keeps its area will hand us a theorem about every normed space there is. So start in three dimensions, where we can look. Here is the unit sphere, and there, drawn around its middle, is its equator. Let it turn once, so you can see it is a ball and not a disc. Now mark off a band around that equator: every point within zero point two radians of it, about twelve degrees of latitude either way. On this sphere the band is nothing special. It holds about twenty percent of the area, and you can see that: most of the surface is out in the two caps. Keep that picture, and raise the dimension. I cannot draw the sphere in dimension one thousand, but I can tell you what happens to this same band up there: it swallows essentially all of the area, and the two caps, which look like most of the sphere here, hold almost nothing at all. The reason is a picture we can draw. Slice the sphere by latitude, and ask how much area sits at each one. In dimension n the answer is proportional to sine of theta, raised to the power n minus one. For our sphere that exponent is just one, and the profile is this gentle mound. Now raise the exponent. At n equals eleven the mound sharpens. At n around fifty it is a spike, standing on the equator. The area has nowhere else to go: any latitude away from the equator has sine less than one, and a number below one, raised to the fiftieth power, is nothing. Here are the guarantees, for the very same band. On our sphere, dimension two, it holds about twenty percent. By dimension one hundred it already holds at least seventy two percent. By dimension one thousand, at least ninety nine point nine nine nine nine nine percent. And by ten thousand, everything except one part in ten to the eighty six. A band you would barely notice on a globe is, up there, the whole sphere. That is concentration of measure, and two questions come with it. Why the equator, of all places, and what guarantees those numbers? Keep one eye on where this is heading: the same phenomenon will, at the end, buy Dvoretzky's theorem. But first the why, and the why is an inequality about caps. A cap is the sphere's own disc. Pick a point x zero, and take everything within spherical distance R of it. There it is, poured around the pole. Let it turn, so you can see the cap lying on the surface, and there is its formula. Now grow it. Take any set A and fatten it by epsilon: everything within distance epsilon of A. That is the epsilon extension, written A sub epsilon, and for a cap it means the cap plus this collar. Growing a set costs area, and the bill is paid along the boundary: the longer the edge, the more the collar adds. So here is Levy's question. Among all sets of one fixed area, which shape, when you fatten it by epsilon, grows the least? Watch what happens to a bad candidate. Here is the same total area, split into two half-size caps, one at each pole. Nothing about the area changed. What changed is the edge: this set has two boundary circles where the single cap had one. Now fatten both sides by the same epsilon. The single cap grows one collar. The split set grows two, and two collars around two circles is more new area than one. Same area in, more area out: splitting was a bad trade. And that is the pattern in general. Wiggly sets, stretched sets, scattered sets: at a fixed area, every one of them carries at least as much boundary as the cap, and grows at least as fast. The cap is the minimizer. Say it precisely. If A and the cap C cover the same fraction of the sphere, then for every positive epsilon, the extension of A is at least as big as the extension of C. That is Levy's spherical isoperimetric inequality, and it is the engine under everything that follows. One more reading before we use it. The cap is solving the same problem the disc solves in the plane and the ball solves in space: least boundary for the area enclosed. On the sphere, least boundary becomes least growth. Now let us spend it. Here is the payoff move, and it is short. Take any set A covering exactly half the sphere: half is where a set and its complement balance, and it is the case that powers everything. Which cap has measure one half? The one of radius pi over two: a hemisphere. So the hemisphere is the shape to beat. Let it turn once, so that blue reads as half of a surface rather than as a lid laid across one. Fatten the hemisphere by epsilon and you get this: everything above the equator, plus a collar of angular width epsilon below it. By the isoperimetric inequality, whatever A actually was, its extension covers at least as much of the sphere as this does. So what does the grown hemisphere miss? Only this: the cap of radius pi over two minus epsilon, around the opposite pole. Call it B. The extension covers one minus the measure of B, and everything now rides on how small B is. And B is exactly the kind of region the spike killed. Every point of it sits at latitude at least epsilon away from the equator, where the sine to the n minus one profile has already collapsed. Integrate that profile and the measure of B comes out at most a constant times e to the minus c n epsilon squared. Put the chain together. Any set of measure one half, grown by epsilon, covers all of the sphere except an exponentially small remainder. The constants C and c are universal: they depend on nothing, not the dimension, not the set, not epsilon. That is the concentration inequality. How small is exponentially small? Let me plot the escaping mass, e to the minus n epsilon squared, against the dimension. Here it is for epsilon equal to zero point two, and above it, decaying slower, the thinner margin zero point one. Now ride the steep one. At n around fifty the escaping mass is already near a tenth. By one hundred and fifty it is a quarter of one percent, and off the right edge, in dimension one thousand, it is ten to the minus eighteen. And compare the two curves: halving epsilon quarters the exponent, so the thin margin decays four times slower. The square on epsilon is real. This behavior earned the spheres a title. A sequence of spaces satisfying exactly this bound, uniformly in n, is called a normal Levy family, and the spheres are the founding example. The name matters because the mechanism travels: any family with the bound inherits everything we do next. Concentration so far is about sets. The version that gets used is about functions. Take any real function f on the sphere that is Lipschitz with constant one, so it moves by at most the distance you moved, and let M be its median: half the sphere sits at or below M. Now grow that half. Every point of the extension is within epsilon of somewhere f was at most M, and a Lipschitz function cannot climb faster than distance. So on the whole extension, f is at most M plus epsilon, and the extension is nearly everything. Run the same argument from above, with the set where f is at least M, and the mirror bound comes out. Put the two together: apart from an exception of size at most twice the old one, every point of the sphere has f within epsilon of the median. There it is: essentially the entire sphere lands in this little window. Read that slowly, because it is strange. A Lipschitz function on a high-dimensional sphere is, for every practical purpose, a constant. You may design f as cleverly as you like; the geometry flattens it. Milman's insight was that this is not a curiosity but a tool. Here is the tool at work, first where we can see it. This is the cube, and everything that follows is one flat cut through its centre, taken two different ways. Let it turn first, so you know where its corners are. Cut it straight across, parallel to a face. The knife goes in flat, like this, and the face it leaves behind is a square. That much you would have guessed without me. Now tilt the knife, until it stands square to the long diagonal of the cube, the line joining one corner to the corner furthest away from it. Same cube, same centre, different angle. And the face it leaves is a hexagon. A regular one, with six equal sides, cut out of a body built entirely from squares. This is the step nobody believes until they watch it. So turn it again, and follow the rim. It crosses six of the cube's twelve edges, and it crosses every one of them at its midpoint. Six midpoints, six equal sides. Now lay the two sections flat and measure them. Draw the biggest circle that fits inside each one, and the smallest that wraps around it. For the square, the inner circle has radius one and the outer reaches the corners at root two, so the outer is forty one percent bigger than the inner. The hexagon is measurably rounder. Its inner circle is the larger of the two, while its outer circle is the same root two, because its corners are still edge midpoints of the same cube. The ratio drops to two over root three, about one point one five. Same cube, better direction, and the section has forgotten most of the cube's corners. In dimension n the same cube has central sections round to within any tolerance you name, and, more surprisingly, you do not have to hunt for the direction. The norm of the cube, like any norm, is a Lipschitz function on the sphere, so it concentrates at its median. On a random subspace of the right dimension it is nearly constant, and nearly constant is nearly round. That is Dvoretzky's theorem, in the sharp form Milman proved with exactly this concentration. For every tolerance epsilon there is a constant c of epsilon, and every n-dimensional normed space, no matter how jagged its unit ball, contains a subspace of dimension at least c of epsilon times log n on which the norm is within one plus epsilon of Euclidean. And the proof is this lecture run in reverse. Caps grow slowest, so sets of half measure grow to everything, so Lipschitz functions sit at their medians, so norms are constant on random subspaces, so round sections exist. One inequality about the oldest shape on the sphere, pushed four steps, reaches every norm in every dimension. Log n is small, and that is honest: the cube itself shows you cannot do better than logarithmic in general. But it is not zero, and that is the miracle. In high dimension, roundness is not the exception. It is what is left when there is nowhere else for the measure to go.","watch":{"version":1,"scenes":[{"title":"Where the Area Hides","start":0,"end":157.78227083333334,"objects":{"band":"a Surface [yellow] drawn in sph (function=<function>, opacity=0.6)","c11":"a FunctionPlot [green] labelled \"n = 11\" drawn in prof (function=<function>)","c2":"a FunctionPlot [blue] labelled \"n = 2\" drawn in prof (function=<function>)","c51":"a FunctionPlot [red] labelled \"n = 51\" drawn in prof (function=<function>)","caption":"a Tex [text] that says \"Within $0.2$ radians of the equator: about a fifth of $S^2$.\"","card":"a Title that says \"High-Dimensional Geometry — Concentration of Measure on the Sphere\"","dest":"a Tex [text] that says \"Where this is heading: Dvoretzky's theorem, that high-dimensional convex bodies have almost round slices.\"","eq_line":"a Line [gray] labelled \"upright(\"equator\")\" drawn in prof (start=(1.5707963267948966, 0.0), end=(1.5707963267948966, 1.04), dashed=True)","globe":"a Sphere [blue] drawn in sph (opacity=0.25)","head_band":"a Heading that says \"A Band Around the Equator\"","head_lives":"a Heading that says \"Where the Area Lives\"","head_numbers":"a Heading that says \"The Same Band, Dimension by Dimension\"","prof":"an Axes (x_range=(0.0, 3.141592653589793), y_range=(0.0, 1.08), include_ticks=False)","prop_math":"a Math [text] that says \"$upright(\"area near latitude\") thin theta prop sin^(n - 1) theta$\"","sph":"an Axes3D (x_range=(-1.6, 1.6), y_range=(-1.6, 1.6), z_range=(-1.6, 1.6))","tbl":"a Table [text] that says \"Dimension $n$ Share inside the band 2 about 20% 100 at least 72% 1,000 at least 99.99999% 10,000 $1 - 10^(-86)$\" (rows=(('Dimension $n$', 'Share inside the band'), ('2', 'about 20%')…, header=True)"},"beats":[{"start":0,"say":"Take a sphere. Not the one in front of you, but its cousin in a thousand dimensions. Almost everything you know about the round one quietly fails up there, and it fails in a useful direction: by the end of this lecture, one fact about where a sphere keeps its area will hand us a theorem about every normed space there is.","live":[],"does":[[0,"card is shown on the screen, written out."],[1.5,"card: enter:write-left-to-right."],[18.2745,"card is hidden from the screen — left the board."]]},{"start":19.4745,"say":"So start in three dimensions, where we can look. Here is the unit sphere, and there, drawn around its middle, is its equator. Let it turn once, so you can see it is a ball and not a disc.","live":null,"does":[[19.4745,"head_band is shown on the screen, written out."],[19.4745,"sph is shown on the screen, written out."],[23.956,"globe is shown on the screen, written out."],[28.925,"sph turns in its own slot."]]},{"start":32.845,"say":"Now mark off a band around that equator: every point within zero point two radians of it, about twelve degrees of latitude either way. On this sphere the band is nothing special. It holds about twenty percent of the area, and you can see that: most of the surface is out in the two caps.","live":["sph","head_band","globe"],"does":[[33.913,"band is shown on the screen, written out."],[44.827,"caption is shown on the screen, written out."]]},{"start":50.813500000000005,"say":"Keep that picture, and raise the dimension. I cannot draw the sphere in dimension one thousand, but I can tell you what happens to this same band up there: it swallows essentially all of the area, and the two caps, which look like most of the sphere here, hold almost nothing at all.","live":["sph","caption","head_band","globe","band"],"does":[[62.726000000000006,"sph turns in its own slot."],[66.5805,"caption is hidden from the screen — left the board."],[66.5805,"head_band is hidden from the screen — left the board."],[66.5805,"sph is hidden from the screen — left the board."],[66.5805,"globe is hidden from the screen — sph left the board."],[66.5805,"band is hidden from the screen — sph left the board."]]},{"start":67.7805,"say":"The reason is a picture we can draw. Slice the sphere by latitude, and ask how much area sits at each one. In dimension n the answer is proportional to sine of theta, raised to the power n minus one. For our sphere that exponent is just one, and the profile is this gentle mound.","live":[],"does":[[67.7805,"head_lives is shown on the screen, written out."],[67.7805,"prof is shown on the screen, written out."],[71.31,"eq_line is shown on the screen, written out."],[76.86,"prop_math is shown on the screen, written out."],[86.171,"c2 is shown on the screen, drawn."]]},{"start":87.53750000000001,"say":"Now raise the exponent. At n equals eleven the mound sharpens. At n around fifty it is a spike, standing on the equator. The area has nowhere else to go: any latitude away from the equator has sine less than one, and a number below one, raised to the fiftieth power, is nothing.","live":["prof","prop_math","head_lives","eq_line","c2"],"does":[[90.89299999999999,"c11 is shown on the screen, drawn."],[94.979,"c51 is shown on the screen, drawn."],[107.867,"head_lives is hidden from the screen — left the board."],[107.867,"prof is hidden from the screen — left the board."],[107.867,"eq_line is hidden from the screen — prof left the board."],[107.867,"c2 is hidden from the screen — prof left the board."],[107.867,"c11 is hidden from the screen — prof left the board."],[107.867,"c51 is hidden from the screen — prof left the board."],[107.867,"prop_math is hidden from the screen — left the board."]]},{"start":109.06700000000001,"say":"Here are the guarantees, for the very same band. On our sphere, dimension two, it holds about twenty percent. By dimension one hundred it already holds at least seventy two percent.","live":[],"does":[[109.06700000000001,"head_numbers is shown on the screen, written out."],[109.88000000000001,"tbl is shown on the screen, written out."],[114.45400000000001,"tbl is shown on the screen, written out."],[115.928,"tbl (the \"about 20%\" part) is emphasized."],[118.459,"tbl is shown on the screen, written out."],[120.18900000000001,"tbl (the \"about 20%\" part) is no longer emphasized."],[120.18900000000001,"tbl (the \"at least 72%\" part) is emphasized."]]},{"start":122.22850000000001,"say":"By dimension one thousand, at least ninety nine point nine nine nine nine nine percent. And by ten thousand, everything except one part in ten to the eighty six. A band you would barely notice on a globe is, up there, the whole sphere.","live":["head_numbers"],"does":[[123.52900000000001,"tbl is shown on the screen, written out."],[124.63200000000002,"tbl (the \"at least 72%\" part) is no longer emphasized."],[124.63200000000002,"tbl (the \"at least 99.99999%\" part) is emphasized."],[130.669,"tbl is shown on the screen, written out."],[131.42400000000004,"tbl (the \"$1 - 10^(-86)$\" part) is emphasized."],[131.42400000000004,"tbl (the \"at least 99.99999%\" part) is no longer emphasized."],[134.558,"tbl (the \"$1 - 10^(-86)$\" part) is no longer emphasized."]]},{"start":137.63150000000002,"say":"That is concentration of measure, and two questions come with it. Why the equator, of all places, and what guarantees those numbers? Keep one eye on where this is heading: the same phenomenon will, at the end, buy Dvoretzky's theorem. But first the why, and the why is an inequality about caps.","live":null,"does":[[148.04600000000002,"dest is shown on the screen, written out."],[156.74060416666669,"dest is hidden from the screen — left the board."],[156.74060416666669,"head_numbers is hidden from the screen — left the board."],[156.74060416666669,"tbl is hidden from the screen — left the board."]]}]},{"title":"Caps, Collars, and Levy's Inequality","start":157.78227083333334,"end":292.8135208333333,"objects":{"axesL":"an Axes3D (x_range=(-1.6, 1.6), y_range=(-1.6, 1.6), z_range=(-1.6, 1.6))","axesR":"an Axes3D (x_range=(-1.6, 1.6), y_range=(-1.6, 1.6), z_range=(-1.6, 1.6))","baseL":"a Sphere [gray] drawn in axesL (opacity=0.15)","baseR":"a Sphere [gray] drawn in axesR (opacity=0.15)","capL":"a Surface [blue] drawn in axesL (function=<function>)","capN":"a Surface [blue] drawn in axesR (function=<function>)","capS":"a Surface [blue] drawn in axesR (function=<function>)","cap_math":"a Math [text] that says \"$C = brace.l x in S^n : op(\"dist\")(x, x_0) lt.eq R brace.r$\"","colN":"a Surface [yellow] drawn in axesR (function=<function>, opacity=0.6)","colS":"a Surface [yellow] drawn in axesR (function=<function>, opacity=0.6)","collarL":"a Surface [yellow] drawn in axesL (function=<function>, opacity=0.6)","ext_def":"a Panel that says \"The $epsilon$-extension of $A$ is everything within distance $epsilon$ of it: the set, together with a collar of width $epsilon$ grown along its boundary.\"","ext_math":"a Math [text] that says \"$A_epsilon = brace.l x in S^n : op(\"dist\")(x, A) lt.eq epsilon brace.r$\"","head_cap":"a Heading that says \"Caps and Their Collars\"","head_split":"a Heading that says \"Same Area, More Boundary\"","head_thm":"a Heading that says \"Levy's Isoperimetric Inequality\"","ineq":"a Math [text] that says \"$sigma_n (A) = sigma_n (C) quad arrow.r.double quad sigma_n (A_epsilon) gt.eq sigma_n (C_epsilon)$\"","lblL":"a Tex [text] that says \"One cap\"","lblR":"a Tex [text] that says \"Two half-size caps, same total area\"","thm":"a Panel that says \"Among all $A subset S^n$ with a given measure $sigma_n (A)$, the spherical cap of that measure has the smallest $epsilon$-extension, for every $epsilon > 0$.\"","x0":"a Point [text] labelled \"x_0\" drawn in axesL (location=(0.0, 0.0, 1.0))"},"beats":[{"start":157.78227083333334,"say":"A cap is the sphere's own disc. Pick a point x zero, and take everything within spherical distance R of it. There it is, poured around the pole. Let it turn, so you can see the cap lying on the surface, and there is its formula.","live":[],"does":[[157.78227083333334,"head_cap is shown on the screen, written out."],[157.78227083333334,"axesL is shown on the screen, written out."],[157.78227083333334,"baseL is shown on the screen, written out."],[160.38327083333334,"x0 is shown on the screen, written out."],[162.43827083333335,"capL is shown on the screen, written out."],[168.30127083333335,"axesL turns in its own slot."],[171.63327083333334,"axesL moves to a new place on the board."],[171.63327083333334,"cap_math is shown on the screen, written out."]]},{"start":173.21977083333334,"say":"Now grow it. Take any set A and fatten it by epsilon: everything within distance epsilon of A. That is the epsilon extension, written A sub epsilon, and for a cap it means the cap plus this collar. Growing a set costs area, and the bill is paid along the boundary: the longer the edge, the more the collar adds.","live":["cap_math","axesL","head_cap","baseL","x0","capL"],"does":[[176.04127083333333,"ext_def is shown on the screen, written out."],[182.41427083333335,"ext_math is shown on the screen, written out."],[186.40827083333335,"collarL is shown on the screen, written out."]]},{"start":195.19327083333334,"say":"So here is Levy's question. Among all sets of one fixed area, which shape, when you fatten it by epsilon, grows the least? Watch what happens to a bad candidate.","live":["ext_def","cap_math","ext_math","axesL","head_cap","baseL","x0","capL","collarL"],"does":[[207.76727083333333,"axesL moves to a new place on the board."],[207.76727083333333,"cap_math is hidden from the screen — left the board."],[207.76727083333333,"ext_def is hidden from the screen — left the board."],[207.76727083333333,"ext_math is hidden from the screen — left the board."],[207.76727083333333,"head_cap is hidden from the screen — left the board."]]},{"start":208.96727083333334,"say":"Here is the same total area, split into two half-size caps, one at each pole. Nothing about the area changed. 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If A and the cap C cover the same fraction of the sphere, then for every positive epsilon, the extension of A is at least as big as the extension of C. That is Levy's spherical isoperimetric inequality, and it is the engine under everything that follows.","live":[],"does":[[256.15877083333334,"head_thm is shown on the screen, written out."],[256.76227083333333,"ineq is shown on the screen, written out."],[260.28027083333336,"ineq (the \"sigma_n (A) = sigma_n (C)\" part) is emphasized."],[264.00627083333336,"ineq (the \"sigma_n (A) = sigma_n (C)\" part) is no longer emphasized."],[264.00627083333336,"ineq (the \"sigma_n (A_epsilon) gt.eq sigma_n (C_epsilon)\" part) is emphasized."],[269.60227083333336,"thm is shown on the screen, written out."],[272.4472708333334,"ineq (the \"sigma_n (A_epsilon) gt.eq sigma_n (C_epsilon)\" part) is no longer emphasized."],[273.74727083333335,"A box is drawn around ineq."]]},{"start":275.50877083333336,"say":"One more reading before we use it. The cap is solving the same problem the disc solves in the plane and the ball solves in space: least boundary for the area enclosed. On the sphere, least boundary becomes least growth. Now let us spend it.","live":["ineq","thm","head_thm"],"does":[[291.7718541666667,"head_thm is hidden from the screen — left the board."],[291.7718541666667,"ineq is hidden from the screen — left the board."],[291.7718541666667,"thm is hidden from the screen — left the board."]]}]},{"title":"The Concentration Inequality","start":292.8135208333333,"end":456.4749791666667,"objects":{"axes3":"an Axes3D (x_range=(-1.6, 1.6), y_range=(-1.6, 1.6), z_range=(-1.6, 1.6))","b_note":"a Tex [text] that says \"$H$: a hemisphere. $B$: the cap of radius $pi slash 2 - epsilon$ that its extension still misses.\"","base3":"a Sphere [gray] drawn in axes3 (opacity=0.15)","claim_tex":"a Tex [text] that says \"Take any $A subset S^n$ with $sigma_n (A) = 1 slash 2$.\"","collar":"a Surface [yellow] drawn in axes3 (function=<function>, opacity=0.6)","decay":"an Axes (x_range=(0.0, 150.0), y_range=(0.0, 1.05), x_ticks_every=50.0)","dot":"a PlotPoint [yellow] labelled \"n = 5\" drawn in decay (target='e2', x=<VariableNumber nv = 150.0>)","e1":"a FunctionPlot [blue] labelled \"epsilon = 0.1\" drawn in decay (function=<function>)","e2":"a FunctionPlot [red] labelled \"epsilon = 0.2\" drawn in decay (function=<function>)","escape":"a Surface [red] drawn in axes3 (function=<function>, opacity=0.55)","family_tex":"a Tex [text] that says \"The spheres $(S^n)_(n gt.eq 1)$ are a normal Levy family.\"","head_family":"a Heading that says \"A Normal Levy Family\"","head_half":"a Heading that says \"Half the Sphere, Grown by Epsilon\"","hemi":"a Surface [blue] drawn in axes3 (function=<function>, opacity=0.45)","levy_def":"a Panel that says \"A sequence of metric probability spaces is a normal Levy family when there are constants $C, c > 0$ such that every set of measure $1 slash 2$ satisfies $sigma_n (A_epsilon) gt.eq 1 - C e^(- c n epsilon^2)$.\"","nv":"a VariableNumber (initial_value=5.0, format_spec='.0f')","res":"a Math [text] that says \"$sigma_n (A_epsilon) gt.eq 1 - C thin e^(- c n epsilon^2)$\"","steps":"a Derivation [text] that says \"$sigma_n (A_epsilon) &gt.eq sigma_n (H_epsilon) \\ &= 1 - sigma_n (B) \\ &gt.eq 1 - C thin e^(- c n epsilon^2)$\"","x1":"a Point [text] labelled \"- x_0\" drawn in axes3 (location=(1.2246467991473532e-16, 0.0, -1.0))"},"beats":[{"start":292.8135208333333,"say":"Here is the payoff move, and it is short. Take any set A covering exactly half the sphere: half is where a set and its complement balance, and it is the case that powers everything. Which cap has measure one half? The one of radius pi over two: a hemisphere. So the hemisphere is the shape to beat.","live":[],"does":[[292.8135208333333,"head_half is shown on the screen, written out."],[292.8135208333333,"axes3 is shown on the screen, written out."],[292.8135208333333,"base3 is shown on the screen, written out."],[298.1655208333333,"axes3 moves to a new place on the board."],[298.1655208333333,"claim_tex is shown on the screen, written out."],[309.82252083333333,"hemi is shown on the screen, written out."]]},{"start":314.0450208333333,"say":"Let it turn once, so that blue reads as half of a surface rather than as a lid laid across one.","live":["claim_tex","axes3","head_half","base3","hemi"],"does":[[314.7185208333333,"axes3 turns in its own slot."]]},{"start":320.9490208333333,"say":"Fatten the hemisphere by epsilon and you get this: everything above the equator, plus a collar of angular width epsilon below it. By the isoperimetric inequality, whatever A actually was, its extension covers at least as much of the sphere as this does.","live":null,"does":[[326.6145208333333,"collar is shown on the screen, written out."],[331.28152083333333,"steps is shown on the screen, written out."]]},{"start":338.1860208333333,"say":"So what does the grown hemisphere miss? Only this: the cap of radius pi over two minus epsilon, around the opposite pole. Call it B. The extension covers one minus the measure of B, and everything now rides on how small B is.","live":["claim_tex","axes3","head_half","base3","hemi","collar"],"does":[[339.92752083333335,"escape is shown on the screen, written out."],[345.37252083333334,"x1 is shown on the screen, written out."],[347.2415208333333,"steps is shown on the screen, written out."],[347.6415208333333,"b_note is shown on the screen, written out."]]},{"start":354.66802083333334,"say":"And B is exactly the kind of region the spike killed. Every point of it sits at latitude at least epsilon away from the equator, where the sine to the n minus one profile has already collapsed. Integrate that profile and the measure of B comes out at most a constant times e to the minus c n epsilon squared.","live":["claim_tex","b_note","axes3","head_half","base3","hemi","collar","escape","x1"],"does":[[366.47552083333335,"steps is shown on the screen, written out."]]},{"start":374.05302083333333,"say":"Put the chain together. Any set of measure one half, grown by epsilon, covers all of the sphere except an exponentially small remainder. The constants C and c are universal: they depend on nothing, not the dimension, not the set, not epsilon. That is the concentration inequality.","live":null,"does":[[374.7145208333333,"res is shown on the screen, written out."],[379.10352083333333,"axes3 turns in its own slot."],[384.8735208333333,"A box is drawn around res."],[393.5230208333333,"axes3 is hidden from the screen — left the board."],[393.5230208333333,"base3 is hidden from the screen — axes3 left the board."],[393.5230208333333,"hemi is hidden from the screen — axes3 left the board."],[393.5230208333333,"collar is hidden from the screen — axes3 left the board."],[393.5230208333333,"escape is hidden from the screen — axes3 left the board."],[393.5230208333333,"x1 is hidden from the screen — axes3 left the board."],[393.5230208333333,"b_note is hidden from the screen — left the board."],[393.5230208333333,"claim_tex is hidden from the screen — left the board."],[393.5230208333333,"head_half is hidden from the screen — left the board."],[393.5230208333333,"res is hidden from the screen — left the board."],[393.5230208333333,"steps is hidden from the screen — left the board."]]},{"start":394.7230208333333,"say":"How small is exponentially small? Let me plot the escaping mass, e to the minus n epsilon squared, against the dimension. Here it is for epsilon equal to zero point two, and above it, decaying slower, the thinner margin zero point one.","live":[],"does":[[394.7230208333333,"head_family is shown on the screen, written out."],[394.7230208333333,"decay is shown on the screen, written out."],[406.5185208333333,"e2 is shown on the screen, drawn."],[410.4075208333333,"e1 is shown on the screen, drawn."]]},{"start":411.6235208333333,"say":"Now ride the steep one. At n around fifty the escaping mass is already near a tenth. By one hundred and fifty it is a quarter of one percent, and off the right edge, in dimension one thousand, it is ten to the minus eighteen. And compare the two curves: halving epsilon quarters the exponent, so the thin margin decays four times slower. The square on epsilon is real.","live":["decay","head_family","e2","e1"],"does":[[412.2275208333333,"dot is shown on the screen, written out."],[414.6765208333333,"dot is redrawn as the numbers it depends on change."],[414.6765208333333,"nv ticks to 50.0."],[418.4155208333333,"dot is redrawn as the numbers it depends on change."],[418.4155208333333,"nv ticks to 150.0."]]},{"start":435.6640208333333,"say":"This behavior earned the spheres a title. A sequence of spaces satisfying exactly this bound, uniformly in n, is called a normal Levy family, and the spheres are the founding example. The name matters because the mechanism travels: any family with the bound inherits everything we do next.","live":["decay","head_family","e2","e1","dot"],"does":[[437.37052083333333,"decay moves to a new place on the board."],[437.37052083333333,"levy_def is shown on the screen, written out."],[447.0875208333333,"family_tex is shown on the screen, written out."],[455.4333125,"decay is hidden from the screen — left the board."],[455.4333125,"e2 is hidden from the screen — decay left the board."],[455.4333125,"e1 is hidden from the screen — decay left the board."],[455.4333125,"dot is hidden from the screen — decay left the board."],[455.4333125,"family_tex is hidden from the screen — left the board."],[455.4333125,"head_family is hidden from the screen — left the board."],[455.4333125,"levy_def is hidden from the screen — left the board."]]}]},{"title":"Functions, Slices, and Dvoretzky","start":456.4749791666667,"end":730.3927708333333,"objects":{"band_brace":"a Brace [yellow] labelled \"upright(\"almost all of\") thin S^n\" drawn in vline (x_start=-0.25, x_end=0.25)","claim4":"a Tex [text] that says \"Let $f : S^n arrow.r RR$ be $1$-Lipschitz, with median $M$.\"","closing":"a Text [text] that says \"Milman's proof: a norm is a Lipschitz function on the sphere, so it concentrates at its median; on a random subspace of the right dimension it is nearly constant, and a norm that is nearly constant on a sphere is nearly Euclidean.\"","cube":"a Solid [blue] drawn in solid (upper=<function>, lower=<function>, x_range=(-1.0, 1.0))","figH":"a Figure (x_range=(-1.7, 1.7), y_range=(-1.7, 1.7), aspect=(1, 1))","figS":"a Figure (x_range=(-1.7, 1.7), y_range=(-1.7, 1.7), aspect=(1, 1))","flat_knife":"a Plane [gray] drawn in solid (size=3.0, opacity=0.25)","head_cube":"a Heading that says \"Slices of a Cube\"","head_dvo":"a Heading that says \"Dvoretzky's Theorem\"","head_lip":"a Heading that says \"Lipschitz Functions Cannot Wander\"","head_round":"a Heading that says \"How Round Is the Slice?\"","hex3":"a Surface [red] drawn in solid (function=<function>, opacity=0.45, show_helper_lines=False)","hexP":"a Polygon [blue] drawn in figH (vertices=((1.4142135623730951, 0.0), (0.7071067811865477, 1.224744871391…, fill_opacity=0.25)","inH":"a Circle [green] drawn in figH (radius=1.224744871391589)","inS":"a Circle [green] drawn in figS","kmath":"a Math [text] that says \"$k gt.eq c(epsilon) thin log n$\"","lblH":"a Tex [text] that says \"Diagonal slice: out over in $= 2 slash sqrt(3) approx 1.15$\"","lblS":"a Tex [text] that says \"Straight slice: out over in $= sqrt(2) approx 1.41$\"","med_tick":"a Line [text] labelled \"M\" drawn in vline (start=(0.0, -0.15), end=(0.0, 0.15))","outH":"a Circle [gray] drawn in figH (radius=1.4142135623730951)","outS":"a Circle [gray] drawn in figS (radius=1.4142135623730951)","res4":"a Math [text] that says \"$sigma_n (abs(f - M) lt.eq epsilon) gt.eq 1 - 2 C thin e^(- c n epsilon^2)$\"","solid":"an Axes3D (x_range=(-1.7, 1.7), y_range=(-1.7, 1.7), z_range=(-1.7, 1.7))","sq":"a Polygon [blue] drawn in figS (vertices=((-1, -1), (1, -1), (1, 1), (-1, 1)), fill_opacity=0.25)","square3":"a Surface [green] drawn in solid (function=<function>, opacity=0.45, show_helper_lines=False)","steps4":"a Derivation [text] that says \"$sigma_n (f lt.eq M) &gt.eq 1 slash 2 \\ sigma_n (f lt.eq M + epsilon) &gt.eq 1 - C e^(- c n epsilon^2) \\ sigma_n (f gt.eq M - epsilon) &gt.eq 1 - C e^(- c n epsilon^2)$\"","thm4":"a Panel that says \"For every $epsilon > 0$ there is $c(epsilon) > 0$: every $n$-dimensional normed space contains a subspace of dimension $k gt.eq c(epsilon) log n$ on which the norm is within a factor $1 + epsilon$ of Euclidean.\"","tilt_knife":"a Plane [gray] drawn in solid (normal=(1.0, 1.0, 1.0), size=3.0, opacity=0.25)","vline":"a NumberLine labelled \"f\" (x_range=(-1.0, 1.0), include_ticks=False)"},"beats":[{"start":456.4749791666667,"say":"Concentration so far is about sets. The version that gets used is about functions. Take any real function f on the sphere that is Lipschitz with constant one, so it moves by at most the distance you moved, and let M be its median: half the sphere sits at or below M.","live":[],"does":[[456.4749791666667,"head_lip is shown on the screen, written out."],[456.4749791666667,"vline is shown on the screen, written out."],[464.5439791666667,"vline moves to a new place on the board."],[464.5439791666667,"claim4 is shown on the screen, written out."],[470.23297916666667,"med_tick is shown on the screen, written out."],[473.0659791666667,"steps4 is shown on the screen, written out."]]},{"start":474.8154791666667,"say":"Now grow that half. Every point of the extension is within epsilon of somewhere f was at most M, and a Lipschitz function cannot climb faster than distance. So on the whole extension, f is at most M plus epsilon, and the extension is nearly everything.","live":["claim4","vline","head_lip","med_tick"],"does":[[486.23997916666667,"steps4 is shown on the screen, written out."]]},{"start":492.9694791666667,"say":"Run the same argument from above, with the set where f is at least M, and the mirror bound comes out. Put the two together: apart from an exception of size at most twice the old one, every point of the sphere has f within epsilon of the median. There it is: essentially the entire sphere lands in this little window.","live":null,"does":[[497.7999791666667,"steps4 is shown on the screen, written out."],[500.2719791666667,"res4 is shown on the screen, written out."],[501.79297916666667,"A box is drawn around res4."],[512.1619791666667,"band_brace is shown on the screen, written out."]]},{"start":513.6439791666667,"say":"Read that slowly, because it is strange. A Lipschitz function on a high-dimensional sphere is, for every practical purpose, a constant. You may design f as cleverly as you like; the geometry flattens it. Milman's insight was that this is not a curiosity but a tool.","live":["claim4","res4","vline","head_lip","med_tick","band_brace"],"does":[[532.0804791666667,"claim4 is hidden from the screen — left the board."],[532.0804791666667,"head_lip is hidden from the screen — left the board."],[532.0804791666667,"res4 is hidden from the screen — left the board."],[532.0804791666667,"steps4 is hidden from the screen — left the board."],[532.0804791666667,"vline is hidden from the screen — left the board."],[532.0804791666667,"med_tick is hidden from the screen — vline left the board."],[532.0804791666667,"band_brace is hidden from the screen — vline left the board."]]},{"start":533.2804791666667,"say":"Here is the tool at work, first where we can see it. This is the cube, and everything that follows is one flat cut through its centre, taken two different ways. Let it turn first, so you know where its corners are.","live":[],"does":[[533.2804791666667,"head_cube is shown on the screen, written out."],[533.2804791666667,"solid is shown on the screen, written out."],[537.8079791666667,"cube is shown on the screen, written out."],[543.9379791666667,"solid turns in its own slot."]]},{"start":547.2089791666667,"say":"Cut it straight across, parallel to a face. The knife goes in flat, like this, and the face it leaves behind is a square. That much you would have guessed without me.","live":["solid","head_cube","cube"],"does":[[551.1089791666667,"flat_knife is shown on the screen, written out."],[554.9059791666667,"square3 is shown on the screen, written out."],[558.4239791666666,"flat_knife is hidden from the screen."]]},{"start":559.0239791666667,"say":"Now tilt the knife, until it stands square to the long diagonal of the cube, the line joining one corner to the corner furthest away from it. Same cube, same centre, different angle.","live":["solid","head_cube","cube","square3"],"does":[[559.0239791666667,"square3 is hidden from the screen."],[562.2629791666667,"tilt_knife is shown on the screen, written out."]]},{"start":572.2784791666667,"say":"And the face it leaves is a hexagon. A regular one, with six equal sides, cut out of a body built entirely from squares. This is the step nobody believes until they watch it.","live":["solid","head_cube","cube","tilt_knife"],"does":[[572.2784791666667,"tilt_knife is hidden from the screen."],[573.9269791666667,"hex3 is shown on the screen, written out."]]},{"start":584.5119791666666,"say":"So turn it again, and follow the rim. It crosses six of the cube's twelve edges, and it crosses every one of them at its midpoint. Six midpoints, six equal sides.","live":["solid","head_cube","cube","hex3"],"does":[[585.1159791666666,"solid turns in its own slot."],[595.9824791666666,"head_cube is hidden from the screen — left the board."],[595.9824791666666,"solid is hidden from the screen — left the board."],[595.9824791666666,"cube is hidden from the screen — solid left the board."],[595.9824791666666,"hex3 is hidden from the screen — solid left the board."]]},{"start":597.1824791666667,"say":"Now lay the two sections flat and measure them. Draw the biggest circle that fits inside each one, and the smallest that wraps around it. For the square, the inner circle has radius one and the outer reaches the corners at root two, so the outer is forty one percent bigger than the inner.","live":[],"does":[[597.1824791666667,"head_round is shown on the screen, written out."],[598.8079791666667,"figS is shown on the screen, written out."],[607.2139791666666,"sq is shown on the screen, written out."],[608.1429791666667,"inS is shown on the screen, written out."],[610.2089791666667,"outS is shown on the screen, written out."],[613.8429791666667,"figS moves to a new place on the board."],[613.8429791666667,"lblS is shown on the screen, written out."]]},{"start":616.1614791666667,"say":"The hexagon is measurably rounder. Its inner circle is the larger of the two, while its outer circle is the same root two, because its corners are still edge midpoints of the same cube. The ratio drops to two over root three, about one point one five. Same cube, better direction, and the section has forgotten most of the cube's corners.","live":["lblS","figS","head_round","sq","inS","outS"],"does":[[616.1614791666667,"figH is shown on the screen, written out."],[616.6839791666666,"hexP is shown on the screen, written out."],[619.4699791666667,"inH is shown on the screen, written out."],[622.1289791666667,"outH is shown on the screen, written out."],[631.6609791666667,"lblH is shown on the screen, written out."]]},{"start":638.3209791666667,"say":"In dimension n the same cube has central sections round to within any tolerance you name, and, more surprisingly, you do not have to hunt for the direction. The norm of the cube, like any norm, is a Lipschitz function on the sphere, so it concentrates at its median. On a random subspace of the right dimension it is nearly constant, and nearly constant is nearly round.","live":["lblS","figS","lblH","figH","head_round","sq","inS","outS","hexP","inH","outH"],"does":[[662.7719791666667,"figH is hidden from the screen — left the board."],[662.7719791666667,"hexP is hidden from the screen — figH left the board."],[662.7719791666667,"inH is hidden from the screen — figH left the board."],[662.7719791666667,"outH is hidden from the screen — figH left the board."],[662.7719791666667,"figS is hidden from the screen — left the board."],[662.7719791666667,"sq is hidden from the screen — figS left the board."],[662.7719791666667,"inS is hidden from the screen — figS left the board."],[662.7719791666667,"outS is hidden from the screen — figS left the board."],[662.7719791666667,"head_round is hidden from the screen — left the board."],[662.7719791666667,"lblH is hidden from the screen — left the board."],[662.7719791666667,"lblS is hidden from the screen — left the board."]]},{"start":663.9719791666666,"say":"That is Dvoretzky's theorem, in the sharp form Milman proved with exactly this concentration. For every tolerance epsilon there is a constant c of epsilon, and every n-dimensional normed space, no matter how jagged its unit ball, contains a subspace of dimension at least c of epsilon times log n on which the norm is within one plus epsilon of Euclidean.","live":[],"does":[[663.9719791666666,"head_dvo is shown on the screen, written out."],[665.5159791666667,"thm4 is shown on the screen, written out."],[680.4819791666666,"kmath is shown on the screen, written out."],[685.9849791666667,"A box is drawn around kmath."]]},{"start":687.5369791666667,"say":"And the proof is this lecture run in reverse. Caps grow slowest, so sets of half measure grow to everything, so Lipschitz functions sit at their medians, so norms are constant on random subspaces, so round sections exist. One inequality about the oldest shape on the sphere, pushed four steps, reaches every norm in every dimension.","live":["thm4","kmath","head_dvo"],"does":[[690.1259791666666,"closing is shown on the screen, written out."]]},{"start":710.1254791666668,"say":"Log n is small, and that is honest: the cube itself shows you cannot do better than logarithmic in general. But it is not zero, and that is the miracle. In high dimension, roundness is not the exception. It is what is left when there is nowhere else for the measure to go.","live":["thm4","kmath","closing","head_dvo"],"does":[[729.3511041666667,"closing is hidden from the screen — left the board."],[729.3511041666667,"head_dvo is hidden from the screen — left the board."],[729.3511041666667,"kmath is hidden from the screen — left the board."],[729.3511041666667,"thm4 is hidden from the screen — left the board."]]}]}]},"durationSeconds":730,"chapters":[{"title":"Where the Area Hides","startSeconds":0,"narration":"Take a sphere. Not the one in front of you, but its cousin in a thousand dimensions. Almost everything you know about the round one quietly fails up there, and it fails in a useful direction: by the end of this lecture, one fact about where a sphere keeps its area will hand us a theorem about every normed space there is. So start in three dimensions, where we can look. Here is the unit sphere, and there, drawn around its middle, is its equator. Let it turn once, so you can see it is a ball and not a disc. Now mark off a band around that equator: every point within zero point two radians of it, about twelve degrees of latitude either way. On this sphere the band is nothing special. It holds about twenty percent of the area, and you can see that: most of the surface is out in the two caps. Keep that picture, and raise the dimension. I cannot draw the sphere in dimension one thousand, but I can tell you what happens to this same band up there: it swallows essentially all of the area, and the two caps, which look like most of the sphere here, hold almost nothing at all. The reason is a picture we can draw. Slice the sphere by latitude, and ask how much area sits at each one. In dimension n the answer is proportional to sine of theta, raised to the power n minus one. For our sphere that exponent is just one, and the profile is this gentle mound. Now raise the exponent. At n equals eleven the mound sharpens. At n around fifty it is a spike, standing on the equator. The area has nowhere else to go: any latitude away from the equator has sine less than one, and a number below one, raised to the fiftieth power, is nothing. Here are the guarantees, for the very same band. On our sphere, dimension two, it holds about twenty percent. By dimension one hundred it already holds at least seventy two percent. By dimension one thousand, at least ninety nine point nine nine nine nine nine percent. And by ten thousand, everything except one part in ten to the eighty six. A band you would barely notice on a globe is, up there, the whole sphere. That is concentration of measure, and two questions come with it. Why the equator, of all places, and what guarantees those numbers? Keep one eye on where this is heading: the same phenomenon will, at the end, buy Dvoretzky's theorem. But first the why, and the why is an inequality about caps."},{"title":"Caps, Collars, and Levy's Inequality","startSeconds":157.78227083333334,"narration":"A cap is the sphere's own disc. Pick a point x zero, and take everything within spherical distance R of it. There it is, poured around the pole. Let it turn, so you can see the cap lying on the surface, and there is its formula. Now grow it. Take any set A and fatten it by epsilon: everything within distance epsilon of A. That is the epsilon extension, written A sub epsilon, and for a cap it means the cap plus this collar. Growing a set costs area, and the bill is paid along the boundary: the longer the edge, the more the collar adds. So here is Levy's question. Among all sets of one fixed area, which shape, when you fatten it by epsilon, grows the least? Watch what happens to a bad candidate. Here is the same total area, split into two half-size caps, one at each pole. Nothing about the area changed. What changed is the edge: this set has two boundary circles where the single cap had one. Now fatten both sides by the same epsilon. The single cap grows one collar. The split set grows two, and two collars around two circles is more new area than one. Same area in, more area out: splitting was a bad trade. And that is the pattern in general. Wiggly sets, stretched sets, scattered sets: at a fixed area, every one of them carries at least as much boundary as the cap, and grows at least as fast. The cap is the minimizer. Say it precisely. If A and the cap C cover the same fraction of the sphere, then for every positive epsilon, the extension of A is at least as big as the extension of C. That is Levy's spherical isoperimetric inequality, and it is the engine under everything that follows. One more reading before we use it. The cap is solving the same problem the disc solves in the plane and the ball solves in space: least boundary for the area enclosed. On the sphere, least boundary becomes least growth. Now let us spend it."},{"title":"The Concentration Inequality","startSeconds":292.8135208333333,"narration":"Here is the payoff move, and it is short. Take any set A covering exactly half the sphere: half is where a set and its complement balance, and it is the case that powers everything. Which cap has measure one half? The one of radius pi over two: a hemisphere. So the hemisphere is the shape to beat. Let it turn once, so that blue reads as half of a surface rather than as a lid laid across one. Fatten the hemisphere by epsilon and you get this: everything above the equator, plus a collar of angular width epsilon below it. By the isoperimetric inequality, whatever A actually was, its extension covers at least as much of the sphere as this does. So what does the grown hemisphere miss? Only this: the cap of radius pi over two minus epsilon, around the opposite pole. Call it B. The extension covers one minus the measure of B, and everything now rides on how small B is. And B is exactly the kind of region the spike killed. Every point of it sits at latitude at least epsilon away from the equator, where the sine to the n minus one profile has already collapsed. Integrate that profile and the measure of B comes out at most a constant times e to the minus c n epsilon squared. Put the chain together. Any set of measure one half, grown by epsilon, covers all of the sphere except an exponentially small remainder. The constants C and c are universal: they depend on nothing, not the dimension, not the set, not epsilon. That is the concentration inequality. How small is exponentially small? Let me plot the escaping mass, e to the minus n epsilon squared, against the dimension. Here it is for epsilon equal to zero point two, and above it, decaying slower, the thinner margin zero point one. Now ride the steep one. At n around fifty the escaping mass is already near a tenth. By one hundred and fifty it is a quarter of one percent, and off the right edge, in dimension one thousand, it is ten to the minus eighteen. And compare the two curves: halving epsilon quarters the exponent, so the thin margin decays four times slower. The square on epsilon is real. This behavior earned the spheres a title. A sequence of spaces satisfying exactly this bound, uniformly in n, is called a normal Levy family, and the spheres are the founding example. The name matters because the mechanism travels: any family with the bound inherits everything we do next."},{"title":"Functions, Slices, and Dvoretzky","startSeconds":456.4749791666667,"narration":"Concentration so far is about sets. The version that gets used is about functions. Take any real function f on the sphere that is Lipschitz with constant one, so it moves by at most the distance you moved, and let M be its median: half the sphere sits at or below M. Now grow that half. Every point of the extension is within epsilon of somewhere f was at most M, and a Lipschitz function cannot climb faster than distance. So on the whole extension, f is at most M plus epsilon, and the extension is nearly everything. Run the same argument from above, with the set where f is at least M, and the mirror bound comes out. Put the two together: apart from an exception of size at most twice the old one, every point of the sphere has f within epsilon of the median. There it is: essentially the entire sphere lands in this little window. Read that slowly, because it is strange. A Lipschitz function on a high-dimensional sphere is, for every practical purpose, a constant. You may design f as cleverly as you like; the geometry flattens it. Milman's insight was that this is not a curiosity but a tool. Here is the tool at work, first where we can see it. This is the cube, and everything that follows is one flat cut through its centre, taken two different ways. Let it turn first, so you know where its corners are. Cut it straight across, parallel to a face. The knife goes in flat, like this, and the face it leaves behind is a square. That much you would have guessed without me. Now tilt the knife, until it stands square to the long diagonal of the cube, the line joining one corner to the corner furthest away from it. Same cube, same centre, different angle. And the face it leaves is a hexagon. A regular one, with six equal sides, cut out of a body built entirely from squares. This is the step nobody believes until they watch it. So turn it again, and follow the rim. It crosses six of the cube's twelve edges, and it crosses every one of them at its midpoint. Six midpoints, six equal sides. Now lay the two sections flat and measure them. Draw the biggest circle that fits inside each one, and the smallest that wraps around it. For the square, the inner circle has radius one and the outer reaches the corners at root two, so the outer is forty one percent bigger than the inner. The hexagon is measurably rounder. Its inner circle is the larger of the two, while its outer circle is the same root two, because its corners are still edge midpoints of the same cube. The ratio drops to two over root three, about one point one five. Same cube, better direction, and the section has forgotten most of the cube's corners. In dimension n the same cube has central sections round to within any tolerance you name, and, more surprisingly, you do not have to hunt for the direction. The norm of the cube, like any norm, is a Lipschitz function on the sphere, so it concentrates at its median. On a random subspace of the right dimension it is nearly constant, and nearly constant is nearly round. That is Dvoretzky's theorem, in the sharp form Milman proved with exactly this concentration. For every tolerance epsilon there is a constant c of epsilon, and every n-dimensional normed space, no matter how jagged its unit ball, contains a subspace of dimension at least c of epsilon times log n on which the norm is within one plus epsilon of Euclidean. And the proof is this lecture run in reverse. Caps grow slowest, so sets of half measure grow to everything, so Lipschitz functions sit at their medians, so norms are constant on random subspaces, so round sections exist. One inequality about the oldest shape on the sphere, pushed four steps, reaches every norm in every dimension. Log n is small, and that is honest: the cube itself shows you cannot do better than logarithmic in general. But it is not zero, and that is the miracle. In high dimension, roundness is not the exception. It is what is left when there is nowhere else for the measure to go."}]}}