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Verbitsky }} \renewcommand{\@evenhead}{\@oddhead} \renewcommand{\@oddfoot}{\hfil\thepage\hfil} \renewcommand{\@evenfoot}{\@oddfoot}} \pagestyle{verbit} \setlength\paperheight {10in}% \setlength\paperwidth {13.5in} \setlength{\textwidth}{0.8\paperwidth} \setlength{\textheight}{0.8\paperheight} \setlength{\pdfpageheight}{\paperheight} \setlength{\pdfpagewidth}{\paperwidth} \addtolength{\topmargin}{-20mm} \addtolength{\leftmargin}{-25mm} \addtolength{\rightmargin}{-25mm} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Lemma, sublemma, corollary, proposition, theorem, % % definition,example defined there: % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcounter{section} \newcounter{Mycounter}[section] \newcounter{lemma}[section] \setcounter{lemma}{0} \renewcommand{\thelemma}{\noindent{Lemma \thesection.\arabic{lemma}}} \newcommand{\lemma}{% \setcounter{lemma}{\value{Mycounter}} \refstepcounter{lemma} \stepcounter{Mycounter} {\bf \green LEMMA:\ }} \newcounter{claim}[section] \setcounter{claim}{0} \renewcommand{\theclaim}{\noindent{Claim \thesection.\arabic{claim}}} \newcommand{\claim}{% \setcounter{claim}{\value{Mycounter}} \refstepcounter{claim} \stepcounter{Mycounter} {\bf \green CLAIM:\ }} \newcounter{corollary}[section] \setcounter{corollary}{0} \renewcommand{\thecorollary}{\noindent{Corollary \thesection.\arabic{corollary}}} \newcommand{\corollary}{% \setcounter{corollary}{\value{Mycounter}} \refstepcounter{corollary} \stepcounter{Mycounter} {\bf \green COROLLARY:\ }} \newcounter{theorem}[section] \setcounter{theorem}{0} \renewcommand{\thetheorem}{\noindent{Theorem \thesection.\arabic{theorem}}} \newcommand{\theorem}{% \setcounter{theorem}{\value{Mycounter}} \refstepcounter{theorem} \stepcounter{Mycounter} {\bf \green THEOREM:\ }} \newcounter{conjecture}[section] \setcounter{conjecture}{0} \renewcommand{\theconjecture}{\noindent{Conjecture \thesection.\arabic{conjecture}}} \newcommand{\conjecture}{% \setcounter{conjecture}{\value{Mycounter}} \refstepcounter{conjecture} \stepcounter{Mycounter} {\bf \green CONJECTURE:\ }} \newcounter{proposition}[section] \setcounter{proposition}{0} \renewcommand{\theproposition} {\noindent{Proposition \thesection.\arabic{proposition}}} \newcommand{\proposition}{% \setcounter{proposition}{\value{Mycounter}} \refstepcounter{proposition} \stepcounter{Mycounter} {\bf \green PROPOSITION:\ }} \newcounter{definition}[section] \setcounter{definition}{0} \renewcommand{\thedefinition} {\noindent{Definition~\thesection.\arabic{definition}}} \newcommand{\definition}{% \setcounter{definition}{\value{Mycounter}} \refstepcounter{definition} \stepcounter{Mycounter} {\bf \green DEFINITION:\ }} \newcounter{example}[section] \setcounter{example}{0} \renewcommand{\theexample}{\noindent{Example \thesection.\arabic{example}}} \newcommand{\example}{% \setcounter{example}{\value{Mycounter}} \refstepcounter{example} \stepcounter{Mycounter} {\bf \green EXAMPLE:\ }} \newcounter{remark}[section] \setcounter{remark}{0} \renewcommand{\theremark}{\noindent{Remark \thesection.\arabic{remark}}} \newcommand{\remark}{% \setcounter{remark}{\value{Mycounter}} \refstepcounter{remark} \stepcounter{Mycounter} {\bf \green REMARK:\ }} \newcounter{observation}[section] \setcounter{observation}{0} \renewcommand{\theobservation}{\noindent{Question \thesection.\arabic{observation}}} \newcommand{\observation}{% \setcounter{observation}{\value{Mycounter}} \refstepcounter{observation} \stepcounter{Mycounter} {\bf \green OBSERVATION:\ }} \newcounter{question}[section] \setcounter{question}{0} \renewcommand{\thequestion}{\noindent{Question \thesection.\arabic{question}}} \newcommand{\question}{% \setcounter{question}{\value{Mycounter}} \refstepcounter{question} \stepcounter{Mycounter} {\bf \green QUESTION:\ }} \newcommand{\exercise}{% {\bf \green EXERCISE:\ }} \newcommand{\proof}{% {\bf \green Proof:\ }} \newcommand{\pstep}{% {\bf \green Proof. Step 1:\ }} \begin{document} \setcounter{page}{1} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{center} {\Large\bf Symplectic geometry\\[15mm] \small lecture 10: Proof of Gromov's Non-Squeezing Theorem} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \\[14mm] {\small Misha Verbitsky } \\[20mm] {\tiny\bf HSE, room 306, 16:20, \\[2mm] October 6, 2021 } \end{center} \newpage {\bf \blue Almost complex structures (reminder)} {\bf\green DEFINITION:} Let $M$ be a smooth manifold. An {\bf \blue almost complex structure} is an operator $I:\; TM \arrow TM$ which satisfies $I^2 = - \Id_{TM}$. The eigenvalues of this operator are $\pm \1$. The corresponding eigenvalue decomposition is denoted $TM=T^{0,1}M\oplus T^{1,0}(M)$. {\bf\green DEFINITION:} An almost complex structure is {\bf \blue integrable} if $\forall X,Y \in T^{1,0}M$, one has $[X,Y]\in T^{1,0}M$. In this case $I$ is called {\bf \blue a complex structure operator}. A manifold with an integrable almost complex structure is called {\bf \blue a complex manifold}. {\bf\green THEOREM:} (Newlander-Nirenberg)\\ {\bf \purple This definition is equivalent to the usual one.} \definition Let $(M, \omega)$ be a symplectic manifold, and $I$ an almost complex structure. We say that {\bf \blue $I$ is compatible with the symplectic structure} if $g(x, y):= \omega(Ix, y)$ for some Riemannian form $g$. {\bf \green THEOREM 1:} Let $(M, \omega)$ be a manifold equipped with a non-degenerate skew-symmetric 2-form. Then {\bf \red the space $C$ of almost complex structures compatible with $\omega$ is contractible.} \newpage {\bf \blue Calibrations (reminder)} \newcommand{\comass}{\operatorname{\sf comass}} \definition (Harvey-Lawson, 1982)\\ Let $W\subset V$ be a $p$-dimensional subspace in a Euclidean space, and $\Vol(W)$ denote the Riemannian volume form of $W \subset V$, defined up to a sign. For any $p$-form $\eta \in \Lambda^p V$, let {\bf\blue comass} $\comass(\eta)$ be the maximum of $\frac{\eta(v_1, v_2, ..., v_p)}{|v_1||v_2|...|v_p|}$, for all $p$-tuples $(v_1, ..., v_p)$ of vectors in $V$ and {\bf\blue face} be the set of planes $W\subset V$ where $\frac\eta {\Vol(W)}=\comass(\eta)$. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \definition\label{_precalibra_Definition_} A {\bf\blue precalibration} on a Riemannian manifold is a differential form with comass $\leq 1$ everywhere. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \definition A {\bf\blue calibration} is a precalibration which is closed. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \definition Let $\eta$ be a $k$-dimensional precalibration on a Riemannian manifold, and $Z\subset M$ a $k$-dimensional subvariety (we always assume that the Hausdorff dimension of the set of singular points of $Z$ is $\leq k-2$, because in this case a compactly supported differential form can be integrated over $Z$). We say that $Z$ is {\bf\blue calibrated by $\eta$} if at any smooth point $z\in Z$, the space $T_zZ$ is a face of the precalibration $\eta$. \newpage {\bf \blue Calibrations and minimal submanifolds (reminder)} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \remark Clearly, for any precalibration $\eta$, one has \[ \Vol(Z) \geq \int_Z\eta,\ \ \ \ \ (*) \] where $\Vol(Z)$ denotes the Riemannian volume of a compact $Z$, and the equality happens iff $Z$ is calibrated by $\eta$. If, in addition, $\eta$ is closed, the number $\int_Z\eta$ is a cohomological invariant. Then, {\bf \purple the inequality (*) implies that $Z$ minimizes the Riemannian volume in its homology class.} \definition A subvariety $Z$ is called {\bf \blue minimal} if for any sufficiently small deformation $Z'$ of $Z$ in class $C^1$, one has $\Vol(Z')\geq \Vol(Z)$. \remark {\bf \red Calibrated subvarieties are obviously minimal.} \newpage {\bf \blue Pseudoholomorphic curves (reminder)} \definition Let $(M,J)$ be an almost complex manifold, $(\Sigma, I)$ a Riemann surface, and $\phi:\; \Sigma \arrow M$ an $I$-holomorphic map, that is, a smooth map with $D\phi(Ix)= J(D\phi(x))$. Then $\phi(\Sigma)$ is called {\bf \blue a pseudo-holomorphic curve}, or {\bf \blue a $J$-holomorphic curve}. \theorem {\bf \blue (Wirtenger's inequality):} \\ Let $(M, I, \omega)$ be an almost K\"ahler manifold. {\bf \purple Then $\frac{1}{2}\omega$ is a calibration which calibrates pseudo-holomorphic curves.} \proof Let $g_S$ be the Riemannian volume form on $S$, and $x, y\in T_sS$ be orthogonal vectors of length 1. Then $g_S(x, y)=1$ and $\omega(x, y)= g(x, Iy)\leq 1$, and the equality is realized if and only if $x=Iy$, by Cauchy-Bunyakovsky-Schwarz inequality. \endproof \corollary {\bf \red Pseudoholomorphic curves are minimal.} \newpage {\bf \blue Symplectic capacity and the pseudoholomorphic curves} {\bf \green THEOREM 2:} Let $M=\C P^1 \times T^{2n}$ be the product of $\C P^1$ and a torus, equipped with the standard symplectic structure, and $J$ a compatible almost complex structure. {\bf \red Then for any $x\in M$ there exists a pseudo-holomorphic curve $S$ homologous to $\C P^1 \times \{m \}$ and passing through $x$.} This theorem implies Gromov's non-squeezing theorem. \theorem {\bf \blue (Gromov)} {\bf \red Symplectic capacity of a symplectic cylinder $\Cyl_1$ is equal to $\pi$.} {\bf \green Scheme of the proof:} We map $\Cyl_1= B_1^2 \times \R^{2n-2}$ to $M=\C P^1 \times T^{2n-2}$ in such a way that the disk $B_1^2$ is bijectively mapped to $\C P^1 \backslash \infty$, and $\R^{2n-2}$ is mapped to $\R^{2n-2}/\Z^{2n-2} = T^{2n-2}$. \newpage {\bf\blue Proof of Gromov's theorem} \theorem {\bf \blue (Gromov)}\\ {\bf \red Symplectic capacity of a symplectic cylinder $\Cyl_1$ is equal to $\pi$.} \pstep Let $f_1:\; B_r \arrow \Cyl_1$ be a symplectic embedding, $r>1$, and $I$ the usual (flat) almost complex structure on $B_r\subset \C^{n+1}$. Consider the manifold $M=\C P^1 \times T^{2n}$, equipped with the standard symplectic structure, and let $f_2:\; \Cyl_1 \arrow \C P^1 \times T^{2n}$ be a symplectic map taking $\Cyl_1=\Delta \times \R^{2n}$ to $\C P^1 \times T^{2n}$ applying the $\Z^{2n}$ quotient on the second argument and the natural symplectomorphism $\Delta\tilde \arrow \C P^1 \backslash \infty$ on the first argument. {\bf \green Step 2:} Choose the lattice $\Z^{2n}\subset \R^{2n}$ in such a way that its fundamental domain contains $f_1(B_r)$. Then the composition $f_1 \circ f_2$ gives a symplectic embedding $B_r \arrow M=\C P^1 \times T^{2n}$. We obtained that {\bf \purple Gromov's non-squeezing theorem is deduced from the following result.} \newpage {\bf \blue Proof of Gromov's theorem (2)} \theorem Let $M= \C P^1 \times T^{2n}$ be equipped with the standard symplectic form, with the symplectic volume of $\C P^1$ equal to $\pi$, and $\phi:\; B_r \arrow M$ a symplectic embedding. {\bf \red Then $r\leq 1$.} \pstep Choose a flat complex structure and the flat Hermitian metric on $B_r$. Denote by $g_0$ the corresponding Hermitian metric on $\phi(B_r)$. Then $g_0$ can be extended to a Riemannian metric $g_1$ on $M$ such that $g_0=g_1$ in a ball $\phi(B_{r-\epsilon})$, for some $\epsilon$ such that $r-\epsilon >1$. The operation $g_1(\cdot, \cdot) \arrow g_1(B_1\cdot, \cdot)$ constructed in the proof of Theorem 1, gives a metric $g$ compatible with the symplectic structure on $M$ and coinciding with $g_0$ in a ball $\phi(B_{r-\epsilon})$. {\bf \purple Replacing $B_r$ by $B_{r-\epsilon}$, we can add to the assumptions of the theorem the following assumption.} {\bf \green There exists a compatible almost complex structure such that uts restriction to $\phi(B_r)$ is equal to the standard complex structure on $B_r\subset \C^{n+1}$.} \newpage {\bf\blue Proof of Gromov's theorem (3)} \theorem Let $M= \C P^1 \times T^{2n}$ equipped with the standard symplectic form, with the symplectic volume of $\C P^1$ equal to $\pi$, and $\phi:\; B_r \arrow M$ a symplectic embedding. {\bf \purple Assume that there exists a compatible almost complex structure such that its restriction to $\phi(B_r)$ is equal to the standard complex structure on $B_r\subset \C^{n+1}$.} {\bf \red Then $r\leq 1$.} {\bf \green Step 2:} Let $x\in M$ be the image of the center of $B_r$, and $S\subset M$ the pseudo-holomorphic curve which passes through $x$ by Theorem 2. Then $\pi=\int_S \omega_M \geq \int_{\phi^{-1}(S)}\omega$, where $\omega_M$ is the symplectic form on $M$, and $\omega$ the symplectic form on $B_r$. Since $S$ is pseudo-holomorphic, {\bf \purple $\int_{\phi^{-1}(S)}\omega_{B_r}$ is the Riemannian volume of its intersection with $\phi(B_r)$.} {\bf \green Step 3:} We obtained a complex curve $D:= \phi^{-1}(S)$ passing through 0 in a ball $B_r$ with flat Riemannian metric and the standard complex structure, with the Riemannian volume $\Vol(D)\leq \pi$. Applying the homothety, we obtain a properly embedded complex disk in the ball $B$ of radius 1, passing through 0 and with area $r^{-1} \pi$. {\bf \purple For any $r>1$, this is impossible, as follows from the following statement,} proven later today. \proposition Let $D\subset B_1$ be a closed complex disk in a unit ball $B_1 \subset \C^n$, with $0\in D$. {\bf \red Then $\Vol(D) \geq \pi$.} \newpage {\bf \blue Monotonicity formula (1)} \proposition Let $D\subset B_1$ be a complex curve (that is, a closed 1-dimensional complex subvariety) in a unit ball $B_1 \subset \C^n$, with $0\in D$. {\bf \red Then $\Vol(D) \geq \pi$.} We deduce it from the {\bf \blue monotonicity formula.} \lemma {\bf \blue (monotonicity formula)}\\ Let $(B, \eta)$ be a unit ball $B_1(0)\subset \R^n$ equipped with a calibration $\eta\in \Lambda^k B_1$ with constant coefficients, and the standard flat metric, and $\Delta\subset B$ a calibrated $k$-dimensional submanifold passing through 0. We assume that $\Delta$ is immersed to $B$ with its boundary in such a way that $\6\Delta$ is mapped to $\6 B$. {\bf \red Let $\Delta_t$ be the intersection of $\Delta$ with a ball $B_t(0)$ of radius $t$. Then the function $t\arrow t^{-k} \Vol(\Delta_t)$ is non-decreasing.} Clearly, $\lim\limits_{t\rightarrow 0} \Vol(\Delta_t)= \pi t^2$ (on a very small scale, any smooth disk is flat). {\bf \purple Then the monotonicity lemma gives $\Vol\Delta \geq \pi$.} \newpage {\bf \blue Monotonicity formula (2)} \lemma {\bf \blue (monotonicity formula)}\\ Let $(B, \eta)$ be a unit ball equipped equipped with a calibration $\eta\in \Lambda^k B_1$ with constant coefficients, and $\Delta\subset B$ a calibrated $k$-dimensional submanifold passing through 0. {\bf \red Let $\Delta_t$ be the intersection of $\Delta$ with a ball $B_t(0)$ of radius $t$. Then the function $t\arrow t^{-k} \Vol(\Delta_t)$ is non-decreasing.} \pstep Let $C(\6\Delta)\subset B$ be the cone obtained as a union of all intervals connecting $0$ and points of $\6\Delta\subset \6 B$. Then \[ \Vol C(\6\Delta)\geq \int_{C(\6\Delta)}\eta=\int_\Delta\eta=\Vol\Delta. \] The first equality follows from the Stokes' theorem, because $\6\Delta= \6 C(\6\Delta)$, and the first inequality holds because $\eta$ is a calibration. {\bf \green Step 2:} The formula for the cone volume gives $\Vol C(\6\Delta)=\frac 1 k l(\6\Delta)$, where $l(\6\Delta)$ is the area of the boundary $\6\Delta$. {\bf \purple This implies $\Vol\Delta\leq \frac 1 k l(\6\Delta)$.} \newpage {\bf \blue Monotonicity formula (3)} \lemma {\bf \blue (monotonicity formula)}\\ Let $(B, \eta)$ be a unit ball equipped equipped with a calibration $\eta\in \Lambda^k B_1$ with constant coefficients, and $\Delta\subset B$ a calibrated $k$-dimensional submanifold passing through 0. {\bf \red Let $\Delta_t$ be the intersection of $\Delta$ with a ball $B_t(0)$ of radius $t$. Then the function $t\arrow t^{-k} \Vol(\Delta_t)$ is non-decreasing.} {\bf \green Step 2:} We have obtained $\Vol\Delta\leq \frac 1 k l(\6\Delta)$. {\bf \green Step 3:} The same argument applied to $\Delta_t$ gives $t^{-k}\Vol\Delta_t\leq \frac 1 {kt^{k-1}} l(\6\Delta_t)$. (homothety maps the $k$-dimensional Riemann volume $\mu$ to $t^k \mu$). On the other hand, $\frac d {dt} \Vol\Delta_t\geq l(\6\Delta_t)$, because the volume of the strip of $\Delta$ between $\6\Delta_t$ and $\6\Delta_{t+\epsilon}$ is bounded by $\epsilon l(\6\Delta_{t+\epsilon})$. This gives \[ \Vol\Delta_t \leq \frac t {k} l(\6\Delta_t) \leq \frac t {k}\frac d {dt} \Vol\Delta_t. \] {\bf \green Step 4:} Let $f(t):= \Vol\Delta_t$. The last formula of Step 3 gives $f(t) \leq \frac t k f'(t)$, hence \[ \frac d{dt} t^{-k}f(t) = t^{-k}f'(t)- kt^{-k-1}f(t) \geq t^{-k}f'(t)-t^{-k}f'(t)=0. \] \endproof \end{document}