主 题: “Boundness” of some groups of geometric origin
报告人: Prof. D. Burago (Pennsylvania State University)
时 间: 2006-12-08 下午 2:30 - 3:30
地 点: 理科一号楼 1114（数学所活动）
We say that a group is {\\it bounded} if it is bounded (as a metric
>space) with respect to any bi-invariant metric. One can see that a
>group $G$ is unbounded if it admits a q-norm, that is a function
>$q: G \\to \\R$ such that the following three properties hold:
>
>1. $q$ is quasi-subadditive: $\\sup (q(ab) - q(a)-q(b)) < \\infty$;
>
>2. $q$ is quasi-conjugacy-invariant: $\\sup|q(b^{-1}ab) - q(b)|
>< \\infty$;
>
>3. $q$ is unbounded.
>
>In particular, bounded groups do not admit non-trivial
>quasi-morphisms. There are however examples of unbounded finitely
>generated groups that do not admit non-trivial quasi-morphisms.
>For instance, $SL(2,\\Z)$ carries an abundance of quasi-morphisms
>and hence is unbounded. In contrast, $SL(n,\\Z)$ is bounded for
>$n\\ge 3$. Furthermore, $SL(n,R)$ is bounded for all $n$.
>
>One can prove that an abelian group is bounded iff it
>is finite. On the other hand, if the commutator norm on $[G, G]$
>is unbounded, then so is $G$.
>
>By our definition some evidently "small" groups are unbounded:
>i.e. $S^1$. However, most of examples have the following property:
>there exists a finite subset of elements whose conjugates generate
>the whole group. We say that such groups are {\\it finitely
>c-generated}. If a group is {\\it not} finitely c-generated, it
>maybe natural to require that q-norms are {\\it bounded on
>compacts}, to rule out fuzzy example like $S^1$ (which is
>infinite-dimensional over $\\Z$, and hence admits q-norms that are
>highly discontinuous).
>
>\\begin{definition}
>\\label{c-generates} We say that a set $K\\subset G$ {\\it
>c-generates} $G$ in $N$ steps (where $N$ is a positive integer or
>infinity) if every element $h\\in G$ can be represented as a
>product $h={\\tilde h_1} {\\tilde h_c}\\dots {\\tilde h_n}$, where
>$n\\le N$ and each $\\tilde h_i$ is conjugate to some element $h_i
>\\in K$: ${\\tilde h_i}=\\alpha_i h_i \\alpha_i^{-1}$, $\\alpha_i \\in
>G$. Hence a finitely c-generated group is c-generated by a finite
>subset (not necessarily in finitely many steps).
>\\end{definition}
>
>The following simple example is of key importance:
>
>\\begin{example}
> if a group is c-generated in infinitely many steps,
>one can define a norm $q_K$: $q_K(b)$ is the length of a shortest
>word representing $b$ and such that each letter is a conjugate to
>an element from $K$.One can show that for any q-norm $q$ there is
>a constant $c$ such that $q \\leq cq_K+c$; hence, in a sense, $q_K$
>is a {\\it maximum} norm.
>
>
>Furthermore, if a group is c-generated by a finite subset $K$ in
>finitely many steps, then $G$ is bounded.
>\\end{example}
>
>The main result of this work is the following theorem:
>
>\\begin{theorem}
>1. The identity components $Diff_0^{comp}(\\R^n)$ of the groups
>$Diff^{comp}(\\R^n)$ of all compactly supported diffeomorphisms of
>$\\R^n$ are bounded.
>
>2. For every manifold $M$, the identity component
> $Diff_0^{comp}(M \\times \\R)$ of the group of
>compactly supported diffeomorphisms of $M \\times \\R$, is bounded.
>In particular, the group of compactly supported diffeomorphisms of
>the open annulus is bounded.
>
>3. The identity components $Diff_0(S^n)$ of the groups $Diff(S^n)$
>of all diffeomorphisms of $S^n$ are bounded.
>
>4. The identity component $Diff_0(M^3)$ of the groups $Diff(M^3)$
>of the diffeomorphisms of a 3-dimensional compact manifold $M$ is
>bounded.
>
>5. The group $PL=PL^{comp}(\\R)$ of compactly supported
>piecewise-linear homeomorphisms of $\\R$
>\\end{theorem}
>
>The proof motivates the following definition.
>
>\\begin{definition}
>Let $M$ be smooth a manifold. A subset $D \\in M$ along with a
>diffeomorphisms $\\phi: B^n \\to D \\subset M$ is said to be a {\\it
>disc in $M$}. We say that a diffeomorphism is {\\it elementary} if
>its support lies in some disc in $M$.
>
>Let $f \\in Dif_0^{comp}(M)$. We define the {\\em complexity} $L(f)$
>of $f$ to be the smallest $k$ such that $f$ can be represented as
>a product of $k$ elementary diffeomorphisms.
>\\end{definition}
>
>In a sense, $L$ is a "universal candidate" for a q-norm:
>$Diff_0^{comp}(M)$ is unbounded iff $L$ is unbounded. Furthermore,
>for every q-norm $q$ on $Diff_0^{comp}(M)$, there is a positive
>constant $c$ such that $q \\leq cL$.
>
>Surprisingly, we do not know a single manifold with unbounded
>$Diff_0^{comp}(M)$. Furthermore, we even could not answer the
>following striking, intriguing and seemingly elementary questions:
>
>{\\em Open Problem}: Does there exist a constant $C$ such that
>every diffeomorphism of the two-dimensional torus isotopic to the
>identity can be represented as a product of no more than $C$
>elementary diffeomorphisms (diffeomorphisms supported in discs)?
>In other words, is $L$ bounded on $Diff_0(T^2)$? What about
>$Diff_0^{comp}(\\R^2\\#RP^2)$, the group of compactly supported
>diffeomorphisms of the (open) M\\"obius strip?
