Geodesic metrics that admit dilatation at each point - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T12:51:34Z http://mathoverflow.net/feeds/question/39960 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/39960/geodesic-metrics-that-admit-dilatation-at-each-point Geodesic metrics that admit dilatation at each point Dmitri 2010-09-25T17:19:01Z 2010-11-08T18:02:56Z <p>Consider the class of geodesic metrics $g$ on manifolds, that have the following property: for each point $x$ there exists a neighbourhood $U_x$ and a smooth vector field $v_x$ in $U_x$ that vanishes at $x$ and whose flow (for small time) dilatates $g$ by a constant factor. Let us call such metrics <em>dilatatable</em>.</p> <p>An obvious example is provided by an Euclidean $\mathbb R^n$, the flow of the field $\sum_i x_i \frac{\partial}{\partial x_i}$ dilatates the Euclidean metric by a constant factor. More generally one can take any Banach space. I would like to make a guess about the structure of such metrics in general.</p> <p><strong>Guess.</strong> Suppose $g$ on $M^n$ is dilatatable. Then there exists a triangulation of $M^n$<br> such that the restriction of the metric $g$ to each simplex if flat with respect to the flat structure on the simplex, and $g$ is flat on the complement to the union of all co-dimension $2$ simplexes. </p> <p>The <strong>first question</strong> is the following: was such class of metrics considered somewhere and is this guess correct? Are there obvious counterexamples?</p> <p>Second part of the question is about examples. It is not hard to construct an example of such a metric, if we don't require $M^n$ to be a smooth manifold. Namely, we can take any polyhedral metric on $M^n$, i.e. glue $M^n$ from a union of Euclidean simplexes (glue the boundaries by isometries). Then for each point there is a conical neighbourhood, and obviously we can always scale this neighbourhood by the radial field emanating from $x$. So now comes the</p> <p><strong>Second question.</strong> Take a topological manifold $M^n$ of dimension $n&lt;7$ with such a polyhedral metric. It is known then that such a manifold has a smooth structure (because a PL structure in dimension up to $6$ always defines a unique smooth structure). Is it possible to chose this smooth structure in such a way, that the polyhedral metric is dilatatable for the smooth structure?</p> <p>The answer to this question is positive for $n=2$, but I don't know already what happen for $n=3$. At the same time, there are non-trivial examples in higher dimensions, coming from complex geometry. For example one can quotient some complex tori $\mathbb T^n$ by a finite group of isometries to get $\mathbb CP^n$, the obtained polyheral metric on $\mathbb CP^n$ is dilatatable with respect to the canonical complex (and hence smooth) structure on $\mathbb CP^n$.</p> http://mathoverflow.net/questions/39960/geodesic-metrics-that-admit-dilatation-at-each-point/39969#39969 Answer by Sergei Ivanov for Geodesic metrics that admit dilatation at each point Sergei Ivanov 2010-09-25T19:43:52Z 2010-09-25T20:00:03Z <p>Concerning the first question: you description is incomplete, even in the homogeneous case.</p> <p>There are homogeneous geodesic metrics that admit smooth families of dilatations but are not made of flat Banach metrics. In particular, some <a href="http://en.wikipedia.org/wiki/Sub-Riemannian_manifold" rel="nofollow">Carnot-Caratheodory metrics</a> are.</p> <p>For example, consider the Heisenberg group $H$, which can be thought of as $\mathbb R^3$ equipped with the following group law: $$(x,y,z)\cdot(x',y',z') = (x+x',y+y',z+z'+x'y) .$$ Observe that for every $t\in\mathbb R$, the map $\phi_t:(x,y,z)\mapsto (e^tx,e^ty,e^{2t}z)$ is a group homomorphism, and these maps form a smooth 1-parameter group of diffeomorphisms (and hence a flow generated by a smooth vector field).</p> <p>Consider a left-invariant two-dimensional distribution $V\subset TH$ spanned by left-invariant vector fields $X$ and $Y$ whose values at $(0,0,0)$ equal $\partial/\partial x$ and $\partial/\partial y$, respectively. Equip this distribution with a left-invariant Euclidean metric. The distribution is completely non-integrable, so we get a Carnot-Caratheodory metric on $H$. Observe that $\phi_t$ maps $X$ to $e^tX$ and $Y$ to $e^tY$, hence it is a $e^t$-dilatation of the Carnot-Caratheodory metric.</p> <p>The Carnot-Caratheodory metric is very different from Banach metrics. For example, its Hausdorff dimension equals 4.</p> http://mathoverflow.net/questions/39960/geodesic-metrics-that-admit-dilatation-at-each-point/45334#45334 Answer by Marius Buliga for Geodesic metrics that admit dilatation at each point Marius Buliga 2010-11-08T18:02:56Z 2010-11-08T18:02:56Z <p>Relative to comments by Sergei Ivanov and Bill Thurston, maybe this line of research concerning "metric spaces with dilations" or "dilation structures" provides a precise answer, more general than Berestovskii result. See <a href="http://arxiv.org/abs/1007.2362" rel="nofollow">this introduction</a> and dig into the biblio. </p> <p>Concerning examples related to Carnot-Caratheodory geometry and nilpotent groups (precisely: "Carnot groups"), they appear naturally as models of the (metric) tangent space to a point in a space with dilations. </p> <p>If you stand to read a more algebraic account, see <a href="http://arxiv.org/abs/0907.1520" rel="nofollow">emergent algebras</a>, where it is proven that this is not really a metric induced phenomenon. </p>