User malte - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T03:19:36Z http://mathoverflow.net/feeds/user/20557 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110709/a-riemannian-manifold-with-finitely-many-closed-contractible-geodesics A riemannian manifold with finitely many closed contractible geodesics Malte 2012-10-25T22:20:21Z 2013-05-12T08:17:31Z <p>By a closed geodesic, I mean a smooth periodic geodesic $\mathbb{R} \rightarrow (M,g)$. I will consider them up to geometric distinction. This means that any two closed geodesics are equivalent if they have the same image in $(M,g)$. </p> <p>Manifolds with constant curvature $\leq 0$, by Cartan's theorem, cannot have any closed contractible geodesics, and every riemannian metric on $S^2$ has infinitely many closed geodesics (for $n\geq 3$, the analogous theorem for $S^n$ is not known). Moreover, if the sequence of Betti numbers of the loops space $\Omega(M)$ is unbounded and $M$ is simply-connected, then $(M,g)$ contains infinitely many (contractible) closed geodesics.</p> <p><strong>Are there any known examples of riemannian manifolds with finitely and positively many closed contractible geodesics (or even just closed geodesics)?</strong></p> <p>There is a theorem associated with Gromov asserting that the word problem of $\pi_1 M$ is solvable if there is a metric $g$ on $M$ with only finitely many contractible closed geodesics. I was wondering if there are any non-trivial examples for this theorem.</p> http://mathoverflow.net/questions/128819/riemannian-manifolds-with-small-geodesics-and-bounded-curvature Riemannian manifolds with small geodesics and bounded curvature Malte 2013-04-26T11:55:17Z 2013-04-27T03:12:42Z <p>Let $(M,g)$ be a compact riemannian manifold with sectional curvature $|K_g| \leq 1$. A lemma due to Klingenberg asserts that then either the injectivity radius $i_g \geq \pi$ or $(M,g)$ contains a geodesic loop $\gamma$ of length $L(\gamma) = 2i_g$.</p> <p>Let $n = 2$. Suppose for a moment that $i_g &lt; \pi$ and $M$ is orientable. Note that $\gamma$ is automatically simple. (I am note sure how often this occurs. An example, however, is a thin long cylinder with two balloon-like 2-spheres attached.) Then $\gamma$ cuts $M$ into two pieces. </p> <p>Intuition says that these two pieces should be rather large, as the manifold cannot curve together very quickly due to the curvature condition. By another Klingenberg lemma, known as the "Long Homotopy Lemma", any null-homotopy $\gamma_t$ of $\gamma_0 = \gamma$ must sweep through some curve $\gamma_{t_0}$ of length $L(\gamma_{t_0}) \geq \pi$. (This lemma holds for all $n$ and non-orientable manifolds. It is Exercise 1 in Chapter 10 of do Carmo's text book.)</p> <p>To some extend, this is the kind of theorem I am looking for. It asserts that away from $\gamma$, $(M,g)$ must be slightly larger (on the scale of the curvature bound) than near $\gamma$. However, I'm looking for some result describes the geometry of $M$ away from $\gamma$ in a "more global" fashion. For instance: Is there any estimate of the diameter or the volume of the pieces that $M$ is cut into? </p> <p>In view of the examples given by horse with no name, one should assume that $M$ has $g(M) \neq 0$ (or that $M$ has some other non-vashing characteristic class in the case that $n > 2$). </p> <p>As pointed out in the comments and Anton Petrunin's answer, $M$ should either be a sphere or $\gamma$ must also be assumed to cut $M$ in half.</p> http://mathoverflow.net/questions/118693/is-the-volume-functional-contiunuous-for-compact-manifolds-with-lower-bounds-on-v Is the volume functional contiunuous for compact manifolds with lower bounds on volume? Malte 2013-01-12T02:01:16Z 2013-01-12T14:04:18Z <p>Let $n\in\mathbb{N}$ Is the volume functional continuous on the set of isometry classes of compact riemannian $n$-manifolds with volume $\geq \varepsilon$\ (with respect to Gromov--Hausdorff distance)?</p> <p>Without the volume bound, a collapsing torus gives a counterexample. But it seems that this is the only singularity. There are a number of weird results on the semi-continuity of the volume functional. For instance, every metric on $S^3$ is the limit of a metric with volume converging to $0$. I don't know what happens to the curvature of these metrics.</p> http://mathoverflow.net/questions/103087/how-many-quotients-can-a-finitely-generated-group-have-or-how-many-bundles-over-a How many quotients can a finitely generated group have or how many bundles over aspherical spaces does a fixed total space support? Malte 2012-07-25T10:11:47Z 2012-08-16T23:12:45Z <p>Consider $M^3_{pq}$, a torus bundle over $S^1$ with fundamental group the HNN extension generated by three generators $x,y,z$ satisfying the relations $\quad [x,y], \quad x^z = x^p \quad$ and $y^z = y^q$. By the exact homotopy sequence, $M^3_{pq}$ is aspherical.</p> <p>Suppose there is a closed manifold $E^{n+3}$ that is the total space of a $T^n$ bundle over infinitely many $M^3_{pq}$ satisfying $p\neq q$. As $M_{pq}^3$ is aspherical, so is $E$. Hence, this can only be the case if $\pi_1 M_{pq}$ is a quotient of $\pi_1 E$ by the fundamental group $\pi_1T^n$; note that $\pi_1E$ and $n$ may not depend on $(p,q)$. It seems unlikely that one can obtain an infinite amount of mutually non-isomorphic, non-abelian groups as quotients from a finitely generated group by only identifying elements that commute. However, I have failed to prove this.</p> <p>This begs the general question: How many exact sequences with non-abelian groups $H$<br> $$0 \rightarrow Z^n \rightarrow G \rightarrow H \rightarrow e$$ can a finitely presented group $G$ admit for fixed $n$? What are possible obstructions, aside from the kernel of $G\rightarrow H$ having to be abelian?</p> http://mathoverflow.net/questions/99247/preissmann-and-byers-theorems/99258#99258 Answer by Malte for Preissmann and Byers Theorems Malte 2012-06-10T20:53:54Z 2012-07-21T01:28:24Z <p>1) To construct a manifold with no solvable fundamental group, take for example a finite unsolvable group $G$ and embed it into $SU(n)$ as a discrete subgroup. This embedding is obtained by realizing $G$ as a subgroup of the permutations $S(G)$ of the set $G$, then note $S(G)$ is isomorphic to a subgroup of $SU(n)$ via monomial matrices ($n=|G|$). The quotient $SU(n)/G$ carries a natural manifold structure with fundamental group $G$. Note these examples are compact. Taking product with $R$, one obtains non-compact examples.</p> <p>2) For manifolds with cyclic subgroups of finite index, consider products of real projective spaces $RP^n$ or lens spaces $S^{2n+1}/Z_p$ with manifolds with finite fundamental group.</p> <p>To combine 1) and 2), take products.</p> http://mathoverflow.net/questions/102542/historical-question-fiber-bundles/102770#102770 Answer by Malte for Historical question: fiber bundles Malte 2012-07-20T21:26:05Z 2012-07-20T21:26:05Z <p>The first paragraph of the following article hints that Steenrod, Whitehead, Chern and Sun indepentently arrived at the fact that -- provided that the universal bundle exists -- it classifies principal $G$ bundles. (I don't have a subscription so I can't check the references for details, in particular the date of publication...)</p> <p>S.-T. Hu: The Equivalence of Fiber Bundles, Ann. Math. (2) 53 1953, <a href="http://www.jstor.org/discover/10.2307/1969542?uid=3737864&amp;uid=2129&amp;uid=2&amp;uid=70&amp;uid=4&amp;sid=49501340154377" rel="nofollow">http://www.jstor.org/discover/10.2307/1969542?uid=3737864&amp;uid=2129&amp;uid=2&amp;uid=70&amp;uid=4&amp;sid=49501340154377</a></p> <p>A partial result (the fibre is $S^n$) was proved earlier by Steenrod (I believe you can find the general result in Steenrod's book on fibre bundles, which was published in 1951). Milnor, however, seems to be the first one to establish the existence of universal bundles.</p> http://mathoverflow.net/questions/102173/second-homotopy-group-of-graph-manifolds Second Homotopy Group of Graph Manifolds Malte 2012-07-13T19:41:56Z 2012-07-14T16:31:37Z <p>A graph manifold is a closed 3-manifold $M$ that admits a finite collection of disjoint embedded tori $\mathcal{T}$ so that $M \setminus \mathcal{T}$ is a disjoint union of Seifert fibred spaces (i.e. spaces admitting a foliation with a circle as fibre).</p> <p>My interest in this class of manifolds lies in its characterization through the methods of geometrization: A 3-manifold is a graph manifold if and only if its simplicial norm vanishes.</p> <p>Question: What is known about the second homotopy group of graph manifolds? I am particulary interested in an answer to the following question: How many graph manifolds have finite second homotopy group?</p> <p>The only hint I could find was the following result due to D. Hume (cf. <a href="http://arxiv.org/abs/1112.0263v2" rel="nofollow">http://arxiv.org/abs/1112.0263v2</a>): The universal cover of any graph manifold quasi-isometrically embeds in the product of three metric trees.</p> http://mathoverflow.net/questions/98107/metric-deformations-from-non-negative-to-positive-curvature Metric Deformations from Non-Negative to Positive Curvature Malte 2012-05-27T12:39:56Z 2012-05-27T16:06:13Z <p>Is it possible to deform the metric $g$ of a closed Riemannian manifold $(M,g)$ satisfying $\mathrm{Ricci}(M,g) > 0$ and $\mathrm{sec}(M,g) \geq 0$ to a metric $g_1$ satisfying $\mathrm{sec}(M,g_1) > 0$?</p> <p>I understand that this question is ludicrous, at best (for instance, an affirmative would prove that $S^n\times S^m$ always carries a metric of positive sectional curvature). I'm assuming that there is a counterexample, but I can't seem to figure it out. Most examples of manifolds with positive Ricci curvature that I know can't admit metric of non-negative sectional curvature for topological reasons, for instance the Sha--Yang construction of metrics with positive Ricci curvature on connected sums of $S^n \times S^m$.</p> <p>I was wondering about this in view of the following two results: The first, due to Aubin and Ehrlich, asserts that a metric $g$ with $\mathrm{Ricci}(M,g) \geq 0$ everywhere and $\mathrm{Ricci}(M,g) > 0$ somewhere can be deformed into a metric $g_1$ with $\mathrm{Ricci}(M,g_1) > 0$ everywhere. The second, due to Gao--Yau, asserts that if $\mathrm{Ricci}(M,g) > 0$ everywhere, then $g$ can be deformed into a metric $g_1$ with $\mathrm{Ricci}(M,g) > 0$ everywhere and $\mathrm{sec}(M,g) > 0$ somewhere. </p> <p>The Gao--Yau result is a local solution to the intial question. </p> <p>As a related question: Suppose the Ricci curvature of $(M,g)$ is pinched. Does this change the answer to the question if the pinching constant is particulary small or would it seem that pinching plays no role in this? Is there any theory of pinched Ricci curvature? </p> <p>T. Aubin, Metriques riemannienes et courbure. Diff. Geom., 4:383--424, 1970</p> <p>P. Ehrlich, Metric deformations of curvature. Geom. Ded., 5:1--23, 1976</p> <p>L. Gao, S.-T. Yau, The existence of negatively Ricci curved metrics on three manifolds. Invent. Math., 85:637--652, 1986</p> http://mathoverflow.net/questions/92616/does-normalized-ricci-flow-on-surfaces-yield-a-bundle Does normalized Ricci flow on surfaces yield a bundle? Malte 2012-03-29T22:04:10Z 2012-03-31T14:49:32Z <p>As is well known, the normalized Ricci flow is defined for all $t>0$ on compact surfaces, and every metric on a compact surfaces converges to a metric constant curvature if $X \neq S^2$ (at least I can't find a reference that asserts that same result for $X=S^2$; B. Chow's "Ricci flow on the 2-sphere" only shows that metrics of positive Gaußian curvature converge to constant curvature metrics). This is somewhat related to <a href="http://mathoverflow.net/questions/47981/ricci-flow-with-surgery-in-dimension-2" rel="nofollow">this</a>.</p> <p>One thus has a map $\mathcal{R}(X) \rightarrow T_X$ from the Riemannian moduli space to the Teichmüller space $T_X$ of constant curvature metrics associating to $g\in \mathcal{R}(X)$ its limit $g^\ast$ under the normalized Ricci flow. Note that the fibres of this map are convex.</p> <p>Question: Is this map a fibre bundle?</p> http://mathoverflow.net/questions/91994/recognizing-the-4-sphere-and-the-adjan-rabin-theorem Recognizing the 4-sphere and the Adjan--Rabin theorem Malte 2012-03-23T10:49:20Z 2012-03-23T19:33:32Z <p>The problem of recognizing the standard $S^n$ is the following: Given some simplicial complex $M$ with rational vertices representing a closed manifold, can one decide (in finite time) if $M$ is homeomorphic to $S^n$. </p> <p>For $n=1$, this is obvious, and for $n=2$, one can solve it by computing $\chi(M)$. A solution for $n=3$ is due to </p> <p>J.H. Rubinstein. An algorithm to recognize the 3-sphere. In Pro- ceedings of the International Congress of Mathematicians, vol- ume 1, 2, pages pp. 601–611, Basel, 1995. Birkhäuser.</p> <p>By a theorem of S.P. Novikov, the problem is unsolvable if $n\geq 5$. The idea is the following: By the Adjan--Rabin theorem, there is a sequence of super-perfect groups $\pi_i$ for which the triviality problem is unsolvable. Now construct homology spheres $\Sigma_i$ with fundamental groups $\pi_i$. If one can decide which of the $\Sigma_i$ are standard spheres, then one can solve the triviality problem for the fundamental groups.</p> <p>Question: Is the recognition problem for $S^4$ solvable?</p> <p>The problem with this proof of S.P. Novikov's theorem is that there is no result that asserts that for any given super-perfect group $\pi$ there is a homology $4$-sphere satisfying $\pi_1(\Sigma) = \pi$. However, Kervaire has proved that every perfect group with the same amount of generators and relators may be realized as the fundamental group of a homology $4$-sphere.</p> <p>Thus the question: Is there an improved Adjan--Rabin theorem that asserts the existence of a sequence of perfect groups $\pi_i$ with the same amount of generators and relators, the triviality problem of which is unsolvable?</p> http://mathoverflow.net/questions/91604/does-regularity-of-the-boundary-imply-interior-sphere-condition/91732#91732 Answer by Malte for Does regularity of the boundary imply interior sphere condition Malte 2012-03-20T16:12:01Z 2012-03-20T16:12:01Z <p>I think the remark on the curvature of the boundary of $\Omega$ might give some insight into this problem. Assume that $\Omega \subseteq \mathbb{R}^2$ has a $C^2$ boundary curve. Then its curvature is bounded from above by $\varepsilon > 0$. This implies that for any point of the curve, there is a osculating circle of radius $R \leq 1/\varepsilon$.</p> <p>The tube lemma should imply that there is some sort of $\delta$-collar around the boundary curve (using the normal bundle of the curve). Outside the $\delta$-collar, every point $x\in \Omega$ is contained in $B(x,\delta)$. After taking $\delta &lt; 1/\varepsilon$, inside the collar, every point is contained in an osculating circle of radius $\delta$.</p> <p>I assume this argument should work (after some refinement) for more complicated boundaries (say, if $\Omega$ is an annulus) and in higher dimensions using the Riemannian curvature.</p> http://mathoverflow.net/questions/91462/thomas-clausens-puzzle/91468#91468 Answer by Malte for Thomas Clausen's puzzle Malte 2012-03-17T14:31:07Z 2012-03-17T14:31:07Z <p>The equality $(e^z)^w = e^{zw}$ does not hold for $z,w \in \mathbb{C}$ in general, so the error lies between lines three and four.</p> http://mathoverflow.net/questions/91377/the-category-of-posets/91381#91381 Answer by Malte for The category of posets Malte 2012-03-16T16:10:06Z 2012-03-16T16:10:06Z <p>Abstract simplicial complexes happen to be posets, and to every abstract simplicial complex, one may associate a topological space, its geometric realization. This is as functor $\mathrm{Pos} \rightarrow \mathrm{Top}$. (asc's are defined by the property that if $A \in \Delta$ and $B\subseteq A$, then $B \in \Delta$)</p> <p>Also, you might want to take a look into the theory of Bruhat--Tits buildings. Basically, one associates to a simple algebraic group a certain simplicial complex $\Delta(G)$. However, I have one tried to figure out if that association is a functor without any success since given up.</p> <p>Both example's seem to be more geometric/topological/algebraic than really category theoretical (viz. more focused on the objects than the functor), but maybe still instructive. </p> http://mathoverflow.net/questions/90862/what-are-some-examples-of-ingenious-unexpected-constructions/90932#90932 Answer by Malte for What are some examples of ingenious, unexpected constructions? Malte 2012-03-11T21:04:26Z 2012-03-11T21:04:26Z <p>1) Alexander's horned sphere, disproving the Jordan curve theorem in high dimensions.</p> <p>2) Maybe not a watershed construction as some others named, but certainly high up on the "awesome" list: Higman's universal group.</p> http://mathoverflow.net/questions/85595/relation-between-combinatorial-manifolds-and-pl-manifolds Relation between combinatorial manifolds and PL manifolds Malte 2012-01-13T16:58:30Z 2012-01-26T16:14:52Z <p>In </p> <p>W. W. Boone, W. Haken, and V. Poenaru, On Recursively Unsolvable Problems in Topology and Their Classification, Contributions to Mathematical Logic (H. Arnold Schmidt, K. Schütte, and H. J. Thiele, eds.), North-Holland, Amsterdam, 1968.</p> <p>a combinatorial manifold is defined as a simplicial complex with the property that the star of every vertex is combinatorially equivalent to the standard $n$-simplex. (two simplicial complexes are combinatorially equivalent if they possess linear subdivisions, the associated abstract simplicial complexes of which are isomorphic) This is equivalent to the condition that the link of every vertex be a combinatorial $(n-1)$-sphere (=boundary of the standard $n$-simplex) if the underlying manifold has no boudnary.</p> <p>However, in </p> <p>A. Ranicki (ed.), The Hauptvermutung Book, K-Monographs in Mathematics, vol. 1, Kluwer Academic Publishers, Dordrecht, Boston, London, 2010.</p> <p>on page 4, this condition is used to define the term </p> <p>combinatorial manifold (or PL manifold)''.</p> <p>I find this very weird; a PL manifold should be defined as a topological manifold with a maximal atlas of homeomorphisms with PL coordinate changes (and I know a lot of authors who use this definition).<br> The obvious question now is: Is a simplicial complex, the vertices of which have $S^{n-1}$ as link, the same as a topological manifold with a maximal atlas of homeomorphisms piecewise linear coordinate changes? Of course, Ranickis nomenclature implies that it does.</p> <p>Obviously, the condition on the links can be used to construct such an atlas. However, the converse puzzles me, as it seems to be equivalent to the question if every manifold with a maximal PL atlas admits a triangulation.</p> <p>If anyone could point me to an article where this problem is addressed, I would be thrilled. </p> <p>Best regards, Malte</p> http://mathoverflow.net/questions/128819/riemannian-manifolds-with-small-geodesics-and-bounded-curvature Comment by Malte Malte 2013-04-26T12:45:27Z 2013-04-26T12:45:27Z Thank you, Rbega. This, of course, doesn't take the length of $\gamma$ into account. (One should expect the volume of $\Omega_i$ to increase as $i_g \rightarrow 0$. http://mathoverflow.net/questions/124840/does-a-riemannian-manifold-with-bounded-geometry-admit-an-isometric-proper-embedd Comment by Malte Malte 2013-03-18T10:11:25Z 2013-03-18T10:11:25Z On the off-chance of making a fool of myself: By &quot;open&quot;, do you mean &quot;open and complete&quot;? Otherwise, gettig a proper embedding should be impossible. On the other hand, isn't proper automatically satisfied once you require completeness? http://mathoverflow.net/questions/118693/is-the-volume-functional-contiunuous-for-compact-manifolds-with-lower-bounds-on-v/118694#118694 Comment by Malte Malte 2013-01-12T14:05:22Z 2013-01-12T14:05:22Z Thank you! This means that the volume functional is continuous on the set of riemannian manifolds with upper diameter bound and lower bound on Ricci curvature. Good enough, anyway. http://mathoverflow.net/questions/110709/a-riemannian-manifold-with-finitely-many-closed-contractible-geodesics/110768#110768 Comment by Malte Malte 2012-10-27T14:18:44Z 2012-10-27T14:18:44Z I figured out the problem here. The result in Igor Rivin's post is (almost) the Ljusternik-Schnirelmann theorem, which asserts that an ellipsoid with axes very close to the round sphere has exactly three <i>simple</i> closed gedesics (namely, the intersections of the ellipsoid and the coordinate planes). Moreover, it seems that Calabi once conjectured that there are no compact mf. with finitely many closed geodesics. (This does not mean, however, that there are no non-simply-connected mf. with only finitely many contractible closed geodesics). I guess the question has no answer (yet). http://mathoverflow.net/questions/110709/a-riemannian-manifold-with-finitely-many-closed-contractible-geodesics/110768#110768 Comment by Malte Malte 2012-10-26T20:51:24Z 2012-10-26T20:51:24Z This looks/sounds convincing. However, if this were true, it would contradict the theorem I mentioned above. The theorem is due to Franks-Bangert and asserts that <i>every</i> riemannian metric on $S^2$ has infinitely many prime closed geodesics. The &quot;prime&quot; relation is the same as the geometric distinction: $c_0$ is prime if it is not a multiple of another closed geodesic. http://mathoverflow.net/questions/110709/a-riemannian-manifold-with-finitely-many-closed-contractible-geodesics Comment by Malte Malte 2012-10-26T09:17:02Z 2012-10-26T09:17:02Z They are counted up to geometric distinction, i.e. any two closed geodesics are equivalent if they have the same image in $(M,g)$. http://mathoverflow.net/questions/103087/how-many-quotients-can-a-finitely-generated-group-have-or-how-many-bundles-over-a/103093#103093 Comment by Malte Malte 2012-07-26T21:33:27Z 2012-07-26T21:33:27Z I apologise for being inconvenient. My interest in this question really comes from the example given in the question (this is why I added &quot;centre-free&quot; to the list of properties). http://mathoverflow.net/questions/103087/how-many-quotients-can-a-finitely-generated-group-have-or-how-many-bundles-over-a Comment by Malte Malte 2012-07-26T21:31:11Z 2012-07-26T21:31:11Z Hmm... could you elaborate on that, Misha? Just to clarify: I am considering surface bundles that are obtained by taking the mapping torus of some diffeomorphism of the surface. In this case, the surface is a torus and the diffeomorphism is the composition of two Dehn twists. In general, I would get an element of $GL(2,Z)$. But where is the problem with having $p,q &gt; 1$? http://mathoverflow.net/questions/103087/how-many-quotients-can-a-finitely-generated-group-have-or-how-many-bundles-over-a/103093#103093 Comment by Malte Malte 2012-07-25T16:04:12Z 2012-07-25T16:04:12Z Thank you for the answer, Mark. However, I believe that the question becomes harder if one requires that $H$ be centre-free (as is the case for $H = \pi_1 M_{pq}$, or similar families of surface bundles with fixed fibre genus). http://mathoverflow.net/questions/103087/how-many-quotients-can-a-finitely-generated-group-have-or-how-many-bundles-over-a Comment by Malte Malte 2012-07-25T10:41:14Z 2012-07-25T10:41:14Z Thank you, Henry. I implicitly used that $E$ is aspherical to obtain the short exact sequence in the third paragraph, but I have pointed it out more clearly now. http://mathoverflow.net/questions/102173/second-homotopy-group-of-graph-manifolds/102181#102181 Comment by Malte Malte 2012-07-14T09:28:28Z 2012-07-14T09:28:28Z Thank you very much, Agol! http://mathoverflow.net/questions/99506/blackbox-theorems/99511#99511 Comment by Malte Malte 2012-06-14T16:19:54Z 2012-06-14T16:19:54Z 1) This is correct. I don't see the problem, however... 1b) By &quot;easy&quot; you mean &quot;compact case&quot;? 2) True. It just popped into my mind because of this quote by Gromov, who claimed that going through the proof of NET is more than worth the time. 3)+4) There's alot more to the topic of differential topology than WET. For instance, characteristic classes would make a nice topic for a course. But of course I might have just underestimated the importance of WET. http://mathoverflow.net/questions/99488/the-ds-which-appears-in-an-integral-with-respect-to-arclength-is-not-a-1-form/99500#99500 Comment by Malte Malte 2012-06-13T20:26:20Z 2012-06-13T20:26:20Z More generally, given a regular curve, i.e. an immersion $\gamma: [a,b] \rightarrow (M,g)$, one has the pullback metric $\gamma^\ast g$. The volume of $([a,b],\gamma^\ast g)$ is the length of the curve $\gamma$, which is $\int g(\dot\gamma(s),\dot\gamma(s)) ds. However,$ds$here denotes the volume element of$[a,b]$with the standard euclidean metric from the real line. http://mathoverflow.net/questions/99488/the-ds-which-appears-in-an-integral-with-respect-to-arclength-is-not-a-1-form Comment by Malte Malte 2012-06-13T20:15:29Z 2012-06-13T20:15:29Z How do you even consider$ds$as a function of the tangent bundle? The title suggests that you mean the expression$ds$in an integral such as$\int \dot\gamma(s)ds$, right? This, however, is not &quot;the same&quot;$ds$that people sometimes use to describe the local expression of a Riemannian metric, as in$ds^2 = dx^2 + dy^2\$. Of course, if you mean the latter, then you indeed just have a family of euclidean norms, as the definition of Riemannian metric suggests. http://mathoverflow.net/questions/99340/kleins-protocols-a-window-into-our-mathematical-past Comment by Malte Malte 2012-06-12T15:21:35Z 2012-06-12T15:21:35Z This is handwritten in S&#252;tterlin. I believe zou'll probably have a hard time finding anyone below the age of 60 who can fluently read S&#252;tterlin. I don't get it though: Your aim is just a concise list of topic covered, or full translations?