User thomas kragh - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T16:11:49Z http://mathoverflow.net/feeds/user/4500 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/26461/are-the-stiefel-whitney-classes-of-the-tangent-bundle-determined-by-the-mod-2-coh Are the stiefel-Whitney classes of the tangent bundle determined by the mod 2 cohomology? Thomas Kragh 2010-05-30T15:32:47Z 2013-05-05T08:03:50Z <p>Let $G=\mathbb{Z}/2\mathbb{Z}$. Let $f\colon L \to N$ be a smooth map of connected smooth <strong>closed</strong> $n$-dimensional manifolds such that the induced map</p> <p>$f^* \colon H^*(N,G) \to H^*(L,G)$</p> <p>is an isomorphism.</p> <p><strong>Question</strong>: Are the pull back of the Stiefel-Whitney classes of the tangent bundle of $N$ the Stiefel-Whitney classes of the tangent bundle of $L$?.</p> <p>This is in fact true for the first Stiefel-Whitney class by considering coverings and degrees, but what about the higher degree classes?</p> <p><strong>Motivation</strong>: This came up because (relative) spin is important in defining Floer homology with $\mathbb{Z}$ coefficients. So I am in fact mostly interested in the following sub-question.</p> <p><strong>Question</strong>: In particular what about the second Stiefel-Whitney class in the case where both $N$ and $L$ are also assumed to be oriented? and if the answer is negative: what extra conditions do I need to make it positive?</p> <p>The idea is that I apriori have to use $G$ coefficients, but can prove that it is a $G$-cohomology equivalence, and want to use that to start the argument over again with other coefficients, but for that I need this property of the second Stiefel-Whitney class. </p> <p>This sub-question and the relation to Floer homology is related to orientations in real $K$-theory and delooping in the following sense: take a map $h\colon X \to U/O$ by delooping we get a map $\Omega h \colon \Omega X \to \Omega U/O \simeq \mathbb{Z}\times BO$ which classifies a virtual bundle over the loop space of $X$. This bundle is oriented iff the original map composed with the canonical map $U/O \to BO$ classified a virtual $0$-dimensional bundle with vanishing second Stiefel-Whitney class. This is the main point of why orientations in Floer homology is initimitely linked with spin! In the case of a Lagrangian sub-manifold $L\subset T^*N$ the difference of the tangent bundles precisely defines such a map $L \to U/O$ ($U(n)/O(n)$ classifies Lagrangians in $\mathbb{R}^{2n}$) such that the composition to $BO$ classifies the virtual bundle $TN-TL$. So in fact you may add this lifting property as an extra condition to the sub-question if you like, and then I would lose no generality. I believe that this condition implies that all the relative Prontryagin classes vanishes, which may be helpfull.</p> <p>ADDED: in light of the answer, all this motivation and these extra possible assumptions are not important nor relevant for the actual question.</p> http://mathoverflow.net/questions/52396/are-there-topological-restrictions-to-the-existence-of-almost-quaternionic-struct/52436#52436 Answer by Thomas Kragh for Are there topological restrictions to the existence of almost quaternionic structures on compact manifolds? Thomas Kragh 2011-01-18T21:33:28Z 2011-01-20T18:18:36Z <p>It seems to me that if I understood the comments to my comment correctly that the map</p> <p>$$\mathrm{Sp}(1) \times \mathrm{Sp}(n) \to \mathrm{SO}(4n)$$</p> <p>induced by right unit quarternionic multiplaction on $\mathbb{H}^n$ of the left factor and right matrix multiplication on $\mathbb{H}^n$ of the left factor has kernel $\{ \pm 1\}$. Since the source is simply connected it must lift to the spin group. So we have a map</p> <p>$$\mathrm{Sp}(1) \times \mathrm{Sp}(n) \to \mathrm{Spin}(4n)$$.</p> <p>Covering the map</p> <p>$$\mathrm{Sp}(1)\mathrm{Sp}(n) \to \mathrm{SO}(4n)$$</p> <p>Since the covering fiber is $\mathbb{Z}/2\mathbb{Z}$ and we can check that after taking the functor $B$ both fibers are $K(\mathbb{Z}/2\mathbb{Z},1)$-spaces we see that</p> <p>$$\begin{matrix} B(\mathrm{Sp}(1)\times \mathrm{Sp}(n)) &amp; \longrightarrow &amp; B(\mathrm{Spin}(4n)) \\ \downarrow &amp;&amp; \downarrow \\ B(\mathrm{Sp}(1)\mathrm{Sp}(n)) &amp; \longrightarrow &amp; B(\mathrm{SO}(4n)) \end{matrix}$$</p> <p>is homotopy cartesian.</p> <p>So if $M$ is spinable <em>and</em> has an almost Quarternionic structure it means that its classifying map lifts to $B(\mathrm{Sp}(1) \times \mathrm{Sp}(n))$</p> <p>Edit: The conclusion (which is now removed) was wrong, but at least it seems to simplify the picture when $M$ is spin.</p> <p><strong>Added</strong>: For spheres $S^{4n}$ we may use the above on the $4n$th homotopy group and deloop. This implies that if we had a quartenionic structure on $S^{4n}$ we would have that the image of the map</p> <p>$$\pi_{4n-1}(\mathrm{Sp}(1)\times \mathrm{Sp}(n) ) \to \pi_{4n-1} (\mathrm{SO}(4n))$$</p> <p>contains the image of the map $\mathbb{Z} \cong \pi_{4n-1}(\Omega S^{4n}) \to \pi_{4n-1}(\mathrm{SO}(4n)) \cong \mathbb{Z}\times \mathbb{Z}$ (*) induced by the delooping of the classifying map for the tangent bundle of $S^{4n}$.</p> <p>We know that not having <strong>an almost hypercomplex</strong> structure implies that the image of $\pi_{4n-1}(\mathrm{Sp}(n)) \to \pi_{4n-1} (\mathrm{SO}(4n))$ never contains this image, and since $\pi_{4n-1}(\mathrm{Sp}(1))$ is torsion for $n\geq 2$ the above map can not do so either for $n\geq 2$.</p> <p>(*) $\pi_{4n-1}(\mathrm{SO}(4n)) \cong \mathbb{Z}\times\mathbb{Z}$ follows WHEN $n\geq 4$ from the paper </p> <p>Barratt, M. G.; Mahowald, M. E. The metastable homotopy of O(n). Bull. Amer. Math. Soc. 70 1964 775-760.</p> <p>I think this is true in general. Indeed, it is true for $n=1$ where the above is not a contradiction because there $\pi_3(\mathrm{Sp}(1))\cong \mathbb{Z}$. Andrei pointed out in a comment that this is also true for $n=1,2$.</p> http://mathoverflow.net/questions/27367/good-reference-for-homology-of-k-mathbbz-2n/27456#27456 Answer by Thomas Kragh for Good reference for homology of $K(\mathbb{Z}, 2n)$? Thomas Kragh 2010-06-08T12:31:17Z 2010-06-08T14:29:37Z <p>I am not sure if this qualifies as a simple argument, but to me it is very nice. I did not get the reference in the comments above - so sorry if this is close to that. I dont have an exact reference since this is pieced together from several places.</p> <p>Let $A$ be any abelian group. Let $X=X(A)$ be the simplicial set defined by:</p> <p>$X_p = \{$colorings of the $n$ faces of the standard $p$ simplex by elements in $A$ such that when restricted to any $n+1$ face the sum with alternating sign of the colorings of the faces are $0\}$</p> <p>The face and degeneracy maps are defined by restriction and pull-back (the latter introduces 0's when an $n$ face is degenerate.</p> <p>with this $X_p$ is a single point when $p &lt; n$ (a single empty coloring), $X_{n}=A$, $X_{n+1}$ is "relations", and $X_p$ for $p>n+1$ is given by its restrictions to the $n+1$ faces. I.e. it is $n+(1 or 2?)$ co-skeletal. This means that its geometric realization $\lvert X \rvert$ is a $K(A,n)$.</p> <p>In fact this is a simplicial group by adding the colorings, and $X(A)\times X(A)$ is isomorphic as simplicial groups to $X(A\oplus A)$ by definition.</p> <p>To simplify we now use $A=\mathbb{Z}$ and to understand this product on the $\mathbb{Z}$ span of the simplices, I will assume some familiarity with the Eilenberg-Zilber operator, which appears naturally in the product of simplicial sets. Using this we describe the power map</p> <p>(1) $H_n(X(A),A)^{\otimes k} \to H_{kn}(X(A)\times\cdots\times X(A),A)=H_{kn}(X(A^{\oplus k}),A) \to H_{kn}(X(A),A)$</p> <p>$H_n(X(A),A)$ is a single $A$ and it is generated by the $n$ simplex $\alpha$ colored by $1$, the tensor product $\alpha^{\otimes k}$ can when mapped to the middle term of (1) be written as the sum</p> <p>$\sum_{\sigma \in S(k,n)}$ sgn$(\sigma) \beta_{\sigma}$</p> <p>where $S(k,N)$ is the permutations on $\{1,\dots,kn\}$ which preserves the order of the $k$ sequences SEQ$\strut_i=\{in+1,\dots,(i+1)n\}, i=0,k-1$ (a generalized shuffle), and $\beta_{\sigma}$ is the associated product of degenerations defined by</p> <p>(2) $\beta_{\sigma}=(\sigma_1^* \alpha)\times \cdots \times (\sigma_k^* \alpha)$,</p> <p>where $\sigma_i$ is the order preserving surjective map from $\{0,\dots,kn\}$ to $\{0,\dots,n\}$ defined by increasing one precisely when a number is in the image of $\sigma$ restricted to SEQ$\strut_i$.</p> <p>In the case of $n$ even it is clear that the sign of the permutations $\sigma$ which simply permutes the sequences SEQ$\strut_i$ (without intertwining them) has sign 1 and thus we may act on the $\sigma$'s in the sum by these with out changing the sign, this action, however, permutes the factors in (2), but when mapped to the last factor in (1), they are the same, so the image of the sum in (1) is divisible by $k!$.</p> <p>To see injectivity of the product one can see that the power map is injective on rational homology by using a Hopf-algebra argument as in section 3.C of Hatchers book.</p> http://mathoverflow.net/questions/26203/floer-homology-and-status-of-the-arnold-conjecture Floer homology and status of the Arnold conjecture. Thomas Kragh 2010-05-27T22:07:06Z 2010-06-03T19:56:32Z <p>The Arnold conjecture on a closed symplectic manifold $(M,\omega)$ says in the weakest version that for a non-degenerate Hamiltonian there are at least $k$ 1-periodic orbits where $k$ is the sum of the beti numbers of $M$. It is easy to show that one can assume that $\omega$ is integral, so I do so in the following.</p> <p>On wikipedia it says that the Arnold conjecture is solved in many cases using Floer homology. </p> <p>However, I was given the impression that this version of the Arnold conjecture has been solved in all cases, but are scattered around in several papers - due to several different complications.</p> <p><strong>Question</strong>: What is the current status of this weak Arnold conjecture precisely, and what are the refferences for these results?</p> <p>Added: I know most of the details of the monotone case, so I am mostly interested in the more exotic cases.</p> http://mathoverflow.net/questions/26823/area-enclosed-by-x4-y4-1/26828#26828 Answer by Thomas Kragh for Area enclosed by x^4 + y^4 = 1 Thomas Kragh 2010-06-02T14:54:18Z 2010-06-02T16:25:59Z <p>I think this question smells of homework, but another answer, which to me totally obscures the geometric nature of the question has been posted, and I feel that this justifies the following answer (even if the question is closed):</p> <p>The $l^p$ norms $\lvert(x,y)\rvert_p = (\lvert x\rvert^p+\lvert y \rvert^p)^{1/p}$ are norms and satisfies that if $\lvert(x,y)\rvert_p=1$ and $q>p$ then $\lvert(x,y)\rvert_q\leq 1$. So the unit "circles" of which you want to find the area grows.</p> <p>It is also a fact that $\lvert (x,y) \rvert_p \to \max (\lvert x\rvert,\lvert y\rvert)$ as $p\to \infty$. So the unit circles converges to the square which is the boundary of $[-1,1]\times [-1,1]$. This implies by monotone convergence theorem that your integral converges to 1. Because the entire square has area 4.</p> http://mathoverflow.net/questions/25866/smooth-sections-of-smooth-fiber-bundles/25885#25885 Answer by Thomas Kragh for smooth sections of smooth fiber bundles Thomas Kragh 2010-05-25T14:29:28Z 2010-05-25T14:29:28Z <p>The answer is: Yes (at least for finite dimensional manifolds).</p> <p>In fact you only need that the fiber is contractible not smoothly contractible. Take any continuous section $s_0 \colon B \to E$. cover $B$ by open sets $U_i$ such that the bundle is trivializable over each $U_i$, also make sure that the closure of each $U_i$ is compact and that the cover is locally finite.</p> <p>Furthermore, give $E$ any complete Riemannian structure. This provides each fiber $E_x$ with an induced Riemannian structure, which is also complete. We use this to define the obvious supremum distance between any two sections of $E$ over any sub-space of $B$. We may also construct a continuous map $r \colon E \to \mathbb{R}_+$ such that the ball of radius $r(x)$ and center $x$ in each fiber is geodesically convex. </p> <p>The construction now goes in 2 steps:</p> <p>1) local construction: for each $i$ we may find a smooth section $s \colon U_i \to E$ such that the supremum distance defined above is smaller than $r$ on all the points of $S_0$ restricted to $U_i$. This is easy and follows from smooth approximation of any function from $U_i \to F$ defining a section $U_i \to U_i \times F$.</p> <p>2) global construction: use a partion of unity to get a global construction. This is now possible because we were carefull enough to create the smooth local sections such that they lie in a geodesically convex neighborhood.</p> <p>While finishing this Andrew Stacey posted a similar answer, but it seemed a waste not to poste this also. Especially since we are focussing on different details.</p> http://mathoverflow.net/questions/25568/real-interpretations-of-discontinuities-in-floer-homology Real interpretations of Discontinuities in Floer homology Thomas Kragh 2010-05-22T10:53:18Z 2010-05-23T10:08:52Z <p>This question is motivated by the answer in <a href="http://mathoverflow.net/questions/25442/hamiltonian-displaceability-of-tori-in-symplectic-balls" rel="nofollow">this</a> question (you dont have to read it to understand the following).</p> <p>I am not that proficient in calculating Floer homology, and I held back on answering and just commenting on the question because of the following.</p> <p>Let $L$ be the equator on $S^2$ then it is obvious that no "Hamiltonian" symplectomorphism can take $L$ to another Lagrangian not intersecting $L$. Indeed, $L$ cuts $S^2$ into two equally sized pieces and it is not difficult from there.</p> <p>It is quite a different thing to assert that the intersection FLoer homology $FH(L,L)$ is non-trivial, which is used in the solution to the problem (together with a kunneth formula).</p> <p>The thing that initially bugged me about this was that arbitrarily close to $L$ there are Langrangians whose Floer homology intersection is $0$ both with $L$ and themselves. I have later come to realize that I myself when considering exact Lagrangians in cotangent bundles have encountered this discontinuity frequently, and should really not be surprised. Here the exactness is very important for non-triviallity. However, for me this comes up in a different way because I am considering finite reductions of the loop spaces using Chaperons broken geodesics (as Viterbo does).</p> <p>I, however, realized that the discontinuity does not seem that bad and this leads me to my question:</p> <p><strong>Question</strong>: Is there a natural Floer homology with real coefficients, which resolves these discontinuities in the sense that the differentials are continuous.</p> <p>A little more motivation: In the above case it seems that with coefficients in the novikov ring $\mathbb{Z}[t]$ we can realize the Floer complex with two generators, and then for all Lagrangians not dividing $S^2$ into two equally volumed parts the differential is an isomorphism. But if the Lagrangians are both of that particular type the differential is $0$. This could be described continuously with real coefficients as:</p> <p>$d \colon \mathbb{R}[t] \to \mathbb{R}[t]$</p> <p>where $d$ is the differential from the $\mathbb{Z}[t]$ coefficient complex multiplied by the scalar $(A_1-A_2)^2+(A_1'-A_2')^2$ where $A_i$ are the two volumes for the first Langrangian and $A_i'$ are the volumes for the second Lagrangian.</p> http://mathoverflow.net/questions/24378/equality-vs-isomorphism-vs-specific-isomorphism/24439#24439 Answer by Thomas Kragh for Equality vs. isomorphism vs. specific isomorphism Thomas Kragh 2010-05-12T21:38:07Z 2010-05-12T21:38:07Z <p>How $X \times Y$ is isomorphic to $Y \times X$ in a category with products should not be ignored. To illuminate this think a little more general:</p> <p>Keeping track of isomorphisms are essential in symmetric monoidal categories:</p> <p>In a monoidal category you have a product $\oplus$ and coherency isomorphisms relating the associativity of the product $c_{A,B,C} \colon A\oplus (B \oplus C) \to (A \oplus B) \oplus C$ satisfying the pentagon relation. You also have a unit (zero) object and other isomorphisms satisfying some relations.</p> <p>There is a theorem saying that one can always replace this by an equivalent monoidal category where all these isomorphisms are identities (including the ones for the unit making the unit strict). So really this tells you that you dont need these isomorphisms to define them up to equivalence.</p> <p>However, for symmetric monoidal categories, which also have a "commutativity" isomorphism you cannot always do this, and so it is essential to keep track of the isomorphisms to ensure full generality.</p> <p>This is ture also for categorical products and coproducts, and more generally for compatibility betweem objects defined by different universal properties in a category. E.g. the eaxmple $X \times Y$ is isomorphic to $Y\times X$ mentioned at the beginning. This is more than keeping track of the automorphisms of objects (which is also very important, but mentioned in other answers) - it is remembering certain important isomorphisms depending on some objects.</p> http://mathoverflow.net/questions/24229/maps-of-loop-spaces-with-infinity-bounded-differential Maps of loop spaces with infinity-bounded differential. Thomas Kragh 2010-05-11T09:58:29Z 2010-05-11T12:56:38Z <p>I am currently working with loop spaces of manifold and finite dimensional manifolds approximating these and the following comes up very naturally:</p> <p>In the following piece-wise smooth means smooth on each set of a closed covering, implying continuity and boundedness of 1-sided derivatives.</p> <p>For any closed Riemannian manifold $M$ define $\Lambda M$ as the space of piece-wise smooth maps from $S^1=I/\{0,1\}$ to M. define the energy of $\gamma \in \Lambda M$ by</p> <p>$E(\gamma)=\int_{S^1} \mid\mid \gamma'(t)\mid\mid^2dt$</p> <p>This follows Milnors book on Morse theory and the tangent space of $\Lambda M$ at $\gamma$ is defined to be piece-wise smooth tangent fields along $\gamma$ (WARNING: see comments by Andrew Stacey). Remark: with this definition $\gamma'$ may not be a tangent vector since it can be discontinuous. We may define the supremums norm on the tangent space by</p> <p>$\mid\mid {\partial \gamma} \mid\mid_\infty = \sup_{t\in S^1} \mid\mid {\partial \gamma(s)}\mid\mid$</p> <p>for any $\partial \gamma \in T\Lambda M$.</p> <p>I define a <strong>bounded differentiable</strong> function $F\colon \Lambda M \to \Lambda N$ by the following criteria:</p> <p>$E(F(\gamma)) \leq C_F E(\gamma)$ and $\mid\mid F_*(\partial \gamma)\mid\mid_\infty \leq C_F\mid \mid\partial \gamma\mid\mid_\infty$</p> <p>for some $C_F>0$. Here $F_*$ is assumed to be well-defined using variations. Since I assume that $M$ and $N$ are closed this constant may depend on the Riemannian structures, but the notion does not. These arise e.g. as loops of differentiable maps $f\colon M\to N$, but I need them in their generality.</p> <p><strong>Question:</strong> Has anybody seen this notion of boundedness or maybe a similar local definition used anywhere?</p> <p><strong>Motivation:</strong> To begin with I felt this was an unnatural mix of $L^2$ and $L^\infty$, but working with these on the following spaces have made them feel much more natural: Define $\Lambda^\beta M$ as the space of loops with energy less than $\beta$, and define $\Lambda_r^\beta M$ as the space of piece-wise geodesics each piece parametrized by an interval of length $1/r$ with total energy less than $\beta$. If $\beta/r$ is small enough then this is a manifold given by the endpoints of the geodesics (see Milnors book on Morse theory). The above conditions are very suited for transfering arguments back and forth between $\Lambda^\beta M$ and $\Lambda_r^\beta M$ since the supremum norm is compatible with evaluations at points, but also the inclusions $\Lambda_r^\beta M \to \Lambda M$ for small $\beta/r$ is compatible. I need a lot of lemmas regarding these (e.g. existence of homotopy through such maps, when one have a continuous homotopy between two such which stayes constant outside a set of compact homotopy type) and if some one has already worked on this it can help me greatly.</p> <p>One could ask why use the energy in the first place why not define $\Lambda_r^\beta M$ using length. The reasons is that the energy is more natural to use in the setting I am looking at, which is often the case since the energy is modelled on $L^2$ which is the nicest $L^p$ space.</p> http://mathoverflow.net/questions/23710/does-this-approach-for-the-poincare-conjecture-work/23842#23842 Answer by Thomas Kragh for Does this approach for the Poincare conjecture work? Thomas Kragh 2010-05-07T10:33:28Z 2010-05-07T10:33:28Z <p>I also Had a quick look (maybe a little less quick), and although I very much like the other answer, which illustrates that it may be difficult to fix, I may have found a more specific error, which may be more helpfull as an answer to the question.</p> <p>Firstly, I am a little confused as to what constitutes a stratification. I see two possibilities:</p> <p>1) The one which is actually defined which allows the following stratification: $S_1=S_2=D^2\times [0,1]$ and they are glued along a closed disc in the <strong>interior</strong> of $D^2\times {1}$ of $S_1$ and the same disc in the <strong>interior</strong> of $D^2 \times {0}$ in $S_2$.</p> <p>2) The one which I think is implied at some points: $S_i$ and $S_{i+1}$ may only be identified such that $U(S_i) \cup L(S_{i+1})$ is in fact a sub-surface in the 3-manifold.</p> <p>I will describe my problems related to both definitions:</p> <p>In the proof of prop 5.8 parts (2-3-4) he attaches "3-cell"s (I would write 3-disc as to avoid confusion with CW complex attachments of cells, or attach both a 2-cell and a 3-cell) $W$, and extends the stratification.</p> <p>If we work under definition 2) above then this seems generally impossible because you would often also have to attach it at the top of $S_{i+1}$ to get the extra surface assumption in 2).</p> <p>If we work under definition 1) above then this doesn't even make the new $F_{i+1}$ a surface in the simple example described above.</p> http://mathoverflow.net/questions/22860/do-rational-numbers-admit-a-categorification-which-respects-the-following-dualit/23112#23112 Answer by Thomas Kragh for Do rational numbers admit a categorification which respects the following "duality"? Thomas Kragh 2010-04-30T14:35:45Z 2010-04-30T14:35:45Z <p>Another way of categorifying $\mathbb{Q}_+$ is as follows:</p> <p>Let $\mathcal{F}$ be the category of finite non-empty sets and bijections. This is a symmetric monoidal category with respect to $\times$. Let $\mathbb{N}$ be viewed as a symmetric monoidal category, such that the objects are $\mathbb{N}$ and we only have identity morphism. The product is given by the usual product $\cdot$ in $\mathbb{N}$.</p> <p>Then there is an obvious symmetric monoidal functor $G \colon \mathcal{F} \to \mathbb{N}$, which simply counts the elements in the set. Taking Grayson Quillen construction on both sides, we get a symmetric monoidal functor:</p> <p>$G' \colon \mathcal{F}^{-1}\mathcal{F} \to \mathbb{Q}_+$,</p> <p>whic may be viewed as a categorification (note that this is also the $\pi_0$ functor). The Grayson Quillen construction may be thought of as a categorified version of the Grothendeick construction. Objects in $\mathcal{F}^{-1}\mathcal{F}$ are pairs of objects in $\mathcal{F}$ which I suggestively write as $A/B$</p> <p>If we only use the skeletal category of $\mathcal{F}$ given by objects $s_n=\{1,\dots,n\}$ for each $n\in \mathbb{N}$ we get that $G$ is an isomorphism on the set of objects. This is not true for $G'$, because we have implicitly divided out by isomorphisms in the image (i.e. $p/q=np/(nq)$ in $\mathbb{Q}_+$ not just isomorphic objects as they are in the Grayson Quillen construction on $\mathbb{N}$). We cannot do this on the domain of $G'$ because $s_p/s_q$ is NOT isomorphic to $s_{np}/s_{nq}$. However, there are many maps from $s_p/s_q$ to $s_{np}/s_{nq}$, which identifies them when taking $\pi_0$.</p> <p>I should note that I believe (but am not sure) that this can be made <strong>bi</strong>-monoidal by taking the category of finite sets and bijections as a bimonoidal category and localizing with respect to $\times$ over the full subcategory of finite and non-empty sets.</p> http://mathoverflow.net/questions/22567/diameter-of-a-circle-in-an-embedded-riemannian-manifold/22589#22589 Answer by Thomas Kragh for Diameter of a circle in an embedded Riemannian manifold Thomas Kragh 2010-04-26T11:44:23Z 2010-04-26T12:35:52Z <p>The answer to the first is NO.</p> <p>As mentioned by another answer you can use Do Carmo to prove a local result of the type: at points where the normal curvature is non-zero the 0-directions defines a foliation of straight lines. </p> <p>With this one can prove that embedding $\mathbb{R}²$ into $\mathbb{R}^3$ preserves the diameter of circles, but one can construct a counter example in the case of a piece of paper:</p> <p>Take a equilateral triangle on your table with points barely out side of the piece of paper. Then since these sides do not meet inside the piece you may fold (using very sharp foldings) the piece of paper up at the sides of the triangle such that out side of a nighborhood of the triangle the paper lies in planes which are perpendicular to the table. Note that this can not be extended to all of $\mathbb{R}^2$ because the folding lines interset! Then shrinking the circumscribed sphere slightly of the triangle to lie in the paper - this is folded up so that the diameter becomes strictly less.</p> <p>For the second question I really dont see $\pi_1$ any where. What I see is the following: for me it is natural to defined the distance on a Riemmanian manifold $X$ as:</p> <p>$d(x,y) = \inf_{\gamma} \textrm{len}(\gamma)$</p> <p>where the infimum is over curves $\gamma$ from $x$ to $y$, and len($\gamma$) is the length of the curve. This is <strong>intrinsic</strong> and gives the same diameters for the piece of paper, but when you remove an open set it becomes different, because the curves has to avoid this set.</p> <p>Since isometric embeddings preserve lengths of curves you get with this definition</p> <p>$d(Fx,Fy) \leq d(x,y)$</p> http://mathoverflow.net/questions/22327/partition-of-r-into-midpoint-convex-sets/22338#22338 Answer by Thomas Kragh for Partition of R into midpoint convex sets Thomas Kragh 2010-04-23T12:18:43Z 2010-04-23T12:18:43Z <p>Yes if you assume AC:</p> <p>With AC let $\{v_\alpha\}$ be a $\mathbb{Q}$-basis for $\mathbb{R}$ then the following two sets satisfies your property:</p> <p>$A = \{q_1v_{\alpha_1}+\cdots+q_nv_{\alpha_n} \mid q_i \in \mathbb{Q} , \sum q_i \geq 0 \}$</p> <p>and </p> <p>$B = \{q_1v_{\alpha_1}+\cdots+q_nv_{\alpha_n} \mid q_i \in \mathbb{Q} , \sum q_i &lt; 0 \}$</p> <p>So in fact these are $\mathbb{Q}$ convex (in the obvious sense).</p> http://mathoverflow.net/questions/22036/pullbacks-in-category-of-sets-and-partial-functions/22046#22046 Answer by Thomas Kragh for Pullbacks in Category of Sets and Partial Functions Thomas Kragh 2010-04-21T12:14:57Z 2010-04-21T12:41:41Z <p>Pullbacks exists but are not what you describe.</p> <p>The answer is as follows:</p> <p>The category $\mathcal{C}$ of sets and partial functions is equivalent to the category of based sets and based functions, by sending the set $A$ to $A$ disjoint union a base point $*$ and sending $f$ to the obvious based function which sends everything on which $f$ was not defined to the base-point.</p> <p>The pullback in based sets are well-known and for example the product $\times$ in based sets translates back through this equivalence to $\mathcal{C}$ and becomes:</p> <p>$A \times_\mathcal{C} B \approx (A \times_{\textrm{set}} B) \sqcup_{\textrm{set}} A \sqcup_{\textrm{set}} B$</p> <p>From the purely sets and partial function point of view this is also explainable. Indeed, any morphism from $Z$ to this product is given by a choice for each point in $Z$ of either: a point in $A$ and a point in $B$, or a point in $A$, or a point $B$, or nothing.</p> http://mathoverflow.net/questions/20712/do-surjective-degree-zero-maps-exist Do "surjective" degree zero maps exist? Thomas Kragh 2010-04-08T12:11:50Z 2010-04-09T01:25:25Z <p>Is there a map $f\colon X \to Y$ of closed, connected, smooth and orientable $n$-dimensional manifolds such that the degree of $f$ is 0 but $f$ is not <strong>homotopic</strong> to a non-surjective map?</p> <p><strong>Added</strong>: The motivation is: There is a "mild version" of the Nearby Langrangian conjecture stating: any exact Lagrangian manifold $X \to T^*Y$ has non-zero degree when composed with the projection $T^*Y \to Y$. It is known that the map is always surjective. I am looking at a <strong>possible</strong> inbetween stating that the map cannot be homotoped to a non-surjective map.</p> http://mathoverflow.net/questions/20144/if-the-second-derivative-of-a-function-on-mathbb-rn-is-everywhere-nondegenera/20223#20223 Answer by Thomas Kragh for If the second derivative of a function on $\mathbb R^n$ is everywhere nondegenerate, does it follow that the first derivative is an injection? Thomas Kragh 2010-04-03T10:16:57Z 2010-04-03T10:16:57Z <p>The counter example given in the comments by Brian Conrad (and Dylan Thurston) is very nice. However, it oscillates wildly at $\infty$, and I believe that if you assume some nice properties at $\infty$ you will get a possitive answer.</p> <p>Translating $f$ by a linear function does not change the assumptions on $f$ and so the claim is equivalent to the fact that all such $f$ has no or a unique critical points.</p> <p>If we add assumptions such that e.g. the Conley index (or homotopy index) is well-defined and a sphere then this modified claim would follow. Indeed, if two or more critical points existed they would by assumptions be non-degenerate with same Morse index and a small pertubation would yield a Conley index which is a vedge of two or more spheres - a contradiction.</p> http://mathoverflow.net/questions/19929/is-there-a-general-statement-about-structures-on-spheres-relating-to-division-alg is there a general statement about structures on spheres relating to division algebras? Thomas Kragh 2010-03-31T10:22:59Z 2010-03-31T16:48:48Z <p>It is classical to take a division algebra over $\mathbb{R}$ and defining an H-space structure on the unit spheres by restricting and normalizing.</p> <p>There are commutative division algebras of dimension 1 and 2 leading to commutative products on $S^0$ and $S^1$ identifying them as Eilenberg-MacLane spaces - Or if we forget some structure as an $E_{\infty}$-spaces.</p> <p>The associative division algebras $\mathbb{H}$ defines an associative product on $S^3$, which is also a Lie-group, but forgetting some structure it is an $A_\infty$-space.</p> <p>There division algebra $\mathbb{O}$ defines an $A_2$ structure on $S^7$, which is not $A_\infty$ (is it $A_3$?).</p> <p>As is well known it is possible to prove that no other spheres has $A_2$ structure.</p> <p><strong>Question:</strong> Is there a heiraki of structures below $A_2$ yet related such that $S^{15}$ has this structure, but $S^{31}$ does not?</p> <p>Remark: A heiraki below $A_2$ could be that $A_2=D_\infty$ for some definition of structures $D_n$, analagous to $E_1$ being $A_\infty$. </p> <p><strong>Question:</strong> Is there an even more general definition of "lower" structures and a statement about all spheres (including possibly non-trivial structures on even-dimensional spheres)?</p> http://mathoverflow.net/questions/19822/a-problem-concerning-two-symmetric-matrices/19825#19825 Answer by Thomas Kragh for A problem concerning two symmetric matrices Thomas Kragh 2010-03-30T12:53:25Z 2010-03-30T12:53:25Z <p>Counter example:</p> <p>take the diagonal matrices:</p> <p>$A=D(2,1,1,0,0)$ and $B=D(0,0,1,2,0)$</p> <p>If you want the multiplicity to match (thinking of $X \cup Y$ is a "multiset" union) then it is easy to create an inductive argument proving the assertion.</p> http://mathoverflow.net/questions/19644/what-is-the-definition-of-canonical/19724#19724 Answer by Thomas Kragh for What is the definition of "canonical" ? Thomas Kragh 2010-03-29T13:07:39Z 2010-03-29T13:07:39Z <p>Vague definition of canonical:</p> <p>Let $X$ and $Y$ be collections (often sets) for which assumptions has been made (has been given structures and/or are related somehow). A function $f\colon X \to Y$ is canonical if it is given by a rule using only the already given structure.</p> <p>This explains the relation to the greek word rule (kanon). The precise meaning of the above are open for enterpretation: how much structure can the rule itself contain (maybe this can be made precise)! This "definition" somewhat contradicts many of the other answers, which for some reason is under the impression that canonical implies unique (or almost unique), which in my point of view is very wrong since different rules may define different maps. E.g. if we let $X$ be the objects in the category of abelian groups and $Y$ the morphisms then the definitions makes all the group homomorphisms $A \to A$ given by multiplication with an element in $\mathbb{Z}$ canonical, which to me is not a problem.</p> <p>Usually when there is an especially simple rule it is often assumed without mentioning that this is the rule defining the function. E.g. most will understand the following:"there is a canonical endemorphism of any object in a category". This emphasizes the multiplication with 1 above as somehow speciel or "more canonical" than the rest. This is simply because the rule works in much greater generality and is shorter.</p> <p>Usually if a rule is very simple the function will have nice properties. E.g. simply rules in category theory often define functors, natural transformations, e.t.c. This leeds many people to confuse the notion of canonical with "something behaving nicely".</p> <p>I am somewhat puzzled by the use of the word uniform in one of the answers. The nature of the word uniform is "of the same form" and relates more to symmetries and things looking the same every where. This often leeds to canonical maps, since a choice at one point can sometimes be extended to a choice at every point. Please someone comment on this since maybe this is just a use of the word I have not seen before!</p> http://mathoverflow.net/questions/19328/what-is-enough-to-conclude-that-something-is-a-cw-complex-part-ii What is enough to conclude that something is a CW complex (part II)? Thomas Kragh 2010-03-25T17:09:09Z 2010-03-25T18:38:48Z <p>A while ago I asked a <a href="http://mathoverflow.net/questions/18505/what-is-enough-to-conclude-that-something-is-a-cw-complex" rel="nofollow">question about recoqnizing CW complexes</a> and got an extremely nice and concrete answer. However, I am still interested in a more general treatment of this and therefore pose the following closely related question:</p> <p>Assume that $X$ is an $n−1$ dimensional finite CW complex. and assume that $X'$ is given as a a set by the disjoint union of $X$ and a single open cell $e$ of dimension $n$. I.e. $e$ is an open subspace of $X'$ homeomorphic to the open $n$ disc (and of course $X$ is homeomorphic to the complement). Also assume that $X'$ is compact Hausdorff.</p> <p><strong>Question</strong>: Are there some natural topological conditions to further put on $X'$ such that it follows that $X'$ is in fact a CW-complex with $X$ as a sub-CW complex and $e$ a single new cell?</p> <p>Remark: The fact that $X$ is such is equivalent to whether or not one can modify the homeomorphism of $e$ to the open unit disc such that the inverse extends to the closed unit disc.</p> <p>Ideas on answers (but I have no proofs or counter examples in any of the cases):</p> <p>1) $X'$ is locally contractible.</p> <p>2) $X'$ has the homotopy type of a CW complex.</p> <p>3) The Combination of 1) and 2).</p> <p>4) $X'$ is homeomorphic to a CW complex</p> <p>The last is weird, but it is not clear to me that even this is enough!</p> <p>Any ideas, counter examples, or references?</p> http://mathoverflow.net/questions/19303/good-example-of-a-non-continuous-function-all-of-whose-partial-derivatives-exist/19304#19304 Answer by Thomas Kragh for Good example of a non-continuous function all of whose partial derivatives exist Thomas Kragh 2010-03-25T13:45:41Z 2010-03-25T13:45:41Z <p>The standard example I have seen is: $f(x,y)=\frac{2xy}{x^2+y^2}$.</p> http://mathoverflow.net/questions/18701/existence-of-fermi-coordinates-on-a-riemannian-manifold/18727#18727 Answer by Thomas Kragh for Existence of Fermi coordinates on a Riemannian manifold Thomas Kragh 2010-03-19T11:36:00Z 2010-03-19T11:36:00Z <p>It is not enough to be within the injective radius. Let $(M,g)$ be an $(n+1)$-dimensional manifold and $\gamma\colon (-a,a) \to M$ a geodesic. One way to define Fermi coordinates</p> <p>$\phi \colon (-a,a)\times D^n_{\epsilon} \to M$</p> <p>is by</p> <p>$\phi(t,x_1,\dots,x_n)$ = exp$\strut_{\gamma(t)}(x_1v_1(t)+\cdots + x_nv_n(t)$,</p> <p>where $D^n_{\epsilon}$ is the open disc with radius $\epsilon$ and $v_i(s)$ is the $i^{\textrm{th}}$ vector of an othonormal frame of the orthogonal complement of $\gamma'(t) \in T_{\gamma(t)} M$, which is given by parallel transporting $v_i(0)$ back and forth to all of $\gamma$.</p> <p>If we let $\delta>0$ be so small that any geodesic triangle with all sides less than $\delta$ cannot have two right angles (this exists locally, and may have to be strictly smaller than the injective radius) then if $2a$ and $\epsilon$ is smaller than this $\delta$ then $\phi$ will be injective and in fact a diffeomorphisms onto its image. To prove injectivity assume it is not then you get a geodesic triangle with two right angles - contradiction.</p> <p>To the concrete question: The answer is NO. Knowing bounds at the point $p$ is never enough to get even an injective radius depending on these bounds, and certainly not this $\delta$.</p> <p>HOWEVER: If you know concrete bounds on all of $M$ or within a concrete radius of $p$ then this is possible. I.e. you $f(h)$ does not exist, but if you assume the bounds in a radius depending also on $h$ it does.</p> http://mathoverflow.net/questions/18505/what-is-enough-to-conclude-that-something-is-a-cw-complex What is enough to conclude that something is a CW complex? Thomas Kragh 2010-03-17T16:55:19Z 2010-03-17T19:41:36Z <p>This question was something I considered when looking into CW-structures on Grassmannians, but I found no general treatment of this in the literature:</p> <p>Question: Assume that $X$ is an $n-1$ dimensional finite CW complex. and assume that $X'$ is given as a a set by the disjoint union of $X$ and a single open cell $e$ of dimension $n$. I.e. $e$ is an open subspace of $X'$ homeomorphic to the open $n$ disc (and of course $X$ is homeomorphic to the complement). Assume also that $X'$ is compact Hausdorff. Is $X'$ homeomorphic to a CW complex given by attaching a single $n$ cell to $X$?</p> <p>Remark: Maybe I am missing some obvious counter example!</p> http://mathoverflow.net/questions/18464/what-are-the-correct-axioms-for-a-weakly-associative-monoidal-functor/18481#18481 Answer by Thomas Kragh for What are the correct axioms for a "weakly associative monoidal functor"? Thomas Kragh 2010-03-17T11:48:01Z 2010-03-17T11:48:01Z <p>Firstly: A functor that messes up the associator but preserves product is NOT strictly monoidal, and from the following discussing I believe that in general it cannot be rectified into one that is.</p> <p>This is not an idea on weakning the assumptions of being an associator, and is thus not a direct answer to the question, but it is an analogy putting the question into another context, which to me at least clarifies it.</p> <p>In my point of view a very good way of looking at these things are on the classifying space of the categories.</p> <p>Any monoidal structure defines an $A^\infty$ product on the classifying space (i.e. the type of product you get on a loop space, but without inverse).</p> <p>Any lax-monoidal functor defines a map of $A^\infty$ spaces, i.e. it preserves the product and the underlying homotopy associativity structure both up to homotopy. The same is true for a strong monoidal functor. A strict moniodal functor preserves the product and the structure on the nose.</p> <p>The sort of functor you are talking about preserves the product but not the underlying homotopy associativity structure, and this makes it a little weird.</p> <p>I have no concrete example for associativity, but I do have one for commutativity. I.e. symmetric monoidal functors and $E^\infty$ structures, and I believe this can provide an example for associativity as well if it is delooped a couple of times.</p> <p>Ex: Let $C$ be the topological groupoid with objects $\mathbb{Z}$ and morfisms only automorphisms aut$(n) = S^1$ as a topological abelian group (meaning that composition is the usual product). The monoidal product $\oplus$ is the following:</p> <p>on objects: $n\oplus m = n+m$</p> <p>on morphisms: $a\oplus b = a\cdot b$</p> <p>This is somewhat trivial and $B C$ is just the product $\mathbb{Z} \times K(\mathbb{Z},2)$ in the category of $E^\infty$ spaces.</p> <p>However, I have said nothing about associators or symmetry isomorphisms because the product is both associative and commutative, BUT we can in fact give it a non-trivial such structure:</p> <p>Let $C'$ be the same category with the same monoidal product but the natural coherency isomorphisms are not all assumed to be the identity. indeed, the symmetry</p> <p>$\gamma_{n,m}^{\oplus} \colon n\oplus m \to m\oplus n$</p> <p>are defined to be $\gamma_{n,m}^{\oplus} = (-1)^{nm}$, but all others are identities.</p> <p>One can check that this defines a symmetric monoidal category and the map $C \to C'$ given by the identity is a product preserving functor, but it is NOT \emph{symmetric} monoidal. Furthermore, in the category of $E^\infty$ spaces $BC'$ is not the trivial product $BC$ and delooping a number of times (3 I think) produces a space which is not a product of Eilenberg Maclane spaces. So no symmetric monoidal equaivalence exists between $C$ and $C'$.</p> <p>In fact one can prove that there are precisely two extensions in the category of $E^\infty$ spaces and $E^\infty$ maps of $K(\mathbb{Z},n)$ and $K(\mathbb{Z},n+2)$ (the third homology group of the Eilenberg maclane spectrum $H\mathbb{Z}$ is $\mathbb{Z}/2\mathbb{Z}$), and one can check that these two examples are precisely those two.</p> <p>I am not sure if this helps with the concrete example of which I am not that familiar. My idea was that maybe there is an obstruction to getting something more than product preserving (i.e. the $A^\infty$ structure may be different).</p> http://mathoverflow.net/questions/18319/can-a-continuous-nowhere-differentiable-function-have-specified-shape-at-every/18345#18345 Answer by Thomas Kragh for Can a continuous, nowhere differentiable function have specified "shape" at every point? Thomas Kragh 2010-03-16T09:02:09Z 2010-03-16T09:15:58Z <p>Assume WLOG that $\phi(x)>0$ when $x>0$. Since the limit described exists for all $x$ in the source of $f$. We get for any $x$ the bound:</p> <p>$f(x+\delta)-f(x) \leq C\phi(\delta)$</p> <p>for $0 &lt; \delta &lt; \delta_0$ for some $C,\delta_0>0$ which may depend on $x$.</p> <p>diving by $\delta$ we get by the assumptions on $\phi$ that</p> <p>$\underline{\lim}_{\delta \to 0} ( \frac{f(x+\delta) - f(x)}{\delta}) \leq 0$</p> <p>This is one the four derivatives of $f$, and proposition 2 chapter 5 in Real Analysis by H.L. Royden states that if $f$ is continuous then it is (non-strictly) decreasing. Similar for increasing. So $f$ is constant.</p> http://mathoverflow.net/questions/127303/what-is-known-about-the-strong-arnold-conjecture Comment by Thomas Kragh Thomas Kragh 2013-05-25T13:24:17Z 2013-05-25T13:24:17Z Hi Nate. The reference you give seems (to me) to prove the cup length conjecture in those case. Ie that the number of fixed points is greater than or equal to the cup length (or something stronger). Does it mention anything about Morse functions? Or does that somehow follow from that? http://mathoverflow.net/questions/100612/contractible-open-sets Comment by Thomas Kragh Thomas Kragh 2012-06-25T18:21:48Z 2012-06-25T18:21:48Z @Mariano: You can make a spiral construction in $\mathbb{R}^2$ such that that the inverse of a value (say 1) is the unit circle and some curves in the unit disc spiraling out towards the unit circle, but never getting there. You can put the unique minimum at 0 taking the value 0. The inverse image of 1 would be connected but not path connected. http://mathoverflow.net/questions/95151/from-microlocal-to-local Comment by Thomas Kragh Thomas Kragh 2012-04-25T20:37:11Z 2012-04-25T20:37:11Z The integral does not seem to depend on $u$? Are you missing a $u(y)$? http://mathoverflow.net/questions/89798/is-s2-x-s4-a-complex-manifold Comment by Thomas Kragh Thomas Kragh 2012-02-29T19:46:07Z 2012-02-29T19:46:07Z I thought this question was resolved for $S^6$ a while ago (after the question you link to was posed and answered)? Am I wrong? http://mathoverflow.net/questions/18319/can-a-continuous-nowhere-differentiable-function-have-specified-shape-at-every/18345#18345 Comment by Thomas Kragh Thomas Kragh 2011-05-27T15:12:11Z 2011-05-27T15:12:11Z Wildly oscilating refers to $\phi(x)/x$ being both close to zero and unbounded for very small $x$. The assumption WLOG refers to the fact that the limit would not exist if $\phi$ were not strictly positive or negative in a small $(0,\epsilon)$ (assuming continuity). http://mathoverflow.net/questions/55297/codimension-of-measurable-sets/55316#55316 Comment by Thomas Kragh Thomas Kragh 2011-02-13T17:42:57Z 2011-02-13T17:42:57Z @Simon: You didn't fix it you still have the two equivalent sets Jim described. You should rather look at all subsets $J \subset I$ such that $i\notin J$ for some fixed $i\in I$. http://mathoverflow.net/questions/51367/are-there-cohomology-classes-on-a-hyperkahler-manifolds-which-pull-back-to-the-st Comment by Thomas Kragh Thomas Kragh 2011-02-05T05:13:33Z 2011-02-05T05:13:33Z In fact this is Bens calculation for the other Lagrangian the zero section $\mathbb{R}^N$. http://mathoverflow.net/questions/51367/are-there-cohomology-classes-on-a-hyperkahler-manifolds-which-pull-back-to-the-st Comment by Thomas Kragh Thomas Kragh 2011-02-04T20:31:28Z 2011-02-04T20:31:28Z @Eric: The tangent bundle of the ENTIRE conormal bundle (which is the actual Lagrangian in question) is trivial. So this is not a counter example to the original question. http://mathoverflow.net/questions/52396/are-there-topological-restrictions-to-the-existence-of-almost-quaternionic-struct/52436#52436 Comment by Thomas Kragh Thomas Kragh 2011-01-21T16:44:18Z 2011-01-21T16:44:18Z I realized that non-surjectivity of the map on $\pi_{4n}$ is detected by the Euler class. That is - the Euler class evaluates to 0 on the image. So any manifold with non-trivial euler class restricted to $\pi_{4n}$ cannot have a quaternionic structure for $n\geq 2$. This of course does not help with $\mathbb{C}P^{2n}$. http://mathoverflow.net/questions/52396/are-there-topological-restrictions-to-the-existence-of-almost-quaternionic-struct/52436#52436 Comment by Thomas Kragh Thomas Kragh 2011-01-19T00:28:25Z 2011-01-19T00:28:25Z I put this into the answer, and also added the extra stable $\math{Z}$ in $\pi_{4n-1}(SO(4n))$ which I had forgotten (I replaced $\mathbb{Z}$ with $\mathbb{Z}\times\mathbb{Z}$. Sorry for the numerous edits. http://mathoverflow.net/questions/52396/are-there-topological-restrictions-to-the-existence-of-almost-quaternionic-struct/52436#52436 Comment by Thomas Kragh Thomas Kragh 2011-01-18T23:44:21Z 2011-01-18T23:44:21Z Ups - My reference for $\pi_{4n-1}(SO(4n))$ only works for $n\geq 4$. So have not settled $S^8$ and $S^12$. http://mathoverflow.net/questions/52396/are-there-topological-restrictions-to-the-existence-of-almost-quaternionic-struct/52436#52436 Comment by Thomas Kragh Thomas Kragh 2011-01-18T23:20:27Z 2011-01-18T23:20:27Z Hope my addition makes the point clear - and that there are no errors :). http://mathoverflow.net/questions/52396/are-there-topological-restrictions-to-the-existence-of-almost-quaternionic-struct Comment by Thomas Kragh Thomas Kragh 2011-01-18T19:20:55Z 2011-01-18T19:20:55Z What do you mean by $Sp(1)Sp(n)$? since $Sp(1)$ is a sub-Lie-group of $Sp(n)$ this is with the obvious definition simply $Sp(n)$ gain. http://mathoverflow.net/questions/27367/good-reference-for-homology-of-k-mathbbz-2n/27515#27515 Comment by Thomas Kragh Thomas Kragh 2010-06-09T06:29:06Z 2010-06-09T06:29:06Z I really meant mine on the chain level - i write that $\alpha$ is a simplex generating (meant to say a simplex representing a generator). http://mathoverflow.net/questions/27367/good-reference-for-homology-of-k-mathbbz-2n Comment by Thomas Kragh Thomas Kragh 2010-06-08T11:14:32Z 2010-06-08T11:14:32Z @algori: Do you have a more precise reference?