# Tag Info

7

Yes, for any group $G$ there exists a (path connected) $K(G,1)$, and there are non-trivial groups with trivial abelianization.

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I think this answers your question, but I'm not entirely sure. Hopefully some bits of it are useful to you--let me know! I think, first and foremost, you have a bit of confusion regarding the two uses of the word 'differential' here. On one hand, when you speak of the differentials as $g=\dim H^0(X,\Omega^1_{X/\mathbb{C}})$ you're speaking of algebraic ...

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Regarding question 1, even two non-self-intersecting curves that enclose the same sets of points with winding number 1 might not be homotopic:

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Without non-self-intersection, the first statement is already false for the plane minus one point, and the second statement is already false for the plane minus two points. In fact, the fundamental group of $\mathbb{R}^2$ minus $k$ points is the free group on $k$ generators (so free homotopy classes of loops correspond to conjugacy classes in the free ...

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A group whose abelianization is trivial is called perfect, and there are many such groups. In particular, any nonabelian finite simple group is perfect, so $A_5$, for example, is perfect. $A_5$ is in fact the smallest nontrivial perfect group. So any space with fundamental group $A_5$ is an example. A famous example which almost has fundamental group $A_5$ ...

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This is exactly the type of space that Poincare constructed to show that homology was not enough to distinguish three-manifolds from the three-sphere. He took a dodecahedron and glued opposite faces with a minimal clockwise twist. The resulting space is a homology sphere --- it has the homology groups of $\Bbb S^3$, but has nontrivial $\pi_1$.

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By the Universal coefficient theorem (for cohomology) and the assumption the $M$ is simply connected, we have $$H^2(M)=\operatorname{Hom}(H_2(M),\mathbb{Z})\oplus \operatorname{Ext}(H_1(M),\mathbb{Z})= \operatorname{Hom}(H_2(M),\mathbb{Z})$$ which is torsion free, hence free abelian (because it is finitely generated). By Poincaré duality, $H^2(M)\cong ... 4 The answers are, broadly speaking, "yes": homology is a property that's invariant within a homoeomorphism class, i.e., if$X$and$Y$are homeomorphic, they end up having the same homology groups, and the same is true if they're merely homotopy equivalent. These claims can be proved (with limitations) for simiplicial objects, but the proofs are not very ... 4 The first statement is not true, even for the plane minus finitely many (even one) point. Assuming we know that$\pi_1(\mathbb{R}^2\setminus 0) = \mathbb{Z}$, generated by loops around$0$, then the double winding around$0$and the single winding around$0$both contain the same "pole" but are not homotopic. 3 Yes there exists such a surjective map! Actually for any$n\gt 0$there exists a surjective continuous map from the$1$-simplex, aka the unit interval$[0,1]$, to$\mathbb R^n$or$S^n$: these are examples of the Peano phenomenon of space-filling curves. The most general result in that area is the Hahn-Mazurkiewicz theorem: Every second-countable, ... 3 Spanier's Algebraic Topology, Theorem 6.2.17 (p. 296): Let$U$be an orientation over [a commutative ring]$R$of an$n$-manifold$X$and let$(A, B)$be a compact pair in$X$[i.e.,$A \subseteq X$is compact and$B \subseteq A$is closed in$A$]. Then for all$q$and$R$-modules$G$, there is an isomorphism $$\bar{\gamma}_U: H_q(X - B, X - A; G) ... 3 It is called De Rham complex and the cohomology it induces is called De Rham cohomology. U can be replaced by any smooth manifold. It is naturally isomorphic to other cohomology theories with \mathbb{R}-coefficients; the exterior product of forms corresponds to the cup product in singular/simplicial cohomology. Note also that for U\subseteq ... 2 Differential Forms in Algebraic Topology by Bott and Tu has many examples that you could present if you want. A User's Guide to Spectral Sequences by McCleary has also tons and tons of examples. Here are some examples off the top of my head: The Serre SS for computing the homology of \Omega S^n ("easy"), of BU_n and U_n ("hard"). In homological ... 2 Recall that c_1(\xi) lives in H^2(\mathbb{C}P^1), so when you write c_1(\xi)=-1, you actually mean c_1(\xi) is the negative of the generator of H^2(\mathbb{C}P^1). The first 1 in c(\xi), on the other hand, is the generator of H^0(\mathbb{C}P^1). As for the geometric reasoning, the top Chern class (highest non-zero Chern class) of a complex ... 2 I am not surprised that you find this a difficult area to get into! One has to some extent to appreciate the historical background, which is nicely covered in the books "History of Topology" Edited IM James, particularly the article by Lefshetz, and also in the book by Dieudonné. The early articles on topology used terms like cycles modulo boundary, but ... 2 I'm going to echo Incnis Mrsi's comment: The first concept you describe is called homotopy equivalence of spaces. Two spaces are then homotopy equivalent if there are compositions of continuous maps X \overset{f}{\to} Y \overset{g}{\to} X homotopic to \operatorname{id}_X and Y \overset{g}{\to} X \overset{f}{\to} Y homotopic to \operatorname{id}_Y. As ... 2 If \omega and \tau are closed forms (which you have not used), then$$d(\omega \wedge \tau) = d\omega \wedge \tau + (-1)^{p-1}\omega \wedge d\tau = 0,$$hence \omega \wedge \tau is also closed, hence it corresponds to a class in cohomology. Let us see that this class depends only on the classes of \omega and \tau. Let us suppose that \omega = ... 2 Let's start with 1-cycles. The canonical example of a 1-cycle is an oriented simple closed curve \gamma in the 1-skeleton of the simplicial complex, subdivided into 0-cells and oriented 1-simplices. Notice that each 0-simplex in the simplicial complex has exactly the same number of incoming oriented 1-simplices as it has outgoing oriented 1-simplices: ... 2 Yes this is correct. It obviously suffices to show that \tilde H_{k+1}(\Sigma X) = \tilde H_{k}(X) for all k. To show this consider the Mayer Vietors Sequence for suitable neighborhoods of the two cones C_1X,C_2X which are glued together to obtain the suspension. Using homotopy equivalences you will get the following exact sequence:$$ \cdots \to ... 2 All the cellular and simplicial homology groups of any (triangulable) space are isomorphic. For a proof see e.g. Hatcher. 2 There are cases when simplicial homology can't be applied but cellular homology can, for example in the case of a compact manifold that can't be triangulated (which exist in dimensions greater than or equal to 4). 2 The quotes on the "submanifold" are important. It's hard to visualize this case, because the curves$\alpha$and$\beta$intersect. When the curves don't intersect then things are a bit easier to see. For example, consider the$2$-dimensional torus. Two disjoint longitudes on the torus are homologous, and this is geometrically justified because the ... 1 Neither$K(ku)$nor tmf are complex orientable, so neither$K(ku)$nor tmf are elliptic cohomology theories in the strict sense.$K(ku)$"is" an elliptic cohomology theory in the looser sense that it "detects$v_2$-periodic phenomena" (although I can't elaborate too much on what this means), and tmf "is" an elliptic cohomology theory in the looser sense that ... 1 Yes; in fact$f$is an isomorphism onto$H^{\bullet}(BU(1)^n)^{S_n}$. This is an aspect of the splitting principle for complex vector bundles. 1 On one hand, $$C_1=d(C_0)\oplus d(C_0)^{\perp}=d(C_0)\oplus\ker d^*$$ On the other hand, $$C_1=d^*(C_2)\oplus d^*(C_2)^{\perp}=d^*(C_2)\oplus\ker d$$ Since$d(C_0)\subseteq\ker d$, this implies$\ker d=\ker d\cap\left(d(C_0)\oplus\ker d^*\right)=d(C_0)\oplus\left(\ker d\cap\ker d^*\right)$,$\mathrm{H}^1(S^2)=\ker d/ d(C_0)\simeq\ker d\cap\ker d^*$and ... 1 One of the good properties of Chern classes is that if$E$is a complex vector bundle of rank$r$on a manifold$X$and$s$is a global section of$E$that intersects the zero-section transversely, then$c_r(E)$is exactly the cohomology class in$H^{2r}(X,\mathbf Z)$of the submanifold$Z(s)=\{x \in X \mid s(x)=0\}$. As @fixedp says, the "1" here refers ... 1 Just to have this down as an answer. Yes, such a$d$does exist. Note since$\mathcal{O}(1)$(the canonical very ample bundle on$\mathbb{P}^n$) is ample, we know that there exists$d\geqslant 0$such that$\mathcal{F}\otimes\mathcal{O}(d)$is globally generated. Suppose that$h^0(\mathcal{F}\otimes\mathcal{O}(d))=0$. Then, by the fact that ... 1 I think the 2nd quest is still open. By$s(x)s'(x)^{-1} \in \ker(p)=\text{im}(i)$and injectivity of$i$there is a unique$g(x) \in A$such that$i(g(x))=s(x)s'(x)^{-1}$. This defines a function$g:G \to A$. Before we can deduce the formula in part (ii) of the question, it's important to note that the action of$G$on$A$is defined by$i(g\cdot a) := ...

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