# Tag Info

7

There are many more 3-manifolds that arise as the boundary of compact contractible 4-manifolds than just those, including some Brieskorn spheres like $\Sigma(2,3,5)$. The keyword you want is Mazur manifold. It is my impression that, and I would be surprised if it weren't true, there is not much known about what 3-manifolds arise as the boundary of a Mazur ...

5

The Hodge star is an isomorphism and you've transported the differential along that isomorphism, so you just get de Rham cohomology with the indices backwards. A more fun thing to do is to study both differentials at the same time, which leads to Hodge theory.

4

The Yoneda lemma says that any natural transformation such as $a$ between two representable functors ($H^i(-, \Bbb Z/2\Bbb Z)$ and $H^{i+k}(-, \Bbb Z/2\Bbb Z)$ in this case) is determined uniquely by maps between the representing objects. In our case this is a map $K(\Bbb Z/2\Bbb Z, i) \to K(\Bbb Z/2\Bbb Z, i+k)$, which corresponds precisely to the element ...

4

If $X \to Y$ is a finite Galois cover with Galois group $G$ then it's known that the induced map $H^{\bullet}(Y, \mathbb{Q}) \to H^{\bullet}(X, \mathbb{Q})$ is injective and induces an isomorphism $$H^{\bullet}(Y, \mathbb{Q}) \cong H^{\bullet}(X, \mathbb{Q})^G.$$ (This has nothing to do with $X$ and $Y$ being compact or manifolds.) So here the answer is ...

3

Let me write $B^n A$ for $K(A, n)$. The Yoneda lemma implies that $[B^n \mathbb{Z}, B^{np} \mathbb{Z}]$ is precisely the set of natural transformations $H^n(-, \mathbb{Z}) \to H^{np}(-, \mathbb{Z})$. An obvious candidate for such a natural transformation is the $p^{th}$ cup power $$H^n(-, \mathbb{Z}) \ni \alpha \mapsto \alpha^p \in H^{np}(-, \mathbb{Z})$$ ...

3

Consider the pair $(\Bbb T, A)$ where $A \subset \Bbb T$ is one of the longitudal circles in $\Bbb T$. We thus have the long exact sequence $$\cdots \to H_{k+1}(\Bbb T) \to H_{k+1}(\Bbb T, A) \stackrel{\partial}{\to} H_k(A) \to H_k(\Bbb T) \to H_k(\Bbb T, A) \stackrel{\partial}{\to} H_{k-1}(A) \to \cdots$$ $(\Bbb T, A)$ has homotopy extension property, as ...

3

Let $x \in S^n$ not be in $A$. Then $A \subset S^n \setminus \{x\} \cong \Bbb R^n$, and this is contractible. The the map $H_n(A) \to H_n(S^n)$ factors through $H_n(A) \to H_n(\Bbb R^n) \to H_n(S^n)$, and must therefore be zero. So the sequence $0 \to H_n(S^n) \to H_n(S^n,A)$ is exact, and $H_n(S^n,A)$ is nontrivial as desired.

3

An element of $H_0(X)$ is basically just a point. Take any $a\in A\,,$ then $a\in X$ generates $H_0(X)$ since any other point $x\in X$ can be connected to $a$ since $X$ is path connected, hence $x-a$ is a boundary, ie. $x-a\equiv 0\,,$ hence $x\equiv a\,.$

3

Well, if you know already that $M$ is orientable, then there is $\eta$ a nowhere zero $n$-form on $M$. Let $C = \int_M \eta \in \mathbb R$. Now for any $[\omega] \in H^n(M)$, consider $D:= \int_M \omega$. This is nonzero. Then $[\frac{D}{C} \eta] = [\omega]$ as $$\int_M \frac DC \eta = D = \int_M \omega .$$ Note that $\frac DC \eta$ is nowhere ...

3

In general it really depends on how well you understand $f$ and its action on the cohomology of $X$. Here are two nice special cases where you only need to understand the action of $f$ on a single cohomology group and then everything else is determined by the cup product. Tori Let $\Gamma$ be a lattice in $\mathbb{R}^n$. (You can take $\Gamma = ... 2 A very good, more recent introduction to Khovanov homologies was written by Dolotin and Morozov. Introduction to Khovanov Homologies. I. Unreduced Jones superpolynomial. Introduction to Khovanov Homologies. II. Reduced Jones superpolynomials. Introduction to Khovanov Homologies. III. A new and simple tensor-algebra construction of Khovanov-Rozansky ... 2 First off, yes, you're right: contractibility is a very strong hypothesis, and is overkill if all you want to know is that closed$k$-forms are exact. You have heard about de Rham cohomology. Let me quickly recall what it is all about. Given a manifold$M$, the space$\Omega^k(M)$is the (vector) space of$k$-forms on$M$. You have the linear map$d_k : ...

2

The Mayer-Vietoris sequence goes $\cdots \rightarrow H^{n - 1}(U) \oplus H^{n - 1}(V) \rightarrow H^{n-1}(U \cap V) \rightarrow H^{n}(M) \rightarrow H^{n}(U) \oplus H^{n}(V) \rightarrow 0$. Note that since $V \cong \mathbb{R}^{n}$ we have $H^{n}(V) = 0$. We also already know that $H^{n-1}(U \cap V)$ and $H^{n}(M)$ are both $\mathbb{R}$, so in particular they ...

2

The first relation: take any vector space with subspaces $C, Z\supseteq B$. The natural surjection $Z/B \to (Z+C)/(B+C)$ has kernel $(Z\cap C +B)/B$ which is isomorphic to $Z\cap C /B\cap C$ so the isomorphism you want follows from the first isomorphism theorem. $\partial^0$ is the map formed as the direct sum of the maps $C_{dp}/C_{d,p-1}\to ... 2 By the second isomorphism theorem, the condition that$HK=G$, or probably in your case$H+K=G$(if we're using additive notation) is sufficient. 2 As Stefan Hamcke has indicated in his comments, one can obtain$S^2/A$step-by-step by first pinching the equator to a point to get$S^2 \vee S^2$and then pinching the pole to the wedged point to have$S^2 \vee S^2 \vee S^1$. This is homotopy equivalent to our space. Hence,$\widetilde{H_n}(S^2/A) \cong H_n(S^2 \vee S^1 \vee S^1)$which is$\Bbb Z^2$when ... 2 Let$a,b\in A\,.$If$[a]_X=[b]_X\,,$then$[a]_X-[b]_X$is a boundary, hence there is a path in$X$connecting$a$to$b\,,$so that$a,b$lie in the same path connected component of$X\,.$Since each path connected component of$X$contains at most one path connected component of$A\,,a,b$can also be connected by a path in$A\,,$hence we also have ... 2 One place to look at is Topospaces (The Topology Wiki), which was suggested by a now-deleted user in a now-deleted answer. The description says This is a pre-alpha stage topology wiki primarily managed by Vipul Naik, a Ph.D. in Mathematics at the University of Chicago. We have over 400 articles including some material in basic point-set topology. As ... 2 Presumably, you mean$\sigma_{i_0,\ldots,i_p} = -\sigma_{i_0,\ldots,i_{q-1},i_{q+1},i_q,i_{q+2},\ldots,i_p}$. Let us consider the case$p=1$. Then we want to show that$\sigma_{i_0,i_1}=-\sigma_{i_1,i_0}$for any$(i_0,i_1)$. But$\delta\sigma = 0$means that for all$i$, we have $$\sigma_{i_1,i}|_{V_i}-\sigma_{i_0,i}|_{V_i}+\sigma_{i_0,i_1}|_{V_i} = 0,$$ ... 2 The category$\mathcal{O}$is closed under quotienting, submodules and finite direct sums, but not under extensions. Hence$Y$need not be in$\mathcal{O}$in general (see here). 1 You are right that an extra degree is needed in the hypotheses. Conceptually, you don't expect the zero-th groups of something contractible to vanish, because you should have the homology of a point. Note that$H_0(P_*)$is only zero if$P_1 \to P_0$is onto. If there existed a$D_{-1}: 0 \to P_0$satisfying$D_{-1}d_0 + d_1D_0 = \operatorname{Id}$, then ... 1 There are two reasonable ways to define Cech cohomology, and they give the same answer in end. The exact answer to your question depends on which route you take. Let$X$be a topological space. Let$U_i$be an open cover, with the index$i$running through some set$I$. For$i_0$,$i_1$, ...,$i_p \in I$, write$U_{i_0 i_1 \cdots i_p}$for$U_{i_0} \cap ...

1

You can always construct a non vanishing volume form, e.g. by choosing a Riemannian metric. Call this $\eta$. Since $H^n_{dR}(M)$ is a one-dimensional vector space, any nonzero class $[\omega]$ is a multiple of $[\eta]$, say, $[\omega] = c[\eta]$, $c\neq 0$. So $[\omega] = [c\eta]$ and you have a nowhere vanishing representative of $[\omega]$.

1

I think your question is a little ambiguous so I will answer both potential questions. 1 An inclusion of spaces does not induce an inclusion of groups. As pointed out in the comments you can include a space with non-trivial homology into a space with trivial homology (i.e. same homology as a point). That would imply that there has to be an induced map in ...

1

Let $x_1x_2\ldots x_n$ denote the element in $H^{n}(\Bbb R P^\infty \times \dots \times \Bbb R P^\infty; \Bbb Z/ 2\Bbb Z)$ that I am assuming is the image of the generator of $H^n(K(\Bbb Z/2\Bbb Z, n); \Bbb Z/ 2\Bbb Z)$ Instead of trying to compute $\operatorname{Sq}^I(x_1x_2\ldots x_n)$ completely, if you just want to demonstrate a monomorphism consider ...

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