New answers tagged

0

By simple computations, one checks $AB=BA$, $AC=CA$, $CB=BCA$ is a presentation. So in the abelianization $A=1$, $BC=CB$ the abelianized group is $\bf Z^2$. Note that $dy, dz$ are invariant by $G$ and form a base for the de Rham co-homology


1

I would leave a comment but don't have the reputation: check out http://www.math.udel.edu/~lazebnik/papers/dioph2.pdf, which also looks to have several useful references (in particular http://www.math.tamu.edu/~rojas/kannanbachemhermitesmith79.pdf for algorithms to compute Smith normal form).


0

So we can actually build up the answer inductively here: let's label the simplices of $\Delta_2$ as $v_0, v_1, v_2$, and the corresponding vertices of the top and bottom faces of $\Delta_2 \times I$ as $t_0, t_1, t_2, b_0, b_1, b_2$. So let's define $P$ on $C_0(\Delta_2)$. Here $\partial P + P\partial = \partial P$ (as $\partial$ vanishes on 0-simplices) ...


1

Here's a map that goes backwards. If $M$ is equipped with a hermitian metric, then $\Omega^{p,q}(M)$ is an inner product space and $\bar{\partial} : \Omega^{p,q-1}(M) \to \Omega^{p,q}(M)$ has an adjoint $\bar{\partial}^* : \Omega^{p,q}(M) \to \Omega^{p,q-1}(M)$ satisfying $\langle\bar{\partial}\alpha, \beta\rangle = \langle\alpha, ...


0

Yes, see Tyler Lawson's answer here http://mathoverflow.net/a/10983 This is for free abelian groups but the same argument works in your case.


4

There's a little abuse of notation. First, $M$ has to be compact for all this to make sense; it's probably written somewhere before in the paper. When $M$ is compact, orientable, and connected, $H^{2d}(M;\mathbb{R})$ is isomorphic to $\mathbb{R}$, and the choice of an orientation basically amounts to the choice of an isomorphism $H^{2d}(M;\mathbb{R}) \cong ...


3

Here is the construction of a (co)homology theory for lattices which yields the Chech (co)homology in the case of topological spaces when applied to the lattice of open subsets. Recall that the Chech theory works with open coverings. An analogue of an open covering in a lattice $L$ is a subset $C$ of $L$ satisfying the property that the join of the elements ...


0

In intuitive terms, results about $\mathbb{Z}$ are the most general. They are in a sense arithmetic with all the subtleties that implies. Results about $\mathbb{Q}$ or $\mathbb{R}$ or really any field are much cruder. The arithmetic subtleties are lost and all considerations of torsion are erased. Of course in some situations this is just what is needed. ...


5

The 2-sphere with 4 points deleted is the plane with three points deleted which retracts to the bouquet of three circles, so its fundamental group is the free group generated by three elements. Its universal cover is the plane.


0

First of all, since $\pi = \pi\circ\sigma$, if $\beta$ is a $k$-form on $\Bbb RP^n$, then $\alpha = \pi^*\beta = (\pi\circ\sigma)^*\beta = \sigma^*(\pi^*\beta) = \sigma^*\alpha$. Conversely, we know that $\pi$ is a local diffeomorphism. So, given a point $[x]\in\Bbb RP^n$, choose a neighborhood $U$ of $x\in S^n$ so that $\pi|_U\colon U\to \pi(U)$ is a ...


1

Your proof of invariance of domain doesn't work, because you have given no justification that $\tilde{f}$ is proper. In fact, it probably won't be proper. Indeed, if (as you suggest) you replace $\epsilon$ by $\epsilon+\xi$ for some $\xi>0$, then $\tilde{f}$ will definitely not be proper, since $\psi^{-1}(\overline{B(\psi(y), \epsilon)})$ is a compact ...


1

No. Consider the 1-form $d\theta$ on the unit circle, and the two 1-simplices $$ f:[0, 1] \to R : x \mapsto (\cos 2\pi x, \sin 2\pi x) \\ g:[0, 1] \to R : x \mapsto (\cos 4\pi x, \sin 4\pi x) $$ These have the same image (the whole circle), but the integrals are $2\pi$ and $4\pi$ respectively. As a simpler example, consider the two maps $f$ and $g$ of ...


2

It's not correct to assume that the vanishing of the first cohomology implies every closed curve must be a boundary. The curve $BA$ you mention, for instance, is not one, as would show up in the fundamental group or in first homology with integer coefficients, though not in the first cohomology. With real coefficients, $BA$ represents zero in cohomology ...


2

Fundamental Classes Let $Y$ be a connected, closed, orientable $k$-dimensional manifold, then $H_k(Y; \mathbb{Z}) \cong \mathbb{Z}$. The choice of a generator of $H_k(Y; \mathbb{Z})$ is equivalent to a choice of orientation. One often denotes the generator by $[Y]$; we call this a fundamental class of $Y$. If we consider $Y$ with it's other orientation, the ...


3

Let $e_1, \cdots, e_n$ be a local basis of $E$ so that $\nabla e_i (x) = 0$ at $x\in M$. Using $1.$, the dual basis $e_1^*, \cdots, e_n^*$ also has $\nabla e_i^* (x) = 0$. Using this basis, we have locally $$\text{Id} = \sum_{i=1}^n e_i^* \otimes e_i,$$ thus $\nabla \text{Id}(x) = 0$ by $2.$. Since $x\in M$ is arbitrary, we have $\nabla \text{Id} = 0$.


7

This is false, and somewhat surprising. There is a counterexample due to Milnor and Barrat . The space is a union of a nested sequence of spheres embedded in $\mathbb{R}^3$ and it has infinitely many non-zero singular homology groups. The nesting of the spheres is similar as in the Hawaiian earring. However, the result is true if one imposes local niceness ...


1

The Mayer-Vietoris sequence is defined as the long exact sequence $$H_*(U \cap V) \to H_*(U) \oplus H_*(V) \to H_*(X)$$ where the first map is defined by $(i_*, j_*)$ and the second map is defined by $k_* - l_*$ where $i, j$ are the inclusion maps of $U \cap V$ inside $U$ and $V$ and $k, l$ are the inclusions of $U$ and $V$ inside $X$. If $X$ is the Klein ...


1

You're having difficulty proving this because the statement is wrong. As stated, the condition that there exists a compact $C'\subseteq C$ such that $H^p(C')=0$ is true for every space $X$, since you can always take $C'=\emptyset$. I would guess that the intended statement should have $C\subseteq C'$ rather than $C'\subseteq C$. In fact, with this ...


1

I) You can't find such $D_3$. You have correctly proved that if you can, the integral is zero. If you could find such a $D_3$, the surface would be null-homologous; but there are plenty of non-null surfaces (eg $\Bbb{CP}^1 \subset \Bbb{CP}^2$). II) Your proof only works if $A$ is defined on both sides of the curve, so that $F=dA$ globally. But if this were ...


-1

These classes are not always representable, but classes of $H_l(M,Z/2)$ are representable. Positive multiple of elements of $H_l(M,Z)$ can be represented by smooth manifolds, this is shown in the thesis of Rene Thom.


2

The reason for this is the following fact: Theorem: Let $N\to M$ be a submanifold and $[N]$ the homology class it defines in $M$. Then in $M$ the Thom class of the normal bundle is dual to $[N]$. (see e.g. this exercise in Milnor Stasheff) Use this together with the map $[M,K(G,i)] \to H^i(M;G)$ which is precisely given by pulling back the unique ...


3

A major issue is the multiplicative structure that is around in cohomology. This allows you to distinguish spaces, which have the same homology. As an example, the $X:=\mathbb CP^2$ and $Y:=S^2\vee S^4$. Then both $X$ and $Y$ are CW-complexes with one cell in dimensions $0$, $2$, and $4$. Hence in both cases the homology with integral coefficients is ...


5

There are several elementary points that can be made : On the other hand, why consider only homology ? I don't see why one would be more natural than the other (actually for me cohomology seems more natural because I'm used to subjects where cohomology natuarally appears). There are cohomology theories that are clearly useful and natural and are related to ...


0

Given $A^{\bullet} \xrightarrow{f}B^{\bullet}$ such that $f$ is nullhomotopic we have by hypothesis $$f=d^{B}s+sd^{A}$$ in particular $$f_n=d_{n-1}^{B}s+sd_n^{A}$$ Applying the universal property of the kernel we can induce a map $f^{k}_n:\operatorname{Ker}(d^{A}_n) \to \operatorname{Ker}(d^{B}_n) $ with the property $$ f_n \circ k^{A}_n= k^{B}_n \circ ...


2

Interior here means the interior with respect to the lowest-dimensional Euclidean space in which the simplex can be embedded. Technically, the standard-$n$-simplex $\Delta^n$ is the convex hull of the unit vectors $e_0,\dots,e_n$ in $\Bbb R^{n+1}$, with the subspace topology. That is $$ \textstyle \Delta^n = \{ (t_0,\dots,t_n) \mid t_i\ge0, \sum_i t_i = 1 ...


1

When computing homology, it is not just the chain complexes, but also the boundary maps that are important. Because $A$ and $B$ are included in different ways in the simplicial complex structure of $X$ in the problem you references, the boundary maps will be different, so the homology groups will also be different. There is also a theorem, which you are ...


2

You are perfectly right that the bilinear form $\sigma: H^{2n}(M; \mathbb{Z}) \times H^{2n}(M; \mathbb{Z}) \to \mathbb{Z}$ is defined on the $\mathbb Z$-module $H^{2n}(M; \mathbb{Z})$, which is certainly not a vector space nor even a free abelian group. And symmetric bilinear forms on $\mathbb Z$-modules are a notoriously difficult subject where nothing so ...


2

No such groups exist. Suppose $\operatorname{Ext}(A,\mathbb{Z})=\operatorname{Hom}(A,\mathbb{Z})=0$. First, note that the functor $\operatorname{Ext}(-,\mathbb{Z})$ turns injections into surjections (this follows from the long exact sequence and the fact that $\operatorname{Ext}^2$ always vanishes for abelian groups). So we must have ...


2

A map is given explicitly by composing the diagonal map $x \rightarrow (x,x)$ with quotienting by a point. Let $\alpha_1 \in H^n(S^n \vee S^n)$ send the generator of the homology of the the "first" sphere $\sigma_1$ to $1$ and $\alpha_2$ send the generator of the second sphere $\sigma_2$ to $1$. Under the diagonal map the generator of $H_n(S^n)$ gets sent to ...


1

In general, with respect to a particular choice of basis, a symmetric bilinear form on a finite dimensional vector space can be represented by a symmetric diagonal matrix. Sylvester's law of inertia says that the number of positive, negative, and zero entries on the diagonal are independent of the choice of basis. In the case that the bilinear form is ...


2

Consider $\mathbb{RP}^3$, the real projective space. It is a compact, connected, orientable manifold of dimension $3$. Its cohomology is given by: $$H^p(\mathbb{RP}^3;\mathbb{Z}) = \begin{cases} \mathbb{Z} & p = 0 \\ 0 & p = 1 \\ \mathbb{Z}/2\mathbb{Z} & p = 2 \\ \mathbb{Z} & p = 3 \\ 0 & p > 3 \end{cases}$$ Then for the nonzero ...


3

Consider the map $ u : H^n_c(M) \to \mathbb{R}$ defined setting $$u: \ [ \omega ] \mapsto \int_{M} \omega \ .$$ This is well defined because of Stokes theorem, and is injective because of your assumption. So (under your assumption) either $H^n_c(M)=0$ or $\mathbb{R}$. On the other hand $H^n_c(M)\not=0$ since $u$ is non-zero (working in local coordinates, ...


1

(1) If $\omega =fdx$, then define $g(t):=\int_{-\infty}^t\omega$ Then $ g'(t)=f(t)$ so that $dg=\omega$ (2) $H^1_c({\bf R})={\bf R}$ : If $\omega$ is $1$-form on compact support on ${\bf R}$, then $d\omega =0$ And let $\int_{\bf R}\omega =C$ If $\omega'$ is another form of compact support with $\int_{\bf R} \omega'=C$, then $$ \int_{\bf R} \omega ...


0

Neither $\theta_{1}$ nor $\theta_{2}$ is a (global, continuous) function on the torus, only on the universal cover. Integrating "$d\theta_{i}$" over a closed path parallel to the $\theta_{i}$ axis gives $1$, while the integral of an exact form would be $0$.


0

Alexander duality says that: If $K$ is a compact, locally contractible, nonempty, proper subspace of $S^n$, then $$\tilde{H}_i(S^n-K;\mathbb{Z})\cong \tilde{H}^{n-i-1}(K;\mathbb{Z})$$ for all $i$. For a proof see Hatcher Theorem 3.44.


2

The boundary operator $\partial \colon H_k(X)\to H_{k-1}(A\cap B$ in Mayer Vietoris for $X=A\cup B$ is defined as follows. Given a $k$-cycle $x$ in X, write it as $[x]=[a+b]$ where $a$ and $b$ are chains supported in $A$ and $B$ respectively. Then $\partial a$ is contained in $A\cap B$, and is the image of the MV boundary operator. For the sphere, the ...


3

The problem is that it is not true that the canonical map $i:\mathbb{Z}[[x]]\otimes\mathbb{Q}\to\mathbb{Q}[[x]]$ is an isomorphism. Any power series of the image of $i$ must have only finitely many different denominators on its coefficients, since an element of $\mathbb{Z}[[x]]\otimes\mathbb{Q}$ is a sum involving only finitely many elements of ...


5

It depends on the ring. Consider $R = \mathbb{Q}$, for example. Then there is no continuous map $S^1 \to S^1$ inducing the graded ring morphism $\Lambda_\mathbb{Q}(\xi_1) \to \Lambda_\mathbb{Q}(\xi_1)$ sending $\xi_1$ to $\frac{1}{2} \xi_1$. Indeed, every map $S^1 \to S^1$ is homotopic to one of the maps $z \mapsto z^n$ for some $n \in \mathbb{Z}$, and the ...


2

When $R=\Bbb Z$ the answer to your question is no: Since $T^n$ is a $K(\pi_1(T^n),1)$, every group endomorphism of $\pi_1(T^n)$ is induced by a map $T^n\to T^n$. (See for example Prop 1B.9 in Hatcher). Therefore, any endomorphism of $\pi_1(T^n)^{ab}=H_1(T^n)=H^1(T^n)$ is also induced by a map of spaces. Since $H^*(T^n)\cong \bigwedge_{\Bbb Z}\left[ ...


1

$\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}$I'm not sure if this is an answer to the question, since it does refer to $H_{\ast}(M, \ZZ)$, but I think it sheds some interesting light on why the problem is hard. We begin with a flawed proof attempt. Write $\delta$ for the diagonal map $M \to M \times M$, and $\pi_1$ and $\pi_2$ for the projections from $M \times ...


1

Hochschild (co)homology is not a "topological invariant", because it doesn't even deal with topological spaces at all. Not all things called "homology" are about topology. In this case, Hochschild (co)homology is an invariant of associative algebras and their bimodules. Hochschild homology (resp. cohomology) takes as input an associative algebra $A$ and an ...



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