Tag Info

New answers tagged

0

As explained in another answer here, simplicial homology can be regarded as a special case of cellular homology; but the standard definition of the latter and establishment of its properties requires the main facts on singular homology. For a short discussion of some history of algebraic topology, and some anomalies, relevant to this question see this ...


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

All the cellular and simplicial homology groups of any (triangulable) space are isomorphic. For a proof see e.g. Hatcher.


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) ...


0

It seems to me the you have already shown everything. Let $SX$ be the set of n-simplexes of $X$, you have a surjective 2:1 map $S\widetilde X\to SX$. Then, $C_n(X,\mathbb{Z}_2)=\oplus_{SX}\mathbb{Z}_2$, and the map $ \oplus_{S\widetilde X} \mathbb{Z}_2\to \oplus_{SX} \mathbb{Z}_2$ is the one induced by $S\widetilde X\to SX$. This immediately implies the ...


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, ...


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 ...


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

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 ...


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 ...


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: ...


5

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 $\pi_1$.


5

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 $A_5$ ...


7

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


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 ...


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

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 ...


0

Consider the homotopy cofiber sequence induced by the quotient map $S^n \to \mathbb RP^n$, that's just attaching a disk along the map, and so we obtain $S^n \to \mathbb RP^n \to \mathbb RP^{n+1} $. Now, on homology or cohomology long exact sequences, we can see by knowing that the quotient map induces the trivial map on homology/cohomology, or by knowing the ...


5

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 ...


6

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


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.


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 ...


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.


0

A functor $F:\mathcal{C}\to\mathcal{D}$ takes a composition $f\circ g\mapsto F(f)\circ F(g)$. Hence, the functor $H_\ast$ takes a sequence of inclusions $A\hookrightarrow B\hookrightarrow C$ to a sequence of maps $F(A)\to F(B)\to F(C)$. As Qiaochu says, "that's what it to be a functor".


0

It suffices to know the comultiplication induced by $\mu$ on cohomology. For degree reasons the unique possibility is $$\Delta(\alpha) = \alpha \otimes 1 + 1 \otimes \alpha$$ and from here you are in the realm of a purely Hopf-algebraic calculation: you want to apply the "$n^{th}$ comultiplication" $\Delta_n : H \to H^{\otimes n}$ (where $H$ is the Hopf ...


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 ...


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 ...


5

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 ...


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 ...


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 ...


0

I don't have access to the paper, but I think I can help: I imagine that $I_m$ refers to an $m$-cell (or equivalently up to homotopy, an $m$-simplex). What's going on here is that $Y$ is filling the role of the descending manifold, for homological purposes, it seems. Being finite-dimensional, $Y$ intersected with the ball is an $m$-ball, and $\tau$ is ...


0

There is another nice answer to the first statement, For $d\gg 0$ we have $$p_\mathcal{F}(d)= h^0 \mathcal{F}(d)$$ where $p_\mathcal{F}(d)$ is the Hilbert polynomial. So the cohomology groups for very large twists should be nonzero because Hilbert polynomial can't have infinitely many roots. and the proof of the second claim is as the answer above.


7

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 ...


0

A triangulated space means: a space (which admits at least one triangulation) together with a fixed triangulation of it. This means, you consider this fixed triangulation, which gives you e.g. a finite set of $n$-vertices, which become a basis of $C_n(X)$ after ordering them. Similar situation: Oriented manifold or orientable manifold. In the latter we just ...


0

Note that $\phi_A$ is defined up to an exact form form $d\alpha$ and $$\int_U (\phi_A + d\alpha) \wedge \omega = \int_U \phi_A \wedge \omega + \int_U d(\alpha\wedge w)$$ and $$\int_U d(\alpha\wedge \omega) = \int_{\partial U} \alpha \wedge \omega$$ So the LHS of your equation is not well defined if $\partial U \neq \emptyset$. If you consider 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) := ...


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

If you have two sections $s$, $s':G\to E$, then for all $g\in G$ we have that $s(g)$ and $s'(g)$ are two elements of $E$ with the same image in $G$: it follows that there exists a unique $t(g)\in A$ such that $s(g)=t(g)s'(g)$. Figure out what consequences this has in terms of the factor sets corresponding to $s$ and $s'$.


1

The easiest way, for me at least, is first understand the dual of each abelian group in the complex, then construct the dual to the boundary operator. Let $G$ be an abelian group. The dual to $G$ is just the group whose elements are homomorphisms $G\to\mathbb{Z}$. Note that if $G$ is freely generated by the set $\{g_i|i\in I\}$, then a homomorphism ...



Top 50 recent answers are included