# Tag Info

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No. Even Chern numbers are not topologically invariant — an example was given by Borel and Hirzebruch in 1959 (Characteristic classes and homogeneous spaces, §24.11).

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...it turns out there are non-trivial high dimensional smooth knots $S^n\subset S^{n+2}$ such that $\pi_1$ of the knot complement $X$ is infinite cyclic. The inclusion of the meridian $S^1 \subset X$ is a homology isomorphism, a $\pi_1$ isomorphism but is never a weak equivalence since by a result of Levine if it were then the knot would be ...

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Hint (for the first question): Since the group acts freely, for any point $x \in S$, the elements $g \cdot x$ are pairwise distinct for $g \in G$. Because the action is continuous, this means that for every $g$, there are neighborhoods $U_g$ and $V_g$ respectively of $x$ and $g \cdot x$ that are disjoint. Now $g$ is finite so $\bigcap_{g \in G} U_g$ is a ...

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Hint: Show that the quotient map $S \to S/G$ is a finite-sheeted covering to deduce that $S/G$ is a closed surface. Also, an interesting example to consider is the action $\mathbb{Z}_2 \curvearrowright \mathbb{S}^2$ whose quotient is $\mathbb{R} \mathbb{P}^2$.

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The key is that since the action of $I^*$ on $S^3$ is free, the fundamental group (via the usual theory of covering spaces) should be easy enough to compute. As for homology, this is a purely algebraic problem. What can you say about the commutator subgroup of $I^*$? (In this case, I would look up the notion of a 'perfect' group)

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You may embed your oriented closed surface into $\mathbb R^3$ in the usual way. Then the normal bundle is trivial, and of course the tangent bundle of $\mathbb R^3$ is trivial, so that the same is true of its restriction to $\Sigma$. Thus we find that indeed, $$T\Sigma \oplus \text{ rank one trivial (i.e. normal bundle to \Sigma in \mathbb R^3) }$$ $$... 2 I don't have a complete answer for you, but I can say a bit. It's a general theorem that if we have a covering space p:\tilde{X}\rightarrow X with \tilde{X} simply-connected and X path-connected and locally path-connected, then the group of deck transformations G is isomorphic to \pi_1(X). (For example, see Hatcher p.71 prop. 1.39.) All the ... 0 I suppose you want to know where (a,b) travels during the homotopy from f_0 to f_1? If f_0 is an injective path, so it has no self-intersections, the answer is f_1(f_0^{-1}(a,b)). If f_0 is not injective, (a,b) might have more than one preimage and theres no unique answer. Consider the following homotopic curves: 1 There is a way around using Van Kampen. In a sense it is a dirty cheat, since the core idea of the argument is essentially the same as the proof of the Van Kampen theorem. The idea is to use the supplied hint. By Sard's theorem, you can apply a small homotopy leaving the base-point fixed to any map S^1 \to K^2, ensuring the new map misses some point in ... 1 You should convince yourself with a drawing that the Klein bottle K minus a point (not your base point) deformation retracts onto the wedge of circles S^1\vee S^1 generated by the two loops \alpha and \beta. If \gamma is a loop in K, it is based-homotopic to a loop that misses a point \mathrm{pt} (necessarily other than the base point). Use the ... 1 Letting \gamma(t) = e^{2 \pi i t} be a loop (with base point 1+0i) representing the generator for \pi_1(C^*), you get the presentation$$\pi_1(C^*/H) = \langle \gamma,\psi \,\, | \,\, \psi \gamma \psi^{-1} = \gamma^{-1}\rangle $$1 Your proof works fine, but you should say that p\psi is homotopic rel endpoints or path homotopic to a constant path and, since a path homotopy lifts to a path homotopy, ψ is path homotopic to a constant map, too, which implies a=b. With little more effort, we can show that if a loop \phi at b_0 in B is in the image p_*(\pi_1(E,e_0)) (which is ... 0 Let f be a continuous map from S^1\longrightarrow S^1. It induces a map on \pi_1; that is a map f_*: \pi_1(S^1)\cong \mathbb Z \rightarrow \pi_1(S^1)\cong \mathbb Z . Now the only group homomorphism from \mathbb Z to \mathbb Z is multiplication by an integer k_f. This k_f is the degree of f. The fact that (f\circ g)_*=f_*\circ g_* is a ... 1 In general, if two maps are homotopic, then their induced maps conincide: For a map f:(X,x_0)\rightarrow (Y,y_0) (that is a continuous map X\rightarrow Y with f(x_0)=y_0) we denote its induced map \pi_1(X,x_0)\rightarrow \pi_1(Y,y_0) by f_*. Claim: If f,g:(X,x_0)\rightarrow (Y,y_0) are homotopic relative to \{x_0\}, then f_*=g_*. ... 7 To a topologist, topological groups are interesting in their own right. The group structure actually gives us interesting topological structure, too! One interesting fact is that the fundamental group (an important topological invariant) of a topological group is Abelian, a fact that spectacularly fails to be true in general - any group can be the ... 5 Topological groups also arise out of purely algebraic situations. Here are just two of the most common examples. In number theory one considers the rings of p-adic numbers \mathbb Z_p for primes p. I won't get into their construction except to note that their topology is built out of purely algebraic considerations, and that \mathbb Z_p is in fact a ... 5 Converted from comment to answer: Topological groups are used to study continuous symmetries, and are used very frequently in modern theoretical physics. Local symmetries form the basis for gauge theories - QED is an abelian gauge theory with the symmetry group U(1), quantum chromodynamics, is a gauge theory with the action of the SU(3) group, and the ... 12 The reason that <concept> is important is that their are enough examples of <concept> that justifies making the definition. So I will answer this by giving some examples. The first example that we come across is the real numbers, (\mathbb{R}, +) (by extension all the \mathbb{R}^n). We see that the map sub:\mathbb{R}^2\to \mathbb{R} ... 0 Basically you can construct the surface of genus g, by gluing a disk \mathbb{D}^2 to a wedge of 2g circles, \bigvee\limits_{i=1}^{2g} \mathbb{S}_i^1. First we think the border of \mathbb{D}^2 as 2g-sided polygon, and we put labels in each side a_1,\ b_1,\ a_1,\ldots,\ a_g,\ b_g, and arrows as in the picture: Then we label each copy ... 1 You could use the fact that a covering map with finitefibers is a perfect map, i.e. a surjective and closed map with compact fibers. Closedness of p is shown as follows: Let C\subset E be closed, b\notin p[C]. Then p^{-1}(b)=\{b_1,...,b_n\} is a finite subset of E-C. There is an evenly covered neighborhood U of b whose preimage is the union of ... 1 You haven't used that B is compact. Hint: project your open cover of E to an open cover of B... 1 There are two finite 2-dimensional complexes A, B which are not homotopy-equivalent but are homology-equivalent, i.e., there exists a continuous map$$ f: A\to B $$inducing isomorphism of fundamental groups and homology groups. See the last paragraph (page 522) of this paper, the actual example is due to Dunwoody. Edit 1: The homology equivalence part ... 2 Consider any subset E_m of your abelian group of size m. Then since the group is abelian, |E_m^n| is no more than the number of ways you can choose m non-negative integers to sum up to n, which is{{n+m-1} \choose {m-1}} which is no more than than (n+m-1)^{m-1}. Since the exponent is a constant, you will get \lim_{n \to \infty}(n+m-1)^{(m-1)/n} ... 0 I have not covered homotopy category yet so cannot use it I'm afraid. However I have found the solution I think... Consider the following$$x\circ h\circ f \circ k\circ y$$We know that x\circ h\circ f\sim 1_{X} so if we substitute this into the above we get that$$x\circ h\circ f \circ k\circ y\sim 1_{X}\circ k\circ y\sim k\circ y$$Similarly we know ... 0 This question has a nice answer on MO: this answer by Reid Barton, as mentioned by Bruno Stonek in the comments. The Whitehead tower has most of the properties I was looking for, too; thanks to Piotr Pstrągowski for mentioning this in the comments as well. 0 Another example would be the interplay between Möbius strip, Klein bottle and real projective plane. They motivate the study of orientability and are fun to visualize, especially how one transform into another. 2 The most basic such map is:$$x\mapsto \frac x{\|x\|}$$The homotopy is:$$h(x,t)=\frac{x}{(1-t)+t\|x\|}$$3 \DeclareMathOperator{\cls}{cls}Let \epsilon : C_0(X) \to \mathbb{Z} be defined by \epsilon(\sum m_i x_i) = \sum m_i. This is clearly a homomorphism, and it vanishes on boundaries (easy check), so it yields \epsilon : H_0(X) \to \mathbb{Z}. Suppose that \cls(\gamma) is a generator of H_0(X) and let n = \epsilon(\cls(\gamma)). Let x \in X be ... 1 There is another way of putting this question. The cyclic group C_2 of order 2 acts on the circle S^1 by conjugation z \mapsto \bar{z}. The fundamental group of the circle at 1 is \mathbb Z and the induced action on \mathbb Z is n \mapsto -n; the quotient of \mathbb Z by this action is cyclic of order 2, which is the wrong answer, as ... 0 An interesting read that would give you considerable insight into some natural applications of mathematics in physics would be the three volume set Group Theory in Physics by John F. Cornwell. I think it is fair to say these contain some calculus which is probably not covered by the other beautiful books mentioned in answers here. In a similar vein, you ... 1 They are just saying that if A=\bigcup_nA_n, then \ell^2(A)\simeq\bigoplus \ell^2(A_n). The direct sum has to be understood as a direct sum of Hilbert spaces, i.e. an \ell^2-sum. The left regular representation of \Lambda leaves each direct summand invariant. So we can think of it acting that way on \ell^2(\Gamma) (as a block-diagonal operator). ... 1 Your space is homotopic to half circle \{ e^{i\theta} : \theta\in[0,\pi]\} which is contractible. We can write down the contraction$$ H(\theta,t) = e^{i \min\{\theta,\pi(1-t)\}}. $$Fundamental group of contractible space is trivial. 1 Let U = S^n\setminus A and V = S^n\setminus B then by hypothesis U and V cover S^n so we can apply Mayer-Vietoris:$$ H_1(S^n) \overset{\partial_*}\longrightarrow H_0(U\cap V) \overset{(i_*,j_*)}{\longrightarrow} H_0(U)\oplus H_0(V) \overset{g}{\longrightarrow} H_0(S^n) \longrightarrow 0 $$If we assume that n\geq 2 then H_1(S^n) = \{0\}, ... 2 Am I right in saying that any two lifts of A to an endomorphism of F_2 differ by some inner automorphism of F_2? No, using a free basis F_2 = \langle a,b \rangle, the map given by f(a)=a, f(b) = a b^3 a^3 b^{-1} a^{-4} b^{-1} is not an inner automorphism but it induces the identity on \mathbb{Z}^2, as does the identity automorphism ... 1 Yes, this is a standard notation. 0 I suggest that you read two of the early papers which gave the first interesting examples of infinite dimensional bounded cohomology. In the first paper Brooks computes 2-D bounded cohomology of free groups, and in the second paper Brooks and Series compute 2-D bounded cohomology of surface groups. Brooks, Robert. "Some remarks on bounded cohomology". ... 0 Intuitively, a disc is simply connected because any loop can always be shrunk to a point. Now if we take a point away from the interior of the disc, this is no longer the case, since when when we shrink the loop it gets caught in the hole we just created. This is not the case when we remove a point from the boundary of D^2, that is, S^1. 1 Hint: The map$$\psi:\quad {\mathbb C}\to{\mathbb C}^*/G,\qquad w\mapsto e^w/_\sim$$is a covering of your Riemann surface R:={\mathbb C}^*/G. Two points w, w' map onto the same point of R iff they are equivalent modulo the lattice generated by \omega_1:=\log 2, \>\omega_2:=2\pi i. 1 I'm assuming the ordinary Euclidean topology. First note that the boundary of D^2 is (homeomorphic to) S^1 (not S^2, mind you). If a point belongs to the interior of D^2  (which you can embed in \mathbb R^2  as the unitary disc) you may as well assume it is the origin; build up the map  H : [0,1] \times D^2 \to D^2  sending$$ (t,x) \mapsto ...

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I don't if I am right, but I will give it a try. This is what I know- (1) $\displaystyle f_{*} (<\alpha>) = <f \circ \alpha>$ where $\alpha$ is a loop based at $1$ in $S^1$. (2) The exponential mapping $\pi: \mathbb{R} \rightarrow S^1, x \mapsto e^{2\pi ix}$ and the path $p_n(s) = ns, 0\leq s\leq1, n \in \mathbb{Z}$, joining $0$ to $n$ in ...

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Yes, because when you do GNS your state becomes a vector state. So, if in the direct sum you put all zeroes and the cyclic vector for your $w_\phi$ in its corresponding component, you get a vector that implements your state.

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First define the auxiliary function $f:I\times I\longrightarrow I$ given by $$f(s, t)=(1-s)t+s.$$ Define $H:C(X)\times I\longrightarrow C(X)$ setting $$H([x, t], s)=[x, f(s, t)].$$ It is easy to check $H$ is homotopy from $I_{C(X)}$ to a constant. Obs: You must take care with the function $f$ to be chosen. You must assure $f$ assumes its values on $I$. For ...

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Do you know Weibel's note on the history of Homological Algebra? See page 17: The familiar "five-lemma" occurs for the first time on p. 16 of [ES]. (Its companion, the "snake lemma", first appeared in [CE].) Nobody who looked at Cartan-Eilenberg book has been disappointed. It's an evergreen, no matter if it is 60 years old! :) [ES]: S. Eilenberg and ...

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Think at $\mathbb{R}P^2$ as the quotient of the 2-sphere by the antipodal relation. Then think at a closed band around the equator (precisely, as if you think at the band between the tropics :)), under the previous identification this band gives rise to a Möbius Band, right? Then the complement in the 2-sphere are two open disks ("shortened hemispheres"), ...

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In order to prove that $\tilde f^*(E)$ is isomorphic to $S^1 \times \mathbb{R}$ you need to construct a map $\tilde f' : S^1 \times \mathbb{R} \to E$ having the property that its restriction to the fiber $(S^1 \times \mathbb{R})_z$ is a linear isomorphism between $(S^1 \times \mathbb{R})_z$ and $E_{z^3}$. For this purpose you cannot use the exact same ...

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You can indeed use other objects instead of a closed interval. The important thing to make things work is when considering a homotopy between two morphisms $f,g:X\to Y$ is to replace $Y$ by a cylinder object for it. A cylinder object for $Y$ is an a space (and more generally the situation can be considered in any Quillen model category, so in fact the ...

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Note that the homomorphisms of $\mathbb Z$ are all of the form $n\mapsto kn$ for some $k\in\mathbb Z$, and are completely determined by the fact that $1$ maps to $k$. You want to show that each of these can be realized as as the induced homomorphism of a map $\varphi_k:S^1\to S^1$. Intuitively, the isomorphism $\pi_1(S^1)\to \mathbb Z$ counts how many times ...

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$Out(G)$ is a quotient group of $Aut(G)$, so if we find a well-defined map to $Aut(G)$, there's a well-defined surjection onto $Out(G)$ from there. Well $f^* : G \to G : p \to f\circ p$ is a well defined bijection of $G$. Further $f^* (pq) = f \circ (pq) = (f\circ p)(f\circ q)$, prove that! Thus $f^*$ is in $Aut(G)$ and so there's a well-defined map $f ... 2 I think the answer is yes: if$X$is a connected finite 2-dimensional CW-complex with$\pi_1(X)$free, then$X$is homotopy equivalent to a wedge of 1-spheres and 2-spheres. This is stated in the paragraph spanning the first and second pages of Trees of Homotopy Types of Two-Dimensional CW-Complexes by Dyer and Sieradski, available here: ... 3 If$Y=A\times I \cup X\times\{0\}$is closed in$X\times I$, then$Y\cap (X\times\{1\})$is closed in$X\times\{1\}$. But this set is just$A\times\{1\}$, and if$A\times\{1\}$is closed in$X\times\{1\}$, then$A$must be closed in$X\$.

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