# Tag Info

5

There is the long exact sequence in homology of the pair $(X, X \setminus p)$ (everything is with $\mathbb{Z}$-coefficients unless indicated otherwise): $$0 \to H_n(X \setminus p) \to H_n(X) \to H_n(X, X \setminus p) \to H_{n-1}(X \setminus p) \to H_{n-1}(X) \to 0$$ (recall that the local homology groups satisfy, by excision, $H_n(X, X \setminus p) = ... 4 Yes. First, recall that a suspension has no nontrivial cup products. By Poincaré duality, it follows that the integral cohomology of$M$is torsion in degrees$1$through$n-1$, and that the same is true of the integral homology. From universal coefficients we know that$H_{n-1}$is torsion-free, which here implies that it's trivial. By Poincaré duality ... 3 I'll prove the statement I have in my comment. That any CW-complex with the homology of$S^n$suspends to a space homotopy equivalent to$S^{n+1}$for$n\geq 1$. If$X$is such a CW-complex, by Mayer-Vietoris,$SX$has the homology of$S^{n+1}$. As$X$is$(0)$-connected,$\pi_1(SX)=0$by the Freudenthal suspension theorem. By applying Hurewicz repeatedly, ... 3 Consider the long exact sequence of the pair$(M,\partial M)$and notice that Lefschetz duality gives an isomorphism$H_k(M)\cong H^{n-k}(M,\partial M)\cong H_{n-k}(M,\partial M)$, using that$M$is orientable and the coefficients are over a field. Consider the truncated exact sequence$H_m(\partial M)\to H_m(M)\to H_m(M,\partial M) \to H_{m-1}(\partial ...

3

While the tangent spaces $T_pM$ are all distinct abstract vector spaces, if $M$ has dimension $n$, then they are all isomorphic to $\mathbb{R}^n$, even though there is a priori no "canonical" isomorphism $T_p M \cong \mathbb{R}^n$. On a Riemannian manifold, parallel transport gives a way of associating to a curve $\gamma(t) \subset M$ in the manifold ...

3

Since $M$ is orientable, without boundary and connected, we have $H_0(M;\mathbb{Z}) = H_3(M;\mathbb{Z}) = \mathbb{Z}$. Using Poincaré duality, you know that the torsion subgroup of $H_2(M;\mathbb{Z})$ is isomorphic to the torsion subgroup of $H_{3-2-1}(M;\mathbb{Z}) = H_0(M;\mathbb{Z}) = \mathbb{Z}$ and thus is zero. Using Poincaré duality again, you know ...

2

For a quick and dirty explanantion that offers no insight (as requested), take a $3\times 3$ permutation matrix and draw a star in every unoccupied spot that is not above (in the same column) or to the right (in the same row) of any $1$. Put $0$s everywhere else. That's the Schubert cell corresponding to the permutation matrix. For example, for the ...

2

Yes, this is true. The natural map $H_n(X,\mathbb{Z})\otimes \mathbb{Z}_p\to H_n(X,\mathbb{Z}_p)$ is an isomorphism for $X=S^n$ and $X=M$. We thus have the following commutative diagram, where the vertical maps are isomorphisms: $$\require{AMScd} \begin{CD} H_n(S^n,\mathbb{Z})\otimes\mathbb{Z}_p @>{f_*\otimes 1}>> H_n(M,\mathbb{Z})\otimes ... 2 [This argument is stolen from the end of this answer (which handles the case q=1).] Let 1\leq i\leq n-1 and \alpha\in H_i(M,\mathbb{Z}). By Poincare duality, there exists \beta\in H^{n-i}(M,\mathbb{Z}) such that \alpha=z\cap\beta. Since H^{n-i}(S^n,\mathbb{Z})=0, f^*(\beta)=0. Thus 0=f_*(i_n\cap ... 2 Every odd-dimensional manifold has vanishing Euler characteristic, so$$0 = \chi(M) = b_0 - b_1 + b_2 - b_3.$$We have b_0 = 1, and b_0 = 0 since M is nonorientable. Hence, b_1 > 0 and thus H_1(M, \mathbb{Z}) is infinite. 2 You can talk about homomorphic and antiholomorphic functions, but a general complex-valued smooth function will not be the sum of a holomorphic function with an antiholomorphic one (for example, |z|^2 on \mathbb{C}). Also, the constant functions are both holomorphic and antiholomorphic. On the other hand, complex-valued smooth 1-forms do decompose ... 2 The transfer homomorphism of the orientation cover, composed with the pushforward, is multiplication by 2. The image is free abelian since the (n-1)-integral homology of the orientation cover is by my answer here, so the only torsion in H_{n-1}(X, \mathbb{Z}) is 2-torsion. Applying the universal coefficient theorem for H_n with \mathbb{Z}/2 ... 1 It should mean M*M = M\times M/ \sim, where (x, y) \sim (y, x). -- John Ma To confirm that the answer by John Ma is correct, I looked up another paper which gives this definition of symmetric product just before citing the aforementioned Milnor's paper about it. 1 So you have to show that F_{\ast}(X_p) : C^{\infty}_{F(p)}(N) \rightarrow \mathbb{R} is a linear map verifying a Leibniz rule, i.e. for all f,g \in C^{\infty}_{F(p)}(N) we have$$ F_{\ast}(X_p)(fg) = g(p)(F_{\ast}(X_p)f) + f(p)(F_{\ast}(X_p)g)  So let $f, g \in C_{F(p)}^\infty(N)$, et $\lambda \in \mathbb{R}$. We have \begin{eqnarray*} F_*(X_p) ...

1


1

Let $N$ be the neighbourhood of $g(t)$ in $M$. As $N$ is open and $g$ is continuous, $g^{-1}(N)$ is an open subset of $[a, b]$. As $t \in g^{-1}(N)$ and $g^{-1}(N)$ is open, there is $\delta > 0$ such that $(t - \delta, t + \delta) \subseteq g^{-1}(N)$. As such, $g((t - \delta, t + \delta)) \subseteq g(g^{-1}(N)) \subseteq N$.

1

Take $X = Y = \Bbb R$, and let $f:X\to Y$ be given by $f(x) = x^3$. Then $f$ is a $C^\infty$ bijection with continuous inverse $f^{-1}(y) = \sqrt[3]y$. However, the derivative of $f^{-1}$ fails to exist at $0$, so $f^{-1}$ is not $C^\infty$.

1

Usually Schubert cells are constructed using the structure theory of semisimple Lie groups (in this case $SL(3,\mathbb C)$), but for this case, one can give an explicit description as follows. Start by fixing one flag, say the standard one $\mathbb C\subset\mathbb C^2\subset\mathbb C^3$. This will be the unique Schubert-cell of dimension $0$. The further ...

1

The definition of Differentiable function on smooth manifold, does not depend on the choice of local chart. Since if $f:M\to \mathbb{R}$ be Differentiable at $p\in M$ and $(U,\varphi)$ be the corresponding local chart. For each local chart $(V,\psi)$ which $p\in V$, we have $p\in W=U\cap V$ and $f\circ \psi^{-1}|_{\psi(W)}=f\circ\varphi^{-1}\circ\varphi ... 1 A closed orientable$n$-manifold is called homology sphere if it has homology (or equivalently cohomology) of a sphere. Note that for a closed connected orientable 3-manifold we have always$H_0M\cong \mathbb Z \cong H_3M$and also$H_2M \cong H^1M \cong Hom(H_1M,\mathbb Z)$. You see that$H_1M=0\$ forces it to be a homology sphere.

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