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9

Yes, it holds for manifolds with boundary. One way to see this is to note that if $M$ is a smooth manifold with boundary, then the inclusion map $\iota\colon \text{Int}\ M\hookrightarrow M$ is a smooth homotopy equivalence (Thm. 9.26 in ISM), which therefore induces isomorphisms on de Rham cohomology (Thm. 17.11). Since $\iota$ also induces isomorphisms ...


4

$H^0(X;G)$ is a direct product of as many copies of $G$ as path-connected components $X$ has. Your computation is correct.


3

The calculation, once again, looks correct, noting that since your resolving sheaves (the powers of the cotangent sheaf) are acyclic on this affine, you can just takes global sections instead of the more complicated hypercohomological approach. How you visualize what you get for $H^1$ is somewhat clear--and similar to your picture. Namely, you have for each ...


3

Holomorphic functions are always orientation-preserving, so this is not the reflection you're used to from real linear algebra. The map has degree $1$. If you want to think about it in real coordinates (e.g., looking at the induced map on $S^{2n+1}$), you'll see that it is a composition of two real reflections.


3

Tate writes in his article in Cassels and Frolich that the cohomology calculations have their origin in genus theory (going back --- at least --- to Gauss's Disquitiones). I haven't gone that far back, but I have read parts of Hilbert's Zahlbericht, where e.g. his Theorem 90 appeared. (The theorems are labelled in order throughout the book.) If you look at ...


3

It's just a matter of arithmetic, I think. Observe that the $n$-simplex has $\binom{n+1}{r+1}$ $r$-simplices as faces. Thus, the Euler characteristic of the simplicial chain complex of the $k$-skeleton is $$\sum_{r = 0}^{k} (-1)^r \binom{n+1}{r+1} = 1 + (-1)^k \binom{n}{k+1}$$ and as you say, its homology is concentrated in degrees $0$ and $k$, so (assuming ...


2

The cohomology formulation came at the end of a long line of development. Briefly, local class field theory was developed using Brauer groups, and Brauer groups can be interpreted as cohomology groups, then it was seen that one could dispense with the Brauer formalism and deal directly with the cohomology groups, finally the basic maps were interpreted as ...


2

All of these polynomial algebras are known; this is probably very classical material but I don't know a reference. Their generators have cohomological degree $2d$ where $d$ runs over the numbers from this list. For example, $H^{\bullet}(BG_2, \mathbb{Q})$ is a polynomial algebra on generators of degrees $4$ and $12$. You might want to look up some material ...


1

For those who want to compute it all out, if we use the delta-complex below instead of the one I gave in the question we have: Let $\phi \in C^0(K,\mathbb{Z}_2)$ be dual to $v\in C_0(K,\mathbb{Z}_2)$, $\alpha, \beta, \gamma \in C^1(K,\mathbb{Z}_2)$ be the dual elements to $a,b,c\in C_1(K,\mathbb{Z}_2)$, respectively, and $\mu,\lambda\in ...


1

The space $H^1(K) = \mathbb{Z}^2$ is generated by the Poincaré duals $\alpha = A^*$ and $\beta = B^*$ to $A$ and $B$, respectively. (I'm working over $\mathbb{Z}_2$ throughout, so that $H^*(K)$ is actually $H^*(K, \mathbb{Z}_2)$. For reasons of dimension, the only products you need to compute in the ring $H^*(K)$ are $\alpha^2, \alpha \beta$, and $\beta^2$. ...


1

Another German textbook including full details of the geometric proof of Poincaré duality is: Ralph Stöcker, Heiner Zieschang: Algebraische Topologie (1988) Contrary to the claim frequently found in sketches of the argument, the authors stress that the dual ‘cells’ are not in general cells in the topological sense: “Examples and sketches in ...


1

Suppose given a diagram of cochain complexes $$\begin{array} AA & =& A \\ \downarrow{f} & & \downarrow{g} \\ B & \stackrel{h}{\longrightarrow} & C. \end{array} $$ If $h$ is a homotopy equivalence, with homotopy inverse $h'$ also commuting with $f$ and $g$, then the induced map on mapping cones $$\mathrm{cone}(f) ...


1

If I understand your question correctly, and I'm not mistaken, the answer for a non-cyclic, finitely generated free group $F$ is negative. Fix a prime $p$. Clearly $F$ has a non-cyclic group $E$ of order $p^2$ as an image. Suppose $E = \langle a, b \rangle$, and let $K = \langle g \rangle$ be a cyclic group of order $p$. Now $E$ is the set-theoretic union ...


1

For any finite complex $X$ that is embedded in $\mathbb{R}^n$ in a piecewise smooth manner, $X$ has a regular neighborhood which is a compact $n$-manifold with boundary that deformation retracts to $X$. In the case of the standard embedding $S^2 \vee S^2 \subset \mathbb{R}^3$, that regular neighborhood is as described by @DanielRust, and consists of large ...



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