Talk:Betti numbers of 3-manifolds (Ex)

Revision as of 09:07, 7 January 2019 (view source) Maggie Miller (Talk | contribs) (Created page with "<wikitex>; Now let $M$ be a closed, orientable $3$-manifold with $\pi_1(M)=G$. View $M$ as a finite CW complex, with cells in dimensions-$0,1,2,3$, as usual. Attach cells of ...")		Latest revision as of 07:06, 8 January 2019 (view source) Fabian Henneke (Talk | contribs) m
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	Thus, $b_1(G)=\text{rk} H_1(N;\mathbb{Z})=\text{rk}H_1(M;\mathbb{Z})=\text{rk}H_2(M;\mathbb{Z})\ge \text{rk} H_2(N;\mathbb{Z})=b_2(G)$, so $b_1(G)\ge b_2(G)$.		Thus, $b_1(G)=\text{rk} H_1(N;\mathbb{Z})=\text{rk}H_1(M;\mathbb{Z})=\text{rk}H_2(M;\mathbb{Z})\ge \text{rk} H_2(N;\mathbb{Z})=b_2(G)$, so $b_1(G)\ge b_2(G)$.
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Latest revision as of 07:06, 8 January 2019

Now let $<wikitex>; Now let $M$ be a closed, orientable $-manifold with $\pi_1(M)=G$. View $M$ as a finite CW complex, with cells in dimensions-<script type="math/tex">M$ be a closed, orientable $3$-manifold with $\pi_1(M)=G$. View $M$ as a finite CW complex, with cells in dimensions-$0,1,2,3$, as usual. Attach cells of dimension $\ge 3$ to $M$ to obtain $N$, a $K(G,1)$ space. Then $b_1(G)$ is the rank of $H_1(N;\mathbb{Z})$ and $b_2(G)$ is the rank of $H_2(N;\mathbb{Z})$. By construction of $N$, $H_1(N;\mathbb{Z})\cong H_1(M;\mathbb{Z})$ and the rank of $H_2(N;\mathbb{Z})$ is bounded above by the rank of $H_2(M;\mathbb{Z})$. Moreover, $M$ is a $3$-manifold, so Poincaré duality says the ranks of $H_1(M;\mathbb{Z})$ and $H_2(M;\mathbb{Z})$ agree.

Thus, $b_1(G)=\text{rk} H_1(N;\mathbb{Z})=\text{rk}H_1(M;\mathbb{Z})=\text{rk}H_2(M;\mathbb{Z})\ge \text{rk} H_2(N;\mathbb{Z})=b_2(G)$, so $b_1(G)\ge b_2(G)$.

,1,2,3$, as usual. Attach cells of dimension $\ge 3$ to $M$ to obtain $N$, a $K(G,1)$ space. Then $b_1(G)$ is the rank of $H_1(N;\mathbb{Z})$ and $b_2(G)$ is the rank of $H_2(N;\mathbb{Z})$. By construction of $N$, $H_1(N;\mathbb{Z})\cong H_1(M;\mathbb{Z})$ and the rank of $H_2(N;\mathbb{Z})$ is bounded above by the rank of $H_2(M;\mathbb{Z})$. Moreover, $M$ is a $-manifold, so Poincaré duality says the ranks of $H_1(M;\mathbb{Z})$ and $H_2(M;\mathbb{Z})$ agree. Thus, $b_1(G)=\text{rk} H_1(N;\mathbb{Z})=\text{rk}H_1(M;\mathbb{Z})=\text{rk}H_2(M;\mathbb{Z})\ge \text{rk} H_2(N;\mathbb{Z})=b_2(G)$, so $b_1(G)\ge b_2(G)$. M be a closed, orientable $3$-manifold with $\pi_1(M)=G$. View $M$ as a finite CW complex, with cells in dimensions-$0,1,2,3$, as usual. Attach cells of dimension $\ge 3$ to $M$ to obtain $N$, a $K(G,1)$ space. Then $b_1(G)$ is the rank of $H_1(N;\mathbb{Z})$ and $b_2(G)$ is the rank of $H_2(N;\mathbb{Z})$. By construction of $N$, $H_1(N;\mathbb{Z})\cong H_1(M;\mathbb{Z})$ and the rank of $H_2(N;\mathbb{Z})$ is bounded above by the rank of $H_2(M;\mathbb{Z})$. Moreover, $M$ is a $3$-manifold, so Poincaré duality says the ranks of $H_1(M;\mathbb{Z})$ and $H_2(M;\mathbb{Z})$ agree.

Thus, $b_1(G)=\text{rk} H_1(N;\mathbb{Z})=\text{rk}H_1(M;\mathbb{Z})=\text{rk}H_2(M;\mathbb{Z})\ge \text{rk} H_2(N;\mathbb{Z})=b_2(G)$, so $b_1(G)\ge b_2(G)$.