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user156937
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Let $\tilde{M}\to M$ be a finite covering of closed 3-manifolds.

Is it possible that $\beta_1(\tilde{M})-1> [\pi_1(M):\pi_1(\tilde{M})]\beta_1(M)$$\beta_1(\tilde{M})-1> [\pi_1(M):\pi_1(\tilde{M})](\beta_1(M)-1)>0$?

Here, $\beta_1(X)=\dim_{\mathbb{Q}}H_1(X;\mathbb{Q})$.

From the Reidemeister-Schreier method, $\operatorname{rank}(\pi_1(\tilde{M}))-1\leq [\pi_1(M):\pi_1(\tilde{M})](\operatorname{rank}\pi_1(M)-1)$.

Here, $\operatorname{rank}(G)$ is the least possible number of generators of $G$. By definition, $\operatorname{rank}(\pi_1(X))\geq \beta_1(X)$.

I'm mainly interested in the 3-manifold case though the question does make sense for general $n$-dimensional manifolds.

Edit: For a psychological reason, I wantknow that there is a slightly strongerhomology 3-sphere which has a finite cover which is not a homology 3-sphere. So I changed the inequality.

Let $\tilde{M}\to M$ be a finite covering of closed 3-manifolds.

Is it possible that $\beta_1(\tilde{M})-1> [\pi_1(M):\pi_1(\tilde{M})]\beta_1(M)$?

Here, $\beta_1(X)=\dim_{\mathbb{Q}}H_1(X;\mathbb{Q})$.

From the Reidemeister-Schreier method, $\operatorname{rank}(\pi_1(\tilde{M}))-1\leq [\pi_1(M):\pi_1(\tilde{M})](\operatorname{rank}\pi_1(M)-1)$.

Here, $\operatorname{rank}(G)$ is the least possible number of generators of $G$. By definition, $\operatorname{rank}(\pi_1(X))\geq \beta_1(X)$.

I'm mainly interested in the 3-manifold case though the question does make sense for general $n$-dimensional manifolds.

Edit: For a psychological reason, I want a slightly stronger inequality.

Let $\tilde{M}\to M$ be a finite covering of closed 3-manifolds.

Is it possible that $\beta_1(\tilde{M})-1> [\pi_1(M):\pi_1(\tilde{M})](\beta_1(M)-1)>0$?

Here, $\beta_1(X)=\dim_{\mathbb{Q}}H_1(X;\mathbb{Q})$.

From the Reidemeister-Schreier method, $\operatorname{rank}(\pi_1(\tilde{M}))-1\leq [\pi_1(M):\pi_1(\tilde{M})](\operatorname{rank}\pi_1(M)-1)$.

Here, $\operatorname{rank}(G)$ is the least possible number of generators of $G$. By definition, $\operatorname{rank}(\pi_1(X))\geq \beta_1(X)$.

I'm mainly interested in the 3-manifold case though the question does make sense for general $n$-dimensional manifolds.

Edit: I know that there is a homology 3-sphere which has a finite cover which is not a homology 3-sphere. So I changed the inequality.

[Edit removed during grace period]; added 77 characters in body
Source Link
user156937
  • 541
  • 2
  • 8

Let $\tilde{M}\to M$ be a finite covering of closed 3-manifolds.

Is it possible that $\beta_1(\tilde{M})-1> [\pi_1(M):\pi_1(\tilde{M})](\beta_1(M)-1)$$\beta_1(\tilde{M})-1> [\pi_1(M):\pi_1(\tilde{M})]\beta_1(M)$?

Here, $\beta_1(X)=\dim_{\mathbb{Q}}H_1(X;\mathbb{Q})$.

From the Reidemeister-Schreier method, $\operatorname{rank}(\pi_1(\tilde{M}))-1\leq [\pi_1(M):\pi_1(\tilde{M})](\operatorname{rank}\pi_1(M)-1)$.

Here, $\operatorname{rank}(G)$ is the least possible number of generators of $G$. By definition, $\operatorname{rank}(\pi_1(X))\geq \beta_1(X)$.

I'm mainly interested in the 3-manifold case though the question does make sense for general $n$-dimensional manifolds.

Edit: For a psychological reason, I want a slightly stronger inequality.

Let $\tilde{M}\to M$ be a finite covering of closed 3-manifolds.

Is it possible that $\beta_1(\tilde{M})-1> [\pi_1(M):\pi_1(\tilde{M})](\beta_1(M)-1)$?

Here, $\beta_1(X)=\dim_{\mathbb{Q}}H_1(X;\mathbb{Q})$.

From the Reidemeister-Schreier method, $\operatorname{rank}(\pi_1(\tilde{M}))-1\leq [\pi_1(M):\pi_1(\tilde{M})](\operatorname{rank}\pi_1(M)-1)$.

Here, $\operatorname{rank}(G)$ is the least possible number of generators of $G$. By definition, $\operatorname{rank}(\pi_1(X))\geq \beta_1(X)$.

I'm mainly interested in the 3-manifold case though the question does make sense for general $n$-dimensional manifolds.

Let $\tilde{M}\to M$ be a finite covering of closed 3-manifolds.

Is it possible that $\beta_1(\tilde{M})-1> [\pi_1(M):\pi_1(\tilde{M})]\beta_1(M)$?

Here, $\beta_1(X)=\dim_{\mathbb{Q}}H_1(X;\mathbb{Q})$.

From the Reidemeister-Schreier method, $\operatorname{rank}(\pi_1(\tilde{M}))-1\leq [\pi_1(M):\pi_1(\tilde{M})](\operatorname{rank}\pi_1(M)-1)$.

Here, $\operatorname{rank}(G)$ is the least possible number of generators of $G$. By definition, $\operatorname{rank}(\pi_1(X))\geq \beta_1(X)$.

I'm mainly interested in the 3-manifold case though the question does make sense for general $n$-dimensional manifolds.

Edit: For a psychological reason, I want a slightly stronger inequality.

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user156937
  • 541
  • 2
  • 8

The first Betti number of a finite covering space of a closed 3-manifold

Let $\tilde{M}\to M$ be a finite covering of closed 3-manifolds.

Is it possible that $\beta_1(\tilde{M})-1> [\pi_1(M):\pi_1(\tilde{M})](\beta_1(M)-1)$?

Here, $\beta_1(X)=\dim_{\mathbb{Q}}H_1(X;\mathbb{Q})$.

From the Reidemeister-Schreier method, $\operatorname{rank}(\pi_1(\tilde{M}))-1\leq [\pi_1(M):\pi_1(\tilde{M})](\operatorname{rank}\pi_1(M)-1)$.

Here, $\operatorname{rank}(G)$ is the least possible number of generators of $G$. By definition, $\operatorname{rank}(\pi_1(X))\geq \beta_1(X)$.

I'm mainly interested in the 3-manifold case though the question does make sense for general $n$-dimensional manifolds.