Betti number of nonnegative Ricci curvature and positive scalar curvature closed 3-mfd Suppose that $M^3$ is a closed 3-manifold with nonnegative Ricci curvature and positive scalar curvature, I think $b_1(M^3)\leq 1$. Is this right and is there a quick cut proof?
 A: By taking the orientable cover if necessary, we assume that $M$ is orientable (Note that the Betti number is not changed after taking the orientable cover). The Bochner formula for one form is 
$$\tag{1} \Delta_d \alpha = -\nabla^*\nabla \alpha + \text{Ric}(\alpha),$$
for all $\alpha$. Let $\alpha$ be a harmonic one form, then integrating the above formula gives
$$ 0= \int_M \|\nabla \alpha\|^2 + \text{Ric} (\alpha, \alpha) \ge \int_M \|\nabla \alpha\|^2 .$$
Thus $\alpha$ is parallel, that is $\nabla \alpha = 0$. Also using $(1)$ again, $$\tag{2} \text{Ric}(\alpha)  = 0$$
on $M$. If there are two nontrivial linearly independent harmonic one forms $\alpha_1, \alpha_2$, then 
$$\alpha_3 = *(\alpha_1\wedge \alpha_2)$$ 
is also a harmonic one form as $\Delta_d$ commutes with Hodge star $*$. Using $(2)$, we have
$$\text{Ric}(\alpha_i, \alpha_j) = 0 $$
on $M$, for $i, j = 1,2, 3$. Thus $M$ has zero Ricci curvature as $\{\alpha_1, \alpha_2, \alpha_3\}$ is linearly independent and $\dim M = 3$. Hence $M$ does not have positive scalar curvature. As a result, there cannot be two linearly independent harmonic one forms on $M$ and so $b_1(M) \le 1$ by Hodge theory. 
