Is there any Bochner Technique type for two forms?

Question

One of the beautiful classical results in DG is the Bochner Technique.

Theorem (Bochner, 1948). If $(M, g)$ is compact and has $\rm Ric\geq 0$, then every harmonic $1$-form is parallel.

I want to know is there any similar results for two forms together non-negative Ricci curvature and then some estimate for second Betti number (If am right)?

Answer 1

I am not sure if it is what you want, but we do have a similar result, under a stronger hypothesis:

Theorem (D. Meyer, S. Gallot). Let $(M,g)$ be a closed Riemannian manifold with nonnegative curvature operator. Then all harmonic form on $M$ are parallel.

The proof is simply Bochner technique in general. For a general tensor $T$, we have the Bochner formula $$\frac{1}{2}\Delta|T|^2=|\nabla T|^2-\langle\nabla^*\nabla T,T\rangle\geq0,$$ where $\nabla^*$ is the formal adjoint of $\nabla$. Thus to generalize Bochner's theorem, we only need to ensure that $\langle\nabla^*\nabla T,T\rangle\leq0$.

Define the Weitzenböck curvature operator $\operatorname{Ric}$ acting on any tensor $T$ by $$(\operatorname{Ric}(T))(X_1,\ldots,X_n):=\sum_{i,j}(R(E_j,X_i)T)(X_1\ldots,X_{i-1},E_j,X_{i+1}\ldots,X_n),$$ where $E_j$ is an orthonormal frame. For $c>0$, define the Lichnerowicz Laplacian $\Delta_L:=\nabla^*\nabla+c\operatorname{Ric}$ acting on any tensor. To show that $\langle\nabla^*\nabla T,T\rangle\leq0$, it suffices that $T\in\ker\Delta_L$ and $\langle\operatorname{Ric}(T),T\rangle\geq0$. It remains to do two things. First, find $c$ such that $\Delta_L$ is geometrically meaningful. Second, analyze when $\operatorname{Ric}(T)$ is nonnegative.

The latter is ensured if the curvature operator is nonnegative. For the former, we need the following Weitzenböck formula stating that, on $k$-forms, the Hodge Laplacian is just the Lichnerowicz Laplacian with $c=1$: $$(d\delta+\delta d)\omega=\nabla^*\nabla\omega+\operatorname{Ric}(\omega)\quad\text{for $k$-form }\omega.$$ Then our proof is complete.

In fact, using more analysis, we can obtain Betti number estimates even when the curvature operator $\mathfrak{R}$ is only bounded below. Explicitly, if $\mathfrak{R}\geq-k$ for $k\geq 0$, and $\operatorname{diam}M\leq D$, then we have the following estimate for the Betti numbers: $$b_\ell(M)\leq\binom{n}{\ell}e^{C(n,kD^2)},$$ and $\lim_{\varepsilon\to0}C(n,\varepsilon)=0$.

I am no expert in Riemannian geometry, and I learned all of the above from Chapter 9 of the book Riemannian Geometry (Third Edition) by Peter Petersen. By the way, this book is both well-written and very informative!