Let $(M,g\_0)$ be a closed $n$-dimensional Riemannian manifold. Let $1<k<n$ be fixed, and let $\Delta\_{g\_0}:\Omega^k(M) \to \Omega^k(M)$ be the $g\_0$-Laplacian. Let $H^k\_{g\_0}=\text{ker} \Delta\_{g\_0}$. 



> 
> Suppose $g\_{\epsilon}$ is close to $g\_0$ in the $C^1$ sense. Is it true that $H^k\_{g\_0}$ is "close" to $H^k\_{g\_{\epsilon}}$ in some sense?
> 
> 
> 


**Here is one way of quantifying what does it mean for $H^k\_{g\_0}$ to be close $H^k\_{g\_{\epsilon}}$:** (I guess there are other ways, this seemed natural to me).


Take the sup norm on $\Omega^k(M)$ (where the pointwise norm is the one induced by $g\_0$), and let $S$ be the unit sphere of $\Omega^k(M)$ w.r.t this norm. Define $S\_0=H^k\_{g\_0} \cap S, S\_{\epsilon}=H^k\_{g\_{\epsilon}} \cap S.$ Set


$$ d(H^k\_{g\_{\epsilon}},H^k\_{g\_{0}}):=d\_H(S\_{\epsilon},S\_0)$$
where $d\_H$ is the [Hausdorff distance](https://en.wikipedia.org/wiki/Hausdorff_distance) of $H^k\_{g\_{\epsilon}},H^k\_{g\_{0}}$ inside $\Omega^k(M)$.



> 
> Is it true that if $g\_{\epsilon}$ is close to $g\_0$ then $H^k\_{g\_{\epsilon}}$ is close to $H^k\_{g\_{0}}$ in this sense?
> 
> 
> 


I am not particularly fussed on this specific proximity measure, so results using 
other notions of distance are also welcomed.



Does the space of harmonic forms change continuously with the metric?
I think the answer is positive.


Let $D$ be the subspace of smooth **closed** $k$-forms on $M$. Equip $D$ with the supremum- $C^1$ norm:
$$
\| \omega \|\_{C^1,sup}:=\max\{ \|\omega\|\_{sup}, \|T\omega\|\_{sup} \},
$$


All the norms are w.r.t $g\_0$.


Let $\delta\_g$ be the codifferential of $d$ w.r.t the metric $g$. We consider $\delta\_g$ as as family of bounded linear operators, where $g \to g\_0$ in $C^1$.


$$
\delta\_{g}:(D,\| \cdot \|\_{C^1,sup}) \to (\Omega^{k-1}(M),\| \cdot \|\_{sup})
$$


The image of $\delta\_g$ is **closed:** Indeed, by the Hodge theorem
$$ \delta\_g(\Omega^k(M)) \supseteq \delta\_g(D) \supseteq \delta\_g\big(\text{Image}(d)\big) = \delta\_g(\Omega^k(M))=(\text{Image}(d)\oplus H\_g)^{\perp}$$ 


so $ \delta\_g(D)= \delta\_g(\Omega^k(M)) $ is closed in $\Omega^k(M)$ w.r.t the $L^2$ metric induced by $g$ hence (since $M$ is compact) also w.r.t to the the $L^2$ metric induced by $g\_0$. This implies it's closed w.r.t the uniform norm $\|\cdot\|\_{sup}$.


Also, it is easy to see that


$$\| \delta\_{g}-\delta\_{g\_0} \|\_{op} \le C(g\_0) \|g-g\_0\|\_{C^1,sup}, $$


so $\delta\_{g} \to \delta\_{g\_0}$ when $g \to g\_0$ in $C^1$.


Thus, we have a family of operators with closed images which converge to a limiting operator, where all the operators in the family have finite-dimensional kernels of the same dimension. This implies that "the kernels converge": we can choose bases for $H\_g=\ker \delta\_g$ which converge to a basis of $H\_{g\_0}=\ker \delta\_{g\_0}$.


The full argument for that claim is [here](https://math.stackexchange.com/a/2892319/104576).


