# Is there an upper bound on dimension of kernel of elliptic operator for a fixed closed manifold M

Assume that $M$ is a smooth closed manifold and $E,F$ are fixed smooth vector bundles over $M.$

Is there a number $C,$ such that for any elliptic operator $\mathcal{D}:\Gamma(E)\to\Gamma(F)$ $$\dim\ker\mathcal{D}\leqslant C.$$

For simplicity, we may assume that $\Gamma(E)=\Gamma(F)=C^\infty(M).$

• I think $\prod_{k=1}^n, (d^2+k^2)$ on $S^1$ is another counter example. But do you have an example of such manifold? What about if we assume some restriction on the principal simple and requre the bounded ness of index? – Ali Taghavi Jun 13 '17 at 20:34

There are, at least, many examples where this is false. For instance, Hitchin showed (I think this was his thesis) that there is a sequence of metrics $\{g_k\}_{k=1}^{\infty}$ on the three sphere such that the kernel of the Dirac operator associated to $g_{k}$ has dimension at least $k.$ It is stated as a conjecture in these notes http://www.mathematik.uni-regensburg.de/ammann/talks/11BerlinSFB.pdf that this is a generic phenomena for spin manifolds of dimension at least three.

In the case where $E=F=M\times \mathbb{R},$ I don't have any examples in mind off the top of my head, but I could imagine one could construct a family of metrics $\{g_{k}\}$ and a family of functions $f_{k}:M\rightarrow \mathbb{R}$ such that the sequence of elliptic operators \begin{align} D_{k}=\Delta_{g_{k}} + f_{k} \end{align} has arbitrarily large kernel.

With this in mind, it's hard to imagine a circumstance where what you are asking is true. A better question might be: for some sensible elliptic operator that depends on a parameter, like the Dirac operator on a spin manifold depending on a Riemannian metric, can you find sensible geometric bounds on the Riemannian metric under which this kernel has a uniform bound.

• I purpously posted this question in such naive way, because I was unable to came up with any exaples. Thank you for this neat answer. – Fallen Apart Feb 18 '17 at 21:59

Here is another simple counterexample: every holomorphic vector bundle over a Riemann surface gives rise to an elliptic operator (the $\bar\partial$-operator) whose kernel is the space of holomorphic sections. Two vector bundles over the same compact surface are isomorphic as smooth complex vector bundles if and only if the have the same rank and the same degree. Now take the Riemann sphere $P^1$ and a line bundle $L\to P^1$ of degree $d\geq 1$ and the rank $2$ vector bundle $$E=L\oplus L^*\to P^1.$$ $E$ has degree $0$ for every $d$ but the dimension of holomorphic sections, i.e. the dimension of the kernel is $d+1$.

On the other hand, there is a class of elliptic operators for which such an estimate exists: every linear elliptic differential operator $$D\colon\Gamma(M,L)\to\Gamma(M,\tilde L)$$ of order 1 between complex line bundles $L,\tilde L$ over a surface $M$ is given as the $\bar\partial$-operator of some holomorphic line bundle (with respect to an appropriate Riemann surface structure). The dimension of the kernel is then restricted by $$\dim\ker D\leq \mathrm{deg}(L)+1$$ because we can always produce a $(k-1)$th order zero at a point for some section in a $k$-dimensional space of holomorphic sections of a line bundle.

• Excuse me Sebastian, but I am afraid that I cannot understand the phrase starting with "On the other hand", could you please edit it? – Paul-Benjamin Feb 20 '17 at 10:22
• Sorry for the late reply: What I mean is that in general, there is no such estimate, but in special situations (like line bundles over compact Riemann surfaces and elliptic operators of order 1), there can be such an estimate. – Sebastian Jun 14 '17 at 6:33