We consider the following simple fact about matrices. Then we try to generalize it in the context of smooth manifolds;


Let $L$ be the collection of all $n \times n$ real matrices $A=(a\_{ij})$ with the following property:


$$\sum\_{i=1}^{n} a\_{ij}=0$$ for every fixed $j$.


Obviousely $L$ is a Lie algebra.([As I have already learned from Qiaochu Yuan in another MO post](https://mathoverflow.net/questions/225811/a-particular-lie-algebra-l-n-and-various-lie-groups-whose-lie-algebra-is-i/225815))


Moreover the linear map $X \mapsto AX$ has non trivial kernel.


These simple facts can be modelized in an infinite dimensional manner.(note that a vector $X\in\mathbb{R}^{n}$ can be considered as a function on a finite ($n$ pointed)set $M$ equiped with discrete counting measure and a matrix is a function on $M\times M$. Now the product $AX$ has an integral representation if we replace $\sum$ by the integral sign. That is we read the expersion $\sum a\_{ij}x\_{j}$ in the integral form $\int\_{M} a\_{ij}x\_{j}$ where the integration is based on the normalized counting measure.


Now we state our questions as generalization of the above simple fact about matrices.


Assume that $M$ is a compact orientable manifold or a Lie or topological group. So $M$ has a natural measure, correspond to volum form or the invariant metric or Haar measure.
Assume that $g: M \times M \to \mathbb{R}$ is a smooth function which satisfies $$\int\_{M} g(x,y)dx=0\;\;\;\;\;(1) $$ for all $y \in M$.



> 
> Does the linear map $A$ on $C^{\infty} (M)$ has nontrivial kernel?
>  $$A(f)(x)=\int\_{M} g(x,y)f(y)dy$$
> 
> 
> 


Note that for topological groups we consider continuous functions, since smoothness is meaningles.


For our next question, we assume that $M$ is a symplectic manifold, so $M \times M$ has a natural symplectic structure. 
Let $L$ be the space of all smooth functions on $M\times M$ which satisfy the equation (1).



> 
> Is $L$ closed under Poisson bracket?
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> 
> 



Infinite dimensional version of a simple fact on certain singular matrices
For the first question, the answer is not necessarily. 


**Very rough idea**: The rank-nullity theorem doesn't always hold on infinite dimensional spaces. 


**Rough idea**: Let the operator $A$ be defined on $L^2(M)$ be a injective mapping such that its range does not include the constant function. More precisely, since $M$ is compact we can enumerate its eigenvalues (of the Laplacian) increasing with multiplicity as $\lambda\_i$, with $\lambda\_0 = 0$ corresponding to the constants. Now let $\psi:\overline{\mathbb{N}}\to \overline{\mathbb{N}}$ such that $\psi$ is injective and such that the range of $\psi$ does not contain $0$. Then defining $A$ as the map that sends the $i$th eigenspace to $\psi(i)$th eigenspace will provide a counterexample. 


**Realization**: In practice to guarantee smoothness it is easier to not keep $A$ an isometry. Take $M = \mathbb{S}^1$ for simplicity. Let 


$$ \phi\_-(x) := \sum\_{k < 0} 2^{-|k|} e^{ik x} $$


The series is absolutely convergent and in fact defines a $C^\infty$ function. Similarly we define


$$ \phi\_+(x) := \sum\_{k \geq 0} 2^{-|k|} e^{ikx} $$


Define your function $g$ by 


$$ g(x,y) = \phi\_-(x-y) + e^{ix} \phi\_+(x-y) $$


It is easy to check that $\int\_{0}^{2\pi} g(x,y) ~\mathrm{d}x = 0$ for any fixed $y$. But the operator $f(x) \mapsto \int g(x,y) f(y) ~\mathrm{d}y$ has no nontrivial kernel. 


