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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)

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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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.

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  • $\begingroup$ Is it possible that you have the sign wrong in the definition of $\phi_-$ (i.e. $2^{-k} \mapsto 2^k$)? $\endgroup$ – kosta Jul 1 '16 at 18:02
  • $\begingroup$ @kosta: it is entirely possible. Thanks, fixed. $\endgroup$ – Willie Wong Jul 1 '16 at 18:35
  • $\begingroup$ @WillieWong thanks a lot for your answer. $\endgroup$ – Ali Taghavi Jul 2 '16 at 20:47
  • $\begingroup$ @WillieWong I am sorry if my question is elementary: But are not you using some thing as Mercer Theorem?If yes, Is your g a symmetric function? $\endgroup$ – Ali Taghavi Feb 25 '17 at 19:20
  • $\begingroup$ @AliTaghavi: (1) I don't know Mercer Theorem. Perhaps it is using the same idea. (The idea is elementary, so likely appears in many places.) (2) Do you mean symmetric in $x$ and $y$? Then obviously no: if it were symmetric, then $\int g(x,y) f(y) dy = 0$ for $f$ being any constant function, which basically is the entire opposite of the construction. $\endgroup$ – Willie Wong Feb 26 '17 at 3:04
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Willie Wong answered the general case, and I'd like to give a counterexample for the symplectic case:

Let $M=S^2$, the standard 2-sphere embedded in $\mathbb{R}^3$ as the unit sphere, with the standard symplectic form corresponding to the Euclidean metric on $\mathbb{R}^3$. Take on $(M \times M, \Omega = \omega \oplus - \omega)$ the functions \begin{align} f(z,w) &= z_3\cdot F(w), \\ g(z,w) &= z_3 \cdot G(w) \end{align}

both of which are in $L$. Now, $\{ f, g \} = \Omega(\mathrm{sgrad}(g), \mathrm{sgrad}(f)) =: \Omega'(\mathrm{d}f, \mathrm{d}g)$. Then,

\begin{align} \{f,g\} &= (\omega' \oplus (-\omega')) \left( F(w)\mathrm{d} z_3 + z_3 \mathrm{d} F(w), G(w)\mathrm{d} z_3 + z_3 \mathrm{d} G(w) \right) \\ &= F(w)G(w) \omega'(\mathrm{d}z_3,\mathrm{d}z_3) - z_3^2 \omega'(\mathrm{d} F(w),\mathrm{d} G(w)) \\ &= z_3^2 \{G,F\}(w) =:h(z,w) \end{align}

Now, unless $\{G,F\} \equiv 0$, $h$ will not be in $L$.

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  • $\begingroup$ thank you so much for your answer.I think you mean $\Omega(X_{f},X_{g}$ where the later are hamiltonian vector fields. Moreover what is $\Omega '$. How a two form act one forms df and dg. $\endgroup$ – Ali Taghavi Jul 2 '16 at 18:18
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    $\begingroup$ @AliTaghavi $X_f \equiv \mathrm{sgrad}(f)$ - just different notation. A symplectic form $\omega$ gives an isomorphism between $TM$ and $T^*M$ by $\eta \leftrightarrow \iota_X \omega$. Define $\omega'(\eta,\sigma) := - \omega(X,Y)$ where $\eta = \iota_X \omega$ and $\sigma = \iota_Y \omega$. I may have messed up the sign somewhere. $\endgroup$ – kosta Jul 2 '16 at 18:34
  • $\begingroup$ Thank you again for your very interesting answer. I am sorry that I can not accept two answers, simultaneously. meta.mathoverflow.net/questions/1491/… $\endgroup$ – Ali Taghavi Aug 10 '16 at 9:31
  • $\begingroup$ where did you use the minus sign in $\Omega= \omega\oplus -\omega$? $\endgroup$ – Ali Taghavi Feb 25 '17 at 10:32
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    $\begingroup$ I think the same argument works for $\omega \oplus \omega$. am I right? $\endgroup$ – Ali Taghavi Feb 25 '17 at 10:34

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