A Lie algebra associated to a foliation(A kind of saturation of foliations)

Question

Inspired by the Lie algebra discussed in this answer, we consider the following Lie agebra associated to a given foliation:

Let $\mathcal{F}$ be a nontrivial foliation of a manifold $M$ tangent to integrable subbundle $D$ of $TM$. We define the following Lie algebra of vector fields on $M$:

$$A_{\mathcal{F}}=\{Y\in \chi^{\infty}(M)\mid [X,Y]\;\; \text{is tangent to $\mathcal{F}$} \text{ for all } X\in \Gamma (D)\} $$

In fact $A_{\mathcal{F}}$ is the idealizer of the Lie algebra $L_{\mathcal{F}}$ of vector fields on $M$ which are tangent to the foliation.

First Question: Is there a foliation $\mathcal{F}$ for which $A_{\mathcal{F}}=L_{\mathcal{F}}$?

The second question:Is it true to say that the dimension of $\{Y(p)\mid Y \in A_{\mathcal{F}}\}\subseteq T_pM$ is independent of Choosing $p\in M$?

If the answer of the second question is yes, then $A_{\mathcal{F}}$ defines an integrable distribution $D'$ containing the initial distribution $D$. It generates a foliation $\mathcal{F}'$ which would be defined as saturation of $\mathcal{F}$.

Is there an example of this situation such that $dim(\mathcal{F}) <dim({\mathcal{F}')<dim(M)}$?

Remark: As we see in the linked question, when we have a $1$-dimensional foliation $\mathcal{F}$ tangent to a non vanishing vector field $X$ on a surface $M$ with volume form $\omega$ , then the Lie algebra $A_{\mathcal{F}}$ is equal to $$\{Y\mid [X,Y]=fX,\;\text{for some }f\in C^{\infty}(M)\}=\{Y\in X^{\infty}(M)\mid X.\omega(X,Y)=Div X\omega(X,Y)\}$$.

Added: According to the comment of Bertram Arnold we add the following question:

Is it true to say that there is an open dense subset $U\subset M$ with the following two properties:

1) For every $x\in U,\; \{V(x)|V\in A_{\mathcal{F}}\}=T_x M \}$.

2)$U$ and $M\setminus U$ are $\mathcal{F}$- saturated.

Then it seems that $M\setminus U$ is a characyteristic set in the sence that it is invariant under every leaf preserving diffeomorphism.

Answer 1

1. First question

The answer to your first question is yes: consider $D=TM$. In other words, the foliation $\mathcal F$ has a unique leaf, which is $M$ itself. Hence $A_{\mathcal F}=L_{\mathcal F}=\mathfrak{X}(M)$.

2. $A_{\mathcal F}$ are not sections of a bundle on $M$

As Bertram Arnold says, $A_{\mathcal F}$ is not stable under multiplication by functions in $C^\infty(M)$. But it is stable under multiplication by functions $f$ that are constant along the leaves (functions on $M/\mathcal F$, so to say).

3. A bit of symplectic geometry

There is a nice symplectic interpretation of the geometric setup you describe. You can skip it and go directly to the next paragraph if you don't know anything about symplectic geometry. Consider the cotangent space $T^*M$, which is a symplectic manifold. Functions on this are generated by functions on $M$ and vector fields, and the corresponding Poisson bracket is defined by: $$ \{X,f\}=X\cdot f\quad\mathrm{and} \quad\{f,g\}=0\,. $$ On can then consider the ideal $\mathcal I_D$ generated by $\Gamma(D)$. Geometrically this is the ideal of functions vanishing on the zero locus of the induced dual map $$ T^*M\longrightarrow D^* $$ This zero locus is a coisotropic submanifold, and thus one can perform symplectic reduction. Algebraically (i.e. on the level of functions), it amounts to consider the quotient $NP(\mathcal I_D)/\mathcal I_D$ of the Poisson normalizer $NP(\mathcal I_D)$ of the ideal by the ideal itself. Reduced spaces are in general quite singular. But, morally speaking, the reduced space shall be thought of as $T^*(M/\mathcal F)$, the cotangent to the leaf space $M/\mathcal F$.

4. What is $A_{\mathcal F}/L_{\mathcal F}$ ?

One can see from 2 and 3 above that $A_{\mathcal F}/L_{\mathcal F}$ is the Lie algebra of vector fields on $M/\mathcal F$. If the leaf space happens to be a manifold, then $A_{\mathcal F}/L_{\mathcal F}$ will be the space of sections of a vector bundle on $M/\mathcal F$ (the tangent bundle of $M/\mathcal F$), and thus so will be $A_{\mathcal F}$. Hence, in this case, the answer to your second question will be yes.

5. Last question

I would guess the answer to your last question is no. This is because having $\{V(x)|V\in A_{\mathcal F}\}=T_xM$ for every $x\in U$ means that $D_{|U}=TM_{|U}$. Indeed, if $V\in L_{\mathcal F}$ is non-zero at $x$, then we chose a function $f$ such that $(V\cdot f)(x)=1$ and we have $$ V'(x)=[V,fV'](x)-f(x)[V,V'](x)\in D_x $$ for every vector field $V'$. Hence $D_{|U}=TM_{|U}$.

Thus, if $U$ is moreover dense ($U$ being non-empty and $M$ connected is actually sufficient), then you get that $D=TM$.