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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. Indeed, from what I explained above, having $\{V(x)|V\in A_{\mathcal F}=T_xM\}$ for $x\in U$ means that $U$ is included in a leaf. But if your foliation is regular (being tangent to an integrable distribution $D$) then $D_{|U}=TU$, and thus if $U$ is moreover dense, then you get that $D=TM$.

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 $D=TM$.

The answer to your first question is yes: consider $D=TM$.

Let me add a few comments that may help.

a) 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).

b) 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 $M/\mathcal F$.

c) One can see from a) and b) that $A_{\mathcal F}$ is the Lie algebra of vector fields on $M/\mathcal F$. If the leaf space happens to be a manifold, then $A_{\mathcal F}$ will be the space of sections of a vector bundle on $M/\mathcal F$ (the tangent bundle of $M/\mathcal F$) and then the answer to your second question will be yes.

I would guess the answer to your last question is no. Indeed, from what I explained above, having $\{V(x)|V\in A_{\mathcal F}=T_xM\}$ for $x\in U$ means that $U$ is included in a leaf. But if your foliation is regular (being tangent to an integrable distribution $D$) then $D_{|U}=TU$, and thus if $U$ is moreover dense, then you get that $D=TM$.

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. Indeed, from what I explained above, having $\{V(x)|V\in A_{\mathcal F}=T_xM\}$ for $x\in U$ means that $U$ is included in a leaf. But if your foliation is regular (being tangent to an integrable distribution $D$) then $D_{|U}=TU$, and thus if $U$ is moreover dense, then you get that $D=TM$.

The answer to your first question is yes: consider $D=TM$.

Let me add a few comments that may help.

a) 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).

b) 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 $M/\mathcal F$.

c) One can see from a) and b) that $A_{\mathcal F}$ is the Lie algebra of vector fields on $M/\mathcal F$. If the leaf space happens to be a manifold, then $A_{\mathcal F}$ will be the space of sections of a vector bundle on $M/\mathcal F$ (the tangent bundle of $M/\mathcal F$) and then the answer to your second question will be yes.

I would guess the answer to your last question is no. Indeed, from what I explained above, having $\{V(x)|V\in A_{\mathcal F}=T_xM$ for $x\in U$ means that $U$ is included in a leaf. But if your foliation is regular (being tangent to an integrable distribution $D$) then $D_{|U}=TU$, and thus if $U$ is moreover dense, then you get that $D=TM$.

The answer to your first question is yes: consider $D=TM$.

Let me add a few comments that may help.

a) 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).

b) 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 $M/\mathcal F$.

c) One can see from a) and b) that $A_{\mathcal F}$ is the Lie algebra of vector fields on $M/\mathcal F$. If the leaf space happens to be a manifold, then $A_{\mathcal F}$ will be the space of sections of a vector bundle on $M/\mathcal F$ (the tangent bundle of $M/\mathcal F$) and then the answer to your second question will be yes.

I would guess the answer to your last question is no. Indeed, from what I explained above, having $\{V(x)|V\in A_{\mathcal F}=T_xM\}$ for $x\in U$ means that $U$ is included in a leaf. But if your foliation is regular (being tangent to an integrable distribution $D$) then $D_{|U}=TU$, and thus if $U$ is moreover dense, then you get that $D=TM$.