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I am studying the basics of geometrical control theory, and I am struggling with some concepts. At the moment I am studyng the concept of distribution.

So far I have understood that a distribution is a law that associates to each point $x$ a subspace of the teangent space of $x$ :

$$\Delta : x \rightarrow \Delta (x)\subset T_x\mathbb{R}^{n}$$

but I cannot grasp the concept. For example, if I consider a distribution:

$$\Delta (x)=\begin{pmatrix} x_1 & 1\\ x_1x_3 &x_1 \\ 0 &0 \end{pmatrix}$$

if I consider the definition, it should associate to each point a subspace, but what does it mean?

Maybe each column of the distribution if a vector, and so a collection of vectors defines a subspace? This that I just said is just a reasoning I did, so I am not sure.

Moreover, I have studied that a distribution is given by a set of independet vectors:

$$\Delta (x)=\operatorname{span}[f_1(x),....,f_n(x)]$$

which I think is has to be true, otherwise they won't define a space. But I am also confused by the fact that each vectors is associated to a point, so if I take each vector by itself, I have a space with more vectors associated to points.

After this, the notes of my professor start describing costant rank distributions and integrable distributions, which are hard to understand for me at this point, considering that I have not clear the concept of distribution.

Can somebody please help me?

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    $\begingroup$ It's just a subbundle of your tangent bundle, nothing more, one thing to note: rank of the subbundle is necessarily less than of equal to $m$( which is dimension of your underlying smooth manifold). And the "each point" really comes from the tangent space, because to each point $p \in M$ of the manifold there exists a tangent space $T_pM$ which is isomorphic (vector space isomorphism) to $\mathbb{R}^{\text{dim} M}$ $\endgroup$
    – Siddhartha
    Apr 27, 2020 at 16:13

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A distribution $D$ with rank $k\leq n$ of a $n-$dimensional smooth manifold $M$ is a $k$-dimensional subspace of the tangent bundle $TM$. It can be locally generated by $k$ independent tangent vectors, which in general means that it can be generated by $k$ independent vector fields $X_1,...,X_k$, namely $$X_1\wedge...\wedge X_k\neq 0\,\text{on }M.$$

For example if $M=\mathbb{R}^n$, with $n>2$ as in your question, a 2 dimensional distribution can be the one generated by $\partial_x,\partial_y$, two independent vector fields.

An interesting property of distributions is the integrability one. Saying that a distribution $D$ is integrable means that there exstists a submanifold $N\subset M$ with tangent spaces which everywhere coincide with the element $D_x$ of the distribution $D$.

Here comes an interesting theorem, Froebenius one, which says that a distribution $D$ spanned by $X_1,...,X_k$, is (completely) integrable if and only if it is closed with respect to the Lie bracket, i.e. $$ [X_i,X_j]=X_iX_j-X_jX_i\in D,\quad\forall i,j=1,...,k. $$

To think about distributions I suggest you to start with one dimensional ones, which are nothing else than tangent vector fields. For example on $\mathbb{R}^n$ a one dimensional distribution is $D=\langle\partial_x\rangle$. One dimensional distributions are integrable if they are smooth enough (by existence and uniqueness theorem) because as the one dimensional submanifold of $\mathbb{R}^n$ you take the integral curve of the vector field generating the distribution.

A similar thing can be done for a general smooth manifold and this concept can be extended to higher rank distributions.

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