We consider the usual Riemannian metric on $S^{n}$. Its corresponding LC connection gives us a distribution on $TS^{n}$. Is this distribution completely nonintegrable?


In general, what type of curvature criterions can be used to prove that certain Levi Civita connection gives a completely non integrable distributation?



Is this distribution completely non integrable?
Here is a more complete answer to your question, which describes exactly which vectors in $TM$ can be joined by $L$-curves. (I changed $H$ to $L$ since I have to use $H$ for holonomy.)


Recall that, when $(M,g)$ is a Riemannian manifold (assumed connected and simply connected for simplicity), there is a unique horizontal plane field $L$ on $TM$ (i.e., $L$ is a complement to the kernel of $\pi':T(TM)\to TM$) with the property that a curve $\gamma:[a,b]\to TM$ is tangent to $L$ if and only if it is the $\nabla^g$-parallel translation of $\gamma(a)\in T\_{\pi(\gamma(a))}M$ along $\pi\circ\gamma:[a,b]\to M$, where $\nabla^g$ is the Levi-Civita connection of $g$.


Fix a point $x\in M$ and let $H\_x\subset\mathrm{SO}(T\_xM,g\_x)$ be the holonomy group of $\nabla^g$ at $x$. Any $H\_x$-invariant function $p\_x:T\_xM\to\mathbb{R}$ then extends uniquely to a function $p:TM\to\mathbb{R}$ that is constant along all $L$-curves. For example, the function $g\_x:T\_xM\to\mathbb{R}$ extends uniquely to $g:TM\to\mathbb{R}$, and this function is constant along all $L$-curves, since $\nabla^g$-parallel translation preserves the $g$-norm of tangent vectors.


Conversely, it is not hard to show, using Berger's classification of Riemannian holonomy groups, that, if two vectors $v\in T\_xM$ and $w\in T\_yM$ have equal $p$-values for all such "$L$-constant" functions $p:TM\to\mathbb{R}$, then there is an $L$-curve that joins them.


To see why, remember that there is a decomposition $T\_xM=V\_1\oplus\cdots\oplus V\_m$ into nontrivial subspaces $V\_i$ for $1\le i\le m$ and a product decomposition
$$
H\_x = H\_1\times \cdots \times H\_m
$$
where $H\_i$ acts trivially on $V\_j$ for $j\not=i$ and irreducibly on $V\_i$. Moreover, each $H\_i$ is isomorphic to the holonomy group of an irreducible Riemannian manifold $(M\_i,g\_i)$ of dimension $d\_i = \dim V\_i\ge 1$, where the isomorphism is induced via some isometric isomorphism $T\_{x\_i}M\_i\simeq V\_i$.


Now, by Berger's classification, either $H\_i$ acts transitively on the unit sphere in $V\_i$ (in which case the ring of $H\_i$-invariant polynomials is generated by $g\_i$ and so has algebraic dimension $r\_i=1$) or else $(M\_i,g\_i)$ is a Riemannian symmetric space of rank $r\_i\ge1$. In either case, there exists a ring $R\_i$ of $H\_i$-invariant polynomials on $V\_i$ with algebraic dimension $r\_i$ such that the polynomial functions in $R\_i$ have the $H\_i$-orbits as their simultaneous level sets.


Putting all of this together, we see that there is a finitely generated ring $R\_x$ of $H\_x$-invariant polynomials on $T\_xM$ such that two vectors $v$ and $w$ in $T\_xM$ can be joined by an $L$-curve if and only if they have the same $p\_x$-value for all $p\_x\in R\_x$. 


In the specific case of $S^n$, it is known that, for *any* Riemannian metric $g$ on $S^n$, the holonomy group is $H\_x = \mathrm{SO}(T\_xM,g\_x)$ for all $x\in S^n$. Thus, when $n>1$, it follows that any two tangent vectors to $S^n$ can be joined by an $L$-curve if and only if they have the same $g$-length.


