I know the familiar differences between tight and overtwisted contact structures. For example, each homotopy class of plane-bundles on a three-manifold has an overtwisted representative but tight contact structures are less common. Also a tight contact structure gives a genus bound on an embedded closed surface and therefore can tell things about the topology of the ambient manifold. My question is what were the motivations behind distinguishing between the two originally? Is there possibly an intuition which makes the two fundamentally different and then proving the above differences comes as a consequence or it was these differences which led to distinguishing between the two? In other words I am trying to see the difference between the two without appealing to the statements like the above.
[Math] Tight vs. overtwisted contact structure
3-manifoldscontact-geometry
Related Solutions
Well, here is what I can say. Perhaps this will answer some of your questions about $S^{2n+1}$ at least.
Suppose that $G/H = S^{2n+1}$ where $n>0$ and that the action of $G$ on $S^{2n+1}$ is effective and preserves a contact structure on $S^{2n+1}$.
By a result of Montgomery (Simply connected homogeneous spaces, PAMS 1950), $G$ has a compact subgroup that acts transitively on $S^{2n+1}$ (and preserves the contact structure), and this implies that a maximal compact subgroup $U\subset G$ acts transitively on $S^{2n+1}$ with compact stabilizer $K = U\cap H$, so that $S^{2n+1} = U/K$ where $U$ preserves the given contact structure on $S^{2n+1}$. Without loss of generality, we can assume that $U$ is connected, which implies that $K$ is connected as well.
By results of Borel, it follows that $U$ has an embedding into $\mathrm{SO}(2n{+}2)$ for which $K = U\cap \mathrm{SO}(2n{+}1)$ (i.e., $U$ acts as a transitive group of isometries of $S^{2n+1}$ endowed with its standard metric of constant sectional curvature $+1$). Examining Borel's list of the possibilities, one sees that the connected compact subgroup $U\subset \mathrm{SO}(2n{+}2)$ acts transitively on $S^{2n+1}$ and preserves a contact structure if and only if $U$ is conjugate in $\mathrm{SO}(2n{+}2)$ to one of the following subgroups $$ \mathrm{U}(n{+}1),\quad \mathrm{SU}(n{+}1),\quad \mathrm{Sp}\bigl(\tfrac{n+1}2\bigr)\cdot S^1,\quad \mathrm{Sp}\bigl(\tfrac{n+1}2\bigr). $$ (The latter two cases only happen when $n$ is odd.) The first three subgroups preserve a unique contact structure, namely the contact structure defined by the $1$-form $\xi$ on $S^{2n+1}$ defined by $\xi(v) = \mathrm{d}r(Jv)$, where $J:\mathbb{C}^{n+1}\to \mathbb{C}^{n+1}$ is the complex structure map and $r = |z|^2$ is the squared Hermitian norm. The fourth subgroup preserves a $2$-sphere of contact structures, namely, one identifies $\mathbb{C}^{n+1}$ with $\mathbb{H}^{(n+1)/2}$ (thought of as column vectors of height $\tfrac12(n{+}1)$ with quaternion entries) and uses the same formula as before, but now, one allows $J$ to be scalar multiplication (on the right) by any unit imaginary quaternion. Upon conjugating by an element of the subgroup $\mathrm{Sp}(1)\subset \mathrm{SO}(2n{+}2)$ consisting of multiplication on the right by a unit quaternion, any two of these contact structures can be identified, so that each of these homogeneous contact structures in the fourth case are homogeneously isometric to the contact structure identified in the first three cases.
Thus, there are really only four cases to consider: When the group $G$ contains, as identity component $U$ of its maximal compact, one of the four groups listed above, and that subgroup acts on $S^{2n+1}$ preserving a metric of constant sectional curvature $+1$.
This is a classification problem that can be worked out. Though I haven't done it myself, there is a routine method to do this.
For example, when $U = \mathrm{U}(n{+}1)$, one could have, in addition to $G=\mathrm{U}(n{+}1)$, that $G = \mathrm{Sp}(n{+}1,\mathbb{R})$, the symplectic transformations of $\mathbb{R}^{2n+2}$, or $G=\mathrm{SU}(n{+}1,1)$, the CR-autmorphisms of $S^{2n+1}$ as a CR-manifold. (There might be others; I haven't checked.)
Best Answer
As mentioned in the previous answer, the first use of the overtwisted/tight dichotomy is most certainly Bennequin's Theorem that there are non-isomorphic contact structures on $\mathbb{R}^3$ and $\mathbb{S}^3$, a landmark result.
However, the relevance of this dichotomy goes now far beyond this. As you probably know, contact topology has a Darboux theorem: locally, all contact structures are the same, isomorphic to the standard contact structure on $\mathbb{R}^3$. So, all of them are "locally tight", and having an overtwisted disc must be a global condition. A global way of distinguishing objects which are locally the same is tremendously important, here it can be thought of as some analogue of Gromov's non-squeezing theorem in symplectic geometry.
Moreover, while the classification of overtwisted structure has been achieved quite early by Eliashberg (Inventiones 1989), the tight contact structures happened to be very rich (see e.g. Giroux, Inventiones 1999 and its introduction -- in French) : certain manifold have only one of them, but infinitely many are shown to bear infinitely many non-isomorphic tight contact structure in the paper cited. The relation with symplectic fillings is one more indication of the relevance of this dichotomy. One can now consider that the tight contact structure are the one to study, as overtwisted ones are pretty well understood.
Note that the relevance of this dichotomy has recently been more firmly established in higher dimensions too by Niederkrüger, Massot and Wendl (Inventiones 2013).
About Bennequin's theorem, I can not explain its proof, although I learned one in my graduate years. I cannot give a complete account of it any more, but I can give some flavor. Consider the usual cylindrically symmetric version of the standard contact structure, and decompose $\mathbb{R}^3$ as a trivial open book (the vertical line through the origin is the binding, and each half plane it bounds is a page). Assume there is a closed horizontal curve(in the sense of the contact structure) which bounds a disc tangent to the structure, which you assume in general position, and look at the way this disc intersects the pages. In each page the non-transversal points are vertices of a graph, so you get combinatorial objects to work with. That's pretty much what I remember, but there are several proofs in the literature (at least one by Giroux, but likely to be written in French; it is probably not a waste of time to learn enough French to read mathematics if you are planning to work in contact topology). In fact, a friend of mine working in this area once told me that it was almost a duty for anyone working in this area to find its own proof of Bennequin's theorem. You should be ale to locate several proofs through the literature, but I am too far from this field to help you there.