Let $M$ be a smooth manifold. The classical construction is the tangent bundle $TM$. What does the tangent groupoid $GM$ give me that this construction doesn't, and why is it useful in non-commutative geometry?

Heuristically, the tangent groupoid, which actually is a bundle too, thickens the tangent bundle with approximations to it; instead of being over $M$, it is over $M\mathbb{R}$, and the fibre over zero is the standard tangent bundle.

To construct it, we first note two simpler constructions:

a. Any bundle $E \rightarrow M$ can be considered as a groupoid $G$ with object space $M$ and morphisms $G(a,a)=E_a$ with the obvious composition and all other hom-spaces empty.

b. The pair groupoid $M \times M$ on $M$ has object space $M$ and morphism space $M^2$, with composition $(a,b)(b,c):=(a,c)$ (with all others empty).

Then given a manifold $M$, its tangent groupoid $GM$ is a disjoint union of groupoids $G_t M$ for $t$ in the real line; and where $G_0 M$=$TM$ considered as a groupoid and $G_t M=M \times M$

Now the object space of $GM$ is the disjoint union of the object space of each fibre which is $M$, for each $t$ in the real line. This means we can identify the object space with $M \mathbb{R}$.

We give it a weak topology so that fibres spaces away from the fibre over zero are seen as spaces of approximate tangent vectors, that is:

for $ f \in C^{\infty}M$ we take the weakest topology (the one with the fewest open sets) such that the following are continuous:

a. $(X,m,0) \rightarrow Xf$

b. $(m,n,t) \rightarrow \frac{(fn-fm)}{t}$

natureis quantum? $\endgroup$ – Qiaochu Yuan Jul 9 '12 at 0:13naturallyreach for when I want to think about tangents. Classical is a better & right word. $\endgroup$ – Mozibur Ullah Jul 9 '12 at 0:45