TITLE: What's the intution behind defining the cotangent sheaf as $\Delta^\ast(\mathscr{I}/\mathscr{I}^2)$?
QUESTION [15 upvotes]: This definition seems to be given all over the place (e.g. Hartshorne II.8, Vakil 21.2.20, Wikipedia, McKernan's lecture notes from MIT), and never with any explanation as to why the map $\Delta : X \to X \times X$ should have any relation to relative differentials.  Is this purely some kind of formal trick?  Is there an intuitive meaning to it?  Could it work with other maps than $\Delta$, but $\Delta$ is just a convenient choice for some reason?

REPLY [2 votes]: Let me maybe try to interpret in a naive fashion. When I say "smooth", I mean a reasonable varying way in your context (smooth, holomorphic, being polynomial, etc.), and all arguments below are local but eventually you glue these smoothly.
What are functions on $X\times X$? If you give a function $f(\cdot, \cdot)$ on $X\times X$ and a point $p$ on $X$, $f(\cdot, p)$ gives a function on X. Hence I regard functions on $X\times X$ as smooth varying functions on X parametrized by points in X. Now I look at their Taylor series $f(x, p)=\sum a_{i}(p)(x-p)^{i}$ and then what you wanted is precisely $a_{1}(p)$ . If I view $a_{1}(p)(x-p)$ again as a section of $\mathcal{O}_{X\times X}$, first of all at $p$ it better be zero(because as a series at p it doesn't have constant term), namely $a_{1}(x-p)$ vanishes on the diagonal $i: \Delta\subset X\times X$. Let $I$ be ideal of $\Delta$ then it is a section of $I$. But on the other hand you don't care about higher terms, so you kill them, by looking at $I/I^{2}$. Now recall that how we produced these is by looking at Taylor expansion of $f(\cdot, p)$ centered at $p$, so at $p$ you only care about $a_{1}(p)$(but not other $a_{1}(q)$), this gives you a section on $X$. This is precisely given by $i^{*} (I/I^{2})$.
Now back to an example, given a function $f \in \mathcal{O}_{X}$, what is $\mbox{d}f\in \Omega_{X}$? Let's first view it as a function on $X\times X$, where at each $p$, we associate the function $f$, namely $F(x,p)=f(x)$ for any $p$. How do I get linear term of Taylor expansion? First I need to kill constant term by modifying $F(x,p)=f(x)-f(p)$, then I kill higher terms by looking at its image in $I/I^{2}$. If you look at construction in Matsumura, then our $f(x)-f(p)$ is precisely his $f\otimes 1- 1\otimes f$, which we call it $\mbox{d}f$. 
Now a general differential form is just locally the above example, then you glue them smoothly, this justifies that our initial construction is correct.