# Apparently the conormal sheaf of the diagonal constructs differentials … how do I plug in vectors?

Let me specify the construction I am thinking about, then I have a specific question to ask at the end.

We are given a morphism of schemes $f: X \to Y$ (let's just say separated to avoid something else I am confused about). Then $\Delta : X \to X \times_Y X$ is a closed immersion, and we can construct the conormal sheaf on $X$ as $\Delta^*(I)$, where $I$ is the ideal sheaf cutting out $\Delta : X \to X \times_Y X$.

Let $p_1$ and $p_2$ denote the two projection maps from $X \times_Y X \to X$.

Given a section $f \in \Gamma(U,O_X)$, we can build a section of $I$ on $X \times_Y X$ by $p_1^*(f) - p_2^*(f)$.

I am pretty confused about the construction (though I can understand the details of the following proofs in Ravi that show it is naturally / universally isomorphic on affines to the sheafification of the Kaehler differential construction).

If this is supposed to be a differential, how can I plug in tangent vectors to it?

I'm talking calculus here, hopefully nothing fancy... we are pulling back (polynomial) functions on $\mathbb{R}$ via the projections $\mathbb{R}^2 \to \mathbb{R}$ and taking their difference - the tricky spot is that one has to do something in pulling it back to a sheaf on $X$. But I guess there is a calculus style way to interpret this, and I am missing that piece of intuition. One can try to push forward tangent vectors from the diagonal (it's a closed immersion), and then try to plug them into $p_1^*(f) - p_2^*(f)$. How can this be done? Is my question clear?

As for your example with polynomials: Let me just introduce some notation: Let's call Functions($\mathbb{R}^2)=\mathbb{R}[x,y]$, Functions$(\mathbb{R})=\mathbb{R}[x]$. And the projections, in coordinates, are $p_1: \mathbb{R}^2 \rightarrow \mathbb{R}, (x,y) \mapsto x$, $p_2: \mathbb{R}^2 \rightarrow \mathbb{R}, (x,y) \mapsto y$. The conormal sequence involving the module of Kahler differentials is then: \begin{align} 0 \rightarrow I/I^2 \rightarrow \mathbb{R}[x,y]/I^2 \rightarrow \mathbb{R}[x,y]/I \rightarrow 0 \end{align} Where $I$ is the ideal generated by polynomials $f(x,y)$ that vanish on the diagonal, i.e. with $f(x,x)=0$. $p_1^*(I/I^2)$ is what we call $\Omega$, the $\mathbb{R}[x]$-module of differentials. Since $x-y \in I$, $\mathbb{R}[x,y]/I = \mathbb{R}[x]$. Given a function $f(x) \in \mathbb{R}[x]$, this construction of the de Rham differential gives $df =p_1^*f(x)-p_2^*f(x) \text{ mod }I^2=f(x)-f(y) \text{ mod }I^2$. This can be thought of as $f(x)-f(y)$ where $x-y$ is small (read $< \epsilon$) since $(x-y)^2=0$. After applying $p_{1*}$ this becomes $f(x)-f(0) \text{ mod } x^2$ which is the linear term of $f(x)$!
• Oh, and I'd like to add one thing. The de Rham differential is characterized by a universal property: namely $\mathcal{Hom}_{O_X}(\Omega_{X/S},F)=\mathcal{Der}_S (O_X, F)$. If you take $F= O_X$, you get $\Omega_{X/S}^*=\mathcal{Hom}_{O_X}(\Omega_{X/S}, O_X)=\mathcal{Der}_S(O_X, O_X)$. So sections of the tangent sheaf are identified with derivations. – This account was hacked May 4 '16 at 12:00