# Kähler differential and higher derivations (geometric interpretation of diagonal here)

I am studing Kähler differentials and I tried to understand the geometric motivation behind these settings. What I do not understand is the role which plays the diagonal in all these theory. The cotangent sheaf is later defined in terms of the diagonal map. Why is this geometrically interesting? I tried to write a short introduction to Kähler differential to make the geometric nature more available, but I do not now if it make sense. Here it is:

Differential $1$-forms are linear transformations $\omega_{p}:T_p X\mapsto K$ assigning an element in $K$ to a tangent vector of the tangent space $T_p X$ of a point $p\in X$ in some differential manifold $X$. Differential $1$-forms can be viewed as \textit{infinitesimal} direction vectors $\triangle p$. This means in physical terms, that the scalar $\omega_p(\triangle t)\in K$ with $\triangle t$ a tangent vector represents the \textit{work} required to move from $x_i$ to $x_{i+1}$ with $p\in (x_i,x_{i+1})$ along some curve. In other words, differential forms are cotangent vectors over some field $K$, which give information about the work which is locally required to move along some curve. However, they can be generalized and captured by sheaf theory. For this attempt, we observe first that $\triangle p$ is related to Taylor expansions. Indeed, let $f$ be a smooth function, that is $f\in C^{\infty}(\mathbb{C})$, on a differential manifold $X$ and let $\mathfrak{J}$ be the ideal of smooth functions vanishing at the point $p\in X$. The zero order part of the Taylor series of a smooth function $f$ is the value of $f$ at the point $p$, let us say, $f(p)=c$, so that $f-c\in\mathfrak{J}$. Now the first order derivatives of $f-c$ correspond to the first order terms in the Taylor series and these are given by the image of $f$ in $\mathfrak{J}/\mathfrak{J}^2$. Let us denote this map by $d(f)$ with $d: \mathcal{O}_X\rightarrow \mathfrak{J}/\mathfrak{J}^2$ and where $\mathcal{O}_X$ denotes the ring of smooth functions on $X$. Moreover, if $f$ is constant, this means a fixed vector, then $d(f)=0$. Another important input is that $\triangle p$ is required to be non zero, as there is no direction available for the zero vector. But $\triangle p=0$ is satisfied if and only if the two endpoints of the tangent vector $\triangle p$ are choosen to be the same, which happens if and only if the point $p\in X$ correspond to an element on the diagonal of $X\times X$. So we demand that we just consider elements in $X\times X$ vanishing on the diagonal (or in the complement of the diagonal).

Summarizing up, my two questions are the following:

1) which geoemtric interpretation have the diagonal in these context?

2) Higher derivations seemed to me a generalization of Kähler differentials, but what is their motivation or geometric nature (analogy to differential geometry), since I cannot see any connection between Kähler differentials and higher derivations ?

I believe that you are interested to the algebraic geometry setting: if yes then I suggest you Vakil's FOAG (available at http://math.stanford.edu/~vakil/216blog/FOAGdec2915public.pdf); exactly the chapters 12 (sections 1, 2, 3 and 6) and 21 (sections from 1 to 5).

This is an old question, but the question may have interest for others on this site.

Question 1. "Which geometric interpretation have the diagonal in these context?"

Answer. If $$\phi:A \rightarrow B$$ is a map of commutative rings and $$I:=ker(m) \subseteq B\otimes_A B$$ is the ideal of the diagonal, we get a closed immersion

D1. $$\Delta:X:=Spec(B) \rightarrow Spec(B\otimes_A B):=X\times_S X.$$

The sheafication $$\Omega:=\Omega^1_{X/S}$$ of the module of Kahler differentials $$\Omega^1_{B/A}$$ is a sheaf of $$\mathcal{O}_{X\times_S X}$$-modules with support on $$\Delta(X)$$. The two projection maps $$p,q:X\times_S X\rightarrow X$$ induce a left and right $$\mathcal{O}_X$$-module structure on $$\Omega$$, hence $$\Omega$$ is a sheaf of left and right $$\mathcal{O}_X$$-modules with the property that $$a(\omega b)=(a\omega)b$$ for all local setions $$a,b,\omega$$. The following formula holds:

D2. $$a\omega = \omega a.$$

hence the left and right structure coincide. There is an isommorphism of $$B$$-modules

D3. $$Hom_B(\Omega^1_{B/A},B) \cong Der_A(B)$$

where $$Der_A(B)$$ is the set of $$A$$-linear maps $$\delta:B\rightarrow B$$ with $$\delta(ab)=a\delta(b)+\delta(a)b$$. At the level of sheaves you get an ismorphism

D4. $$\Omega^* \cong \Theta_{X/S}$$

hence the dual of the module of Kahler differentials is the "tangent bundle" $$\Theta_{X/S}$$.

Let $$k\subseteq A$$ be a map of commutative unital rings. A "higher derivation of $$A/k$$" is constructed as follows: Let $$B:=A[t]/(t^{n+1})$$ and let $$D:A \rightarrow B$$ be a map of $$k$$-algebras. It follows

D5. $$D(a)=\sum_{i\geq 0}D_i(A)t^i$$

where $$D_i:A \rightarrow A$$ are $$k$$-linear maps and $$D_k(ab)=\sum_{i+j=k}D_i(a)D_j(b)$$. The map

D6. $$D_1: A\rightarrow A$$

is a derivation, but the other maps $$D_i$$ are not. See Matsumura's book "Commutative ring theory" for some applications to the study of local rings.

Question: "Higher derivations seemed to me a generalization of Kähler differentials, but what is their motivation or geometric nature (analogy to differential geometry), since I cannot see any connection between Kähler differentials and higher derivations?"

Answer: If $$k$$ is a field of characteristic zero it follows for any derivation $$\delta \in Der_k(A)$$ we get a higher derivation $$D^n(\delta)$$

$$D^n(\delta)(\sum_{i=0}^n a_it^i):=\sum_{i\geq 0}\frac{1}{i!}\delta^i(a_i)t^i$$

where $$\delta^i:=\delta \circ \cdots \circ \delta.$$ This construction and it's relation to smoothness is studied in Matsumuras book Chapter 27.

They also go under the name "Hasse-Schmidt derivations". In some cases (over a field of charateristic zero) they are related to the notion "flat connection". You may find some results on this on the arXiv preprint server.