Intuition/Heuristic behind I/I^2 definition of Kähler differentials

Hello,

this one has always been mysterious to me. The Kähler differentials $\Omega_{A/k}$ are definined, by the universal property $$Der_k(A,M)=A-Mod(\Omega_{A/k},M)$$ so for $M=A$ we get that $\Omega_{A/k}$ is the cotangent space of $spec(A)$. (or a relative version of it if k is no field).

There are two constructions of Kähler Differentials I know. The first one is $$\Omega_{A/k}=\langle df : \text{relations satisfied by any derivation} \rangle$$ I think I sort of understand this one, it says that the differential of a function just contains enough information to extract the derivation of the function out of it. And this is what a section into cotangent space should be. Something that contains just enough information to pair it with a vector-field into a function. The other construction is $$\Omega_{A/k}=I/I^2$$ Where $I$ is the Ideal of functions vanishing on the diagonal in $spec(A)\times_{spec(k)} spec(A)$.

More geometrically it says sections into cotangent space=functions vanishing on the diagonal mod higher order. But still I don't think I understand this equality on an intuitive level. Can someone explain the heuristic behind this equality? Or maybe explain $\Omega_{A/k}=I/I^2$ from another intuitive viewpoint?

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4 Answers

Think of the Taylor expansion of a smooth function on ${\mathbb R}^n$. Let $I$ be the ideal of smooth functions vanishing at a given point $x_0$, then the zero order part of the Taylor series of a smooth function $f$ gives just the value of $f$ at the point. If we subtract this constant from $f$, we land in the ideal $I$. Now the first order derivatives of $f$ correspond to the first order terms in the Taylor series abd these are given by the image of $f$ in $I/I^2$.

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Thanks for the answer. What you write makes sense to me, however what I don't understand is why I in the definition is the ideal of the diagonal. –  Jan Weidner Mar 13 '10 at 17:56

Start with $Spec(A)$ being a vector space $V$, and let $V'$ be the subspace $V\times 0$ complementary to the diagonal $V_\Delta$. Then the ideal $I$ is functions on $V\times V$ vanishing on the diagonal, which we can think of as functions on $V'$ vanishing at $0$, indexed by points of $V_\Delta$. As Anton Deitmar explained, $I/I^2$ is then linear functions on $V'$, indexed by points of $V_\Delta$. Identifying $V' = V = V_\Delta$ we get $V \times V^*$.

Then this analysis works locally at smooth points of a general $Spec(A)$.

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Thanks, for your answer, I didn't understand everything, but now I figuered out how to think about the differentials. –  Jan Weidner Mar 14 '10 at 20:49

In calculus we teach that if $x = x_0 + \Delta x,$ then $f(x) = f(x_0) + f'(x_0) \Delta x.$ In other words, the derivative of $f$ at $x_0$ tells us the scaling factor for the change in $f$ when $x_0$ moves by an infinitesimal amount $\Delta x$. What does infinitesimal mean? Well, we are ignoring higher order terms, i.e. terms in $\Delta x^n$ for $n \geq 2,$ i.e. we are working not at the level $\Delta x =0$, which says $x = x_0$, i.e. that we are on the diagonal, but at the next level $\Delta x^2 = 0,$ which is working modulo $I^2$. (Working modulo $I$ is the same as setting $\Delta x = 0$.)

This is written in one variable, but works for any number of variables.

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Thank you for your answer! –  Jan Weidner Mar 14 '10 at 20:52

There are a number of definitions of the module of Kähler differentials that are proven to be equivalent in these notes. Your question is discussed on page 20 near the bottom.

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