What are Kähler differentials, really?
Given a commutative ring \(A\) and an \(A\)-algebra \(B\), one can define a \(B\)-module \(\Omega_{B/A}\), called the module of Kähler differentials. It is often described as the commutative-algebraic analogue of differential \(1\)-forms, but it is not clear from the definition why this is so. In this article, we reconsider the definition of differential \(1\)-forms on manifolds and explain why the definition of Kähler differentials is their natural analogue.
Definition of Kähler differentials
Let \(A\) be a commutative ring, and let \(B\) be a commutative \(A\)-algebra. In other words, suppose that we are given a ring homomorphism \(A\longrightarrow B\). Let \(I\) be the ideal of \(B\otimes_A B\) defined by
\[ I=\operatorname{Ker}(\mu\colon B\otimes_AB\longrightarrow B;\ b\otimes b'\longmapsto bb') \]The module of Kähler differentials \(\Omega_{B/A}\) is then defined by
\[ \Omega_{B/A}=I/I^2 \]At this point, the definition does not look much like differential \(1\)-forms.
A common conceptual description of Kähler differentials uses a universal property. Given a \(B\)-module \(M\), an \(A\)-linear map \(D\colon B\longrightarrow M\) satisfying
\[ D(bb')=bD(b')+b'D(b) \]is called an \(A\)-derivation. With \(\Omega_{B/A}\) defined as above, define \(d\colon B\longrightarrow \Omega_{B/A}\) by
\[ b\longmapsto [1\otimes b-b\otimes 1] \in I/I^2 \]One can check that \(d\) is an \(A\)-derivation. Moreover, \(d\colon B\longrightarrow \Omega_{B/A}\) is an initial object in the category of \(A\)-derivations. That is, for every \(A\)-derivation \(D\colon B\longrightarrow M\), there is a unique \(B\)-linear map \(\varphi\colon \Omega_{B/A}\longrightarrow M\) such that \(D=\varphi\circ d\). This property characterizes \(\Omega_{B/A}\) uniquely up to isomorphism.
Since derivations abstract the Leibniz rule for differentiation, this characterization does suggest something differential, but it still leaves two questions: why is \(d\) a derivation, and why can the universal derivation be constructed as \(I/I^2\)?
From commutative rings to schemes
The starting point for a reinterpretation is to replace the commutative rings by their corresponding schemes.
\[ S=\operatorname{Spec} A,\qquad X=\operatorname{Spec} B \]Let us use this notation. Then \(X\) is a scheme over \(S\), and the ring \(B\) can be regarded as the ring of functions on \(X\). The ring homomorphism \(\mu\colon B\otimes_A B\longrightarrow B\) corresponds to the diagonal morphism
\[ \Delta\colon X\longrightarrow X\times_SX \]Therefore, \(I\) can be regarded as the ideal of functions on \(X\times_SX\) that vanish on the diagonal. With this interpretation in place, we compare \(I/I^2\) with ordinary differential \(1\)-forms to see why it can be regarded as their algebraic counterpart.
Rethinking cotangent spaces
Let \(M\) be a smooth manifold. A differential \(1\)-form on \(M\) is a smooth section of the cotangent bundle of \(M\): it assigns to each point \(p\in M\) an element of the cotangent space \(T^*_pM\), and does so smoothly. The cotangent space \(T^*_pM\) is the dual vector space of the tangent space \(T_pM\). Since \(T_pM\) is the vector space of differential operators at \(p\), an element of the cotangent space describes a linear response to differential operators at \(p\).
Write \(C^\infty(M)\) for the ring of smooth functions on \(M\), and let \(I_p\) be the ideal of functions that vanish at \(p\in M\). For \(f\in C^\infty(M)\), the total derivative of \(f\) at \(p\),
\[ (df)_p\colon T_pM\longrightarrow \mathbb{R};\qquad v\longmapsto v(f) \]defines an element of the cotangent space \(T^*_pM\). This gives a surjective linear map
\[ d_p\colon C^\infty(M)\longrightarrow T^*_pM \]Adding a constant does not change the total derivative, so this map factors as
\[ C^\infty(M)\xrightarrow{f\longmapsto f-f(p)} I_p\xrightarrow{d_p|_{I_p}} T^*_pM \]The linear map \(d_p|_{I_p}\colon I_p\longrightarrow T^*_pM\) is surjective, so \(T^*_pM\) can be regarded as a quotient of \(I_p\). In other words, \(T^*_pM\) can be viewed as
\[ \boxed{ \text{\(I_p\) modulo equality of total derivatives at \(p\)} } \]This is another way to view the cotangent space.
What does the equivalence relation given by equality of total derivatives at \(p\), or equivalently the kernel of \(d_p|_{I_p}\), look like? If \(f,g\in I_p\), then \(f(p)=g(p)=0\), and the Leibniz rule shows that the total derivative of \(fg\) at \(p\) is zero. Thus \(I_p^2\subseteq \operatorname{Ker}(d_p|_{I_p})\). In fact, this inclusion is an equality.
Lemma. We have \(I_p^2=\operatorname{Ker}(d_p|_{I_p})\).
Proof. The preceding paragraph proves that \(I_p^2\subseteq \operatorname{Ker}(d_p|_{I_p})\). Conversely, suppose that \(f\in I_p\) and \((df)_p=0\), and choose local coordinates \(x_1,\ldots,x_n\) centered at \(p\). By Hadamard's lemma, in a neighborhood of \(p\) we can write
\[ f=\sum_i x_i g_i \]Differentiating this identity at \(p\) gives \(g_i(p)=\partial f/\partial x_i(p)=0\), so another application of Hadamard's lemma gives
\[ g_i=\sum_j x_jh_{ij},\qquad f=\sum_{i,j}x_ix_jh_{ij} \]Choose bump functions \(\chi,\eta\) supported in this coordinate neighborhood such that \(\chi=1\) near \(p\) and \(\eta=1\) on \(\operatorname{supp}\chi\). Extend \(\eta x_i\) and \(\chi x_jh_{ij}\) by zero to functions on all of \(M\). These functions belong to \(I_p\), and the sum of their products is \(\chi f\), so \(\chi f\in I_p^2\). Moreover, \(1-\chi,f\in I_p\), so \((1-\chi)f\in I_p^2\), and hence \(f=\chi f+(1-\chi)f\in I_p^2\). \(\square\)
The lemma gives an isomorphism between the cotangent space and \(I_p/I_p^2\):
\[ \boxed{ T^*_pM\cong I_p/I_p^2 } \]Under this isomorphism, the total derivative \(d_p\colon C^\infty(M)\longrightarrow T^*_pM\) becomes
\[ d_p\colon C^\infty(M)\longrightarrow I_p/I_p^2;\qquad f\longmapsto \bigl[f-f(p)\bigr] \]This is already beginning to resemble the module of Kähler differentials.
Rethinking differential \(1\)-forms
Continue to let \(M\) be a smooth manifold. In the preceding section, we described the cotangent space \(T^*_pM\) algebraically for a fixed point \(p\in M\). We now let \(p\) vary and seek a similar algebraic description of the space \(\Omega(M)=\Gamma(M,T^*M)\) of differential \(1\)-forms.
In the preceding section, we considered the functions on \(M\) that vanish at \(p\). Since \(p\) will now vary, we work on the product manifold \(M\times M\). For each \(p\in M\), one may think of the copy \(\{p\}\times M\) of \(M\). For a smooth function \(h\in C^\infty(M\times M)\), the condition
\[ \forall p\in M,\qquad h(p,{-})\in I_p \]is equivalent to \(h\) belonging to the ideal
\[ I=\bigl\{h\in C^\infty(M\times M)\mid \forall x\in M,\ h(x,x)=0\bigr\}. \]Moreover, if \(h\in I^2\), then \(h(p,{-})\in I_p^2\), so for each \(p\in M\) there is a natural linear map
\[ I/I^2\longrightarrow I_p/I_p^2;\quad [h]\longmapsto \bigl[h(p,{-})\bigr]. \]Combining these maps as \(p\) varies gives a map
\[ I/I^2\longrightarrow \Omega(M). \]A suitable globalization of the proof of the lemma in the preceding section shows that this map is an isomorphism; we omit the details. Thus the differential \(1\)-forms on a smooth manifold admit an algebraic description parallel to the definition of Kähler differentials:
\[ \boxed{ \Omega(M)\cong I/I^2. } \]Let us interpret this isomorphism more intuitively. An element of \(I\) is a function of a pair \((x,y)\in M\times M\) that vanishes when \(x=y\). Taking the quotient by \(I^2\) amounts, in local coordinates, to discarding terms of order at least two in \(y-x\). It retains only the first-order behavior as \(y\) approaches \(x\) and discards all other information. The result is a linear response to tangent vectors, which is precisely a differential \(1\)-form. As a rough slogan, the isomorphism expresses the idea that
\[ \boxed{ \begin{gathered} \text{a differential \(1\)-form is a linear function of the difference}\\ \text{between two infinitesimally close points} \end{gathered} } \]This is the viewpoint expressed by the isomorphism.
The differential map
We now ask how to express the differential map \(d\colon C^\infty(M)\longrightarrow \Omega(M)\) algebraically. Given a function \(f\in C^\infty(M)\), consider the function on \(M\times M\) defined by
\[ (x,y)\longmapsto f(y)-f(x). \]This function clearly vanishes on the diagonal, so it belongs to \(I\). For each \(p\in M\), consider the natural map
\[ I/I^2\longrightarrow I_p/I_p^2\cong T^*_pM. \]Under this map, \(\bigl[f(y)-f(x)\bigr]\) is sent as follows:
\[ \bigl[f(y)-f(x)\bigr]\longmapsto \bigl[f-f(p)\bigr]\longmapsto (df)_p. \]Thus \(\bigl[f(y)-f(x)\bigr]\in I/I^2\) corresponds to the differential \(1\)-form \(df\in \Omega(M)\). Consequently, the differential map \(d\colon C^\infty(M)\longrightarrow \Omega(M)\) is described algebraically by
\[ d\colon C^\infty(M)\longrightarrow I/I^2;\qquad f\longmapsto \bigl[f(y)-f(x)\bigr]. \]For a real-valued function of one variable \(f\colon \mathbb{R}\longrightarrow \mathbb{R}\), differentiation can be written as
\[ f(y)-f(x) = f'(x)(y-x)\mod(y-x)^2. \]The formula above is precisely a generalization of this identity.
Finally, recall the universal derivation that characterizes the module of Kähler differentials:
\[ d\colon B\longrightarrow \Omega_{B/A};\qquad b\longmapsto [1\otimes b-b\otimes 1]. \]This is exactly the algebraic translation of the preceding description of \(d\). Indeed, \(1\otimes b\) is the pullback of the function \(b\) along the second projection
\[ \operatorname{pr}_2\colon X\times_S X \longrightarrow X. \]It is therefore analogous to the function \((x,y)\longmapsto f(y)\) on \(M\times M\). Similarly, \(b\otimes 1\) corresponds to \((x,y)\longmapsto f(x)\). In general, however,
\[ C^\infty(M\times M)\not\cong C^\infty(M)\otimes_{\mathbb R}C^\infty(M), \]so this analogy does not produce an isomorphism between \(\Omega_{C^\infty(M)/\mathbb R}\) and \(\Omega(M)\).