Affine Schemes, Dual Numbers, and Tangent Spaces

Derivations of Affine Schemes via Dual Numbers

Let $k$$k$ be a field, and let

$$X=\mathrm{Spec}A$$

be an affine $k$$k$-scheme.

The basic idea is:

$$\text{a derivation is a first-order infinitesimal lift.}$$

The infinitesimal object we use is the algebra of dual numbers. For every commutative $k$$k$-algebra $R$$R$, define

$$R[\epsilon ]:=R[t]/(t^2).$$

We write the image of $t$$t$ as $\epsilon$$\varepsilon$, so

$${\epsilon }^2=0.$$

There is a natural projection

$${\pi }_R:R[\epsilon ]\to R,\epsilon \mapsto 0.$$

The tangent functor of $X$$X$ is defined by

$$TX(R):=X(R[\epsilon ]).$$

Since

$$X(R)={\mathrm{Hom}}_{k\text{-alg}}(A,R),$$

we have

$$TX(R)=X(R[\epsilon ])={\mathrm{Hom}}_{k\text{-alg}}(A,R[\epsilon ]).$$

The projection $R[\epsilon ]\to R$$R[\varepsilon]\to R$ induces a natural map

$$TX(R)\to X(R).$$

Thus a tangent vector should be a point of $X(R[\epsilon ])$$X(R[\varepsilon])$ lying over a chosen point of $X(R)$$X(R)$.

Tangent Vectors at an $R$$R$-Point

Let

$$x\in X(R)$$

be an $R$$R$-point of $X$$X$. Equivalently, $x$$x$ is a $k$$k$-algebra homomorphism

$$x^{\mathrm{\#}}:A\to R.$$

A tangent vector at $x$$x$ is a lift

$$\tilde{x}:A\to R[\epsilon ]$$

such that

$${\pi }_R\circ \tilde{x}=x^{\mathrm{\#}}.$$

So we have a diagram

$$\begin{array}{ccc}A& \stackrel{\tilde{x}}{\to }& R[\epsilon ]\\ & {\searrow }_{x^{\mathrm{\#}}}& \downarrow {\pi }_R\\ & & R.\end{array}$$

Since ${\pi }_R\circ \tilde{x}=x^{\mathrm{\#}}$$\pi_R\circ\widetilde{x}=x^\#$, every element $\tilde{x}(a)$$\widetilde{x}(a)$ must have the form

$$\tilde{x}(a)=x^{\mathrm{\#}}(a)+\epsilon D(a)$$

for a unique element $D(a)\in R$$D(a)\in R$.

Thus a lift $\tilde{x}$$\widetilde{x}$ is the same data as a function

$$D:A\to R$$

such that

$$\tilde{x}(a)=x^{\mathrm{\#}}(a)+\epsilon D(a).$$

Now we ask when $\tilde{x}$$\widetilde{x}$ is a $k$$k$-algebra homomorphism.

Since $\tilde{x}$$\widetilde{x}$ is additive, $D$$D$ must be additive. Since $\tilde{x}$$\widetilde{x}$ is $k$$k$-linear, $D$$D$ must be $k$$k$-linear. The main condition comes from multiplication.

We need

$$\tilde{x}(ab)=\tilde{x}(a)\tilde{x}(b).$$

The left-hand side is

$$\tilde{x}(ab)=x^{\mathrm{\#}}(ab)+\epsilon D(ab).$$

The right-hand side is

$$(x^{\mathrm{\#}}(a)+\epsilon D(a))(x^{\mathrm{\#}}(b)+\epsilon D(b)).$$

Since ${\epsilon }^2=0$$\varepsilon^2=0$, this becomes

$$x^{\mathrm{\#}}(a)x^{\mathrm{\#}}(b)+\epsilon (x^{\mathrm{\#}}(a)D(b)+D(a)x^{\mathrm{\#}}(b)).$$

Comparing the coefficient of $\epsilon$$\varepsilon$, we get

$$D(ab)=x^{\mathrm{\#}}(a)D(b)+D(a)x^{\mathrm{\#}}(b).$$

This is precisely the Leibniz rule, where $R$$R$ is regarded as an $A$$A$-module via

$$x^{\mathrm{\#}}:A\to R.$$

We denote this $A$$A$-module by $R_x$$R_x$.

Therefore

$$\boxed{{\displaystyle T_{x}X(R)\cong {\mathrm{Der}}_k(A,R_x).}}$$

In words:

$$\boxed{{\displaystyle \text{a tangent vector at }x=\text{a derivation }A\to R\text{ relative to }x.}}$$

Derivations as Infinitesimal Lifts

The previous computation can be summarized as follows.

A lift

$$\tilde{x}:A\to R[\epsilon ]$$

over $x^{\mathrm{\#}}:A\to R$$x^\#:A\to R$ is necessarily of the form

$$\tilde{x}=x^{\mathrm{\#}}+\epsilon D.$$

This is a $k$$k$-algebra homomorphism if and only if

$$D(ab)=x^{\mathrm{\#}}(a)D(b)+D(a)x^{\mathrm{\#}}(b).$$

Hence

$$\boxed{{\displaystyle \{\text{lifts }A\to R[\epsilon ]\text{ over }{x}^{\mathrm{\#}}\}\cong {\mathrm{Der}}_k(A,R_x).}}$$

This is the dual-number definition of derivations.

The derivation is not an extra object placed on top of the geometry. It is exactly the first-order part of a map to dual numbers.

The Global Version

There is also a global version.

Take $R=A$$R=A$ and let

$$x^{\mathrm{\#}}={\mathrm{id}}_A:A\to A.$$

Then a global vector field on $X=\mathrm{Spec}A$$X=\operatorname{Spec}A$ is a section

$$s:A\to A[\epsilon ]$$

of the projection

$$A[\epsilon ]\to A.$$

That is,

$$\pi \circ s={\mathrm{id}}_A.$$

Every such section has the form

$$s(a)=a+\epsilon D(a).$$

The condition that $s$$s$ is a $k$$k$-algebra homomorphism is exactly

$$D(ab)=aD(b)+D(a)b.$$

Therefore

$$\boxed{{\displaystyle {\mathrm{Der}}_k(A,A)\cong \{\text{sections }A\to A[\epsilon ]\text{ of }A[\epsilon ]\to A\}.}}$$

Geometrically,

$${\mathrm{Der}}_k(A,A)$$

is the module of global vector fields on the affine scheme $X$$X$.

Square-Zero Extensions

The dual-number construction is a special case of a more general construction.

Let $M$$M$ be an $A$$A$-module. Define the square-zero extension

$$A\oplus M$$

with multiplication

$$(a,m)(b,n)=(ab,an+bm).$$

Then

$$(0,m)(0,n)=(0,0),$$

so $M$$M$ is a square-zero ideal inside $A\oplus M$$A\oplus M$.

There is a projection

$$p:A\oplus M\to A,(a,m)\mapsto a.$$

A section

$$s:A\to A\oplus M$$

of $p$$p$ must have the form

$$s(a)=(a,D(a)).$$

The condition that $s$$s$ is multiplicative is

$$s(ab)=s(a)s(b).$$

The left-hand side is

$$s(ab)=(ab,D(ab)).$$

The right-hand side is

$$(a,D(a))(b,D(b))=(ab,aD(b)+bD(a)).$$

Thus

$$D(ab)=aD(b)+bD(a).$$

Therefore

$$\boxed{{\displaystyle {\mathrm{Der}}_k(A,M)\cong \{\text{sections }A\to A\oplus M\text{ of }A\oplus M\to A\}.}}$$

When $M=A$$M=A$, we have

$$A\oplus A\cong A[\epsilon ],$$

by identifying

$$(a,b)\leftrightarrow a+\epsilon b.$$

So the usual dual-number picture is the case $M=A$$M=A$.

Relation with Kähler Differentials

The module of Kähler differentials ${\mathrm{\Omega }}_{A/k}$$\Omega_{A/k}$ represents derivations.

That is, for every $A$$A$-module $M$$M$, there is a natural isomorphism

$${\mathrm{Der}}_k(A,M)\cong {\mathrm{Hom}}_A({\mathrm{\Omega }}_{A/k},M).$$

The universal derivation is

$$d:A\to {\mathrm{\Omega }}_{A/k}.$$

Every derivation

$$D:A\to M$$

factors uniquely as

$$A\stackrel{d}{\to }{\mathrm{\Omega }}_{A/k}\stackrel{\varphi }{\to }M.$$

For a point

$$x^{\mathrm{\#}}:A\to R,$$

we get

$$T_{x}X(R)\cong {\mathrm{Der}}_k(A,R_x)\cong {\mathrm{Hom}}_A({\mathrm{\Omega }}_{A/k},R_x).$$

Equivalently, after base change to $R$$R$,

$$T_{x}X(R)\cong {\mathrm{Hom}}_R(x^{\ast }{\mathrm{\Omega }}_{A/k},R),$$

where

$$x^{\ast }{\mathrm{\Omega }}_{A/k}={\mathrm{\Omega }}_{A/k}{\otimes }_{A}R.$$

So the dual-number definition and the Kähler differential definition are two expressions of the same object.

Example: Affine Space

Let

$$X={\mathbb{A}}_k^n=\mathrm{Spec}k[x_1,\dots ,x_n].$$

An $R$$R$-point of $X$$X$ is a $k$$k$-algebra homomorphism

$$k[x_1,\dots ,x_n]\to R.$$

It is determined by elements

$$r_1,\dots ,r_n\in R,$$

where

$$x_i\mapsto r_i.$$

A lift to $R[\epsilon ]$$R[\varepsilon]$ has the form

$$x_i\mapsto r_i+\epsilon v_i.$$

There are no equations to satisfy. Hence every choice of

$$v_1,\dots ,v_n\in R$$

gives a tangent vector.

Therefore

$$T_x{\mathbb{A}}_k^n(R)\cong R^n.$$

This matches the usual fact that affine $n$$n$-space has tangent space of rank $n$$n$ at every point.

Example: A Hypersurface

Let

$$X=\mathrm{Spec}k[x_1,\dots ,x_n]/(f).$$

An $R$$R$-point of $X$$X$ is a tuple

$$r=(r_1,\dots ,r_n)\in R^n$$

such that

$$f(r_1,\dots ,r_n)=0.$$

A tangent vector at $r$$r$ is a lift

$$x_i\mapsto r_i+\epsilon v_i.$$

This lift must satisfy the equation

$$f(r_1+\epsilon v_1,\dots ,r_n+\epsilon v_n)=0$$

in $R[\epsilon ]$$R[\varepsilon]$.

Expanding to first order gives

$$f(r+\epsilon v)=f(r)+\epsilon \sum _{i=1}^n\frac{\partial f}{\partial x_i}(r)v_i.$$

Since $r$$r$ is already a point of $X$$X$, we have

$$f(r)=0.$$

Therefore the tangent condition is

$$\sum _{i=1}^n\frac{\partial f}{\partial x_i}(r)v_i=0.$$

So

$$T_{r}X(R)=\{v\in R^n|\sum _{i=1}^n\frac{\partial f}{\partial x_i}(r)v_i=0\}.$$

This is the usual Jacobian equation for the tangent space.

Example: Several Equations

Let

$$X=\mathrm{Spec}k[x_1,\dots ,x_n]/(f_1,\dots ,f_m).$$

An $R$$R$-point is a tuple

$$r=(r_1,\dots ,r_n)$$

such that

$$f_j(r)=0$$

for all $j$$j$.

A tangent vector is a tuple

$$v=(v_1,\dots ,v_n)\in R^n$$

such that

$$x_i\mapsto r_i+\epsilon v_i$$

still satisfies all equations modulo ${\epsilon }^2$$\varepsilon^2$.

For each $j$$j$, we get

$$f_j(r+\epsilon v)=f_j(r)+\epsilon \sum _{i=1}^n\frac{\partial f_j}{\partial x_i}(r)v_i.$$

Since $f_j(r)=0$$f_j(r)=0$, the tangent condition is

$$\sum _{i=1}^n\frac{\partial f_j}{\partial x_i}(r)v_i=0$$

for every $j$$j$.

Hence

$$T_{r}X(R)=\{v\in R^n|J_f(r)v=0\},$$

where $J_f(r)$$J_f(r)$ is the Jacobian matrix

$$J_f(r)={(\frac{\partial f_j}{\partial x_i}(r))}_{j,i}.$$

Thus the dual-number definition recovers the familiar Jacobian tangent space.

Example: The Node

Let

$$X=\mathrm{Spec}k[x,y]/(xy).$$

This is the union of the two coordinate axes.

Consider the origin

$$x=0,y=0.$$

A tangent vector at the origin over $k$$k$ is given by

$$x\mapsto \epsilon a,y\mapsto \epsilon b.$$

The equation is

$$xy=0.$$

After lifting, we get

$$(\epsilon a)(\epsilon b)={\epsilon }^{2}ab=0.$$

So there is no linear condition on $a$$a$ and $b$$b$.

Thus

$$T_{0}X\cong k^2.$$

At the origin, the tangent space is two-dimensional, even though each branch is one-dimensional. This is one way singularities appear in the tangent-space picture.

Example: The Cusp

Let

$$X=\mathrm{Spec}k[x,y]/(y^2-x^3).$$

At the origin, a tangent vector is

$$x\mapsto \epsilon a,y\mapsto \epsilon b.$$

The equation becomes

$$(\epsilon b{)}^2-(\epsilon a{)}^3=0.$$

Since

$${\epsilon }^2=0,$$

both terms vanish. Hence there is again no linear condition.

So

$$T_{0}X\cong k^2.$$

The cusp is a curve, but its Zariski tangent space at the singular point is two-dimensional.

Away from the singular point, the Jacobian condition cuts out a one-dimensional tangent space.

Relation with Lie Algebras of Affine Algebraic Groups

Now suppose $G$$G$ is an affine algebraic group.

As a functor,

$$G:{\mathbf{CAlg}}_k\to \mathbf{Grp}.$$

The tangent functor is

$$TG(R)=G(R[\epsilon ]).$$

The projection $R[\epsilon ]\to R$$R[\varepsilon]\to R$ gives

$$TG(R)\to G(R).$$

The Lie algebra functor is the fiber over the identity:

$$\mathrm{Lie}(G)(R)=\ker (G(R[\epsilon ])\to G(R)).$$

This is a special case of the affine scheme construction.

For a general affine scheme $X$$X$, the tangent space at $x$$x$ is a fiber:

$$T_{x}X(R)=\{\tilde{x}\in X(R[\epsilon ])\mid \tilde{x}mod\epsilon =x\}.$$

For an affine group $G$$G$, the identity point gives a distinguished fiber. Because $G(R[\epsilon ])\to G(R)$$G(R[\varepsilon])\to G(R)$ is a group homomorphism, this fiber is a kernel:

$$T_{e}G(R)=\ker (G(R[\epsilon ])\to G(R)).$$

Thus

$$\boxed{{\displaystyle \mathrm{Lie}(G)=T_{e}G.}}$$

The Lie algebra of an affine algebraic group is the tangent space at the identity.

The Main Slogan

The dual-number method says:

$$\boxed{{\displaystyle \text{to differentiate algebraic geometry, map into }R[\epsilon ].}}$$

A tangent vector is a first-order deformation

$$x+\epsilon v.$$

A derivation is the coefficient of $\epsilon$$\varepsilon$ in a first-order lift

$$a\mapsto x(a)+\epsilon D(a).$$

The Leibniz rule is not imposed artificially. It is exactly the condition that the lift is multiplicative.

So the conceptual summary is:

$$\boxed{{\displaystyle \text{Derivations are first-order algebra maps into square-zero extensions.}}}$$

For affine schemes,

$$T_{x}X(R)\cong {\mathrm{Der}}_k(A,R_x).$$

For affine algebraic groups,

$$\mathrm{Lie}(G)(R)=\ker (G(R[\epsilon ])\to G(R)).$$

These are the same idea: infinitesimal geometry is encoded by maps to dual numbers.