Vector space structure of the Zariski tangent space

Question

Let $X$ be a $k$-scheme and $p\in X(k)$. The Zariski tangent space $T_p X$ is usually defined as being the $k$-vector space $\hom_k(\mathfrak{m}_p/\mathfrak{m}_p^2,k)$. In general, this coincides as a set with $$\widetilde{T_p X}:=\{f\in X(k[\varepsilon])\:|\: f((x))=p\},$$ where $k[\varepsilon]$ is the ring of dual numbers. I wonder how can we describe the $k$-vector space structure on $\widetilde{T_p X}$ inherited from $T_p X$ under this identification.

The action of $k$ seems to arise in this way: the morphism $k[\varepsilon]\to k[\varepsilon]$ given by $\varepsilon\mapsto a\varepsilon$, where $a\in k$, induces a morphism $\operatorname{Spec}(k[\varepsilon])\to \operatorname{Spec}(k[\varepsilon])$ and so $a$ acts on $\widetilde{T_p X}$ by precomposition. Now, I'm not sure if this action coincides with the one inherited by $T_p X$ nor I know how to describe addition of tangent vectors.

Also, if $X$ is a group scheme with identity $p$, then we can describe $\widetilde{T_p X}$ set-theoretically as the kernel of the natural morphism (of groups) $X(k[\varepsilon])\to X(k)$. Naturally, this kernel is a group. I wonder if the group operation coincides with the vector sum.

Answer 1

The functor-of-points perspective is very useful here in giving a clean answer to your question. When $X$ is a scheme and $R$ is a ring, write $X(R)=\mathrm{Hom}_{\mathbf{Sch}}(\mathrm{Spec}(R), X)$. We have a morphism of sets $X(k[\epsilon]/\epsilon^2)\to X(k)$ induced by the morphism of rings $k[\epsilon]/\epsilon^2\to k$ induced by mapping $\epsilon\mapsto 0$. The set-theoretic preimage over some $p\in X(k)$ is your $\widetilde{T_pX}$. We will equip this set with the structure of a $k$-vector space.

First we will define addition on $\widetilde{T_pX}$. In preparation, we want to define a map $$ +\colon X(k[\epsilon]/\epsilon^2)\times_{X(k)}X(k[\epsilon]/\epsilon^2)\to X(k[\epsilon]/\epsilon^2). $$ Now, $$ X(k[\epsilon]/\epsilon^2)\times_{X(k)}X(k[\epsilon]/\epsilon^2) \simeq X(k[\epsilon]/\epsilon^2\otimes_kk[\epsilon]/\epsilon^2)\simeq X(k[\epsilon_1,\epsilon_2]/(\epsilon_1^2,\epsilon_2^2)). $$ Thus to define the map $+$, it's enough to define a map on rings. Take the map on rings given by $\epsilon_1\mapsto\epsilon$ and $\epsilon_2\mapsto\epsilon$. The map $+$ gives a morphism over $X(k)$ that is vector addition in the tangent space.

Scalar multiplication works similarly, along the lines you had thought: Given an element $a\in k$, we get a morphism of $k$-algebras $k[\epsilon]/\epsilon^2\to k[\epsilon]/\epsilon^2$ given by $\epsilon\mapsto a\epsilon$. Applying the functor $X(-)$ gives a morphism $$ X(k[\epsilon]/\epsilon^2)\to X(k[\epsilon]/\epsilon^2) $$ over $X(k)$ (that is, the triangular diagram I would like to be able to draw here commutes).

Edit:

I apologize as these sections I think are worse than the above. The main point is that if $A$ is a $k$-algebra and you fix a morphism $\phi\colon A\to k$ of $k$-algebras, then a morphism $A\to k[\epsilon]/\epsilon^2$ of the form $a\mapsto\phi(a)+\psi(a)\epsilon$ is just the data of a $k$-derivation $\psi$ of $A$.

Below we'll dispense with what happens at least for linear algebraic groups over the field $k$. (Note that affine implies linear and this comparison makes sense only if $G$ is taken over $k$.)

$\widetilde{T_pX}$ and $T_pX$

This is essentially what the other answer discussed. On the level of sets, the isomorphism $\widetilde{T_pX}\to T_pX$ is given as follows. Let $f\in\widetilde{T_pX}$. Then $f$ is a morphism $$ f\colon\mathrm{Spec}k[\epsilon]/\epsilon^2\to X $$ such that the only closed point $(0)$ of the source is sent to $p\in X$. We get a morphism of local rings $$ f^{\#}\colon \mathcal{O}_{X,p}\to k[\epsilon]/\epsilon^2 $$ that must send $\mathfrak{m}_p\to \epsilon k[\epsilon]/\epsilon^2\simeq k$. This is clearly the same as map $\mathfrak{m}_p/\mathfrak{m}_p^2\to k$. Moreovoer, a map $\mathfrak{m}_p/\mathfrak{m}_p^2\to k$ is the same as a $k$-derivation of $\mathcal{O}_{X,p}$, and addition of linear functionals correspdonds to addition of derivations under this bijection. Explicitly, given a linear functional $\lambda$, one defines $d_\lambda(x)=0$ if $x$ is a unit, and $d_\lambda(x)=\lambda(x)$ if $x\in\mathfrak{m}_p$, and this is a $k$-derivation.

As for the map $+$, if we choose an affine open $\mathrm{Spec}A$ of $X$, containing $p$, we see that the sum of two elements of $\widetilde{T_pX}$ is a ring morphism $$ A\to k[\epsilon_1,\epsilon_2]/(\epsilon_1^2,\epsilon_2^2)\to k[\epsilon]/\epsilon^2. $$ By definition of $+$, the derivation this ring morphism encodes (the function giving the coefficient of $\epsilon$) is the sum of the derivations we started with. This is the statement that the the isomorphis of sets above is an isomorphism of abelian groups. Scalar mutliplication is similar.

The case of linear algebraic groups

Let $G$ be a linear (equivalently, affine) algebraic $k$-group, and define $$ \mathrm{Lie}_1(G):=\mathrm{ker}(G(k[\epsilon]/\epsilon^2)\to G(k)). $$ Thus $\mathrm{Lie}_1(G)$ is a group, although a priori it is not obviously an abelian group. As just the kernel of a group homomorphism, it is not a priori a vector space over $k$. We'll try to show that it's isomorphic as a group to your $T_pX$, and then that it has a natural $k$-action that makes this isomorphism one of $k$-vector spaces. Then we'll conclude from the section below. This material is totally standard and in Borel, Springer, Milne, various books of Conrad and co-authors etc, but nowhere did I find the precise statement that I need below (at least, not yet).

Recall that your $T_pX$ is isomorphic as a $k$-vector space to the space of $k$-derivations of $\mathcal{O}_{X,p}$. Indeed, given a morphism $\varepsilon\colon A\to k$ of $k$-algebras, let $\mathfrak{m}=\ker\varepsilon$. Then $\mathrm{Der}_k(A)\to(\mathfrak{m}/\mathfrak{m}^2)^*$ sending $\delta\mapsto\delta|_{\mathfrak{m}}$ is a linear isomorphism.

Define $$\mathrm{Lie}(G):=T_eG.$$

We will define an isomorphism $\mathrm{Lie}_1(G)\to\mathrm{Lie}(G)$ of groups. Indeed, a morphism of rings $A\to k[\epsilon]/\epsilon^2$ is given by an morphism of $A$-algebras $A\to k$ and a $k$-derivation of $A$. For such a morphism to belong to $\mathrm{Lie}_1(G)$ means that the morphism of $k$-algebras is always the counit $\varepsilon\colon A\to k$ of the Hopf algebra $A$. Thus if $g_1$, $g_2$ are in the kernel, their product is $$ g_1g_2\colon A\to A\otimes_kA\to k[\epsilon]/\epsilon^2 $$ and is also in the kernel. Thus the corresponding morphism $A\to k$ is again the counit, and the only question is what the derivation is. Of course, it just has to be the sum of the derivations for the $g_i$, but I'm dense enough to not extract this immediately from the slightly different treatments I listed above.

Answer 2

Question: "I wonder how can we describe the k-vector space structure on TpX˜ inherited from TpX under this identification?"

Answer: Since any point in a scheme is contained in an open affine subscheme and since the tangent space is defined "locally" we may assume $X$ is affine.

Let $I\in X(k)$ with $X:=Spec(A)$ and $A$ a $k$-algebra ($I:=\mathfrak{m}_p$ in your notation). Given any element $\mu\in Hom_k(I/I^2, R/I)$, there is a direct sum decomposition $R/I^2 \cong k \oplus I/I^2$ and a canonical map

$$ \phi(\mu): R/I^2 \rightarrow k[\epsilon]$$

defined by

$\phi(\mu)(a,b):=a+\mu(b)\epsilon$ where $a\in k$ and $b\in I/I^2$. You get a canonical map

$$\phi_{\mu}: R \rightarrow k[\epsilon]$$

defined by

$$\phi_{\mu}(x):=\phi(\mu)(\overline{x}),$$

where $\overline{x}\in R/I^2$.

It seems that if you choose $\mu , \eta \in Hom_k(I/I^2,R/I)$ you will get the map

$$\phi_{\mu +\eta}(a,b):=a+(\mu+\eta)(b)\epsilon.$$

and for any element $u\in k$

$$\phi_{u\mu}(a,b):=a+(u\mu)(b)\epsilon.$$

This construction seems to induce a $k$-module structure on $ker(X(k[\epsilon])\rightarrow X(k))$ compatible with the one on $(I/I^2)^*$. If $X$ is a group scheme you should be able to determine the kernel of $X(k[\epsilon]) \rightarrow X(k)$: There is a canonical map

$$u:X(k[\epsilon]) \rightarrow X(k)$$

induced by the canonical map

$$R \rightarrow k[\epsilon] \rightarrow k$$

and your point $I\in X(k)$. Hence if you assume $\phi\in X(k[\epsilon])$ with $\phi \in ker(u)$ it follows

$$\phi(a):=\overline{a}1 +\phi_1(a)\epsilon$$

where $\overline{a}\in R/I$ is the class of $a$. Given two maps $\phi, \psi\in ker(u)$ you may "add them" and "multiply with a constant $\alpha$":

$$\rho, \eta:R \rightarrow k[\epsilon]$$

defined by

$$\rho(a):= \overline{a}1+(\phi_1+\psi_1)(a)\epsilon$$

and

$$\eta(a):=\overline{a}1+(\alpha \psi_1(a))\epsilon.$$

This should type of reasoning should induce a $k$-module structure on $\widetilde{T_p(X)}:=ker(u)$ giving an isomorphism of $k$-modules

$$T_p(X) \cong \widetilde{T_p(X)}.$$

Example: $Lie(SL(V))$ Let $V:= k\{e_1,e_2\}$ and $SL(V)$ be the group scheme of linear automorphisms of $V$ with determinant $1$, where $k$ is any commutative unital ring. Let

$$R:=k[x_{11}, x_{12},x_{21},x_{22}]/(det(x_{ij})-1).$$

It follows $SL(V):=Spec(R)$. We want to determine $T_e(SL(V))$ - the tangent space at the identity $e$.

By definition $T_e(SL(V)):=ker(SL(V)(k[\epsilon]) \rightarrow SL(V)(k)):=ker(u)$.

A map $\phi \in ker(u)$ is a map on the form

$$\phi: R \rightarrow k[\epsilon]$$

with $\phi(x_{11}):=1+a_{11}\epsilon, \phi(x_{12}):=a_{12}\epsilon, \phi(x_{21}):=a_{21}\epsilon, \phi(x_{22}):=1+a_{22}\epsilon$ with $a_{ij}\in k$.

The map $\phi$ induce a map $\phi^*:k[x_{ij}]\rightarrow k[\epsilon]$ with the property that $\phi^*(det(x_{ij})-1)=0$, and since the composed map $k[x_{ij}]\rightarrow k$ should correspond to the identity matrix, it follows the induced map $\phi$ must satisfy the above conditions. Since the map $\phi$ is well defined it follows $\phi(det(x_{ij})-1)=0$ and this holds iff $a_{11}+a_{22}=0$. Hence $ker(u)$ is the set of matrices $sl(V) :=\{A:=(a_{ij})\in Mat(2,k)$ with $tr(A)=0\}$. Hence

$$Lie(SL(V))\cong sl(V).$$

The following post considers an intrinsic construction of the structure of a Lie algebra on $Lie(SL(V))$ and the universal envloping algebra $U(Lie(SL(V))$ (you must consider the algebra of first order differential operators and the algebra of distributions):

Distributions of a group scheme as differential operators.

If $G:=Spec(R)$ is a group scheme and $I\subseteq G$ is the ideal of the unit element it follows

$$Dist^n(G,e):= Hom_{R/I}(R/I^{n+1}, R/I)$$

has the property that if the base field $k$ is of characteristic zero it follows

$$Dist^n(G,e) \cong U^n(Lie(G))$$

is the $n$'th piece of the canonical filtration of the enveloping algebra $U(Lie(G))$. Hence $Dist^1(G,e)=U^1(Lie(G))\cong k\oplus Lie(G)$. Since $Dist^1(G,e) \cong Diff^1_k(R/I^2,k)$ is the module of first order differential operators from $R/I^2$ to $k$, there is a canonical structure of Lie algebra on $k\oplus Lie(G)$ inducing one on $Lie(G)$. This structure comes from the group-scheme structure on $G$.

Note: The construction is elementary - you must get used to working with coordinate rings and their maximal and prime ideals. In fact: If $R$ is a Hilbert-Jacobson ring it is enough to work with maximal ideals. Any commutative unital ring that is finitely generated over a field or a Dedekind domain is a Hilbert-Jacobson ring.