Motivation behind the definition of Zariski tangent space

Question

Intuitively, I think of tangent space at a point as the set of all points lying in the tangent plane passing through that point.

Here is the definition of the Zariski tangent space:

Let $X$ be an algebraic variety and $p \in X$. The tangent space of $X$ at point $p$ is defined as $$T_pX= \operatorname{Der}_k(O_{X,p},k).$$

How does the above definition match with my intuition? Or more specifically, can someone give a one-one correspondence between $T_pX$ and the set of all points lying in the tangent plane passing through $p$?

Answer 1

The tangent space at $p$ is the space of all directions in which you can take a directional derivative at $p$. Whatever "directional derivative" means, it should only depend on the germ at $p$, so it's a function on $\mathcal{O}_{X, p}$. And it should be linear and obey the Leibniz rule, so it's a derivation. These conditions turn out to be enough to give a notion of tangent space that agrees with intuition (e.g. you can compute the Zariski tangent space to a variety cut out by various polynomials and it will be the thing you think it is).

Answer 2

$\def\A{\mathbb{A}} \def\sI{\mathscr{I}} \def\ord{\operatorname{ord}} \def\sO{\mathcal{O}} \def\der{\operatorname{Der}} \def\m{\mathfrak{m}}$Let's examine the classical situation and redefine the tangent space in more geometric terms. Let $k$ be an algebraically closed field, $X\subset\A_k^n$ be an algebraic subset. Let $a=(a_1,\dots,a_n)\in X$ be a point. Let $\sI(X)\subset k[x_1,\dots,x_n]$ be the ideal of polynomials vanishing on $X$. Given $f\in\sI(X)$ and a vector $\vec{v}\in k^n$, the polynomial $f(a+t\vec{v})$ vanishes at $t=0$. Given a ring $A$ and a polynomial $g\in A[t]$, its vanishing order, denoted $\ord g$, is the maximum integer $e$ such that $t^e$ divides $g$.

Definition 1. Let $\vec{v}\in k^n$.

• The intersection multiplicity of $X$ with the line $l=\{a+\lambda\vec{v}\mid \lambda\in k\}$ is the number $$ \mu_a(X,l)=\min_{f\in\sI(X)}\ord_tf(a+t\vec{v})\geq 1. $$

• The vector $\vec{v}$ is tangent to $X$ at $a$ if $\mu_a(X,l)\geq 2$.

We define the tangent space of $X$ at $a$, denoted $T_aX$, to be the set of vectors $\vec{v}\in k^n$ tangent to $X$ at $a$. Then $T_aX$ is a $k$-vector subspace of $k^n$ [A, Proposition 5]. For each $\vec{v}\in k^n$, we define the directional derivative at $a$ in the direction $\vec{v}$

\begin{align*} \partial_{\vec{v}}:k[x_1,\dots,x_n]&\to k\\ f&\mapsto\left.\frac{d}{dt}f(a + t\vec{v})\right|_{t=0}. \end{align*} In [A] it is shown:

• $\partial_{\vec{v}}$ is a $k$-derivation.
• If $\vec{v}\in T_aX$, then $\partial_{\vec{v}}f=0$ for all $f\in\mathscr{I}(X)$ and therefore we get a $k$-derivation $\partial_{\vec{v}}:\sO_{X,a}\to k$.

Proposition 2. The map $\vec{v}\mapsto\partial_{\vec{v}}$ induces an isomorphism $T_aX\cong\der_k(\sO_{X,a},k)$ as $k$-vector spaces.

Remark 3. The polynomial ring $A=k[x_1,\dots,x_n]$ has a canonical graded ring structure $A=\bigoplus_{i\geq 0}A_i$, where $A_i$ is the zero polynomial and the homogeneous polynomials of degree $i$. Pushingforward this graded structure along the automorphism $f\mapsto f(x-a)$ of $A$ gives a graded structure $A=\bigoplus_{i\geq 0}B_i$, where $B_i=\{\lambda(x_1-a_1)^{r_1}\cdots(x_n-a_n)^{r_n}\mid\lambda\in k,r_1+\cdots+r_n=i\}$.

The following proof is [B, Proposición 5.4], translated to English by me.

Proof. Given a $k$-derivation $D\in\der_k(\sO_{X,a},k)$, define $\vec{v}_D$ to be the vector $(D(x_1-a_1),\dots,D(x_n-a_x))$. Then $D\mapsto\vec{v}_D$ is $k$-linear. Let's see it is the inverse to $\vec{v}\in T_aX\mapsto\partial_{\vec{v}}$. Firstly, we see that $\vec{v}_D$ is tangent to $X$ at $a$. Let $f\in\sI(X)$. We have to show that the polymomial $g(t)=f(a+t\vec{v}_D)$ has a zero of multiplicity $\geq 2$ at $t=0$. On the one hand, $g(0)=f(a)=0$. Thus, $\ord g\geq 2$ if and only if $g'(0)=0$. By the multivariate chain rule for polynomials [ref], $g'(t)=\sum_{i=1}^n\partial_if(a+t\vec{v}_D)D(x_i-a_i)$, whence $g'(0)=\sum_{i=1}^n\partial_if(a)D(x_i-a_i)=D\left(\sum_{i=1}^n\partial_i f(a)(x_i-a_i)\right)=D(f)=0$, for $\sum_{i=1}^n\partial_i f(a)(x_i-a_i)$ is the component of degree $1$ of $f$ (in the graded structure $\bigoplus_{i\geq 0}B_i$ of Remark 3), and thus the difference $f-\sum_{i=1}^n\partial_i f(a)(x_i-a_i)$ is in $\m_{X,a}^2\subset\sO_{X,a}$ (and $D$ vanishes on $\m_{X,a}^2$ by Leibniz's rule).

If we apply $\partial_{\vec{v}}$ to $x_i-a_i$, we obtain the derivative at $t=0$ of $(a_i+tv_i)-a_i=tv_i$, which is $v_i$. Hence $\vec{v}_{\partial_\vec{v}}=\vec{v}$ for $\vec{v}\in T_aX$. Conversely, given $D\in\der_k(\sO_{X,a},k)$, we need to show $\partial_{\vec{v}_D}=D$. Firstly, note that these two derivations are equal when evaluated at $x_i-a_i$: the scalar $\partial_{\vec{v}}(x_i-a_i)$ is the $i$-th component of $\vec{v}$ for all $\vec{v}$, so $\partial_{\vec{v}_D}(x_i-a_i)$ is the $i$-th component of $\vec{v}_D$, which by definition is $D(x_i-a_i)$. On the other hand, by Leibniz's rule (for quotients), a derivation $\sO_{X,a}\to k$ is completely determined by its values at $x_i-a_i$, $i=1,\dots,n$ (expand a polynomial in the graded structure $\bigoplus_{i\geq 0}B_i$ of Remark 3). Thus $\partial_{\vec{v}_D}=D$. $\square$

Let $D\in\der_k(\sO_{X,a},k)$. By Leibniz's rule, $D$ vanishes at $\m_{X,a}^2$. Hence it defines a $k$-linear map $\m_{X,a}/\m_{X,a}^2\to k$.

Proposition 4. This map induces an isomorphism $\der_k(\sO_{X,a},k)\cong\operatorname{Hom}_k(\m_{X,a}/\m_{X,a}^2,k)$ as $k$-vector spaces.

Combination of this result with Proposition 2 means that the tangent space $T_aX$ is the Zariski tangent space.

Proof. The result is actually more general, and it is proven in [C, Proposition 4.29]. $\square$

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References

[A] Mathematics Stack Exchange, Tangent space definition in Algebraic Geometry

[B] A. Rojas León, Notas y ejercicios de Geometría Algebraica (archived in the Wayback Machine)

[C] J. S. Milne, Algebraic Geometry (archived in the Wayback Machine)

Answer 3

First use GAGA to pass to complex manifolds. Then use base change to pass to a real manifold....Now, given any smooth embedding (use the Whitney embedding theorem) of your manifold in Euclidean space, assume that the tangent surface at p exists and does everything you want it to. The one-to-one correspondence you ask for, which was not given by the other answer, is:

Given a vector starting at p and lying in the tangent surface, consider the geodesic line lying in your manifold, passing through p, and whose tangent vector in the usual Euclidean sense is the given vector. Problem: to define a derivation of the space of germs of functions on your manifold at p. Hint: First extend your function to the entire ambient space, smoothly. Then use the ordinary directional derivative.
Remark: this is not canonical, but you didn't ask for a canonical isomorphism.

Second remark: you didn't specify whether the field has characteristic zero or not...I assume it does. Third remark: what you wrote down isn't actually the definition of the Zariski tangent space. The definition of the Zariski tangent space is $m_p/m_p^2$ where $m_p$ is the ideal of all algebraic functions which vanish at p.

Answer 4

There is a one-to-one correspondence, here is a simple example from elementary differential geometry :

let $\Gamma:\mathbb{R}\to \mathbb{R}^2$ be a curve in $\mathbb{R}^2$, the derivative $\frac{d\Gamma}{dt}|_p$ of $\Gamma$ at some point $p$ on the curve is a vector tangent to the curve at this point, so it can be the basis vector for the tangent space at $p$ by spanning it so : $$T_p\Gamma=\left\{ v\frac{d\Gamma}{dt}\Big{|}_p,~v\in\mathbb{R} \right\}$$
now we can forget about $\Gamma$ and define it using only the operator $\frac{d}{dt}|_p$ and nothing will be effected: $$T'_p\Gamma=\left\{ v\frac{d}{dt}\Big{|}_p,~v\in\mathbb{R} \right\} $$

you can see that the old and the new $T_p\Gamma$ are isomorphic so you can use any one of them as the tangent space