Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $V$ be a non-singular affine variety in $\mathbb{C}^n$. $V$ can be regarded as a complex manifold. Let $p = (a_1,\dots,a_n) $ be a point of $V$. Let $\mathcal{O}_p$ be the local ring of $V$ at $p$. A tangent vector $v$ at $p$ is a derivation $\mathcal{O}_p \rightarrow \mathbb{C}$, i.e. a $\mathbb{C}$-linear map $v$ such that $v(fg) = v(f)g(p) + f(p)v(g)$ for $f, g \in \mathcal{O}_p$. Let $T_p$ be the set of tangent vectors at $p$. We regard $T_p$ as a vector space over $\mathbb{C}$ in the obvious way.

On the other hand, we can define a tangent space at $p$ as follows. Let $f_1,\dots f_r$ be defining polynomials for $V$. Let $L_i$ be the hyperplane defined by $\sum_k \frac{\partial f_i}{\partial x_k}(p)(x_k - a_k) = 0$. Let $S_p = \bigcap_i L_i$.

Are $T_p$ and $S_p$(or rather the vector space attached to it) canonically isomorphic?

share|improve this question
add comment

2 Answers 2

up vote 3 down vote accepted

Yes, there is a canonical isomorphism of $\mathbb C$-vector spaces $$i:S_{X,p}\stackrel {\cong}{\to} T_{X,p}=Der(\mathcal O_{X,p}, \mathbb C)$$ If one starts with a vector $v=(v_1,...,v_n)\in S_p\subset \mathbb C^n$ (so that $\sum_k \frac{\partial f_i}{\partial x_k}(p)v_k = 0$), the isomorphism $i$ associates to it the derivation $i(v)=\partial _v$ defined on $\mathcal O_{X,p}$ by the formula $$ \partial_v(g)=\sum_k \frac{\partial g}{\partial x_k}(p)v_k $$ where $g\in \mathcal O_{X,p}$ is an arbitrary local function.
The inverse isomorphism is given by $$i^{-1}: Der(\mathcal O_{X,p},\mathbb C) \stackrel {\cong}{\to} S_{X,p} :\partial \mapsto (\partial x_1,...,\partial x_n) $$

Edit
The vector space of derivations is also isomorphic to Zariski's tangent space $(\frak m_p/\frak m^2_p)^*$, defined via the maximal ideal $\mathfrak m_p\subset \mathcal O_{X,p}$.
The isomorphism is $$ Der(\mathcal O_{X,p} ) \stackrel {\cong}{\to} (\frak m_p/\frak m^2_p)^*:\partial \mapsto \overline{\partial} $$ where $\overline{\partial} (g \;\text {mod} \;m^2_p)=\partial (g)$ for $g\in \frak m_p$.

Second Edit
The following remark may be of some interest, since it does not seem to be addressed in Algebraic Geometry books:
If $X\subset \mathbb A^n_k$ is an affine algebraic variey and if $p\in X$, we may consider the maximal ideal $M_p\subset \mathcal O(X)$ of global functions vanishing at $p$.
We may also consider, as we already did, the maximal ideal $\mathfrak m_p\subset \mathcal O_{X,p}$ of germs of functions regular ay $p$ and vanishing at $p$.
We then have a natural $k$-linear map $ M_p/M_p^2 \to \mathfrak m_p/\mathfrak m_p^2 $ and the slightly surprising but pleasant fact is that this linear map is an isomorphism.

share|improve this answer
    
Dear Georges, could you explain why it is an isomorphism? –  Makoto Kato Nov 18 '12 at 11:20
    
Dear Makoto, you can look up the proofs either in Shafarevich's Basic Algebraic Geometry, vol.1, Chapter II, Section 1.3 or in in Mezzetti's online notes, section 13. –  Georges Elencwajg Nov 18 '12 at 12:19
add comment

Lemma 1 Let $A$ be a local algebra over a field $k$. Let $\mathfrak{m}$ be the unique maximal ideal of $A$. Suppose the canonical homomorphism $k \rightarrow A/\mathfrak{m}$ is an isomorphism. Let $f \in A$. There exists a unique $c \in k$ such that $f \equiv c$ (mod $\mathfrak{m}$). We denote this $c$ by $f(0)$. We call a $k$-linear map $v\colon A \rightarrow k$ a derivation if $v(fg) = v(f)g(0) + f(0)v(g)$ for $f, g \in A$. Let $Der(A, k)$ be the set of derivations. We regard $Der(A, k)$ as a vector space over $k$ in the obvious way. Let $v \in Der(A, k)$. If $f, g \in \mathfrak{m}$, $v(fg) = v(f)g(0) + f(0)v(g) = 0$. Hence $v(\mathfrak{m}^2) = 0$. Hence $v$ induces a $k$-linear map $\bar v\colon \mathfrak{m}/\mathfrak{m}^2 \rightarrow k$. Hence we get a $k$-linear map $\psi\colon Der(A, k) \rightarrow (\mathfrak{m}/\mathfrak{m}^2)^*$. Then $\psi$ an isomorphism.

Proof: This is proved here

Lemma 2 Let $k$ be a field. Let $L_i = \sum_{k = 1}^{n} a_{ik} x_k, i = 1,\dots, r$ be linear polynomials in $k[x_1,\dots,x_n]$. Let $T$ be the vector subspace $\{v \in k^n| L_i(v) = 0$ for all $i\}$ of $k^n$. Let $W$ be the vector subspace of $k^n$ generated by $L_i(e_1,\dots,e_n)$ for all $i$, where $\{e_1,\dots,e_n\}$ is the canonical basis of $k^n$. Then $T$ is canonically isomorphic to $(k^n/W)^*$.

Proof: An element of $(k^n/W)^*$ is identified with a linear map $f\colon k^n \rightarrow k$ such that $f(W) = 0$. Such a map $f$ is uniquely determined by the condition $\sum_{k = 1}^{n} a_{ik} f(e_k) = 0, i = 1,\dots, r$. Hence the assertion follows. QED

Proposition Let $A = k[x_1,\dots,x_n]$ be the polynomial ring over $k$. Let $I$ be an ideal of $A$ generated by $F_1,\dots,F_r$. Let $B = A/I$. Let $p = (a_1,\dots, a_n)$ be a point of $k^n$ such that $F_i(p) = 0$ for all $i$. Let $M = (x_1 - a_1,\dots,x_n - a_n)$ be the maximal ideal of $A$. Let $\mathfrak{m} = M/I$. Let $L_i = \sum_k \frac{\partial F_i}{\partial x_k}(p)x_k$ for $i = 1, \dots, r$. Let $T$ be the vector subspace $\{v \in k^n| L_i(v) = 0$ for all $i\}$ of $k^n$. Then $T$ is canonically isomorphic to $Der(B_{\mathfrak{m}},k)$.

Proof: $\mathfrak{m}/\mathfrak{m}^2$ is canonically isomorphic to $M/(I + M^2)$. $M/M^2$ is a $k$-vector space with a basis $\bar x_1,\dots,\bar x_n$, where $\bar x_i = x_i$ (mod $M^2$). Let $F \in I$. Since $F(p) = 0$, $F = \sum_k \frac{\partial F}{\partial x_k}(p)(x_k - a_k) + \cdots$. Hence $(I + M^2)/M^2$ is a vector subspace of $M/M^2$ generated by $L_i(\bar x_1, \dots,\bar x_n)$ for all $i$. Hence the assertion follows from Lemma 1 and Lemma 2. QED

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.