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In page 205 of "Algebraic Curves, Algebraic Manifolds and Schemes" by Shokurov and Danilov, the tangent space $T_x X$ of an affine variety $X$ at a point $x \in X$ is defined as the subspace of $K^n$, where $K$ is the underlying field, such that $\xi \in T_x X$ if $(d_x g)(\xi)=0$ for any $g \in I$, where $I$ is the ideal of $K[T_1,\cdots,T_n]$ that defines $X$ and by definition $(d_x g)(\xi)=\sum_{i=1}^n \frac{\partial g}{\partial T_i}(x) \xi_i$, where partial derivatives are formal. So far so good. Next, it is mentioned, that if $f:X \rightarrow Y$ is a morphism of affine varieties, then we obtain a well-defined map $d_x f : T_x X \rightarrow T_{f(x)} Y$. How is this mapped defined and why is it well-defined?

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4 Answers 4

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This is explained in Shafarevich, p.88. Let $f(x)=y.$ The intrinsic and natural way to define this map is to use the isomorphisms $T_xX\cong (\mathfrak m_x/\mathfrak m_x^2)^*$ and $T_yY\cong(\mathfrak m_y/\mathfrak m_y^2)^*$ and the fact that $f$ defines a pullback on regular functions $f^*:k[Y]\to k[X].$ The map $f^*$ induces a map $\mathfrak m_y\to\mathfrak m_x$ which factors to give $d_x^*:\mathfrak m_y/\mathfrak m_y^2\to\mathfrak m_x/\mathfrak m_x^2.$

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Thanks for the answer. Why is the kernel of the map $m_y \rightarrow m_x/{m_x^2}$ equal to $m_y^2$? I can see that $m_y^2$ must be in the kernel, how about the reverse direction? –  Manos Nov 16 '12 at 20:27
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I think it doesn't need to be the whole kernel; as long as $f^*(\mathfrak m_y^2)\subseteq\mathfrak m_x^2$ we will have a well defined map between the quotients. –  Andrew Nov 16 '12 at 20:28
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Got it, the map need not be injective. Great. –  Manos Nov 16 '12 at 20:29

If $X\subset K^n$ and $Y\subset K^m$ are affine subvarieties , the map $f:X\to Y$ is the restriction of some polynomial map $F:K^n\to K^m: x\mapsto (F_1(x),...,F_m(x))$, where the $F_i$'s are polynomials $F_i\in K[T_1,...,T_n]$.
The map $d_x f : T_x X \rightarrow T_{f(x)} Y$ is the restriction to $T_x(X)$ of the linear map given by the Jacobian $$d_xF=Jac(F)(x)=(\frac {\partial F_i}{\partial X_j}(x)):K^n\to K^m$$ [The subspace $T_xX \subset K^n$ is the set of solutions of the humongous (but extremely redundant!) system of linear equations $\Sigma \frac {\partial g}{\partial X_j}(x)\xi_j=0$ where $g$ runs through $I(X)$]
The only thing to check is that we have in $K^m$ : $$(d_xf)(T_xX)\subset T_y(Y) $$
This means that we must show that $$(d_yh)(d_xf(v))=0 \quad (?)$$ for alll $v\in T_xX$ and all $h\in I(Y)$.
This follows from the following two facts:
a) For all $h\in I(Y)$ we have $h\circ F\in I(X)$, since $F$ maps $X$ into $Y$.
b) Functoriality of the differential: $d_x(h\circ F)=d_yh\circ d_xF$ .

And now if $v\in T_xX$ we can write $$(d_yh)(d_xf(v))=(d_yh)(d_xF(v))=d_x(h\circ F)(v)=0$$ since $h\circ F\in I(X)$ . We have thus proved $(?)$

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Brilliant answer. In your "b) functoriality of the differential", y=f(x)? –  Manos Nov 16 '12 at 21:09
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Dear @Manos, "functoriality" is a pretentious way of saying that the differential of a composition of regular maps is the composition of the differentials, exactly as in our beloved good old calculus... –  Georges Elencwajg Nov 16 '12 at 21:17

Let $y = f(x)$. Then f induces a map on stalks, which in turn induces a map on cotangent spaces:

$$\mathfrak{m}_{y,Y}/\mathfrak{m}_{y,Y}^2 \rightarrow \mathfrak{m}_{x,X}/\mathfrak{m}_{x,X}^2$$

Taking the dual over $K$ yields the map on tangent spaces.

To put this into context, your definition of the tangent space makes use of the explicit basis

$$\frac{\partial}{\partial T_i} \bigg|_x$$

That is, partial derivation with respect to the variable $T_i$, followed by evaluation at $x$.

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There is a good analogy between tangent spaces of differentiable manifolds and those of affine varieties. By the proposition of my answer to this question, $T_x$ is canonically isomorphic to $Der(\mathcal{O}_{X, x},K)$, where $\mathcal{O}_{X,x}$ is the local ring of $X$ at $x$. The morphism $f\colon X \rightarrow Y$, induces a morphism $f^{\#}_x\colon \mathcal{O}_{Y, f(x)} \rightarrow \mathcal{O}_{X, x}$ of $K$-local algebras. Hence $f^{\#}_x$ induces a linear map $d_xf\colon Der(\mathcal{O}_{X, x},K) \rightarrow Der(\mathcal{O}_{Y, f(x)},K)$, i.e. $d_x f\colon T_x X \rightarrow T_{f(x)}Y$.

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