6
$\begingroup$

Let $X$ be an irreducible algebraic variety over some field $k$. It is well known, that if $X$ is smooth and of dimension $d$, then the tangent bundle of $X$ is smooth, irreducible and of dimension $2d$. There is no difficulty in defining the tangent bundle $T_X$ in the non-smooth case. It is defined as the (global) Spec of the symmetric algebra of the sheaf of Kähler differentials $\Omega^1_{X|k}$. In fact this definition works for an arbitrary morphism of schemes $X\to S$. (See EGA IV$_4$ 16.5.12.1.)

All the irreducible components of $T_X$ come from points of $X$. See Prop. 2.2 in this paper for a precise statement. The generic point of $X$ always yields an irreducible component of $T_X$ of dimension $2d$, but typically there will be other components.

I would like to know if there is an example such that $T_X$ has an irreducible component of dimension strictly smaller than $2d$.

Note that this will not be possible if $X\subset\mathbb{A}^n_k$ is a complete intersection: If the ideal of $X$ is generated by $f_1,\ldots,f_{n-d}$ then $T_X$ is described by the $2(n-d)$ equations $f_1,\ldots,f_{n-d},d(f_1),\ldots,d(f_{n-d})$ in $\mathbb{A}^{2n}_k$. (Cf. Example 1.5 in the above paper.)

$\endgroup$

1 Answer 1

4
$\begingroup$

How about the union in $\mathbb{A}^9$ of the three $5$-planes, $\Lambda_1 = \text{span}(\mathbf{e}_1,\mathbf{e}_6,\mathbf{e}_7,\mathbf{e}_8,\mathbf{e}_9)$, $\Lambda_2=\text{span}(\mathbf{e}_2,\mathbf{e}_4,\mathbf{e}_5,\mathbf{e}_8,\mathbf{e}_9)$ and $\Lambda_3=\text{span}(\mathbf{e}_3,\mathbf{e}_4,\mathbf{e}_5,\mathbf{e}_6,\mathbf{e}_7)$? The Zariski tangent space at the origin is the entire $9$-dimensional affine space. Each pair $\Lambda_i$, $\Lambda_j$ is contained in a hyperplane $H_{i,j} = Z(x_k)$, where $k$ is the missing index. So a tangent vector at the origin contained in none of the three hyperplanes $H_{1,2}=Z(x_3)$, $H_{1,3}=Z(x_2)$, $H_{2,3}=Z(x_1)$ is not a "limit" of nearby tangent vectors, i.e., the Zariski tangent space of the origin is an irreducible component.

Of course this algebraic set is reducible. But since this problem is a formal local problem, I can make a "global" irreducible variety which is formally locally isomorphic to the algebraic set above (i.e., it is not "unibranch").

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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