Suppose that $X$ is an algebraic variety with divisor $D$ such that the logarithmic tangent bundle $\mathcal{T}_X(-\log D)$ is locally free. Suppose moreover that $\iota\colon Y\to X$ is a regular embedding of varieties with $Y$ smooth so that the normal bundle $\mathcal{N}_{Y/X}$ is also locally free and the natural map $\iota^\ast\mathcal{T}_X\to \mathcal{N}_{Y/X}$ is surjective.

Question: Are there natural geometric conditions on $D$ and $Y$ that guarantee that the restriction of the map $\iota^\ast \mathcal{T}_X\to \mathcal{N}_{Y/X}$ to a map $\iota^\ast\mathcal{T}_X(-\log D)\to \mathcal{N}_{Y/X}$ remains surjective? Moreover if the answer is yes, does it appear in the literature?

I suspect that the answer is along the lines that the map is surjective when $D$ and $Y$ intersect transversely for some suitable notion of transverse intersection in this context but I'd be grateful for something precise.

My motivation comes from the analogous question where 'algebraic variety' is replaced in my first sentence by 'rigid analytic variety' but I'm guessing I'm more likely to get an answer as stated and that it will translate easily. I don't mind assuming that $X$ is smooth as well as $Y$.


1 Answer 1


I guess my answer will be in the category of complex manifolds, therefore we definitely need to assume that $X$ is smooth. I am not that familiar with this language, so I apologize for any stupid mistakes coming from this.

Assume that $D$ is simple normal crossing and that $Y$ is smooth. Denote by $X_0 = X \setminus D$, $X_1 = D \setminus \text{Sing}(D)$, and inductively $X_{k+1}$ is the non-singular part of $\text{Sing}(\overline{X_k})$. Then the sections of $\mathcal{T}_X(- \log D)$ are those holomorphic vector fields on $X$, which are tangent to all strata $X_k$. I think a sufficient criterion for surjectivity is:

Y intersects all the strata $X_k$ transversally (in the sense $T_p Y + T_p X_k = T_p X$ for all $p \in Y \cap X_k$)

Indeed, to check whether the morphism $i^* \mathcal{T}_X(- \log D) \to \mathcal{N}_{Y/X}$ of locally free sheaves on $Y$ is surjective, it suffices to check in the fibres. At a point $p \in Y \cap X_k$ the divisor $D$ has equation $z_1 z_2 \ldots z_k =0$ in local coordinates and a local basis of $\mathcal{T}_X(- \log D)$ is given by $$z_1 \frac{\partial}{\partial z_1}, \ldots, z_k \frac{\partial}{\partial z_k}, \frac{\partial}{\partial z_{k+1}} \ldots, \frac{\partial}{\partial z_{\dim X}}.$$ The assumption, that $Y$ intersects $X_k$ transversally means that $T_p Y$ together with $\frac{\partial}{\partial z_{k+1}}|_p \ldots, \frac{\partial}{\partial z_{\dim X}}|_p$ span the vector space $T_p X$, thus the sections $\frac{\partial}{\partial z_{k+1}} \ldots, \frac{\partial}{\partial z_{\dim X}}$ restrict to generators of the fibre of $\mathcal{N}_{Y/X}$ at $p$.

If we drop the assumption that $Y$ is smooth, we might be able to phrase the transversality condition in terms of the matrix of derivatives of a regular sequence $x_1, \ldots, x_l$ defining $\iota$ along tangent directions to $X_k$ having rank $l$. [For this see also the discussion in the comments.]

  • $\begingroup$ Thanks. This looks helpful. I don't have time today to think further about it but will do so tomorrow. One quick question: are you overloading $k$ when you say that 'At a point $p\in Y\cap X_k$ the divisor $D$ has equation $z_1\ldots z_k=0$ in local coordinates'? $\endgroup$ Feb 16, 2016 at 13:56
  • $\begingroup$ No, these k should be the same. My reference is Section 2 of the paper arxiv.org/abs/math/0210263 . From what is contained there, it should follow that we can choose the coordinates around $p$ such that $D$ is given by the equation $z_1, \ldots, z_k$. $\endgroup$
    – JoS
    Feb 16, 2016 at 14:14
  • $\begingroup$ I see. Thanks for that reference. I'm not sure I understand where you are using $Y$ is smooth as a stronger statement than the statement that $\iota^\ast\mathcal{T}_X\to \mathcal{N}_{Y/X}$ is itself surjective. $\endgroup$ Feb 16, 2016 at 20:29
  • $\begingroup$ If $Y$ is not smooth, it is not clear what $T_p Y$ means (at least it is not a vector space of dimension $\dim Y$). So I guess here one must be careful how to phrase the transversality condition. However, I think that the proposal at the end of the answer should work, because then tangent vectors $v_1, \ldots, v_l$ to $X_k$ with the matrix of differentials $(\partial_{v_i} x_j)_{i,j=1, \ldots, l}$ having full rank should restrict to a basis of $\mathcal{N}_{Y/X}|_p$. $\endgroup$
    – JoS
    Feb 17, 2016 at 7:33
  • $\begingroup$ To see this last claim you might want to take a look at Proposition 21.2.16 of Vakil's notes math.stanford.edu/~vakil/216blog/FOAGdec2915public.pdf , where the conormal sheaf of a regular embedding is described explicitly. $\endgroup$
    – JoS
    Feb 17, 2016 at 10:04

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