$\DeclareMathOperator{\Bl}{\operatorname{Bl}}$It is known that if we have a projective variety $X$ and a projective smooth subvariety $Y$ then the exceptional divisor $E \subset \Bl_{Y}X$ of the blow-up of $X$ along $Y$ is the projectivization of the normal bundle $N_{Y|X}$. In particular points in $E$ parametrizes lines (directions) normal to $Y$.

My question now is the following: suppose for simplicity that $X= \mathbb{P}^3$ and $Y=\ell$ is a line. If we have a point $p \in \ell$ and a smooth curve $C \subset \mathbb{P}^3$ such that $\mathbb{T}_pC=\ell$, then if $$\nu:\Bl_{\ell}\mathbb{P}^3 \rightarrow \mathbb{P}^3$$ what is the intersection $\widetilde{C} \cap E$, where $\widetilde{C}$ is the strict transform of $C$ under $\nu$?

In general, if I have a curve tangent to the locus that I'm blowing up, where does its "direction" go if the exceptional locus parametrize only normal directions?

Thanks in advance.


1 Answer 1


If $C \subset X$ is a smooth curve and $p \in C$ there is a unique plane (so-called osculating plane) $$ T^2_pC \subset T_pX $$ in the tangent space $T_pX$ to $X$ at $p$ such that $C$ any element of $T_p^\vee X = \mathfrak{m}_p/\mathfrak{m}^2_p$ vanishing on $T^2_pC$ vanishes to order 2 on $C$ at $p$. Of course, it contains the tangent line $T_pC$. So, if $T_pC = \ell$ then this plane defines a normal direction to $\ell$ at $p$; the corresponding point of $E$ is the intersection point of $\tilde{C}$ with $E$.

  • $\begingroup$ Thank you very much. The only thing that is not clear to me is why the osculating plane tells exactly the "right" normal direction to choose for C at p. $\endgroup$
    – gigi
    Oct 14, 2020 at 19:04
  • 4
    $\begingroup$ The easiest thing is to look at local coordinates (taking into account that if you have a parametric representation for the curve then the osculating plane is spanned by the "velocity" and "acceleration" vectors). $\endgroup$
    – Sasha
    Oct 14, 2020 at 19:05

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