# Strict transform of a tangent curve under blow-up

$$\DeclareMathOperator{\Bl}{\operatorname{Bl}}$$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

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 $$E$$.

• 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. – gigi Oct 14 '20 at 19:04
• 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). – Sasha Oct 14 '20 at 19:05