# Holomorphic vector fields on $\mathbb{P}^n$ that extend to the blow up

Hi, these are three questions regarding extendability of holomorphic vector fields on complex projective space to its blow up along a subvariety. Let $\mathbb{P}^n$ be the complex projective space, with homogeneous coordinates of a point $p\in\mathbb{P}^n$ $$p=[z_0:\ldots:z_n]$$

1) suppose we blow up the point $p_0=[1:0:\ldots:0]$, which are the global holomorphic vector fields of $\mathbb{P}^n$ that extend to the blow up?

I made some computations in the affine chart $\lbrace z_0\neq0 \rbrace$ and modulo (fatal) errors i obtain that my vector fields are of the form $$l_0\frac{\partial}{\partial z_0}+\sum_{i=1}^n c_iz_i\frac{\partial}{\partial z_i}$$ with $l_0\in\mathbb{C}[z_0,\ldots,z_n]$ homogeneous of degree 1 and $c_i\in\mathbb{C}$. So vector fields that extend to the blow up span a vector subspace of dimension $2n$ of $H^0(T\mathbb{P}^n)$. Am i right?

2)In a similar fashion i find that if i blow up a k-codimensional variety of type $$\lbrace z_k=\ldots=z_n=0 \rbrace$$ with $k>1$ my vector fields span a vector subspace of dimension $n(k+2)$ of $H^0(T\mathbb{P}^n)$. Again, am i right?

3) If i blow up a smooth curve of sufficiently high degree/genus do i still have holomorphic vector fields that extend to the blow up?

Thank you in advance.

## 1 Answer

1) No. There are many more vector fields. The vector fields you are looking for are precisely those which vanish at $p_0$. Since $h^0( \mathbb P^n, T \mathbb P^n) = (n+1)^2 -1$ and you are imposing $n$ linearly independent conditions you should get $n^2 + n - 1$ vector fields. In homegeneous coordinates they can be written as $$l_0 \frac{\partial}{\partial z_0} + \sum_{i=1}^n \sum_{j=1}^n c_{ij} z_i \frac{\partial}{\partial z_j} .$$ Notice that the homogeneous vector field $\sum_{i=0}^n z_i \frac{\partial}{\partial z_i}$ does not contribute to the counting since it corresponds to the zero vector field on $\mathbb P^n$.

2) You should get $k^2 + (n+1-k) n -1$ vector fields. As pointed out in the comments the result remains unchanged if we swap $k$ and $n-k+1$. The point is that $PGL(n+1)$ acts on $\mathbb P^n$ as well as on the dual projective space $\check{\mathbb P}^n$. If we look at the subgroup preserving a linear subspace of codimension $k$ on $\mathbb P^n$, then the natural action on $\check{\mathbb P}^n$ will preserve its dual: a linear subspace of codimension $n-k+1$.

3) You will get a holomorphic vector field on the blow-up if and only if the curve is left invariant by the original vector field.

If your vector field vanishes on a linear subspace of dimension at least two then every curve contained in it will be invariant by the vector field. Thus on the blow-up along these curves you will still get holomorphic vector fields.

If instead you restrict your attention to vector fields with zero set of dimension at most one then while you can find curves of arbitrarily high degree invariant by this class of vector fields (think on orbits of $\mathbb C^*$-actions on $\mathbb P^n$), they all have (geometric) genus zero.

• I believe your statement about blowing up of curves is incorrect. If the curve is contained in a linear subspace which is in the zero locus of the vector field, then the vector field can extend to the blowing up. Oct 31, 2011 at 17:12
• @Jason: This seems right. Any curve contained in a dimension-$k$ linear subspace should have at least $(n+1-k)n-1$ of the vector fields from the whole subspace. In the formula of $2$, the answer does not change when $k$ and $n-k$ are swapped. Is there some sort of duality at work here? Oct 31, 2011 at 18:23
• @Jason: Thanks for catching that. You are of course right. Oct 31, 2011 at 18:24
• @Will: I have edited my answer to reflect your comment. Thanks. Oct 31, 2011 at 18:54