Terracini loci and a codimension one Alexander-Hirschowitz theorem

Edoardo Ballico, Maria Chiara Brambilla, Claudio Fontanari
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Abstract

The Terracini locus $\mathbb{T}(n, d; x)$ is the locus of all finite subsets $S \subset \mathbb{P}^n$ of cardinality $x$ such that $\langle S \rangle = \mathbb{P}^n$, $h^0(\mathcal{I}_{2S}(d)) > 0$, and $h^1(\mathcal{I}_{2S}(d)) > 0$. The celebrated Alexander-Hirschowitz Theorem classifies the triples $(n,d,x)$ for which $\dim\mathbb{T}(n, d; x)=xn$. Here we fully characterize the next step in the case $n=2$, namely, we prove that $\mathbb{T}(2,d;x)$ has at least one irreducible component of dimension $2x-1$ if and only if either $(d,x)=(6,10)$ or $(d,x)=(4,4)$ or $d\equiv 1,2 \pmod{3}$ and $x=(d+2)(d+1)/6$.
特雷西尼位置和一维亚历山大-赫肖维兹定理
Terracini 所在地 $\mathbb{T}(n, d; x)$ 是心数为 $x$ 的所有有限子集的所在地$S \subset \mathbb{P}^n$ ,使得 $\langle S \rangle =\mathbb{P}^n$, $h^0(\mathcal{I}_{2S}(d))>0$,并且 $h^1(\mathcal{I}_{2S}(d))>0$.著名的亚历山大-赫肖维兹定理对 $\dim\mathbb{T}(n, d; x)=xn$ 的三元组$(n,d,x)$ 进行了分类。在这里,我们完全描述了$n=2$情况下的下一步,即我们证明了$\mathbb{T}(2,d;x)$至少有一个维数为$2x-1$的不可还原分量,当且仅当$(d,x)=(6,10)$或$(d,x)=(4,4)$或$d\equiv 1,2 \pmod{3}$且$x=(d+2)(d+1)/6$。
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