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Let $\mathbb{P}^N$ be the projective space parametrizing $n\times n$ non-zero matrices modulo scalar multiplication, and let $\mathbb{P}^M\subset\mathbb{P}^N$ be the subspaces of symmetric matrices.

In $\mathbb{P}^N$ we have a stratification $X_1\subset X_2\subset \dots \subset X_r \subset \dots \subset \mathbb{P}^N$ where a general point of $X_r$ corresponds to a rank $r$ matrix. Such stratification restricts to the corresponding stratification $Y_r = X_r\cap \mathbb{P}^M$ of the space of symmetric matrices.

Assume there exists a polynomial of degree $d$ on $\mathbb{P}^M$ having multiplicity $m_r$ along $Y_r$ for all $r$. Can we affirm then that there exists a polynomial of degree $d$ on $\mathbb{P}^N$ having multiplicity $m_r$ along $X_r$ for all $r$ ?

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I don't have a neat answer, but what I want to write is too long to be comment. First of all, if $I_X$ is the homogeneous (saturated) ideal of an integral subscheme of $\mathbb{P}^N$ (say $X$), then the $m$-th symbolic power of $I_X$ twisted by $\mathcal{O}(d)$ (which I will denote by $I^{[m]}_X(d)$) is precisely the ideal of polynomials of degree $d$ which vanish with multiplicity at least $m$ on $X$.

\strikeout {I looked a bit around on Google, and it seems known that if $X$ is arithmetically Cohen-Macaulay (meaning that its coordinate ring is Cohen-Macaulay), then the symbolic powers of $I_X$ agree with the regular powers of $I_X$.} : not sure if that is true, probably not correct...

So basically your question can be formulated as follows:

Let $P$ a non-zero vector which lies in the intersection of the $I_{Y_r}^{[m_r]}(d)$, can we find a non-vector in the intersection of the $I_{X_r}^{[m_r]}(d)$?

I found this paper by Sullivant: https://arxiv.org/pdf/math/0608542.pdf where he gives a quite detailed description of the graded generators of $I_{Y_r}^{[m_r]}$ and $I_{X_r}^{[m_r]}$ (corollary $5.7$ and Theorem $5.8$). As a consequence, it seems that we get:

$$I_{X_r}^{[m_r]}(d) \otimes \left(\mathbb{C}[X_0,\ldots,X_N]/I_{\mathbb{P}^M}\right) = I_{Y_r}^{[m_r]}(d).$$

In particular, one has a surjection:

$$I_{X_r}^{[m_r]}(d) \rightarrow I_{Y_r}^{[m_r]}(d) \rightarrow 0,$$

which looks pretty close to what you want.

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