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Algebraic varieties, stacks, sheaves, schemes, moduli spaces, complex geometry, quantum cohomology.

47 votes
5 answers
4k views

Is the determinant equal to a determinant?

Let $\det_d = \det((x_{i,j})_{1 \leq i,j\leq d})$ be the determinant of a generic $d \times d$ matrix. Suppose $k \mid d$, $1 < k < d$. Can $\det_d$ be written as the determinant of a $k \times k$ mat …
Zach Teitler's user avatar
  • 6,237
12 votes

What is the largest tensor rank of $n \times n \times n$ tensor?

For tensors in $\mathbb{R}^3 \otimes \mathbb{R}^3 \otimes \mathbb{R}^3$ or in $\mathbb{C}^3 \otimes \mathbb{C}^3 \otimes \mathbb{C}^3$, the maximum rank is $5$. See Bremner, Hu, On Kruskal's theorem t …
Zach Teitler's user avatar
  • 6,237
10 votes
Accepted

2x2-determinantal representations of cubic curves

Let $(\ell_1,\ell_2)$ be the ideal of a point in $V(f)$, so $f \in (\ell_1,\ell_2)$.
Zach Teitler's user avatar
  • 6,237
10 votes
Accepted

When is an algebra defined by generators and relations finite-dimensional and satisfies Poin...

Say $I = (f_1,\dotsc,f_l)$ is the ideal generated by the $f_i$. The $f_i$ are homogeneous; let’s add an assumption that none of the $f_i$ are constant (degree zero). The following conditions are equiv …
Zach Teitler's user avatar
  • 6,237
9 votes
Accepted

Is the Grassmannian contained in a Plücker hyperplane?

The Grassmannian in its Plücker embedding spans the space. The space $\mathbb{P}^{\binom{n}{k}-1}$ of alternating tensors is spanned by the simple wedges (also called decomposable) $v_1 \wedge \dotsb …
Zach Teitler's user avatar
  • 6,237
9 votes
Accepted

A different notion of a decomposable symmetric tensor (besides Veronese)

Your $\vee$ is essentially multiplication of polynomials. The variety of tensors $x_1 \vee \dotsb \vee x_m$ corresponds to polynomials that factor as products of linear factors. Points of the (project …
Zach Teitler's user avatar
  • 6,237
9 votes
Accepted

Is a flattening rank a lower bound for the border rank?

Yes, the flattening rank is a lower bound for border rank. First note that flattening rank is a lower bound for rank. If $T$ is a decomposable tensor (simple tensor, rank one tensor) then every flatt …
Zach Teitler's user avatar
  • 6,237
8 votes
Accepted

Linear spaces secant to Veronese varieties

Here is an answer in terms of power sum decompositions of polynomials. A point $p \in \mathbb{P}^9$ corresponds to a homogeneous polynomial $P$ of degree $3$ in $3$ variable, defining a plane cubic. P …
Zach Teitler's user avatar
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7 votes
Accepted

Is the Waring rank homogeneous polynomials sub-multiplicative?

You asked: Is $W(P \otimes Q) \leq W(P) W(Q)$ for two homogeneous polynomials $P$ and $Q$? First, I think you have to be a little bit careful about the difference between tensor product and mult …
Zach Teitler's user avatar
  • 6,237
7 votes
Accepted

Waring rank of monomials, and how it depends on the ground field

The answer to question 1 is affirmative. There are several lower bounds in various papers. I'll take the idea from https://arxiv.org/abs/1503.08253 (Buczyński and myself, "Some examples of forms of hi …
Zach Teitler's user avatar
  • 6,237
7 votes

Tensor rank of anti-symmetric tensor

The tensor ranks of determinants and permanents are currently not known. In the $3 \times 3$ case it is known: the $3 \times 3$ determinant has tensor rank $5$ and the $3 \times 3$ permanent has tenso …
Zach Teitler's user avatar
  • 6,237
7 votes

Sum of a big divisor and a pseudo-effective divisor

Theorem 2.2.26 of Lazarsfeld's Positivity in Algebraic Geometry, 2004, is the following: Theorem: The big cone is the interior of the pseudoeffective cone and the pseudoeffective cone is the closure …
Zach Teitler's user avatar
  • 6,237
7 votes
Accepted

Classification of curves whose tangent spaces are large

If $Z$ is a pure one-dimensional closed subscheme of degree $d$ in projective space, with $d$-dimensional tangent space at every closed point, then $Z$ is supported on a line. To see this, let $H$ be …
Zach Teitler's user avatar
  • 6,237
6 votes
Accepted

Very general quartic hypersurface in $\mathbb{P}^3$ has Picard group $\mathbb{Z}$

That is the Noether-Lefschetz theorem. Searching online should find plenty of results in web pages and lecture notes. If you want a published source, how about: Mark Green, A new proof of the explicit …
Zach Teitler's user avatar
  • 6,237
6 votes
1 answer
1k views

If some powers of polynomials are linearly independent, does it imply higher powers are also...

Let $P_1,\dotsc,P_k$ be polynomials. Assume they are pairwise non-proportional (i.e., any two of them are linearly independent). Suppose $N$ is a power such that $P_1^N,\dotsc,P_k^N$ are linearly inde …
Zach Teitler's user avatar
  • 6,237

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