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Most tensor subspaces of low dimension have rank-1 defining equations

Let $V_1,ldots , V_k$ be vector spaces of dimensions $n_1,ldots , n_k$ over a field of characteristic zero.

Consider the rational map
$
newcommand{PP}{mathbb{P}}
newcommand{bs}{boldsymbol}
DeclareMathOperator{Gr}{Gr}
DeclareMathOperator{im}{im}
$

$$
Phi : left[PP (V_1^*) times cdots times PP (V_k^*)right]^j dashrightarrow Gr (n_1 cdots n_k – j, , V_1otimes cdots otimes V_k )$$

defined by
$$
left([alpha_1^i], ldots , [alpha_k^i]right)_{1le ile j} mapsto { T mid (alpha_1^i otimes cdots otimes alpha_k^i) (T) =0, , , 1le i le j }.
$$

Counting dimensions, we expect that the map is dominant when
$$
j , (n_1 + n_2 cdots +n_k – k) ge j , left( n_1 cdots n_k – j right)
$$

$$
Rightarrow , , , j ge n_1 cdots n_k – (n_1 + n_2 cdots +n_k – k) .
$$

To see that our expectation is correct in the case where equality holds, it is enough to show generically finite fibers. Let
$$
L = Phi ( bs{alpha}) =
biglangle
alpha_1^1 otimes cdots alpha_k^1, ,
ldots , ,
alpha_1^j otimes cdots alpha_k^j
bigrangle^perp ,
$$

so that $PP (L^perp)$ is a $(j-1)$-secant of the Segre variety
$$
Sigma_{n_1-1, ldots, n_k-1}subset PP (V_1 otimes cdots otimes V_k ),$$

which has dimension $(n_1 + n_2 cdots +n_k – k).$ Since the dimensions of $PP (L^perp)$ and $Sigma_{n_1-1, ldots, n_k-1}$ are complementary, the classical "trisecant lemma" implies for generic $bs{alpha}$ that
$$PP (L^perp ) cap Sigma _{n_1-1, ldots, n_k-1} = { [alpha_1^1 otimes cdots alpha_k^1], ,
ldots , ,
[alpha_1^j otimes cdots alpha_k^j] }. $$

(See eg. Prop 1.3.3 in Russo’s "Geometry of Special Varieties." The statement assumes characteristic $0,$ though I’m not sure if it’s necessary for me.)

Thus, if $Phi (bs{alpha} ) = Phi (bs{alpha}’)$ for $bs{alpha}$ generic, then $bs{alpha }$ equals $bs{alpha}’$ after permuting some of the $j$ factors in the domain.

I discovered the above argument for the case where $k=2$ and $(n_1, n_2) = (3,3),$ in a context apparently unrelated to tensors, and was surprised that it generalized with little effort. I am thus wondering if it is a well-known fact or if there is some simpler explanation.

A seemingly more difficult question I am also interested in: describe the constructible set $im Phi .$

MathOverflow Asked on January 5, 2022

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