Is the asymptotic rank of a tensor bounded by (naive) border rank?

Question

Write $R(T)$ for the rank of an order-$3$ tensor $T \in \mathbb C^{m \times n \times p}$ over the complex numbers. If $T' \in \mathbb {C}^{m' \times n' \times p'}$ is another such tensor then let $T \otimes T' \in \mathbb{C}^{mm' \times nn' \times pp'}$ denote their tensor product (aka Kronecker product). It is straightforward to verify that $R(T \otimes T') \le R(T) R(T')$, and from this it follows that the sequence $R(T^k)^{1/k}$ converges to its infimum. The asymptotic rank of $T$ is $$\tilde R(T) = \lim_{k \to \infty} R(T^k)^{1/k} = \inf_{k \ge 1} R(T^k)^{1/k}.$$

For example, if $T = m_2$ is the structure tensor for $2 \times 2$ matrix multiplication then $\tilde R(T) = 2^\omega$ where $\omega$ is the exponent of matrix multiplication. It would therefore be putting it mildly to say that $\tilde R(T)$ is not easy to calculate.

My question is actually about the border rank of tensors. The literature agrees universally that this is denoted $\underline R(T)$, but seems to vary wildly on what it's actual definition is. Here are some possibilities:

1. According to some sources, including Wikipedia, border rank is the minimal $r$ such that $T$ is a limit of tensors of rank at most $r$. Let's call this $\underline R^\text{naive}$. Wikipedia has a basic example of a tensor with $R(T) = 3$ but $\underline R^\text{naive}(T) = 2$.

2. That cannot possibly be the "right" definition for an arbitrary field since there is no natural topology in general. I have seen at least one source therefore define border rank to be the least $r$ such that $T$ is in the Zariski closure of the set of tensors of rank at most $r$. Call this $\underline{R}^\text{Zariski}$.

3. Most of the literature that actually carefully defines and proves things (such as the book Algebraic Complexity Theory, Chapter 15) defines border rank in terms of algebraic expressions over $\mathbb C[\epsilon]$ of the type $\epsilon^d T + O(\epsilon^{d+1}) = (\text{rank-$r$ tensor})$ for some integer $d$. Let's call this definition $\underline R^\text{original}(T)$.

Note that $\underline R^\text{Zariski} \le \underline R^\text{naive} \le \underline R^\text{original}$.

Question 1: Are these things really different? What are good examples?

Anyway, the crucial inequality relating border rank (at least in its original definition) to asymptotic rank is $\tilde R(T) \le \underline R^\text{original}(T)$. However, it seems to be implicit in some of the more recent literature that this holds for $R^\text{naive}$ (see for example the intro to https://arxiv.org/abs/1112.6007). Is this true?

Question 2: Is it true that $\tilde R \le \underline R^\text{naive}$?

Of course, we could ask the same question for $\underline R^\text{Zariski}$, which would be even better.

Answer 1

It is true that over $\mathbb{C}$ (and over every algebraically closed field) we have $\underline{R}^{\mathrm{Zariski}}(T) = \underline{R}^{\mathrm{original}}(T)$.

The standard reference is the doctoral thesis of Alder "Grenzrang und Grenzkomplexität aus algebraischer und topologischer Sicht", but it is not accessible. A proof is also available in §20.6 of the book "Algebraic complexity theory" by Bürgisser, Clausen, Shokrollahi, and a stronger statement can be found in the following paper: T. Lehmkuhl, T. Lickteig. "On the order of approximation in approximative triadic decompositions of tensors". Theor. Comp. Sci. 66(1):1–14.

The idea is that the set $X_r$ of tensors of rank at most $r$ is constructible and therefore contains a subset $Y_r$ open in the Zariski closure $\overline{X}_r$. If $x \in \overline{X}_r \setminus X_r$, then, taking an intersection of $\overline{X}_r$ with a general affine subspace of appropriate dimension containing $x$, we can get an algebraic curve $C \subset \overline{X}_r$ such that $x \in C$ and $C \cap Y_r$ is open in $C$. Then we take a normalization of this curve to get rid of possible singularity at $x$, which gives a local parametrization of $C$ at $x$ equivalent to the required expression with $\varepsilon$.