I hope this is appropriate for the site. I am reading the paper "The analytic rank of a tensor" [S. Lovett, Discrete Analysis (2019), #7, 10 pp.] and am a bit confused in one of the applications sections. In the paper, a tensor is defined to be a multilinear map from $V^d\to {\bf F}$, where ${\bf F}$ is a field and $V$ is an $n$-dimensional vector space over ${\bf F}$.
However, later on, it is mentioned that to solve the cap-set problem, we define a tensor $T$ which captures the problem structure, then bound the largest independent set in $T$. This is a subset $A\subseteq [n]$ such that for all $i_1,\ldots,i_d\in A$, the coefficient $T_{i_1,\ldots,i_d}$
of the tensor is nonzero if and only if $i_1 = \cdots = i_d$.
A cap set is a subset $A\subseteq {\bf F}_3^n$ such that $x+y+z\ne 0$ whenever $x,y,z\in A$ are pairwise distinct.
From what I've seen, given $A\subseteq {\bf F}_3^n$,
the "tensor" for the cap-set problem is
$$T(x,y,z) = \cases{1, & if $x=y=z$ and $x\in A$;\cr 0, & otherwise.}$$
If $A$ is a cap set, then this definition makes $A$ an independent set in $T$. However, I don't think that $T$ actually is a tensor, since if $x=y=z\in A$, then
$$f(x+x, y, z) = 0 \ne 2 = 1 + 1 = f(x,y,z) + f(x,y,z).$$
My conclusion is that I'm considering the wrong choice of $T$, but I can't seem to figure out the tensor alluded to by the author.
Best Answer
The cap-set tensor is a tensor $V^3 \to \mathbb F_3$ where $V$ is $\mathbb F_3^{ \mathbb F_3^n}$, not simply $\mathbb F_3^n$. It is defined by
$$T (e_a,e_b,e_c) = \begin{cases} 1 & \textrm{if }a+b+c = 0 \\ 0 & \textrm{if } a+b+c \neq 0 \end{cases}$$
and extended to arbitrary vectors by linearity.
The independent sets in this case are subsets of the index set $\mathbb F_3^n$, not subsets of the vector space $\mathbb F_3^{ \mathbb F_3^n}$.