Suppose $A\in\mathbb{R}^{n^k}$ is a $k$-dimensional tensor with $n$ elements along each dimension. Morover suppose $u_1,u_2,\dots,u_k\sim\text{Unif}(\pm1)^n$ are $n$ dimensional vectors with each of their components drawn iid from Rademacher distribution. More generally, you can think of their components as zero-mean, unit-variance iid sub-Gaussian random variables. Moroever, let $U:=u_1\otimes u_2 \otimes \dots \otimes u_k$ be a rank-1 random tensor constructed from these vectors. Finally, let $X=\langle A, U \rangle$, in which by $\langle, \rangle$ we simply mean the inner product of tensor if we flatten them as vectors.

Question: What is the concentration of $X^2$ around its mean? Does $X$ have exponentially light tails? It is easy to show that $\mathbb{E}[X^2]$ is simply the sum of squares of all elements of $A$, using linearity of expectation and the fact that $u_i$s have zero-mean unit variance elements (the proof given at the end of the post). Therefore, if $X^2$ has a distribution with exponentially light tails, the product would in effect sketch the sum of squares of elements of tensor $A$.

**Derivation of $\mathbb{E} X^2$:** $$\mathbb{E} X^2 = \mathbb{E}\sum_{i_1,j_1, \dots,i_k,j_k} a_{i_1, \dots,i_k} a_{j_1,\dots,j_k} \Pi_{m=1}^k (u_{m,i_m} u_{m,j_m})$$
Because of linearity of expectation and independence of $u_i$ vectors from each other we have:

$$\mathbb{E} X^2 = \sum_{i_1,j_1, \dots,i_k,j_k} a_{i_1, \dots,i_k} a_{j_1,\dots,j_k} \Pi_{m=1}^k \mathbb{E}(u_{m,i_m} u_{m,j_m})$$ Now, for all the terms that have $i_m\neq j_m$ based on indep $$\mathbb{E}(u_{m,i_m} u_{m,j_m}) = \mathbb{E}u_{m,i_m} \mathbb{E}u_{m,j_m} = 0\cdot 0 = 0$$ Therefore, only terms that satisfy $i_m=j_m$ for all $m\in[k]$ remain, and for those terms we have: $$\Pi_{m=1}^k\mathbb{E}(u_{m,i_m} u_{m,j_m}) = \Pi_{m=1}^k\mathbb{E}(u_{m,i_m})^2 = \Pi_{m=1}^k\text{Var}(u_{m,i_m})=\Pi_{m=1}^k 1 = 1$$ And therefore: $$\mathbb{E} X^2 = \sum_{i_1,j_1, \dots,i_k,j_k} a_{i_1, \dots,i_k}^2$$ which is the result we wanted.