Suppose we are given an $n$-dimensional order-3 symmetric tensor $T \in (\mathbb{R}^n)^{\otimes 3}$ that is the sum of $r$ random rank-1 terms. The problem of recovering the rank-1 components is possible in principle when $r \lesssim n^2$ but polynomial-time algorithms are only known in the regime $r \ll n^{3/2}$. Similar "statistical-computational gaps" occur in many high-dimensional inference tasks, and in recent years there has been a flurry of work on explaining the apparent computational hardness in these problems by proving lower bounds against restricted (yet powerful) models of computation such as statistical queries (SQ), sum-of-squares (SoS), and low-degree polynomials (LDP). However, no such prior work exists for tensor decomposition, largely because its hardness does not appear to be explained by a "planted versus null" testing problem. We consider a model for random order-3 tensor decomposition where one component is slightly larger in norm than the rest (to break symmetry), and the components are drawn uniformly from the hypercube. We resolve the computational complexity in the LDP model: $O(\log n)$-degree polynomial functions of the tensor entries can accurately estimate the largest component when $r \ll n^{3/2}$ but fail to do so when $r \gg n^{3/2}$. This provides rigorous evidence suggesting that the best known algorithms for tensor decomposition cannot be improved, at least by known approaches. A natural extension of the result holds for tensors of any fixed order $k \ge 3$, in which case the LDP threshold is $r \sim n^{k/2}$.

翻译：假设我们给定一个 $n$ 维三阶对称张量 $T \in (\mathbb{R}^n)^{\otimes 3}$，它是 $r$ 个随机秩-1项的和。理论上，当 $r \lesssim n^2$ 时，恢复秩-1分量是可能的，但多项式时间算法仅在 $r \ll n^{3/2}$ 的范围内已知。类似的“统计-计算差距”出现在许多高维推理任务中，近年来，大量工作通过针对受限（但强大）的计算模型（如统计查询（SQ）、平方和（SoS）和低阶多项式（LDP））证明下界，来解释这些问题的明显计算困难。然而，此前对于张量分解并无此类研究，主要是因为其困难似乎无法通过“植入与零假设”的检验问题来解释。我们考虑一个随机三阶张量分解模型，其中一个分量在范数上略大于其余分量（以打破对称性），且所有分量均匀地从超立方体中抽取。我们解决了LDP模型中的计算复杂度：当 $r \ll n^{3/2}$ 时，张量条目的 $O(\log n)$ 度多项式函数能准确估计最大分量，但当 $r \gg n^{3/2}$ 时则失效。这提供了严格证据，表明已知最佳的张量分解算法无法改进，至少无法通过现有方法改进。该结果可自然推广到任意固定阶数 $k \ge 3$ 的张量，此时LDP阈值为 $r \sim n^{k/2}$。