If we have a symmetric tensor, $\phi \in S^d V$, we can think of the tensor as a polynomial. Namely, with a polarization identity given by: $$ \bar{\phi}(x_1, \cdots, x_d) = \frac{1}{k!} \sum_{I \subset [k], I \not= \emptyset} (-1)^{k - |I|} \phi\left( \sum_{i \in I} x_i \right) $$

Define symmetric tensor rank to be the smallest $r$ such that $\phi = v_1^d + \cdots + v_r^d$ for some $v_i \in V$ and define symmetric border rank to be the smallest $r$ such that $\phi$ is the limit of symmetric tensors of symmetric tensor rank $r$.

Then, consider a partially polarized form of $\phi$. Namely, $\phi \in S^s V \otimes S^{d-s} V$. If we think of $S^s V \otimes S^{d-s} V$ as the space of linear maps from $S^s V^\ast \rightarrow S^{d-s} V$, then the mapping is: $$ \phi_{s, d-s} : (\alpha^1, \ldots, \alpha^s) \mapsto \bar{\phi}(\alpha^1, \ldots, \alpha^s, \cdot, \ldots, \cdot)$$

Prove that if the symmetric border rank of $\phi$ is less than or equal to $k$, then the tensor rank of $\phi_{s, d-s}$ is less than or equal to $k$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.