Timeline for A different notion of a decomposable symmetric tensor (besides Veronese)
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Sep 29, 2020 at 0:43 | history | became hot network question | |||
| Sep 29, 2020 at 0:19 | vote | accept | Ben | ||
| Sep 29, 2020 at 0:17 | answer | added | Zach Teitler | timeline score: 9 | |
| Sep 28, 2020 at 21:37 | comment | added | Ben | $x_1 \vee x_2=x_1 \otimes x_2 + x_2 \otimes x_1$ is often tensor rank 2, but clearly has rank 1 with respect to the notion I am asking about. | |
| Sep 28, 2020 at 19:22 | comment | added | Abdelmalek Abdesselam | The $x_1\vee\cdots\vee x_m$ is the tensor rank, whereas the $x^{\otimes m}$ is the Waring rank. | |
| Sep 28, 2020 at 19:20 | comment | added | Ben | Good point, so the $x_1 \vee \dots \vee x_m$ rank is at most the tensor rank (and also the Waring rank). I'm pretty sure that all three of these ranks can be different, though. | |
| Sep 28, 2020 at 19:17 | comment | added | Abdelmalek Abdesselam | symmetrize post decomposition | |
| Sep 28, 2020 at 19:15 | comment | added | Ben | @AbdelmalekAbdesselam I thought Comon's conjecture was that the tensor rank and Waring rank of a symmetric tensor agree? For tensor rank, the decomposables are $x_1 \otimes \dots \otimes x_m$. For Waring rank, the decomposables are $x^{\otimes m}$. Neither are $x_1 \vee \dots \vee x_m$. | |
| Sep 28, 2020 at 19:12 | comment | added | Abdelmalek Abdesselam | The notions of rank and (border rank) for the two definitions are related and were conjectured to be the same (Comon Conjecture). However, this has been disproved arxiv.org/abs/1705.08740 | |
| Sep 28, 2020 at 17:32 | comment | added | Ben | It's worth noting that $\mathbb{C}^n \otimes \mathbb{C}^m$ is isomorphic to the vector space of $n \times m$ matrices, with the elements of the form $v \otimes w$ mapping to the set of rank-one matrices. | |
| Sep 28, 2020 at 17:30 | comment | added | Ben | Let's stick to two spaces: $V \otimes W$ is defined as the free vector space spanned by $(v,w)$ for all $v \in V$ and $w \in W$, quotiented by relations like $(v,w+u)=(v,w)+(v,u)$ that make $(\cdot, \cdot)$ into a bilinear map. This quotient space is $V \otimes W$ and the equivalence class of elements of the form $(v,w)$ is referred to as $v \otimes w$. See en.wikipedia.org/wiki/Tensor_product | |
| Sep 28, 2020 at 17:23 | comment | added | DCM | Re. $m$ and $n$: no that's a typo. Sorry! Re. what I'm getting at: how do you define the '$m$-factors' of an arbitrary element of the tensor product? (they aren't all of the form $x_1\otimes \dots \otimes x_m$). Note: this is possibly me just being ignorant, but it's possibly worth clarifying. | |
| Sep 28, 2020 at 17:17 | comment | added | Ben | Not sure what you mean... Have you replaced $m$ with $n$ for a reason? en.wikipedia.org/wiki/Tensor_product#Tensor_powers_and_braiding | |
| Sep 28, 2020 at 17:14 | comment | added | DCM | This is probably me being dumb, but... how does one define the $m$-factors of $\sum_{i\in I} x_{i_1}\otimes\dots \otimes x_{i_n}$ for an arbitrary finite $I$? | |
| Sep 28, 2020 at 17:10 | comment | added | Ben | @DCM Yes, thanks! I'm not sure what the ambiguity is with $(\mathbb{C}^n)^{\otimes m}$. I mean the $m$-fold tensor product of $\mathbb{C}^n$, as a vector space. | |
| Sep 28, 2020 at 17:10 | comment | added | DCM | Ah - you fixed the $S_m$ :) | |
| Sep 28, 2020 at 17:08 | history | edited | Ben | CC BY-SA 4.0 |
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| Sep 28, 2020 at 17:07 | comment | added | DCM | Could you clarify what you mean by $(\mathbb{C}^n)^{\otimes m}$ here? (I'm not sure I know what the "m-factors" of a generic element of $(\mathbb{C}^n)^{\otimes m}$ are, but perhaps I'm just naive). Also, I think the $S_n$ in your definition of $x_1\vee \dots \vee x_m$ should be an $S_m$. | |
| Sep 28, 2020 at 16:37 | history | asked | Ben | CC BY-SA 4.0 |