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Sep 6, 2013 at 20:24 vote accept George Bergman
Sep 6, 2013 at 19:20 comment added Martin Brandenburg This is a nice proof. Although it seems to be algebraic, I think it is geometric in disguise. Perhaps someone can formulate this proof coordinate-free, and/or in terms of the the intersection of $\mathbb{P}(V) \subseteq \mathbb{P}(A \otimes B)$ with $\mathbb{P}(A) \times \mathbb{P}(B) \hookrightarrow \mathbb{P}(A \otimes B)$?
Sep 6, 2013 at 19:16 history edited Martin Brandenburg CC BY-SA 3.0
1) fixed a typo, 2) replaced twice $M \in V \mapsto$ by $V \ni M \mapsto$, because M (not V) gets mapped.
Sep 6, 2013 at 16:57 history edited Clément de Seguins Pazzis CC BY-SA 3.0
added 41 characters in body
Sep 6, 2013 at 16:50 history answered Clément de Seguins Pazzis CC BY-SA 3.0