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Dec 20, 2011 at 7:14 answer added Chi-Keung Ng timeline score: 0
Dec 15, 2011 at 5:59 answer added Chi-Keung Ng timeline score: 0
Nov 22, 2010 at 15:19 comment added HenrikRüping let all groups be trivial and let $R$ be the rationals. Then $2\otimes 2\otimes 2...$ is not in the image, which is generated by $1\otimes 1\otimes ...$ and these two elements are linear independent: Consider the multilinear form $f(x_1,....):=\prod x_i$, if almost all $x_i$ are $1$,$0$ otherwise.
Nov 22, 2010 at 15:17 answer added Todd Trimble timeline score: 1
Nov 22, 2010 at 15:05 comment added Thomas Nikolaus How do we see that the product to the infinite tensor product is not surjective (or better say bijective)?
Nov 22, 2010 at 14:37 comment added Martin Brandenburg (See mathoverflow.net/questions/11767/infinite-tensor-products for what Henrik means by the infinite tensor product here. There $g_1 \otimes g_2 \otimes ...$ makes sense, in contrast to the usual infinite tensor product of algebras.)
Nov 22, 2010 at 13:56 history asked HenrikRüping CC BY-SA 2.5