Timeline for Iterated Reduced Tensor Power of Graded Vector spaces
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 22, 2013 at 18:15 | comment | added | Nevermind | ok. I see now . | |
| Jul 22, 2013 at 17:39 | comment | added | David E Speyer | To make darij's comment more explicit: Let $V_1 = \mathbb{R}$ and all other $V_0=0$. Then the graded pieces of $\bar{T}(V)$ have dimensions $(0,1,1,1,1,\dots)$ and the graded pieces of $\bar{T}(\bar{T}(V))$ have dimensions $(0,1,2,4,8,16, \dots)$. This simply isn't true. | |
| Jul 22, 2013 at 16:11 | comment | added | Nevermind | I'm not sure I understand this.You mean if the underlying field of the vector space is finite? In that case ok, I have to rewrite the question, because I'm only interested in $\mathbb{R}$-vector spaces. | |
| Jul 22, 2013 at 15:52 | answer | added | Peter Michor | timeline score: 3 | |
| Jul 22, 2013 at 15:51 | comment | added | darij grinberg | Assume that each homogeneous component of $V$ is free of finite rank over $\mathbb Z$. Assume also that the $0$-th homogeneous component of $V$ is $0$. Then, if $H\left(W\right)$ denotes the Hilbert series of a graded $\mathbb Z$-module $W$ whose homogeneous components are free, then we have $H\left(\overline T\left(V\right)\right) = \frac{H\left(V\right)}{1-H\left(V\right)}$. From this, it should be pretty easy to see that in general, $H\left(\overline T\left(V\right)\right) \neq H\left(\overline T\left(\overline T\left(V\right)\right)\right)$. | |
| Jul 22, 2013 at 15:05 | history | asked | Nevermind | CC BY-SA 3.0 |