Timeline for $G$-invariant bilinear maps
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Mar 29, 2018 at 6:23 | comment | added | Michiel Van Couwenberghe | @JeremyRickard I think I just answered my own question. Since the Weyl module itself is $\mathbb{Z}$-free, the same is true for the zeto weight space since it is a submodule of a free module over a PID. So yes, I think that I want to assume that $M$ is $\mathbb{Z}$-free. | |
| Mar 29, 2018 at 5:24 | comment | added | Michiel Van Couwenberghe | @JeremyRickard To be more precise, I am especially interested in the case where $M$ is the zero weight space of a Weyl module $V(\lambda)$ of a linear algebraic group over $\mathbb{Z}$, considered as a module for the Weyl group. Do you know whether this is $\mathbb{Z}$-free? | |
| Mar 28, 2018 at 15:30 | history | edited | YCor |
edited tags
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| Mar 28, 2018 at 15:02 | comment | added | LSpice | @MichielVanCouwenberghe, you can take $N$ to be the quotient of $M$ by its torsion submodule. | |
| Mar 28, 2018 at 14:49 | comment | added | Michiel Van Couwenberghe | @JeremyRickard I am not really sure. Probably, if that is the only way that I can say something useful. How can we check whether $M$ is $\mathbb{Z}$-free? Or can you always find a $\mathbb{Z}$-free $N$ such that $N \otimes \mathbb{C} \cong M \otimes \mathbb{C}$? | |
| Mar 28, 2018 at 14:36 | comment | added | Jeremy Rickard | Do you want to assume $M$ is $\mathbb{Z}$-free? Otherwise it could have a large summand annihilated by $\text{char}(k)$ and $M\otimes k$ would have little to do with $M\otimes\mathbb{C}$. | |
| Mar 28, 2018 at 14:35 | history | edited | Michiel Van Couwenberghe | CC BY-SA 3.0 |
corrected spelling
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| Mar 28, 2018 at 14:34 | comment | added | Michiel Van Couwenberghe | @LSpice Yes, that is correct. By $M \otimes M \otimes k$, I really mean the tensor product of the representation $M \otimes_\mathbb{Z} k$ with itself. It could also have written $M \otimes_{\mathbb{Z}G} M \otimes_{\mathbb{Z}G} kG$. | |
| Mar 28, 2018 at 13:53 | comment | added | LSpice | Is it correct that $M \otimes k$ means $M \otimes_{\mathbb Z} k$, but $M \otimes M \otimes k$ means $M \otimes_{\mathbb ZG} M \otimes_{\mathbb Z} k$? | |
| Mar 28, 2018 at 13:49 | history | asked | Michiel Van Couwenberghe | CC BY-SA 3.0 |