Timeline for Structure of the variety of $n$-tuples of $m \times m$ matrices with zero product
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Jun 29, 2019 at 21:26 | comment | added | Will Sawin | @Libli I don't think Chow motives are defined for singular spaces, and if they were I don't think the motive of a stratified space would be the sum of the motives of the strata. No, I don't think one can get the dimensions just from this stratification with no further information. | |
| Jun 29, 2019 at 21:22 | comment | added | S. carmeli | @WillSawin right, thanks! | |
| Jun 29, 2019 at 21:21 | comment | added | Libli | @WillSawin : The maps I use in my argument are all locally trivial in the Zariski topology, so I think we can conclude that the Chow motive of $X$ is given by $\sum_{k=0}^{n} \mathbb{L}^{\otimes (m-k)k} \otimes [\mathrm{GL}_k] \otimes [\mathrm{Gr}(k,m)]^{\otimes 2}$. The cellular decomposition of the Grassmannian is well-known and I guess the motive of $\mathrm{GL}_k$ should be well understood as well. Do you think that we can derived the dimension of the etale cohomology groups with compact support of $X$ from this motivic decomposition? | |
| Jun 29, 2019 at 18:54 | comment | added | Will Sawin | @S.carmeli I don't think is is true. In the case of $n=2$, described by Libli, $X_k$ is just $GL_m$, which does not admit a cell decomposition. | |
| Jun 29, 2019 at 18:44 | comment | added | S. carmeli | @WillSawin it seems that in this case your stratification can be refined to an algebraic cell-decomposition of $X$, hence in particular it supports no differentials for purity reasons, isn't it? This is because every locally-closed strata is the total space of a vector bundle over a product of Grassmannians, which itself admit a cell decomposition. or maybe i'm completely wrong here? | |
| S Jun 29, 2019 at 11:26 | history | suggested | anon | CC BY-SA 4.0 |
Changed link to the "official" version of Milne's notes.
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| Jun 29, 2019 at 11:23 | review | Suggested edits | |||
| S Jun 29, 2019 at 11:26 | |||||
| Jun 29, 2019 at 11:21 | history | edited | Libli | CC BY-SA 4.0 |
added 190 characters in body
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| Jun 28, 2019 at 23:04 | comment | added | Will Sawin | This method works to count points for higher $n$ as we can stratify by the tuple $\operatorname{rank} (A_1 \dots A_i)$. We can also use it to write down an excision spectral sequence computing the compactly supported cohomology groups , but it is not obvious what the differentials are, so this can only naively give upper bounds on the Betti numbers. | |
| Jun 28, 2019 at 23:02 | comment | added | Will Sawin | @Libli Your edit is not correct because, without nonsingularity, the factorization is not unique. | |
| Jun 28, 2019 at 22:44 | history | edited | Libli | CC BY-SA 4.0 |
typos
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| Jun 28, 2019 at 22:10 | comment | added | Libli | @sawdada : non-singularity is useless to get the dimensions of the etale cohomology groups starting from the Zeta function of $X$. I have edited my answer | |
| Jun 28, 2019 at 22:08 | history | edited | Libli | CC BY-SA 4.0 |
add comments on the singularity hypothesis
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| Jun 28, 2019 at 2:54 | comment | added | Zhiyu | As $X$ may not be smooth (so weights may be different), I don't see why this can compute the dimension of the cohomology group. And can we simplify the formula? | |
| Jun 27, 2019 at 19:41 | history | edited | Libli | CC BY-SA 4.0 |
typos
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| Jun 27, 2019 at 19:32 | history | answered | Libli | CC BY-SA 4.0 |