Timeline for Quadratic algebras and Koszul algebras
Current License: CC BY-SA 4.0
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Dec 10, 2018 at 15:44 | history | edited | Mare | CC BY-SA 4.0 |
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| Dec 10, 2018 at 11:12 | answer | added | Vladimir Dotsenko | timeline score: 3 | |
| Dec 10, 2018 at 10:52 | history | edited | Mare | CC BY-SA 4.0 |
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| Dec 10, 2018 at 10:37 | history | edited | YCor |
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| Dec 10, 2018 at 10:36 | comment | added | YCor | @JohnPalmieri In your context (quotient of the tensor algebra on $V$) it seems that both conditions are equivalent, and this holds more generally in a graded quotient $B$ of $A$ (graded in nonnegative integers) if the image of $A_0\otimes A_2$ and $A_2\otimes A_0$ in $B_2$ are both contained in the image of $A_1\otimes A_1$; this is satisfied in the context of the question (linked Def 1.2.2), if I understand properly. | |
| Dec 10, 2018 at 10:11 | history | edited | Mare | CC BY-SA 4.0 |
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| Dec 10, 2018 at 10:07 | comment | added | Mare | @JulianKuelshammer No, I dont assume $A$ to be finite dimensional. For example it could apply to give that the global dimenion of the polynomial algebra in n variables in n or that the global dimension of an arbitrary non-trivial quiver algebra (possibly infinite dimensional) is equal to one. | |
| Dec 10, 2018 at 10:05 | comment | added | Julian Kuelshammer | @Mare Do you assume $A$ to be finite dimensional and of finite global dimension? (Otherwise $B$ is infinite dimensional and of infinite Loewy length and then intuitively I would think it is possible to construct counterexamples). | |
| Dec 9, 2018 at 19:48 | comment | added | Mare | @JohnPalmieri I use the definition of quadratic as in definition 1.2.2. of ams.org/journals/jams/1996-9-02/S0894-0347-96-00192-0 . | |
| Dec 9, 2018 at 18:18 | comment | added | John Palmieri | @Mare: you may be right. Does quadratic mean that the relations are in degree 2 or that the relations are in $V \otimes V$, where the algebra is a quotient of the tensor algebra on $V$? | |
| Dec 9, 2018 at 17:02 | comment | added | Mare | @JohnPalmieri Im not very experienced with this, but when one generator has degree larger than 1, then the relations seem to be non-quadratic or? | |
| Dec 9, 2018 at 16:21 | comment | added | John Palmieri | Suppose $A$ is an exterior algebra on two generators, so $B$ is polynomial on two generators. The equality in question 2 will hold. If both generators are in degree 1, $A$ is Koszul, but if they are in different degrees, it is not. | |
| Dec 9, 2018 at 15:31 | comment | added | Benjamin Steinberg | If you are assuming $A$ is finite dimensional and the grading is by path length of the quiver then the first equality is always true in question 1. I am not sure what conditions you need for the radical of the Ext algebra to be generated in degree one. | |
| Dec 9, 2018 at 14:42 | history | edited | Mare | CC BY-SA 4.0 |
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| Dec 9, 2018 at 14:35 | history | asked | Mare | CC BY-SA 4.0 |