Timeline for Proving the graded structure of the tensor algebra from only the universal property
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Mar 16, 2023 at 11:41 | comment | added | LSpice | @Zerox, re, I think that your proposal may be related to @TheoJohnsonFreyd's idea. I think that is a clear way of seeing what's going on, but proving your version seems to involve something like constructing $\mathrm T^nV$ "the usual way" as a source of those multilinear maps anyway. | |
| Mar 16, 2023 at 5:50 | comment | added | Zerox | Doesn't the graded structure come from that, to get the subalgebra in $A$ generated by $f(V)$, we multiply elements in $f(V)$ with each other, resulting in images of some $R$-multilinear maps such as $f(u)f(v)$? | |
| Mar 16, 2023 at 5:41 | history | edited | LSpice | CC BY-SA 4.0 |
Correcting a typo, and speculating on cheating
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| Mar 16, 2023 at 4:27 | history | edited | LSpice | CC BY-SA 4.0 |
Incorporating comments, and confessing ignorance
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| Mar 16, 2023 at 3:08 | comment | added | sudgy | I found a solution of my own: mathoverflow.net/a/442828/394901 | |
| Mar 15, 2023 at 21:59 | comment | added | LSpice | @sudgy, OK, if instead of maps $V \to R$, or even maps $V \to R/I$ as I suggested, we use maps $V \to A$ for all $R$-algebras $A$ to define the grading, then we have a correct definition of the grading. But, right now, I can only see how to prove that by knowing something about candidate $R$-algebras … for example, it would be easy if we could use the existing tensor algebra. I am still hoping to manufacture a non-circular candidate. | |
| Mar 14, 2023 at 18:16 | comment | added | LSpice | @sudgy, hmm$\DeclareMathOperator\T{T}\DeclareMathOperator\Hom{Hom}$. I agree that that seems to be a counterexample, and that all I showed is that, with this (wrong) definition, $\bigoplus \T^nV \to \T V$ is a surjection. In this case, I think that things could be fixed by replacing $\Hom(V^*, R[x])$ by the direct sum of all $\Hom(V^*, (R/I)[x])$ over all prime ideals $I$, but it's not clear to me whether that works in general. | |
| Mar 14, 2023 at 17:58 | comment | added | sudgy | I know this is a bit old now, but I think I've found a counterexample where this doesn't work. Consider the module $\mathbb Z/2 \mathbb Z$ over $\mathbb Z$. Then every linear functional is zero, which I think implies that in this definition, everything has every grade. | |
| Oct 4, 2021 at 3:54 | vote | accept | sudgy | ||
| Mar 16, 2023 at 3:07 | |||||
| Oct 2, 2021 at 19:12 | history | edited | LSpice | CC BY-SA 4.0 |
Mild clarification; finite-dimensional -> finitely generated
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| Oct 1, 2021 at 20:41 | history | answered | LSpice | CC BY-SA 4.0 |