Timeline for What's so special about the forgetful functor from G-rep to Vect?
Current License: CC BY-SA 4.0
14 events
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| S Dec 16, 2019 at 18:02 | history | edited | Amir Sagiv | CC BY-SA 4.0 |
Reformatted symbols with LaTeX for ease of reading.
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| S Dec 16, 2019 at 18:02 | history | suggested | Dat Minh Ha | CC BY-SA 4.0 |
Reformatted symbols with LaTeX for ease of reading.
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| Dec 16, 2019 at 17:06 | review | Suggested edits | |||
| S Dec 16, 2019 at 18:02 | |||||
| Sep 9, 2010 at 15:35 | vote | accept | Theo Johnson-Freyd | ||
| Sep 8, 2010 at 21:48 | comment | added | Theo Johnson-Freyd | Isomorphic to my last comment: your point above was that $k^G$ and $k'$ aren't isomorphic as algebras. That they are both commutative doesn't make any difference. | |
| Sep 8, 2010 at 21:48 | comment | added | Theo Johnson-Freyd | No, wait, what I said is still false. They're isomorphic as functors, but not as monoidal functors. The point is that a monoidal functor is a functor along with some extra structure, required to satisfy a condition. It is then symmetric if it satisfies another condition --- symmetric monoidal functors are not extra data on top of monoidal functors (symmetric monoidal structures on a category are extra data). Similarly, a monoidal natural transformation is a natural transformation satisfying some condition. But once it satisfies this, being symmetric is not an extra condition. | |
| Sep 8, 2010 at 21:31 | comment | added | Theo Johnson-Freyd | Oh, great. Thanks. So this is the same thing that David Jordan pointed out, which is that the symmetric structure is important too. They are isomorphic as monoidal functors. | |
| Sep 8, 2010 at 21:29 | comment | added | Theo Johnson-Freyd | But maybe this is a "Law of Small Numbers" problem. | |
| Sep 8, 2010 at 21:20 | comment | added | Jacob Lurie | The isomorphism you describe is not an isomorphism of symmetric monoidal functors. To see this, let $k^G$ denote the regular representation of $G$, regarded as commutative algebra via pointwise multiplication. If F: Rep(G) -> Vect(k) is any symmetric monoidal functor, then $F(k^G)$ will inherit the structure of a commutative algebra. In the above example, $F(k^G)$ is the space of $G$-invariants in $k'^{G}$, where $G$ is acting both by permuting the factors and by Galois symmetries.Evaluation at the identity element of $G$ determines an isomorphism of this algebra with $k'$,which is not $k^G$. | |
| Sep 8, 2010 at 21:12 | comment | added | Jacob Lurie | In this context, a $G$-torsor means a $k$-scheme with an action of $G$, which is ($G$-equivariantly) isomorphic to $G$ over some extension field of $k$. So if $k'$ is a Galois extension of k with Galois group $G$, then the spectrum of $k'$ is a $G$-torsor (since $k' \otimes_k k' \simeq k^G$ by virtue of the Galois assumption). | |
| Sep 8, 2010 at 21:08 | comment | added | Theo Johnson-Freyd |
So, I'm going to think out loud while I do the simplest example. Let $k=\mathbb R$, $k'=\mathbb C$, $G=\{\pm 1\}$, then $G\text{-rep}=\text{Vect}[\epsilon]/(\epsilon^2=1)$, so that $\epsilon$ is the nontrivial one-dimensional G-rep, and as a G-rep $\mathbb C=1+\epsilon$. Then your functor is $(a+b\epsilon)\mapsto((a+b\epsilon)(1+\epsilon))^{\epsilon=0}=(a+b+\epsilon(a+b))^{\epsilon=0}=(a+b)$, which is isomorphic object-by-object to the forgetful functor $(a+b\epsilon)\mapsto(a+b)$. Moreover, there's not enough room for morphisms to be interesting. I think these functors are isomorphic.
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| Sep 8, 2010 at 20:43 | comment | added | Theo Johnson-Freyd | Interesting! But this is not something I'm used to. In non-algebraic-geometry land, all G-torsors for a given group G are isomorphic, almost by definition. How is the definition different in affine-algebraic land that allows for non-isomorphic torsors-over-k? | |
| Sep 8, 2010 at 20:12 | history | edited | Jacob Lurie | CC BY-SA 2.5 |
added 478 characters in body
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| Sep 8, 2010 at 19:22 | history | answered | Jacob Lurie | CC BY-SA 2.5 |