Timeline for Is super-vector spaces a "universal central extension" of vector spaces?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 3, 2012 at 12:58 | vote | accept | André Henriques | ||
| Dec 10, 2011 at 16:54 | comment | added | André Henriques | @Todd: $\mathbb Z$-graded vector spaces admits a functor to $sVect$. So, maybe, what I should have said it that $sVect$ is (weakly?) terminal among extensions of $Vect$. Here's by "weakly terminal" I mean that I only care about the ∃ part of the def of terminal, but not about the uniqueness part. | |
| Dec 10, 2011 at 15:50 | comment | added | Todd Trimble | Sorry, I'm still not getting it. Why isn't $\mathbb{Z}$-graded spaces (with a sign change when you permute homogeneous elements of odd degree) such a non-trivial extension? | |
| Dec 10, 2011 at 13:50 | answer | added | Vladimir Dotsenko | timeline score: 1 | |
| Dec 9, 2011 at 17:51 | comment | added | André Henriques | Maybe "universal central extension" is a bit misleading. It was suggested to me by Alexandru Chirvasitu that $sVect$ might be more like an "algebraic closure" of $Vect$. | |
| Dec 8, 2011 at 22:10 | answer | added | Chris Schommer-Pries | timeline score: 21 | |
| Dec 8, 2011 at 20:28 | comment | added | André Henriques | Konrad: it's an analogy. A central extension of a group $G$ is a map $\tilde G\to G$. Here, the situation is different: $sVect$ admits an inclusion functor $Vect \to sVect$. So, you see, the arrow goes the other way. I wouldn't know how to answer your question, but I think that it's safe to say that $sVect$ is twice as big as $Vect$. That's why I mentioned $\mathbb Z/2$ in my question. | |
| Dec 8, 2011 at 20:05 | comment | added | Konrad Waldorf | $sVect$ is an extension of $Vect$ by what category, exactly? | |
| Dec 8, 2011 at 18:22 | comment | added | André Henriques | All the symmetric monoidal extensions that you name are split: they admit a symmetric monoidal functor back to $Vect$ (the functor takes a representation to its underlying vector space). In that sense they're not "non-trivial extensions". | |
| Dec 8, 2011 at 17:46 | comment | added | Angelo | I am sure I don't understand the question, but $Vect$ has lots of symmetric monoidal extensions (for example, representations of some affine group scheme), that are not contained in $sVect$. | |
| Dec 8, 2011 at 17:39 | history | asked | André Henriques | CC BY-SA 3.0 |