3} M$

Current License: CC BY-SA 4.0

24 events

when toggle format	what		by	license	comment
Jul 27, 2023 at 23:12	vote	accept	Noah B
Jul 27, 2023 at 14:15	comment	added	Noah B		@DavidRoberts That all makes sense! Thank you so much!
Jul 27, 2023 at 13:33	answer	added	Tom Goodwillie		timeline score: 3
Jul 27, 2023 at 7:39	comment	added	David Roberts♦		If so, take $(i,j) = (-n,-m)+\phi(k)$, so that $(i,j)\cdot ((n,m), x) := (\phi(k), x) = ((0,0),k\cdot x)$. Then the quotient by this action means that $[((i,j),x)] = [((0,0),k\cdot x)]$ in the quotient space, for every $k\in \mathbb{Z}/3$. Thus in particular $[((0,0),x)] = [((0,0),k\cdot x)]=[((i,j),x)]$. So the $(\mathbb{Z}/3)^2$ factor is killed off, and the left-over quotient gives us $S^5/(\mathbb{Z}/3)$, which given the appropriate action on the sphere, is the lens space.
Jul 27, 2023 at 7:32	comment	added	David Roberts♦		If $((n,m), k\cdot x) \simeq ((n,m)+\phi(k),x)$, where $n,m,k \in \mathbb{Z}/3,\ x\in S^5$, then I guess $(i,j)\in (\mathbb{Z}/3)^2$ acts on $(\mathbb{Z}/3)^2\times_{\mathbb{Z}/3} S^5$ by $(i,j)\cdot ((n,m), x) := ((i,j)+(n,m), x)$?
Jul 27, 2023 at 6:42	comment	added	Noah B		@DavidRoberts Cool, thank you! For the $S^5$ case, will our space $((\mathbb{Z}_3)^2 \times_{\mathbb{Z}_3} S^5)/(\mathbb{Z}_3)^2$ be the lens space $L_3^5$?
Jul 27, 2023 at 6:37	comment	added	David Roberts♦		@NoahB OK, based on the "check", I would assume that what is meant is what I wrote in my comment.
Jul 27, 2023 at 6:32	history	edited	Noah B	CC BY-SA 4.0	added 51 characters in body
Jul 27, 2023 at 6:32	comment	added	Noah B		@DavidRoberts I have not seen a construction like this either. Given Theorem 1.2 on page 2 from the cited paper, it seems like the case of $S^1 \times S^1$ is allowed, or any product of odd dimensional spheres for that matter. My check is coming from the remarks preceding Theorem 1.2.
Jul 27, 2023 at 6:12	comment	added	David Roberts♦		@NoahB I'm not sure. I've not seen a lens space built from a torus like that! Was that meant to be $S^n\times S^n$ for $n\gt1$?
Jul 26, 2023 at 3:08	comment	added	Noah B		@DavidRoberts Is this definition compatible with the check I included?
Jul 26, 2023 at 2:18	history	edited	Noah B	CC BY-SA 4.0	added 140 characters in body
Jul 26, 2023 at 0:52	history	edited	David Roberts♦	CC BY-SA 4.0	link and tag
Jul 26, 2023 at 0:50	comment	added	David Roberts♦		It might also mean mod out by the anti-diagonal action, so that you are identifying $(gn,p)\sim (g,np)$, rather than $(gn,np)\sim (g,p)$; the former is more like a tensor product than the latter, and it is what one would use for making an associated bundle. Here the associated bundle is one with fibre $(\mathbb{Z}/3)^2$ over the lens space, rather than the $\mathbb{Z}/3$-bundle given by the quotient map.
Jul 25, 2023 at 23:49	comment	added	Noah B		That’s interesting. The lens space $L^5_3$ I assume.
Jul 25, 2023 at 23:45	comment	added	Ryan Budney		One lens space. One of the $\mathbb{Z}/3$ actions permutes the three copies of $S^5$, the other does the lens space action on them, individually.
Jul 25, 2023 at 23:43	comment	added	Noah B		@RyanBudney Thanks for the response. As a follow up, what would $((\mathbb{Z}/3)^2\times_{\mathbb{Z}_3} S^5)/(\mathbb{Z}_3)^2$ be? Just three lens spaces?
Jul 25, 2023 at 23:23	comment	added	Ryan Budney		This means you take the Cartesian product and mod out by the diagonal action. i.e. it is three copies of $S^5$. The action of $(\mathbb{Z}/3)^2$ basically just permutes the three factors, and does the lens space action, independently. If you look at Tom Dieck's book on transformation groups (equivariant homotopy theory, etc) you will see this is standard notation in the field.
Jul 25, 2023 at 23:11	comment	added	Vik78		@NoahB I don't know, but if you know how to show that I would be very interested to see it
Jul 25, 2023 at 22:17	comment	added	Noah B		@Vik78 This is what I was thinking too. Then we would think of $(\mathbb{Z}/3)^2\times_{\mathbb{Z}/3} S^5$ as a disjoint union of lens spaces?
Jul 25, 2023 at 22:06	comment	added	Vik78		This is a guess, but I think it's a sort of "tensor". You have groups $H \subseteq G$ and $H$ acting on a set $X$. The object you want is the Cartesian product $G \times X$ modulo the relations $(gh, x) \sim (g, hx)$ for all $g, h, x$. The $G$–action acts on the first coordinate.
Jul 25, 2023 at 22:04	history	edited	Noah B	CC BY-SA 4.0	added 70 characters in body
Jul 25, 2023 at 22:02	comment	added	Alec Rhea		A blind guess from category theory: it is the pullback of $(\mathbb{Z}/3)^2$ and $S^5$ along the relevant morphisms into $\mathbb{Z}/3$.
Jul 25, 2023 at 21:50	history	asked	Noah B	CC BY-SA 4.0

toggle format