I guess they're just defining the sections on the slice $Y\times\{t_0\}$
They don't say it's a global section
Although...maybe I do have a real question. What does the fibers being Euclidean spaces have to do with extending the section to those open sets (also in the last paragraph)
I think you can just repeat the usual proof for smoothly extending a vector field
@0celo7 This is continuity of transition functions, me thinks. They are just the identity matrix on that particular slice, so if you perturb it a bit, it's impossible it would change much.
@Alan Is there the smallest function which is $C^1$ and bigger than $|x|$? I think it is not. And I posted this as an answer. (Of course, I might have made a mistake.)
My first answer in vector-lattices tag. I think that vector lattices and Riesz spaces are interesting topic. Some time ago I wanted to start studying it, I even decided which book(s) to use. But I never get around to it :-(
It's where my thesis advisor wants me studying, just shifting from student mode into research mode
I'll bring it up with him soon.
also tried to work out the algebra behind his very first exercise, and got lost. For $x,y,z\in E_+$, show that $(z-x)^+ + (y-x)^+ +(2x- x\lor y$)^+\ge 0$
sweird, the last part fdidn't render right...should have been the positive part of the last term, and the whole sum is greater than or equal to 0..and all we know is that each of the terms is greater than or equal to 0 to start with
@ItachÃUchiha That strongly depends on the context. But in the context of matrix polynomials (like Cayley-Hamilton theorem and similar topics), this is precisely what you do.
@MartinSleziak, The person who asked this question is stater on meta. He started with me after 1.5 year. I've answered genuinely, but some one voting down my answer and I'm disappointing with this.
As I said, I do not see a problem with that question. And I do not see the main purpose of the post is to single out you personally.
@MithleshUpadhyay If you feel that it would be better to remove your name from the post, simply leave a comment there. You can point him to some older discussions on meta about naming specific users, like this one: What's the deal with naming names?
And even though you are saying that they made the post to blame you, I see praise in the post: "The edits are definitely time consuming and are meant for good."
Hi all, I have a super noob question. I totally forget how to solve this: I have $2x^2+2yx-1=0$ and $2y^2+2xy-1=0$. Could someone give me a hint about the method how to solve this?
@MithleshUpadhyay I think you have been around long enough to know that a few downvotes are nothing to worry about. (This is true especially about meta.)
I did not downvote your post, but since you asked what might be reasons to downvote it: I guess users who do not like edits for purposes of "badge-hunting" could use downvote to make their opinion on this clear.
It's also worth noting that downvotes on meta means that someone disagreed with your opinion, in contrast to the main site, where it means that someone thought you wrote a poor quality post
@MithleshUpadhyay If I understand the faq correctly the question has to be closed first for regular users to have even possibility to vote to delete: meta.stackexchange.com/questions/5221/…
So at the moment, only moderators can delete that particular question.
"Users with reputation ≥ 10k (more precisely, the moderator-tools privilege; 2k on beta sites) can vote to delete questions that have been closed/on-hold for 48 hours."
"Users with reputation ≥ 20k (more precisely, the trusted-user privilege; 4k on beta sites) are not subject to the 48-hour waiting period for deleting closed questions with a score of −3 or lower. They may also delete answers of score −1 or lower, unless they are accepted."
@BalarkaSen Suppose we have a smooth function $f:\Bbb R^n\to\Bbb R^m$. Embed $\Bbb R^n$ into $\Bbb R^{n+k}$ in the usual manner. Then $\tilde f:\Bbb R^{n+k}\to\Bbb R^m$ with $\tilde f(x_1,\dotsc,x_n, x_{n+1},\dotsc,x_{n+k})=f(x_1,\dotsc,x_n)$ is a smooth extension.
I think you just have to apply that in each trivialization, and done.
they're kind of weird at first, and I always need to look back to the specific wording of the C and W conditions when I need them, but I can appreciate the structure they impose
An equivalent definition of CW complexes is that it's a space $X$ with a decomposition into a chain of subspaces $X^0 \subset X^1 \subset \cdots X^n \subset X^{n+1} \subset \cdots$ (i.e., $X = \cup X^i$), with $X^{n+1}$ obtained from gluing a bunch of $(n+1)$-cells to $X^n$.
@BalarkaSen: let $S$ be a compact hyperbolic surface and $f: S \to S$ be an isometry. denote by $\phi_p$ the homomorphism from MCG(S) to the automorphism group of H_1(S, Z/pZ). if $f$ is not the identity, then $\phi_p([f])$ is not the identity for $p \geq 3$. do you know whether this theorem has a name?
@BalarkaSen: do you know anything about residually finite/finitely approximable rings? I have a finitely generated ring $A \subset \mathbb{R}$ and the author claims that "it is well know that such a ring is finitely approximable", i.e. for any nonzero $r \in A$, there is an ideal $I$ such that $A/I$ is finite and $r \notin I$.
well, planarity of a graph is an important notion, so like, what are the obstructions (if any) to embedding a 2-complex in $\mathbb{R}^3$? an $n$-complex in $\mathbb{R}^{n+1}$? what about embedding into other manifolds of dimension $n+1$?
It's a relation between homology and cohomology of $X$ and $S^n - X$ respectively, where $X$ is a nice enough subspace of $S^n$.
Locally contractible subspace, I think. But that's satisfied here because you probably want to embed your simplicial complex in a piecewise-linear way.
well, sure. topologically people usually talk about edges of a graph embedding being jordan arcs, and straight line or piecewise linear embeddings are special cases
so I dunno if that's full enough generality, although I'm pretty sure there's a theorem that says any planar graph has a linear embedding, for example
This Q has a related Q, but an answer to both Qs is given only with a comment. This is unsatisfying because the core of the question obviously keeps recurring.
