I don't really understand your question. What is the classification of 1D complex manifolds?
I know a classification for 2D real compact manifolds and you can see which of these are complex manifolds, but $\Bbb C \setminus \{0\}$ is not compact
you can always ask about the universal cover of a connected Riemann surface, though. In the case of $\Bbb C - \{0\}$, the universal cover is given by $\exp: \Bbb C \to \Bbb C- \{0\}$
$p:E\to B$ is a covering map if it's surjective and if for each $b\in B$, you can find some neighborhood $V$ such that $p^{-1}(V)$ is a disjoint union of open sets such that the restriction of $p$ to any of them is a homeomorphism
hey everyone :) I would love to get some help on this generating function question. All the details are written on here. I do understand the first case, but the second case (in which there is significance to the order) I don;t. Thank you in advance, and sorry for my poor english math.stackexchange.com/questions/2595727/…
@Daminark Hm, no, I mean take an annulus $U$ in $\Bbb C$ and define the covering map $U \to \Bbb C$ by $z \mapsto z^2$. That seems to be a perfectly fine non-surjective covering map to me.
I guess I don't see why playing with surjectivity is a non-dumb thing to do at all
@Daminark Ah! Hatcher's definition is that $p : Y \to X$ is a covering space if for each $x \in X$ there is a neighborhood $U$ of $x$ such that $p^{-1}(U)$ is disjoint union of a collection (possibly empty) of open sets of $Y$ on which $p$ restricts as a homeomorphism.
Now THAT does simply if $X$ is (path) connected, $p$ is surjective
It's a good exercise to figure out why!
(Also why my example above is not a covering space in this definition)
Really tho the fact that this fucker has anything to do with $S^1\vee S^1$ feels like I'm about to end up on "EPIC PRANKS COMPILATION 2018 (GONE BADLY)"
Good, yes. More generally, if $\Gamma = \Gamma(G, S)$ is the Cayley graph of $G$ wrt some choice of generators $S = \{g_1, \cdots, g_n\}$, define the action of $G$ on $\Gamma$ by $g \cdot "x" = "gx"$ where $"y"$ is the point on the Cayley graph corresponding to the element $y \in G$.
Like, dammit Balarka all we have to do is send a chat transcript instead of all these videos about people whose lives were ruined by drugs to discourage kids
Yo for a sec I was conflicted about starring that. But now that I'm not gonna be the first
Let $A \in\mathbb{R}^{n\times n}$, $n\geq 3$ be a matrix with $n+1$ elements $1$ and the remaining elements are $0$. I want to show that $\det (A)\in \{-1, 0, 1\}$ and each of these $3$ possible values can occur. Do I have to use induction?
Hey, if I have a vector u = [2, 4, 8] how can I calculate a matrix such that Mu=[.25, .5, 1] in other words, how can I divide all the elements by the last element via a matrix?
Let $\{a_{k,K}\}$ and $\{b_{k,K}\}$ be real sequences with the following properties (perhaps not all of them will be needed for the result):
$0\leq a_{k,K} \leq b_{k,K}$ for all $k$, $K\in\Bbb{N}$.
$a_{k,K}=0$ for $k>K$, but $b_{k,K}$ is nonzero for all $k$, $K\in\Bbb{N}$.
$a_{k,K} \sim b_{k,K}...
@Startec, dividing by a single number is a quite basic operation, so you could think simply of $u/8$, and then if you necessarily want a matrix we can add the identity matrix as $Iu=u$ for any $u$. Also you could take a general $M$ and write down the result; you'd see that there are infinitely many $M$ that satisfy the equation. It's even simpler as the rows of $M$ don't interact. For example, the first row of M could be (1/8,0,0), (0,1/16,1/16), etc.
Hey, I am trying to do some integration. But I am stuck at some point and I am not sure if everything is correct so far. Does anyone see a mistake or has a an idea how to go on ?
lol an linear algerbra class topology a more advance analysis class and a more advanced number theory class only doing 4 cause i almost died last semester but i am sitting in on a Galois theory class as well
also got and advisor to do a summer reasearch project that i dont understand
Let $\{a_{k,K}\}$ and $\{b_{k,K}\}$ be real sequences with the following properties (perhaps not all of them will be needed for the result):
$0\leq a_{k,K} \leq b_{k,K}$ for all $k$, $K\in\Bbb{N}$.
$a_{k,K}=0$ for $k>K$, but $b_{k,K}$ is nonzero for all $k$, $K\in\Bbb{N}$.
$a_{k,K} \sim b_{k,K}...
