Is that the one where you want a set closed under a bunch of operations?
> In point-set topology, Kuratowski's closure-complement problem asks for the largest number of distinct sets obtainable by repeatedly applying the set operations of closure and complement to a given starting subset of a topological space.
> During World War II, he gave lectures at the underground university in Warsaw, since higher education for Poles was forbidden under German occupation.
> World War II saw the cultivation of underground education in Poland. Secretly conducted education prepared scholars and workers for the postwar reconstruction of Poland and countered German and Soviet threats to eradicate Polish culture.
Let $R_X$ be the simplicial complex with vertex set $X$, where, for every $S\subseteq X$ such that there exists a $y\in Y$ with, for all $x\in S$, $xRy$ (aka $(x,y)\in R$), then you fill in a simplex with vertices $S$
So see how there are 5 columns, and the simplicial complex on the left has 5 vertices
That last column represents the vertex that's not connected to anything
Columns 2, 3, and 4 represent the vertices with the filled in triangle because they're all represented by row 4
Both simplicial complexes look like "two disconnected pieces; one (on top) homotopic to a circle, one (below) homotopic to a point", so you can see how they're homotopic to each other
They each have a piece with a loop and a piece with no loops
@LucasHenrique The theory of representation of Compact abelian groups does use topology to a certain extent. But a more geometrical framework can be found in Kirillov theory, where one uses very symplectic geometric techniques of analysing coadjoint orbits to understand group representations. Fascinating ideas!
There's also something called representation of quivers and it has plenty of connections with ideas coming from BPS states in string theory but I know nothing but words here.
I was right about the integral, you'll be happy to know.. and after wasting the entire day figuring out where I went wrong, I will now finally go to sleep
hey is there anyone, who knows a lot about BV spaces?
If I take $f,g \in BV({[0,1]}^2)$ then $fg \in BV([0,1])$. In particular for $f_1,g_1 \in BV([0,1])$, the mapping $h: [0,1]^2 \rightarrow [0,1], f_1(x)g_1(y)$ is in $BV({[0,1]}^2)$. Now I wonder if mappings like $h$ are dense in $BV({[0,1]^2})$ ...
If I take $f,g \in BV({[0,1]}^2)$ then $fg \in BV([0,1])$. In particular for $f_1,g_1 \in BV([0,1])$, the mapping $h: [0,1]^2 \rightarrow \mathbb{},(x,y) \mapsto f_1(x)g_1(y)$ is in $BV({[0,1]}^2)$. Now I wonder if mappings like $h$ are dense in $BV({[0,1]^2})$ ...
fixed that
I don't want a clear answer, unless it is in the literature, but I wonder if someone can give me any tips hints or ideas.
@EdwardEvans the other day I spent 15 minutes trying to figure out why property A implies property B because this was being used implicitly in the paper, only to realize that property A was actually defined as property B+other stuff and I only remembered about the other stuff
I had something called $V(K)$ defined as the span of some guys of the form $v - \pi(k)v$ and was wondering why $f(v - \pi(k)v) = 0$ for all such guys means that $f(V(K)) = 0$
@love_sodam What is the image in $F_2$ of the covering-induced homomorphism?
Note that this is a normal covering space. So the group of deck transformations will be $\pi_1(S^1 \vee S^1)/\text{im } p_*$, where $p$ is covering map.
Write down, $\text{im } p_*$. That should give you all the relators for $S_3$.
@BalarkaSen I am thinking something very wrong but why isn't the deck group for a normal cover same as the symmetric group of the fiber?
What I forgot is that the deck transformation group isn't the same as the action of $\pi_1$ on fibers, since the deck transformation group corresponds to a proper subgroup of $\pi_1$ acting on the fibers.
There are non-regular covers where $\pi_1$ can act by the full symmetric group, but as Alessandro says as soon as the cover is degree >2 a regular cover cannot act as the symmetric group on the fibers
Yeah this is a subtle difference. The deck transformation group is quotient of a normalizer of a subgroup of $\pi_1$ if it's not regular in the first place
Ah yeah nevermind. Take the following beautiful example: Consider the cover of $S^1 \vee S^1$ corresponding to $\langle a \rangle$ (fundamental group is $F_2 = \langle a, b \rangle$). Then my guy above has NO deck transformations; it looks like a two copies of the Cayley graph of $F_2$ coming out of a point, with a self-loop at that point
This is actually bad news for me because it means my "proof" of the polynomials puzzle I gave @Thorgott is not a proof. The puzzle might be wrong altogether.
Yeah man I love setting up the same machine in confusing language to save you some time calculating $\pi_1(S^1)$
Let $X$ be an arbitrary space and $X \ast X$ the join of $X$ with itself. If G-SvK is so useful, tell me what $\pi_1(X \ast X)$ is; I still don't know the answer
