Say $G_n$ is a sequence of groups, and $G_n=\pi_1(X_n)$ where $X_n$ is a sequence of spaces
Form the "Hawaiian bouquet" of the $X_n$ by joining them at their basepoints and saying any neighborhood of the basepoint must engulf all but finitely many $X_n$
Call that $X$
Then is $\pi_1(X)$ determined by the $G_n$?
If so, I'd call it the "Hawaiian product" of the groups
I wonder if $\pi_2$ of the "spherical Hawaiian earring" (subset of R^3 that's smaller and smaller spheres put together at a basepoint) is the same as $\pi_1$ of the original
She's a leftist trans woman who does hot takes on many politically incorrect topics
I need to learn the proof that the minimal genus surface representing the homology class $n[\Bbb{CP}^1]\in H_2(\Bbb{CP}^2)$ in $\Bbb{CP}^2$ is in fact $(n-1)(n-2)/2$
The genus coming from Riemann-Roch
But apparently it requires Seiberg-Witten theory so that'd take a couple years
@Akiva Wait, $SO(3)$ is covered by $SU(2)$ which is isometric to $S^3$ so has isometry group $SO(4)$ which has dimension $10$, and isometries of $SO(3)$ lift to those of $SU(2)$ so $\dim \text{Isom}(SO(3)) \leq 10$ better hold, no
Please help me with this problem of Conic Sections
If $\alpha, \beta, , \gamma and \delta $ be the eccentric angles of the four points of intersection of the ellipse and any circle, prove that $$ \alpha + \beta + \gamma +\delta$$ is an even multiple of $\pi$
Hi, I'm trying to solve this problem that exploits the implicit function theorem. The second point, however, to me is a headache. i.sstatic.net/sRkre.png
I seeeee, in the theory of modular forms it's an equivalence class of cusps and the cohomology theory is called "Eichler cohomology" and is defined as $H^1_P(\Gamma, M) := \operatorname{ker}\left\lbrace H^1(\Gamma, M) \to \prod_{\mathfrak{s}}H^1(\Gamma_\mathfrak{s}, M)\right\rbrace$ where $\Gamma$ is a congruence subgroup of SL_2(Z) and $M$ is a representation of $\Gamma$ and $\Gamma_\mathfrak{s}$ are stabilisers of the cusps $\mathfrak{s}$ rofl
It feels very nice when I hear someone is working on Set Theory, Topology or other pure mathematical topics than those technocrats and computer science things.
@AkivaWeinberger Your intuitive argument that $[\ell_1, \ell_2][\ell_2, \ell_3]\cdots$ is nontrivial in $H_1(\Bbb H)$ can actually be made precise as follows (I should have realized this immediately but I was extremely sleep deprived). If it was trivial, then it would belong to $[\pi_1(\Bbb H), \pi_1(\Bbb H)]$.
So it's a finite product of commutators in $\Bbb H$. Take the maximal $n$ such that $\ell_n$ appears in that word.
That means this word $[\ell_1, \ell_2][\ell_2, \ell_3]\cdots$ is homotopic to a word in a finite rank free subgroup in $\pi_1(\Bbb H)$ generated by the first $n$ circles.
There should be a quick argument to show this is impossible.
@AkivaWeinberger Oh I lost track there, scratch the above. Here's what it is: Say it's a product of $n$ commutators in $\Bbb H$ (doesn't make sense to take the maximal $n$ such that $\ell_n$ appears, it can be commutators of infinite-length words).
Chop off every letter after $\ell_{n+3}$. Then $[\ell_1, \ell_2][\ell_2, \ell_3]\cdots$ gets sent to $[\ell_1, \ell_2][\ell_2, \ell_3]\cdots[\ell_n, \ell_{n+1}][\ell_{n+1}, \ell_{n+2}] \in F_{n+2}$, which is a product of $n+1$ commutators. But since this word in $\pi_1(\Bbb H)$ is equal to $n$ commutators after chopping off we also get it's equal to product of $n$ commutators in $F_{n+2}$.
Essentially because $\pi_1(\Bbb H)$ sits inside $\varprojlim F_k$, so the sequence of words in every free group are consistent under the "chop off the last letter map" $F_{k+1} \to F_k$
I guess it's not super clear to me that commutator-length should be an invariant
In $F_n = \langle x_1, y_1, \cdots, x_n, y_n \rangle$, $[x_1, y_1][x_2, y_2]\cdots[x_n, y_n]$ is not a product of less than $2n$ commutators because if that was so then if say $[x_1, y_1]\cdots[x_n, y_n] = [a_1, b_1]\cdots[a_m, b_m]$, then you'll get a non-nullhomotopic map $\Sigma_{2m} \to \Sigma_{2n}$
@MikeMiller Yeah got it. $\Sigma_g \to \Sigma_h$ for $g < h$ has to be non-surjective on $H_1$ so non-injective on $H^1$. You get some $\alpha \in H^1(\Sigma_h)$ which pulls back to $0$ in $\Sigma_g$
Take it's dual, which cups to $1$ in $\Sigma_h$ but pullback cups to $0$
@AlessandroCodenotti This is standard. Suppose $M_f$ deformation retracts to the $X$ end. Then run the homotopy long exact sequence on $(M_f, X)$ to get $\pi_* X \to \pi_* M_f \to \pi_* (M_f, X)$. Since $M_f$ def rets to $X$ the first map is an isomorphism, forcing $\pi_* (M_f, X) = 0$.
Some argument I forget here. Damn I am forgetting all of topology
@Shootforthemoon: You mean to test? If you have a critical point $a$, the second-degree T.P. at $a$ might tell you. But what happens with something like $f(x)=x^3$ or $x^4$ at $0$?
Oh ok assume everything is cellular. $X, Y$ are CW complexes, $f : X \to Y$ is a cellular map. So $X$ is a subcomplex of $M_f$ obtained from attaching cells. $\pi_*(M_f, X) = 0$ implies you can collapse the cells of $M_f$ attached to $X$ by leaving it fixed along the boundaries
That gives a deformation retraction, cell-by-cell, of $M_f$ to $X$
And for non-cellular maps you have to jiggle a little to make stuff cellular. This bit I really don't remember
@AkivaWeinberger Anyway, I think it suffices to take the word $w = [\ell_1, \ell_2][\ell_3, \ell_4] \cdots$ in $H_1(\Bbb H)$. $S(w) = [\ell_3, \ell_4]\cdots$ which is homologous to $w$ of course. And $w \neq 0$ because if it is zero, then $w \in [\pi_1(\Bbb H), \pi_1(\Bbb H)]$ thereby a product of $n$ commutators, but then chop of all the letters $\ell_{n+3}, \ell_{n+4}, \cdots$ to get $[\ell_1, \ell_2] \cdots [\ell_{n+1}, \ell_{n+2}] = \text{product of n commutators}$ in a free group $F_{n+2}$
That's not possible because of the surface and degree argument above
@TedShifrin Thanks! Well, in those cases for $f(x)=x^3$ we have a change in concavity, while for $x^4$ we have no changes and a positive concavity, right?
