$S^1$ is the only one dimensional closed manifold so $\Bbb R\Bbb P^1$ needs to be isomorphic to it, even though the construction you wrote above should be a much better justification than this abstract nonsense
@TedShifrin it did well! This first year (especially the second semester) was challenging, and both linear algebra and vector calculus were very fun. I plan to study (very) introductory algebraic geometry with my linear algebra teacher next year. Also, I'll be studying Galois theory until August 2020, and I can surely say that it's one of the coolest subjects I know 'till now.
connected and compact are genuine adjectives associated to manifolds, without-boundary is a clarifier, and non-Hausdorff and non-second countable are abominations
so thats curious. I have $S^1 \rightarrow S^1 \lor S^1 \rightarrow S^1$ but the "net action" is still homeomorphic (where $S^1 \rightarrow S^1 \lor S^1$ is not, in particular the circle and the figure 8 have different fundamental groups..)
oh, there was a planar geometry problem which came up on chat earlier. solving it with analytic geometry was easy enough but I wonder if there's a more geometric argument.
Yeah, because for any open subset of R^n there is a further open sub-subset which is homeomorphic to R^n. But as Ted would tell you, this would be bad for complex analytic manifolds @Alessandro
@JoeShmo A space can have quotients homeomorphic to the whole space, this should not be surprising, here you're factoring such a quotient through an intermediate space by doing it in two steps
suppose I have an ellipse and I draw two perpendicular lines through its center. each intersects the ellipse twice. let the distances from the origin to those intersections be $r_1,r_2$
Or even varieties, @Alessandro. A variety is locally biregular to an open subset of a closed subset of A^n, not locally biregular to a closed subset of A^n nor all of A^n
it's not bad in analytic geometry: in polar coordinates, one has $r^{-2} = a^{-2}+(b^{-2}-a^{-2})\sin^2\theta$ where $a,b$ are the semi-major/minor axes
So is there any way to get it out of just the sum of the distances to the foci is constant @Semiclassic? I'll allow differentiating to prove a function is constant.
@Eric: I volunteered to do a bunch of prep work one evening ahead, but I don't think they'll trust me to do much — cook ground beef, soak and cook beans, etc.
So let's say I have a f.g. group with $n$ generators. This come with an $F_n$ action for free, write it as $G=F_n/H$ and let $F_n$ act on the cosets by multiplication. Everything makes sense so far?
So now I want to obtain $\mathrm{Cay}(G)$ as the quotient of an action of $F_n$ on a topological space (maybe $\mathrm{Cay}(F_n)$? I'm not sure what I should use here)
@TedShifrin fyi, the original problem was actually a bit stronger than what I've said: You take the two intercepts, draw the line segment between them, and then draw the perpendicular from said segment to the origin
at which point the claim is that that distance doesn't change
@Alessandro Cay(G) is indeed a quotient of Cay(F_n), given by pinching the subgraphs spanned by the vertices contained in a single H-coset in F_n, and then stacking togather all the edges of the same color (i.e, corresponding to a specific generator in your original generating set in F_n) going between two subgraphs corresponding to two different cosets.
To be precise this describes $Cay(G; \overline{S})$ as a quotient of $Cay(F_n, S)$ where $\overline{S}$ is the induced generating set coming from $S$ by taking image under the quotient map $F_n \to G$
Todhunter's version: "The perpendicular from the centre on a straight line joining the ends of perpendicular diameters of an ellipse is of constant length."
just to be clear, the idea is: work in terms of the stretched circle parametrization, and use that to suss out how 90 degree angles in the ellipse get mapped to the circle
@ted the other cute implication of this business, though it's totally obvious, is that if you vary your choice of perpendicular lines then the point D traces out a circle
@TedShifrin I think it's true because of geometry. For example, normalizers of elements of $\pi_1(\Sigma_g)$ for $g > 1$ are cyclic, because you take a geodesic representative, lift it upstairs in $\Bbb H^2$ to get a geodesic line, and the normalizer becomes all elements of $\pi_1(\Sigma_g)$ which stabilizes the geodesic is cyclic, because the hyperbolic isometries which fix a geodesic (aka fix endpoints of the geodesic) are cyclic
@Alessandro Ohhh you can cook up some HNN extension
I think normalizer of cyclic subgroups of a hyperbolic group in general are virtually cyclic by the way
I don't actually think essential curves are essential (wink). I can get a geodesic representative without making it collapse to the cusps (just not a minimal geodesic), and then lift to get a geodesic line in $\Bbb H^2$
I think nothing fails
All you need is subgroup of $\text{PSL}_2(\Bbb R)$ fixing two points on the boundary is cyclic, and then the normalizer is intersection of this cyclic subgroup with your group (in this case $F_2$) which is also cyclic
@Alessandro So yeah if this is fine, any 2-generated 1-word non-cyclic group is quotient of F_2 by F_1
And gives you an example
Ugh, no. You have to quotient by the normal closure, not normalizer of the subgroup generated by the relators.
That can be huge. Normal closure of $\langle a \rangle$ in $\langle a, b \rangle$ is $\langle b^n a b^{-n}\rangle$, infinite cyclic.
@AlessandroCodenotti OK: If $H$ is a finite rank free normal subgroup of $F_2$, then $F_2/H$ will actually be a finite group. Consider the covering space $X$ of the figure 8 corresponding to $H$. This is a graph with finitely many cycles since $H$ is finitely generated. If $X \to S^1 \vee S^1$ was infinite-sheeted, so take a fiber which contained a point $x$ in some cycle of $X$.
You can find another point $y$ in the fiber which is not contained in any of the cycles of $X$ by finitude of cycles. But then there is no graph automorphism of $X$ taking $x$ to $y$, which contradicts the fact that this is a regular cover (deck transformations are simplicial here).
This implies it's a finite sheeted cover, so $F_2/H$ is a finite group
Take any infinite 2-generated group which is not 1-generated, which is isomorphic to $F_2/H$ by choosing a presentation. $H$ is not isomorphic to any finite rank free group, it has to be infinite rank.
So all the examples we spoke of, normal closure of $\langle a^3b^{-2}\rangle$, $\langle aba^{-1}b^{-1} \rangle$, etc, are infinite rank
Does the symmetric group $S_n$ contain subgroups of all possible (that is, dividing $n!$) orders? It's true for $S_1,S_2,S_3,S_4$; I am not sure about $S_5$.
How do I find $[\mathbb{Q}(\sqrt{2}, \sqrt{-5}) : \mathbb{Q}]$? I'm pretty sure it's 2. If I show that $[\mathbb{Q}(\sqrt{2}, \sqrt{-5}) : \mathbb{Q}(\sqrt{2})] = 1$ I'm done.
The problem is that IDK how I can judge what's the order of the min poly of $\sqrt{-5}$ over $\mathbb{Q}(\sqrt{2})$
That should work out to be $4$. If $[\mathbb{Q}(\sqrt{2},\sqrt{-5})\colon\mathbb{Q}(\sqrt{2})]=1$, that would mean $\sqrt{-5}\in\mathbb{Q}(\sqrt{2})$, but $\sqrt{-5}$ is imaginary, whereas $\mathbb{Q}(\sqrt{2})\subseteq\mathbb{R}$.
