You need to use change-of-basis and rotate to make your plane a standard coordinate plane where you know how to parametrize a circle. Either that, or you need an orthonormal basis for the plane (through the origin), @Leaky. Either way, you need some linear algebra.
idk if I have a good justification, but it took a long time to come up with the right generalizations that captured how continuity behaved, it wasnt easy to come up with @maimslap so I don't know if we should be able to armchair reason why it is the right definition
@PVAL @MikeM: Leaky is asking an interesting question. What is the least $n$ for which we can embed the torus in $\Bbb R^n$ and get an ambient isotopy from the identity to the map that switches the orientation on (any) tangent space?
You can complete the square in the quadratic stuff and see what it is. But you can't just throw away the third-order stuff unless you're a physicist and are lying.
@TedShifrin bonsoir, s'il vous plait comment trouver l'équation de la droite tangente à un cercle de rayon 2 et de centre 0=(0,0) et passe par le point $A=(3,0)$
@Leaky: Rotate simultaneously in the $x_1x_2$- and $x_3x_4$-planes. But now you have to think about how you change embeddings of the torus, regardless.
Problem: Let $X = \prod_{n \in \Bbb{N}} X_i$. If each $X_i$ has a countable subset $A_i$, then so does $X$. Observations (sadly no proof yet): First, $\prod A_i$ doesn't work, since the countable product of countable sets is uncountable. Second, letting $A \subseteq X$ denote a dense subset, it's clear that writing $A$ as some cartesian $\prod D_i$ won't work, since $\prod \overline{D}_i = \overline{A} = \prod X_i$ and therefore the $D_i$ are dense; but this won't work because
the only dense sets we are given are the $A_i$, and their product is uncountable. At this point, I am not sure what to do. I was thinking maybe the canonical projections $\{\pi_k \}$ might help, since they are open maps and therefore $\overline{\pi^{-1}_k(D_k)} = \pi_k^{-1}(\overline{D}_k)$
Oh, well, I'm not 100% sure about that. I was trying to do a path in $SO(n)$ and then argue that we could look at the $2$-plane path of our original tangent frame (and normal frame) and induce embeddings of the tori by modding out those $2$-planes. I didn't think it through carefully, @Balarka. That's why we need you.
@Balarka: I was definitely using triviality of the tangent bundle to reduce to pointwise considerations.
@user193319: You need the Cantor diagonalization trick.
@TedShifrin if you cut the torus open to form a square, and then glue it back to the emmersion state, you find that you inverted one loop and preserved the other
What automorphism? The isotopy doesn't need to give any automorphism. The identity map is isotopic to the antipodal map through immersions S^2 --> R^3. The identity and the antipodal map are NOT isotopic as maps S^2 --> S^2
@Ted Holonomic approximation theorem says exactly that the space of immersions $M^2 \to \Bbb R^3$ is weak homotopy equivalent to $\hom(TM, T\Bbb R^3)$ I believe.
So yeah for the torus this is obvious because the tangent bundle is trivial (hence the classifying map is nullhomotopic); I can homotope the derivatives of the two immersions
@Ted @Leaky Ah, what is nice about this is that my thing immediately says any embedding of $S^{n-1}$ everts in $\Bbb R^n$ if and only if $\pi_{n-1} SO(n)$ is zero. That forces $n = 1, 3, 7$
@TedShifrin about showing that $SO(4)$ is path connected (regardless of the topology problem). I tried it like this: any matrix in $SO(4)$ is the product of an even number of hyperplane reflections (and the number is uniformly bounded by some integer $2n$) The map which sends a nonzero vector $v \in \Bbb R^4$ to the reflection through the orthogonal complement of $v$ is continuous (there's some formula in terms of the dot product. )
Thus the subset $R$ of $O(4)$ consisting of all hyperplane reflections is path-connected. Continuity of matrix multiplication gives that $SO(4)$ is path-connected as the image of the map $R^{2n} \to SO(4), (R_1, \dots, R_{2n}) \mapsto R_1 \cdot \dots \cdot R_{2n}$
Say I have a group $G$ of order $p^nq$. By the first Sylow theorem there is a subgroup of $G$, let's call it $S$, of order $p^n$. Furthermore let $f: G \to G, g \mapsto g^{p^m}$ and let $G^{p^m}$ be the image of $f$. I want to show that for every $m > n, S \cong \frac{G}{G^{p^m}}$.
My guess is ...
Say I have a group $G$ of order $p^nq$. By the first Sylow theorem there is a subgroup of $G$, let's call it $S$, of order $p^n$. Furthermore let $f: G \to G, g \mapsto g^{p^m}$ and let $G^{p^m}$ be the image of $f$. I want to show that for every $m > n, S \cong \frac{G}{G^{p^m}}$.
My guess is ...
Holomorphic injections must have non-zero derivative, right? My heuristic reason for believing is that if you have non-zero derivative, then you write $f(z) = f(z_0) + (z-z_0)^ng(z)$ where $g(z_0) \ne 0$, but this $(z-z_0)^n$ business is not injective
Say I have a group $G$ of order $p^nq$. By the first Sylow theorem there is a subgroup of $G$, let's call it $S$, of order $p^n$. Furthermore let $f: G \to G, g \mapsto g^{p^m}$ and let $G^{p^m}$ be the image of $f$. I want to show that for every $m > n, S \cong \frac{G}{G^{p^m}}$.
My guess is ...
