For a smooth manifold $M$ with $p \in M$ and $v \in T_pM$ and $f \in C^{\infty}(M)$, $v$ 'acting' on $f$, just means the derivation $v$ taking $f$ as its input right? So $v(f) = c$ for some $c \in \mathbb{R}$ since $v : C^{\infty}(M) \to \mathbb{R}$
Hi, $$\text{Let }f\in C([0,1]), g\in C^1([0,1]) \text{ such as } f\geq g \text{ and increasing } g. \\\text{Is there the existence a sequence of function in }C^1([0,1]) \text{ minus by } g \text{ and having the uniform limit }f?$$
Let $f, g: \Bbb S^1 \to X$ be two loops, and $\varphi, \psi : \Bbb S^1 \times I \to X$ be two homotopies of $f$ and $g$. Must there be a homotopy $\Bbb S^1 \times I^2 \to X$ between $\varphi$ and $\psi$? If the universal cover exists then the answer is certainly yes, because the universal cover is loops quotient homotopy and is simply connected, and the homotopy required is a path between the two points in the universal cover. What if the universal cover doesn't exist?
Now that I think about it without category theory I would've never understood why free groups, and free vector spaces and free modules all popped up everywhere
It's as if mathematicians were all trying to give me this free stuff I didn't want
@Semiclassical already noticed that it is sufficient to consider only $k$ from $1$ through $n$, because for $k$ larger than $n$ $A^{k+i}$ can be written as a linear combination of lower powers of $A$ by Cayley-Hamilton
which suggests induction, but it also suggests proof by contradiction
but it's only extra credit so i probably wont get to it, but im sharing it because ti's interesting
@KasmirKhaan if you want to train your proving skills: for a group $G$ and a normal subgroup $N \trianglelefteq G$, prove that $G/N$ is abelian iff $[G,G] \subseteq N$
@Balarka as I said $\displaystyle \lim_{\longleftarrow} D_{n} \cong \hat{\Bbb{Z}} \rtimes \mathbb{Z}/2\mathbb{Z}$ and $D_\infty = \Bbb{Z} \rtimes \mathbb{Z}/2\mathbb{Z}$, so basically you take the embedding $\Bbb{Z} \to \hat{\Bbb Z}$ on the first component and the identity on $\Bbb Z/ 2\Bbb Z$
@LeakyNun not entirely obvious, no. I can prove it by messing around with an explicit construction of projective limits in the category of groups, but I guess that's not really pretty
@BalarkaSen i do accually, if you seen for the first time 1 apple, would you know if it was small or big ? you need at least 2 thing to be able to compare
I would say that is because of the Peano axioms on $\Bbb N$.
More accurately, the "successor" function
Successor of each natural number $n$ is defined to be $S(n) = n+1$
The fundamental working principle of induction is to rig the successor function to prove a statement for all natural numbers knowing only for $n = 1$ (or $0$, whatever $\Bbb N$ starts with)
There are "mathematical induction" techniques that are defined for more general sets than naturals, or the integers, however... there are things called "real induction", which is induction on the real numbers? I have no idea how it works
@BalarkaSen There's a couple of typesetting errors here and there, but it should be overall understandable. I typeset all of that on a single night in rage, so it can only be so good :P
@MatheinBoulomenos @TedShifrin I will show that there is a path from $I$ to any matrix $A \in SO(n)$:
Now, $A$ can be diagonalized over the complex numbers (follows from a simple argument by induction where you reduce the dimension of the space by one each time by taking the orthogonal complement)
@MatheinBoulomenos @TedShifrin Now, let $(\lambda,v)$ be an eigenpair. Then, $\|\lambda\|^2 \langle v, v \rangle = \langle \lambda v, \lambda v \rangle = \langle Av, Av \rangle = \langle v, v \rangle$, so $\|\lambda\|^2=1$. Now, pair each non-real eigenvalue with its conjugate, and on the diagonal matrix, change $\begin{bmatrix} e^{i\theta} & 0 \\ 0 & e^{-i\theta} \end{bmatrix}$ to $\begin{bmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{bmatrix}$.
@MatheinBoulomenos @TedShifrin now, since the determinant is $1$, the real eigenvalues have product $1$, so the number of $-1$ is even, so they can still be paired up and there $\theta = \pi$. Now, the path is given by changing each $R_{\theta}$ block to $R_0$ simultaneously.
@TedShifrin I just think that certain matrix computations (with actual numbers) are among the most dull things I did in college. I don't like following an algorithm. I had fun computing Galois groups etc.
Actually the other direction should be too, if you have the right subnormal series you can just extend it to a composition series and the quotients have to remain abelian, right @Mathein?
@EricSilva Well, to be fair, you have to have a very high IQ to understand how to make a Rick and Morty meme. Without a good understanding of theoretical physics the harsh realization of the memes coming to real life and the realism of the elitism would be completely lost to the memer....
I think they're like all small but vocal interest groups. The incel. The Bernie-bro. The red-piller. The algebraist. They're out there, but hard to find.
World is a strange place. Everyone seems to think the other person is a sheeple, and that person himself is thought as a sheeple by a person watching both, and ad infinitum to hell
