I want to know whether those "may be false" cases in the following statement is only when one of the a or b is zero: "We must be careful working with congruences. Some properties we may expect to be true are not valid. For example, if ac ≡ bc (mod m), the congruence a ≡ b (mod m) may be false."
@Niing In any case, this happens when m is not prime
The main problem is zero divisors. 2*3 = 0 mod 6, for example, despite the fact that neither 2 nor 3 are 0 mod 6
So add 2*whatever to both sides and you get another example (in my above example, whatever=2)
Another example: 3*5 = 0 mod 15. Add 3*7 to both sides, we get 3*12 = 3*7 mod 15
despite the fact that 12 is not 7 mod 15
ac ≡ bc (mod m) if c is coprime to m
(meaning that c and m share no factors other than 1)
(Re: my book) And then he does a bunch of calculations with it, which I have no intuition for because I have no intuition for the weird object he just defined
and it turns out that it has nice properties, by magic
Consider a coordinate neighborhood of a point in your manifold with coordinates $(x^1, \cdots, x^n)$
Then $R(x^i, x^j) = \nabla_{x^i} \nabla_{x^j} - \nabla_{x^j} \nabla_{x^i}$
This is like the Lie bracket of the operators $\nabla_{x^i}$ and $\nabla_{x^j}$
You can draw the same square picture
If you feed $R_{ij}:=R(x^i, x^j)$ a vector field $X$ on your manifold, it spits another vector field obtained after parallel transporting $X$ along an infinitisimal square $"x^i x^j (x^i)^{-1} (x^j)^{-1}"$
Think of it in analogy with Lie derivative. Notice that the Jacobi formula for the Lie bracket implies $\mathcal{L}_X \mathcal{L}_Y - \mathcal{L}_Y \mathcal{L}_X - \mathcal{L}_{[X, Y]} = 0$
(Because $\mathcal{L}_X Y = [X, Y]$, so the identify above, when fed to a vector field $Z$, implies $[X, [Y, Z]] - [Y, [X, Z]] - [[X, Y], Z] = 0$ i.e., $[X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y]] = 0$ - Jacobi identity)
But that tensor is nonzero for an affine connection in general. (In fact, it precisely measures flatness of the connection)
Hey: anyone have a good recommendation for a computer algebra system (at no cost) which can handle finite rings and their free modules? And preferably quotients of those free modules?
@AkivaWeinberger Look, it's the same idea as in Lie bracket. If $\pi^X_t$ and $\pi^Y_t$ are the parallel transports along $X$ and $Y$ for time $t$, then $R(X, Y)Z$ is $\partial_t \partial_s (\pi^X_s)^{-1} (\pi^Y_t)^{-1}\pi^X_s \pi^Y_t Z|_{t = 0, s = 0}$
@AkivaWeinberger The formal analogy between the Lie bracket and the connection is this. If $X$ is a vector field and $\gamma^X(t)$ are the flowlines of $X$ then $\mathcal{L}_X Y$ is "derivative of $Y$ along the flowline $\gamma^X(t)$ at $t = 0$".
On the other hand if $X$ is a vector field and $\pi^X(t)$ is the parallel transport defined by $X$ (which is an isomorphism between $T_{X(t_1)} M$ and $T_{X(t_2)} M$), then $\nabla_X Y$ is the "derivative of $Y$ along the parallel transport $\pi^X(t)$ defined by $X$"
(The information of the parallel transport is purely encoded in the path-space $\mathcal{P} M$. Parallel transport is a map $\mathcal{P} M \to \text{GL}(n, \Bbb R)$ sending a path $\gamma$ in $M$ starting at $x_0$ to the isomorphism $T_{\gamma(0)} M \to T_{\gamma(1)} M$, satisfying some coherence properties)
How about a map which preserves the grading on the filtrations
@EricSilva I used them a few days ago to prove that if $M$ is a negatively curved complete Riemannian manifold then it has a unique geodesic representative in every free homotopy class of loops.
I feel like half the time the big curvature tensor shows up cause I tried to commute a thing, the other contracted boiz are a lot more geometrical to me
@gian I had in mind just order-preservation. If $X = \bigcup X_i$, $Y = \bigcup Y_i$ are filtered spaces then require a map $f : X \to Y$ to satisfy $f(X_i) \subset Y_i$.
Not entirely sure if that's the right morphism in this category but feels like it
@EricSilva riemannian geometry is just lie algebras on the large
@Secret We can bring up the issue with the SE team, and then after that it's kind of case-dependent what if anything gets done about it. But by convention we usually leave that sort of detailed investigation to the mods of the site that "owns" the chat room.
I guess for now, we can try to bug the math mods to contact the SE team, and get them to look into the issue, cause recently that overstarring is escalating and many users found it disruptive
@usukidoll It's kinda measuring how badly the function fails to have an antiderivative near that point. Analytic functions, and functions like $\frac1{z^2}$, have antiderivatives, so the the residue is $0$. $\frac1z$ doesn't have an antiderivative near zero, so we give it the residue $1$.
The main part is that, if something has an antiderivative, then its integral around a loop is gonna be $0$.
'Cause the loop starts and ends at the same point ('cause it's a loop), so if we call that point $a$, then $\int_a^af(z)dz$ will be $F(a)-F(a)=0$
If there is no antiderivative we can't do that, that's why the integral of $\frac1z$ in a loop around the origin isn't zero
but in a sense, the only way to get the integral to be anything other than zero is if there's a $\frac1z$ hiding in the function.
well, the issue is that it has been going on for years, and most importantly, it is escalating recently and I think more than 50% of the users are being disrupted by it
unrelated; it's funny how Riemann curvature tensor of a symmetric connection is $\nabla_X \nabla_Y - \nabla_Y \nabla_X - \nabla_{\nabla_X Y - \nabla_Y X}$
I see the scalar curvature $R$ as an indicator of how a manifold curves locally (the easiest example is for a $2$-dimensional manifold $M$, where the $R=0$ in a point means that it is flat there, $R>0$ that it makes like a hill and $R<0$ that it is a saddle point).
Are there analogous interpreta...
Let $E$ be dense subset of metric space $X$, and let $f$ be a continuous function from $E$ to metric space $Y$. Does there always exist a continuous extension of $f: X\to Y$?
@Zee That does not seem correct. I mean, take $E = Y = \mathbb{Q}$ and the identity function. There is no way to extend this to a continuous function on all of the reals, is there?
@TobiasKildetoft I just saw this example mentioned in my readings, and i wonder why we can't extend this. The reading says'If there were an extension $f$ to mapping from $X$ into $Y$, there would be two extensions from $f$ to mappings from $X$ to $X$, contradicting uniqueness of extension' I wonder what two extensions would be.
Another way to phrase this is that continuous maps are uniquely determined by their value on a dense subset (when we put the right condition on the target space)
Someone help me to prove this theorem in differential equations:
Suppose that $f$ is continuous, bouded and lipschitziane on $x$ in $[\underline{t}-2\overline{T},\underline{t}+2\overline{T}]\times \overline{B}(\underline{x},2R)$ for all $\overline{T}>0$ and $R>0.$ Then there exists $T\in]0,\overline{T}]$ such that, for each $(t_0,x_0)\in[\underline{t}-\overline{T},\underline{t}+\overline{T}]\times \overline{B}(\underline{x},R),$ the maximal solution for the Cauchy problem will be defined on an interval which containes $[t_0-T, t_0+T].$
note that $\frac 1 {in}=-\frac i n$ so, $\sum_{n\neq 0}-\frac i n(-1)^{n+1}[\cos(n\theta)+i\sin(n\theta)]$ $\sum_{n\neq 0}\frac 1 n(-1)^{n+1}[-i\cos(n\theta)+\sin(n\theta)]$
It gets more complicated if you had a random variable that had a continuous and a discrete part. For instance, you could define X by first flipping a coin. If it comes up heads, then X=1; if it comes up tails, pick a random number between 0 and 1.
@Abcd I was eating my lunch :) so here it is : if $n$ is pos: we get $-i \cos(\theta)+\sin(\theta)-\frac 1 2[-i\cos(2\theta)+\sin(2\theta)]+\frac 1 3[-i\cos(3\theta)+\sin(3\theta)]... $ if $n$ is neg: we get $(-1)[-i \cos(\theta)-\sin(\theta)]+\frac 1 2[-i\cos(2\theta)-\sin(2\theta)]-\frac 1 3[-i\cos(3\theta)-\sin(3\theta)]... =i \cos(\theta)+\sin(\theta)-\frac 1 2[i\cos(2\theta)+\sin(2\theta)]+\frac 1 3[i\cos(3\theta)+\sin(3\theta)]...$ If I add these together: $2\sin(\theta)-\sin(2\theta)+\frac 2 3 \sin(3\theta)...=2\sum_1^\infty (-1)^{n+1}\frac {\sin(n\theta)} n$
@Bright if you take a measure-theory based course on probability then (as I understand it) then these issues get treated more carefully. But otherwise you’re probably safe to assume the r.v.s are either discrete or continuous
@Abcd I don’t see an immediate answer myself. But I’d start the problem by letting $z=\cos(\pi/8)+i\sin(\pi/8)$
