Yeah so it'd make sense to write $S_n$ to not confuse it with that, like you can just let the superscript mean sphere and subscript mean symmetric group
Im looking for examples of algorithms $A,B,C,D$ such that :
$A$ halts If $B$ does not halt.
$C$ does not halt If $D$ halts.
$A,B,C,D$ are not algorithms that halt on all input.
Also they are not algorithms that loop forever on all input.
I am aware that the general halting problem is undecid...
I have another question How can I integrate $∫ (x- 1/2x)^2$ I re-wrote it as $∫(x-(2x)^-1)^2$ then letting u equal $(x-(2x^-1) $ i get $du=(1+2x^-2)dx$ and $dx = 1/(x-2x^-2)du $ ∫ 1/3(U)^3* 1/(x-2x^-2)du $ nothing cancels
Im looking for examples of algorithms $A,B,C,D$ such that :
$A$ halts If $B$ does not halt.
$C$ does not halt If $D$ halts.
$A,B,C,D$ are not algorithms that halt on all input.
Also they are not algorithms that loop forever on all input.
I am aware that the general halting problem is undecid...
Let $X,Y$ be algorithms that accept an ordered set of positive integers as imput.
What are examples of $X$ halts IFF $Y$ does not halt ?
Im aware that the general halting problem is undecidable.
If I require the trivial bundle to be isomorph to $M\times\Bbb R^n$, does it fix my connection? My goal is namely to understand, which step in the trivialisation does fix my connection.
No, @quallenjäger, only if you require compatibility with a metric or some other such thing do you get a unique connection. Otherwise there are truly uncountably many.
@TedShifrin Then, for example for a Lie group $G$, can I define a parallel transport by the left translation $L_g$ to induce an unique flat connection on the Lie group?
You can't do either part of the assignment? Why don't you take the very simplest $L$-structure you can think of, say a one-element structure, and figure out whether $\phi$ is true or false there. Depending on what you find out, call that structure $A$ or $B,$ and you've got half of the problem solved! — bof6 mins ago
@quallenjäger: Declaring a particular frame to be covariant constant will give vanishing Christoffel symbols for that frame. I didn't necessarily know that's what you were doing. I guess it is clear that if you fix one basis and declare it "constant," then any other left-invariant basis will be as well.
This is not the natural connection on the Lie group when you want to think of a natural metric on it.
In particular, this connection has lots of torsion (and the Riemannian connection has zero torsion).
@TedShifrin First I thought I could roll the curve onto Lie group without specifying the connection. But I think it might be not realistic as I have to transport the tangent over the manifold.
I haven't thought about this rolling stuff too much, other than for a few exercises I wrote for my notes. I'm not sure how much your need the connection for.
Maybe if I have some free time tomorrow I'll think about it.
@TedShifrin The approach I was choosing is to parallel transport the tangent vector of the curve onto the manifold and then require that the transported vector "is the same" as the tangent vector of the rolling curve.
let me use this as a springboard to strongly encourage everyone (who needs to grade things) to use Gradescope: https://gradescope.com/ it's really fantastic. it saves tons of time, the students get a much clearer picture of the rubric and where/why they lost points, you can grade at home in pajamas instead of staying at the department til 1am with your fellow graders...
I know chem has a bit of a problem as well since so many people don't care about teaching. And the sad part is, the one person who has been banned from teaching undergrad classes is the one who really tries hard and loves teaching but is just bad at it
Consider $\Bbb{R}^2$ with the usual topology. Let $X$ be a subset of $\Bbb{R}^2$. If for every $a \in X$ and $v \in \Bbb R^2$ there exists a d>0, such that $a+vt\in X$, for every $0\leq t<d$, then X is open.
I suppose this theorem is wrong as the choice of the radius of the open ball $d$ is depe...
Im looking for examples of algorithms $A,B,C,D$ such that :
$A$ halts If $B$ does not halt.
$C$ does not halt If $D$ halts.
$A,B,C,D$ are not algorithms that halt on all input.
Also they are not algorithms that loop forever on all input.
I am aware that the general halting problem is undecid...
