@Daminark idk I think there are few things more concrete than the tangent bundle. I think Schlag has a penchant for trying to get students to understand things the way a 19th century mathematician would, and I don't fully understand why he does this
"Although it is widely reported that the title refers to the sound of urine tinkling in a chamber pot, this is a later Joycean embellishment ..." - referring to Joyce's "Chamber Music"
@Eric: $\Bbb P^n$ is the space of lines through the origin. The tautological line bundle on it is the bundle whose fiber at $\ell$ is precisely the $1$-dimensional subspace of $\Bbb R^{n+1}$ that $\ell$ is. Hence tautology.
I have no clue if that's at all a good/correct way to think about it but from that one time when my physics TA did vector bundles (which was rather abstract), that's the vibe I got
The difficulty is that said intuition doesn't generalize in the ways that it needs to.
I mainly have uses of bundles etc. in quantum mechanics in mind.
e.g. the evolution of a wavefunction in a Hilbert space as some kind of Hermitean line bundle? I'm sure it's straightforward once one writes it out, but hearing it my brain just deflates a bit.
Right, @Eric, although I would call the subspace $\Lambda$ or $\xi$ instead of $\ell$ :D
So if you think about Gauss mappings to spheres, projective spaces, or (for non-hypersufaces) to Grassmannians, you can see that the tautological bundles pull back to the tangent bundle.
right when I did the bootcamp thing last summer I lectured both on the moving frames proof and the normal proof and seeing Gauss-Bonnet just fall out was the moment I was like "whoa this is more than just a change in notation"
But evidently there are cases where doing that doesn't give back the original eigenvector. Which I guess is kinda like doing parallel transport on a sphere and not ending up with the same vector you start with.
How would I go about solving a spherical (right) triangle in the ambigous case? I am given two adjacent side lengths and the angle between them which is right as follows:
B
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I'm looking at one of my student's reports, and they say that the predicted vs. experimental value was 8.81 vs. 7.42.
That's not a great agreement, and they say so. They go on to give a decent reasoning for why the latter could be off by a certain amount---say, 1. (they'd measured 1 for something, so that's not pulled out of nowhere)
But then, when they want to use this to update the experimental value, they subtract it to get 6.4 :/
Prove or disprove that there exists a sequence of moves independent of the state of the Rubik's cube you can make that eventually solves all starting positions
"...there is no sequence of moves that will swap a single pair of pieces or rotate a single corner or edge cube. Thus there are twelve possible sets of reachable configurations, sometimes called "universes" or "orbits", into which the Cube can be placed by dismantling and reassembling it."
I agree that the wording of that last line is weird, but the rest of the paragraph makes it obvious.
Particularly since it explicitly says that, if you were to disassemble the cube and reassemble it randomly, there'd be a 1/12 chance of it being solvable.
I think you guys are talking past each other and looking at different questions. I think by any state Simply means one the solvable state (that is you scrambled a standard rubics cube by moves)
Well the rubics group is finite, so choose a sequence of moves that moves to every position, no matter your starting point at some point you will hit the identity (the solved cube)
I think it's better to pose it the other way around: Is there a specific prescription of moves that, starting from the initial state, takes you through every single state in the solvable orbit?
Now, as I just stated it, I think it's trivial. You can construct the Cayley graph of that group; it's large but finite, so there's definitely a path on that graph which visits all sites.
The term "God's number" is sometimes given to the graph diameter of Rubik's graph, which is the minimum number of turns required to solve a Rubik's cube from an arbitrary starting position (i.e., in the worst case). Rokicki et al. (2010) showed that this number equals 20. This computation used a bank of computers at Google and required a total of 35 CPU-years.
" Every solver of the Cube uses an algorithm, which is a sequence of steps for solving the Cube. One algorithm might use a sequence of moves to solve the top face, then another sequence of moves to position the middle edges, and so on. There are many different algorithms, varying in complexity and number of moves required, but those that can be memorized by a mortal typically require more than forty moves.
One may suppose God would use a much more efficient algorithm, one that always uses the shortest sequence of moves; this is known as God's Algorithm. The number of moves this algorithm would take in the worst case is called God's Number. At long last, God's Number has been shown to be 20. "
Zach: If I haven't made a mistake, it won't be a right triangle, but the vector going from the origin to the point on the circle of radius 2 and the vector from the point on the circle of radius 1 to the point on the circle of radius 3 seem to be orthogonal!
I thought of an interesting problem concerning the Rubik's cube, and it is as follows:
Does there exist a sequence of moves that eventually returns any solvable Rubik's cube to the solved state?
That is, can we solve all solvable Rubik's cubes without caring about its current state by using...
I thought of an interesting problem concerning the Rubik's cube, and it is as follows:
Does there exist a sequence of moves that eventually returns any solvable Rubik's cube to the solved state?
That is, can we solve all solvable Rubik's cubes without caring about its current state by using...
@PaulPlummer worldbuilding.SE gets questions about wormholes drilled through spacetime by energy momentum and time warping happening from zooming in and out of them on international space stations
How would I go about solving a spherical (right) triangle in the ambigous case? I am given two adjacent side lengths and the angle between them which is right as follows:
B
/|\
/ | \
/ | \
/ | \
/ | ...
I thought of an interesting problem concerning the Rubik's cube, and it is as follows:
Does there exist a sequence of moves that eventually returns any solvable Rubik's cube to the solved state?
That is, can we solve all solvable Rubik's cubes without caring about its current state by using...
