Lecture 6: Fields (International Winter School on Gravity and Light 2015)

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The WE-Heraeus International Winter School on Gravity and Light

The WE-Heraeus International Winter School on Gravity and Light

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@tuneerchakraborty5836
@tuneerchakraborty5836 9 жыл бұрын
"Imagining is dangerous because you may imagine something different from me."
@Alkis05
@Alkis05 7 жыл бұрын
This guy deserved a better cameraman. He didn't aim the camera on the professor during the whole intuitive explanation of what is a bundle.
@IndianHeathen1982
@IndianHeathen1982 8 жыл бұрын
Extraordinary! Distilling hard concepts into a form that is relatively easy to understand!!
@travellcriner6849
@travellcriner6849 8 жыл бұрын
I'm definitely a huge fan of these lectures and the amazing strength of intuition utilized. My hunger to learn is awakened by these lectures. I have to say though, I'm not a huge fan of the "noses" analogy. I think it hinders the talk of dimensionality especially around 29:00 .
@antoniolewis1016
@antoniolewis1016 7 жыл бұрын
I'd like to keep my nose, thanks.
@luke001
@luke001 8 жыл бұрын
I thought that the tangent spaces TpM for all point p in a manifold M were automatically disjoint, ie they dont share any elements.
@travellcriner6849
@travellcriner6849 8 жыл бұрын
You're correct, although some elements of TpM may look like elements of TqM. I believe the point here is to emphasize that they're not the same even though they look the same. I'll explain: 1) Recall our definition of an element of TpM, namely a velocity v_r,p(f) where r is a curve on M through p and f is in C^inf(M). I like, in this respect, to think of C^inf(M) as a collection of all possible landscapes over M. Then v_r,p tells us the change in height we'd feel over any inputted landscape taking the particular curve r through p. So really, v_r,p is a collection of rises/falls. 2) It's conceivable that TpM and TqM may appear to have some shared collection of rises/falls. Even if that's the case, we understand these collections (the velocities) are to be thought of as totally different -- for one is only relevant to the point p and the other to the point q. 3) Imagine I wanted to create a set containing a '1' for every star in the observable universe. You might try unioning lots of copies of {1} together. But {1}U...U{1} = {1}, because sets don't care for repetitions. To get around this, we may secretly add a bit of information about the star into each of the 1's and publicly refer to the unions as disjoint unions. I'm pretty sure this is the idea behind him stressing disjoint union. I would appreciate if someone with more expertise confirmed or correct this idea, and I hope I helped as well.
@MrAkashvj96
@MrAkashvj96 8 жыл бұрын
Yes you do not have to worry about disjoint condition in this case because it comes implicitly.
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