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this helped me so much! cannot thank you enough!!!

Lie groups are dual to lie algebras. Vectors (contravariant) are dual to co-vectors (covariant) -- Dual bases. Riemann curvature is dual, upper indices are dual to lower indices, contravariant is dual to covariant. Injective is dual to surjective synthesizes bijection or isomorphism. "Always two there are" --Yoda. Sine is dual to cosine or dual sine -- the word co means mutual and implies duality.

This is amazing, thank you so much! (context: trying to draw a ellipsis from a covariance matrix)

Subgroups are dual to subfields -- the Galois correspondence. Reducible is dual to irreducible -- non reducible is dual to non irreducible! Null homotopic is dual to non null homotopic. Symmetry (waves, Bosons) is dual to anti-symmetry (particles, Fermions) -- quantum duality. "Always two there are" -- Yoda. "The negation of the negation synthesizes the positive" -- Hegel. Positive is dual to negative -- numbers or electric charge.

Injective is dual to surjective synthesizes bijective or isomorphism (duality). Antipodal points identify for the rotation group SO(3) -- north poles are dual to south poles (magnets). "Always two there are" -- Yoda. Spinors -- you have to travel around a mobius loop twice to get back to your original position.

5:10

Convergence (syntropy) is dual to divergence (entropy). Teleological physics (syntropy) is dual to non teleological physics (entropy). Ellipsoids are dual to hyperboloids. "Always two there are" -- Yoda.

8:37 I'm not 100% sure but I think it might be that iλ∈(i Lie(T))* since λ:Lie(T)->ℝ but I might be being overly pedantic by pointing that out.

Not enough schools teach students the fact the "complex plane" is really a real algebra for a bivector (plus scalar). Bivectors (unit bivectors, or more precisely unit rotors) generate rotations in _any_ space dimension whatsoever. Every bivecotr plane defines a complex structure, henceforward geometer should really just talk about "bivector structure" which is more linguistically meaningful, and ditch the term "complex". Scalars are ... well... dilations, of course. So the branch cuts and Riemann surfaces obtained thereby have these periodicities because they are an obscure way to encode rotational symmetry (also Lorentz boosts for a Lorentzian bivector/rotor). More complicated functions could be thought of as more generalized bivector transformations, the Riemann curvature tensor, for example, is best thought of as a bivector valued function of a bivector and a spacetime position. All Lie groups are spin groups (generated by bivectors). geometry.mrao.cam.ac.uk/1993/01/lie-groups-as-spin-groups/

This great material suffers hugely from volume being way too low. I have to jack up the computer volume to the max just to make out barely what you're saying, and as long as there's no ambient noise in my area.

Well done Jonathan. Thanks

14:46 I think ℂ^3 should be something like (ℂ^3)^ ⊗^3

@15:00 I don't really get how you comput the derivative of the left side of this equation, can you elaborate on that?

15:20: I was having great difficulty showing that any complex line was a ℂ subrepresentation of the irreducible representation of ℂ^n. I did some digging and found Schur’s Lemma. I found that I was able to show that this was a subrepresentation R|ℂ=π1∘R∘ι Where π1:ℂ^n->ℂ was the projection and ι:ℂ->ℂ^n was the embedding. Then f:ℂ->ℂ^n defined by f(z)=ι(mz) was the U(1)-linear map required to show isomorphism by Schur’s Lemma since R∘f=f∘R|ℂ, this would prove that ℂ=ℂ^n, hence n=1. So I found it is not so much that any complex line is a sub-rep, it is more so that scaling the number and taken with embedding and projection would make a sub-rep and prove isomorphism by Schur’s Lemma. (Unless I missed something or am overcomplicating things).

Thank you very much for this video and for the whole playlist, which I have enjoyed a lot! I liked that you ended the series by providing examples. In particular, you addressed the point I raised in a comment to some earlier video about the existence of a universal cover. The universal cover of the figure 8 is exactly the example I tried to describe in that comment. I had wrongly named it “free space” (basically, I was looking at it as a free tree construction, if that makes sense). It’s kind of funny because the figure 8 is what corresponds to the free group on two generators. While its universal cover corresponds to the trivial group (which could also be seen as a free group, I guess, but without generators). Back to the main point. At the end of the video, you hinted at the fact that there are spaces that don’t admit a universal cover. I thought about several options (using variants of the topologist’s sine curve, one of the weirdest spaces I know). But so far I haven’t been able to find a good counter-example. I will do some research then. Thanks again!

Are there any path-connected, simply path-connected spaces, which do not admit a simply connected universal cover? I may be wrong, but it seems to me like any loop can be unfolded to get some “free space” (I put it in quotes, cause I’m not sure if the term is accurate - but, hopefully, it should be understandable). For instance, take the figure 8. The middle point is a cross. So, it can be unfolded into an infinite tree, each node of which has 4 edges, connecting it to 4 other nodes. The covering map would then map every node in this infinite tree to the single node of the figure 8. Two of the four incident edges would be mapped to one loop (but going in opposite directions), while the other two to the other loop (again, each one following one direction of travel). It’s not hard to see that this space is simply connected and the map is a covering map. Of course, it’s just one example. But it seems to have some degree of generality, which is why it makes me wonder whether the requirement that X admits a universal cover is needed or not..

Let me see if I understand covering isomorphisms. As far as I can tell, the difference between a homeomorphism and a covering isomorphism lies in the presence of the covering maps. So, whenever I have a homeomorphism F : X_1 -> X_2 and a covering map p_1 : X_1 -> Y, I can make a new covering map p_2 = p_1 \circ F^{-1} from X_2 to Y, such that F : (X_1, p_1) -> (X_2, p_2) is a covering iso between them. Moreover, given p_1 and F, p_2 is uniquely defined as above, as follows from the definition of covering transformation. However, a different iso F’ would give rise to a different covering map p_2’. I believe this reasoning should be correct.

Thanks again for this video (I love your content, as you can probably tell). I have two questions: 1) In the first theorem, shouldn’t we also assume that Y_1 is path-connected and locally path-connected? I guess it may be implied by the fact that p_1 is itself a covering map (so that any path in the base space can be lifted to a path in the covering space and this should be enough, somehow). But shouldn’t we justify it? 2) As the video shows, a covering transformation “decomposes” a covering map as the composition of two covering maps (possibly under certain conditions, see comment above). Can we also build covering maps by composing covering maps? This seems trivially true to me. I believe it should follow from the definition of covering map (and we shouldn’t even need to impose any constraints on any of the spaces). But I might be wrong, of course.

I’m not sure I understand why we need to use the inclusion condition on the push-forwards to prove well-definedness of the lifting. Couldn’t we just lift the loop from X to Y? Wouldn’t the two halves of the lifted loop be themselves liftings of the two corresponding halves of the X loop (by definition of path lifting - along with path composition)? This would imply, in particular, that the inclusion condition is true of every path-connected, locally path-connected T, covering map p : Y -> X and continuous function f : T -> X. What am I missing?

As always, thanks for the cool video! In the proof that the quotient map is a covering map, we are using the group G to index the copies of the quotient space. So, we are implicitly endowing it with the discrete topology. In the cases of G equal Z or Z/n, that’s just its natural topology. Are there any non-trivial examples of properly discontinuous (continuous) group actions, where the group in question isn’t normally given the discrete topology (e.g. where G=R)?

The big face reveal! Nice face.

The volume is too soft. Hard to hear you

😂😂😂😂😂😂😅😅😅😅😅😅😅😅😅

Thanks for providing notes along with each video! Just to spell it out (and please correct me if I’m wrong), the two elementary nbhds that cover the figure 8 are the figure 8 without the cross point and an "open cross" around that point (whose pre-image would be two open crosses around the two cross points). In particular, we can’t have two branches of the cross joined, or we would lose continuity of the map h: p^{-1}(U) -> U x {0,1} (where U is the open cross). You ask: “Are there any more 2-to-1 covers of the figure 8?”. How about, two disconnected figure 8’s?

I love this stuff, thanks for the video! I have a couple of questions about the relations between generators, which I recap here for convenience (trying my best not to add any errors): 1) \sigma_i \sigma_j = \sigma_j \sigma_i (when | j - i | >= 2) 2) \sigma_i \sigma_{i+1} \sigma_i = \sigma_{i+1} \sigma_i \sigma_{i+1} And that’s fine. My question is about the case i=j. Initially, I was tempted to think that \sigma_i would be its own inverse, since applying \sigma_i twice would simply swap the same two points twice. But it can’t be, cause the group elements are the (homotopy classes of) paths in (unordered) configuration space. So, if you make the i-th point go behind the (i+1)-th and then do the same thing again, the result is the two points going around each other in a loop. But that can’t be equivalent to the identity (two straight strands), can it? Am I correct to assume that \sigma_i^{-1} would consist of making the (i+1)-th point go behind the i-th point instead? (i.e. right strand goes behind the left, which is the opposite of what the \sigma_i’s do in your video)

That’s crazy! I’m thinking about the implications for orientable surfaces with genus g>1. And I’m sure there are even weirder examples that can be drastically simplified using Van Kampen's theorem. Beautiful!

I was a little puzzled by the definition of mapping torus. Due to the presence of the [0,1] interval, I immediately thought of it as a homotopy from id_X to \phi, even it never says that in the video. And, indeed, it doesn’t have to be. In particular, it isn’t necessarily continuous (when seen as a deformation of id_x into \phi along [0,1]). I guess what fooled me was the fact that \phi itself is continuous. The example of the Klein bottle is what made me confront my misconception. I’m just putting it out there, just in case.. and please correct me if I’m wrong.

Thanks again for another great video! I’m wondering if the definition of HEP could be simplified by letting X=Y, F=id_X and A x [0,1] -h-> A (where Y on the rhs became A). It seems to be what you’re actually using in this particular proof and it would require one quantifier less (for all F …). I’m not sure if it would be equivalent to your definition in general though. In other words, is every homotopy of type A x [0,1] -h-> B always decomposable into a homotopy of type A x [0,1] -h-> A, followed by some A -f-> B ?

I would like to see an example of a quotient map which is not open

In the video, after he states the theorem, he says that it doesn't apply to the example, because [0, 2\pi) is not compact. But, since F is a bijection, I was tempted to check if its inverse satisfied the conditions. After all, S^1 seems compact to me. Similarly, [0, 2\pi) seems to meet the criteria for Hausdorffness… However, it is continuity that fails in this case, I believe. In particular, [0, \theta) \subset [0, 2\pi) is open, but it's pre-image isn’t. Any thoughts?

Please reply Sir, which is the reference book

Thank you for the beautiful videos! I think that the formula may not be correct. At t=(1-s)/2, the argument of alpha should be 0 (for all s). However, the 2t part cancels with s-1 and the -st is generally negative (except when s=0 or s=1).

Good

Does this help with number theory?!! How can it be used?! For example, if I wanted to prove Fermat's Last theorem, how could it be used?

There are jumps which aren't clear for me (and possibly for the audience). I really thought we were going to talk about the fact that if we shrink the fast winding large loop into the point, then it must, at some R, intersect zero. And the fact that a constant polynomial different from zero would always result in a point P(x) different from zero. I really don't see where homotopy is used here? Anyone can explain? Thanks.

31:19 This is not in the video, but in the written version of the presentation found on the professor's website. Here he wrote: this implies that uv is in the normal subgroup generated by the amalgamated relations. I don't quite understand how one reaches this conclusion, especially when we are talking about u_1v_1...u_nv_n. Anybody could help, please?

Aren’t u the one who “owned” Shapiro?

Quit screwing around and get to your damn point. Your presentation deserves a thumbs down.

very nice video :) youre very good at explaining, thank u!

If Poob has a million fans, then I am one of them. If Poob has ten fans, then I am one of them. If Poob has only one fan then that is me. If Poob has no fans, then that means I am no longer on earth. If the world is against Poob, then I am against the world.

Thank you for making these lectures public - they're great to learn from. I got a different interpolation - alpha( (2/(1+s)) t - (2/(1+s) -1)), hopefully doesn't matter as long as there is at least one homotopy then they're equivalent, right?

Thank you

thanks for the video. 9:54 det(..) = t^2-t-1

@quentinmerrit loserrrrrr

Just finished this playlist!❤ Thanks

Thank you, I've been struggling to understand this in my lecture notes (y1 math) 👍 This made the summation part clearer :DD

（u,v）should be written as u+v

why killing form can be viewed as an inner product? Normally, y needs to be transposed tr(xy^T) to make it positive definite.

May I ask what tool you are using to make these videos? Thank you.

hope you can add representation of sp(2n) and its application on mechanics.