In terms of mathematics and ability to clearly present it,Prof. Schuller is the best of all lecturers I have encountered at University or on the internet.
Excellent lecture! I remember watching this a few years ago and not understanding most of it. I've watched this at least 4-5 times now, I think. It's kind of unbelievable what persistence can achieve.
Glad I'm not the only one who's had to watch and re-re-re-watch several times before it clicks. Hats off to the folks who get it after watching the lecture only once. For us mere mortals, the persistence makes it all the more satisfying when it finally clicks. Totally worth the struggle because this is the some of the most beautiful math, tying so many other concepts together in such an elegant way.
At about 40 min into Lecture 13, the push forward should be "g _lower star," not g_ upper star (which would denote the pullback). O/wise, these are superb lectures. Schuller makes things very clear by taking the time to motivate what he is doing and why he is doing it.
At 56:30, there is a mistake in assuming that any function pulls out. Only constant functions can be pulled out. So, L(G) is not a C-infinity(G) module but it is a R-vector space.
Yeah, this threw me for a loop too. I think in the second and third alternative definitions of left invariant vector fields we shouldn't be translating the term on the right hand sides by g at all. Considering the real line with addition as the Lie group and taking d/dx as our left-invariant vector field, we obviously can't scale it by an arbitrary C-infinity function and still have it be left invariant.
I believe it is C-inf(G) module. Correct me if I am wrong. let f,j be C-inf functions on G, X a vector field on G and l_g the left translation map. Then f.X is also a vector field on G and (l_g fX)j = fX (j o l_g) , by definition of push forward map but rhs = f.X(j o l_g) = f.(l_g X)j So we have (l_g fX)j = f.(l_g X)j The function pulls out and hence its C-inf linear.
As it turns out, L(G) is not a C-inf submodule; you skipped over some steps and when you do the derivation more thoroughly you see why it doesnt work (apart from the example in the first comment, of course). Maybe this will help, Let k be a C-inf function on G, and let X be a left invariant vector field on G. C-inf multiplication (at the point p) on the set of vector fields is defined by (kX)(p) := k(p)X(p). Now, in order to have a left invariant vector field, the push forward (of the map l_g, call it L_g due to technical limitations) must satisfy L_g(X(h)) = X(gh) So, we must show that (kX) is also a left invariant vector field for it to be a submodule. Let Y = (kX), then L_g(Y(h))=L_g(k(h)X(h)) = k(h)L_g(X(h)) = k(h)X(gh). But, if Y was left invariant then L_g(Y(h)) should equal Y(gh) = (kX)(gh) = k(gh)X(gh), but it doesnt. Thus, only constant functions that aren't dependent on its input can be pulled out.
On other videos lectures I have seen him reference his notes but not often. He is a great explainer like Feynman was. Dr. Schuller is a bright lecturer.
Awesome lecture/live performance! He has everything in his head that he doesn't need to see any notes. He only needs to "display" the run-through of the ideas and the key points.
1:01:35 can be alternatively formulated with (ii) being the Alternativity condition, where [y,y] = 0, ∀y ∈ G. Then we get (ii) and (i) ⇒ anti-symmetry. To see this, just expand [a + b, a + b] = 0 = [a,b] + [b,a]. **adding in case anyone is confused by this formulation. Schuller's version is equivalent.
In fact it must be formulated like this. The anti symmetry condition is not enough to characterize a Lie algebra if the given Lie algebra is a K - vector space with K being a field of characteristic 2.
L(G) can't be a C^\infty(G)-module. To see this, let X \in L(G) (not null) and consider f\in C^\infty(G) such that it vanishes only at the identity, that is: f(e)=0 but f(g) eq 0, for every g eq e. Then, Y = f X would be left-invariant, not null and yet Y_e = 0. However, this last equality requires Y to be the null vector field, which is a contradiction.
For this to work you would need to show that there exists such a function that satisfies f(e)=0 but f(g) eq 0, for every g eq e which is also smooth, one cannot just claim an existence of such a function. As a specific example if G = S^1 then it would be quite difficult to justify such a function existing.
Exactly. Simply proving the isomorphism implies that if X_e = Y_e, then X_g = Y_g, for all g. But of course if L(G) is C^\infty module, then multiplying X_g by a function f(g) that's equal to 1 at the identity e will lead to a different vector field, but one that corresponds to the same vector at the origin as X_e. This contradicts isomorphism.
36:24 regarding the explanation about push-forward and pull-back in this part of the video, shouldn't the smooth map we use to define the pull-back also be at least injective? Reason being that if 2 ponts in T*M can get to the same point, say n, in T*N, when we construct h*n we must cover the entire T*M, but at the same time we can not allow h*n to point to 2 points in T*M as it would no longer be a map.
I had exactly the same question you asked but then I realized that it is not necessary that h is injective. Well, to be more precise, there are two kinds of push-forward and pull-back: the "pointwise", where you map tangent or cotangent spaces at a point to tangent or cotangent spaces at another point, and then there is the "global", where you map the whole vector or covector field. The pointwise push-forward of tangent vectors does not require h to be injective, as you are mapping T_p M to T_h(p) N and there is no problem in doing so. However, if you want to push-forward a whole vector field, you need to have a tangent vector at every point of N, which is not possible if some q in N is not q=h(p) for some p in M. Now, the interesting part is that the pull-back behaves kind of opposite. Let q be in N and w a cotangent vector at q (that is, and element of T*_q N). Suppose we want to pointwise pull back w. If h is not surjective, where would you pull w back? (Note the difference for the pointwise push-forward, where given v in T_p M, you can always push it forward to T_h(p) N). Moreover, if h is not injective and q = h(p1) = h(p2), then w should be pulled back to T*_p1 M or to T*_p2 M? In this sense, it is true what you said about h having to be surjective and injective so that we can pull-back pointwise. However, if we want to pull back a whole covector field (call it w), the important part is to find at every p in M a covector h_* w which is associated with a covector at some point of N, and this is always possible as we can always define the action of h_* w on a vector v (at p of course) by the pointwise push-forward (h_* w)(v) := w(h*v) (where w is evaluated at h(p)). I know this may be confusing, but it is an important to understand these subtle details. In other words, when you want to pull-back pointwise, you start with a covector at a point q in N and want to associate it to a covector at a unique point p in M, which cannot be possible if h is not bijective. But for the "global" pull-back, you start with a point p in M and want to find a covector at this point which can be constructed from the (already available) covector field w on N, which can be done as in the previous paragraph. Note that for the "global" pull-back, we do not make use of the "pointwise" pull-back, but of the "pointwise" push-forward, which does not require h to be injective nor surjective. Only when h is bijective, both the pointwise and global perspectives coincide and one can be obtained from the other. This is an important point that I think Prof. Schuller omitted in the previous lectures. Hope this helps.
Hey comments! If, for maps in general, the push-forward of a vector field is ill-defined, how could the definition of the pull-back of the n-form make use of the push-forward on its arguments (last lecture)?
In the definition of pull back for covector fields, we only push vectors forward point-wise. Therefore we don't care whether p and q are mapped to the same point under h; we just push a vector at p forward and operate on it with the covector at h(p) and do the same for q. It doesn't matter for the well-definition of the pull back that the vector at p and the (different) vector at q are operated on by the same covector at h(p) = h(q).
The camera zooming in on the chapter number is too funny to me. It's like whoever is in charge of it is doing a double take or wants to let us know that it shouldn't be chapter 4 (but chapter 5).
Yes, they are. Every left-invariant vect. field can be entirely reconstructed from its value at the identity, $X_e$, by application of (the push-forward of) the left-translation, that is: $X_g = (L_g)_* X_e$. By its definition, $L_g$ depends smoothly on $g$ and therefore, so does $X_g$.
@@brandonwillnecker8060 I know, that's why I asked. There was a comment in the lecture that implied otherwise, I don't remember what exactly. It's a 2 year old comment.
All of modern physics uses them to some degree. From the standard model to condensed matter to optics. Also appears in the fascinating intersections between quantum field theory and geometry: both in knot theory and enumerative geometry of curves on moduli spaces. It's as fundamental and necessary as calculus. But (except for here) never taught nearly as well. A typical physics class will just shotgun the representation theory of SU(N) and SO(3,1) at you and have you muck around with commutators. You memorize terms like adjoint representation but have zero intuition for what's going on. These lectures are a god-send by making the geometric intuitions thst are otherwise buried in dense notation in grad math books clear and accessible.
Every quantum operator is a member of a Lie Algebra, the Shcrödinger equation explicitly has a solution that derivative of a group element (acting on something) = Lie algebra member (acting on something)
I get stuck on the word manifold. I am autistic and think in very concrete terms - so when I see the word manifold - I get stuck because the only manifold is an engine manifold. Does it mean surface ? Or topological surface. Smoothness means that you can keep differentiating it ?
A smooth manifold is a C^∞-manifold, i.e. an infinitely continuously differentiable manifold. And a differentiable manifold is a manifold where chart transitions are differentiable.
So, what is a Lie group??! A group is a set of elements, a blob operation, neutral and inverse elements. Lie group is a group with ... (fill in blanks). What does "equipped with ordinary diferentiable manifold" means?
The manifold part requires that each element of a Lie group changes continuously with some parameters in R^n. For example, consider the group of all operators representing the rotation in 2D space. Each element can be written as exp(-i a J_z), where J_z = -i(x d/dy - y d/dx) [d means partial derivative] and 'a' is a real number specifying the angle of rotation. This 'a' is the aforementioned parameter in R^1, and the element changes continuously with 'a'.
Not knowing much about Lie algebras already this lecture was mostly torture with abstract and empty letter constructs, even though he tries very well to make the subject understandable, and I liked most of the previous lectures. But it was just simply too abstract for my taste. In the end I understand now less about Lie algebra than I thought I did. The only time I understood what’s going on is if he does handwaving and appeals to geometric intuition. There is really very little left of that here. He could have talked much more about flows on manifolds and what you can do with Lie groups/algebra that acts on these.
Actually, I’m watching this very lecture again 4 years later, and I must say that it is very well worth it. It is still challenging but I understand it now much better, meaning that it all makes very much sense. I still have to watch it one more time to get some of the technical details clearly in my head, so I can apply it practically.