So if anyone was wondering about the proof at 16:07. The key point is that d\omega is a TENSOR. Therefore, d\omega(X,Y) depends only on the values of X and Y at the point you evaluate at (say it is p\in P). Saying this again, we can smoothly change X,Y at all points but p and it WILL NOT change the value of d\omega(X,Y) at p. Now, the most general vertical vector field -- X -- will be generated by different Lie algebra elements at different points. Suppose X at p is generated by A (\in the Lie algebra). From the argument before, we can smoothly change the value of X everywhere but at p, so that it equals the vector field generated by A everywhere, and this will not change the value of d\omega(X,Y). This is what gives us the 'constant' vertical fields considered in the proof.
Excellent lectures! I've already revisited several of the lectures, several times. I have a quick question/doubt regarding connections and curvature and torsion of connections. It is a well known result that if a connection abla on a smooth manifold M is torsion free and has everywhere vanishing (Riemann) curvature, then for every m\in M there is a local chart (U,x), with m\in U, such that the Chistoffel symbols \Gamma^i_{jk} of abla all vanish over U. On the other hand, in a principal G-bundle P (with base manifold M) with a connection 1-form \omega, it holds that if D\omega = 0 everywhere on P, then for every m\in M, there is a local section \sigma:U\to P (with m\in U) such that the Yang-Mills field \sigma^*\omega vanishes everywhere on U. There is no mention in here of the Torsion, it could even not being defined at all. I guess in the first case, the stronger requirement of the torsion to vanish comes from requiring the particular section into the frame bundle making the Christoffel symbols vanishing, be a coordinate induced section.
A follow up of my own comment: In the context of the frame bundle of a smooth manifold M with a connection 1-form \omega having vanishing curvature, let \sigma:U -> LM be a section such that \sigma^*\omega = 0. The section \sigma can be viewed as n (=dimM) smooth vector fields over U being C^\infty(M)-linearly independent. On the other hand, for fixed (a^1, ..., a^n) \in R^n, define the GL(n,R)-equivariant function over LU, F: LU -> R^n as F(\sigma(m) g) := g^{-1} (a^1, ..., a^n). This function corresponds to a section on the associated tangent bundle \tau: U -> TU given by \tau(m):= a^i \sigma(m)_i, where \sigma(m)_i denotes the i-th vector of the frame \sigma(m). In addition, \sigma^*DF = d\sigma^*F + \sigma^*\omega F=0, because \sigma^*F(m) = F(\sigma(m)) = (a^1, ..., a^n). Now, this vanishing of the covariant exterior derivative implies that the vector field X given by X_m:= a^i \sigma(m)_i is parallelly transported along every (smooth) curve in U, or equivalently, abla_Y X = 0 for every vector field over U. Since this holds for every constant tuple (a^1, ..., a^n), it follows in particular that abla_{\sigma_i} \sigma_j = 0 for every pair of indices (i,j). From this, the torsion is given by T(\sigma_i,\sigma_j) = abla_{\sigma_i} \sigma_j - abla_{\sigma_j} \sigma_i - [\sigma_i , \sigma_j] = [\sigma_j , \sigma_i]. By Frobenius theorem, this shows explicitly that the (non-vanishing of the) torsion is the obstruction for any section rendering a vanishing Yang-Mills field from the connection 1-form, to derive from a coordinate system.
So if anyone was wondering what is happening at 21:30 -- how does a vector X act on a Lie algebra valued function B??? -- the answer is that it is missing an ingredient. The way to correctly do it, according to Wikipedia en.wikipedia.org/wiki/Vector-valued_differential_form, is to define (d\omega)(X,Y):=(d\omega^{a})(X,Y)e_a where e_a is a basis of the Lie algebra, and where \omega^a is a REAL VALUED 1-form that is defined as (\omega^a)(X):=(\epsilon^a)(\omega(X)) where \epsilon^a is the standard DUAL basis to e_a. Since \omega^a is a real valued 1-form, the exterior derivative d\omega^a is completely OK. One can check that this definition of d\omega is independent of the basis e_a that is chosen. Then the statement that (X^{A})(\omega(X^{B}))=0 mentioned at 21:30 should actually read: ( (X^{A})((\omega^a)(X^B)) )e_a. Now the action of X^A is on a real valued function, which is pacifying -- and the reason it is zero is because (\omega^a)(X^B):=(\epsilon^a)(\omega(X^B))=(\epsilon^a)(B) is a real valued CONSTANT function on P.
I think that actually what he said is correct. It's just a substitution of the word "number" for the word "Lie algebra element". I mean, when we define at a point p an ordinary 1-form w and a vector X, we have: w(X)=a, where a is a number. Now, in this case, we say at a point p: w(X)=A where A is a Lie algebra element. So, if we are dealing with a section in Gamma(TP), when we go from point to point, we are goint from vector to vector. Thus, from the definition of the connection, we actually have a Lie algebra valued function as you said (defined on P). To fully understand the situation, we should go to the formal definition of directional derivative, but maybe it is too much extending... PD: in any case, I think that you wrote missleadingly: " (d\omega)(X,Y):=(d\omega^{a})(X,Y)e_a ". I guess it should be: (d\omega)(X,Y):=(d\omega)_{ab}(X,Y)\epsilon^{a}\epsilon^{b} since it is a 2-form.
Ok - so in a proof one has to wonder -> When both sides are proven equal because they come to the same value as in A=A that seems cool, but I always feel suspect in the case where it is 0=0 where we are showing both sides are null. In the A=A case A-A=0, but in the 0=0 or null case well 0 already equals 0 so there is no 0-0 = 0 I am no mathematician, but is just feels sort of like we are missing something by just proving both sides of the equation are null. [Great lectures!]
54:40 How can we have that TM is isomorphic to P_V as associated bundles? Are we considering TM as the associated bundle correspoinding to fiber R^d and principal GL(d)-bundle LM? If so then because associated bundle morphisms can only be defined between two associated G-bundles (that is, with the same underlying Lie group) shouldn't we require P to be a principal GL(d)-bundle?
Good point. I have only ever seen this isomorphism defined as an isomorphism of associated vector bundles. However, we can define a principal bundle map between a principal G bundle and a GL(V) bundle by replacing the equivariance condition with u(p\lhd g) = u(p)\lhd ho(g) where ho is the representation of G. Then can we define an associated bundle map between principal bundles with different Lie groups by \tilde{u}([p,v]) = [[u(p), v]] , where [p,v] = [p',v'] when p' = p\lhd g and v' = g^{-1} hd v for some g\in G, and [[a, v]] = [[a', v']] whenever a'=a\lhd ho(g) and v' = ho(g^{-1}) hd v for some ho(g)\in GL(V)?
At 56:30 , shouldn't the soldering form go from the sections of TP to the set of smooth functions from LM to R^d, since the map theta_e maps a vector at the point e on LM to a point in R^d? If think that's also what the definition says, theta is a V-valued 1-form on P and therefore should map a vector field on P to a smooth function on P.
@Wizardking I think it depends on the structure of the manifold, It is always a GL bunle but if there is a Riemannian metric it can be made a O(n) bundle, if the mfd is orientable even SO(n), or U and SU for complex mfds and maybe Sp for syplectic
Actually that makes a ton of sense because in the definition given there (which roughly corresponds to Herr Schuller's point-wise definition) it uses the differential of the projection map, which corresponds to the push-forward of the projection map.
yes! say that you are considering the 2-sphere S^2 (pictured in R^3) as the base manifold and its frame bundle L(S^2) (which is nontrivial) as the principal bundle. the pullback of the curvature on L(S^2) to S^2 is the Riemann tensor with its 4 components. If you contract components of the Riemann tensor then you will get the 2-component Ricci tensor R_{ij}. This is the familiar curvature in which you feed two vectors at a point and it gives you a number to characterise how curved it is. as schuller was saying, this is just how you get the 'field strength' by pulling back the curvature on the overall principle bundle.
@@kashu7691 Thanks for your long comments. It does make sense to me now and it is really helpful. Although I don’t fully understand, I will learn this soon.
I love his lectures, but man he needs to learn what a lie algebra valued differential form actually is. For those who don't know, it maps, not to the lie algebra g (in this case T_eG), but to the space of sections of the bundle P x g, where g is the lie algebra.
I am still confused about this. He said any vertical vector field was of the form X^A in this lecture which would make it a map into TeG since the sections would just have constant value A on the fiber. However I don't see why this is true. One could have a smooth map A(p) instead of A and still have that the induced vector field Z_p=X^(A(p))_p is a smooth vertical vector field.
@@vivalibertasergovivitelibe4111 perhaps look into a good principal bundle book like Tu’s Differential Geometry : connections, curvature, and characteristic classes
