Lecture 7: Connections (International Winter School on Gravity and Light 2015)

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Strong rec Dale · 2024-12 · Physics Stack Exchange · on “How can I make the notion that a basis is fixed in time with respect to itself more precise?”

Here is the correct wikipedia: [LINK] (mathematics) And I really like Prof. Schuller's lecture series, here is the one where he introduces connections [this video] (the whole series is well done)

Recommends Shirish · 2020-07 · Mathematics Stack Exchange · on “Standard version of covariant derivative properties”

[Throughout we're considering the intrinsic version of the covariant derivative. The extrinsic version isn't of any concern.] I'm having trouble reconciling different versions of the properties to be satisfied by the covariant derivative. Essentially $\nabla$ sends $(p,q)$ -tensors to $(p,q+1)$ -tensors. I'll write down the required properties for $\nabla$ from the two sources. This lecture (relevant timestamp linked) [this video] If $X$ is a vector field, $\nabla_Xf=Xf$ , for a scalar field $f$ $\nabla_X(T+S)=\nabla_XT+\nabla_XS$ $\nabla_X(T(\omega,Y))=(\nabla_XT)(\omega,Y)+T(\nabla_X\omega,Y)+T(\omega,\nabla_XY)$ $\nabla_{fX+Z}\ T=f\nabla_XT+\nabla_ZT$ Core principles of special and general rel

Recommends Shirish · 2020-07 · Physics Stack Exchange · on “Problem with extending a vector field's action on another vector field”

I'm following Schuller's lectures on gravity and light (this question specifically concerns this part of the video [this video] ). Let $M$ be a smooth manifold with a smooth vector field $X$ , and let $f\in C^{\infty}(M)$ be a smooth function on $M$ . $X$ is a $C^{\infty}(M)\to C^{\infty}(M)$ map so that $f\mapsto Xf$ , where $(Xf)(p)$ is the directional derivative of $f$ in the direction of $X(p)$ . This much is clear. The lecturer indicates that the covariant derivative is introduced to extend the notion of derivatives of tensor fields w.r.t. a vector field, whereas $X$ only covers the notion of derivatives of scalar fields w.r.t. a vector field. So I tried to consider how $XY^j$ would look lik

Recommends Shirish · 2020-08 · Mathematics Stack Exchange · on “Calculation of covariant derivative being chart-dependent”

Ref. Schuller's lecture 7 on gravity and light ( derivation starts at this timestamp [this video] ). I'm watching a lecture that introduces connection coefficients - specifically the part of the video showing for the first time how to calculate covariant derivative. Relevant timestamp here [this video] , the calculation is being done for $\nabla_XY$ in the particular chart $(U,x)$ . In short, it goes like this (let $X,Y$ be smooth vector fields on a smooth manifold $M$ ): $$\nabla_XY=X^i\nabla_i(Y^m\partial_m)=X^i(\nabla_iY^m)\partial_m+X^iY^m(\nabla_i\partial_m)$$ Regarding the $\nabla_i\partial_m$ term, the lecturer says: What is the result? It's a vector field because this ( $\partial_m$ ) is a vec

Recommends suncup224 · 2020-09 · Mathematics Stack Exchange · on “Equivalence of definitions for covariant derivative”

I understand that there seems to be many similar sounding questions on stackexchange but none of them seem to be able to address my confusion. First of all, I am using the definition of covariant derivative found here [this video] (e.g. 21 min). Specifically, define $\nabla$ to be a map taking a smooth vector field $X$ and $(p,q)$ -tensor $T$ to another $(p,q)$ -tensor $\nabla_XT$ such that the following 4 axioms hold: $\nabla_Xf = Xf$ for all smooth functions $f\in C^{\infty}(M)$ $\nabla_X(T+S) = \nabla_XT+\nabla_XS$ for tensors of the same ranks $\nabla_X (T(\omega,\dots,Y,\dots))=(\nabla_XT)(\omega,\dots,Y,\dots) + T(\nabla_X\omega,\dots,Y,\dots)+\dots+T(\omega,\dots,\nabla_XY,\dots)+\dots$ $\

Recommends Clemens Bartholdy · 2021-05 · Mathematics Stack Exchange · on “What kind of connection is the directional derivative?”

Given a scalar function $f(X)$ as a function of coordinates I can write it's directional derivative in direction $\hat{u}$ as: $$ D_{\hat{u} } f = \nabla f \cdot \hat{u}$$ This is pretty easy to think about and understand. I recently learned about the concept of connections which is supposed to 'abstract' the idea of directional derivatives from $(0,0)$ tensor fields (scalar fields to a general $(p,q)$ tensor field. Here's the definition I learned: A connection $\nabla$ on a smooth manifold $(M,\theta,A)$ is a map that takes a pair consisting of a vector field and a $(p,q)$ tensor field $T $ and sends that to a $(p,q)$ tensor field $\nabla_x T$ From 20:15 [this video] From what I understand, ther

Recommends Clemens Bartholdy · 2022-01 · Mathematics Stack Exchange · on “How do you reconstruct the surface knowing the connection coefficients?”

In this lec-7 on Connections by Frederic Schuller, the connection is introduced as an extension of the directional derivative of scalar fields to $(p,q)$ tensor field. The initial axioms the connection obeys is same as all that which the directional derivative for scalar field obeys. About 35:07 [this video] , he says that the remaining freedom of how the directional derivative generalizes is determined by the Christoffel symbols At 6:30 of lec-8 [LINK] , it is said that the choice of christoffel symbols , fixes our how our manifold looks in space $(M,O , A, \nabla)$ . The point I am confused is at 46:26 [LINK] , he concludes the Christoffels of the regular round sphere are given as: $$ \Gamma_{2

Recommends LGin94 · 2022-04 · Mathematics Stack Exchange · on “Inconsistency in the definition of the connection coefficients”

I am new to general relativity and I am currently facing an apparent inconsistency in the definition of the connection coefficients. Some references I've been consulting (e.g. the lecture notes by S. Carroll or those by D. Tong) define them as follows: $$ \nabla_{\partial_i}\partial_j=\Gamma^k_{ij}\partial_k $$ where $\{\partial_i\}$ are the chart-induced basis vector fields. Some other references, however (e.g. these beautiful lectures [this video] by prof. Schuller) define them with the lower indices swapped: $$ \nabla_{\partial_i}\partial_j=\Gamma^k_{ji}\partial_k $$ I know that for a torsion-free connection the coefficients are symmetric in the lower indices, so it doesn't really matter which

Recommends Clemens Bartholdy · 2022-05 · Mathematics Stack Exchange · on “Confusion on the intuition of a connection”

At 6:27 of this lecture [LINK] by Frederic Schuller, it is remarked that only when we fix the connection (by providing Christoffel's) that we get the specific shape of the sphere that we usually imagine, and, perhaps a choice of a different connection would have ended up in a sphere: [LINK] This is all good and fine. But then I saw this video by Eigen Chris [LINK] , and 17:55 [LINK] of it, he says that one can set all the Christoffels while on the sphere to zero to get another connection named the boring connection: [LINK] Now this is contradictory with the Schuller lecture's to me because of two points: In Schuller's lectures, it is found that the connection fixes the shape, but here Eigen

Recommends Ben H · 2024-02 · Physics Stack Exchange · on “How to find the double covariant derivative of a general vector?”

I find it very hard to disentangle such expressions without writing out the tensors completely. See this lecture by Frederic Schuller [this video] to see how to take (and define) the covariant derivative. Using the method specified by Schuller, we can write out each term much more explicitly. For example, the first term on the right-hand side is: \begin{equation}\begin{split} \partial_\mu \left( \nabla_\nu V^{\rho} \right) &= \frac{\partial}{\partial x^{\mu}} \left[ \left( \nabla_\nu V \right)^\rho \right]\\ &= \frac{\partial}{\partial x^{\mu}} \left[ dx^\rho \left( \nabla_{\frac{\partial}{\partial x^\nu}} V \right) \right]\\ &= \frac{\partial}{\partial x^{\mu}} \left\{ dx^\rho \left[ \nabla_{\fr