Tensor Calculus 3a: The Covariant Basis
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange users link it when asking how a coordinate system's basis vectors come from differentiating a position vector function.
Recommended by 1 people across 1 different months, 2015 to 2022.
What it is
A university-style lecture from Pavel Grinfeld's Tensor Calculus course that introduces the covariant basis, obtained by differentiating the position vector with respect to curvilinear coordinates. It is part of a series on tensor calculus tied to his textbook.
What people said
Recommends Clemens Bartholdy · 2022-03 · Mathematics Stack Exchange · on “The axiomatic approach to covariant derivative vs seeing it as derivative of a parameterized position vector”
In this lecture [LINK] by Frederic schuller, he introduces the the covariant derivative axiomatically: A connection $\nabla$ on a smooth manifold $(M,O,A)$ is a map that takes a pair consisting of a vector (field) $X$ and a $(p,q)$ - Tensor field $T$ and sends them to a $(p,q)$ tensor field $\nabla_X T$ following these axioms: For a scalar function $\nabla_X f = Xf$ where $Xf$ denotes the directioncal derivative of $f$ in the direction of $X$ $\nabla_X (T+S) =\nabla_X T + \nabla_X S$ $\nabla_X T( \omega,Y) = (\nabla_X T)(\omega,Y) + T(\nabla_X \omega, Y) + T(\omega, \nabla_X Y)$ (Leibniz law) $\nabla_{fX+Z} T = f \nabla_X T+ \nabla_Z T$ Later in the lecture prof. Schuller remarks that for th
Cites Barkimedes · 2015-12 · Mathematics Stack Exchange · on “How can vector functions define coordinate systems?”
I was watching a lecture [this video] in which the professor states that by taking the derivative of a vector function, one can find the basis of the coordinate system being studied. My question is, how can a vector function define a coordinate system? Does it not need the coordinate system in the first place so that we can study it? I know vectors are not the numerical representations we assign to them, they are geometric entities of their own, and a vector valued function goes around every point in a vector space and assigns it a vector, but how can this function work if its arguments are described by the coordinate system? What is the explanation that makes this seemingly circular logic go awa
Cites Chill2Macht · 2016-10 · Mathematics Stack Exchange · on “Are coordinate systems just rank 1 covariant tensor fields?”
Question: Given, the Euclidean plane $\mathcal{P}$ , with the point $O \in \mathcal{P}$ (called "the origin"). 1. Is a Cartesian coordinate system for this plane centered at $O$ just an orthonormal basis for the cotangent space of the Euclidean point at $O$ , i.e. an orthonormal basis for $(T_O \mathcal{P})^*$ ? 2. Are vectors with tail at $O$ in the Euclidean plane $\mathcal{P}$ just elements of the tangent space at $O$ , i.e. $v \in T_O \mathcal{P}$ ? And is a vector $w$ with tail at point $A \in \mathcal{P}$ just an element $w \in T_A \mathcal{P}$ ? 3. That any point can be given a unique coordinate is just a reflection of the fact that $$\mathcal{P} \simeq (T_O\mathcal{P})^* $$ as vector