Tensor Calculus 6b: The Covariant Derivative

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Recommends Clemens Bartholdy · 2022-05 · Mathematics Stack Exchange · on “Why do we not need Christoffel to differentiate contravariant component of a vector?”

I figured it out now. The answer was in this video [this video] . Firstly, the most important thing to understand is the way the Covariant derivative acts on a vector and the way the derivative with coordinate acts is different. In the simplified case when were everything is calculated in $\mathbb{R^3}$ , calculating the rate of change of the vector by the partials and by that of the connection is the same. In the calculation I have shown above, we can the Christoffel popped out of the partial of the basis. Suppose we instead hit the vector with a connection: $$ \nabla_{\alpha} (V) = \nabla_{\alpha} (V^i \vec{ Z_i}) = \nabla_{\alpha}(V^i) \vec{Z_i} + V^i \nabla_{\alpha} (\vec{Z_i})$$ Now the inte

Cites vb628 · 2020-09 · Mathematics Stack Exchange · on “Covariant Derivative of Basis Vectors”

Consider the standard covariant derivative of Riemannian Geometry (torsion free with metric compatibility) in the $\frac{\partial}{\partial x^i}$ direction. Application to a vector field will be denoted $\nabla_i \vec{v} $ . For the purposes of this question, I will restrict myself to flat space (namely the plane). Many introductory sources initially define the Christoffel Symbols by the relationship $$\frac{\partial \vec{\mathbf{e_i}}}{\partial x^j}=\Gamma^k_{ij}\vec{\mathbf{e_k}}$$ where $\vec{\mathbf{e_i}} = \frac{\partial}{\partial x^i}$ . The covariant derivative is then derived quite simply for contravariant and covariant vector fields as being $$\nabla_i \vec{v} =\bigg( \frac{\partial

Cites glS · 2021-09 · Mathematics Stack Exchange · on “Is there always a connection such that $\nabla_i\mathbf e_j=0$ for some basis $\{\mathbf e_i\}_i$?”

Consider a smooth manifold $M$ , with a local basis $\{\mathbf e_i\}_{i=1}^n$ . Can a connection $\nabla$ such that $\nabla_i\mathbf e_j=0$ for all $i,j$ always be defined? For context, the question comes from watching some lecture videos on youtube ( e.g. this one [this video] ) explaining tensor calculus notation. In there, the speaker defines the covariant derivative in such a way that $\nabla_i\mathbf e_j=0$ for all $i,j$ , so that we have $$\partial_i \mathbf V = (\partial_i V^k + \Gamma^k_{~ij}V^j)\mathbf e_k \equiv (\nabla_i V^k)\mathbf e_k, \qquad \mathbf V\equiv V^j\mathbf e_j,$$ defining $\nabla_i V^k\equiv \partial_i V^k + \Gamma^k_{~ij}V^j$ . I think the expression $\nabla_i V^k$ can