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 Apr19 awarded Notable Question Apr14 accepted Curves and tangent vectors in a manifold setting Apr14 accepted Showing $T$ equivalent to linear map Apr14 comment Non-affinely parametrized geodesics Perhaps it is so simple that I can just define $f$ to be that quantity? Apr14 asked Non-affinely parametrized geodesics Apr6 asked Showing $T$ equivalent to linear map Mar17 comment Curves and tangent vectors in a manifold setting Many thanks for your reply! I think I understand but let me apply it to a simpler case and then see if my thinking is right. Suppose we have some function $Y = f(\mathbf{x}(t))$ which I can write as $Y = f \circ \mathbf{x} \circ t$. Then $$\frac{df(\mathbf{x}(t))}{dt} = \sum_{i=1}^n \left(\frac{\partial f}{\partial x^i}\right)|_{\mathbf{x}(t)} \frac{d x^i(t)}{dt} = \sum_{i=1}^n \left(\frac{\partial f(\mathbf{x}(t))}{\partial x^i(t)}\right) \frac{d x^i(t)}{dt}?$$ Mar17 revised Curves and tangent vectors in a manifold setting edited body Mar17 asked Curves and tangent vectors in a manifold setting Mar14 awarded Notable Question Mar12 comment Deriving Ricci identity for co-vector fields Just a quick comment here, in the above you write $2\nabla_{[a} \nabla_{b]} (X^c \lambda_c) = 0\,\,\,(1)$ and then say when we apply this to the third display we get a term $$2(\nabla_{[a}\nabla_{b]}X^c)\lambda_c\,\,\,(2)$$ If the brackets in $(1)$ mean that the object $\nabla_{[a}\nabla_{b]}$ is acting on $X^c \lambda_c$ then why don't we write $(2)$ as $$2\nabla_{[a}\nabla_{b]}(X^c)\lambda_c\,\,\,(2)$$? This is more of a question of trying to understand the notation here I think about when something is acting on something or when it denotes components. Thanks :) Mar12 comment Connections and Ricci identity Ah so you mean $\nabla_c f$ is notation for the $c$th component of $\nabla f$ not $\nabla_c$ acting on $f$ which would be denoted by $\nabla_c (f) = e_c(f)$? Mar12 accepted Deriving Ricci identity for co-vector fields Mar12 comment Deriving Ricci identity for co-vector fields Thanks for this method :) I forgot to say though that I managed to obtain the result from my last display in my post. Just as I said, by interchanging the indices on the pair $\nabla_a \nabla_b$. Mar11 comment Connections and Ricci identity Thanks! so when I do that I get $R^a_{[bcd]}\nabla_a f=0$ and since $f$ was arbritary, I get the result. Is there anything in particular wrong with what I had done in my first attempt, where I did more work with the christoffel symbols? Mar11 comment Connections and Ricci identity So I can write $$2\nabla_{|a}\nabla_{b|} \lambda_c = -R^d_{\,\,cab}\lambda_d \Rightarrow 2\nabla_{|a}\nabla_{b|} \nabla_c f = -R^d_{\,\,cab} \nabla_d f$$ I could then take another covariant derivative wrt $e$ but I am not sure how to progress really. I think the aim is to show $R^{d}{\,\,cab}$ has bianchi symmetry in the indices $c,a,b$. I am new to this material so please bear with me :) Mar10 asked Connections and Ricci identity Mar10 asked Deriving Ricci identity for co-vector fields Mar9 accepted Vanishing of the Riemann tensor Mar8 comment Vanishing of the Riemann tensor Many thanks! The only symmetry I have seen is that $R^{i}_{\,jkl} = -R^i_{\,jlk},$ which looks like will reduce the number of component computations to $2^4/2!$