Recent content by Vanille
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Undergrad Confusion about Dual Basis Vectors: Why are these two relationships equal?
\textbf{e}^i are the reciprocal basis. Ok, now I've understood my mistake; the key is to see the scalar product between a vector and a covector as a tensor of rank (1,1), that's the application of a covector to a vector or vice versa; while the scalar product between two vectors as a tensor of...- Vanille
- Post #11
- Forum: Special and General Relativity
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Undergrad Confusion about Dual Basis Vectors: Why are these two relationships equal?
Yes sorry, my bad! i was focused on dual basis and i missed the partial derivative of the vector component! Also, you're right again; it's better using the notation of the affine connection \nabla. So am i right? i think there's an abuse of notation. For the first expression...- Vanille
- Post #9
- Forum: Special and General Relativity
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Undergrad Confusion about Dual Basis Vectors: Why are these two relationships equal?
right,so i think that my mistake is an abuse of notation... in fact in the first case the operator between \omega^i and \textbf{e}_j is a bilinear operator on \textbf{V}^*\times \textbf{V} whose properties resemble the properties of an inner product. In the second case, i use the inner product...- Vanille
- Post #3
- Forum: Special and General Relativity
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Undergrad Confusion about Dual Basis Vectors: Why are these two relationships equal?
Hello all! I've just started to study general relativity and I'm a bit confused about dual basis vectors. If we have a vector space \textbf{V} and a basis \{\textbf{e}_i\}, I can define a dual basis \{\omega^i\} in \textbf{V}^* such that: \omega^i(\textbf{e}_j) = \delta^i_j But in some pdf and...- Vanille
- Thread
- Basis Dual Dual basis Geometry Linear algebra Tensor algebra Vector analysis
- Replies: 14
- Forum: Special and General Relativity