Mathematics Asked on January 1, 2022
I don’t know why contravariant and covariant vector field are named as such. contravariant literally means going against changing, or changing in the opposite way, covariant literally means changing with something in the same way.
For example, vector field and its dual are (in classical terminology) seems contravariant and covariant vector field of the first order. I don’t see how vector field and its dual bear the literal meanings of the two words. Wiki says that they have something to do with their relations to basis vector, but I still can’t figure it out.
Is there any reason or historical explanation for that naming?
—
I see so
covariant vector is an element in dual space, it’s transformation of or action on vector, but (for first order tensor this transformation of vector is inner product so) it can also be regarded as a transposed vector, and visualized in the same space (but it’s not in the same space) as the upenn link does. The key is to recognize the covariant vector in the above $e^j e_i$ Is not ‘vector‘ $e^j$ but $e^j$ as a transformation or transposed vector (but we need to realize that only when it act on a vector it’s covariant vector).
there are two dual: Space of covariant vector, dual space. Its basis can be any proper set of covariant vectors; dual vector, the special basis of dual space $e^j$ that satisfy that its action on basis of vector space produces $delta_{i, j}$.
formally contravariant means inverse of basis transformation matrix: after changing basis by transformation matrix A, the same vector (=components * basis) equals (components transformed by inverse transformation $A^{-1}$ times $A$ transformed basis, i.e. the new basis $e’_i$ ), the result is easier to understand from perspective of matrix operation as insertion of $A^{-1}A$. For dual basis it’s similar. Transformation $B$ of dual basis that satisfies ‘duality’ requirement needs to be inverse $A^{-1}$ of A too,—that is $delta_{i, j}= e^j e_i = e’^j e’_i = (e^j B)(A e_i)$, where $e’^j , e’_i$ are transformed basis, so $B= A^{-1}$—so transformation of components of dual basis (the components can be easily understood if we view covariant vector as a transposed vector) seems to be just $A$, in this sense it’s covariant.
The notions "contravariant" and "covariant" originally apply to vectors in a vector space $V$ and its dual $V^*$. This transfers of course to vector fields on a manifold $M$. A contravariant vector field is a section $M to TM$ of the tangent bundle, a covariant vector field is a section $M to T^*M$ of the cotangent bundle.
Now see my answer to Why is tensor from a vector space covariant, not contravariant?
Answered by Paul Frost on January 1, 2022
3 Asked on October 15, 2020 by simplex1
1 Asked on October 15, 2020 by user830531
1 Asked on October 14, 2020 by daron
0 Asked on October 14, 2020 by saul-rojas
algebra precalculus rational numbers rationality testing real analysis
0 Asked on October 14, 2020 by oddly-asymmetric
4 Asked on October 13, 2020 by dhruv-agarwal
4 Asked on October 13, 2020 by doctor-reality
0 Asked on October 11, 2020 by dfnu
1 Asked on October 10, 2020 by thomasmart
1 Asked on October 10, 2020 by ricky_nelson
compactness proof explanation real analysis solution verification uniform continuity
1 Asked on October 9, 2020 by subbota
1 Asked on October 9, 2020 by michael-morrow
2 Asked on October 8, 2020
functional analysis grassmannian hilbert spaces mathematical physics optimization
1 Asked on October 8, 2020 by nx37b
1 Asked on October 8, 2020 by fdez
lebesgue integral lebesgue measure measure theory riemann integration
0 Asked on October 7, 2020 by aspiringmathematician
1 Asked on October 7, 2020 by ashids
3 Asked on October 6, 2020 by rosita
0 Asked on October 5, 2020 by emre-yolcu
Get help from others!
Recent Questions
Recent Answers
© 2023 AnswerBun.com. All rights reserved. Sites we Love: PCI Database, MenuIva, UKBizDB, Menu Kuliner, Sharing RPP, SolveDir