Timeline for Why Is the Dual Basis Mathematically Unavoidable?
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| 22 hours ago | answer | added | Man-I-Fold | timeline score: 0 | |
| 2 days ago | answer | added | Nicolas Malebranche | timeline score: 1 | |
| Nov 20 at 15:14 | comment | added | Lee Mosher | One quick summary to the (very good) comments and answers so far: you write "... if we already have a basis for a vector space ...", but that is a very BIG $\Large{IF}$. | |
| Nov 19 at 17:10 | comment | added | Filip Milovanović | See also this answer to a similar question, it's well-written and I think it might provide some insight. | |
| Nov 19 at 14:44 | history | became hot network question | |||
| Nov 19 at 14:40 | history | reopened |
Mikhail Katz Harish Chandra Rajpoot T_T Daniele Tampieri Dominique |
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| Nov 19 at 8:43 | review | Reopen votes | |||
| Nov 19 at 14:40 | |||||
| Nov 18 at 23:42 | history | closed |
Ben Steffan Anne Bauval Dietrich Burde Sebastiano pyridoxal_trigeminus |
Opinion-based | |
| Nov 18 at 23:27 | answer | added | Andreas Blass | timeline score: 10 | |
| Nov 18 at 21:13 | answer | added | Felix Benning | timeline score: 6 | |
| Nov 18 at 20:57 | comment | added | Malady | While for finite dimensional vector spaces, a space is isomorphic to its dual, for finite dimensional vector bundles or infinite dimensional vector spaces, this no longer holds. | |
| Nov 18 at 20:12 | comment | added | Eric Towers | Can be difficult to understand in (locally) flat spaces. In Misner, Wheeler, and Thorne's Gravitation, the duals are introduced to give local coordinates on non-flat space because parallel transport "along vectors" doesn't work the way your intuition on flat spaces makes you expect it to. | |
| Nov 18 at 19:55 | answer | added | KCd | timeline score: 20 | |
| Nov 18 at 19:27 | comment | added | Deane | The need for a dual vector space and a dual basis arises even in physics, because the gradient of a scalar function does not transform, under a change of coordinates, like a vector. Physicists noticed this and call the gradient a "pseudovector". However, if you define the gradient of a function as a dual vector, then all of a sudden everything becomes easy and natural. | |
| Nov 18 at 19:18 | comment | added | Deane | It probably is possible to avoid defining and using the dual vector space. Just as it is possible to do abstract linear algebra using only $\mathbb{R}^n$ and not defining an abstract vector space. However, this makes everything unnecessarily complicated. An important theme in modern math is, given a natural collection of spaces (such as vector spaces or manifolds), the spaces of scalar functions on a space as well as the space of maps from one space to another are fundamental and powerful concepts. The dual vector space is simply an example of this. | |
| Nov 18 at 19:18 | comment | added | Brian Moehring | I'm not even convinced "mathematically unavoidable" is distinct from "convenient" for a human pursuit of mathematics. | |
| Nov 18 at 19:17 | review | Close votes | |||
| Nov 18 at 23:43 | |||||
| Nov 18 at 18:55 | comment | added | Ted Shifrin | Suppose your vector space has an inner product. You want to compute this inner product in terms of coordinates coming from a non-orthonormal basis. | |
| Nov 18 at 18:52 | history | asked | Aurora Borealis | CC BY-SA 4.0 |