Timeline for Invariant characterization of isometric embeddings
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 22, 2016 at 18:49 | vote | accept | Asaf Shachar | ||
| Sep 7, 2016 at 5:39 | answer | added | darij grinberg | timeline score: 1 | |
| Sep 6, 2016 at 9:05 | comment | added | Asaf Shachar | @darijgrinberg I forgot to direct my previous comment to you. I have added details regarding your question (You can see the comments above, and the modified question). Would you care to write your computations? (I only succeeded to prove identiy $(*)$ and got stuck in demonstrating $(**)$). | |
| Sep 5, 2016 at 22:30 | comment | added | Asaf Shachar | see in particular conditions $(3),(3)'$. Of course, the orientations themselves are not supposed to play a truly fundamental role, when considering embeddings of a space in a higher dimensional space, since then there is no meaning to the word "orientation-preserving" embedding. Perhaps this whole discussion can be done without explicit reference to them, but they certainly seem useful as auxiliary objects. | |
| Sep 5, 2016 at 22:27 | comment | added | Asaf Shachar | Good question. I have an idea, but I am not entirely sure yet. The point is that in the case of $\dim V=\dim W$, the identity $(*)$ immediately implies that $A$ is an isometry if and only if the condition $ \Det A \neq 0,\Det A \cdot A= \Cof A$ is satisfied. I came up with an analogous condition in the case where the dimensions are different. To prove my generalized condition works, I need an analogous identity to replace $(*) \, \Cof A \circ A^T= \Det A \cdot \operatorname{Id}_W$. This generalized identity is what I was asking about. I have added many details explaining my goals exactly, | |
| Sep 5, 2016 at 22:23 | history | edited | Asaf Shachar | CC BY-SA 3.0 |
Direction of the question changed, I added lots of motivation and details, and clarified the goals and the main question marks left.
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| Sep 5, 2016 at 20:19 | comment | added | darij grinberg | If my (fairly simple) computations are true, then $(**)$ holds always (with signs chosen appropriately). What does this have to do with isometric embeddings and orientations? | |
| Sep 5, 2016 at 13:18 | history | edited | Asaf Shachar | CC BY-SA 3.0 |
Motivation for the question was added.
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| Sep 4, 2016 at 3:43 | comment | added | Allen Knutson | Annoying point, irrelevant to the question above: your definition of orientation would put only one (not two) on 0-dimensional spaces. A better definition is a continuous function from bases $\to \{+,-\}$ satisfying obvious conditions. Note that one if $A = V\oplus W$ then one wants orientations on any two to determine one on the third, even if $W$ is 0-dimensional. | |
| Sep 3, 2016 at 19:58 | comment | added | Asaf Shachar | An orientation on a vector space $V$ is a choice of equivalence class of bases for $V$, where two bases are said to be equivalent if the transition matrix between them has a positive determinant. (There are other equivalent definitons -en.wikipedia.org/wiki/Orientation_(vector_space)). An oriented vector space is a space together with a specific orientation on it. (There are only 2 possible options). | |
| Sep 3, 2016 at 19:40 | comment | added | David Handelman | What's an oriented inner product space? | |
| Sep 3, 2016 at 14:33 | history | asked | Asaf Shachar | CC BY-SA 3.0 |