Timeline for Does this formula have a rigorous meaning, or is it merely formal?
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Feb 3, 2011 at 18:17 | comment | added | Deane Yang | Let me restate Greg's last sentence in plain English: If you demand that exactly one row of vectors in $|\cdots|$, then $|\cdot|$ is well-defined on a standard vector space and nothing new is needed. If, however, you want to allow more than one row of vectors, then $|\cdot|$ is still well-defined but takes values in the exterior algebra. | |
| Feb 3, 2011 at 7:19 | vote | accept | Dick Palais | ||
| Feb 3, 2011 at 7:17 | comment | added | Harry Gindi | @Mariano: Yes, but it's also got lots of extensions! That's what my comment was about. | |
| Feb 3, 2011 at 6:55 | comment | added | Mariano Suárez-Álvarez | @Harry, it is just the dual numbers. The idea goes back, more or less, to Lagrange and friends! | |
| Feb 3, 2011 at 6:35 | history | edited | Greg Kuperberg | CC BY-SA 2.5 |
added 1783 characters in body; added 2 characters in body
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| Feb 3, 2011 at 4:34 | comment | added | Harry Gindi | This is a beautiful construction and is the basis for a very powerful construction of Quillen called the stabilization, or the tangent category (by others. I think Quillen also called it the "abelianization". | |
| Feb 3, 2011 at 0:31 | comment | added | François G. Dorais | Theo: I think the point of the question is to expand on should... | |
| Feb 3, 2011 at 0:11 | comment | added | Theo Johnson-Freyd | Even better, one should work over $\Lambda^\bullet V = $ the free commutative (in the super sense) ring generated by V in degree 1. | |
| Feb 2, 2011 at 21:53 | comment | added | Dick Palais | @Sergei: ??? Sorry to be so dense, but could you expand on your comment a bit. I'm not sure what you mean by "the same dot product trick with the second row". If you mean take the dot product of the original formula with $B$ then you get $0 = 0$ so you obviously must mean something else. | |
| Feb 2, 2011 at 21:13 | comment | added | Sergei Ivanov | If you define vector to vector as zero and try the same dot product trick with the second row, you would get a wrong answer. | |
| Feb 2, 2011 at 20:59 | history | answered | Greg Kuperberg | CC BY-SA 2.5 |