Timeline for What is the canonical isomorphism between the tensor products of the top exterior powers associated to exact sequences of vector spaces?
Current License: CC BY-SA 3.0
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| Apr 29, 2012 at 14:03 | comment | added | Liviu Nicolaescu | @Dmitry That is true, but in applications one is given either a basis of $X$ (this happens when one computes torsions of based acyclic complexes) or an orientation of $X$, i.e., a basis of $X$ defined up to a positive multiple. Then you run into a bit of trouble with the conventions. | |
| Apr 29, 2012 at 12:37 | comment | added | Dmitri Pavlov | If X is a line, then det(X)=X and det(X*)=X*, but there is no canonical isomorphism between a line and its dual (e.g., a line bundle over a holomorphic manifold need not be isomorphic to its dual). A more natural statement would be (det(X))*=det(X*). The canonical pairing between det(X) and det(X*) can be defined using the canonical pairing between ΛX and Λ(X*). | |
| Apr 29, 2012 at 9:51 | comment | added | Liviu Nicolaescu | Yes, and unfortunately, this cannot be helped. Think of doing intersection of two odd dimensional cycles of complementary dimensions in an even dimensional manifold. | |
| Apr 29, 2012 at 7:47 | comment | added | Alberto Abbondandolo | Thanks for the link to your book. If I understand well, the notion of "weighted line" allows $\Lambda^{\max}(X)$ to "remember" the dimension of the space $X$ and this helps when dealing with sign issues. | |
| Apr 29, 2012 at 1:51 | history | edited | Liviu Nicolaescu | CC BY-SA 3.0 |
added 4 characters in body
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| Apr 28, 2012 at 23:37 | history | answered | Liviu Nicolaescu | CC BY-SA 3.0 |