Timeline for Is there a different construction of "the" tensor product of two modules?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Jul 12, 2012 at 20:26 | answer | added | wildildildlife | timeline score: 5 | |
| Jun 14, 2012 at 19:27 | comment | added | Yongheng Zhang | So far there are two intersecting lines of thought: one is to start a different set theoretic construction directly and the other is to use the tensor-hom adjointness and possibly the double dual isomorphism under finiteness assumption to express the tensor product in a variety of different ways. Both supplement and advance my understanding of $\bigocross$ greatly. | |
| Jun 14, 2012 at 15:52 | comment | added | user2035 | For arbitrary vector spaces, you can take the subspace of the space of bilinear forms $V^*\times W^*\to k$ generated by $v\otimes w$ (defined in the obvious way), $v\in V,w\in W$. | |
| Jun 14, 2012 at 12:59 | comment | added | Konrad Waldorf | For finite-dimensional vector spaces, you can also take $Hom(V^{*},W)$ together with the bilinear map $\tau$ which is $\tau(v,w)(\lambda) := \lambda(v)w$. | |
| Jun 14, 2012 at 8:47 | answer | added | Elemer E Rosinger | timeline score: 0 | |
| Jun 14, 2012 at 7:16 | comment | added | darij grinberg | Mariano: only if the vector spaces are finite-dimensional. | |
| Jun 14, 2012 at 0:59 | comment | added | S. Carnahan♦ | Tom's suggestion works in ZFC if you restrict your pairs to have rank less than any sufficiently large limit ordinal. | |
| Jun 13, 2012 at 21:10 | comment | added | Tom Goodwillie | Well, you could consider the set of all pairs $(P,B:M\times N\to P)$ where $P$ is a $k$-module and $B$ is bilinear, and then define the tensor product as a certain submodule of the product, over all such pairs, of all of the modules $P$ (or rather you could use a modified version of this construction that does not violate the rules of set theory). | |
| Jun 13, 2012 at 20:41 | comment | added | Will Sawin | For $\mathbb Z$-modules, or abelian groups, it's the dual to the space of bilinear maps $M \times N \to S^1$. | |
| Jun 13, 2012 at 20:38 | comment | added | user1437 | The tensor product is usually constructed in the manner you mentioned. It is usually defined, on the other hand, via the universal property: briefly, any billinear map $M\times N\to P$ lifts to a unique map $M\otimes N\to P$; this is the universal property, and by this any construction of the tensor product is isomorphic to any other. | |
| Jun 13, 2012 at 20:03 | comment | added | Mariano Suárez-Álvarez | For vector spaces, one can construct $V\otimes W$ as the dual space of the vector space of bilinear maps $V\times W\to\text{base field}$. | |
| Jun 13, 2012 at 20:00 | comment | added | Steven Gubkin | You don't only want to construct an object, but also a projection map. The universal property says any such gadgets doing the right thing will all be isomorphic. So I am not sure what more you can really ask... | |
| Jun 13, 2012 at 19:51 | history | asked | Yongheng Zhang | CC BY-SA 3.0 |