3 ms·
Crikey, I hope I never have to talk to that mathematician! That's a terse, unintuitive definition that isn't very helpful unless you're already familiar with th
by soVeryTired 8y ago
Crikey, I hope I never have to talk to that mathematician! That's a terse, unintuitive definition that isn't very helpful unless you're already familiar with the concepts. (Also maybe you meant linear maps into bilinear?)
Reminds me of the time an algebraist mentioned to me that he was working on profinite group theory. I asked what a profinite group was, and he immediately replied 'an inverse limit of an inverse system', with no follow up. Well thanks buddy, that really opened my eyes.
- chongli 8y agoMath is just a much deeper topic than most others. The things people do in research level math can take a really long time to explain to a lay person because of the many layers of abstraction involved.
- soVeryTired 8y agoIt is a very deep and specialised topic. However, there are ways to convey intuition to a 'mathematically mature' audience, and there are quick definitions that are correct but unenlightening. I much prefer the former :)
- ginnungagap 8y agoNo, it turns bilinear maps into linear one! If you have three R-modules (one can read K-vector spaces if unfamiliar with modules) N,M,P and a bilinear map N×M→P then there is a unique linear map N⊗M→P compatible with the map N×M→N⊗M which is part of the structure of a tensor product. (What's really going on here in fancy terms is the so called Hom-Tensor adjunction because the _⊗M functor is adjoint to the Hom(M,_) functor, but just thinking about bilinear and linear maps is much clearer)
- soVeryTired 8y agoOk fair enough, I think I get you.