Equivalent definitions of the tensor product.

Question

The way tensor products are usually defined in algebra is by it's universal property with a proof that a tensor product of two modules always exists. However in my differential geometry course we used a different definition. We defined it as the set of all bilinear maps $V_1^*\times V_2^*\to K$ where $V_1$ and $V_2$ are $K$ vector spaces. The pure tensors in this case are maps

$v_1\otimes v_2(\psi_1, \psi_2):=\psi_1(v_1)\psi_1(v_2)$

To get all the awesome results from algebra on tensor product's I've proven that in this case the two definitions are indeed equivalent.

However for the proof I've used the fact that bases of $V_1$ and $V_2$ exist, which makes the proof effectively trivial. I was wondering if it were possible to extend this construction on modules that are not necessarily free. My guess would be no, considering the fact that this method could be used instead of the quite lengthy method for constructing the tensor module in algebra.

This may help answer your question. – i like math Sep 22 '22 at 20:46 — i like math, Sep 22 '22 at 20:46

score 2 · Accepted Answer · answered Sep 22 '22 at 20:48

2

Your suspicion is correct: the definition given in your differential geometry class works for finitely generated free modules but not in general. There is a more general definition found in many algebra textbooks which is actually, in my opinion, more intuitive than the one you quote, but it requires more algebraic background. See for example https://kconrad.math.uconn.edu/blurbs/linmultialg/tensorprod.pdf

answered Sep 22 '22 at 20:48

Michael Cohen

801

Thank you for your answer! – Boxonix Sep 22 '22 at 20:55

Equivalent definitions of the tensor product.

1 Answers1