3
$\begingroup$

Take two vector spaces $V$ and $W$ over $\mathbb{C}$. We want $V {\otimes}_{\mathbb{C}} W$. As far as I know, we form this tensor product via quotienting the larger space $V \times W$ by the space spanning the linear combinations $$(v_1 + v_2, w_1)-(v_1, w_1)-(v_2, w_1)\:\:\:\:\:(1)$$ $$(v_1, w_1 + w_2)-(v_1, w_1)-(v_1, w_2)\:\:\:\:\:(2)$$ $$\lambda (v_1, w_1)-(\lambda v_1, w_1)\:\:\:\:\:(3)$$ $$\lambda (v_1, w_1)-(v_1, \lambda w_1)\:\:\:\:\:(4)$$

The confusion is about combinations (3) and (4) and what they imply. My understanding is that the scalars come from $\mathbb{C}$ so I can do this:

For $z \in \mathbb{C}$, $$z(v \otimes w) = zv \otimes w = v \otimes zw \:\:\:\:\:(*)$$

But someone told me this cannot happen.

I need to take tensor products of vector spaces and I was trying to apply the tensor product of modules from Dummit and Foote's Abstract Algebra (3rd Ed). Since vector spaces are modules, I thought it would work the same way.

I couldn't follow their explanation at the time for why (*) isn't true for vector spaces (if that's what they were even getting at). Can anyone help clear this up?

$\endgroup$
3
  • 11
    $\begingroup$ Whoever told you that they were wrong, that's perfectly valid. Also, the space you're quotienting isn't $V \times W$ but the free vector space on $V \times W$. $\endgroup$ Commented Sep 18, 2022 at 16:31
  • $\begingroup$ As a hint that something’s wrong, if $\dim V=m$ and $\dim W=n$, then $\dim (V\times W)= m+n < mn = \ dim(V\otimes W)$ when $m\ge 2$ and $n> 2$. $\endgroup$ Commented Sep 18, 2022 at 23:31
  • $\begingroup$ If $R$ is the subspace generated by relations (1) thru (4) of, $F(V\times W)$, the free vector space, then $V\otimes_{\mathbb F}W=F(V\times W)/R$, for each field $\mathbb F$, over where $V$ and $W$ are vector spaces. Right? $\endgroup$
    – janmarqz
    Commented Sep 21, 2022 at 14:49

1 Answer 1

1
$\begingroup$

Think $v\otimes w$ as the coset $$v\otimes w=(v,w)+R$$ where $R$ is the mentioned subspace, then the relations (3), (4) mean $$z(v_1,w_1)+R=(z v_1,w_1)+R=(v_1,z w_1)+R,$$ which is your relation $(*)$.

$\endgroup$
1
  • $\begingroup$ Remember that two cosets $(v,w)+R=(u,t)+R$ if and only if $(v,w)-(u,t)\in R$ $\endgroup$
    – janmarqz
    Commented Sep 20, 2022 at 20:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .