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In Gowers' explanation of tensor products, there is a passage where he writes, with vector spaces $V$ and $W$:

For every pair $(v,w)$ in $V\times W$, regard $[[v,w]]$ as a meaningless symbol. We can define a rather large vector space $Z$ by taking formal linear combinations of these symbols. By that I mean that $Z$ consists of all expressions of the form $$a_1[[v_1,w_1]]+ a_2[[v_2,w_2]]+...+ a_n[[v_n,w_n]]$$ with obvious definitions for addition and scalar multiplication.

I am a little confused by what he means at the end. When he talks about addition and scalar multiplication, I am wondering if he means the following: $$a_1[[v_1,w_1]]+a_2[[v_2,w_2]]=[[a_1v_1+a_2v_2,a_1w_1+a_2w_2]]$$ But this also seems incorrect because then there would seemingly be no difference between $Z$ and $V\times W$ other than the fact that instead of $(\cdot,\cdot)$ we write $[[\cdot,\cdot]]$.

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No, $Z$ really is just the vector space whose basis is the set of all symbols $[[v,w]]$. $a[[v,w]]$ can be simplified no further, and there's no combination of elements when you take sums, in particular $[[v+v',w]]$ doesn't split into a sum of two elements. That comes later when we take the quotient to create the tensor product. Even $[[0,0]]\neq 0$, $0$ is the element $0[[v,w]]$ for any $v,w$, so for example $[[0,0]]$ and $2[[0,0]]$ are different elements.

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  • $\begingroup$ Just to make sure I'm understanding you correctly, in this case $[[v+v',w]]$ and $[[v,w]]+[[v',w]]$ are different elements, and they are only made to identify with each other when the quotient is taken, which explicitly sets all symbols of the form $[[v+v',w]]-[[v,w]]-[[v',w]]$ to zero. $\endgroup$
    – Vasting
    Commented Mar 27, 2020 at 19:33
  • $\begingroup$ @Vasting That is correct. $\endgroup$ Commented Mar 27, 2020 at 19:41

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