# What to call the elements of a tensor product.

What does one call the first (or second) factor of an element of a tensor product? For example, if $V,W$ are vector spaces, and $v \in V$, $w \in W$, with $v \otimes w \in V \otimes W$, how would one refer to $v$? First tensor factor?

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I think any denomination other than $v$ is confusing. The proof is that you are asking here, so it is hardly standard. –  Andrea Ferretti Mar 9 '10 at 23:07
Tensorand? I've heard something like that before. –  tkr Mar 9 '10 at 23:08
Call it the first projection of a representative of the tensor $v\otimes w=\otimes((v,w))$, so $(v,w)\in \otimes^{-1}(v\otimes w)\subset V\times W$. –  Harry Gindi Mar 9 '10 at 23:14
@ Andrea: But I've got a collection of about 116 $v \otimes w$'s, and I really don't want to refer to them individually. I need a way to say "Take all the first tensor factors and consider the ideal they generate". –  Aston Smythe Mar 9 '10 at 23:18
The single element $v\otimes w$ is often called a "pure tensor", so you could talk about the "first component(s)/factor(s) of the pure tensors". "First tensor factor" seems okay to me, since you'll have to define whatever locution you choose anyway. –  Arturo Magidin Mar 9 '10 at 23:26

This issue is similar to what someone faces when dealing with a polynomial expression $$c_n\alpha^n + c_{n-1}\alpha^{n-1} + \cdots + c_1\alpha + c_0$$ where $\alpha$ actually satisfies an equation of degree smaller than $n$. Logically speaking such expressions can be written in multiple ways (consider a quartic polynomial expression in $\sqrt{3}$), but nobody has a problem speaking about the $i$th term in the expression.
Just do the same thing when you write down an elementary tensor $v_1 \otimes v_2 \otimes \cdots \otimes v_k$: call $v_i$ the $i$th component (or $i$th term, or perhaps even the $i$th factor). Now comes an issue of who your audience is (which you didn't indicate). If your audience is experts, then it would be clear to your audience that whatever you're doing with $v_i$ is eventually leading to some well-defined result in terms of the tensor itself, so there's nothing more to say.