Let $W$ be a vector space over $\mathbb R$ with basis $\{a,b\}$. Consider the tensor
$$a \otimes a \otimes a – b \otimes b \otimes a + a \otimes b \otimes b + b \otimes a \otimes b.$$
a) Show that this tensor cannot be presented as sum of two decomposable tensors.
b) Show that such presentation is possible in complexification of $W$.
My only attempt was to write an equation
$$(\alpha a + \beta b)\otimes (\gamma a + \delta b) \otimes (\epsilon a + \zeta b) + (\alpha' a + \beta' b)\otimes (\gamma' a + \delta' b) \otimes (\epsilon' a + \zeta' b) =$$
$$= a \otimes a \otimes a – b \otimes b \otimes a + a \otimes b \otimes b + b \otimes a \otimes b,$$
and from equality of coefficients deduce a system of 8 equalities and 12 variables. I think I can brute force it some way, but I was wondering if there are any other approaches.
Best Answer
I'll start by stating some common theorems with tensor products.
1. Preliminaries
Theorem 1 (Zero-sum)
Let $U$,$V$ be two vector spaces on $F$, if for $n$ linearly independent vectors $x_1,x_2,...,x_n \in U$, there are $n$ vectors $v_1,...,v_n \in V$ such that: $$x_1 \otimes v_1 +x_2 \otimes v_2+...+x_n \otimes v_n= 0 $$ then $v_1=v_2=...=v_n=0$ $\square$
Theorem 2 (Span)
Let $U$,$V$ be two vector spaces on $F$. $W$ is a subspace of $V$.
Let $ t $ be a tensor in $U \otimes W$ and if for $n$ linearly independent vectors $x_1,x_2,...,x_n \in U$, there are $n$ vectors $v_1,...,v_n \in V$ such that: $$t=x_1 \otimes v_1 +x_2 \otimes v_2+...+x_n \otimes v_n $$ then $\text{span}(v_1,...,v_n) \subset W$
$\square$
Remark 3: We will use mainly theorem 2, the proof of theorem 2 is based on theorem 1. Furthermore, we only need this theorem in the special case where $\dim W=2$
2. Main answer
Assume there are vectors $u_1 , u_2, u_3,v_1,v_2,v_3 \in W$ such that: $$ a \otimes a \otimes a - b \otimes b \otimes a + a \otimes b \otimes b + b \otimes a \otimes b= u_1\otimes u_2 \otimes u_3+u_1\otimes u_2 \otimes u_3 =:t$$ On one hand, we see that: $$ t \in W\otimes( \underbrace{\text{span}( u_2\otimes u_3, v_2 \otimes v_3)}_{=:U})$$ On the other hand, $$ t= a \otimes ( a \otimes a +b \otimes b)+ b\otimes ( -b \otimes a + a \otimes b) $$ Then by theorem 2, we imply that $a \otimes a +b \otimes b$ and $ -b \otimes a + a \otimes b $ lie in $U$.
By theorem 1, none of two above vector is null (and clearly they are independent) , whereas $\dim(U)\le 2$.
Therefore, $\dim(U)=2$ and $ U = \text{span}( a \otimes a +b \otimes b , -b \otimes a + a \otimes b ) $
So there are two elements $p,q \in F$ ( $F$ is the field on which we define $W$, in this case it is either $\mathbb{R}$ or $\mathbb{C}$) such that: $$p (a \otimes a +b \otimes b ) +q( -b \otimes a + a \otimes b ) =u_2 \otimes u_3$$
Or , $$ a \otimes( pa +qb) +b \otimes ( bp-qa) = u_2 \otimes u_3$$ Again, by theorem $2$ ( or by any other straightforward reasoning), as $a,b$ are independent, $pa+qb$ and $bp-qa$ cannot be independent ( as $\dim( \text{span}(u_3)) = 1$)
As $a,b$ are independent, the two above vector are linearly dependent iff $p^2+q^2=0$ This cannot occur in $\mathbb{R}$ unless $p=q=0$ which also leads to a contradiction.
Hence the conclusion for $\mathbb{R}$.
The conclusion for $\mathbb{C}$ (if true) is just a result from some simple calculations.
Remark Indeed, many arguments I presented above can be omitted or shortened given that we are familiar with the tensor product. I just tried to show this point as clear as possible.