How to calculate the outer product in GR?

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Haorong Wu
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Homework Statement
Show that the scalar product of two vectors is not altered as they are both Fermi-Walker transported along a curve ##C##
Relevant Equations
If a curve is timelike, ##u## is its tangent vector, and ##a=\nabla_u u=Du/d \tau##, then a vector ##V## is said to be Fermi-Walker transported along ##u## if $$ \nabla_u V=(u \otimes a- a \otimes u)\cdot V .$$
I will post the answer here, part of which I do not follow.

Let the two vectors be ##x## and ## y##. Then the Fermi-walker transport law reads$$\nabla_u x=(u \otimes a -a \otimes u) \cdot x$$ $$\nabla_u y=(u \otimes a- a \otimes u)\cdot y ,$$ where ##u## is the tangent vector to the curve ##c## and ##a=\nabla_u u##. Using the product rule we evaluate the change in the dot product along the curve $$\begin{align} \nabla_u ( x \cdot y)&= (\nabla_u x)\cdot y+ x \cdot(\nabla_u y) \nonumber \\& =(a \cdot x) (u \cdot y) -(a \cdot y) (u \cdot x) +(u \cdot x) (a \cdot y) -(a \cdot x) (u \cdot y) =0 . \nonumber \end{align}$$ The scalar product is unaltered.

I do not follow the outer-product part. I know that I should multiply two terms together if they are in the same space. However, in this problem, I do not know how to determin which term belongs to which space. It seems, sometimes ##x## and ##u## are in the same space, and sometimes ##x## and ##a## are in the same space. I am stuck.
 
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By definition, let ##S:= u \otimes a - a \otimes u##, then ##S(\alpha, \beta) = (u \cdot \alpha)(a \cdot \beta) - (a \cdot \alpha)(u \cdot \beta)##
 
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