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Consider a vector space $V$ (with dimension $n+1$ and elements $v$) on which a (commutative and associative) "product" $\odot$ taking $V\odot V\rightarrow V$ is defined, and an $1$ element $v_0$ exists: $v\odot v_0=v$ for all $v\in V$. $\odot$ is now completely defined by choosing a basis $v_m$ $(0\le m\le n)$ ($v_0$ is predefined - hands off - you can't change it) and giving the "structure constants" $A^{ij}_k$ $(0\le i,j,k\le n)$ which are free parameters (except those with $i*j=0$ who are predefined by $v_0$ being the $1$ element): $v_i\odot v_j=\sum_k A^{ij}_k*v_k$.
After you defined the $A^{ij}_k$, you are free to chose a convenient other basis. First question: I have $n(n+1)$ free parameters to do so. Just from counting, I can use them to have $v_k\odot v_k=v_0$ for all $k$ and use up exactly all "gauge" freedom this way. My linear algebra is rusty - can I really? Or are there $A^{ij}_k$ sets where this fails?
Second question. Define an "outer product" $\otimes$ (with is distributive over vector sums, and follows $(v_i\otimes v_j)\odot v_k:=v_i\otimes (v_j\odot v_k)$ and $v_i\odot (v_j\otimes v_k):=(v_i\odot v_j)\otimes v_k$) and a quantity $S=\sum_i\sum_j a_{ij}*v_i \otimes v_j$. $S$ should be an "eigenvector": $S\odot v_k=v_k\odot S$ for all $k$. Express the allowed set of $a_{ij}$ in terms of the $A^{ij}_k$. (This is trivial for small n, where I do it with diagrams and by hand - in fact this is knot/graph theory in disguise as always when I ask :-) But a closed formula would be nice.)

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3 Answers 3

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Assuming you want $(A+B) ⊙ V = A⊙V + B⊙V$ (if not things get a lot more complicated , you might as well ignore the vector structure having no way to link the addition and product) , the product , being bi-linear , can be described as a tensor (essentially n square matrices $M_p$, n*n each , such that if $V_1⊙V_2 =C $ , then $< V_1 | M_k | V2 > = C_k$ , the k'th component of $C$ . ) . I like to imagine them as a cube .

A⊙V = V⊙A means that each of these matrices will be symmetric . The second constraint is $v\odot v_0=v$ . Linear algebra ensures we can chose a convenient form for $v_0$ , so let $v_0= (1,0,0 ,0...) $ . We have $< v_0 | M_k | B > =B_k $ , the k'th component of $B$ . This constrains the first column of $M_k$ to be (0,0..,0,1,0..0) where the k'th component is 1. By symmetry , the first row is also constrained . So the only freedom we have is choosing the parameters of $n \hspace{5 mm} (n-1)*(n-1)$ symmetric matrices , in total $n*n*(n-1)$ parameters for a fixed $v_0$ . Since $v_0$ can be chosen arbitrarily from a space of n dimentions , we have $(n + n*n*(n+1))$ degrees of freedom for the product operator . Hope that helps .

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The answer to your first question is - generally, it is not possible. For instance, let $V=\mathbb{C}$ viewed as a 2-dimensional vector space over $\mathbb{R}$, with the usual product of complex numbers. What you are asking for is an $\mathbb{R}$-basis of $\mathbb{C}$ consisting of elements which square to $1$. But all such complex numbers are $1$ and $-1$, and you cannot build a basis from these.

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Let $k$ be your base field, then consider the ring $k[\epsilon]/\epsilon^{n+1}$. The elements of the ring such that $v^2=1$ are just $v=\pm1$. These elements do not span the ring as a vector space over $k$.

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I am confused - for your $v$, we have $v^2=1\pm 2f\epsilon^{(n+1)/2}$, which is neither $1$ (probably that's what you meant, at least that's what the OP was asking for) nor 0. –  Vladimir Dotsenko May 6 '12 at 16:24
    
Ah, you are correct. That was a stupid mistake. Luckily it just makes my point simpler. –  Will Sawin May 6 '12 at 16:54
    
Indeed. You might want to delete "for f in the ring" as well, to make it not only simple but not confusing ;-) –  Vladimir Dotsenko May 6 '12 at 16:56
    
I have to leave SOMETHING to keep my readers on their toes. –  Will Sawin May 6 '12 at 20:11
1  
Yes. That'd be "$v^2=0$" which is still there :-) –  Vladimir Dotsenko May 6 '12 at 23:41

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