Vector "product" diagonalization - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T13:29:53Z http://mathoverflow.net/feeds/question/95994 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95994/vector-product-diagonalization Vector "product" diagonalization Hauke Reddmann 2012-05-04T15:17:02Z 2012-05-07T00:42:02Z <p>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): $v_i\odot v_j=\sum_k A^{ij}_k*v_k$.<br> 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 <em>counting</em>, 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?<br> 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.)</p> http://mathoverflow.net/questions/95994/vector-product-diagonalization/96018#96018 Answer by Will Sawin for Vector "product" diagonalization Will Sawin 2012-05-04T20:09:30Z 2012-05-07T00:42:02Z <p>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 $k$.</p> http://mathoverflow.net/questions/95994/vector-product-diagonalization/96139#96139 Answer by Dan Fodor for Vector "product" diagonalization Dan Fodor 2012-05-06T15:37:53Z 2012-05-06T15:37:53Z <p>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 $&lt; V_1 | M_k | V2 > = C_k$ , the k'th component of $C$ . ) . I like to imagine them as a cube . </p> <p>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 $&lt; v_0 | M_k | B > =B_k$ , 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 .</p> http://mathoverflow.net/questions/95994/vector-product-diagonalization/96143#96143 Answer by Vladimir Dotsenko for Vector "product" diagonalization Vladimir Dotsenko 2012-05-06T16:29:09Z 2012-05-06T16:29:09Z <p>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.</p>