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| − | + | In this case <math>n</math> represents the number of points and <math>m</math> represents the dimension of the vector space the points lie in. So for <math>n</math>=4 the matrix <math>\bold{D}</math> constructed below is 3 x 3. Thus there are <math>\binom{3}{2} \times \binom{3}{2} = 9</math> (<math>2 \times 2</math>)-minors of <math>\bold{D}</math>. By the construction of <math>\bold{D}</math>, there is a certain symmetry to the matrix(this symmetry occurs for all <math>n\ge{4}</math>). For this reason, with <math>n</math>=4, the polynomials given by the (<math>2 \times 2</math>)-minors are not all unique. In fact, there are only 6 distinct polynomials. Below is what I'm thinking may be the case in general. | |
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| − | be the matrix | + | |
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| + | Idea: For <math>2\le{m}\le{n-2}</math>, let <math>D_{i,j}</math> be indeterminates <math>(1\le{i}<j\le{n})</math> and let | ||
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| + | <math>\bold{D}</math> = (<math>D_{i,j} - D_{i,n} - D_{j,n}</math>)<math>_{i,j=1,...,n-1}</math> | ||
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| + | be the matrix where we set <math>D_{i,i}</math> := 0 and <math>D_{i,j}</math> := <math>D_{j,i}</math> for <math>i>j</math>. Set <math>k</math> := <math>n - m</math>. Then there are <math>\binom{\binom{n-1}{k-1}+1}{2}</math> distinct (<math>m \times m</math>)-minors of <math>\bold{D}</math>. | ||
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| + | I'm thinking that the number of distinct minors of <math>\bold{D}</math> is actually the number of polynomials of <math>n\times m</math> variables that we are looking for to determine constructibility of the <math>n</math>-point configurations. My reasoning for this is that in the algorithm presented in the proof of Theorem 1.6, we set <math>F:=F_1F_2</math> where <math>F(d_{1,2},...,d_{n-1,n})=f(P_1,...,P_n)\ne 0</math>. Here, I'm thinking that <math>F_1</math> can only be one polynomial. <math>F_2</math> on the other hand can be any (<math>m \times m</math>)-minor of <math>\bold{D}</math>. If I'm right, then the number of distinct minors of <math>\bold{D}</math> is the same number of distinct polynomials <math>F</math>, and thus the same for <math>f</math>. | ||
Revision as of 18:47, 11 June 2009
In this case $ n $ represents the number of points and $ m $ represents the dimension of the vector space the points lie in. So for $ n $=4 the matrix $ \bold{D} $ constructed below is 3 x 3. Thus there are $ \binom{3}{2} \times \binom{3}{2} = 9 $ ($ 2 \times 2 $)-minors of $ \bold{D} $. By the construction of $ \bold{D} $, there is a certain symmetry to the matrix(this symmetry occurs for all $ n\ge{4} $). For this reason, with $ n $=4, the polynomials given by the ($ 2 \times 2 $)-minors are not all unique. In fact, there are only 6 distinct polynomials. Below is what I'm thinking may be the case in general.
Idea: For $ 2\le{m}\le{n-2} $, let $ D_{i,j} $ be indeterminates $ (1\le{i}<j\le{n}) $ and let
$ \bold{D} $ = ($ D_{i,j} - D_{i,n} - D_{j,n} $)$ _{i,j=1,...,n-1} $
be the matrix where we set $ D_{i,i} $ := 0 and $ D_{i,j} $ := $ D_{j,i} $ for $ i>j $. Set $ k $ := $ n - m $. Then there are $ \binom{\binom{n-1}{k-1}+1}{2} $ distinct ($ m \times m $)-minors of $ \bold{D} $.
I'm thinking that the number of distinct minors of $ \bold{D} $ is actually the number of polynomials of $ n\times m $ variables that we are looking for to determine constructibility of the $ n $-point configurations. My reasoning for this is that in the algorithm presented in the proof of Theorem 1.6, we set $ F:=F_1F_2 $ where $ F(d_{1,2},...,d_{n-1,n})=f(P_1,...,P_n)\ne 0 $. Here, I'm thinking that $ F_1 $ can only be one polynomial. $ F_2 $ on the other hand can be any ($ m \times m $)-minor of $ \bold{D} $. If I'm right, then the number of distinct minors of $ \bold{D} $ is the same number of distinct polynomials $ F $, and thus the same for $ f $.
