How should one find complete sets of invariants for vectors whose elements can be permuted by a given group?












4














Disclaimer / Introduction



I am a physicist by training who hasn't taken courses in invariant theory. I hope my description of my question that doesn't mis-use the terminology in `invariant theory' -- but in case it does I have provided a concrete example in less mathematical terminology to try to make clear what I actually meant!



Attempt at mathematical statement:



Notation



Suppose I have an $n$-dimensional vector space $V$ over a field $F$ which may (at your preference) be taken to be either the reals or the complex numbers. I also have a group $G$ which permutes (some of) the components of the vectors $vec x$ in $V$. In all the cases I am interested in, $G$ is a proper subgroup of the $S_n$, by which I mean that it does not include all possible $(n!)$ permutations of the $n$ elements of each $vec x$. We will assume that $G$ is generated by $m$ generators $g_1,ldots,g_min G$.



We new define an equivalence relation $sim$ in the vector space by: $$(vec xsim vec y) Leftrightarrow (exists gin G text{s.t.} vec x = gvec y). $$



The problem itself



Given any concretely specified $G$, $n$, $F$ and $V$ as defined above, I wish to learn how I can construct a set of $n$ functions (let's call them $f_1,f_2,ldots, f_n$) which each map $Vrightarrow F$ in such a way that:
$$( f_i(vec x) = f_i(vec y) forall iin{1,2,ldots,n})Leftrightarrow(vec x sim vec y).$$



[I think these might be called "indicator functions" or "complete sets of invariants", in that they unambiguously indicate to which equivalence class any $vec x$ belongs. Though perhaps I am mis-using those terms.]



A very simple example (Example 1)



Suppose $n=3$, and an arbitrary element of $V$ is denoted $vec x=(a,b,p)$. Assume that $G$ is generated by a single generator $g$ which exchanges the first and second elements of $x$ ($aleftrightarrow b$). I.e. $$g=left( begin{array}{ccc}
0 & 1 & 0 \
1 & 0 & 0 \
0 & 0 & 1
end{array}right)$$
(Note that in more complicated examples $G$ could have more than one generator!)



In the above case, the three functions:
$$
f_1(vec x) = a+b
\
f_2(vec x) = ab
\
f_3(vec x) = p
$$

is a set of functions $f_i$ which satisfies my requirements, but in contrast
$$
f_1(vec x) = a+b
\
f_2(vec x) = a-b
\
f_3(vec x) = p
$$

is a bad set of functions which does not meet the requirement, and
$$
f_1(vec x) = a+b+p
\
f_2(vec x) = ab+ap+bp
\
f_3(vec x) = abp
$$
does not work either.



A more complicated example (Example 2)



Suppose $n=6$, and an arbitrary element of $V$ is denoted $vec x=(a,b,p,q,x,y)$. Assume that $G$ is generated by a single generator $g$ which exchanges the first and second elements of $x$ ($aleftrightarrow b$) at the same time as exchanging the third and fourth elements ($pleftrightarrow q$). I.e. $$g=left( begin{array}{cccccc}
0 & 1 & 0 & 0 & 0 & 0 \
1 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 1 & 0 \
0 & 0 & 0 & 0 & 0 & 1
end{array}right)$$
(Note that in more complicated examples $G$ could have more than one generator!)



In the above case, the six functions:
$$
f_1(vec x) = p+q
\
f_2(vec x) = a+b
\
f_3(vec x) = pq-ab
\
f_4(vec x) = bp+aq
\
f_5(vec x) = x
\
f_6(vec x) = y
$$

is a set of functions $f_i$ which satisfies my requirements.



An example showing why I am not interested in the case where $G=S_n$



If $G$ were $S_6$, the symmetric group on six objects having all possible $6!$ elements, then symmetric polynomials would (trivially) be a valid set of $f_i$:
$$
f_1 = a+b+p+q+x+y \
f_2 = ab+ap+aq+ax+ay+bp+bq+bx+by+pq+px+py+qx+qy+xy \
f_3 = abp+abq+cdots+qxy \
f_4 = abpq+cdots+pqxy \
f_5 = abpqx+abpqy+abpxy+abqxy+apqxy+bpqxy\
f_6 = abpqxy.
$$

The above sets are easy to construct, but do not answer the question when $G$ is not $S_n$ (see my examples above).



Notes



The limitations of my own method of finding functions $f_i$



Alas: the way (which I have not described) that I generated a valid set of functions for Example 2 does not allow me to work with any $G$ containing an element which contains three or more transpositions. I cannot therefore presently write down a set of functions which meets my requirements if the generator $g$ above is replaced by, say
$$g'=left( begin{array}{cccccc}
0 & 1 & 0 & 0 & 0 & 0 \
1 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 & 1 & 0
end{array}right).$$
I can't believe that there is no way of answering such marginally more complicated cases, though, so I assume that any invariant-theory professional will have an armoury of tools much better than mine, and can point me to books and or resources that will tell me how to turn any $G$ that's a permutation group into a set of functions $f_i$ that meet my needs.










share|cite|improve this question




















  • 1




    I don't understand Example 1. You have $f(a,b,p)=p+q$. Where did $q$ come from?
    – Gerry Myerson
    Nov 25 at 11:14










  • Oops - in example 1 I transposed the roles of (p,q) with (a,b). Thanks to your spotting this I have now corrected that typo in example 1.
    – KesterKester
    Nov 25 at 20:12












  • @GerryMyerson Thank you for spotting this typo in my example 1. I have fixed it. I transposed the roles of (p,q) with (a,b).
    – KesterKester
    Nov 25 at 20:38










  • For Example 2, would the four functions $p+q,pq,a+b,ab$ do as well as the four you give?
    – Gerry Myerson
    Nov 25 at 21:54






  • 1




    Dear @GerryMyerson : I don't think $(f1,f2,f3,f4)=(p+q,pq,a+b,ab)$ is valid for Example 2, as it is too permissive. Ignoring $x,y$ it puts $(a,b,p,q)$ in the same equivalence class as $(a,b,q,p)$ and $(b,a,p,q)$ and $(b,a,q,p)$, whereas actually $(a,b,p,q)$ should only be in the same equivalence class as $(b,a,q,p)$ in Example 2.
    – KesterKester
    Nov 25 at 22:09


















4














Disclaimer / Introduction



I am a physicist by training who hasn't taken courses in invariant theory. I hope my description of my question that doesn't mis-use the terminology in `invariant theory' -- but in case it does I have provided a concrete example in less mathematical terminology to try to make clear what I actually meant!



Attempt at mathematical statement:



Notation



Suppose I have an $n$-dimensional vector space $V$ over a field $F$ which may (at your preference) be taken to be either the reals or the complex numbers. I also have a group $G$ which permutes (some of) the components of the vectors $vec x$ in $V$. In all the cases I am interested in, $G$ is a proper subgroup of the $S_n$, by which I mean that it does not include all possible $(n!)$ permutations of the $n$ elements of each $vec x$. We will assume that $G$ is generated by $m$ generators $g_1,ldots,g_min G$.



We new define an equivalence relation $sim$ in the vector space by: $$(vec xsim vec y) Leftrightarrow (exists gin G text{s.t.} vec x = gvec y). $$



The problem itself



Given any concretely specified $G$, $n$, $F$ and $V$ as defined above, I wish to learn how I can construct a set of $n$ functions (let's call them $f_1,f_2,ldots, f_n$) which each map $Vrightarrow F$ in such a way that:
$$( f_i(vec x) = f_i(vec y) forall iin{1,2,ldots,n})Leftrightarrow(vec x sim vec y).$$



[I think these might be called "indicator functions" or "complete sets of invariants", in that they unambiguously indicate to which equivalence class any $vec x$ belongs. Though perhaps I am mis-using those terms.]



A very simple example (Example 1)



Suppose $n=3$, and an arbitrary element of $V$ is denoted $vec x=(a,b,p)$. Assume that $G$ is generated by a single generator $g$ which exchanges the first and second elements of $x$ ($aleftrightarrow b$). I.e. $$g=left( begin{array}{ccc}
0 & 1 & 0 \
1 & 0 & 0 \
0 & 0 & 1
end{array}right)$$
(Note that in more complicated examples $G$ could have more than one generator!)



In the above case, the three functions:
$$
f_1(vec x) = a+b
\
f_2(vec x) = ab
\
f_3(vec x) = p
$$

is a set of functions $f_i$ which satisfies my requirements, but in contrast
$$
f_1(vec x) = a+b
\
f_2(vec x) = a-b
\
f_3(vec x) = p
$$

is a bad set of functions which does not meet the requirement, and
$$
f_1(vec x) = a+b+p
\
f_2(vec x) = ab+ap+bp
\
f_3(vec x) = abp
$$
does not work either.



A more complicated example (Example 2)



Suppose $n=6$, and an arbitrary element of $V$ is denoted $vec x=(a,b,p,q,x,y)$. Assume that $G$ is generated by a single generator $g$ which exchanges the first and second elements of $x$ ($aleftrightarrow b$) at the same time as exchanging the third and fourth elements ($pleftrightarrow q$). I.e. $$g=left( begin{array}{cccccc}
0 & 1 & 0 & 0 & 0 & 0 \
1 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 1 & 0 \
0 & 0 & 0 & 0 & 0 & 1
end{array}right)$$
(Note that in more complicated examples $G$ could have more than one generator!)



In the above case, the six functions:
$$
f_1(vec x) = p+q
\
f_2(vec x) = a+b
\
f_3(vec x) = pq-ab
\
f_4(vec x) = bp+aq
\
f_5(vec x) = x
\
f_6(vec x) = y
$$

is a set of functions $f_i$ which satisfies my requirements.



An example showing why I am not interested in the case where $G=S_n$



If $G$ were $S_6$, the symmetric group on six objects having all possible $6!$ elements, then symmetric polynomials would (trivially) be a valid set of $f_i$:
$$
f_1 = a+b+p+q+x+y \
f_2 = ab+ap+aq+ax+ay+bp+bq+bx+by+pq+px+py+qx+qy+xy \
f_3 = abp+abq+cdots+qxy \
f_4 = abpq+cdots+pqxy \
f_5 = abpqx+abpqy+abpxy+abqxy+apqxy+bpqxy\
f_6 = abpqxy.
$$

The above sets are easy to construct, but do not answer the question when $G$ is not $S_n$ (see my examples above).



Notes



The limitations of my own method of finding functions $f_i$



Alas: the way (which I have not described) that I generated a valid set of functions for Example 2 does not allow me to work with any $G$ containing an element which contains three or more transpositions. I cannot therefore presently write down a set of functions which meets my requirements if the generator $g$ above is replaced by, say
$$g'=left( begin{array}{cccccc}
0 & 1 & 0 & 0 & 0 & 0 \
1 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 & 1 & 0
end{array}right).$$
I can't believe that there is no way of answering such marginally more complicated cases, though, so I assume that any invariant-theory professional will have an armoury of tools much better than mine, and can point me to books and or resources that will tell me how to turn any $G$ that's a permutation group into a set of functions $f_i$ that meet my needs.










share|cite|improve this question




















  • 1




    I don't understand Example 1. You have $f(a,b,p)=p+q$. Where did $q$ come from?
    – Gerry Myerson
    Nov 25 at 11:14










  • Oops - in example 1 I transposed the roles of (p,q) with (a,b). Thanks to your spotting this I have now corrected that typo in example 1.
    – KesterKester
    Nov 25 at 20:12












  • @GerryMyerson Thank you for spotting this typo in my example 1. I have fixed it. I transposed the roles of (p,q) with (a,b).
    – KesterKester
    Nov 25 at 20:38










  • For Example 2, would the four functions $p+q,pq,a+b,ab$ do as well as the four you give?
    – Gerry Myerson
    Nov 25 at 21:54






  • 1




    Dear @GerryMyerson : I don't think $(f1,f2,f3,f4)=(p+q,pq,a+b,ab)$ is valid for Example 2, as it is too permissive. Ignoring $x,y$ it puts $(a,b,p,q)$ in the same equivalence class as $(a,b,q,p)$ and $(b,a,p,q)$ and $(b,a,q,p)$, whereas actually $(a,b,p,q)$ should only be in the same equivalence class as $(b,a,q,p)$ in Example 2.
    – KesterKester
    Nov 25 at 22:09
















4












4








4







Disclaimer / Introduction



I am a physicist by training who hasn't taken courses in invariant theory. I hope my description of my question that doesn't mis-use the terminology in `invariant theory' -- but in case it does I have provided a concrete example in less mathematical terminology to try to make clear what I actually meant!



Attempt at mathematical statement:



Notation



Suppose I have an $n$-dimensional vector space $V$ over a field $F$ which may (at your preference) be taken to be either the reals or the complex numbers. I also have a group $G$ which permutes (some of) the components of the vectors $vec x$ in $V$. In all the cases I am interested in, $G$ is a proper subgroup of the $S_n$, by which I mean that it does not include all possible $(n!)$ permutations of the $n$ elements of each $vec x$. We will assume that $G$ is generated by $m$ generators $g_1,ldots,g_min G$.



We new define an equivalence relation $sim$ in the vector space by: $$(vec xsim vec y) Leftrightarrow (exists gin G text{s.t.} vec x = gvec y). $$



The problem itself



Given any concretely specified $G$, $n$, $F$ and $V$ as defined above, I wish to learn how I can construct a set of $n$ functions (let's call them $f_1,f_2,ldots, f_n$) which each map $Vrightarrow F$ in such a way that:
$$( f_i(vec x) = f_i(vec y) forall iin{1,2,ldots,n})Leftrightarrow(vec x sim vec y).$$



[I think these might be called "indicator functions" or "complete sets of invariants", in that they unambiguously indicate to which equivalence class any $vec x$ belongs. Though perhaps I am mis-using those terms.]



A very simple example (Example 1)



Suppose $n=3$, and an arbitrary element of $V$ is denoted $vec x=(a,b,p)$. Assume that $G$ is generated by a single generator $g$ which exchanges the first and second elements of $x$ ($aleftrightarrow b$). I.e. $$g=left( begin{array}{ccc}
0 & 1 & 0 \
1 & 0 & 0 \
0 & 0 & 1
end{array}right)$$
(Note that in more complicated examples $G$ could have more than one generator!)



In the above case, the three functions:
$$
f_1(vec x) = a+b
\
f_2(vec x) = ab
\
f_3(vec x) = p
$$

is a set of functions $f_i$ which satisfies my requirements, but in contrast
$$
f_1(vec x) = a+b
\
f_2(vec x) = a-b
\
f_3(vec x) = p
$$

is a bad set of functions which does not meet the requirement, and
$$
f_1(vec x) = a+b+p
\
f_2(vec x) = ab+ap+bp
\
f_3(vec x) = abp
$$
does not work either.



A more complicated example (Example 2)



Suppose $n=6$, and an arbitrary element of $V$ is denoted $vec x=(a,b,p,q,x,y)$. Assume that $G$ is generated by a single generator $g$ which exchanges the first and second elements of $x$ ($aleftrightarrow b$) at the same time as exchanging the third and fourth elements ($pleftrightarrow q$). I.e. $$g=left( begin{array}{cccccc}
0 & 1 & 0 & 0 & 0 & 0 \
1 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 1 & 0 \
0 & 0 & 0 & 0 & 0 & 1
end{array}right)$$
(Note that in more complicated examples $G$ could have more than one generator!)



In the above case, the six functions:
$$
f_1(vec x) = p+q
\
f_2(vec x) = a+b
\
f_3(vec x) = pq-ab
\
f_4(vec x) = bp+aq
\
f_5(vec x) = x
\
f_6(vec x) = y
$$

is a set of functions $f_i$ which satisfies my requirements.



An example showing why I am not interested in the case where $G=S_n$



If $G$ were $S_6$, the symmetric group on six objects having all possible $6!$ elements, then symmetric polynomials would (trivially) be a valid set of $f_i$:
$$
f_1 = a+b+p+q+x+y \
f_2 = ab+ap+aq+ax+ay+bp+bq+bx+by+pq+px+py+qx+qy+xy \
f_3 = abp+abq+cdots+qxy \
f_4 = abpq+cdots+pqxy \
f_5 = abpqx+abpqy+abpxy+abqxy+apqxy+bpqxy\
f_6 = abpqxy.
$$

The above sets are easy to construct, but do not answer the question when $G$ is not $S_n$ (see my examples above).



Notes



The limitations of my own method of finding functions $f_i$



Alas: the way (which I have not described) that I generated a valid set of functions for Example 2 does not allow me to work with any $G$ containing an element which contains three or more transpositions. I cannot therefore presently write down a set of functions which meets my requirements if the generator $g$ above is replaced by, say
$$g'=left( begin{array}{cccccc}
0 & 1 & 0 & 0 & 0 & 0 \
1 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 & 1 & 0
end{array}right).$$
I can't believe that there is no way of answering such marginally more complicated cases, though, so I assume that any invariant-theory professional will have an armoury of tools much better than mine, and can point me to books and or resources that will tell me how to turn any $G$ that's a permutation group into a set of functions $f_i$ that meet my needs.










share|cite|improve this question















Disclaimer / Introduction



I am a physicist by training who hasn't taken courses in invariant theory. I hope my description of my question that doesn't mis-use the terminology in `invariant theory' -- but in case it does I have provided a concrete example in less mathematical terminology to try to make clear what I actually meant!



Attempt at mathematical statement:



Notation



Suppose I have an $n$-dimensional vector space $V$ over a field $F$ which may (at your preference) be taken to be either the reals or the complex numbers. I also have a group $G$ which permutes (some of) the components of the vectors $vec x$ in $V$. In all the cases I am interested in, $G$ is a proper subgroup of the $S_n$, by which I mean that it does not include all possible $(n!)$ permutations of the $n$ elements of each $vec x$. We will assume that $G$ is generated by $m$ generators $g_1,ldots,g_min G$.



We new define an equivalence relation $sim$ in the vector space by: $$(vec xsim vec y) Leftrightarrow (exists gin G text{s.t.} vec x = gvec y). $$



The problem itself



Given any concretely specified $G$, $n$, $F$ and $V$ as defined above, I wish to learn how I can construct a set of $n$ functions (let's call them $f_1,f_2,ldots, f_n$) which each map $Vrightarrow F$ in such a way that:
$$( f_i(vec x) = f_i(vec y) forall iin{1,2,ldots,n})Leftrightarrow(vec x sim vec y).$$



[I think these might be called "indicator functions" or "complete sets of invariants", in that they unambiguously indicate to which equivalence class any $vec x$ belongs. Though perhaps I am mis-using those terms.]



A very simple example (Example 1)



Suppose $n=3$, and an arbitrary element of $V$ is denoted $vec x=(a,b,p)$. Assume that $G$ is generated by a single generator $g$ which exchanges the first and second elements of $x$ ($aleftrightarrow b$). I.e. $$g=left( begin{array}{ccc}
0 & 1 & 0 \
1 & 0 & 0 \
0 & 0 & 1
end{array}right)$$
(Note that in more complicated examples $G$ could have more than one generator!)



In the above case, the three functions:
$$
f_1(vec x) = a+b
\
f_2(vec x) = ab
\
f_3(vec x) = p
$$

is a set of functions $f_i$ which satisfies my requirements, but in contrast
$$
f_1(vec x) = a+b
\
f_2(vec x) = a-b
\
f_3(vec x) = p
$$

is a bad set of functions which does not meet the requirement, and
$$
f_1(vec x) = a+b+p
\
f_2(vec x) = ab+ap+bp
\
f_3(vec x) = abp
$$
does not work either.



A more complicated example (Example 2)



Suppose $n=6$, and an arbitrary element of $V$ is denoted $vec x=(a,b,p,q,x,y)$. Assume that $G$ is generated by a single generator $g$ which exchanges the first and second elements of $x$ ($aleftrightarrow b$) at the same time as exchanging the third and fourth elements ($pleftrightarrow q$). I.e. $$g=left( begin{array}{cccccc}
0 & 1 & 0 & 0 & 0 & 0 \
1 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 1 & 0 \
0 & 0 & 0 & 0 & 0 & 1
end{array}right)$$
(Note that in more complicated examples $G$ could have more than one generator!)



In the above case, the six functions:
$$
f_1(vec x) = p+q
\
f_2(vec x) = a+b
\
f_3(vec x) = pq-ab
\
f_4(vec x) = bp+aq
\
f_5(vec x) = x
\
f_6(vec x) = y
$$

is a set of functions $f_i$ which satisfies my requirements.



An example showing why I am not interested in the case where $G=S_n$



If $G$ were $S_6$, the symmetric group on six objects having all possible $6!$ elements, then symmetric polynomials would (trivially) be a valid set of $f_i$:
$$
f_1 = a+b+p+q+x+y \
f_2 = ab+ap+aq+ax+ay+bp+bq+bx+by+pq+px+py+qx+qy+xy \
f_3 = abp+abq+cdots+qxy \
f_4 = abpq+cdots+pqxy \
f_5 = abpqx+abpqy+abpxy+abqxy+apqxy+bpqxy\
f_6 = abpqxy.
$$

The above sets are easy to construct, but do not answer the question when $G$ is not $S_n$ (see my examples above).



Notes



The limitations of my own method of finding functions $f_i$



Alas: the way (which I have not described) that I generated a valid set of functions for Example 2 does not allow me to work with any $G$ containing an element which contains three or more transpositions. I cannot therefore presently write down a set of functions which meets my requirements if the generator $g$ above is replaced by, say
$$g'=left( begin{array}{cccccc}
0 & 1 & 0 & 0 & 0 & 0 \
1 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 & 1 & 0
end{array}right).$$
I can't believe that there is no way of answering such marginally more complicated cases, though, so I assume that any invariant-theory professional will have an armoury of tools much better than mine, and can point me to books and or resources that will tell me how to turn any $G$ that's a permutation group into a set of functions $f_i$ that meet my needs.







abstract-algebra polynomials reference-request invariant-theory






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edited Nov 25 at 20:57









user26857

39.2k123983




39.2k123983










asked Nov 24 at 22:17









KesterKester

886




886








  • 1




    I don't understand Example 1. You have $f(a,b,p)=p+q$. Where did $q$ come from?
    – Gerry Myerson
    Nov 25 at 11:14










  • Oops - in example 1 I transposed the roles of (p,q) with (a,b). Thanks to your spotting this I have now corrected that typo in example 1.
    – KesterKester
    Nov 25 at 20:12












  • @GerryMyerson Thank you for spotting this typo in my example 1. I have fixed it. I transposed the roles of (p,q) with (a,b).
    – KesterKester
    Nov 25 at 20:38










  • For Example 2, would the four functions $p+q,pq,a+b,ab$ do as well as the four you give?
    – Gerry Myerson
    Nov 25 at 21:54






  • 1




    Dear @GerryMyerson : I don't think $(f1,f2,f3,f4)=(p+q,pq,a+b,ab)$ is valid for Example 2, as it is too permissive. Ignoring $x,y$ it puts $(a,b,p,q)$ in the same equivalence class as $(a,b,q,p)$ and $(b,a,p,q)$ and $(b,a,q,p)$, whereas actually $(a,b,p,q)$ should only be in the same equivalence class as $(b,a,q,p)$ in Example 2.
    – KesterKester
    Nov 25 at 22:09
















  • 1




    I don't understand Example 1. You have $f(a,b,p)=p+q$. Where did $q$ come from?
    – Gerry Myerson
    Nov 25 at 11:14










  • Oops - in example 1 I transposed the roles of (p,q) with (a,b). Thanks to your spotting this I have now corrected that typo in example 1.
    – KesterKester
    Nov 25 at 20:12












  • @GerryMyerson Thank you for spotting this typo in my example 1. I have fixed it. I transposed the roles of (p,q) with (a,b).
    – KesterKester
    Nov 25 at 20:38










  • For Example 2, would the four functions $p+q,pq,a+b,ab$ do as well as the four you give?
    – Gerry Myerson
    Nov 25 at 21:54






  • 1




    Dear @GerryMyerson : I don't think $(f1,f2,f3,f4)=(p+q,pq,a+b,ab)$ is valid for Example 2, as it is too permissive. Ignoring $x,y$ it puts $(a,b,p,q)$ in the same equivalence class as $(a,b,q,p)$ and $(b,a,p,q)$ and $(b,a,q,p)$, whereas actually $(a,b,p,q)$ should only be in the same equivalence class as $(b,a,q,p)$ in Example 2.
    – KesterKester
    Nov 25 at 22:09










1




1




I don't understand Example 1. You have $f(a,b,p)=p+q$. Where did $q$ come from?
– Gerry Myerson
Nov 25 at 11:14




I don't understand Example 1. You have $f(a,b,p)=p+q$. Where did $q$ come from?
– Gerry Myerson
Nov 25 at 11:14












Oops - in example 1 I transposed the roles of (p,q) with (a,b). Thanks to your spotting this I have now corrected that typo in example 1.
– KesterKester
Nov 25 at 20:12






Oops - in example 1 I transposed the roles of (p,q) with (a,b). Thanks to your spotting this I have now corrected that typo in example 1.
– KesterKester
Nov 25 at 20:12














@GerryMyerson Thank you for spotting this typo in my example 1. I have fixed it. I transposed the roles of (p,q) with (a,b).
– KesterKester
Nov 25 at 20:38




@GerryMyerson Thank you for spotting this typo in my example 1. I have fixed it. I transposed the roles of (p,q) with (a,b).
– KesterKester
Nov 25 at 20:38












For Example 2, would the four functions $p+q,pq,a+b,ab$ do as well as the four you give?
– Gerry Myerson
Nov 25 at 21:54




For Example 2, would the four functions $p+q,pq,a+b,ab$ do as well as the four you give?
– Gerry Myerson
Nov 25 at 21:54




1




1




Dear @GerryMyerson : I don't think $(f1,f2,f3,f4)=(p+q,pq,a+b,ab)$ is valid for Example 2, as it is too permissive. Ignoring $x,y$ it puts $(a,b,p,q)$ in the same equivalence class as $(a,b,q,p)$ and $(b,a,p,q)$ and $(b,a,q,p)$, whereas actually $(a,b,p,q)$ should only be in the same equivalence class as $(b,a,q,p)$ in Example 2.
– KesterKester
Nov 25 at 22:09






Dear @GerryMyerson : I don't think $(f1,f2,f3,f4)=(p+q,pq,a+b,ab)$ is valid for Example 2, as it is too permissive. Ignoring $x,y$ it puts $(a,b,p,q)$ in the same equivalence class as $(a,b,q,p)$ and $(b,a,p,q)$ and $(b,a,q,p)$, whereas actually $(a,b,p,q)$ should only be in the same equivalence class as $(b,a,q,p)$ in Example 2.
– KesterKester
Nov 25 at 22:09












1 Answer
1






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I think I might have an answer for the case that $F$ is the reals.
First construct a polynomial $P(vec X)$ that is a function of an arbitrary point $vec X=(X_1,X_2,ldots,X_n)in V$ as follows
$$ P(vec X) =
| vec X - g_1 vec x |^2
| vec X - g_2 vec x |^2
| vec X - g_3 vec x |^2
cdots
| vec X - g_m vec x |^2
$$
in which $|vec x|^2$ represents a Euclidean norm. This $P$ has roots at $vec x$ and all the places to which $vec x$ can be moved by the action of the group $G$. Because of the way we have constructed $P$, if we were to multiply out all its terms we would find that the collected coefficients of any given term (e.g. the collected coefficients of the $X_1^4 X_2^2 X_5$ term) would be invariant under the action of $G$. The set of all of these coefficients entirely specifies the polynomial, and the polynomial is entirely specified by its roots and its form, so these coefficients are almost the set of functions $f_i$ requested. The only problem is that there are (in general) too many of them. Among the big list, though, exactly $n$ independent ones must exist, since the coefficients capture the same information as exists in the specification of the $n$ roots of the polynomial. A sufficient set of $n$ independent coefficients could be extracted by the following algorithm:



(1) create an empty set $K$.



(2) put all the polynomial coefficients in a set $C$.



(3) choose an element $cin C$ and remove that element from $C$ permanently.



(4) create a set $T=K cup {c}$.



(5) construct a Groebner basis for the polynomials in $T$. If this basis has a length equal to the number of elements in $T$, then replace $K$ with $T$, otherwise leave $K$ unaltered.



(6) If $K$ has fewer than $n$ elements, go to step (3).



(7) If this step is reached, $K$ contains a set of $n$ independent functions $f_i$ which (I hope) answer my question.



At least, I hope so.



I have tried this on a few concrete examples which have all worked out, though that doesn’t prove much.






share|cite|improve this answer





















  • No. This my proposed answer is not right. For "Example 2", the basis it gives me is equivalent to the one suggested by GerryMyerson in one of his comments on my original question. Which means it is no good for the reason I gave then. (Unless I was wrong then too ... )
    – KesterKester
    Nov 28 at 10:08










  • Perhaps things are not as bad as I thought. I had been looking for $n$ functions $f_i$ and so my proposal algorithm above stops in step (6) if K has $n$ elements. However, if I replace step (6) with "If there are still elements in $C$ then go to step (3)" then the final $K$ generated does appear always to contain the desired answer (and some dross). For example 2 it generates 7 functions $f_i$ of which six answer my problem. This leaves open the questions of (a) whether the conjecture is generally true, and (b) how one would find which of the functions in $K$ are (in general) the right $n$.
    – KesterKester
    Nov 28 at 11:02











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I think I might have an answer for the case that $F$ is the reals.
First construct a polynomial $P(vec X)$ that is a function of an arbitrary point $vec X=(X_1,X_2,ldots,X_n)in V$ as follows
$$ P(vec X) =
| vec X - g_1 vec x |^2
| vec X - g_2 vec x |^2
| vec X - g_3 vec x |^2
cdots
| vec X - g_m vec x |^2
$$
in which $|vec x|^2$ represents a Euclidean norm. This $P$ has roots at $vec x$ and all the places to which $vec x$ can be moved by the action of the group $G$. Because of the way we have constructed $P$, if we were to multiply out all its terms we would find that the collected coefficients of any given term (e.g. the collected coefficients of the $X_1^4 X_2^2 X_5$ term) would be invariant under the action of $G$. The set of all of these coefficients entirely specifies the polynomial, and the polynomial is entirely specified by its roots and its form, so these coefficients are almost the set of functions $f_i$ requested. The only problem is that there are (in general) too many of them. Among the big list, though, exactly $n$ independent ones must exist, since the coefficients capture the same information as exists in the specification of the $n$ roots of the polynomial. A sufficient set of $n$ independent coefficients could be extracted by the following algorithm:



(1) create an empty set $K$.



(2) put all the polynomial coefficients in a set $C$.



(3) choose an element $cin C$ and remove that element from $C$ permanently.



(4) create a set $T=K cup {c}$.



(5) construct a Groebner basis for the polynomials in $T$. If this basis has a length equal to the number of elements in $T$, then replace $K$ with $T$, otherwise leave $K$ unaltered.



(6) If $K$ has fewer than $n$ elements, go to step (3).



(7) If this step is reached, $K$ contains a set of $n$ independent functions $f_i$ which (I hope) answer my question.



At least, I hope so.



I have tried this on a few concrete examples which have all worked out, though that doesn’t prove much.






share|cite|improve this answer





















  • No. This my proposed answer is not right. For "Example 2", the basis it gives me is equivalent to the one suggested by GerryMyerson in one of his comments on my original question. Which means it is no good for the reason I gave then. (Unless I was wrong then too ... )
    – KesterKester
    Nov 28 at 10:08










  • Perhaps things are not as bad as I thought. I had been looking for $n$ functions $f_i$ and so my proposal algorithm above stops in step (6) if K has $n$ elements. However, if I replace step (6) with "If there are still elements in $C$ then go to step (3)" then the final $K$ generated does appear always to contain the desired answer (and some dross). For example 2 it generates 7 functions $f_i$ of which six answer my problem. This leaves open the questions of (a) whether the conjecture is generally true, and (b) how one would find which of the functions in $K$ are (in general) the right $n$.
    – KesterKester
    Nov 28 at 11:02
















0














I think I might have an answer for the case that $F$ is the reals.
First construct a polynomial $P(vec X)$ that is a function of an arbitrary point $vec X=(X_1,X_2,ldots,X_n)in V$ as follows
$$ P(vec X) =
| vec X - g_1 vec x |^2
| vec X - g_2 vec x |^2
| vec X - g_3 vec x |^2
cdots
| vec X - g_m vec x |^2
$$
in which $|vec x|^2$ represents a Euclidean norm. This $P$ has roots at $vec x$ and all the places to which $vec x$ can be moved by the action of the group $G$. Because of the way we have constructed $P$, if we were to multiply out all its terms we would find that the collected coefficients of any given term (e.g. the collected coefficients of the $X_1^4 X_2^2 X_5$ term) would be invariant under the action of $G$. The set of all of these coefficients entirely specifies the polynomial, and the polynomial is entirely specified by its roots and its form, so these coefficients are almost the set of functions $f_i$ requested. The only problem is that there are (in general) too many of them. Among the big list, though, exactly $n$ independent ones must exist, since the coefficients capture the same information as exists in the specification of the $n$ roots of the polynomial. A sufficient set of $n$ independent coefficients could be extracted by the following algorithm:



(1) create an empty set $K$.



(2) put all the polynomial coefficients in a set $C$.



(3) choose an element $cin C$ and remove that element from $C$ permanently.



(4) create a set $T=K cup {c}$.



(5) construct a Groebner basis for the polynomials in $T$. If this basis has a length equal to the number of elements in $T$, then replace $K$ with $T$, otherwise leave $K$ unaltered.



(6) If $K$ has fewer than $n$ elements, go to step (3).



(7) If this step is reached, $K$ contains a set of $n$ independent functions $f_i$ which (I hope) answer my question.



At least, I hope so.



I have tried this on a few concrete examples which have all worked out, though that doesn’t prove much.






share|cite|improve this answer





















  • No. This my proposed answer is not right. For "Example 2", the basis it gives me is equivalent to the one suggested by GerryMyerson in one of his comments on my original question. Which means it is no good for the reason I gave then. (Unless I was wrong then too ... )
    – KesterKester
    Nov 28 at 10:08










  • Perhaps things are not as bad as I thought. I had been looking for $n$ functions $f_i$ and so my proposal algorithm above stops in step (6) if K has $n$ elements. However, if I replace step (6) with "If there are still elements in $C$ then go to step (3)" then the final $K$ generated does appear always to contain the desired answer (and some dross). For example 2 it generates 7 functions $f_i$ of which six answer my problem. This leaves open the questions of (a) whether the conjecture is generally true, and (b) how one would find which of the functions in $K$ are (in general) the right $n$.
    – KesterKester
    Nov 28 at 11:02














0












0








0






I think I might have an answer for the case that $F$ is the reals.
First construct a polynomial $P(vec X)$ that is a function of an arbitrary point $vec X=(X_1,X_2,ldots,X_n)in V$ as follows
$$ P(vec X) =
| vec X - g_1 vec x |^2
| vec X - g_2 vec x |^2
| vec X - g_3 vec x |^2
cdots
| vec X - g_m vec x |^2
$$
in which $|vec x|^2$ represents a Euclidean norm. This $P$ has roots at $vec x$ and all the places to which $vec x$ can be moved by the action of the group $G$. Because of the way we have constructed $P$, if we were to multiply out all its terms we would find that the collected coefficients of any given term (e.g. the collected coefficients of the $X_1^4 X_2^2 X_5$ term) would be invariant under the action of $G$. The set of all of these coefficients entirely specifies the polynomial, and the polynomial is entirely specified by its roots and its form, so these coefficients are almost the set of functions $f_i$ requested. The only problem is that there are (in general) too many of them. Among the big list, though, exactly $n$ independent ones must exist, since the coefficients capture the same information as exists in the specification of the $n$ roots of the polynomial. A sufficient set of $n$ independent coefficients could be extracted by the following algorithm:



(1) create an empty set $K$.



(2) put all the polynomial coefficients in a set $C$.



(3) choose an element $cin C$ and remove that element from $C$ permanently.



(4) create a set $T=K cup {c}$.



(5) construct a Groebner basis for the polynomials in $T$. If this basis has a length equal to the number of elements in $T$, then replace $K$ with $T$, otherwise leave $K$ unaltered.



(6) If $K$ has fewer than $n$ elements, go to step (3).



(7) If this step is reached, $K$ contains a set of $n$ independent functions $f_i$ which (I hope) answer my question.



At least, I hope so.



I have tried this on a few concrete examples which have all worked out, though that doesn’t prove much.






share|cite|improve this answer












I think I might have an answer for the case that $F$ is the reals.
First construct a polynomial $P(vec X)$ that is a function of an arbitrary point $vec X=(X_1,X_2,ldots,X_n)in V$ as follows
$$ P(vec X) =
| vec X - g_1 vec x |^2
| vec X - g_2 vec x |^2
| vec X - g_3 vec x |^2
cdots
| vec X - g_m vec x |^2
$$
in which $|vec x|^2$ represents a Euclidean norm. This $P$ has roots at $vec x$ and all the places to which $vec x$ can be moved by the action of the group $G$. Because of the way we have constructed $P$, if we were to multiply out all its terms we would find that the collected coefficients of any given term (e.g. the collected coefficients of the $X_1^4 X_2^2 X_5$ term) would be invariant under the action of $G$. The set of all of these coefficients entirely specifies the polynomial, and the polynomial is entirely specified by its roots and its form, so these coefficients are almost the set of functions $f_i$ requested. The only problem is that there are (in general) too many of them. Among the big list, though, exactly $n$ independent ones must exist, since the coefficients capture the same information as exists in the specification of the $n$ roots of the polynomial. A sufficient set of $n$ independent coefficients could be extracted by the following algorithm:



(1) create an empty set $K$.



(2) put all the polynomial coefficients in a set $C$.



(3) choose an element $cin C$ and remove that element from $C$ permanently.



(4) create a set $T=K cup {c}$.



(5) construct a Groebner basis for the polynomials in $T$. If this basis has a length equal to the number of elements in $T$, then replace $K$ with $T$, otherwise leave $K$ unaltered.



(6) If $K$ has fewer than $n$ elements, go to step (3).



(7) If this step is reached, $K$ contains a set of $n$ independent functions $f_i$ which (I hope) answer my question.



At least, I hope so.



I have tried this on a few concrete examples which have all worked out, though that doesn’t prove much.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 27 at 9:23









KesterKester

886




886












  • No. This my proposed answer is not right. For "Example 2", the basis it gives me is equivalent to the one suggested by GerryMyerson in one of his comments on my original question. Which means it is no good for the reason I gave then. (Unless I was wrong then too ... )
    – KesterKester
    Nov 28 at 10:08










  • Perhaps things are not as bad as I thought. I had been looking for $n$ functions $f_i$ and so my proposal algorithm above stops in step (6) if K has $n$ elements. However, if I replace step (6) with "If there are still elements in $C$ then go to step (3)" then the final $K$ generated does appear always to contain the desired answer (and some dross). For example 2 it generates 7 functions $f_i$ of which six answer my problem. This leaves open the questions of (a) whether the conjecture is generally true, and (b) how one would find which of the functions in $K$ are (in general) the right $n$.
    – KesterKester
    Nov 28 at 11:02


















  • No. This my proposed answer is not right. For "Example 2", the basis it gives me is equivalent to the one suggested by GerryMyerson in one of his comments on my original question. Which means it is no good for the reason I gave then. (Unless I was wrong then too ... )
    – KesterKester
    Nov 28 at 10:08










  • Perhaps things are not as bad as I thought. I had been looking for $n$ functions $f_i$ and so my proposal algorithm above stops in step (6) if K has $n$ elements. However, if I replace step (6) with "If there are still elements in $C$ then go to step (3)" then the final $K$ generated does appear always to contain the desired answer (and some dross). For example 2 it generates 7 functions $f_i$ of which six answer my problem. This leaves open the questions of (a) whether the conjecture is generally true, and (b) how one would find which of the functions in $K$ are (in general) the right $n$.
    – KesterKester
    Nov 28 at 11:02
















No. This my proposed answer is not right. For "Example 2", the basis it gives me is equivalent to the one suggested by GerryMyerson in one of his comments on my original question. Which means it is no good for the reason I gave then. (Unless I was wrong then too ... )
– KesterKester
Nov 28 at 10:08




No. This my proposed answer is not right. For "Example 2", the basis it gives me is equivalent to the one suggested by GerryMyerson in one of his comments on my original question. Which means it is no good for the reason I gave then. (Unless I was wrong then too ... )
– KesterKester
Nov 28 at 10:08












Perhaps things are not as bad as I thought. I had been looking for $n$ functions $f_i$ and so my proposal algorithm above stops in step (6) if K has $n$ elements. However, if I replace step (6) with "If there are still elements in $C$ then go to step (3)" then the final $K$ generated does appear always to contain the desired answer (and some dross). For example 2 it generates 7 functions $f_i$ of which six answer my problem. This leaves open the questions of (a) whether the conjecture is generally true, and (b) how one would find which of the functions in $K$ are (in general) the right $n$.
– KesterKester
Nov 28 at 11:02




Perhaps things are not as bad as I thought. I had been looking for $n$ functions $f_i$ and so my proposal algorithm above stops in step (6) if K has $n$ elements. However, if I replace step (6) with "If there are still elements in $C$ then go to step (3)" then the final $K$ generated does appear always to contain the desired answer (and some dross). For example 2 it generates 7 functions $f_i$ of which six answer my problem. This leaves open the questions of (a) whether the conjecture is generally true, and (b) how one would find which of the functions in $K$ are (in general) the right $n$.
– KesterKester
Nov 28 at 11:02


















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