Does orthogonal-invariance of a differential imply invariance of the function?












2












$begingroup$


Let $U:text{Hom}(mathbb{R}^d,mathbb{R}^d) to mathbb{R}$ be a smooth function .



If $U$ is orthogonally-invariant, i.e. $U(QA)=U(A)$ for every $Q in text{SO}(n),A in text{Hom}(mathbb{R}^d,mathbb{R}^d)$, then



$$ dU_{QA}(QB)=dU_{A}(B) tag{1}$$



Does the converse hold? That is, suppose $U$ is a $C^{infty}$ function whose differential satisfies equation $(1) $ for every $A,Bin text{Hom}(mathbb{R}^d,mathbb{R}^d),Qin text{SO}(n)$. Is it true that $U(QA)=U(A)$ for every $Q in text{SO}(n)$?






lie-groups symmetry orthogonal-matrices symmetric-functions







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$endgroup$












  • $begingroup$
    You don't really need $f$ periodic: it's sufficient for $f$ to be bounded, which follows from the domain of $h$ being compact
    $endgroup$
    – Dap
    Dec 2 '18 at 17:57
















2












$begingroup$


Let $U:text{Hom}(mathbb{R}^d,mathbb{R}^d) to mathbb{R}$ be a smooth function .



If $U$ is orthogonally-invariant, i.e. $U(QA)=U(A)$ for every $Q in text{SO}(n),A in text{Hom}(mathbb{R}^d,mathbb{R}^d)$, then



$$ dU_{QA}(QB)=dU_{A}(B) tag{1}$$



Does the converse hold? That is, suppose $U$ is a $C^{infty}$ function whose differential satisfies equation $(1) $ for every $A,Bin text{Hom}(mathbb{R}^d,mathbb{R}^d),Qin text{SO}(n)$. Is it true that $U(QA)=U(A)$ for every $Q in text{SO}(n)$?






lie-groups symmetry orthogonal-matrices symmetric-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    You don't really need $f$ periodic: it's sufficient for $f$ to be bounded, which follows from the domain of $h$ being compact
    $endgroup$
    – Dap
    Dec 2 '18 at 17:57














2












2








2





$begingroup$


Let $U:text{Hom}(mathbb{R}^d,mathbb{R}^d) to mathbb{R}$ be a smooth function .



If $U$ is orthogonally-invariant, i.e. $U(QA)=U(A)$ for every $Q in text{SO}(n),A in text{Hom}(mathbb{R}^d,mathbb{R}^d)$, then



$$ dU_{QA}(QB)=dU_{A}(B) tag{1}$$



Does the converse hold? That is, suppose $U$ is a $C^{infty}$ function whose differential satisfies equation $(1) $ for every $A,Bin text{Hom}(mathbb{R}^d,mathbb{R}^d),Qin text{SO}(n)$. Is it true that $U(QA)=U(A)$ for every $Q in text{SO}(n)$?






lie-groups symmetry orthogonal-matrices symmetric-functions







share|cite|improve this question











$endgroup$




Let $U:text{Hom}(mathbb{R}^d,mathbb{R}^d) to mathbb{R}$ be a smooth function .



If $U$ is orthogonally-invariant, i.e. $U(QA)=U(A)$ for every $Q in text{SO}(n),A in text{Hom}(mathbb{R}^d,mathbb{R}^d)$, then



$$ dU_{QA}(QB)=dU_{A}(B) tag{1}$$



Does the converse hold? That is, suppose $U$ is a $C^{infty}$ function whose differential satisfies equation $(1) $ for every $A,Bin text{Hom}(mathbb{R}^d,mathbb{R}^d),Qin text{SO}(n)$. Is it true that $U(QA)=U(A)$ for every $Q in text{SO}(n)$?






lie-groups symmetry orthogonal-matrices symmetric-functions




lie-groups symmetry orthogonal-matrices symmetric-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 9:27







Asaf Shachar

















asked Nov 25 '18 at 13:33









Asaf ShacharAsaf Shachar

5,2473941




5,2473941












  • $begingroup$
    You don't really need $f$ periodic: it's sufficient for $f$ to be bounded, which follows from the domain of $h$ being compact
    $endgroup$
    – Dap
    Dec 2 '18 at 17:57


















  • $begingroup$
    You don't really need $f$ periodic: it's sufficient for $f$ to be bounded, which follows from the domain of $h$ being compact
    $endgroup$
    – Dap
    Dec 2 '18 at 17:57
















$begingroup$
You don't really need $f$ periodic: it's sufficient for $f$ to be bounded, which follows from the domain of $h$ being compact
$endgroup$
– Dap
Dec 2 '18 at 17:57




$begingroup$
You don't really need $f$ periodic: it's sufficient for $f$ to be bounded, which follows from the domain of $h$ being compact
$endgroup$
– Dap
Dec 2 '18 at 17:57










1 Answer
1






active

oldest

votes


















1












$begingroup$

The answer is positive. Indeed, define $h:text{SO}(n) to mathbb{R}$, by
$$ h(Q)=U(QA).$$



Then, $dh_Q(tilde B)=dU_{QA}(tilde BA)=dU_{A}(Q^{-1}tilde BA)=dh_I(Q^{-1}tilde B)$,
so
$$ dh_Q = dh_I circ Q^{-1} ,,,text{ for every } Q in text{SO}(n). tag{1}$$



Since $text{SO}(n)$ is compact, $h$ obtains a maximum at some point $P in text{SO}(n)$. Thus, $dh_P=0$, hence by equation $(1)$, $dh_I=0$, and so $dh_Q =0$ for every $Q in text{SO}(n)$. This implies $h$ is constant.



Comment: This argument use the compactness of the symmetry group $text{SO}(n)$ in an essential way; indeed, for non-compact groups $G$, "differential invariance" of a function $G to mathbb{R}$ does not imply it is constant. Take e.g. $G=mathbb{R}$, and $f(x)=x$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice! This old question of mine is somewhat reminiscent of the present one. Maybe your last comment answers it, in the negative.
    $endgroup$
    – Giuseppe Negro
    Dec 10 '18 at 9:47












  • $begingroup$
    Your question is nice indeed, however the answer turns out to be positive.
    $endgroup$
    – Asaf Shachar
    Dec 10 '18 at 11:07











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

The answer is positive. Indeed, define $h:text{SO}(n) to mathbb{R}$, by
$$ h(Q)=U(QA).$$



Then, $dh_Q(tilde B)=dU_{QA}(tilde BA)=dU_{A}(Q^{-1}tilde BA)=dh_I(Q^{-1}tilde B)$,
so
$$ dh_Q = dh_I circ Q^{-1} ,,,text{ for every } Q in text{SO}(n). tag{1}$$



Since $text{SO}(n)$ is compact, $h$ obtains a maximum at some point $P in text{SO}(n)$. Thus, $dh_P=0$, hence by equation $(1)$, $dh_I=0$, and so $dh_Q =0$ for every $Q in text{SO}(n)$. This implies $h$ is constant.



Comment: This argument use the compactness of the symmetry group $text{SO}(n)$ in an essential way; indeed, for non-compact groups $G$, "differential invariance" of a function $G to mathbb{R}$ does not imply it is constant. Take e.g. $G=mathbb{R}$, and $f(x)=x$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice! This old question of mine is somewhat reminiscent of the present one. Maybe your last comment answers it, in the negative.
    $endgroup$
    – Giuseppe Negro
    Dec 10 '18 at 9:47












  • $begingroup$
    Your question is nice indeed, however the answer turns out to be positive.
    $endgroup$
    – Asaf Shachar
    Dec 10 '18 at 11:07
















1












$begingroup$

The answer is positive. Indeed, define $h:text{SO}(n) to mathbb{R}$, by
$$ h(Q)=U(QA).$$



Then, $dh_Q(tilde B)=dU_{QA}(tilde BA)=dU_{A}(Q^{-1}tilde BA)=dh_I(Q^{-1}tilde B)$,
so
$$ dh_Q = dh_I circ Q^{-1} ,,,text{ for every } Q in text{SO}(n). tag{1}$$



Since $text{SO}(n)$ is compact, $h$ obtains a maximum at some point $P in text{SO}(n)$. Thus, $dh_P=0$, hence by equation $(1)$, $dh_I=0$, and so $dh_Q =0$ for every $Q in text{SO}(n)$. This implies $h$ is constant.



Comment: This argument use the compactness of the symmetry group $text{SO}(n)$ in an essential way; indeed, for non-compact groups $G$, "differential invariance" of a function $G to mathbb{R}$ does not imply it is constant. Take e.g. $G=mathbb{R}$, and $f(x)=x$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice! This old question of mine is somewhat reminiscent of the present one. Maybe your last comment answers it, in the negative.
    $endgroup$
    – Giuseppe Negro
    Dec 10 '18 at 9:47












  • $begingroup$
    Your question is nice indeed, however the answer turns out to be positive.
    $endgroup$
    – Asaf Shachar
    Dec 10 '18 at 11:07














1












1








1





$begingroup$

The answer is positive. Indeed, define $h:text{SO}(n) to mathbb{R}$, by
$$ h(Q)=U(QA).$$



Then, $dh_Q(tilde B)=dU_{QA}(tilde BA)=dU_{A}(Q^{-1}tilde BA)=dh_I(Q^{-1}tilde B)$,
so
$$ dh_Q = dh_I circ Q^{-1} ,,,text{ for every } Q in text{SO}(n). tag{1}$$



Since $text{SO}(n)$ is compact, $h$ obtains a maximum at some point $P in text{SO}(n)$. Thus, $dh_P=0$, hence by equation $(1)$, $dh_I=0$, and so $dh_Q =0$ for every $Q in text{SO}(n)$. This implies $h$ is constant.



Comment: This argument use the compactness of the symmetry group $text{SO}(n)$ in an essential way; indeed, for non-compact groups $G$, "differential invariance" of a function $G to mathbb{R}$ does not imply it is constant. Take e.g. $G=mathbb{R}$, and $f(x)=x$.






share|cite|improve this answer









$endgroup$



The answer is positive. Indeed, define $h:text{SO}(n) to mathbb{R}$, by
$$ h(Q)=U(QA).$$



Then, $dh_Q(tilde B)=dU_{QA}(tilde BA)=dU_{A}(Q^{-1}tilde BA)=dh_I(Q^{-1}tilde B)$,
so
$$ dh_Q = dh_I circ Q^{-1} ,,,text{ for every } Q in text{SO}(n). tag{1}$$



Since $text{SO}(n)$ is compact, $h$ obtains a maximum at some point $P in text{SO}(n)$. Thus, $dh_P=0$, hence by equation $(1)$, $dh_I=0$, and so $dh_Q =0$ for every $Q in text{SO}(n)$. This implies $h$ is constant.



Comment: This argument use the compactness of the symmetry group $text{SO}(n)$ in an essential way; indeed, for non-compact groups $G$, "differential invariance" of a function $G to mathbb{R}$ does not imply it is constant. Take e.g. $G=mathbb{R}$, and $f(x)=x$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 9:36









Asaf ShacharAsaf Shachar

5,2473941




5,2473941












  • $begingroup$
    Nice! This old question of mine is somewhat reminiscent of the present one. Maybe your last comment answers it, in the negative.
    $endgroup$
    – Giuseppe Negro
    Dec 10 '18 at 9:47












  • $begingroup$
    Your question is nice indeed, however the answer turns out to be positive.
    $endgroup$
    – Asaf Shachar
    Dec 10 '18 at 11:07


















  • $begingroup$
    Nice! This old question of mine is somewhat reminiscent of the present one. Maybe your last comment answers it, in the negative.
    $endgroup$
    – Giuseppe Negro
    Dec 10 '18 at 9:47












  • $begingroup$
    Your question is nice indeed, however the answer turns out to be positive.
    $endgroup$
    – Asaf Shachar
    Dec 10 '18 at 11:07
















$begingroup$
Nice! This old question of mine is somewhat reminiscent of the present one. Maybe your last comment answers it, in the negative.
$endgroup$
– Giuseppe Negro
Dec 10 '18 at 9:47






$begingroup$
Nice! This old question of mine is somewhat reminiscent of the present one. Maybe your last comment answers it, in the negative.
$endgroup$
– Giuseppe Negro
Dec 10 '18 at 9:47














$begingroup$
Your question is nice indeed, however the answer turns out to be positive.
$endgroup$
– Asaf Shachar
Dec 10 '18 at 11:07




$begingroup$
Your question is nice indeed, however the answer turns out to be positive.
$endgroup$
– Asaf Shachar
Dec 10 '18 at 11:07


















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