Is the orthogonal polar factor the unique retraction $operatorname{GL}_n^+ to operatorname{SO}_n$?
$begingroup$
$newcommand{psym}{text{Psym}_n}$
$newcommand{sym}{text{sym}}$
$newcommand{Sym}{operatorname{Sym}}$
$newcommand{Skew}{operatorname{Skew}}$
$renewcommand{skew}{operatorname{skew}}$
$newcommand{GLp}{operatorname{GL}_n^+}$
$newcommand{SO}{operatorname{SO}_n}$
This might be silly, but I wonder:
Let $F:GLp to SO$ be a continuous retract. Is it true that $F$ must be the orthogonal polar factor, i.e. $F(A)=O$, where $A=OP,O in SO,Pinpsym$. Does anything changes if we assume $F$ is a deformation retract? Or if it is a smooth deformation retract?
algebraic-topology lie-groups matrix-decomposition orthogonal-matrices retraction
asked Dec 20 '18 at 19:56
Asaf ShacharAsaf Shachar
5,79431145
$endgroup$
add a comment |
$begingroup$
$newcommand{psym}{text{Psym}_n}$
$newcommand{sym}{text{sym}}$
$newcommand{Sym}{operatorname{Sym}}$
$newcommand{Skew}{operatorname{Skew}}$
$renewcommand{skew}{operatorname{skew}}$
$newcommand{GLp}{operatorname{GL}_n^+}$
$newcommand{SO}{operatorname{SO}_n}$
This might be silly, but I wonder:
Let $F:GLp to SO$ be a continuous retract. Is it true that $F$ must be the orthogonal polar factor, i.e. $F(A)=O$, where $A=OP,O in SO,Pinpsym$. Does anything changes if we assume $F$ is a deformation retract? Or if it is a smooth deformation retract?
algebraic-topology lie-groups matrix-decomposition orthogonal-matrices retraction
asked Dec 20 '18 at 19:56
Asaf ShacharAsaf Shachar
5,79431145
$endgroup$
add a comment |
$begingroup$
$newcommand{psym}{text{Psym}_n}$
$newcommand{sym}{text{sym}}$
$newcommand{Sym}{operatorname{Sym}}$
$newcommand{Skew}{operatorname{Skew}}$
$renewcommand{skew}{operatorname{skew}}$
$newcommand{GLp}{operatorname{GL}_n^+}$
$newcommand{SO}{operatorname{SO}_n}$
This might be silly, but I wonder:
Let $F:GLp to SO$ be a continuous retract. Is it true that $F$ must be the orthogonal polar factor, i.e. $F(A)=O$, where $A=OP,O in SO,Pinpsym$. Does anything changes if we assume $F$ is a deformation retract? Or if it is a smooth deformation retract?
algebraic-topology lie-groups matrix-decomposition orthogonal-matrices retraction
asked Dec 20 '18 at 19:56
Asaf ShacharAsaf Shachar
5,79431145
$endgroup$
$newcommand{psym}{text{Psym}_n}$
$newcommand{sym}{text{sym}}$
$newcommand{Sym}{operatorname{Sym}}$
$newcommand{Skew}{operatorname{Skew}}$
$renewcommand{skew}{operatorname{skew}}$
$newcommand{GLp}{operatorname{GL}_n^+}$
$newcommand{SO}{operatorname{SO}_n}$
This might be silly, but I wonder:
Let $F:GLp to SO$ be a continuous retract. Is it true that $F$ must be the orthogonal polar factor, i.e. $F(A)=O$, where $A=OP,O in SO,Pinpsym$. Does anything changes if we assume $F$ is a deformation retract? Or if it is a smooth deformation retract?
algebraic-topology lie-groups matrix-decomposition orthogonal-matrices retraction
algebraic-topology lie-groups matrix-decomposition orthogonal-matrices retraction
asked Dec 20 '18 at 19:56
Asaf ShacharAsaf Shachar
5,79431145
asked Dec 20 '18 at 19:56
Asaf ShacharAsaf Shachar
5,79431145
asked Dec 20 '18 at 19:56
Asaf ShacharAsaf Shachar
5,79431145
asked Dec 20 '18 at 19:56
Asaf ShacharAsaf Shachar
5,79431145
asked Dec 20 '18 at 19:56
Asaf ShacharAsaf Shachar
5,79431145
5,79431145
add a comment |
add a comment |
1 Answer
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$begingroup$
Retractions are a very "soft" concept so there is a huge number of those. Take any continuous function $f:PSym_nto SO_n$ and define $F_f(A):=Ocdot f(P)$ for $A=OP$. Then this is clearly a retraction and it even is a defomormation retraction, which is smooth if $f$ is smooth.
answered Dec 21 '18 at 7:46
Andreas CapAndreas Cap
11.4k923
$endgroup$
$begingroup$
Thanks, I was just thinking something similar myself: we can take e.g. $F(A)= Odet A$ or $F(A)= Ofrac{| A|}{sqrt n}$ or variations on these. However, both of these retractions are "proportional" to the orthogonal polar factor. A more interesting question is: are there retractions $GL^+ to SO$ which are not proportional to it? (By your construction, it suffices to find a continuous non constant map $f:Psym to SO$ which is the identity on $SO$. Is it trivial such a map exists?)
$endgroup$
– Asaf Shachar
Dec 21 '18 at 10:35
$begingroup$
In fact, therer are no conditions on the map $f$, so lots of such maps exist
$endgroup$
– Andreas Cap
Dec 23 '18 at 8:51
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
Retractions are a very "soft" concept so there is a huge number of those. Take any continuous function $f:PSym_nto SO_n$ and define $F_f(A):=Ocdot f(P)$ for $A=OP$. Then this is clearly a retraction and it even is a defomormation retraction, which is smooth if $f$ is smooth.
answered Dec 21 '18 at 7:46
Andreas CapAndreas Cap
11.4k923
$endgroup$
$begingroup$
Thanks, I was just thinking something similar myself: we can take e.g. $F(A)= Odet A$ or $F(A)= Ofrac{| A|}{sqrt n}$ or variations on these. However, both of these retractions are "proportional" to the orthogonal polar factor. A more interesting question is: are there retractions $GL^+ to SO$ which are not proportional to it? (By your construction, it suffices to find a continuous non constant map $f:Psym to SO$ which is the identity on $SO$. Is it trivial such a map exists?)
$endgroup$
– Asaf Shachar
Dec 21 '18 at 10:35
$begingroup$
In fact, therer are no conditions on the map $f$, so lots of such maps exist
$endgroup$
– Andreas Cap
Dec 23 '18 at 8:51
add a comment |
$begingroup$
Retractions are a very "soft" concept so there is a huge number of those. Take any continuous function $f:PSym_nto SO_n$ and define $F_f(A):=Ocdot f(P)$ for $A=OP$. Then this is clearly a retraction and it even is a defomormation retraction, which is smooth if $f$ is smooth.
answered Dec 21 '18 at 7:46
Andreas CapAndreas Cap
11.4k923
$endgroup$
$begingroup$
Thanks, I was just thinking something similar myself: we can take e.g. $F(A)= Odet A$ or $F(A)= Ofrac{| A|}{sqrt n}$ or variations on these. However, both of these retractions are "proportional" to the orthogonal polar factor. A more interesting question is: are there retractions $GL^+ to SO$ which are not proportional to it? (By your construction, it suffices to find a continuous non constant map $f:Psym to SO$ which is the identity on $SO$. Is it trivial such a map exists?)
$endgroup$
– Asaf Shachar
Dec 21 '18 at 10:35
$begingroup$
In fact, therer are no conditions on the map $f$, so lots of such maps exist
$endgroup$
– Andreas Cap
Dec 23 '18 at 8:51
add a comment |
$begingroup$
Retractions are a very "soft" concept so there is a huge number of those. Take any continuous function $f:PSym_nto SO_n$ and define $F_f(A):=Ocdot f(P)$ for $A=OP$. Then this is clearly a retraction and it even is a defomormation retraction, which is smooth if $f$ is smooth.
answered Dec 21 '18 at 7:46
Andreas CapAndreas Cap
11.4k923
$endgroup$
Retractions are a very "soft" concept so there is a huge number of those. Take any continuous function $f:PSym_nto SO_n$ and define $F_f(A):=Ocdot f(P)$ for $A=OP$. Then this is clearly a retraction and it even is a defomormation retraction, which is smooth if $f$ is smooth.
answered Dec 21 '18 at 7:46
Andreas CapAndreas Cap
11.4k923
answered Dec 21 '18 at 7:46
Andreas CapAndreas Cap
11.4k923
answered Dec 21 '18 at 7:46
Andreas CapAndreas Cap
11.4k923
answered Dec 21 '18 at 7:46
Andreas CapAndreas Cap
11.4k923
11.4k923
$begingroup$
Thanks, I was just thinking something similar myself: we can take e.g. $F(A)= Odet A$ or $F(A)= Ofrac{| A|}{sqrt n}$ or variations on these. However, both of these retractions are "proportional" to the orthogonal polar factor. A more interesting question is: are there retractions $GL^+ to SO$ which are not proportional to it? (By your construction, it suffices to find a continuous non constant map $f:Psym to SO$ which is the identity on $SO$. Is it trivial such a map exists?)
$endgroup$
– Asaf Shachar
Dec 21 '18 at 10:35
$begingroup$
In fact, therer are no conditions on the map $f$, so lots of such maps exist
$endgroup$
– Andreas Cap
Dec 23 '18 at 8:51
add a comment |
$begingroup$
Thanks, I was just thinking something similar myself: we can take e.g. $F(A)= Odet A$ or $F(A)= Ofrac{| A|}{sqrt n}$ or variations on these. However, both of these retractions are "proportional" to the orthogonal polar factor. A more interesting question is: are there retractions $GL^+ to SO$ which are not proportional to it? (By your construction, it suffices to find a continuous non constant map $f:Psym to SO$ which is the identity on $SO$. Is it trivial such a map exists?)
$endgroup$
– Asaf Shachar
Dec 21 '18 at 10:35
$begingroup$
In fact, therer are no conditions on the map $f$, so lots of such maps exist
$endgroup$
– Andreas Cap
Dec 23 '18 at 8:51
$begingroup$
Thanks, I was just thinking something similar myself: we can take e.g. $F(A)= Odet A$ or $F(A)= Ofrac{| A|}{sqrt n}$ or variations on these. However, both of these retractions are "proportional" to the orthogonal polar factor. A more interesting question is: are there retractions $GL^+ to SO$ which are not proportional to it? (By your construction, it suffices to find a continuous non constant map $f:Psym to SO$ which is the identity on $SO$. Is it trivial such a map exists?)
$endgroup$
– Asaf Shachar
Dec 21 '18 at 10:35
$begingroup$
Thanks, I was just thinking something similar myself: we can take e.g. $F(A)= Odet A$ or $F(A)= Ofrac{| A|}{sqrt n}$ or variations on these. However, both of these retractions are "proportional" to the orthogonal polar factor. A more interesting question is: are there retractions $GL^+ to SO$ which are not proportional to it? (By your construction, it suffices to find a continuous non constant map $f:Psym to SO$ which is the identity on $SO$. Is it trivial such a map exists?)
$endgroup$
– Asaf Shachar
Dec 21 '18 at 10:35
$begingroup$
In fact, therer are no conditions on the map $f$, so lots of such maps exist
$endgroup$
– Andreas Cap
Dec 23 '18 at 8:51
$begingroup$
In fact, therer are no conditions on the map $f$, so lots of such maps exist
$endgroup$
– Andreas Cap
Dec 23 '18 at 8:51
add a comment |
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