Mixing
Does this geometric rigidity condition forces the map to be the identity?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
$newcommandCofoperatornamecof$
$newcommandidoperatornameId$
$newcommandEndoperatornameEnd$
$newcommandGLoperatornameGL$
Let $V$ be a real $d$-dimensional vector space of dimension $d > 6$.
Let $B in GL(bigwedge^2V) $, and suppose that
$$ B(v_1) wedge B(v_2) wedge B(v_3)=v_1 wedge v_2 wedge v_3 in bigwedge^6V tag* $$
for every multi-vectors $v_1,v_2,v_3 in bigwedge^2V $.
Is it true that $B=textId_V$?
If it helps, we can also assume that $B$ maps decomposable elements to decomposable elements.
It is important that $dim V>6$. If $dim V=6$, one can take any $B=bigwedge^2 A$, where $A in GL(V)$ has determinant one.
Edit: This seems to be true for diagonal maps $B$. Thus it also holds for diagonalizable maps (by multiplicativity).
Indeed, suppose $v_i$ is a basis for $ bigwedge^2V $, and that $Bv_i=lambda_i v_i$. Then condition $(*)$ implies $lambda_i_1 lambda_i_2 lambda_i_3=1$ for every distinct triplet $1le i_1,i_2,i_3 le binomd2=dim(bigwedge^2V )$.
Now we think of the diagonal matrix $A=textdiag(lambda_i)$ as a linear map $W to W$, where $W=bigwedge^2V$ is a $binomd2$-dimensional vector space. Then the $3$-th exterior power of $A$, $bigwedge^3 A=textId_W$. Since $3 < dim(W)$, this implies $A=textId$, so $B=textId$ as required.
We might try to generalize this claim for non-diagonalizable maps, perhaps via some density argument (over $mathbbC$), but at the moment I don't see how to do this.
linear-algebra differential-geometry representation-theory determinant exterior-algebra
asked Jul 23 at 10:07
Asaf Shachar
4,5163832
add a comment |Â
up vote
2
down vote
favorite
$newcommandCofoperatornamecof$
$newcommandidoperatornameId$
$newcommandEndoperatornameEnd$
$newcommandGLoperatornameGL$
Let $V$ be a real $d$-dimensional vector space of dimension $d > 6$.
Let $B in GL(bigwedge^2V) $, and suppose that
$$ B(v_1) wedge B(v_2) wedge B(v_3)=v_1 wedge v_2 wedge v_3 in bigwedge^6V tag* $$
for every multi-vectors $v_1,v_2,v_3 in bigwedge^2V $.
Is it true that $B=textId_V$?
If it helps, we can also assume that $B$ maps decomposable elements to decomposable elements.
It is important that $dim V>6$. If $dim V=6$, one can take any $B=bigwedge^2 A$, where $A in GL(V)$ has determinant one.
Edit: This seems to be true for diagonal maps $B$. Thus it also holds for diagonalizable maps (by multiplicativity).
Indeed, suppose $v_i$ is a basis for $ bigwedge^2V $, and that $Bv_i=lambda_i v_i$. Then condition $(*)$ implies $lambda_i_1 lambda_i_2 lambda_i_3=1$ for every distinct triplet $1le i_1,i_2,i_3 le binomd2=dim(bigwedge^2V )$.
Now we think of the diagonal matrix $A=textdiag(lambda_i)$ as a linear map $W to W$, where $W=bigwedge^2V$ is a $binomd2$-dimensional vector space. Then the $3$-th exterior power of $A$, $bigwedge^3 A=textId_W$. Since $3 < dim(W)$, this implies $A=textId$, so $B=textId$ as required.
We might try to generalize this claim for non-diagonalizable maps, perhaps via some density argument (over $mathbbC$), but at the moment I don't see how to do this.
linear-algebra differential-geometry representation-theory determinant exterior-algebra
asked Jul 23 at 10:07
Asaf Shachar
4,5163832
Note that it is quite possible for $v_i$ to be a basis of $Lambda^2(V)$ with $v_i wedge v_j = 0$ for some $i neq j$ so I'm not sure your argument works even for diagonal maps.
– levap
Jul 24 at 10:45
Thanks, your comment seems correct. I was careless.
– Asaf Shachar
Jul 24 at 11:24
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$newcommandCofoperatornamecof$
$newcommandidoperatornameId$
$newcommandEndoperatornameEnd$
$newcommandGLoperatornameGL$
Let $V$ be a real $d$-dimensional vector space of dimension $d > 6$.
Let $B in GL(bigwedge^2V) $, and suppose that
$$ B(v_1) wedge B(v_2) wedge B(v_3)=v_1 wedge v_2 wedge v_3 in bigwedge^6V tag* $$
for every multi-vectors $v_1,v_2,v_3 in bigwedge^2V $.
Is it true that $B=textId_V$?
If it helps, we can also assume that $B$ maps decomposable elements to decomposable elements.
It is important that $dim V>6$. If $dim V=6$, one can take any $B=bigwedge^2 A$, where $A in GL(V)$ has determinant one.
Edit: This seems to be true for diagonal maps $B$. Thus it also holds for diagonalizable maps (by multiplicativity).
Indeed, suppose $v_i$ is a basis for $ bigwedge^2V $, and that $Bv_i=lambda_i v_i$. Then condition $(*)$ implies $lambda_i_1 lambda_i_2 lambda_i_3=1$ for every distinct triplet $1le i_1,i_2,i_3 le binomd2=dim(bigwedge^2V )$.
Now we think of the diagonal matrix $A=textdiag(lambda_i)$ as a linear map $W to W$, where $W=bigwedge^2V$ is a $binomd2$-dimensional vector space. Then the $3$-th exterior power of $A$, $bigwedge^3 A=textId_W$. Since $3 < dim(W)$, this implies $A=textId$, so $B=textId$ as required.
We might try to generalize this claim for non-diagonalizable maps, perhaps via some density argument (over $mathbbC$), but at the moment I don't see how to do this.
linear-algebra differential-geometry representation-theory determinant exterior-algebra
asked Jul 23 at 10:07
Asaf Shachar
4,5163832
$newcommandCofoperatornamecof$
$newcommandidoperatornameId$
$newcommandEndoperatornameEnd$
$newcommandGLoperatornameGL$
Let $V$ be a real $d$-dimensional vector space of dimension $d > 6$.
Let $B in GL(bigwedge^2V) $, and suppose that
$$ B(v_1) wedge B(v_2) wedge B(v_3)=v_1 wedge v_2 wedge v_3 in bigwedge^6V tag* $$
for every multi-vectors $v_1,v_2,v_3 in bigwedge^2V $.
Is it true that $B=textId_V$?
If it helps, we can also assume that $B$ maps decomposable elements to decomposable elements.
It is important that $dim V>6$. If $dim V=6$, one can take any $B=bigwedge^2 A$, where $A in GL(V)$ has determinant one.
Edit: This seems to be true for diagonal maps $B$. Thus it also holds for diagonalizable maps (by multiplicativity).
Indeed, suppose $v_i$ is a basis for $ bigwedge^2V $, and that $Bv_i=lambda_i v_i$. Then condition $(*)$ implies $lambda_i_1 lambda_i_2 lambda_i_3=1$ for every distinct triplet $1le i_1,i_2,i_3 le binomd2=dim(bigwedge^2V )$.
Now we think of the diagonal matrix $A=textdiag(lambda_i)$ as a linear map $W to W$, where $W=bigwedge^2V$ is a $binomd2$-dimensional vector space. Then the $3$-th exterior power of $A$, $bigwedge^3 A=textId_W$. Since $3 < dim(W)$, this implies $A=textId$, so $B=textId$ as required.
We might try to generalize this claim for non-diagonalizable maps, perhaps via some density argument (over $mathbbC$), but at the moment I don't see how to do this.
linear-algebra differential-geometry representation-theory determinant exterior-algebra
asked Jul 23 at 10:07
Asaf Shachar
4,5163832
edited Jul 23 at 14:07
asked Jul 23 at 10:07
Asaf Shachar
4,5163832
asked Jul 23 at 10:07
Asaf Shachar
4,5163832
asked Jul 23 at 10:07
Asaf Shachar
4,5163832
4,5163832
Note that it is quite possible for $v_i$ to be a basis of $Lambda^2(V)$ with $v_i wedge v_j = 0$ for some $i neq j$ so I'm not sure your argument works even for diagonal maps.
– levap
Jul 24 at 10:45
Thanks, your comment seems correct. I was careless.
– Asaf Shachar
Jul 24 at 11:24
add a comment |Â
Note that it is quite possible for $v_i$ to be a basis of $Lambda^2(V)$ with $v_i wedge v_j = 0$ for some $i neq j$ so I'm not sure your argument works even for diagonal maps.
– levap
Jul 24 at 10:45
Thanks, your comment seems correct. I was careless.
– Asaf Shachar
Jul 24 at 11:24
Note that it is quite possible for $v_i$ to be a basis of $Lambda^2(V)$ with $v_i wedge v_j = 0$ for some $i neq j$ so I'm not sure your argument works even for diagonal maps.
– levap
Jul 24 at 10:45
Note that it is quite possible for $v_i$ to be a basis of $Lambda^2(V)$ with $v_i wedge v_j = 0$ for some $i neq j$ so I'm not sure your argument works even for diagonal maps.
– levap
Jul 24 at 10:45
Thanks, your comment seems correct. I was careless.
– Asaf Shachar
Jul 24 at 11:24
Thanks, your comment seems correct. I was careless.
– Asaf Shachar
Jul 24 at 11:24
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2860192%2fdoes-this-geometric-rigidity-condition-forces-the-map-to-be-the-identity%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Note that it is quite possible for $v_i$ to be a basis of $Lambda^2(V)$ with $v_i wedge v_j = 0$ for some $i neq j$ so I'm not sure your argument works even for diagonal maps.
– levap
Jul 24 at 10:45
Thanks, your comment seems correct. I was careless.
– Asaf Shachar
Jul 24 at 11:24