Mixing
Is every basis for $bigwedge^kV$ satisfying a “complementary” property a rescaling of a “standard”...
$begingroup$
This question was inspired by this beautiful answer:
Let $V$ be a $4$-dimensional real vector space. Let $omega_{i_1,i_2}$ ($1 le i_1 < ldots < i_2 le 4$) be a basis for $bigwedge^2V$, where each $omega_{i_1,i_2}$ is decomposable. Suppose the following property holds: For every basis element $omega=omega_{i_1,i_2}$, there is exactly one basis element $tilde omega=omega_{j_1,j_2}$ such that $omega wedge tilde omega neq 0$.
Must $omega_{i_1,i_2}$ be a rescalig of a "standard" basis
for $bigwedge^2V$ induced by a basis of $V$? i.e. does there exist a basis $v_i$ for $V$, such that $omega_{i_1,i_2}=lambda_{i_1,i_2}v^{i_1} wedge v^{i_2}$?
Note that we must allow a possible rescaling of the original basis:
The "complementary" property is scale-invariant, but being a standard basis is not:
Indeed, if $omega^{i_1,ldots,i_k}$ is a standard basis for $bigwedge^kV$, and $lambda_{i_1,ldots,i_k} omega^{i_1,ldots,i_k} $ is also standard, then the $lambda_{i_1,ldots,i_k}$ must be the $k$-minors of some diagonal $d times d$ matrix. In other words, we have $lambda_{i_1,ldots,i_k}=sigma_{i_1}cdot ldotscdotsigma_{i_k}$ for some $sigma_1,ldots,sigma_d in mathbb{R}$. This implies that the $lambda_{i_1,ldots,i_k}$ cannot be chosen freely; there are non-trivial relations.
Thus, the rescalings of "standard" bases which remain standard are restricted.
Comment: The question can be asked for any even $d$, and $k=d/2$. I thought it would be easier to start with the simplest case.
For the interested reader, here is a proof for the rigidity of standard bases:
We shall prove that $lambda_{i_1,ldots,i_k}$ must be the $k$-minors of some diagonal $d times d$ matrix.
Suppose that $ omega^{i_1,ldots,i_k} =v^{i_1} wedge ldots wedge v^{i_k}$ and $lambda_{i_1,ldots,i_k} omega^{i_1,ldots,i_k} =u^{i_1} wedge ldots wedge u^{i_k}$ for some bases $u_i,v_i$ of $V$. Then, we have $text{span}(v_{i_1},dots,v_{i_k})=text{span}(u_{i_1},dots,u_{i_k})$, for every $1 le i_1 < ldots < i_k le d$. This implies that $u_i in text{span}(v_i)$: Indeed,
by switching between $i_k$ and $i_{k+1}$ in
$$text{span}(v_{i_1},dots,v_{i_{k-1}},v_{i_k})=text{span}(u_{i_1},dots,u_{i_{k-1}},u_{i_k}), tag{1}$$ we obtain
$$text{span}(v_{i_1},dots,v_{i_{k-1}},v_{i_{k+1}})=text{span}(u_{i_1},dots,u_{i_{k-1}},u_{i_{k+1}}). tag{2}$$
By intersecting (1) and (2), we deduce that
$$text{span}(v_{i_1},dots,v_{i_{k-1}})=text{span}(u_{i_1},dots,u_{i_{k-1}}). tag{3}$$
In the passage from $(1)$ to $(3)$ we have "removed" the last vectors $v_{i_k},u_{i_k}$. Continuing in this way, we can remove all vectors, until we get to $text{span}(v_{i_1})=text{span}(u_{i_1})$ as required.
differential-geometry multilinear-algebra exterior-algebra real-algebraic-geometry tensor-decomposition
asked Jan 28 at 9:27
Asaf ShacharAsaf Shachar
5,81131145
$endgroup$
add a comment |
$begingroup$
This question was inspired by this beautiful answer:
Let $V$ be a $4$-dimensional real vector space. Let $omega_{i_1,i_2}$ ($1 le i_1 < ldots < i_2 le 4$) be a basis for $bigwedge^2V$, where each $omega_{i_1,i_2}$ is decomposable. Suppose the following property holds: For every basis element $omega=omega_{i_1,i_2}$, there is exactly one basis element $tilde omega=omega_{j_1,j_2}$ such that $omega wedge tilde omega neq 0$.
Must $omega_{i_1,i_2}$ be a rescalig of a "standard" basis
for $bigwedge^2V$ induced by a basis of $V$? i.e. does there exist a basis $v_i$ for $V$, such that $omega_{i_1,i_2}=lambda_{i_1,i_2}v^{i_1} wedge v^{i_2}$?
Note that we must allow a possible rescaling of the original basis:
The "complementary" property is scale-invariant, but being a standard basis is not:
Indeed, if $omega^{i_1,ldots,i_k}$ is a standard basis for $bigwedge^kV$, and $lambda_{i_1,ldots,i_k} omega^{i_1,ldots,i_k} $ is also standard, then the $lambda_{i_1,ldots,i_k}$ must be the $k$-minors of some diagonal $d times d$ matrix. In other words, we have $lambda_{i_1,ldots,i_k}=sigma_{i_1}cdot ldotscdotsigma_{i_k}$ for some $sigma_1,ldots,sigma_d in mathbb{R}$. This implies that the $lambda_{i_1,ldots,i_k}$ cannot be chosen freely; there are non-trivial relations.
Thus, the rescalings of "standard" bases which remain standard are restricted.
Comment: The question can be asked for any even $d$, and $k=d/2$. I thought it would be easier to start with the simplest case.
For the interested reader, here is a proof for the rigidity of standard bases:
We shall prove that $lambda_{i_1,ldots,i_k}$ must be the $k$-minors of some diagonal $d times d$ matrix.
Suppose that $ omega^{i_1,ldots,i_k} =v^{i_1} wedge ldots wedge v^{i_k}$ and $lambda_{i_1,ldots,i_k} omega^{i_1,ldots,i_k} =u^{i_1} wedge ldots wedge u^{i_k}$ for some bases $u_i,v_i$ of $V$. Then, we have $text{span}(v_{i_1},dots,v_{i_k})=text{span}(u_{i_1},dots,u_{i_k})$, for every $1 le i_1 < ldots < i_k le d$. This implies that $u_i in text{span}(v_i)$: Indeed,
by switching between $i_k$ and $i_{k+1}$ in
$$text{span}(v_{i_1},dots,v_{i_{k-1}},v_{i_k})=text{span}(u_{i_1},dots,u_{i_{k-1}},u_{i_k}), tag{1}$$ we obtain
$$text{span}(v_{i_1},dots,v_{i_{k-1}},v_{i_{k+1}})=text{span}(u_{i_1},dots,u_{i_{k-1}},u_{i_{k+1}}). tag{2}$$
By intersecting (1) and (2), we deduce that
$$text{span}(v_{i_1},dots,v_{i_{k-1}})=text{span}(u_{i_1},dots,u_{i_{k-1}}). tag{3}$$
In the passage from $(1)$ to $(3)$ we have "removed" the last vectors $v_{i_k},u_{i_k}$. Continuing in this way, we can remove all vectors, until we get to $text{span}(v_{i_1})=text{span}(u_{i_1})$ as required.
differential-geometry multilinear-algebra exterior-algebra real-algebraic-geometry tensor-decomposition
asked Jan 28 at 9:27
Asaf ShacharAsaf Shachar
5,81131145
$endgroup$
$begingroup$
Perhaps this is another one for @EricWofsey !
$endgroup$
– Jose Brox
Jan 29 at 16:31
1
$begingroup$
I now asked this also at mathoverflow: mathoverflow.net/questions/322135/…
$endgroup$
– Asaf Shachar
Jan 31 at 14:19
add a comment |
$begingroup$
This question was inspired by this beautiful answer:
Let $V$ be a $4$-dimensional real vector space. Let $omega_{i_1,i_2}$ ($1 le i_1 < ldots < i_2 le 4$) be a basis for $bigwedge^2V$, where each $omega_{i_1,i_2}$ is decomposable. Suppose the following property holds: For every basis element $omega=omega_{i_1,i_2}$, there is exactly one basis element $tilde omega=omega_{j_1,j_2}$ such that $omega wedge tilde omega neq 0$.
Must $omega_{i_1,i_2}$ be a rescalig of a "standard" basis
for $bigwedge^2V$ induced by a basis of $V$? i.e. does there exist a basis $v_i$ for $V$, such that $omega_{i_1,i_2}=lambda_{i_1,i_2}v^{i_1} wedge v^{i_2}$?
Note that we must allow a possible rescaling of the original basis:
The "complementary" property is scale-invariant, but being a standard basis is not:
Indeed, if $omega^{i_1,ldots,i_k}$ is a standard basis for $bigwedge^kV$, and $lambda_{i_1,ldots,i_k} omega^{i_1,ldots,i_k} $ is also standard, then the $lambda_{i_1,ldots,i_k}$ must be the $k$-minors of some diagonal $d times d$ matrix. In other words, we have $lambda_{i_1,ldots,i_k}=sigma_{i_1}cdot ldotscdotsigma_{i_k}$ for some $sigma_1,ldots,sigma_d in mathbb{R}$. This implies that the $lambda_{i_1,ldots,i_k}$ cannot be chosen freely; there are non-trivial relations.
Thus, the rescalings of "standard" bases which remain standard are restricted.
Comment: The question can be asked for any even $d$, and $k=d/2$. I thought it would be easier to start with the simplest case.
For the interested reader, here is a proof for the rigidity of standard bases:
We shall prove that $lambda_{i_1,ldots,i_k}$ must be the $k$-minors of some diagonal $d times d$ matrix.
Suppose that $ omega^{i_1,ldots,i_k} =v^{i_1} wedge ldots wedge v^{i_k}$ and $lambda_{i_1,ldots,i_k} omega^{i_1,ldots,i_k} =u^{i_1} wedge ldots wedge u^{i_k}$ for some bases $u_i,v_i$ of $V$. Then, we have $text{span}(v_{i_1},dots,v_{i_k})=text{span}(u_{i_1},dots,u_{i_k})$, for every $1 le i_1 < ldots < i_k le d$. This implies that $u_i in text{span}(v_i)$: Indeed,
by switching between $i_k$ and $i_{k+1}$ in
$$text{span}(v_{i_1},dots,v_{i_{k-1}},v_{i_k})=text{span}(u_{i_1},dots,u_{i_{k-1}},u_{i_k}), tag{1}$$ we obtain
$$text{span}(v_{i_1},dots,v_{i_{k-1}},v_{i_{k+1}})=text{span}(u_{i_1},dots,u_{i_{k-1}},u_{i_{k+1}}). tag{2}$$
By intersecting (1) and (2), we deduce that
$$text{span}(v_{i_1},dots,v_{i_{k-1}})=text{span}(u_{i_1},dots,u_{i_{k-1}}). tag{3}$$
In the passage from $(1)$ to $(3)$ we have "removed" the last vectors $v_{i_k},u_{i_k}$. Continuing in this way, we can remove all vectors, until we get to $text{span}(v_{i_1})=text{span}(u_{i_1})$ as required.
differential-geometry multilinear-algebra exterior-algebra real-algebraic-geometry tensor-decomposition
asked Jan 28 at 9:27
Asaf ShacharAsaf Shachar
5,81131145
$endgroup$
This question was inspired by this beautiful answer:
Let $V$ be a $4$-dimensional real vector space. Let $omega_{i_1,i_2}$ ($1 le i_1 < ldots < i_2 le 4$) be a basis for $bigwedge^2V$, where each $omega_{i_1,i_2}$ is decomposable. Suppose the following property holds: For every basis element $omega=omega_{i_1,i_2}$, there is exactly one basis element $tilde omega=omega_{j_1,j_2}$ such that $omega wedge tilde omega neq 0$.
Must $omega_{i_1,i_2}$ be a rescalig of a "standard" basis
for $bigwedge^2V$ induced by a basis of $V$? i.e. does there exist a basis $v_i$ for $V$, such that $omega_{i_1,i_2}=lambda_{i_1,i_2}v^{i_1} wedge v^{i_2}$?
Note that we must allow a possible rescaling of the original basis:
The "complementary" property is scale-invariant, but being a standard basis is not:
Indeed, if $omega^{i_1,ldots,i_k}$ is a standard basis for $bigwedge^kV$, and $lambda_{i_1,ldots,i_k} omega^{i_1,ldots,i_k} $ is also standard, then the $lambda_{i_1,ldots,i_k}$ must be the $k$-minors of some diagonal $d times d$ matrix. In other words, we have $lambda_{i_1,ldots,i_k}=sigma_{i_1}cdot ldotscdotsigma_{i_k}$ for some $sigma_1,ldots,sigma_d in mathbb{R}$. This implies that the $lambda_{i_1,ldots,i_k}$ cannot be chosen freely; there are non-trivial relations.
Thus, the rescalings of "standard" bases which remain standard are restricted.
Comment: The question can be asked for any even $d$, and $k=d/2$. I thought it would be easier to start with the simplest case.
For the interested reader, here is a proof for the rigidity of standard bases:
We shall prove that $lambda_{i_1,ldots,i_k}$ must be the $k$-minors of some diagonal $d times d$ matrix.
Suppose that $ omega^{i_1,ldots,i_k} =v^{i_1} wedge ldots wedge v^{i_k}$ and $lambda_{i_1,ldots,i_k} omega^{i_1,ldots,i_k} =u^{i_1} wedge ldots wedge u^{i_k}$ for some bases $u_i,v_i$ of $V$. Then, we have $text{span}(v_{i_1},dots,v_{i_k})=text{span}(u_{i_1},dots,u_{i_k})$, for every $1 le i_1 < ldots < i_k le d$. This implies that $u_i in text{span}(v_i)$: Indeed,
by switching between $i_k$ and $i_{k+1}$ in
$$text{span}(v_{i_1},dots,v_{i_{k-1}},v_{i_k})=text{span}(u_{i_1},dots,u_{i_{k-1}},u_{i_k}), tag{1}$$ we obtain
$$text{span}(v_{i_1},dots,v_{i_{k-1}},v_{i_{k+1}})=text{span}(u_{i_1},dots,u_{i_{k-1}},u_{i_{k+1}}). tag{2}$$
By intersecting (1) and (2), we deduce that
$$text{span}(v_{i_1},dots,v_{i_{k-1}})=text{span}(u_{i_1},dots,u_{i_{k-1}}). tag{3}$$
In the passage from $(1)$ to $(3)$ we have "removed" the last vectors $v_{i_k},u_{i_k}$. Continuing in this way, we can remove all vectors, until we get to $text{span}(v_{i_1})=text{span}(u_{i_1})$ as required.
differential-geometry multilinear-algebra exterior-algebra real-algebraic-geometry tensor-decomposition
differential-geometry multilinear-algebra exterior-algebra real-algebraic-geometry tensor-decomposition
asked Jan 28 at 9:27
Asaf ShacharAsaf Shachar
5,81131145
asked Jan 28 at 9:27
Asaf ShacharAsaf Shachar
5,81131145
asked Jan 28 at 9:27
Asaf ShacharAsaf Shachar
5,81131145
asked Jan 28 at 9:27
Asaf ShacharAsaf Shachar
5,81131145
asked Jan 28 at 9:27
Asaf ShacharAsaf Shachar
5,81131145
5,81131145
$begingroup$
Perhaps this is another one for @EricWofsey !
$endgroup$
– Jose Brox
Jan 29 at 16:31
1
$begingroup$
I now asked this also at mathoverflow: mathoverflow.net/questions/322135/…
$endgroup$
– Asaf Shachar
Jan 31 at 14:19
add a comment |
$begingroup$
Perhaps this is another one for @EricWofsey !
$endgroup$
– Jose Brox
Jan 29 at 16:31
1
$begingroup$
I now asked this also at mathoverflow: mathoverflow.net/questions/322135/…
$endgroup$
– Asaf Shachar
Jan 31 at 14:19
$begingroup$
Perhaps this is another one for @EricWofsey !
$endgroup$
– Jose Brox
Jan 29 at 16:31
$begingroup$
Perhaps this is another one for @EricWofsey !
$endgroup$
– Jose Brox
Jan 29 at 16:31
1
1
$begingroup$
I now asked this also at mathoverflow: mathoverflow.net/questions/322135/…
$endgroup$
– Asaf Shachar
Jan 31 at 14:19
$begingroup$
I now asked this also at mathoverflow: mathoverflow.net/questions/322135/…
$endgroup$
– Asaf Shachar
Jan 31 at 14:19
add a comment |
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$begingroup$
Perhaps this is another one for @EricWofsey !
$endgroup$
– Jose Brox
Jan 29 at 16:31
1
$begingroup$
I now asked this also at mathoverflow: mathoverflow.net/questions/322135/…
$endgroup$
– Asaf Shachar
Jan 31 at 14:19