Mixing
Is “being decomposable” preserved under taking a subspace?
$begingroup$
Let $V$ be a vector space over some field, and $W le V$ a vector subspace. Let $1<k<dim V$ be an integer.
Suppose $omega in bigwedge^k W$ is decomposable as an element in $bigwedge^k V$. Is it decomposable as an element in $bigwedge^k W$?
multilinear-algebra exterior-algebra grassmannian tensor-decomposition
asked Feb 3 at 11:53
Asaf ShacharAsaf Shachar
5,79931145
$endgroup$
add a comment |
$begingroup$
Let $V$ be a vector space over some field, and $W le V$ a vector subspace. Let $1<k<dim V$ be an integer.
Suppose $omega in bigwedge^k W$ is decomposable as an element in $bigwedge^k V$. Is it decomposable as an element in $bigwedge^k W$?
multilinear-algebra exterior-algebra grassmannian tensor-decomposition
asked Feb 3 at 11:53
Asaf ShacharAsaf Shachar
5,79931145
$endgroup$
add a comment |
$begingroup$
Let $V$ be a vector space over some field, and $W le V$ a vector subspace. Let $1<k<dim V$ be an integer.
Suppose $omega in bigwedge^k W$ is decomposable as an element in $bigwedge^k V$. Is it decomposable as an element in $bigwedge^k W$?
multilinear-algebra exterior-algebra grassmannian tensor-decomposition
asked Feb 3 at 11:53
Asaf ShacharAsaf Shachar
5,79931145
$endgroup$
Let $V$ be a vector space over some field, and $W le V$ a vector subspace. Let $1<k<dim V$ be an integer.
Suppose $omega in bigwedge^k W$ is decomposable as an element in $bigwedge^k V$. Is it decomposable as an element in $bigwedge^k W$?
multilinear-algebra exterior-algebra grassmannian tensor-decomposition
multilinear-algebra exterior-algebra grassmannian tensor-decomposition
asked Feb 3 at 11:53
Asaf ShacharAsaf Shachar
5,79931145
asked Feb 3 at 11:53
Asaf ShacharAsaf Shachar
5,79931145
asked Feb 3 at 11:53
Asaf ShacharAsaf Shachar
5,79931145
asked Feb 3 at 11:53
Asaf ShacharAsaf Shachar
5,79931145
asked Feb 3 at 11:53
Asaf ShacharAsaf Shachar
5,79931145
5,79931145
add a comment |
add a comment |
1 Answer
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$begingroup$
The answer is positive. Let $w_i$ be a basis for $W$. Write $omega=sum_{1 le i_1 <dots<i_k le dim W)} w_{i_1} wedge ldots wedge w_{i_k}$. Let $v in V setminus W$. Since $w_i,v$ are linearly independent, we can complete them into a basis of $V$. This implies that $omega wedge v=sum_{1 le i_1 <dots<i_k le dim W)} w_{i_1} wedge ldots wedge w_{i_k} wedge v neq 0$.
By assumption, $omega in bigwedge^k W$ is decomposable in $bigwedge^k V$, i.e. $omega=v_1 wedge dots wedge v_k$, for some $v_i in V$. Since $v_1 wedge dots wedge v_k wedge v=omega wedge v neq 0$, we deduce that $v notin {v_1,dots,v_k}$. Since $v in V setminus W$ was arbitrary, we have proved that $V setminus W=W^c subseteq {v_1,dots,v_k}^c$, so $ {v_1,dots,v_k} subseteq W$, so $omega=v_1 wedge dots wedge v_k$ is decomposable in $bigwedge^k W$.
An alternative proof:
$omega$ is decomposable in $bigwedge^k V$, if and only if $V_{omega}={ v in V , | , v wedge omega=0}$ is $k$-dimensioal.
Now, $v in V setminus W Rightarrow v wedge omega neq 0 Rightarrow v notin V_{omega}$, i.e. $W^c subseteq V_{omega}^c$. Thus,
so $V_{omega} subseteq W$, hence
$$ W_{omega}=V_{omega} cap W=V_{omega}$$
is also $k$-dimensional, so $omega$ is decomposable in $bigwedge^k W$.
answered Feb 3 at 11:53
Asaf ShacharAsaf Shachar
5,79931145
$endgroup$
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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$begingroup$
The answer is positive. Let $w_i$ be a basis for $W$. Write $omega=sum_{1 le i_1 <dots<i_k le dim W)} w_{i_1} wedge ldots wedge w_{i_k}$. Let $v in V setminus W$. Since $w_i,v$ are linearly independent, we can complete them into a basis of $V$. This implies that $omega wedge v=sum_{1 le i_1 <dots<i_k le dim W)} w_{i_1} wedge ldots wedge w_{i_k} wedge v neq 0$.
By assumption, $omega in bigwedge^k W$ is decomposable in $bigwedge^k V$, i.e. $omega=v_1 wedge dots wedge v_k$, for some $v_i in V$. Since $v_1 wedge dots wedge v_k wedge v=omega wedge v neq 0$, we deduce that $v notin {v_1,dots,v_k}$. Since $v in V setminus W$ was arbitrary, we have proved that $V setminus W=W^c subseteq {v_1,dots,v_k}^c$, so $ {v_1,dots,v_k} subseteq W$, so $omega=v_1 wedge dots wedge v_k$ is decomposable in $bigwedge^k W$.
An alternative proof:
$omega$ is decomposable in $bigwedge^k V$, if and only if $V_{omega}={ v in V , | , v wedge omega=0}$ is $k$-dimensioal.
Now, $v in V setminus W Rightarrow v wedge omega neq 0 Rightarrow v notin V_{omega}$, i.e. $W^c subseteq V_{omega}^c$. Thus,
so $V_{omega} subseteq W$, hence
$$ W_{omega}=V_{omega} cap W=V_{omega}$$
is also $k$-dimensional, so $omega$ is decomposable in $bigwedge^k W$.
answered Feb 3 at 11:53
Asaf ShacharAsaf Shachar
5,79931145
$endgroup$
add a comment |
$begingroup$
The answer is positive. Let $w_i$ be a basis for $W$. Write $omega=sum_{1 le i_1 <dots<i_k le dim W)} w_{i_1} wedge ldots wedge w_{i_k}$. Let $v in V setminus W$. Since $w_i,v$ are linearly independent, we can complete them into a basis of $V$. This implies that $omega wedge v=sum_{1 le i_1 <dots<i_k le dim W)} w_{i_1} wedge ldots wedge w_{i_k} wedge v neq 0$.
By assumption, $omega in bigwedge^k W$ is decomposable in $bigwedge^k V$, i.e. $omega=v_1 wedge dots wedge v_k$, for some $v_i in V$. Since $v_1 wedge dots wedge v_k wedge v=omega wedge v neq 0$, we deduce that $v notin {v_1,dots,v_k}$. Since $v in V setminus W$ was arbitrary, we have proved that $V setminus W=W^c subseteq {v_1,dots,v_k}^c$, so $ {v_1,dots,v_k} subseteq W$, so $omega=v_1 wedge dots wedge v_k$ is decomposable in $bigwedge^k W$.
An alternative proof:
$omega$ is decomposable in $bigwedge^k V$, if and only if $V_{omega}={ v in V , | , v wedge omega=0}$ is $k$-dimensioal.
Now, $v in V setminus W Rightarrow v wedge omega neq 0 Rightarrow v notin V_{omega}$, i.e. $W^c subseteq V_{omega}^c$. Thus,
so $V_{omega} subseteq W$, hence
$$ W_{omega}=V_{omega} cap W=V_{omega}$$
is also $k$-dimensional, so $omega$ is decomposable in $bigwedge^k W$.
answered Feb 3 at 11:53
Asaf ShacharAsaf Shachar
5,79931145
$endgroup$
add a comment |
$begingroup$
The answer is positive. Let $w_i$ be a basis for $W$. Write $omega=sum_{1 le i_1 <dots<i_k le dim W)} w_{i_1} wedge ldots wedge w_{i_k}$. Let $v in V setminus W$. Since $w_i,v$ are linearly independent, we can complete them into a basis of $V$. This implies that $omega wedge v=sum_{1 le i_1 <dots<i_k le dim W)} w_{i_1} wedge ldots wedge w_{i_k} wedge v neq 0$.
By assumption, $omega in bigwedge^k W$ is decomposable in $bigwedge^k V$, i.e. $omega=v_1 wedge dots wedge v_k$, for some $v_i in V$. Since $v_1 wedge dots wedge v_k wedge v=omega wedge v neq 0$, we deduce that $v notin {v_1,dots,v_k}$. Since $v in V setminus W$ was arbitrary, we have proved that $V setminus W=W^c subseteq {v_1,dots,v_k}^c$, so $ {v_1,dots,v_k} subseteq W$, so $omega=v_1 wedge dots wedge v_k$ is decomposable in $bigwedge^k W$.
An alternative proof:
$omega$ is decomposable in $bigwedge^k V$, if and only if $V_{omega}={ v in V , | , v wedge omega=0}$ is $k$-dimensioal.
Now, $v in V setminus W Rightarrow v wedge omega neq 0 Rightarrow v notin V_{omega}$, i.e. $W^c subseteq V_{omega}^c$. Thus,
so $V_{omega} subseteq W$, hence
$$ W_{omega}=V_{omega} cap W=V_{omega}$$
is also $k$-dimensional, so $omega$ is decomposable in $bigwedge^k W$.
answered Feb 3 at 11:53
Asaf ShacharAsaf Shachar
5,79931145
$endgroup$
The answer is positive. Let $w_i$ be a basis for $W$. Write $omega=sum_{1 le i_1 <dots<i_k le dim W)} w_{i_1} wedge ldots wedge w_{i_k}$. Let $v in V setminus W$. Since $w_i,v$ are linearly independent, we can complete them into a basis of $V$. This implies that $omega wedge v=sum_{1 le i_1 <dots<i_k le dim W)} w_{i_1} wedge ldots wedge w_{i_k} wedge v neq 0$.
By assumption, $omega in bigwedge^k W$ is decomposable in $bigwedge^k V$, i.e. $omega=v_1 wedge dots wedge v_k$, for some $v_i in V$. Since $v_1 wedge dots wedge v_k wedge v=omega wedge v neq 0$, we deduce that $v notin {v_1,dots,v_k}$. Since $v in V setminus W$ was arbitrary, we have proved that $V setminus W=W^c subseteq {v_1,dots,v_k}^c$, so $ {v_1,dots,v_k} subseteq W$, so $omega=v_1 wedge dots wedge v_k$ is decomposable in $bigwedge^k W$.
An alternative proof:
$omega$ is decomposable in $bigwedge^k V$, if and only if $V_{omega}={ v in V , | , v wedge omega=0}$ is $k$-dimensioal.
Now, $v in V setminus W Rightarrow v wedge omega neq 0 Rightarrow v notin V_{omega}$, i.e. $W^c subseteq V_{omega}^c$. Thus,
so $V_{omega} subseteq W$, hence
$$ W_{omega}=V_{omega} cap W=V_{omega}$$
is also $k$-dimensional, so $omega$ is decomposable in $bigwedge^k W$.
answered Feb 3 at 11:53
Asaf ShacharAsaf Shachar
5,79931145
edited Feb 4 at 9:21
answered Feb 3 at 11:53
Asaf ShacharAsaf Shachar
5,79931145
answered Feb 3 at 11:53
Asaf ShacharAsaf Shachar
5,79931145
answered Feb 3 at 11:53
Asaf ShacharAsaf Shachar
5,79931145
5,79931145
add a comment |
add a comment |
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