Mixing
Is every decomposable basis for $bigwedge^kV$ “standard”?
$begingroup$
This is a curiosity:
Let $V$ be a $d$-dimensional real vector space, and let $1<k<d$. Set
Let $omega^{i_1,ldots,i_k}$ be a basis for $bigwedge^kV$, whose elements are all decomposable. Is $omega^{i_1,ldots,i_k}$ necessarily "induced" by a standard basis up to scaling?
i.e. does there exist a basis $v_i$ for $V$, such that $omega^{i_1,ldots,i_k}=lambda_{i_1,ldots,i_k}v^{i_1} wedge ldots wedge v^{i_k}$ for some real scalars $lambda_{i_1,ldots,i_k}$?
representation-theory multilinear-algebra exterior-algebra tensor-decomposition
asked Jan 27 at 7:56
Asaf ShacharAsaf Shachar
5,78431144
$endgroup$
add a comment |
$begingroup$
This is a curiosity:
Let $V$ be a $d$-dimensional real vector space, and let $1<k<d$. Set
Let $omega^{i_1,ldots,i_k}$ be a basis for $bigwedge^kV$, whose elements are all decomposable. Is $omega^{i_1,ldots,i_k}$ necessarily "induced" by a standard basis up to scaling?
i.e. does there exist a basis $v_i$ for $V$, such that $omega^{i_1,ldots,i_k}=lambda_{i_1,ldots,i_k}v^{i_1} wedge ldots wedge v^{i_k}$ for some real scalars $lambda_{i_1,ldots,i_k}$?
representation-theory multilinear-algebra exterior-algebra tensor-decomposition
asked Jan 27 at 7:56
Asaf ShacharAsaf Shachar
5,78431144
$endgroup$
add a comment |
$begingroup$
This is a curiosity:
Let $V$ be a $d$-dimensional real vector space, and let $1<k<d$. Set
Let $omega^{i_1,ldots,i_k}$ be a basis for $bigwedge^kV$, whose elements are all decomposable. Is $omega^{i_1,ldots,i_k}$ necessarily "induced" by a standard basis up to scaling?
i.e. does there exist a basis $v_i$ for $V$, such that $omega^{i_1,ldots,i_k}=lambda_{i_1,ldots,i_k}v^{i_1} wedge ldots wedge v^{i_k}$ for some real scalars $lambda_{i_1,ldots,i_k}$?
representation-theory multilinear-algebra exterior-algebra tensor-decomposition
asked Jan 27 at 7:56
Asaf ShacharAsaf Shachar
5,78431144
$endgroup$
This is a curiosity:
Let $V$ be a $d$-dimensional real vector space, and let $1<k<d$. Set
Let $omega^{i_1,ldots,i_k}$ be a basis for $bigwedge^kV$, whose elements are all decomposable. Is $omega^{i_1,ldots,i_k}$ necessarily "induced" by a standard basis up to scaling?
i.e. does there exist a basis $v_i$ for $V$, such that $omega^{i_1,ldots,i_k}=lambda_{i_1,ldots,i_k}v^{i_1} wedge ldots wedge v^{i_k}$ for some real scalars $lambda_{i_1,ldots,i_k}$?
representation-theory multilinear-algebra exterior-algebra tensor-decomposition
representation-theory multilinear-algebra exterior-algebra tensor-decomposition
asked Jan 27 at 7:56
Asaf ShacharAsaf Shachar
5,78431144
asked Jan 27 at 7:56
Asaf ShacharAsaf Shachar
5,78431144
asked Jan 27 at 7:56
Asaf ShacharAsaf Shachar
5,78431144
asked Jan 27 at 7:56
Asaf ShacharAsaf Shachar
5,78431144
asked Jan 27 at 7:56
Asaf ShacharAsaf Shachar
5,78431144
5,78431144
add a comment |
add a comment |
1 Answer
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$begingroup$
No. For instance, let $d=4$ and $k=2$. Note that every standard basis for $bigwedge^2 V$ (up to scaling) has the property that for any basis element $b$, there is exactly one basis element $c$ such that $bwedge cneq 0$. On the other hand, if ${w,x,y,z}$ is a basis for $V$, then consider the following decomposable basis:
$$wwedge x, wwedge y, wwedge z,xwedge y, (x+y)wedge z,ywedge z.$$
Here, the basis element $b=wwedge x$ has two different $c$ such that $bwedge cneq 0$, namely $c=ywedge z$ and $c=(x+y)wedge z$.
answered Jan 27 at 8:06
Eric WofseyEric Wofsey
190k14216348
$endgroup$
$begingroup$
This is a perfect answer! extremely elegant. Thank you.
$endgroup$
– Asaf Shachar
Jan 27 at 9:43
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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$begingroup$
No. For instance, let $d=4$ and $k=2$. Note that every standard basis for $bigwedge^2 V$ (up to scaling) has the property that for any basis element $b$, there is exactly one basis element $c$ such that $bwedge cneq 0$. On the other hand, if ${w,x,y,z}$ is a basis for $V$, then consider the following decomposable basis:
$$wwedge x, wwedge y, wwedge z,xwedge y, (x+y)wedge z,ywedge z.$$
Here, the basis element $b=wwedge x$ has two different $c$ such that $bwedge cneq 0$, namely $c=ywedge z$ and $c=(x+y)wedge z$.
answered Jan 27 at 8:06
Eric WofseyEric Wofsey
190k14216348
$endgroup$
$begingroup$
This is a perfect answer! extremely elegant. Thank you.
$endgroup$
– Asaf Shachar
Jan 27 at 9:43
add a comment |
$begingroup$
No. For instance, let $d=4$ and $k=2$. Note that every standard basis for $bigwedge^2 V$ (up to scaling) has the property that for any basis element $b$, there is exactly one basis element $c$ such that $bwedge cneq 0$. On the other hand, if ${w,x,y,z}$ is a basis for $V$, then consider the following decomposable basis:
$$wwedge x, wwedge y, wwedge z,xwedge y, (x+y)wedge z,ywedge z.$$
Here, the basis element $b=wwedge x$ has two different $c$ such that $bwedge cneq 0$, namely $c=ywedge z$ and $c=(x+y)wedge z$.
answered Jan 27 at 8:06
Eric WofseyEric Wofsey
190k14216348
$endgroup$
$begingroup$
This is a perfect answer! extremely elegant. Thank you.
$endgroup$
– Asaf Shachar
Jan 27 at 9:43
add a comment |
$begingroup$
No. For instance, let $d=4$ and $k=2$. Note that every standard basis for $bigwedge^2 V$ (up to scaling) has the property that for any basis element $b$, there is exactly one basis element $c$ such that $bwedge cneq 0$. On the other hand, if ${w,x,y,z}$ is a basis for $V$, then consider the following decomposable basis:
$$wwedge x, wwedge y, wwedge z,xwedge y, (x+y)wedge z,ywedge z.$$
Here, the basis element $b=wwedge x$ has two different $c$ such that $bwedge cneq 0$, namely $c=ywedge z$ and $c=(x+y)wedge z$.
answered Jan 27 at 8:06
Eric WofseyEric Wofsey
190k14216348
$endgroup$
No. For instance, let $d=4$ and $k=2$. Note that every standard basis for $bigwedge^2 V$ (up to scaling) has the property that for any basis element $b$, there is exactly one basis element $c$ such that $bwedge cneq 0$. On the other hand, if ${w,x,y,z}$ is a basis for $V$, then consider the following decomposable basis:
$$wwedge x, wwedge y, wwedge z,xwedge y, (x+y)wedge z,ywedge z.$$
Here, the basis element $b=wwedge x$ has two different $c$ such that $bwedge cneq 0$, namely $c=ywedge z$ and $c=(x+y)wedge z$.
answered Jan 27 at 8:06
Eric WofseyEric Wofsey
190k14216348
answered Jan 27 at 8:06
Eric WofseyEric Wofsey
190k14216348
answered Jan 27 at 8:06
Eric WofseyEric Wofsey
190k14216348
answered Jan 27 at 8:06
Eric WofseyEric Wofsey
190k14216348
190k14216348
$begingroup$
This is a perfect answer! extremely elegant. Thank you.
$endgroup$
– Asaf Shachar
Jan 27 at 9:43
add a comment |
$begingroup$
This is a perfect answer! extremely elegant. Thank you.
$endgroup$
– Asaf Shachar
Jan 27 at 9:43
$begingroup$
This is a perfect answer! extremely elegant. Thank you.
$endgroup$
– Asaf Shachar
Jan 27 at 9:43
$begingroup$
This is a perfect answer! extremely elegant. Thank you.
$endgroup$
– Asaf Shachar
Jan 27 at 9:43
add a comment |
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