In which degrees there exist non-decomposable elements in the exterior algebra?
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I am trying to get a better understanding of the concept "decomposable" element in an exterior algebra. Let $V$ be a $d$-dimensional real vector space, and let $1<k<d$.
For which tuples $(k,d)$, $bigwedge^k V $ contains non-decomposable elements?
Here is a partial answer:
First, the answer for $(k,d),(d-k,d)$ is the same, by duality. So, it suffices to assume $k le d/2$.
For even $kle d/2$, there are always non-decomposable elements:
Set $sigma = e_1 wedge dots wedge e_k + e_{k+1} wedge dots wedge e_{2k} $, where $e_1,dots,e_d$ form a basis for $V$.
Then $sigma wedge sigma=(1+(-1)^{k^2})e_1 wedge dots wedge e_{2k}=(1+(-1)^k)e_1 wedge dots wedge e_{2k}$, which for even $k$ becomes $sigma wedge sigma=2e_1 wedge dots wedge e_{2k} neq 0$, so $sigma$ must be non-decomposable.
- For $(1,d)$, (and hence also for $(d-1,d)$) every element is decomposable.
I am not sure what happens for odd $k le d/2$.
differential-geometry symmetry exterior-algebra tensor-decomposition
asked Jan 21 at 13:23
Asaf ShacharAsaf Shachar
5,64631141
$endgroup$
add a comment |
$begingroup$
I am trying to get a better understanding of the concept "decomposable" element in an exterior algebra. Let $V$ be a $d$-dimensional real vector space, and let $1<k<d$.
For which tuples $(k,d)$, $bigwedge^k V $ contains non-decomposable elements?
Here is a partial answer:
First, the answer for $(k,d),(d-k,d)$ is the same, by duality. So, it suffices to assume $k le d/2$.
For even $kle d/2$, there are always non-decomposable elements:
Set $sigma = e_1 wedge dots wedge e_k + e_{k+1} wedge dots wedge e_{2k} $, where $e_1,dots,e_d$ form a basis for $V$.
Then $sigma wedge sigma=(1+(-1)^{k^2})e_1 wedge dots wedge e_{2k}=(1+(-1)^k)e_1 wedge dots wedge e_{2k}$, which for even $k$ becomes $sigma wedge sigma=2e_1 wedge dots wedge e_{2k} neq 0$, so $sigma$ must be non-decomposable.
- For $(1,d)$, (and hence also for $(d-1,d)$) every element is decomposable.
I am not sure what happens for odd $k le d/2$.
differential-geometry symmetry exterior-algebra tensor-decomposition
asked Jan 21 at 13:23
Asaf ShacharAsaf Shachar
5,64631141
$endgroup$
add a comment |
$begingroup$
I am trying to get a better understanding of the concept "decomposable" element in an exterior algebra. Let $V$ be a $d$-dimensional real vector space, and let $1<k<d$.
For which tuples $(k,d)$, $bigwedge^k V $ contains non-decomposable elements?
Here is a partial answer:
First, the answer for $(k,d),(d-k,d)$ is the same, by duality. So, it suffices to assume $k le d/2$.
For even $kle d/2$, there are always non-decomposable elements:
Set $sigma = e_1 wedge dots wedge e_k + e_{k+1} wedge dots wedge e_{2k} $, where $e_1,dots,e_d$ form a basis for $V$.
Then $sigma wedge sigma=(1+(-1)^{k^2})e_1 wedge dots wedge e_{2k}=(1+(-1)^k)e_1 wedge dots wedge e_{2k}$, which for even $k$ becomes $sigma wedge sigma=2e_1 wedge dots wedge e_{2k} neq 0$, so $sigma$ must be non-decomposable.
- For $(1,d)$, (and hence also for $(d-1,d)$) every element is decomposable.
I am not sure what happens for odd $k le d/2$.
differential-geometry symmetry exterior-algebra tensor-decomposition
asked Jan 21 at 13:23
Asaf ShacharAsaf Shachar
5,64631141
$endgroup$
I am trying to get a better understanding of the concept "decomposable" element in an exterior algebra. Let $V$ be a $d$-dimensional real vector space, and let $1<k<d$.
For which tuples $(k,d)$, $bigwedge^k V $ contains non-decomposable elements?
Here is a partial answer:
First, the answer for $(k,d),(d-k,d)$ is the same, by duality. So, it suffices to assume $k le d/2$.
For even $kle d/2$, there are always non-decomposable elements:
Set $sigma = e_1 wedge dots wedge e_k + e_{k+1} wedge dots wedge e_{2k} $, where $e_1,dots,e_d$ form a basis for $V$.
Then $sigma wedge sigma=(1+(-1)^{k^2})e_1 wedge dots wedge e_{2k}=(1+(-1)^k)e_1 wedge dots wedge e_{2k}$, which for even $k$ becomes $sigma wedge sigma=2e_1 wedge dots wedge e_{2k} neq 0$, so $sigma$ must be non-decomposable.
- For $(1,d)$, (and hence also for $(d-1,d)$) every element is decomposable.
I am not sure what happens for odd $k le d/2$.
differential-geometry symmetry exterior-algebra tensor-decomposition
differential-geometry symmetry exterior-algebra tensor-decomposition
asked Jan 21 at 13:23
Asaf ShacharAsaf Shachar
5,64631141
asked Jan 21 at 13:23
Asaf ShacharAsaf Shachar
5,64631141
asked Jan 21 at 13:23
Asaf ShacharAsaf Shachar
5,64631141
asked Jan 21 at 13:23
Asaf ShacharAsaf Shachar
5,64631141
asked Jan 21 at 13:23
Asaf ShacharAsaf Shachar
5,64631141
5,64631141
add a comment |
add a comment |
1 Answer
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The set of decomposable elements forms a smooth projective subvariety of $Bbb{P}(bigwedge^kV)$ of dimension $k(d-k)$, called the Grassmannian. In particular there are non-decomposable elements if and only if
$$k(d-k)<dimBbb{P}(bigwedge^kV)=binom{d}{k}-1,$$
i.e. if and only if $1<k<d-1$.
answered Jan 22 at 13:21
ServaesServaes
25.8k33996
$endgroup$
$begingroup$
Thanks! This is a nice argument. By the way, do you know a nice reference for reading about the Plucker relations? The wikipedia article contains only a "coordinate formulation", but I guess there should be a more invariant (coordinate-free) approach...
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– Asaf Shachar
Jan 22 at 13:23
$begingroup$
I believe EGA I (second edition) has a section on Grassmannians, which is no doubt coordinate-free. I don't have a copy at hand though.
$endgroup$
– Servaes
Jan 22 at 14:53
add a comment |
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$begingroup$
The set of decomposable elements forms a smooth projective subvariety of $Bbb{P}(bigwedge^kV)$ of dimension $k(d-k)$, called the Grassmannian. In particular there are non-decomposable elements if and only if
$$k(d-k)<dimBbb{P}(bigwedge^kV)=binom{d}{k}-1,$$
i.e. if and only if $1<k<d-1$.
answered Jan 22 at 13:21
ServaesServaes
25.8k33996
$endgroup$
$begingroup$
Thanks! This is a nice argument. By the way, do you know a nice reference for reading about the Plucker relations? The wikipedia article contains only a "coordinate formulation", but I guess there should be a more invariant (coordinate-free) approach...
$endgroup$
– Asaf Shachar
Jan 22 at 13:23
$begingroup$
I believe EGA I (second edition) has a section on Grassmannians, which is no doubt coordinate-free. I don't have a copy at hand though.
$endgroup$
– Servaes
Jan 22 at 14:53
add a comment |
$begingroup$
The set of decomposable elements forms a smooth projective subvariety of $Bbb{P}(bigwedge^kV)$ of dimension $k(d-k)$, called the Grassmannian. In particular there are non-decomposable elements if and only if
$$k(d-k)<dimBbb{P}(bigwedge^kV)=binom{d}{k}-1,$$
i.e. if and only if $1<k<d-1$.
answered Jan 22 at 13:21
ServaesServaes
25.8k33996
$endgroup$
$begingroup$
Thanks! This is a nice argument. By the way, do you know a nice reference for reading about the Plucker relations? The wikipedia article contains only a "coordinate formulation", but I guess there should be a more invariant (coordinate-free) approach...
$endgroup$
– Asaf Shachar
Jan 22 at 13:23
$begingroup$
I believe EGA I (second edition) has a section on Grassmannians, which is no doubt coordinate-free. I don't have a copy at hand though.
$endgroup$
– Servaes
Jan 22 at 14:53
add a comment |
$begingroup$
The set of decomposable elements forms a smooth projective subvariety of $Bbb{P}(bigwedge^kV)$ of dimension $k(d-k)$, called the Grassmannian. In particular there are non-decomposable elements if and only if
$$k(d-k)<dimBbb{P}(bigwedge^kV)=binom{d}{k}-1,$$
i.e. if and only if $1<k<d-1$.
answered Jan 22 at 13:21
ServaesServaes
25.8k33996
$endgroup$
The set of decomposable elements forms a smooth projective subvariety of $Bbb{P}(bigwedge^kV)$ of dimension $k(d-k)$, called the Grassmannian. In particular there are non-decomposable elements if and only if
$$k(d-k)<dimBbb{P}(bigwedge^kV)=binom{d}{k}-1,$$
i.e. if and only if $1<k<d-1$.
answered Jan 22 at 13:21
ServaesServaes
25.8k33996
answered Jan 22 at 13:21
ServaesServaes
25.8k33996
answered Jan 22 at 13:21
ServaesServaes
25.8k33996
answered Jan 22 at 13:21
ServaesServaes
25.8k33996
25.8k33996
$begingroup$
Thanks! This is a nice argument. By the way, do you know a nice reference for reading about the Plucker relations? The wikipedia article contains only a "coordinate formulation", but I guess there should be a more invariant (coordinate-free) approach...
$endgroup$
– Asaf Shachar
Jan 22 at 13:23
$begingroup$
I believe EGA I (second edition) has a section on Grassmannians, which is no doubt coordinate-free. I don't have a copy at hand though.
$endgroup$
– Servaes
Jan 22 at 14:53
add a comment |
$begingroup$
Thanks! This is a nice argument. By the way, do you know a nice reference for reading about the Plucker relations? The wikipedia article contains only a "coordinate formulation", but I guess there should be a more invariant (coordinate-free) approach...
$endgroup$
– Asaf Shachar
Jan 22 at 13:23
$begingroup$
I believe EGA I (second edition) has a section on Grassmannians, which is no doubt coordinate-free. I don't have a copy at hand though.
$endgroup$
– Servaes
Jan 22 at 14:53
$begingroup$
Thanks! This is a nice argument. By the way, do you know a nice reference for reading about the Plucker relations? The wikipedia article contains only a "coordinate formulation", but I guess there should be a more invariant (coordinate-free) approach...
$endgroup$
– Asaf Shachar
Jan 22 at 13:23
$begingroup$
Thanks! This is a nice argument. By the way, do you know a nice reference for reading about the Plucker relations? The wikipedia article contains only a "coordinate formulation", but I guess there should be a more invariant (coordinate-free) approach...
$endgroup$
– Asaf Shachar
Jan 22 at 13:23
$begingroup$
I believe EGA I (second edition) has a section on Grassmannians, which is no doubt coordinate-free. I don't have a copy at hand though.
$endgroup$
– Servaes
Jan 22 at 14:53
$begingroup$
I believe EGA I (second edition) has a section on Grassmannians, which is no doubt coordinate-free. I don't have a copy at hand though.
$endgroup$
– Servaes
Jan 22 at 14:53
add a comment |
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