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:




  1. First, the answer for $(k,d),(d-k,d)$ is the same, by duality. So, it suffices to assume $k le d/2$.


  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.




  1. 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







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    $begingroup$


    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:




    1. First, the answer for $(k,d),(d-k,d)$ is the same, by duality. So, it suffices to assume $k le d/2$.


    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.




    1. 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







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      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:




      1. First, the answer for $(k,d),(d-k,d)$ is the same, by duality. So, it suffices to assume $k le d/2$.


      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.




      1. 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







      share|cite|improve this question









      $endgroup$




      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:




      1. First, the answer for $(k,d),(d-k,d)$ is the same, by duality. So, it suffices to assume $k le d/2$.


      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.




      1. 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






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 21 at 13:23









      Asaf ShacharAsaf Shachar

      5,64631141




      5,64631141






















          1 Answer
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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 $1<k<d-1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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











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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 $1<k<d-1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          2












          $begingroup$

          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 $1<k<d-1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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














          2












          2








          2





          $begingroup$

          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 $1<k<d-1$.






          share|cite|improve this answer









          $endgroup$



          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 $1<k<d-1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • $begingroup$
            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


















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