Do complexification and exterior power commute?












1












$begingroup$


Let $V$ be a $d$-dimensional real vector space, and let $1<k<d$.




Are $(bigwedge^k V)^{mathbb{C}}$ and $bigwedge^k (V^{mathbb{C}})$ naturally isomorphic?




They both have the same complex dimension. I guess that one possible map $bigwedge^k (V^{mathbb{C}}) to (bigwedge^k V)^{mathbb{C}}$ is given by
$$ v_1+itilde v_1 wedge dots wedge v_k+itilde v_k to (v_1 wedge dots wedge v_k)+i,,( tilde v_1 wedge dots wedge tilde v_k).$$



Or maybe the correct one is given by "expanding explicitly" $v_1+itilde v_1 wedge dots wedge v_k+itilde v_k$ in the usual way?



I think I might be confusing here with the way that exterior power commutes with direct sum, since we can think on $V^{mathbb{C}}$ as $Voplus iV$, right? (though then we need to "remember" the way we multiply $v+iw$ by complex numbers...



Edit: The map proposed above cannot be the correct one, since
it is not injective. I guess we need to define our map by using a basis (and the basis it induces on the exterior algebra or complexification) and then prove that the map we built is independent of the original basis chosen.



However, maybe there is a basis-free approach? (e.g. using some universal properties etc).






complex-geometry multilinear-algebra exterior-algebra







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let $V$ be a $d$-dimensional real vector space, and let $1<k<d$.




    Are $(bigwedge^k V)^{mathbb{C}}$ and $bigwedge^k (V^{mathbb{C}})$ naturally isomorphic?




    They both have the same complex dimension. I guess that one possible map $bigwedge^k (V^{mathbb{C}}) to (bigwedge^k V)^{mathbb{C}}$ is given by
    $$ v_1+itilde v_1 wedge dots wedge v_k+itilde v_k to (v_1 wedge dots wedge v_k)+i,,( tilde v_1 wedge dots wedge tilde v_k).$$



    Or maybe the correct one is given by "expanding explicitly" $v_1+itilde v_1 wedge dots wedge v_k+itilde v_k$ in the usual way?



    I think I might be confusing here with the way that exterior power commutes with direct sum, since we can think on $V^{mathbb{C}}$ as $Voplus iV$, right? (though then we need to "remember" the way we multiply $v+iw$ by complex numbers...



    Edit: The map proposed above cannot be the correct one, since
    it is not injective. I guess we need to define our map by using a basis (and the basis it induces on the exterior algebra or complexification) and then prove that the map we built is independent of the original basis chosen.



    However, maybe there is a basis-free approach? (e.g. using some universal properties etc).






    complex-geometry multilinear-algebra exterior-algebra







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Let $V$ be a $d$-dimensional real vector space, and let $1<k<d$.




      Are $(bigwedge^k V)^{mathbb{C}}$ and $bigwedge^k (V^{mathbb{C}})$ naturally isomorphic?




      They both have the same complex dimension. I guess that one possible map $bigwedge^k (V^{mathbb{C}}) to (bigwedge^k V)^{mathbb{C}}$ is given by
      $$ v_1+itilde v_1 wedge dots wedge v_k+itilde v_k to (v_1 wedge dots wedge v_k)+i,,( tilde v_1 wedge dots wedge tilde v_k).$$



      Or maybe the correct one is given by "expanding explicitly" $v_1+itilde v_1 wedge dots wedge v_k+itilde v_k$ in the usual way?



      I think I might be confusing here with the way that exterior power commutes with direct sum, since we can think on $V^{mathbb{C}}$ as $Voplus iV$, right? (though then we need to "remember" the way we multiply $v+iw$ by complex numbers...



      Edit: The map proposed above cannot be the correct one, since
      it is not injective. I guess we need to define our map by using a basis (and the basis it induces on the exterior algebra or complexification) and then prove that the map we built is independent of the original basis chosen.



      However, maybe there is a basis-free approach? (e.g. using some universal properties etc).






      complex-geometry multilinear-algebra exterior-algebra







      share|cite|improve this question











      $endgroup$




      Let $V$ be a $d$-dimensional real vector space, and let $1<k<d$.




      Are $(bigwedge^k V)^{mathbb{C}}$ and $bigwedge^k (V^{mathbb{C}})$ naturally isomorphic?




      They both have the same complex dimension. I guess that one possible map $bigwedge^k (V^{mathbb{C}}) to (bigwedge^k V)^{mathbb{C}}$ is given by
      $$ v_1+itilde v_1 wedge dots wedge v_k+itilde v_k to (v_1 wedge dots wedge v_k)+i,,( tilde v_1 wedge dots wedge tilde v_k).$$



      Or maybe the correct one is given by "expanding explicitly" $v_1+itilde v_1 wedge dots wedge v_k+itilde v_k$ in the usual way?



      I think I might be confusing here with the way that exterior power commutes with direct sum, since we can think on $V^{mathbb{C}}$ as $Voplus iV$, right? (though then we need to "remember" the way we multiply $v+iw$ by complex numbers...



      Edit: The map proposed above cannot be the correct one, since
      it is not injective. I guess we need to define our map by using a basis (and the basis it induces on the exterior algebra or complexification) and then prove that the map we built is independent of the original basis chosen.



      However, maybe there is a basis-free approach? (e.g. using some universal properties etc).






      complex-geometry multilinear-algebra exterior-algebra




      complex-geometry multilinear-algebra exterior-algebra






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 24 at 14:32







      Asaf Shachar

















      asked Jan 23 at 12:24









      Asaf ShacharAsaf Shachar

      5,73031142




      5,73031142






















          2 Answers
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          active

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          1












          $begingroup$

          It depends what field you take the exterior power over.



          $Lambda_{mathbb R}(V)otimesmathbb C = Lambda_{mathbb C}(Votimes mathbb C)$ is true and the map is just by rewriting eg $(votimes i)wedge(wotimes 1) = (vwedge w)otimes i$.



          On the other hand if you use the real exterior power on both sides the dimensions won’t be the same. In degree $k$ you get ${2n choose k} neq 2{nchoose k}$. (Not to mention the thing on the right wont be a complex vector space...)






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. Indeed, I meant to take powers over $mathbb{C}$.
            $endgroup$
            – Asaf Shachar
            Jan 28 at 15:07










          • $begingroup$
            Sure, no problem!
            $endgroup$
            – Ben
            Jan 28 at 15:23



















          0












          $begingroup$

          Here is an attempt to write explicitly the isomorphism $(bigwedge^k V)^{mathbb{C}} to bigwedge^k (V^mathbb{C})$. First, we fix some $z in mathbb{C}$. Then we look at the map



          $$ (v_1,ldots,v_k) to zcdot(v_1 otimes 1) wedge ldots wedge (v_k otimes 1).$$



          This map is multilinear and alternating, hence it lifts to an $mathbb{R}$-linear map $phi_z:bigwedge^k V to bigwedge^k (V^mathbb{C})$, given by



          $$ v_1 wedge ldots wedge v_k to z cdot(v_1 otimes 1) wedge ldots wedge (v_k otimes 1).$$



          Now, we define an $mathbb{R}$-bilinear map $bigwedge^k V times mathbb{C} to bigwedge^k (V^mathbb{C})$ given by $(omega,z) to phi_z(omega)$.



          This lifts to an $mathbb{R}$-linear map $T:bigwedge^k V otimes_{mathbb{R}} mathbb{C} = (bigwedge^k V)^{mathbb{C}} to bigwedge^k (V^mathbb{C})$, $T(omega otimes z)= phi_z(omega)$.



          So, to conclude



          $$ Tbig( (v_1 wedge ldots wedge v_k) otimes z big)=z cdot(v_1 otimes 1) wedge ldots wedge (v_k otimes 1).$$



          It is easy to that $T$ is complex-linear isomorphism.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            2 Answers
            2






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            active

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            active

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            1












            $begingroup$

            It depends what field you take the exterior power over.



            $Lambda_{mathbb R}(V)otimesmathbb C = Lambda_{mathbb C}(Votimes mathbb C)$ is true and the map is just by rewriting eg $(votimes i)wedge(wotimes 1) = (vwedge w)otimes i$.



            On the other hand if you use the real exterior power on both sides the dimensions won’t be the same. In degree $k$ you get ${2n choose k} neq 2{nchoose k}$. (Not to mention the thing on the right wont be a complex vector space...)






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks. Indeed, I meant to take powers over $mathbb{C}$.
              $endgroup$
              – Asaf Shachar
              Jan 28 at 15:07










            • $begingroup$
              Sure, no problem!
              $endgroup$
              – Ben
              Jan 28 at 15:23
















            1












            $begingroup$

            It depends what field you take the exterior power over.



            $Lambda_{mathbb R}(V)otimesmathbb C = Lambda_{mathbb C}(Votimes mathbb C)$ is true and the map is just by rewriting eg $(votimes i)wedge(wotimes 1) = (vwedge w)otimes i$.



            On the other hand if you use the real exterior power on both sides the dimensions won’t be the same. In degree $k$ you get ${2n choose k} neq 2{nchoose k}$. (Not to mention the thing on the right wont be a complex vector space...)






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks. Indeed, I meant to take powers over $mathbb{C}$.
              $endgroup$
              – Asaf Shachar
              Jan 28 at 15:07










            • $begingroup$
              Sure, no problem!
              $endgroup$
              – Ben
              Jan 28 at 15:23














            1












            1








            1





            $begingroup$

            It depends what field you take the exterior power over.



            $Lambda_{mathbb R}(V)otimesmathbb C = Lambda_{mathbb C}(Votimes mathbb C)$ is true and the map is just by rewriting eg $(votimes i)wedge(wotimes 1) = (vwedge w)otimes i$.



            On the other hand if you use the real exterior power on both sides the dimensions won’t be the same. In degree $k$ you get ${2n choose k} neq 2{nchoose k}$. (Not to mention the thing on the right wont be a complex vector space...)






            share|cite|improve this answer









            $endgroup$



            It depends what field you take the exterior power over.



            $Lambda_{mathbb R}(V)otimesmathbb C = Lambda_{mathbb C}(Votimes mathbb C)$ is true and the map is just by rewriting eg $(votimes i)wedge(wotimes 1) = (vwedge w)otimes i$.



            On the other hand if you use the real exterior power on both sides the dimensions won’t be the same. In degree $k$ you get ${2n choose k} neq 2{nchoose k}$. (Not to mention the thing on the right wont be a complex vector space...)







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 24 at 16:59









            BenBen

            4,283617




            4,283617












            • $begingroup$
              Thanks. Indeed, I meant to take powers over $mathbb{C}$.
              $endgroup$
              – Asaf Shachar
              Jan 28 at 15:07










            • $begingroup$
              Sure, no problem!
              $endgroup$
              – Ben
              Jan 28 at 15:23


















            • $begingroup$
              Thanks. Indeed, I meant to take powers over $mathbb{C}$.
              $endgroup$
              – Asaf Shachar
              Jan 28 at 15:07










            • $begingroup$
              Sure, no problem!
              $endgroup$
              – Ben
              Jan 28 at 15:23
















            $begingroup$
            Thanks. Indeed, I meant to take powers over $mathbb{C}$.
            $endgroup$
            – Asaf Shachar
            Jan 28 at 15:07




            $begingroup$
            Thanks. Indeed, I meant to take powers over $mathbb{C}$.
            $endgroup$
            – Asaf Shachar
            Jan 28 at 15:07












            $begingroup$
            Sure, no problem!
            $endgroup$
            – Ben
            Jan 28 at 15:23




            $begingroup$
            Sure, no problem!
            $endgroup$
            – Ben
            Jan 28 at 15:23











            0












            $begingroup$

            Here is an attempt to write explicitly the isomorphism $(bigwedge^k V)^{mathbb{C}} to bigwedge^k (V^mathbb{C})$. First, we fix some $z in mathbb{C}$. Then we look at the map



            $$ (v_1,ldots,v_k) to zcdot(v_1 otimes 1) wedge ldots wedge (v_k otimes 1).$$



            This map is multilinear and alternating, hence it lifts to an $mathbb{R}$-linear map $phi_z:bigwedge^k V to bigwedge^k (V^mathbb{C})$, given by



            $$ v_1 wedge ldots wedge v_k to z cdot(v_1 otimes 1) wedge ldots wedge (v_k otimes 1).$$



            Now, we define an $mathbb{R}$-bilinear map $bigwedge^k V times mathbb{C} to bigwedge^k (V^mathbb{C})$ given by $(omega,z) to phi_z(omega)$.



            This lifts to an $mathbb{R}$-linear map $T:bigwedge^k V otimes_{mathbb{R}} mathbb{C} = (bigwedge^k V)^{mathbb{C}} to bigwedge^k (V^mathbb{C})$, $T(omega otimes z)= phi_z(omega)$.



            So, to conclude



            $$ Tbig( (v_1 wedge ldots wedge v_k) otimes z big)=z cdot(v_1 otimes 1) wedge ldots wedge (v_k otimes 1).$$



            It is easy to that $T$ is complex-linear isomorphism.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Here is an attempt to write explicitly the isomorphism $(bigwedge^k V)^{mathbb{C}} to bigwedge^k (V^mathbb{C})$. First, we fix some $z in mathbb{C}$. Then we look at the map



              $$ (v_1,ldots,v_k) to zcdot(v_1 otimes 1) wedge ldots wedge (v_k otimes 1).$$



              This map is multilinear and alternating, hence it lifts to an $mathbb{R}$-linear map $phi_z:bigwedge^k V to bigwedge^k (V^mathbb{C})$, given by



              $$ v_1 wedge ldots wedge v_k to z cdot(v_1 otimes 1) wedge ldots wedge (v_k otimes 1).$$



              Now, we define an $mathbb{R}$-bilinear map $bigwedge^k V times mathbb{C} to bigwedge^k (V^mathbb{C})$ given by $(omega,z) to phi_z(omega)$.



              This lifts to an $mathbb{R}$-linear map $T:bigwedge^k V otimes_{mathbb{R}} mathbb{C} = (bigwedge^k V)^{mathbb{C}} to bigwedge^k (V^mathbb{C})$, $T(omega otimes z)= phi_z(omega)$.



              So, to conclude



              $$ Tbig( (v_1 wedge ldots wedge v_k) otimes z big)=z cdot(v_1 otimes 1) wedge ldots wedge (v_k otimes 1).$$



              It is easy to that $T$ is complex-linear isomorphism.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Here is an attempt to write explicitly the isomorphism $(bigwedge^k V)^{mathbb{C}} to bigwedge^k (V^mathbb{C})$. First, we fix some $z in mathbb{C}$. Then we look at the map



                $$ (v_1,ldots,v_k) to zcdot(v_1 otimes 1) wedge ldots wedge (v_k otimes 1).$$



                This map is multilinear and alternating, hence it lifts to an $mathbb{R}$-linear map $phi_z:bigwedge^k V to bigwedge^k (V^mathbb{C})$, given by



                $$ v_1 wedge ldots wedge v_k to z cdot(v_1 otimes 1) wedge ldots wedge (v_k otimes 1).$$



                Now, we define an $mathbb{R}$-bilinear map $bigwedge^k V times mathbb{C} to bigwedge^k (V^mathbb{C})$ given by $(omega,z) to phi_z(omega)$.



                This lifts to an $mathbb{R}$-linear map $T:bigwedge^k V otimes_{mathbb{R}} mathbb{C} = (bigwedge^k V)^{mathbb{C}} to bigwedge^k (V^mathbb{C})$, $T(omega otimes z)= phi_z(omega)$.



                So, to conclude



                $$ Tbig( (v_1 wedge ldots wedge v_k) otimes z big)=z cdot(v_1 otimes 1) wedge ldots wedge (v_k otimes 1).$$



                It is easy to that $T$ is complex-linear isomorphism.






                share|cite|improve this answer









                $endgroup$



                Here is an attempt to write explicitly the isomorphism $(bigwedge^k V)^{mathbb{C}} to bigwedge^k (V^mathbb{C})$. First, we fix some $z in mathbb{C}$. Then we look at the map



                $$ (v_1,ldots,v_k) to zcdot(v_1 otimes 1) wedge ldots wedge (v_k otimes 1).$$



                This map is multilinear and alternating, hence it lifts to an $mathbb{R}$-linear map $phi_z:bigwedge^k V to bigwedge^k (V^mathbb{C})$, given by



                $$ v_1 wedge ldots wedge v_k to z cdot(v_1 otimes 1) wedge ldots wedge (v_k otimes 1).$$



                Now, we define an $mathbb{R}$-bilinear map $bigwedge^k V times mathbb{C} to bigwedge^k (V^mathbb{C})$ given by $(omega,z) to phi_z(omega)$.



                This lifts to an $mathbb{R}$-linear map $T:bigwedge^k V otimes_{mathbb{R}} mathbb{C} = (bigwedge^k V)^{mathbb{C}} to bigwedge^k (V^mathbb{C})$, $T(omega otimes z)= phi_z(omega)$.



                So, to conclude



                $$ Tbig( (v_1 wedge ldots wedge v_k) otimes z big)=z cdot(v_1 otimes 1) wedge ldots wedge (v_k otimes 1).$$



                It is easy to that $T$ is complex-linear isomorphism.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 28 at 14:49









                Asaf ShacharAsaf Shachar

                5,73031142




                5,73031142






























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