Ytukyg

Suppose that $V$ is a vector space of dimension $2n$, and let $omega in Lambda^2(V)$. Prove that the following two statements are equivalent.

(1) $tilde{omega} : V rightarrow V^*$ defined by $Xmapsto omega(X,cdot)$ is an isomorphism.

(2) $omega^n=omega wedge omega wedge cdots wedge omega in Lambda^{2n}(V)$ is nonzero.

What I have tried is

Suppose that the map $tilde{omega}: Vrightarrow V^*$ defined by $Xrightarrow i_X omega$ is an isomorphism. Then we have the basis $(e^1,f^1,e^2,f^2,dots, e^n,f^n)$ of $V$ such that
begin{align*}
&(1) text{ }omega(e^i,e^j)=omega(f^i,f^j)=0 hspace{3mm}forall i,j\
&(2)text{ } omega(e^i,f^j)=begin{cases} 1 &i=j\ 0 &ineq j end{cases}.
end{align*}

Then observe that

begin{align*}
omegawedge omega wedge cdots wedge omega in Lambda^{2n}(V)
end{align*}

and that

begin{align*}
omegawedge omega wedge cdots wedge omega (e^1,f^1,e^2,f^2,dots, e^n,f^n)=omega(e^1,f^1)wedge omega(e^2,f^2)wedge cdots omega(e^n,f^n)=1neq 0.
end{align*}

Therefore, $omegawedge omega wedge cdots wedge omega $ is nonzero.

Conversely, suppose that $omegawedge omega wedge cdots wedge omega in Lambda^{2n}(V)$ is nonzero. Clearly, note that $tilde{omega}:Vrightarrow V^*$ defined by $tilde{omega}(X)=i_X(omega)$ is a homomorphism since $i_{X+Y}omega(Z)=omega(X+Y,Z)=omega(X,Z)+omega(Y,Z)=i_Xomega(Z)+i_Yomega(Z)$. We need to prove that $tilde{omega}$ is a bijection.
And assume that $i_X(omegawedge omega wedge cdots wedge omega)=0$ for some $Xin Vsetminus {0}$. Then we can build a basis ${X,r^1,r^2,dots, r^{2n-1}}$ of $V$. Since $omegawedge omega wedge cdots wedge omega(X,r^1,r^2,dots,r^{2n-1})=0$ and ${X,r^1,r^2,dots, r^{2n-1}}$ is a basis of $V$, $omegawedge omega wedge cdots wedge omega=0$ and it is a contradiction. Thus, $i_X(omegawedge omega wedge cdots wedge omega)$ is nonzero for any $Xin Vsetminus {0}$. Then note that, for given $Xin Vsetminus {0}$, we have $Yin V$ such that $i_Xomega(Y)=omega(X,Y)neq 0$ (otherwise, for any collection ${Y^1,dots, Y^{2n-1}}subseteq V$, $i_X(omegawedge omega wedge cdots wedge omega)(Y^1,dots, Y^{2n-1})=0wedge omega(Y^{2},Y^{3})wedge dots wedge omega(Y^{2n-2},Y^{2n-1})=0$). Then $ker tilde{omega}=0$, so $tilde{omega}$ is an injection. Then since $V$ and $V^*$ has same dimension, $tilde{omega}$ is a surjection so is a bijection. Therefore, $tilde{omega}$ is an isomorphism.

I mistakenly used the wrong computation of wedge product of alternating tensors.
I am pretty sure that I can adjust a little bit from here to make solid proof. I am still thinking about but seems I am missing something. Any help would be appreciated. Thanks in advance.

I think I figure it out. I posted this solution for future.

Suppose that the map $tilde{omega}: Vrightarrow V^*$ defined by $Xrightarrow i_X omega$ is an isomorphism. Then we have the basis $(e^1,f^1,e^2,f^2,dots, e^n,f^n)$ of $V$ such that
begin{align*}
&(1) text{ }omega(e^i,e^j)=omega(f^i,f^j)=0 hspace{3mm}forall i,j\
&(2)text{ } omega(e^i,f^j)=begin{cases} 1 &i=j\ 0 &ineq j end{cases}.
end{align*}

Let ${epsilon^i,delta^i}$ be a dual basis such that
begin{align*}
&(1) epsilon^i(e^j)=begin{cases}1 &i=j \0 &ineq j end{cases} &(2) delta^i(f^j)=begin{cases}1 &i=j \0 &ineq j end{cases}\
&(3) epsilon^i(f^j)=0 hspace{4mm}forall i,j &(4) delta^i(e^j)=0 hspace{4mm}forall i,j.
end{align*}

Then observe that

begin{align*}
omegawedge omega wedge cdots wedge omega in Lambda^{2n}(V)
end{align*}

and that, since $dim(Lambda^{2n}(V))=bigg(begin{matrix}
2n\2n
end{matrix} bigg)=1$, for some integer $alphain mathbb{R}$,

begin{align*}
omegawedge omega wedge cdots wedge omega (e^1,f^1,e^2,f^2,dots, e^n,f^n)&=alpha(epsilon^1wedge delta^1wedge epsilon^2wedge delta^2wedgecdots epsilon^nwedge delta^n)(e^1,f^1,e^2,f^2,dots, e^n,f^n)\ &=alphasum_{sigma}sgn(sigma)epsilon^1(v_{sigma(1)})delta^1(v_{sigma(2)})cdots epsilon^n(v_{sigma(2n-1)})delta^n(v_{sigma(2n)})\
&=alpha(epsilon^1(e^1)delta^1(f^1)cdots epsilon^n(e^n)delta^n(f^n))\
&=alphaneq 0,
end{align*}
where
begin{align*}
v_i=begin{cases} e^m &text{ if }i=2m-1\ f^m &text{ if } i=2m end{cases} hspace{2cm} forall i in mathbb{Z}cap [1,2n].
end{align*}

Therefore, $omegawedge omega wedge cdots wedge omega $ is nonzero.

Conversely, suppose that $omegawedge omega wedge cdots wedge omega in Lambda^{2n}(V)$ is nonzero. Clearly, note that $tilde{omega}:Vrightarrow V^*$ defined by $tilde{omega}(X)=i_X(omega)$ is a homomorphism since $i_{X+Y}omega(Z)=omega(X+Y,Z)=omega(X,Z)+omega(Y,Z)=i_Xomega(Z)+i_Yomega(Z)$. We need to prove that $tilde{omega}$ is a bijection.

Assume that $i_X(omegawedge omega wedge cdots wedge omega)=0$ for some $Xin Vsetminus {0}$. and $alphain mathbb{R}$ since $dim(Lambda^{2n}(V))=1$. Then we can build a basis ${X,r^1,r^2,dots, r^{2n-1}}$ of $V$. Then, $forall vin V$, $v$ can be represented as linear combination of the elements in basis. Then, for any ${a^1,dots, a^{2n}}subseteq V$, $omegawedge omega wedge cdots wedge omega(a^1,dots,a^{2n})=0$ since $omega$ is multilinear map. It contradicts to our assumption. Therefore, $i_X (omegawedge omega wedge cdots wedge omega)neq 0$ $forall Xin Vsetminus {0}$.

Let $Xin Vsetminus {0}$ be given. Note that
begin{align*}
i_X(omegawedge omega wedge cdots wedge omega)= (i_Xomega)wedge omega wedge cdots wedge omega+cdots + omegawedge omega wedge cdots wedge (i_Xomega)=n{ (i_Xomega)wedge omega wedge cdots wedge omega }neq 0.
end{align*}

It implies that $(i_Xomega)wedge omega wedge cdots wedge omeganeq 0$, thus, we should have $(i_Xomega)neq 0$. Thus, we have $Yin V$ such that $(i_Xomega)(Y)=omega(X,Y)neq 0$. Since $Xin Vsetminus {0}$ was arbitrary, $i_Xomega =0 iff X=0$. Then $ker tilde{omega}=0$, so $tilde{omega}$ is an injection. Then since $V$ and $V^*$ has same dimension, $tilde{omega}$ is a surjection so is a bijection. Therefore, $tilde{omega}$ is an isomorphism.