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How to prove that the definition of exterior product of differential forms is not ambiguous?
$begingroup$
In page 91 of book A Visual Introduction to Differential Forms and Calculus on Manifolds the exterior product of two differential forms $alpha in bigwedge^{r}(mathbb{R}^n)$ and $beta in bigwedge^{s}(mathbb{R}^n)$ is definide by
$$
alpha wedge beta = left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)=sum_{I}sum_{J}a_Ib_J dx^Iwedge dx^J
hspace{2cm}(ast)
$$
where $alpha =sum_{I}a_I dx^I$ and $beta =sum_{J}b_J dx^J$. Here ${dx^I}_{I}$ is the basis of $bigwedge^{r}(mathbb{R}^n)$ that we get ( by wedge product $dx^I=dx^{i_1}wedge ldotswedge dx^{i_k}$) from basis ${dx^1, ldots, dx^n}$ of $(mathbb{R}^n)^ast$ dual of canonical basis ${e_1,ldots,e_n}$ of $mathbb{R}^n$. In the same way, ${dx^J}_{J}$ is the basis of $bigwedge^{s}(mathbb{R}^n)$ that we get ( by wedge product $dx^J=dx^{j_1}wedge ldotswedge dx^{j_ell}$) from basis ${dx^1, ldots, dx^n}$ of $(mathbb{R}^n)^ast$ dual of canonical basis ${e_1,ldots,e_n}$ of $mathbb{R}^n$.
Do Carmo in his book Differential Forms and Applications makes the same definition of exterior product of differential forms.
Now, let's set any base in $ mathbb{R}^n$, say ${y_1,ldots,y_n}$. Let ${dy^I}_{I}$ is the base of $bigwedge^{r}(mathbb{R}^n)$ that we get ( by wedge product $dy^I=dy^{i_1}wedge ldotswedge dy^{i_k}$) from basis ${dy^1, ldots, dy^n}$ of $(mathbb{R}^n)^ast$ dual of basis ${y_1,ldots,y_n}$ of $mathbb{R}^n$. In the same way, let ${dy^J}_{J}$ the basis of $bigwedge^{s}(mathbb{R}^n)$ that we get ( by wedge product $dy^J=dy^{j_1}wedge ldotswedge dy^{j_ell}$) from basis ${dy^1, ldots, dy^n}$ of $(mathbb{R}^n)^ast$ dual of canonical basis ${y_1,ldots,y_n}$ of $mathbb{R}^n$.
Since ${dy^I}_{I}$ and ${dy^J}_{J}$ are basis of $bigwedge^{r}(mathbb{R}^n)$ and $bigwedge^{s}(mathbb{R}^n)$ respectively we can rewrite the differential forms $alpha$ and $beta$ in terms of these basis as
$$
alpha = sum_{I}g_I dy^I hspace{1cm} mbox{and}hspace{1cm} beta=sum_{J} h_{J} dy^J.
$$
It is not clear, at least for me, that
$$
alphawedge beta = left( sum_{I}g_I dy^I right) wedge left( sum_{J} h_{J} dy^J right)= sum_{I}sum_{J} g_Ih_J dy^I wedge dy^J
$$
will be the same differential form that we get in $(ast)$.
Question. How to prove that the definition of exterior product of differential forms is not ambiguous? That is, how to prove that
$$
left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)
=
left( sum_{I}g_I dy^I right) wedge left( sum_{J} h_{J} dy^Jright) mbox{ ? }
$$
Honestly, I have no idea how to proceed with this problem.
differential-forms multilinear-algebra
asked Jan 11 at 18:40
MathOverviewMathOverview
8,71243163
$endgroup$
add a comment |
$begingroup$
In page 91 of book A Visual Introduction to Differential Forms and Calculus on Manifolds the exterior product of two differential forms $alpha in bigwedge^{r}(mathbb{R}^n)$ and $beta in bigwedge^{s}(mathbb{R}^n)$ is definide by
$$
alpha wedge beta = left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)=sum_{I}sum_{J}a_Ib_J dx^Iwedge dx^J
hspace{2cm}(ast)
$$
where $alpha =sum_{I}a_I dx^I$ and $beta =sum_{J}b_J dx^J$. Here ${dx^I}_{I}$ is the basis of $bigwedge^{r}(mathbb{R}^n)$ that we get ( by wedge product $dx^I=dx^{i_1}wedge ldotswedge dx^{i_k}$) from basis ${dx^1, ldots, dx^n}$ of $(mathbb{R}^n)^ast$ dual of canonical basis ${e_1,ldots,e_n}$ of $mathbb{R}^n$. In the same way, ${dx^J}_{J}$ is the basis of $bigwedge^{s}(mathbb{R}^n)$ that we get ( by wedge product $dx^J=dx^{j_1}wedge ldotswedge dx^{j_ell}$) from basis ${dx^1, ldots, dx^n}$ of $(mathbb{R}^n)^ast$ dual of canonical basis ${e_1,ldots,e_n}$ of $mathbb{R}^n$.
Do Carmo in his book Differential Forms and Applications makes the same definition of exterior product of differential forms.
Now, let's set any base in $ mathbb{R}^n$, say ${y_1,ldots,y_n}$. Let ${dy^I}_{I}$ is the base of $bigwedge^{r}(mathbb{R}^n)$ that we get ( by wedge product $dy^I=dy^{i_1}wedge ldotswedge dy^{i_k}$) from basis ${dy^1, ldots, dy^n}$ of $(mathbb{R}^n)^ast$ dual of basis ${y_1,ldots,y_n}$ of $mathbb{R}^n$. In the same way, let ${dy^J}_{J}$ the basis of $bigwedge^{s}(mathbb{R}^n)$ that we get ( by wedge product $dy^J=dy^{j_1}wedge ldotswedge dy^{j_ell}$) from basis ${dy^1, ldots, dy^n}$ of $(mathbb{R}^n)^ast$ dual of canonical basis ${y_1,ldots,y_n}$ of $mathbb{R}^n$.
Since ${dy^I}_{I}$ and ${dy^J}_{J}$ are basis of $bigwedge^{r}(mathbb{R}^n)$ and $bigwedge^{s}(mathbb{R}^n)$ respectively we can rewrite the differential forms $alpha$ and $beta$ in terms of these basis as
$$
alpha = sum_{I}g_I dy^I hspace{1cm} mbox{and}hspace{1cm} beta=sum_{J} h_{J} dy^J.
$$
It is not clear, at least for me, that
$$
alphawedge beta = left( sum_{I}g_I dy^I right) wedge left( sum_{J} h_{J} dy^J right)= sum_{I}sum_{J} g_Ih_J dy^I wedge dy^J
$$
will be the same differential form that we get in $(ast)$.
Question. How to prove that the definition of exterior product of differential forms is not ambiguous? That is, how to prove that
$$
left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)
=
left( sum_{I}g_I dy^I right) wedge left( sum_{J} h_{J} dy^Jright) mbox{ ? }
$$
Honestly, I have no idea how to proceed with this problem.
differential-forms multilinear-algebra
asked Jan 11 at 18:40
MathOverviewMathOverview
8,71243163
$endgroup$
$begingroup$
This is just because of the algebraic properties of $wedge$. Have you tried writing it out explicitly with $alpha = dx^1$ and $beta=dx^2$ for starters?
$endgroup$
– Ted Shifrin
Jan 11 at 18:45
$begingroup$
@TedShifrin In this case supose $alpha, beta bigwedge^1(mathbb{R}^2)$. We will have $alpha= dx^1=g(y_1,y_2)dy^1$, $beta= dx^2=h(y_1,y_2)dy^2$, $alphawedge beta(e_1,e_2)=1$ and $alphawedge beta g(y)h(y)det begin{pmatrix} dy^1(e_1) & dy^1(e_2) \ dy^2(e_1) & dy^2(e_2) end{pmatrix}$. But how can I use this?
$endgroup$
– MathOverview
Jan 11 at 20:12
$begingroup$
First, it's just about pointwise linear algebra. Second, don't do it by evaluating. Write $dy^i = sum a^i_j dx^j$. (In your set-up, you did the reverse. It doesn't matter.)
$endgroup$
– Ted Shifrin
Jan 11 at 21:30
$begingroup$
@TedShifrin It would be something of the type $dy^{I}wedge dy^{J}=dy^{i_1}wedge ldots wedge dy^{i_k}wedge dy^{j_1}wedge ldots wedge dy^{j_ell}=left( sum_{u=1}^{n} a_u^{i_1} dx^u right)wedge ldotswedge left( sum_{u=1}^{n} a_u^{i_k} dx^u right)wedge left( sum_{v=1}^{n} a_v^{j_1} dx^v right) wedge ldotswedge left( sum_{v=1}^{n} a_v^{j_ell} dx^v right);;;;?$
$endgroup$
– MathOverview
Jan 11 at 21:50
add a comment |
$begingroup$
In page 91 of book A Visual Introduction to Differential Forms and Calculus on Manifolds the exterior product of two differential forms $alpha in bigwedge^{r}(mathbb{R}^n)$ and $beta in bigwedge^{s}(mathbb{R}^n)$ is definide by
$$
alpha wedge beta = left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)=sum_{I}sum_{J}a_Ib_J dx^Iwedge dx^J
hspace{2cm}(ast)
$$
where $alpha =sum_{I}a_I dx^I$ and $beta =sum_{J}b_J dx^J$. Here ${dx^I}_{I}$ is the basis of $bigwedge^{r}(mathbb{R}^n)$ that we get ( by wedge product $dx^I=dx^{i_1}wedge ldotswedge dx^{i_k}$) from basis ${dx^1, ldots, dx^n}$ of $(mathbb{R}^n)^ast$ dual of canonical basis ${e_1,ldots,e_n}$ of $mathbb{R}^n$. In the same way, ${dx^J}_{J}$ is the basis of $bigwedge^{s}(mathbb{R}^n)$ that we get ( by wedge product $dx^J=dx^{j_1}wedge ldotswedge dx^{j_ell}$) from basis ${dx^1, ldots, dx^n}$ of $(mathbb{R}^n)^ast$ dual of canonical basis ${e_1,ldots,e_n}$ of $mathbb{R}^n$.
Do Carmo in his book Differential Forms and Applications makes the same definition of exterior product of differential forms.
Now, let's set any base in $ mathbb{R}^n$, say ${y_1,ldots,y_n}$. Let ${dy^I}_{I}$ is the base of $bigwedge^{r}(mathbb{R}^n)$ that we get ( by wedge product $dy^I=dy^{i_1}wedge ldotswedge dy^{i_k}$) from basis ${dy^1, ldots, dy^n}$ of $(mathbb{R}^n)^ast$ dual of basis ${y_1,ldots,y_n}$ of $mathbb{R}^n$. In the same way, let ${dy^J}_{J}$ the basis of $bigwedge^{s}(mathbb{R}^n)$ that we get ( by wedge product $dy^J=dy^{j_1}wedge ldotswedge dy^{j_ell}$) from basis ${dy^1, ldots, dy^n}$ of $(mathbb{R}^n)^ast$ dual of canonical basis ${y_1,ldots,y_n}$ of $mathbb{R}^n$.
Since ${dy^I}_{I}$ and ${dy^J}_{J}$ are basis of $bigwedge^{r}(mathbb{R}^n)$ and $bigwedge^{s}(mathbb{R}^n)$ respectively we can rewrite the differential forms $alpha$ and $beta$ in terms of these basis as
$$
alpha = sum_{I}g_I dy^I hspace{1cm} mbox{and}hspace{1cm} beta=sum_{J} h_{J} dy^J.
$$
It is not clear, at least for me, that
$$
alphawedge beta = left( sum_{I}g_I dy^I right) wedge left( sum_{J} h_{J} dy^J right)= sum_{I}sum_{J} g_Ih_J dy^I wedge dy^J
$$
will be the same differential form that we get in $(ast)$.
Question. How to prove that the definition of exterior product of differential forms is not ambiguous? That is, how to prove that
$$
left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)
=
left( sum_{I}g_I dy^I right) wedge left( sum_{J} h_{J} dy^Jright) mbox{ ? }
$$
Honestly, I have no idea how to proceed with this problem.
differential-forms multilinear-algebra
asked Jan 11 at 18:40
MathOverviewMathOverview
8,71243163
$endgroup$
In page 91 of book A Visual Introduction to Differential Forms and Calculus on Manifolds the exterior product of two differential forms $alpha in bigwedge^{r}(mathbb{R}^n)$ and $beta in bigwedge^{s}(mathbb{R}^n)$ is definide by
$$
alpha wedge beta = left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)=sum_{I}sum_{J}a_Ib_J dx^Iwedge dx^J
hspace{2cm}(ast)
$$
where $alpha =sum_{I}a_I dx^I$ and $beta =sum_{J}b_J dx^J$. Here ${dx^I}_{I}$ is the basis of $bigwedge^{r}(mathbb{R}^n)$ that we get ( by wedge product $dx^I=dx^{i_1}wedge ldotswedge dx^{i_k}$) from basis ${dx^1, ldots, dx^n}$ of $(mathbb{R}^n)^ast$ dual of canonical basis ${e_1,ldots,e_n}$ of $mathbb{R}^n$. In the same way, ${dx^J}_{J}$ is the basis of $bigwedge^{s}(mathbb{R}^n)$ that we get ( by wedge product $dx^J=dx^{j_1}wedge ldotswedge dx^{j_ell}$) from basis ${dx^1, ldots, dx^n}$ of $(mathbb{R}^n)^ast$ dual of canonical basis ${e_1,ldots,e_n}$ of $mathbb{R}^n$.
Do Carmo in his book Differential Forms and Applications makes the same definition of exterior product of differential forms.
Now, let's set any base in $ mathbb{R}^n$, say ${y_1,ldots,y_n}$. Let ${dy^I}_{I}$ is the base of $bigwedge^{r}(mathbb{R}^n)$ that we get ( by wedge product $dy^I=dy^{i_1}wedge ldotswedge dy^{i_k}$) from basis ${dy^1, ldots, dy^n}$ of $(mathbb{R}^n)^ast$ dual of basis ${y_1,ldots,y_n}$ of $mathbb{R}^n$. In the same way, let ${dy^J}_{J}$ the basis of $bigwedge^{s}(mathbb{R}^n)$ that we get ( by wedge product $dy^J=dy^{j_1}wedge ldotswedge dy^{j_ell}$) from basis ${dy^1, ldots, dy^n}$ of $(mathbb{R}^n)^ast$ dual of canonical basis ${y_1,ldots,y_n}$ of $mathbb{R}^n$.
Since ${dy^I}_{I}$ and ${dy^J}_{J}$ are basis of $bigwedge^{r}(mathbb{R}^n)$ and $bigwedge^{s}(mathbb{R}^n)$ respectively we can rewrite the differential forms $alpha$ and $beta$ in terms of these basis as
$$
alpha = sum_{I}g_I dy^I hspace{1cm} mbox{and}hspace{1cm} beta=sum_{J} h_{J} dy^J.
$$
It is not clear, at least for me, that
$$
alphawedge beta = left( sum_{I}g_I dy^I right) wedge left( sum_{J} h_{J} dy^J right)= sum_{I}sum_{J} g_Ih_J dy^I wedge dy^J
$$
will be the same differential form that we get in $(ast)$.
Question. How to prove that the definition of exterior product of differential forms is not ambiguous? That is, how to prove that
$$
left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)
=
left( sum_{I}g_I dy^I right) wedge left( sum_{J} h_{J} dy^Jright) mbox{ ? }
$$
Honestly, I have no idea how to proceed with this problem.
differential-forms multilinear-algebra
differential-forms multilinear-algebra
asked Jan 11 at 18:40
MathOverviewMathOverview
8,71243163
asked Jan 11 at 18:40
MathOverviewMathOverview
8,71243163
edited Jan 15 at 14:31
MathOverview
asked Jan 11 at 18:40
MathOverviewMathOverview
8,71243163
asked Jan 11 at 18:40
MathOverviewMathOverview
8,71243163
asked Jan 11 at 18:40
MathOverviewMathOverview
8,71243163
8,71243163
$begingroup$
This is just because of the algebraic properties of $wedge$. Have you tried writing it out explicitly with $alpha = dx^1$ and $beta=dx^2$ for starters?
$endgroup$
– Ted Shifrin
Jan 11 at 18:45
$begingroup$
@TedShifrin In this case supose $alpha, beta bigwedge^1(mathbb{R}^2)$. We will have $alpha= dx^1=g(y_1,y_2)dy^1$, $beta= dx^2=h(y_1,y_2)dy^2$, $alphawedge beta(e_1,e_2)=1$ and $alphawedge beta g(y)h(y)det begin{pmatrix} dy^1(e_1) & dy^1(e_2) \ dy^2(e_1) & dy^2(e_2) end{pmatrix}$. But how can I use this?
$endgroup$
– MathOverview
Jan 11 at 20:12
$begingroup$
First, it's just about pointwise linear algebra. Second, don't do it by evaluating. Write $dy^i = sum a^i_j dx^j$. (In your set-up, you did the reverse. It doesn't matter.)
$endgroup$
– Ted Shifrin
Jan 11 at 21:30
$begingroup$
@TedShifrin It would be something of the type $dy^{I}wedge dy^{J}=dy^{i_1}wedge ldots wedge dy^{i_k}wedge dy^{j_1}wedge ldots wedge dy^{j_ell}=left( sum_{u=1}^{n} a_u^{i_1} dx^u right)wedge ldotswedge left( sum_{u=1}^{n} a_u^{i_k} dx^u right)wedge left( sum_{v=1}^{n} a_v^{j_1} dx^v right) wedge ldotswedge left( sum_{v=1}^{n} a_v^{j_ell} dx^v right);;;;?$
$endgroup$
– MathOverview
Jan 11 at 21:50
add a comment |
$begingroup$
This is just because of the algebraic properties of $wedge$. Have you tried writing it out explicitly with $alpha = dx^1$ and $beta=dx^2$ for starters?
$endgroup$
– Ted Shifrin
Jan 11 at 18:45
$begingroup$
@TedShifrin In this case supose $alpha, beta bigwedge^1(mathbb{R}^2)$. We will have $alpha= dx^1=g(y_1,y_2)dy^1$, $beta= dx^2=h(y_1,y_2)dy^2$, $alphawedge beta(e_1,e_2)=1$ and $alphawedge beta g(y)h(y)det begin{pmatrix} dy^1(e_1) & dy^1(e_2) \ dy^2(e_1) & dy^2(e_2) end{pmatrix}$. But how can I use this?
$endgroup$
– MathOverview
Jan 11 at 20:12
$begingroup$
First, it's just about pointwise linear algebra. Second, don't do it by evaluating. Write $dy^i = sum a^i_j dx^j$. (In your set-up, you did the reverse. It doesn't matter.)
$endgroup$
– Ted Shifrin
Jan 11 at 21:30
$begingroup$
@TedShifrin It would be something of the type $dy^{I}wedge dy^{J}=dy^{i_1}wedge ldots wedge dy^{i_k}wedge dy^{j_1}wedge ldots wedge dy^{j_ell}=left( sum_{u=1}^{n} a_u^{i_1} dx^u right)wedge ldotswedge left( sum_{u=1}^{n} a_u^{i_k} dx^u right)wedge left( sum_{v=1}^{n} a_v^{j_1} dx^v right) wedge ldotswedge left( sum_{v=1}^{n} a_v^{j_ell} dx^v right);;;;?$
$endgroup$
– MathOverview
Jan 11 at 21:50
$begingroup$
This is just because of the algebraic properties of $wedge$. Have you tried writing it out explicitly with $alpha = dx^1$ and $beta=dx^2$ for starters?
$endgroup$
– Ted Shifrin
Jan 11 at 18:45
$begingroup$
This is just because of the algebraic properties of $wedge$. Have you tried writing it out explicitly with $alpha = dx^1$ and $beta=dx^2$ for starters?
$endgroup$
– Ted Shifrin
Jan 11 at 18:45
$begingroup$
@TedShifrin In this case supose $alpha, beta bigwedge^1(mathbb{R}^2)$. We will have $alpha= dx^1=g(y_1,y_2)dy^1$, $beta= dx^2=h(y_1,y_2)dy^2$, $alphawedge beta(e_1,e_2)=1$ and $alphawedge beta g(y)h(y)det begin{pmatrix} dy^1(e_1) & dy^1(e_2) \ dy^2(e_1) & dy^2(e_2) end{pmatrix}$. But how can I use this?
$endgroup$
– MathOverview
Jan 11 at 20:12
$begingroup$
@TedShifrin In this case supose $alpha, beta bigwedge^1(mathbb{R}^2)$. We will have $alpha= dx^1=g(y_1,y_2)dy^1$, $beta= dx^2=h(y_1,y_2)dy^2$, $alphawedge beta(e_1,e_2)=1$ and $alphawedge beta g(y)h(y)det begin{pmatrix} dy^1(e_1) & dy^1(e_2) \ dy^2(e_1) & dy^2(e_2) end{pmatrix}$. But how can I use this?
$endgroup$
– MathOverview
Jan 11 at 20:12
$begingroup$
First, it's just about pointwise linear algebra. Second, don't do it by evaluating. Write $dy^i = sum a^i_j dx^j$. (In your set-up, you did the reverse. It doesn't matter.)
$endgroup$
– Ted Shifrin
Jan 11 at 21:30
$begingroup$
First, it's just about pointwise linear algebra. Second, don't do it by evaluating. Write $dy^i = sum a^i_j dx^j$. (In your set-up, you did the reverse. It doesn't matter.)
$endgroup$
– Ted Shifrin
Jan 11 at 21:30
$begingroup$
@TedShifrin It would be something of the type $dy^{I}wedge dy^{J}=dy^{i_1}wedge ldots wedge dy^{i_k}wedge dy^{j_1}wedge ldots wedge dy^{j_ell}=left( sum_{u=1}^{n} a_u^{i_1} dx^u right)wedge ldotswedge left( sum_{u=1}^{n} a_u^{i_k} dx^u right)wedge left( sum_{v=1}^{n} a_v^{j_1} dx^v right) wedge ldotswedge left( sum_{v=1}^{n} a_v^{j_ell} dx^v right);;;;?$
$endgroup$
– MathOverview
Jan 11 at 21:50
$begingroup$
@TedShifrin It would be something of the type $dy^{I}wedge dy^{J}=dy^{i_1}wedge ldots wedge dy^{i_k}wedge dy^{j_1}wedge ldots wedge dy^{j_ell}=left( sum_{u=1}^{n} a_u^{i_1} dx^u right)wedge ldotswedge left( sum_{u=1}^{n} a_u^{i_k} dx^u right)wedge left( sum_{v=1}^{n} a_v^{j_1} dx^v right) wedge ldotswedge left( sum_{v=1}^{n} a_v^{j_ell} dx^v right);;;;?$
$endgroup$
– MathOverview
Jan 11 at 21:50
add a comment |
1 Answer
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I do not know if that is the most elegant and direct answer to the question. I would appreciate suggestions for improving it as well as answers that explore a different point of view.
Lema. Let $omegain bigwedge^{r}(mathbb{R}^n)$. Supose that
$sum_{K}a_K du^K$ is the expression of $omega$ in the basis ${du^K}_{K}$ of $bigwedge^{r}(mathbb{R}^n)$ that we get ( by wedge product $du^K=du^{k_1}wedge ldotswedge du^{k_r}$) from basis ${du^1, ldots, du^n}$ of $(mathbb{R}^n)^ast$ dual of basis ${u_1,ldots,u_n}$ of $mathbb{R}^n$.
$sum_{L}b_L dv^L$ is the expression of $omega$ in the basis ${dv^L}_{L}$ of $bigwedge^{r}(mathbb{R}^n)$ that we get ( by wedge product $dv^L=dv^{ell_1}wedge ldotswedge dv^{ell_r}$) from basis ${dv^1, ldots, dv^n}$ of $(mathbb{R}^n)^ast$ dual of canonical basis ${v_1,ldots,v_n}$ of $mathbb{R}^n$.
$c=(c_{kell})_{mtimes m}$ is the basis change matrix that expresses the basis ${u_1,ldots,u_n}$ in terms of the basis ${v_1,ldots,v_n}$ by the following equations $u_k=sum_{ell=1}^{m}c_{kell}v_ell$.
Then
$$
a_K=sum_{I} b_L det(c_{k_qell_p})_{rtimes r}
$$
with $(c_{k_qell_p})_{rtimes r}$ the matrix $r times r$ whose elements are the elenents of matrix $c=(c_{kell})$ whose indices are such that $k_qin K={k_1<ldots<k_r}$ and $ell_pin L={ell_1<ldots<j_r}$.
Proof. Fix $K={k_1<ldots<k_r}$ and set $u_{k_1},ldots, u_{k_p},ldots u_{k_r}in{u_1,ldots, u_n}$. We have
$$
omega(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
=
sum_{I}a_Idu^I(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
=
a_K
$$
On the other hand,
begin{align}
omega(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
=&
sum_{L}b_Ldv^I(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
dv^{ell_1}wedgeldotswedge dv^{ell_q}wedge ldots wedge du^{ell_r}
(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
det( dv^{ell_q}cdot u_{k_p})_{rtimes r}
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
detleft( dv^{ell_q}cdot left( sum_{ell=1}^{n}c_{k_pell}v_ell right)right)_{rtimes r}
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
detleft( c_{k_qell_p}cdot dv^{ell_q} v_{ell_q}right)_{rtimes r}
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
detleft( c_{k_qell_p}cdot 1 right)_{rtimes r}
\
=&
sum_{L}b_{L}
detleft( c_{k_qell_p} right)_{rtimes r}
\
end{align}
Therefore, it follows that
$$
a_K=sum_{I} b_L det(c_{k_qell_p})_{rtimes r}.
$$
Now the demonstration of the question that this post refers to. Since
$$
left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)in
bigwedge^{r+s}(mathbb{R}^n)
$$
there are numbers $a_{K}$, with $K$ running through all ordered sets ${k_1< ldots< k_{r+s}}subset {1,ldots, n}$, such that
$$
left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)
=
sum_{K}a_K dx^K
$$
because ${dx^K}_{K}$ is basis of $bigwedge^{r+s}(mathbb{R}^n)$.For the same reasons given above there are numbers $b_{L}$, with $L$ running through all ordered sets ${ell_1< ldots< ell_{r+s}}subset {1,ldots, n}$, such that
$$
left( sum_{I}g_I dy^I right) wedge left( sum_{J} h_{J} dy^Jright)
=
sum_{L}b_L dy^L.
$$
Fix $K={k_1<ldots< k_{r+s} }$ and $x_{k_1},ldots, x_{k_q}, ldots, x_{r+s}$. It is easy to see that
$$
left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)
(x_{k_1},ldots, x_{k_q}, ldots, x_{r+s})
=
sum_{G}a_G dx^G
(x_{k_1},ldots, x_{k_q}, ldots, x_{k_{r+s}})
=
a_K
$$
Let $c=(c_{kell})_{mtimes m}$ is the basis change matrix that expresses the base ${x_1,ldots,x_n}$ in terms of the base ${y_1,ldots,y_n}$ by the following equations $x_k=sum_{ell=1}^{m}c_{kell}y_ell$. By analogous calculations to what we did in the above demonstration we have
begin{align}
left( sum_{I}g_I dy^I right) wedge left( sum_{J} h_{J} dy^Jright)
(x_{k_1},ldots, x_{k_q}, ldots, x_{k_{r+s}})
=&
sum_{L}b_L dy^L(x_{k_1},ldots, x_{k_q}, ldots, x_{k_{r+s}})
\
=&
sum_{I} b_L det(c_{k_qell_p})_{rtimes r}
end{align}
By lema, we have $a_K=sum_{I} b_L det(c_{k_qell_p})_{rtimes r}$, and it follows that
$$
left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)
hspace{1cm} mbox{ and } hspace{1cm}
left(sum_{I}g_I dy^I right) wedge left( sum_{J} h_{J} dy^Jright)
$$
are equals in all $(r+s)$-tuple of vectors $x_{k_1},ldots, x_{k_q}, ldots, x_{k_{r+s}}$ in basis ${x_1,ldots,x_n}$. By linearity we have that equality holds for all $(r+s)$-tuple of vectors $w_{k_1},ldots, w_{k_q}, ldots, w_{k_{r+s}}$ in $mathbb{R}^n$.
answered Jan 15 at 16:10
MathOverviewMathOverview
8,71243163
$endgroup$
add a comment |
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$begingroup$
I do not know if that is the most elegant and direct answer to the question. I would appreciate suggestions for improving it as well as answers that explore a different point of view.
Lema. Let $omegain bigwedge^{r}(mathbb{R}^n)$. Supose that
$sum_{K}a_K du^K$ is the expression of $omega$ in the basis ${du^K}_{K}$ of $bigwedge^{r}(mathbb{R}^n)$ that we get ( by wedge product $du^K=du^{k_1}wedge ldotswedge du^{k_r}$) from basis ${du^1, ldots, du^n}$ of $(mathbb{R}^n)^ast$ dual of basis ${u_1,ldots,u_n}$ of $mathbb{R}^n$.
$sum_{L}b_L dv^L$ is the expression of $omega$ in the basis ${dv^L}_{L}$ of $bigwedge^{r}(mathbb{R}^n)$ that we get ( by wedge product $dv^L=dv^{ell_1}wedge ldotswedge dv^{ell_r}$) from basis ${dv^1, ldots, dv^n}$ of $(mathbb{R}^n)^ast$ dual of canonical basis ${v_1,ldots,v_n}$ of $mathbb{R}^n$.
$c=(c_{kell})_{mtimes m}$ is the basis change matrix that expresses the basis ${u_1,ldots,u_n}$ in terms of the basis ${v_1,ldots,v_n}$ by the following equations $u_k=sum_{ell=1}^{m}c_{kell}v_ell$.
Then
$$
a_K=sum_{I} b_L det(c_{k_qell_p})_{rtimes r}
$$
with $(c_{k_qell_p})_{rtimes r}$ the matrix $r times r$ whose elements are the elenents of matrix $c=(c_{kell})$ whose indices are such that $k_qin K={k_1<ldots<k_r}$ and $ell_pin L={ell_1<ldots<j_r}$.
Proof. Fix $K={k_1<ldots<k_r}$ and set $u_{k_1},ldots, u_{k_p},ldots u_{k_r}in{u_1,ldots, u_n}$. We have
$$
omega(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
=
sum_{I}a_Idu^I(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
=
a_K
$$
On the other hand,
begin{align}
omega(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
=&
sum_{L}b_Ldv^I(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
dv^{ell_1}wedgeldotswedge dv^{ell_q}wedge ldots wedge du^{ell_r}
(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
det( dv^{ell_q}cdot u_{k_p})_{rtimes r}
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
detleft( dv^{ell_q}cdot left( sum_{ell=1}^{n}c_{k_pell}v_ell right)right)_{rtimes r}
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
detleft( c_{k_qell_p}cdot dv^{ell_q} v_{ell_q}right)_{rtimes r}
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
detleft( c_{k_qell_p}cdot 1 right)_{rtimes r}
\
=&
sum_{L}b_{L}
detleft( c_{k_qell_p} right)_{rtimes r}
\
end{align}
Therefore, it follows that
$$
a_K=sum_{I} b_L det(c_{k_qell_p})_{rtimes r}.
$$
Now the demonstration of the question that this post refers to. Since
$$
left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)in
bigwedge^{r+s}(mathbb{R}^n)
$$
there are numbers $a_{K}$, with $K$ running through all ordered sets ${k_1< ldots< k_{r+s}}subset {1,ldots, n}$, such that
$$
left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)
=
sum_{K}a_K dx^K
$$
because ${dx^K}_{K}$ is basis of $bigwedge^{r+s}(mathbb{R}^n)$.For the same reasons given above there are numbers $b_{L}$, with $L$ running through all ordered sets ${ell_1< ldots< ell_{r+s}}subset {1,ldots, n}$, such that
$$
left( sum_{I}g_I dy^I right) wedge left( sum_{J} h_{J} dy^Jright)
=
sum_{L}b_L dy^L.
$$
Fix $K={k_1<ldots< k_{r+s} }$ and $x_{k_1},ldots, x_{k_q}, ldots, x_{r+s}$. It is easy to see that
$$
left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)
(x_{k_1},ldots, x_{k_q}, ldots, x_{r+s})
=
sum_{G}a_G dx^G
(x_{k_1},ldots, x_{k_q}, ldots, x_{k_{r+s}})
=
a_K
$$
Let $c=(c_{kell})_{mtimes m}$ is the basis change matrix that expresses the base ${x_1,ldots,x_n}$ in terms of the base ${y_1,ldots,y_n}$ by the following equations $x_k=sum_{ell=1}^{m}c_{kell}y_ell$. By analogous calculations to what we did in the above demonstration we have
begin{align}
left( sum_{I}g_I dy^I right) wedge left( sum_{J} h_{J} dy^Jright)
(x_{k_1},ldots, x_{k_q}, ldots, x_{k_{r+s}})
=&
sum_{L}b_L dy^L(x_{k_1},ldots, x_{k_q}, ldots, x_{k_{r+s}})
\
=&
sum_{I} b_L det(c_{k_qell_p})_{rtimes r}
end{align}
By lema, we have $a_K=sum_{I} b_L det(c_{k_qell_p})_{rtimes r}$, and it follows that
$$
left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)
hspace{1cm} mbox{ and } hspace{1cm}
left(sum_{I}g_I dy^I right) wedge left( sum_{J} h_{J} dy^Jright)
$$
are equals in all $(r+s)$-tuple of vectors $x_{k_1},ldots, x_{k_q}, ldots, x_{k_{r+s}}$ in basis ${x_1,ldots,x_n}$. By linearity we have that equality holds for all $(r+s)$-tuple of vectors $w_{k_1},ldots, w_{k_q}, ldots, w_{k_{r+s}}$ in $mathbb{R}^n$.
answered Jan 15 at 16:10
MathOverviewMathOverview
8,71243163
$endgroup$
add a comment |
$begingroup$
I do not know if that is the most elegant and direct answer to the question. I would appreciate suggestions for improving it as well as answers that explore a different point of view.
Lema. Let $omegain bigwedge^{r}(mathbb{R}^n)$. Supose that
$sum_{K}a_K du^K$ is the expression of $omega$ in the basis ${du^K}_{K}$ of $bigwedge^{r}(mathbb{R}^n)$ that we get ( by wedge product $du^K=du^{k_1}wedge ldotswedge du^{k_r}$) from basis ${du^1, ldots, du^n}$ of $(mathbb{R}^n)^ast$ dual of basis ${u_1,ldots,u_n}$ of $mathbb{R}^n$.
$sum_{L}b_L dv^L$ is the expression of $omega$ in the basis ${dv^L}_{L}$ of $bigwedge^{r}(mathbb{R}^n)$ that we get ( by wedge product $dv^L=dv^{ell_1}wedge ldotswedge dv^{ell_r}$) from basis ${dv^1, ldots, dv^n}$ of $(mathbb{R}^n)^ast$ dual of canonical basis ${v_1,ldots,v_n}$ of $mathbb{R}^n$.
$c=(c_{kell})_{mtimes m}$ is the basis change matrix that expresses the basis ${u_1,ldots,u_n}$ in terms of the basis ${v_1,ldots,v_n}$ by the following equations $u_k=sum_{ell=1}^{m}c_{kell}v_ell$.
Then
$$
a_K=sum_{I} b_L det(c_{k_qell_p})_{rtimes r}
$$
with $(c_{k_qell_p})_{rtimes r}$ the matrix $r times r$ whose elements are the elenents of matrix $c=(c_{kell})$ whose indices are such that $k_qin K={k_1<ldots<k_r}$ and $ell_pin L={ell_1<ldots<j_r}$.
Proof. Fix $K={k_1<ldots<k_r}$ and set $u_{k_1},ldots, u_{k_p},ldots u_{k_r}in{u_1,ldots, u_n}$. We have
$$
omega(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
=
sum_{I}a_Idu^I(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
=
a_K
$$
On the other hand,
begin{align}
omega(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
=&
sum_{L}b_Ldv^I(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
dv^{ell_1}wedgeldotswedge dv^{ell_q}wedge ldots wedge du^{ell_r}
(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
det( dv^{ell_q}cdot u_{k_p})_{rtimes r}
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
detleft( dv^{ell_q}cdot left( sum_{ell=1}^{n}c_{k_pell}v_ell right)right)_{rtimes r}
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
detleft( c_{k_qell_p}cdot dv^{ell_q} v_{ell_q}right)_{rtimes r}
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
detleft( c_{k_qell_p}cdot 1 right)_{rtimes r}
\
=&
sum_{L}b_{L}
detleft( c_{k_qell_p} right)_{rtimes r}
\
end{align}
Therefore, it follows that
$$
a_K=sum_{I} b_L det(c_{k_qell_p})_{rtimes r}.
$$
Now the demonstration of the question that this post refers to. Since
$$
left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)in
bigwedge^{r+s}(mathbb{R}^n)
$$
there are numbers $a_{K}$, with $K$ running through all ordered sets ${k_1< ldots< k_{r+s}}subset {1,ldots, n}$, such that
$$
left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)
=
sum_{K}a_K dx^K
$$
because ${dx^K}_{K}$ is basis of $bigwedge^{r+s}(mathbb{R}^n)$.For the same reasons given above there are numbers $b_{L}$, with $L$ running through all ordered sets ${ell_1< ldots< ell_{r+s}}subset {1,ldots, n}$, such that
$$
left( sum_{I}g_I dy^I right) wedge left( sum_{J} h_{J} dy^Jright)
=
sum_{L}b_L dy^L.
$$
Fix $K={k_1<ldots< k_{r+s} }$ and $x_{k_1},ldots, x_{k_q}, ldots, x_{r+s}$. It is easy to see that
$$
left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)
(x_{k_1},ldots, x_{k_q}, ldots, x_{r+s})
=
sum_{G}a_G dx^G
(x_{k_1},ldots, x_{k_q}, ldots, x_{k_{r+s}})
=
a_K
$$
Let $c=(c_{kell})_{mtimes m}$ is the basis change matrix that expresses the base ${x_1,ldots,x_n}$ in terms of the base ${y_1,ldots,y_n}$ by the following equations $x_k=sum_{ell=1}^{m}c_{kell}y_ell$. By analogous calculations to what we did in the above demonstration we have
begin{align}
left( sum_{I}g_I dy^I right) wedge left( sum_{J} h_{J} dy^Jright)
(x_{k_1},ldots, x_{k_q}, ldots, x_{k_{r+s}})
=&
sum_{L}b_L dy^L(x_{k_1},ldots, x_{k_q}, ldots, x_{k_{r+s}})
\
=&
sum_{I} b_L det(c_{k_qell_p})_{rtimes r}
end{align}
By lema, we have $a_K=sum_{I} b_L det(c_{k_qell_p})_{rtimes r}$, and it follows that
$$
left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)
hspace{1cm} mbox{ and } hspace{1cm}
left(sum_{I}g_I dy^I right) wedge left( sum_{J} h_{J} dy^Jright)
$$
are equals in all $(r+s)$-tuple of vectors $x_{k_1},ldots, x_{k_q}, ldots, x_{k_{r+s}}$ in basis ${x_1,ldots,x_n}$. By linearity we have that equality holds for all $(r+s)$-tuple of vectors $w_{k_1},ldots, w_{k_q}, ldots, w_{k_{r+s}}$ in $mathbb{R}^n$.
answered Jan 15 at 16:10
MathOverviewMathOverview
8,71243163
$endgroup$
add a comment |
$begingroup$
I do not know if that is the most elegant and direct answer to the question. I would appreciate suggestions for improving it as well as answers that explore a different point of view.
Lema. Let $omegain bigwedge^{r}(mathbb{R}^n)$. Supose that
$sum_{K}a_K du^K$ is the expression of $omega$ in the basis ${du^K}_{K}$ of $bigwedge^{r}(mathbb{R}^n)$ that we get ( by wedge product $du^K=du^{k_1}wedge ldotswedge du^{k_r}$) from basis ${du^1, ldots, du^n}$ of $(mathbb{R}^n)^ast$ dual of basis ${u_1,ldots,u_n}$ of $mathbb{R}^n$.
$sum_{L}b_L dv^L$ is the expression of $omega$ in the basis ${dv^L}_{L}$ of $bigwedge^{r}(mathbb{R}^n)$ that we get ( by wedge product $dv^L=dv^{ell_1}wedge ldotswedge dv^{ell_r}$) from basis ${dv^1, ldots, dv^n}$ of $(mathbb{R}^n)^ast$ dual of canonical basis ${v_1,ldots,v_n}$ of $mathbb{R}^n$.
$c=(c_{kell})_{mtimes m}$ is the basis change matrix that expresses the basis ${u_1,ldots,u_n}$ in terms of the basis ${v_1,ldots,v_n}$ by the following equations $u_k=sum_{ell=1}^{m}c_{kell}v_ell$.
Then
$$
a_K=sum_{I} b_L det(c_{k_qell_p})_{rtimes r}
$$
with $(c_{k_qell_p})_{rtimes r}$ the matrix $r times r$ whose elements are the elenents of matrix $c=(c_{kell})$ whose indices are such that $k_qin K={k_1<ldots<k_r}$ and $ell_pin L={ell_1<ldots<j_r}$.
Proof. Fix $K={k_1<ldots<k_r}$ and set $u_{k_1},ldots, u_{k_p},ldots u_{k_r}in{u_1,ldots, u_n}$. We have
$$
omega(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
=
sum_{I}a_Idu^I(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
=
a_K
$$
On the other hand,
begin{align}
omega(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
=&
sum_{L}b_Ldv^I(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
dv^{ell_1}wedgeldotswedge dv^{ell_q}wedge ldots wedge du^{ell_r}
(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
det( dv^{ell_q}cdot u_{k_p})_{rtimes r}
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
detleft( dv^{ell_q}cdot left( sum_{ell=1}^{n}c_{k_pell}v_ell right)right)_{rtimes r}
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
detleft( c_{k_qell_p}cdot dv^{ell_q} v_{ell_q}right)_{rtimes r}
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
detleft( c_{k_qell_p}cdot 1 right)_{rtimes r}
\
=&
sum_{L}b_{L}
detleft( c_{k_qell_p} right)_{rtimes r}
\
end{align}
Therefore, it follows that
$$
a_K=sum_{I} b_L det(c_{k_qell_p})_{rtimes r}.
$$
Now the demonstration of the question that this post refers to. Since
$$
left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)in
bigwedge^{r+s}(mathbb{R}^n)
$$
there are numbers $a_{K}$, with $K$ running through all ordered sets ${k_1< ldots< k_{r+s}}subset {1,ldots, n}$, such that
$$
left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)
=
sum_{K}a_K dx^K
$$
because ${dx^K}_{K}$ is basis of $bigwedge^{r+s}(mathbb{R}^n)$.For the same reasons given above there are numbers $b_{L}$, with $L$ running through all ordered sets ${ell_1< ldots< ell_{r+s}}subset {1,ldots, n}$, such that
$$
left( sum_{I}g_I dy^I right) wedge left( sum_{J} h_{J} dy^Jright)
=
sum_{L}b_L dy^L.
$$
Fix $K={k_1<ldots< k_{r+s} }$ and $x_{k_1},ldots, x_{k_q}, ldots, x_{r+s}$. It is easy to see that
$$
left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)
(x_{k_1},ldots, x_{k_q}, ldots, x_{r+s})
=
sum_{G}a_G dx^G
(x_{k_1},ldots, x_{k_q}, ldots, x_{k_{r+s}})
=
a_K
$$
Let $c=(c_{kell})_{mtimes m}$ is the basis change matrix that expresses the base ${x_1,ldots,x_n}$ in terms of the base ${y_1,ldots,y_n}$ by the following equations $x_k=sum_{ell=1}^{m}c_{kell}y_ell$. By analogous calculations to what we did in the above demonstration we have
begin{align}
left( sum_{I}g_I dy^I right) wedge left( sum_{J} h_{J} dy^Jright)
(x_{k_1},ldots, x_{k_q}, ldots, x_{k_{r+s}})
=&
sum_{L}b_L dy^L(x_{k_1},ldots, x_{k_q}, ldots, x_{k_{r+s}})
\
=&
sum_{I} b_L det(c_{k_qell_p})_{rtimes r}
end{align}
By lema, we have $a_K=sum_{I} b_L det(c_{k_qell_p})_{rtimes r}$, and it follows that
$$
left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)
hspace{1cm} mbox{ and } hspace{1cm}
left(sum_{I}g_I dy^I right) wedge left( sum_{J} h_{J} dy^Jright)
$$
are equals in all $(r+s)$-tuple of vectors $x_{k_1},ldots, x_{k_q}, ldots, x_{k_{r+s}}$ in basis ${x_1,ldots,x_n}$. By linearity we have that equality holds for all $(r+s)$-tuple of vectors $w_{k_1},ldots, w_{k_q}, ldots, w_{k_{r+s}}$ in $mathbb{R}^n$.
answered Jan 15 at 16:10
MathOverviewMathOverview
8,71243163
$endgroup$
I do not know if that is the most elegant and direct answer to the question. I would appreciate suggestions for improving it as well as answers that explore a different point of view.
Lema. Let $omegain bigwedge^{r}(mathbb{R}^n)$. Supose that
$sum_{K}a_K du^K$ is the expression of $omega$ in the basis ${du^K}_{K}$ of $bigwedge^{r}(mathbb{R}^n)$ that we get ( by wedge product $du^K=du^{k_1}wedge ldotswedge du^{k_r}$) from basis ${du^1, ldots, du^n}$ of $(mathbb{R}^n)^ast$ dual of basis ${u_1,ldots,u_n}$ of $mathbb{R}^n$.
$sum_{L}b_L dv^L$ is the expression of $omega$ in the basis ${dv^L}_{L}$ of $bigwedge^{r}(mathbb{R}^n)$ that we get ( by wedge product $dv^L=dv^{ell_1}wedge ldotswedge dv^{ell_r}$) from basis ${dv^1, ldots, dv^n}$ of $(mathbb{R}^n)^ast$ dual of canonical basis ${v_1,ldots,v_n}$ of $mathbb{R}^n$.
$c=(c_{kell})_{mtimes m}$ is the basis change matrix that expresses the basis ${u_1,ldots,u_n}$ in terms of the basis ${v_1,ldots,v_n}$ by the following equations $u_k=sum_{ell=1}^{m}c_{kell}v_ell$.
Then
$$
a_K=sum_{I} b_L det(c_{k_qell_p})_{rtimes r}
$$
with $(c_{k_qell_p})_{rtimes r}$ the matrix $r times r$ whose elements are the elenents of matrix $c=(c_{kell})$ whose indices are such that $k_qin K={k_1<ldots<k_r}$ and $ell_pin L={ell_1<ldots<j_r}$.
Proof. Fix $K={k_1<ldots<k_r}$ and set $u_{k_1},ldots, u_{k_p},ldots u_{k_r}in{u_1,ldots, u_n}$. We have
$$
omega(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
=
sum_{I}a_Idu^I(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
=
a_K
$$
On the other hand,
begin{align}
omega(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
=&
sum_{L}b_Ldv^I(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
dv^{ell_1}wedgeldotswedge dv^{ell_q}wedge ldots wedge du^{ell_r}
(u_{k_1},ldots, u_{k_p},ldots u_{k_r})
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
det( dv^{ell_q}cdot u_{k_p})_{rtimes r}
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
detleft( dv^{ell_q}cdot left( sum_{ell=1}^{n}c_{k_pell}v_ell right)right)_{rtimes r}
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
detleft( c_{k_qell_p}cdot dv^{ell_q} v_{ell_q}right)_{rtimes r}
\
=&
sum_{ell_1<ldots<ell_q<ldots <ell_r}
b_{{ell_1<ldots<ell_q<ldots <ell_r}}
detleft( c_{k_qell_p}cdot 1 right)_{rtimes r}
\
=&
sum_{L}b_{L}
detleft( c_{k_qell_p} right)_{rtimes r}
\
end{align}
Therefore, it follows that
$$
a_K=sum_{I} b_L det(c_{k_qell_p})_{rtimes r}.
$$
Now the demonstration of the question that this post refers to. Since
$$
left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)in
bigwedge^{r+s}(mathbb{R}^n)
$$
there are numbers $a_{K}$, with $K$ running through all ordered sets ${k_1< ldots< k_{r+s}}subset {1,ldots, n}$, such that
$$
left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)
=
sum_{K}a_K dx^K
$$
because ${dx^K}_{K}$ is basis of $bigwedge^{r+s}(mathbb{R}^n)$.For the same reasons given above there are numbers $b_{L}$, with $L$ running through all ordered sets ${ell_1< ldots< ell_{r+s}}subset {1,ldots, n}$, such that
$$
left( sum_{I}g_I dy^I right) wedge left( sum_{J} h_{J} dy^Jright)
=
sum_{L}b_L dy^L.
$$
Fix $K={k_1<ldots< k_{r+s} }$ and $x_{k_1},ldots, x_{k_q}, ldots, x_{r+s}$. It is easy to see that
$$
left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)
(x_{k_1},ldots, x_{k_q}, ldots, x_{r+s})
=
sum_{G}a_G dx^G
(x_{k_1},ldots, x_{k_q}, ldots, x_{k_{r+s}})
=
a_K
$$
Let $c=(c_{kell})_{mtimes m}$ is the basis change matrix that expresses the base ${x_1,ldots,x_n}$ in terms of the base ${y_1,ldots,y_n}$ by the following equations $x_k=sum_{ell=1}^{m}c_{kell}y_ell$. By analogous calculations to what we did in the above demonstration we have
begin{align}
left( sum_{I}g_I dy^I right) wedge left( sum_{J} h_{J} dy^Jright)
(x_{k_1},ldots, x_{k_q}, ldots, x_{k_{r+s}})
=&
sum_{L}b_L dy^L(x_{k_1},ldots, x_{k_q}, ldots, x_{k_{r+s}})
\
=&
sum_{I} b_L det(c_{k_qell_p})_{rtimes r}
end{align}
By lema, we have $a_K=sum_{I} b_L det(c_{k_qell_p})_{rtimes r}$, and it follows that
$$
left( sum_{I}a_I dx^Iright) wedge left( sum_{J}b_J dx^Jright)
hspace{1cm} mbox{ and } hspace{1cm}
left(sum_{I}g_I dy^I right) wedge left( sum_{J} h_{J} dy^Jright)
$$
are equals in all $(r+s)$-tuple of vectors $x_{k_1},ldots, x_{k_q}, ldots, x_{k_{r+s}}$ in basis ${x_1,ldots,x_n}$. By linearity we have that equality holds for all $(r+s)$-tuple of vectors $w_{k_1},ldots, w_{k_q}, ldots, w_{k_{r+s}}$ in $mathbb{R}^n$.
answered Jan 15 at 16:10
MathOverviewMathOverview
8,71243163
edited Jan 15 at 17:19
answered Jan 15 at 16:10
MathOverviewMathOverview
8,71243163
answered Jan 15 at 16:10
MathOverviewMathOverview
8,71243163
answered Jan 15 at 16:10
MathOverviewMathOverview
8,71243163
8,71243163
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$begingroup$
This is just because of the algebraic properties of $wedge$. Have you tried writing it out explicitly with $alpha = dx^1$ and $beta=dx^2$ for starters?
$endgroup$
– Ted Shifrin
Jan 11 at 18:45
$begingroup$
@TedShifrin In this case supose $alpha, beta bigwedge^1(mathbb{R}^2)$. We will have $alpha= dx^1=g(y_1,y_2)dy^1$, $beta= dx^2=h(y_1,y_2)dy^2$, $alphawedge beta(e_1,e_2)=1$ and $alphawedge beta g(y)h(y)det begin{pmatrix} dy^1(e_1) & dy^1(e_2) \ dy^2(e_1) & dy^2(e_2) end{pmatrix}$. But how can I use this?
$endgroup$
– MathOverview
Jan 11 at 20:12
$begingroup$
First, it's just about pointwise linear algebra. Second, don't do it by evaluating. Write $dy^i = sum a^i_j dx^j$. (In your set-up, you did the reverse. It doesn't matter.)
$endgroup$
– Ted Shifrin
Jan 11 at 21:30
$begingroup$
@TedShifrin It would be something of the type $dy^{I}wedge dy^{J}=dy^{i_1}wedge ldots wedge dy^{i_k}wedge dy^{j_1}wedge ldots wedge dy^{j_ell}=left( sum_{u=1}^{n} a_u^{i_1} dx^u right)wedge ldotswedge left( sum_{u=1}^{n} a_u^{i_k} dx^u right)wedge left( sum_{v=1}^{n} a_v^{j_1} dx^v right) wedge ldotswedge left( sum_{v=1}^{n} a_v^{j_ell} dx^v right);;;;?$
$endgroup$
– MathOverview
Jan 11 at 21:50