How to prove that the definition of exterior product of differential forms is not ambiguous?












0












$begingroup$


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.










share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















0












$begingroup$


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.










share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












0








0





$begingroup$


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.










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 14:31







MathOverview

















asked Jan 11 at 18:40









MathOverviewMathOverview

8,94543164




8,94543164












  • $begingroup$
    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










  • $begingroup$
    @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












1 Answer
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oldest

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0












$begingroup$

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 $mathbb{R}^n$.






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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 $mathbb{R}^n$.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      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 $mathbb{R}^n$.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        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 $mathbb{R}^n$.






        share|cite|improve this answer











        $endgroup$



        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 $mathbb{R}^n$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 15 at 17:19

























        answered Jan 15 at 16:10









        MathOverviewMathOverview

        8,94543164




        8,94543164






























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