Expression of $n$-form with two charts.
Let $(U , varphi)$ and $(V , psi)$ be two charts on a $n$-dimensional differentiable manifold $M$, with $U cap V neq emptyset$, such that $varphi = (x_1 , ldots , x_n)$ and $psi = (y_1 , ldots , y_n)$. We have two elements in ${Lambda}^n(T_pM)$ ($p in U cap V$):
$$
{(d x_1)}_p wedge ldots wedge {(d x_n)}_p qquad mbox{ and } qquad {(d y_1)}_p wedge ldots wedge {(d y_n)}_pmbox{.}
$$
How can I show that
$$
{(d y_1)}_p wedge ldots wedge {(d y_n)}_p = left(det d {(psi circ {varphi}^{- 1})}_{varphi(p)}right) {(d x_1)}_p wedge ldots wedge {(d x_n)}_p?
$$
I have got the equality
$$
{(d y_1)}_p wedge ldots wedge {(d y_n)}_p = lambda(p) {(d x_1)}_p wedge ldots wedge {(d x_n)}_pmbox{,}
$$
where
$$
lambda(p) = det {left({(d y_i)}_p {left(frac{partial}{partial x_j}right)}_pright)}_{i , j = 1}^nmbox{.}
$$
I am using the notation ${left{{left(frac{partial}{partial x_i}right)}_pright}}_{i = 1}^n$ basis for $T_pM$ (using the chart $(U , varphi)$) and ${{{(d y_i)}_p}}_{i = 1}^n$ dual basis of ${left{{left(frac{partial}{partial y_i}right)}_pright}}_{i = 1}^n$ for $T_p^*M$ ($= {(T_pM)}^*$).
How can I show that $lambda(p) = det d {(psi circ {varphi}^{- 1})}_{varphi(p)}$?
manifolds differential-forms
asked Nov 25 at 15:04
joseabp91
1,243411
add a comment |
Let $(U , varphi)$ and $(V , psi)$ be two charts on a $n$-dimensional differentiable manifold $M$, with $U cap V neq emptyset$, such that $varphi = (x_1 , ldots , x_n)$ and $psi = (y_1 , ldots , y_n)$. We have two elements in ${Lambda}^n(T_pM)$ ($p in U cap V$):
$$
{(d x_1)}_p wedge ldots wedge {(d x_n)}_p qquad mbox{ and } qquad {(d y_1)}_p wedge ldots wedge {(d y_n)}_pmbox{.}
$$
How can I show that
$$
{(d y_1)}_p wedge ldots wedge {(d y_n)}_p = left(det d {(psi circ {varphi}^{- 1})}_{varphi(p)}right) {(d x_1)}_p wedge ldots wedge {(d x_n)}_p?
$$
I have got the equality
$$
{(d y_1)}_p wedge ldots wedge {(d y_n)}_p = lambda(p) {(d x_1)}_p wedge ldots wedge {(d x_n)}_pmbox{,}
$$
where
$$
lambda(p) = det {left({(d y_i)}_p {left(frac{partial}{partial x_j}right)}_pright)}_{i , j = 1}^nmbox{.}
$$
I am using the notation ${left{{left(frac{partial}{partial x_i}right)}_pright}}_{i = 1}^n$ basis for $T_pM$ (using the chart $(U , varphi)$) and ${{{(d y_i)}_p}}_{i = 1}^n$ dual basis of ${left{{left(frac{partial}{partial y_i}right)}_pright}}_{i = 1}^n$ for $T_p^*M$ ($= {(T_pM)}^*$).
How can I show that $lambda(p) = det d {(psi circ {varphi}^{- 1})}_{varphi(p)}$?
manifolds differential-forms
asked Nov 25 at 15:04
joseabp91
1,243411
add a comment |
Let $(U , varphi)$ and $(V , psi)$ be two charts on a $n$-dimensional differentiable manifold $M$, with $U cap V neq emptyset$, such that $varphi = (x_1 , ldots , x_n)$ and $psi = (y_1 , ldots , y_n)$. We have two elements in ${Lambda}^n(T_pM)$ ($p in U cap V$):
$$
{(d x_1)}_p wedge ldots wedge {(d x_n)}_p qquad mbox{ and } qquad {(d y_1)}_p wedge ldots wedge {(d y_n)}_pmbox{.}
$$
How can I show that
$$
{(d y_1)}_p wedge ldots wedge {(d y_n)}_p = left(det d {(psi circ {varphi}^{- 1})}_{varphi(p)}right) {(d x_1)}_p wedge ldots wedge {(d x_n)}_p?
$$
I have got the equality
$$
{(d y_1)}_p wedge ldots wedge {(d y_n)}_p = lambda(p) {(d x_1)}_p wedge ldots wedge {(d x_n)}_pmbox{,}
$$
where
$$
lambda(p) = det {left({(d y_i)}_p {left(frac{partial}{partial x_j}right)}_pright)}_{i , j = 1}^nmbox{.}
$$
I am using the notation ${left{{left(frac{partial}{partial x_i}right)}_pright}}_{i = 1}^n$ basis for $T_pM$ (using the chart $(U , varphi)$) and ${{{(d y_i)}_p}}_{i = 1}^n$ dual basis of ${left{{left(frac{partial}{partial y_i}right)}_pright}}_{i = 1}^n$ for $T_p^*M$ ($= {(T_pM)}^*$).
How can I show that $lambda(p) = det d {(psi circ {varphi}^{- 1})}_{varphi(p)}$?
manifolds differential-forms
asked Nov 25 at 15:04
joseabp91
1,243411
Let $(U , varphi)$ and $(V , psi)$ be two charts on a $n$-dimensional differentiable manifold $M$, with $U cap V neq emptyset$, such that $varphi = (x_1 , ldots , x_n)$ and $psi = (y_1 , ldots , y_n)$. We have two elements in ${Lambda}^n(T_pM)$ ($p in U cap V$):
$$
{(d x_1)}_p wedge ldots wedge {(d x_n)}_p qquad mbox{ and } qquad {(d y_1)}_p wedge ldots wedge {(d y_n)}_pmbox{.}
$$
How can I show that
$$
{(d y_1)}_p wedge ldots wedge {(d y_n)}_p = left(det d {(psi circ {varphi}^{- 1})}_{varphi(p)}right) {(d x_1)}_p wedge ldots wedge {(d x_n)}_p?
$$
I have got the equality
$$
{(d y_1)}_p wedge ldots wedge {(d y_n)}_p = lambda(p) {(d x_1)}_p wedge ldots wedge {(d x_n)}_pmbox{,}
$$
where
$$
lambda(p) = det {left({(d y_i)}_p {left(frac{partial}{partial x_j}right)}_pright)}_{i , j = 1}^nmbox{.}
$$
I am using the notation ${left{{left(frac{partial}{partial x_i}right)}_pright}}_{i = 1}^n$ basis for $T_pM$ (using the chart $(U , varphi)$) and ${{{(d y_i)}_p}}_{i = 1}^n$ dual basis of ${left{{left(frac{partial}{partial y_i}right)}_pright}}_{i = 1}^n$ for $T_p^*M$ ($= {(T_pM)}^*$).
How can I show that $lambda(p) = det d {(psi circ {varphi}^{- 1})}_{varphi(p)}$?
manifolds differential-forms
manifolds differential-forms
asked Nov 25 at 15:04
joseabp91
1,243411
asked Nov 25 at 15:04
joseabp91
1,243411
asked Nov 25 at 15:04
joseabp91
1,243411
asked Nov 25 at 15:04
joseabp91
1,243411
asked Nov 25 at 15:04
joseabp91
1,243411
1,243411
add a comment |
add a comment |
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