Mixing
Change of coordinates on manifolds
$begingroup$
Suppose I have a smooth manifold $M$ of dimension $n$ and smooth charts $(U, phi)$ and $(V, psi)$ on $M$. If we denote the coordinate functions of $phi$ by $(x^i)$ and $psi$ by $left(widetilde{x^i}right)$.
Then for $p in U cap V$ any tangent vector at $p$ can be represented in terms of basis vectors of $T_pU$ or of $T_pV$, which are $frac{partial}{partial x^i}bigg|_{p}$ and $frac{partial}{partial widetilde{x^i}}bigg|_{p}$ respectively.
The relation between the two representations is the following
$$frac{partial}{partial x^i}bigg|_{p} = sum_{j=1}^nfrac{partial widetilde{x^j}}{partial color{red}{x^i}}(phi(p))frac{partial}{partial widetilde{x^j}}bigg|_{p}$$
Now my question is that does highlighted $x^i$ in the denominator on the right hand side of the equation, mean that we're taking the partial derivative of the coordinate function $widetilde{x^i}$ of $psi$ with respect to the standard coordinates on $mathbb{R}^n$?
If so then consider the following example. Let $(x, y)$ denote the standard coordinates on $mathbb{R}^2$ and let $widetilde{x} = x$ and $widetilde{y} = y+x^3$, then $(widetilde{x}, widetilde{y})$ are global smooth coordinates on $mathbb{R}^2$. Then we would get for $p = (1,0)$ $$frac{partial}{partial x}bigg|_{p} = frac{partial}{partial widetilde{x}}bigg|_{p} + 3 frac{partial}{partial widetilde{y}}bigg|_{p}$$ and $$frac{partial}{partial widetilde{x}}bigg|_{p} = frac{partial}{partial x}bigg|_{p} + frac{partial}{partial y}bigg|_{p}.$$
Is that correct?
differential-geometry
asked Jan 17 at 5:24
PerturbativePerturbative
4,37121553
$endgroup$
add a comment |
$begingroup$
Suppose I have a smooth manifold $M$ of dimension $n$ and smooth charts $(U, phi)$ and $(V, psi)$ on $M$. If we denote the coordinate functions of $phi$ by $(x^i)$ and $psi$ by $left(widetilde{x^i}right)$.
Then for $p in U cap V$ any tangent vector at $p$ can be represented in terms of basis vectors of $T_pU$ or of $T_pV$, which are $frac{partial}{partial x^i}bigg|_{p}$ and $frac{partial}{partial widetilde{x^i}}bigg|_{p}$ respectively.
The relation between the two representations is the following
$$frac{partial}{partial x^i}bigg|_{p} = sum_{j=1}^nfrac{partial widetilde{x^j}}{partial color{red}{x^i}}(phi(p))frac{partial}{partial widetilde{x^j}}bigg|_{p}$$
Now my question is that does highlighted $x^i$ in the denominator on the right hand side of the equation, mean that we're taking the partial derivative of the coordinate function $widetilde{x^i}$ of $psi$ with respect to the standard coordinates on $mathbb{R}^n$?
If so then consider the following example. Let $(x, y)$ denote the standard coordinates on $mathbb{R}^2$ and let $widetilde{x} = x$ and $widetilde{y} = y+x^3$, then $(widetilde{x}, widetilde{y})$ are global smooth coordinates on $mathbb{R}^2$. Then we would get for $p = (1,0)$ $$frac{partial}{partial x}bigg|_{p} = frac{partial}{partial widetilde{x}}bigg|_{p} + 3 frac{partial}{partial widetilde{y}}bigg|_{p}$$ and $$frac{partial}{partial widetilde{x}}bigg|_{p} = frac{partial}{partial x}bigg|_{p} + frac{partial}{partial y}bigg|_{p}.$$
Is that correct?
differential-geometry
asked Jan 17 at 5:24
PerturbativePerturbative
4,37121553
$endgroup$
$begingroup$
The coordinate $(x^i)$ on $(U,x^i)$ is not necessarily standard coordinate in $mathbb{R}^n$ unless we chose that to be so.
$endgroup$
– Sou
Jan 17 at 6:16
add a comment |
$begingroup$
Suppose I have a smooth manifold $M$ of dimension $n$ and smooth charts $(U, phi)$ and $(V, psi)$ on $M$. If we denote the coordinate functions of $phi$ by $(x^i)$ and $psi$ by $left(widetilde{x^i}right)$.
Then for $p in U cap V$ any tangent vector at $p$ can be represented in terms of basis vectors of $T_pU$ or of $T_pV$, which are $frac{partial}{partial x^i}bigg|_{p}$ and $frac{partial}{partial widetilde{x^i}}bigg|_{p}$ respectively.
The relation between the two representations is the following
$$frac{partial}{partial x^i}bigg|_{p} = sum_{j=1}^nfrac{partial widetilde{x^j}}{partial color{red}{x^i}}(phi(p))frac{partial}{partial widetilde{x^j}}bigg|_{p}$$
Now my question is that does highlighted $x^i$ in the denominator on the right hand side of the equation, mean that we're taking the partial derivative of the coordinate function $widetilde{x^i}$ of $psi$ with respect to the standard coordinates on $mathbb{R}^n$?
If so then consider the following example. Let $(x, y)$ denote the standard coordinates on $mathbb{R}^2$ and let $widetilde{x} = x$ and $widetilde{y} = y+x^3$, then $(widetilde{x}, widetilde{y})$ are global smooth coordinates on $mathbb{R}^2$. Then we would get for $p = (1,0)$ $$frac{partial}{partial x}bigg|_{p} = frac{partial}{partial widetilde{x}}bigg|_{p} + 3 frac{partial}{partial widetilde{y}}bigg|_{p}$$ and $$frac{partial}{partial widetilde{x}}bigg|_{p} = frac{partial}{partial x}bigg|_{p} + frac{partial}{partial y}bigg|_{p}.$$
Is that correct?
differential-geometry
asked Jan 17 at 5:24
PerturbativePerturbative
4,37121553
$endgroup$
Suppose I have a smooth manifold $M$ of dimension $n$ and smooth charts $(U, phi)$ and $(V, psi)$ on $M$. If we denote the coordinate functions of $phi$ by $(x^i)$ and $psi$ by $left(widetilde{x^i}right)$.
Then for $p in U cap V$ any tangent vector at $p$ can be represented in terms of basis vectors of $T_pU$ or of $T_pV$, which are $frac{partial}{partial x^i}bigg|_{p}$ and $frac{partial}{partial widetilde{x^i}}bigg|_{p}$ respectively.
The relation between the two representations is the following
$$frac{partial}{partial x^i}bigg|_{p} = sum_{j=1}^nfrac{partial widetilde{x^j}}{partial color{red}{x^i}}(phi(p))frac{partial}{partial widetilde{x^j}}bigg|_{p}$$
Now my question is that does highlighted $x^i$ in the denominator on the right hand side of the equation, mean that we're taking the partial derivative of the coordinate function $widetilde{x^i}$ of $psi$ with respect to the standard coordinates on $mathbb{R}^n$?
If so then consider the following example. Let $(x, y)$ denote the standard coordinates on $mathbb{R}^2$ and let $widetilde{x} = x$ and $widetilde{y} = y+x^3$, then $(widetilde{x}, widetilde{y})$ are global smooth coordinates on $mathbb{R}^2$. Then we would get for $p = (1,0)$ $$frac{partial}{partial x}bigg|_{p} = frac{partial}{partial widetilde{x}}bigg|_{p} + 3 frac{partial}{partial widetilde{y}}bigg|_{p}$$ and $$frac{partial}{partial widetilde{x}}bigg|_{p} = frac{partial}{partial x}bigg|_{p} + frac{partial}{partial y}bigg|_{p}.$$
Is that correct?
differential-geometry
differential-geometry
asked Jan 17 at 5:24
PerturbativePerturbative
4,37121553
asked Jan 17 at 5:24
PerturbativePerturbative
4,37121553
asked Jan 17 at 5:24
PerturbativePerturbative
4,37121553
asked Jan 17 at 5:24
PerturbativePerturbative
4,37121553
asked Jan 17 at 5:24
PerturbativePerturbative
4,37121553
4,37121553
$begingroup$
The coordinate $(x^i)$ on $(U,x^i)$ is not necessarily standard coordinate in $mathbb{R}^n$ unless we chose that to be so.
$endgroup$
– Sou
Jan 17 at 6:16
add a comment |
$begingroup$
The coordinate $(x^i)$ on $(U,x^i)$ is not necessarily standard coordinate in $mathbb{R}^n$ unless we chose that to be so.
$endgroup$
– Sou
Jan 17 at 6:16
$begingroup$
The coordinate $(x^i)$ on $(U,x^i)$ is not necessarily standard coordinate in $mathbb{R}^n$ unless we chose that to be so.
$endgroup$
– Sou
Jan 17 at 6:16
$begingroup$
The coordinate $(x^i)$ on $(U,x^i)$ is not necessarily standard coordinate in $mathbb{R}^n$ unless we chose that to be so.
$endgroup$
– Sou
Jan 17 at 6:16
add a comment |
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$begingroup$
The coordinate $(x^i)$ on $(U,x^i)$ is not necessarily standard coordinate in $mathbb{R}^n$ unless we chose that to be so.
$endgroup$
– Sou
Jan 17 at 6:16