Differential of inverse function to a tubular neighborhood
Suppose $S$ is a (regular) compact differentiable surface embedded in $mathbb{R}^3$ so that tubular neighborhoods exist. Consider the diffeomorphism to one of them:
$$F:Stimes(-epsilon,epsilon)rightarrow N_{epsilon}(S)$$
$$F(p,t)=p+tN_p$$
where $N$ is the Gauss map, that is invertible. I have to calculate the differential of its inverse, however I think I don't fully understand how am I supposed to express it in general. I know the inverse function $F^{-1}$ takes a point in the neighborhood to a point of the space that has for its two first coordinates the orthogonal projection onto the surface and for the third one the oriented distance from said surface. Every time I try to calculate the inverse I get stuck in some circular reasoning (for example I know how to express the distance in function of the projection and vice versa but I cannot explicitly write both of them). Any help is greatly appreciated.
differential-geometry surfaces inverse-function inverse-function-theorem
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Suppose $S$ is a (regular) compact differentiable surface embedded in $mathbb{R}^3$ so that tubular neighborhoods exist. Consider the diffeomorphism to one of them:
$$F:Stimes(-epsilon,epsilon)rightarrow N_{epsilon}(S)$$
$$F(p,t)=p+tN_p$$
where $N$ is the Gauss map, that is invertible. I have to calculate the differential of its inverse, however I think I don't fully understand how am I supposed to express it in general. I know the inverse function $F^{-1}$ takes a point in the neighborhood to a point of the space that has for its two first coordinates the orthogonal projection onto the surface and for the third one the oriented distance from said surface. Every time I try to calculate the inverse I get stuck in some circular reasoning (for example I know how to express the distance in function of the projection and vice versa but I cannot explicitly write both of them). Any help is greatly appreciated.
differential-geometry surfaces inverse-function inverse-function-theorem
Differential of the inverse map is just the inverse of the differential of the original map. If you are able to compute $dF$, then it is immediate to compute $dF^{-1}$ just by inverting the differential pointwise.
– cjackal
Jan 4 at 0:40
@cjackal but the inverse of $dF$ in $p$ is the differential of $F^{-1}$ in $F(p)$ right? So for the differential of $F^{-1}$ at x I need to calculate explicitly $F^{-1}(x)$ or no?
– Renato Faraone
2 days ago
You can (with a little bit of differential geometry) compute the differential of $F$ at $(p,t)$ and so, inverting (abstractly), you'll have the differential of $F^{-1}$ at $x=F(p,t)$. You won't see it naturally in the Euclidean coordinates on $Bbb R^3$. You'll see it more naturally in terms of a basis consisting of the principal directions at $p$ and $N_p$.
– Ted Shifrin
2 days ago
add a comment |
Suppose $S$ is a (regular) compact differentiable surface embedded in $mathbb{R}^3$ so that tubular neighborhoods exist. Consider the diffeomorphism to one of them:
$$F:Stimes(-epsilon,epsilon)rightarrow N_{epsilon}(S)$$
$$F(p,t)=p+tN_p$$
where $N$ is the Gauss map, that is invertible. I have to calculate the differential of its inverse, however I think I don't fully understand how am I supposed to express it in general. I know the inverse function $F^{-1}$ takes a point in the neighborhood to a point of the space that has for its two first coordinates the orthogonal projection onto the surface and for the third one the oriented distance from said surface. Every time I try to calculate the inverse I get stuck in some circular reasoning (for example I know how to express the distance in function of the projection and vice versa but I cannot explicitly write both of them). Any help is greatly appreciated.
differential-geometry surfaces inverse-function inverse-function-theorem
Suppose $S$ is a (regular) compact differentiable surface embedded in $mathbb{R}^3$ so that tubular neighborhoods exist. Consider the diffeomorphism to one of them:
$$F:Stimes(-epsilon,epsilon)rightarrow N_{epsilon}(S)$$
$$F(p,t)=p+tN_p$$
where $N$ is the Gauss map, that is invertible. I have to calculate the differential of its inverse, however I think I don't fully understand how am I supposed to express it in general. I know the inverse function $F^{-1}$ takes a point in the neighborhood to a point of the space that has for its two first coordinates the orthogonal projection onto the surface and for the third one the oriented distance from said surface. Every time I try to calculate the inverse I get stuck in some circular reasoning (for example I know how to express the distance in function of the projection and vice versa but I cannot explicitly write both of them). Any help is greatly appreciated.
differential-geometry surfaces inverse-function inverse-function-theorem
differential-geometry surfaces inverse-function inverse-function-theorem
asked Jan 3 at 22:29
Renato Faraone
2,32911627
2,32911627
Differential of the inverse map is just the inverse of the differential of the original map. If you are able to compute $dF$, then it is immediate to compute $dF^{-1}$ just by inverting the differential pointwise.
– cjackal
Jan 4 at 0:40
@cjackal but the inverse of $dF$ in $p$ is the differential of $F^{-1}$ in $F(p)$ right? So for the differential of $F^{-1}$ at x I need to calculate explicitly $F^{-1}(x)$ or no?
– Renato Faraone
2 days ago
You can (with a little bit of differential geometry) compute the differential of $F$ at $(p,t)$ and so, inverting (abstractly), you'll have the differential of $F^{-1}$ at $x=F(p,t)$. You won't see it naturally in the Euclidean coordinates on $Bbb R^3$. You'll see it more naturally in terms of a basis consisting of the principal directions at $p$ and $N_p$.
– Ted Shifrin
2 days ago
add a comment |
Differential of the inverse map is just the inverse of the differential of the original map. If you are able to compute $dF$, then it is immediate to compute $dF^{-1}$ just by inverting the differential pointwise.
– cjackal
Jan 4 at 0:40
@cjackal but the inverse of $dF$ in $p$ is the differential of $F^{-1}$ in $F(p)$ right? So for the differential of $F^{-1}$ at x I need to calculate explicitly $F^{-1}(x)$ or no?
– Renato Faraone
2 days ago
You can (with a little bit of differential geometry) compute the differential of $F$ at $(p,t)$ and so, inverting (abstractly), you'll have the differential of $F^{-1}$ at $x=F(p,t)$. You won't see it naturally in the Euclidean coordinates on $Bbb R^3$. You'll see it more naturally in terms of a basis consisting of the principal directions at $p$ and $N_p$.
– Ted Shifrin
2 days ago
Differential of the inverse map is just the inverse of the differential of the original map. If you are able to compute $dF$, then it is immediate to compute $dF^{-1}$ just by inverting the differential pointwise.
– cjackal
Jan 4 at 0:40
Differential of the inverse map is just the inverse of the differential of the original map. If you are able to compute $dF$, then it is immediate to compute $dF^{-1}$ just by inverting the differential pointwise.
– cjackal
Jan 4 at 0:40
@cjackal but the inverse of $dF$ in $p$ is the differential of $F^{-1}$ in $F(p)$ right? So for the differential of $F^{-1}$ at x I need to calculate explicitly $F^{-1}(x)$ or no?
– Renato Faraone
2 days ago
@cjackal but the inverse of $dF$ in $p$ is the differential of $F^{-1}$ in $F(p)$ right? So for the differential of $F^{-1}$ at x I need to calculate explicitly $F^{-1}(x)$ or no?
– Renato Faraone
2 days ago
You can (with a little bit of differential geometry) compute the differential of $F$ at $(p,t)$ and so, inverting (abstractly), you'll have the differential of $F^{-1}$ at $x=F(p,t)$. You won't see it naturally in the Euclidean coordinates on $Bbb R^3$. You'll see it more naturally in terms of a basis consisting of the principal directions at $p$ and $N_p$.
– Ted Shifrin
2 days ago
You can (with a little bit of differential geometry) compute the differential of $F$ at $(p,t)$ and so, inverting (abstractly), you'll have the differential of $F^{-1}$ at $x=F(p,t)$. You won't see it naturally in the Euclidean coordinates on $Bbb R^3$. You'll see it more naturally in terms of a basis consisting of the principal directions at $p$ and $N_p$.
– Ted Shifrin
2 days ago
add a comment |
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Differential of the inverse map is just the inverse of the differential of the original map. If you are able to compute $dF$, then it is immediate to compute $dF^{-1}$ just by inverting the differential pointwise.
– cjackal
Jan 4 at 0:40
@cjackal but the inverse of $dF$ in $p$ is the differential of $F^{-1}$ in $F(p)$ right? So for the differential of $F^{-1}$ at x I need to calculate explicitly $F^{-1}(x)$ or no?
– Renato Faraone
2 days ago
You can (with a little bit of differential geometry) compute the differential of $F$ at $(p,t)$ and so, inverting (abstractly), you'll have the differential of $F^{-1}$ at $x=F(p,t)$. You won't see it naturally in the Euclidean coordinates on $Bbb R^3$. You'll see it more naturally in terms of a basis consisting of the principal directions at $p$ and $N_p$.
– Ted Shifrin
2 days ago