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How do I prove that the second derivative of a function $f:Mtomathbb{R}$ defined on a surface...
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How do I prove that the second derivative of a function $f:Mtomathbb{R}$ defined on a surface $Msubsetmathbb{R}^n$ is well defined?
If, as says the user Amitai Yuval, the second derivative of a function defined on a $ M $ surface can not be well defined in general what is the explanation for this impossibility?
By a second derivative I mean an application $D^{(2)}f$ that at each point $ p in M $ associates a well-defined bilinear aplication
$$
D^{(2)}f(p):T_pM times T_pM tomathbb{R}.
$$
Let me make a background in a few steps to try to make the question more precise.
Step one. Here $ M $ designates a $C^k$ surface of dimension $ m $ within the Euclidean space $ mathbb{R}^n$, i.e. a set in $mathbb{R}^n$ which satisfies the following properties: (1) there is a family of open sets ${O_i}_{iin I}$ such that $Msubset bigcup_{iin I} O_i$ and (2) for any $U=Mcap O_i$ there are a open set $U_0$ in $mathbb{R}^m$ and a $C^k$ ($kgeq 2$) parametrization $varphi:U_0to U$.
Step two. I'm assuming that, fixed a point $ p in M $, for any two parametrizations $varphi:U_0to Usubset M$ and $psi:V_0to Vsubset M$ such that $pin Vcap U$ it is proved that
$$
(varphi^{-1}circpsi):psi^{-1}(Ucap V)to varphi^{-1}(Ucap V)
quad mbox{ and } quad
( psi^{-1}circ varphi):varphi^{-1}(Ucap V)to psi^{-1}(Ucap V)
$$
are $C^k$ ($kgeq 2$) diffeormorphisms.
Step three. I am also assuming that the tangent plane $T_pM$ is the well defined vector space given by any parameterization $varphi:U_0to Usubset M$ such that $varphi(a)=p$ as
$$
T_pM= Dvarphi(a)(mathbb{R}^m).
$$
By well defined I mean $ Dvarphi(a)(mathbb{R}^m)=Dpsi(b)(mathbb{R}^m)$ for any other parameterization $psi:V_0to Vsubset M$ such that $psi(b)=p$.
Step four. A function $ f: M to mathbb{R}$ is $C^r$ ($1leq r<k$) differentiable at point $p$ if there is a parameterization $varphi:U_0to Usubset M$ with $varphi(a)=p$ such that $fcircvarphi$ is $C^r$ differentiable in $a$. Once the parameter change $(varphi^{-1}circpsi):psi^{-1}(Ucap V)to varphi^{-1}(Ucap V)$ is differentiable it follows that the application $fcircpsi$ is differentiable in $b$ for all parameterization $psi:V_0to Vsubset M$ such that $psi(b)=p$.
Step five If $ f:Mto mathbb{R} $ is $C^r$ $(r>1)$ differentiable at point $ p in M$ then its derivative at that point is the linear transformation $ Df(p):T_pMtomathbb{R} $ defined as follows. Let's take a parameterization $varphi:U_0to Usubset M$ with $varphi(a)=p$. Given a vector $uin T_p M $ there exists unique a vector $ mu in mathbb{R}^m$ such that $u=Df(a)mu$.The derivative of $ f $ at point $ p $ is then simply defined by
$$
Df(p)cdot u =D (fcirc varphi)(a)cdotmu
$$
Step six. The linear transformation of step five is well defined. That is, if $psi:V_0to V$ is any other parameterization with $psi(b)=p$ and $u=Dpsi(b)zeta$ for some vector $zetainmathbb{R}^m$, then
$$
D (fcirc varphi)(a)cdotmu= D (fcirc psi)(b)cdotzeta.
$$
Indeed, $psi=varphicirc(varphi^{-1}circpsi)$ at where $(varphi^{-1}circpsi):psi^{-1}(Ucap V)to varphi^{-1}(Ucap V)$ is a $C^{k}$ diffeomorphism such that $(varphi^{-1}circpsi)(b)=a$. We have
begin{align}
Dvarphi(a)cdot mu=& u \
=& Dpsi(b)cdot zeta \
=& D((varphicirc varphi^{-1})circpsi )(b)cdot zeta\
=& D(varphicirc(varphi^{-1}circpsi))(b)cdot zeta\
=& Dvarphi(a)cdot D(varphi^{-1}circpsi)(b)cdotzeta
end{align}
Since $ D varphi(a)$ is injective we have $mu=D(varphi^{-1}circpsi)(b)cdotzeta$.
Therefore,
begin{align}
D(fcircpsi)(b)zeta =& D(fcirc(varphicircvarphi^{-1})circpsi)(b)cdotzeta \
=& D(fcirc varphicirc(varphi^{-1}circpsi))(b)cdotzeta \
=& D(fcirc varphi)(a) cdot (varphi^{-1}circpsi)(b)cdotzeta \
=& D(fcirc varphi)(a) cdotmu
end{align}
Let us end the background here and return to the question. Below is my attempt to answer the question.
By analogy to the first order derivative the second derivative would work as follows. If $ f:Mto mathbb{R} $ is $C^2$ differentiable at point $ p in M$ then its second derivative $D^{(2)}f$ each point $p$ associates the bilinear transformation $ D^{(2)}f(p):T_pMtimes T_pMtomathbb{R} $ defined as follows. Let's take a parameterization $varphi:U_0to Usubset M$ with $varphi(a)=p$. Given the vectors $u,vin T_p M $ there are a vectors $ mu,nu in mathbb{R}^m$ such that $u=Df(a)mu$ and $v=Df(a)nu$ .The second derivative of $ f $ at point $ p $ is then simply defined by
$$
D^{(2)}f(p)cdot (u,v) =D^{(2)} (fcirc varphi)(a)cdot(mu,nu)
$$
Similarly to the case of the first derivative, the second derivative will be well defined if for any other parameterization $psi:V_0to V$ such that $psi(b)=p$, $Dpsi(b)eta=u$ and $Dpsi(b)zeta=v$ for some $etainmathbb{R}^m$ and for some $zetainmathbb{R}^m$ we have
$$
D^{(2)} (fcirc varphi)(a)cdot(mu,nu)=D^{(2)} (fcirc psi)(b)cdot(eta,zeta)
$$
To show the above equality I tried to imitate step six. We have
begin{align}
D^{(2)} (fcirc psi)(b)cdot(eta,zeta)
=&
D^{(2)} (fcirc(varphicircvarphi^{-1})circ psi)(b)cdot(eta,zeta)
\
=&
D^{(2)} (fcircvarphicirc(varphi^{-1}circ psi))(b)cdot(eta,zeta)
end{align}
But in this last equality I can not use the chain rule.
On the other hand the development of the second derivative $D^{(2)} (fcirc varphi)(a)cdot(mu,nu)$ in terms of the partial derivatives $frac{partial^{2}f}{partial x_ipartial x_j}(a)$ was more productive. Look
begin{align}
D^{(2)} (fcirc varphi)(a)cdot(mu,nu)
=&
left[
frac{partial}{partial nu}
left(
frac{partial}{partial mu} (fcircvarphi)
right)
right](a)
\
=&
left[
frac{partial}{partial nu}
left(
sum_{j=1}^{m}mu_jcdot frac{partial}{partial x_j} (fcircvarphi)
right)
right](a)
\
=&
sum_{i=1}^msum_{j=1}^{m}
nu_icdotmu_jcdot frac{partial^2}{partial x_i partial x_j} (fcircvarphi)(a)
\
end{align}
By $frac{partial}{partial mu} (fcircvarphi)(a)= D(fcirc varphi)(a) cdotmu =u= D(fcircpsi)(b)eta= frac{partial}{partial eta} (fcircpsi)(b)$ we have
begin{align}
D^{(2)} (fcirc varphi)(a)cdot(mu,nu)
=&
left[
frac{partial}{partial nu}
left(
frac{partial}{partial mu} (fcircvarphi)
right)
right](a)
\
=&
left[
frac{partial}{partial nu}
left(
frac{partial}{partial eta} (fcircpsi)
right)
right](b)
\
=&
left[
frac{partial}{partial nu}
left(
sum_{j=1}^{m}eta_jcdot frac{partial}{partial x_j} (fcircpsi)
right)
right](b)
\
=&
sum_{i=1}^msum_{j=1}^{m}
nu_icdoteta_jcdot frac{partial^2}{partial x_i partial x_j} (fcircpsi)(b)
\
=&
D^{(2)}(fcircpsi)(a)cdot(eta,nu)
end{align}
By an entirely analogous calculation and applying the Schwarz theorem we have
$$
D^{(2)}(fcircvarphi)(a)cdot(mu,nu)= D^{(2)}(fcircpsi)(a)cdot(mu,zeta).
$$
The question then becomes the following. How to use the identities below to prove the well definiteness of $ D^{(2)}f$?
begin{align}
D^{(2)} (fcirc varphi)(a)cdot(mu,nu)=&D^{(2)}(fcircpsi)(a)cdot(eta,nu)
\
D^{(2)}(fcircvarphi)(a)cdot(mu,nu)=&D^{(2)}(fcircpsi)(a)cdot(mu,zeta).
end{align}
multivariable-calculus derivatives differential-geometry differential-topology
asked Jan 31 at 17:02
MathOverviewMathOverview
8,95243164
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add a comment |
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How do I prove that the second derivative of a function $f:Mtomathbb{R}$ defined on a surface $Msubsetmathbb{R}^n$ is well defined?
If, as says the user Amitai Yuval, the second derivative of a function defined on a $ M $ surface can not be well defined in general what is the explanation for this impossibility?
By a second derivative I mean an application $D^{(2)}f$ that at each point $ p in M $ associates a well-defined bilinear aplication
$$
D^{(2)}f(p):T_pM times T_pM tomathbb{R}.
$$
Let me make a background in a few steps to try to make the question more precise.
Step one. Here $ M $ designates a $C^k$ surface of dimension $ m $ within the Euclidean space $ mathbb{R}^n$, i.e. a set in $mathbb{R}^n$ which satisfies the following properties: (1) there is a family of open sets ${O_i}_{iin I}$ such that $Msubset bigcup_{iin I} O_i$ and (2) for any $U=Mcap O_i$ there are a open set $U_0$ in $mathbb{R}^m$ and a $C^k$ ($kgeq 2$) parametrization $varphi:U_0to U$.
Step two. I'm assuming that, fixed a point $ p in M $, for any two parametrizations $varphi:U_0to Usubset M$ and $psi:V_0to Vsubset M$ such that $pin Vcap U$ it is proved that
$$
(varphi^{-1}circpsi):psi^{-1}(Ucap V)to varphi^{-1}(Ucap V)
quad mbox{ and } quad
( psi^{-1}circ varphi):varphi^{-1}(Ucap V)to psi^{-1}(Ucap V)
$$
are $C^k$ ($kgeq 2$) diffeormorphisms.
Step three. I am also assuming that the tangent plane $T_pM$ is the well defined vector space given by any parameterization $varphi:U_0to Usubset M$ such that $varphi(a)=p$ as
$$
T_pM= Dvarphi(a)(mathbb{R}^m).
$$
By well defined I mean $ Dvarphi(a)(mathbb{R}^m)=Dpsi(b)(mathbb{R}^m)$ for any other parameterization $psi:V_0to Vsubset M$ such that $psi(b)=p$.
Step four. A function $ f: M to mathbb{R}$ is $C^r$ ($1leq r<k$) differentiable at point $p$ if there is a parameterization $varphi:U_0to Usubset M$ with $varphi(a)=p$ such that $fcircvarphi$ is $C^r$ differentiable in $a$. Once the parameter change $(varphi^{-1}circpsi):psi^{-1}(Ucap V)to varphi^{-1}(Ucap V)$ is differentiable it follows that the application $fcircpsi$ is differentiable in $b$ for all parameterization $psi:V_0to Vsubset M$ such that $psi(b)=p$.
Step five If $ f:Mto mathbb{R} $ is $C^r$ $(r>1)$ differentiable at point $ p in M$ then its derivative at that point is the linear transformation $ Df(p):T_pMtomathbb{R} $ defined as follows. Let's take a parameterization $varphi:U_0to Usubset M$ with $varphi(a)=p$. Given a vector $uin T_p M $ there exists unique a vector $ mu in mathbb{R}^m$ such that $u=Df(a)mu$.The derivative of $ f $ at point $ p $ is then simply defined by
$$
Df(p)cdot u =D (fcirc varphi)(a)cdotmu
$$
Step six. The linear transformation of step five is well defined. That is, if $psi:V_0to V$ is any other parameterization with $psi(b)=p$ and $u=Dpsi(b)zeta$ for some vector $zetainmathbb{R}^m$, then
$$
D (fcirc varphi)(a)cdotmu= D (fcirc psi)(b)cdotzeta.
$$
Indeed, $psi=varphicirc(varphi^{-1}circpsi)$ at where $(varphi^{-1}circpsi):psi^{-1}(Ucap V)to varphi^{-1}(Ucap V)$ is a $C^{k}$ diffeomorphism such that $(varphi^{-1}circpsi)(b)=a$. We have
begin{align}
Dvarphi(a)cdot mu=& u \
=& Dpsi(b)cdot zeta \
=& D((varphicirc varphi^{-1})circpsi )(b)cdot zeta\
=& D(varphicirc(varphi^{-1}circpsi))(b)cdot zeta\
=& Dvarphi(a)cdot D(varphi^{-1}circpsi)(b)cdotzeta
end{align}
Since $ D varphi(a)$ is injective we have $mu=D(varphi^{-1}circpsi)(b)cdotzeta$.
Therefore,
begin{align}
D(fcircpsi)(b)zeta =& D(fcirc(varphicircvarphi^{-1})circpsi)(b)cdotzeta \
=& D(fcirc varphicirc(varphi^{-1}circpsi))(b)cdotzeta \
=& D(fcirc varphi)(a) cdot (varphi^{-1}circpsi)(b)cdotzeta \
=& D(fcirc varphi)(a) cdotmu
end{align}
Let us end the background here and return to the question. Below is my attempt to answer the question.
By analogy to the first order derivative the second derivative would work as follows. If $ f:Mto mathbb{R} $ is $C^2$ differentiable at point $ p in M$ then its second derivative $D^{(2)}f$ each point $p$ associates the bilinear transformation $ D^{(2)}f(p):T_pMtimes T_pMtomathbb{R} $ defined as follows. Let's take a parameterization $varphi:U_0to Usubset M$ with $varphi(a)=p$. Given the vectors $u,vin T_p M $ there are a vectors $ mu,nu in mathbb{R}^m$ such that $u=Df(a)mu$ and $v=Df(a)nu$ .The second derivative of $ f $ at point $ p $ is then simply defined by
$$
D^{(2)}f(p)cdot (u,v) =D^{(2)} (fcirc varphi)(a)cdot(mu,nu)
$$
Similarly to the case of the first derivative, the second derivative will be well defined if for any other parameterization $psi:V_0to V$ such that $psi(b)=p$, $Dpsi(b)eta=u$ and $Dpsi(b)zeta=v$ for some $etainmathbb{R}^m$ and for some $zetainmathbb{R}^m$ we have
$$
D^{(2)} (fcirc varphi)(a)cdot(mu,nu)=D^{(2)} (fcirc psi)(b)cdot(eta,zeta)
$$
To show the above equality I tried to imitate step six. We have
begin{align}
D^{(2)} (fcirc psi)(b)cdot(eta,zeta)
=&
D^{(2)} (fcirc(varphicircvarphi^{-1})circ psi)(b)cdot(eta,zeta)
\
=&
D^{(2)} (fcircvarphicirc(varphi^{-1}circ psi))(b)cdot(eta,zeta)
end{align}
But in this last equality I can not use the chain rule.
On the other hand the development of the second derivative $D^{(2)} (fcirc varphi)(a)cdot(mu,nu)$ in terms of the partial derivatives $frac{partial^{2}f}{partial x_ipartial x_j}(a)$ was more productive. Look
begin{align}
D^{(2)} (fcirc varphi)(a)cdot(mu,nu)
=&
left[
frac{partial}{partial nu}
left(
frac{partial}{partial mu} (fcircvarphi)
right)
right](a)
\
=&
left[
frac{partial}{partial nu}
left(
sum_{j=1}^{m}mu_jcdot frac{partial}{partial x_j} (fcircvarphi)
right)
right](a)
\
=&
sum_{i=1}^msum_{j=1}^{m}
nu_icdotmu_jcdot frac{partial^2}{partial x_i partial x_j} (fcircvarphi)(a)
\
end{align}
By $frac{partial}{partial mu} (fcircvarphi)(a)= D(fcirc varphi)(a) cdotmu =u= D(fcircpsi)(b)eta= frac{partial}{partial eta} (fcircpsi)(b)$ we have
begin{align}
D^{(2)} (fcirc varphi)(a)cdot(mu,nu)
=&
left[
frac{partial}{partial nu}
left(
frac{partial}{partial mu} (fcircvarphi)
right)
right](a)
\
=&
left[
frac{partial}{partial nu}
left(
frac{partial}{partial eta} (fcircpsi)
right)
right](b)
\
=&
left[
frac{partial}{partial nu}
left(
sum_{j=1}^{m}eta_jcdot frac{partial}{partial x_j} (fcircpsi)
right)
right](b)
\
=&
sum_{i=1}^msum_{j=1}^{m}
nu_icdoteta_jcdot frac{partial^2}{partial x_i partial x_j} (fcircpsi)(b)
\
=&
D^{(2)}(fcircpsi)(a)cdot(eta,nu)
end{align}
By an entirely analogous calculation and applying the Schwarz theorem we have
$$
D^{(2)}(fcircvarphi)(a)cdot(mu,nu)= D^{(2)}(fcircpsi)(a)cdot(mu,zeta).
$$
The question then becomes the following. How to use the identities below to prove the well definiteness of $ D^{(2)}f$?
begin{align}
D^{(2)} (fcirc varphi)(a)cdot(mu,nu)=&D^{(2)}(fcircpsi)(a)cdot(eta,nu)
\
D^{(2)}(fcircvarphi)(a)cdot(mu,nu)=&D^{(2)}(fcircpsi)(a)cdot(mu,zeta).
end{align}
multivariable-calculus derivatives differential-geometry differential-topology
asked Jan 31 at 17:02
MathOverviewMathOverview
8,95243164
$endgroup$
1
$begingroup$
You are perfectly right - the second derivative of a function on $M$, as opposed to the first derivative, is not well-defined. It is, by the way, well-defined at critical points of the function in question. Furthermore, it may become well-defined everywhere once we equip $M$ with an affine connection. One way to do this is take the Levi-Civita connection of the Riemannian metric on $M$ induced by the embedding in $mathbb{R}^n$. One can also equip $M$ with many other affine connections.
$endgroup$
– Amitai Yuval
Jan 31 at 18:01
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@AmitaiYuval Excuse me but I have no experience with Riemannian geometry. Could you explain me using the background of the question?
$endgroup$
– MathOverview
Jan 31 at 18:29
$begingroup$
The second derivatives are not well-defined, exactly in the sense you work with. That is if you compute them by means of a coordinate chart just like in $mathbb{R}^m$, the result you get depends on the choice of coordinates. The geometric construction (or data) which enables one to define reasonable high-order derivatives is called an affine connection. You should grab a basic textbook in differential geometry and read about this.
$endgroup$
– Amitai Yuval
Jan 31 at 18:53
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@AmitaiYuval Is there an example of a function defined on a surface that illustrates the impossibility of defining its second derivative? If so, could you put it here?
$endgroup$
– MathOverview
Jan 31 at 19:05
add a comment |
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How do I prove that the second derivative of a function $f:Mtomathbb{R}$ defined on a surface $Msubsetmathbb{R}^n$ is well defined?
If, as says the user Amitai Yuval, the second derivative of a function defined on a $ M $ surface can not be well defined in general what is the explanation for this impossibility?
By a second derivative I mean an application $D^{(2)}f$ that at each point $ p in M $ associates a well-defined bilinear aplication
$$
D^{(2)}f(p):T_pM times T_pM tomathbb{R}.
$$
Let me make a background in a few steps to try to make the question more precise.
Step one. Here $ M $ designates a $C^k$ surface of dimension $ m $ within the Euclidean space $ mathbb{R}^n$, i.e. a set in $mathbb{R}^n$ which satisfies the following properties: (1) there is a family of open sets ${O_i}_{iin I}$ such that $Msubset bigcup_{iin I} O_i$ and (2) for any $U=Mcap O_i$ there are a open set $U_0$ in $mathbb{R}^m$ and a $C^k$ ($kgeq 2$) parametrization $varphi:U_0to U$.
Step two. I'm assuming that, fixed a point $ p in M $, for any two parametrizations $varphi:U_0to Usubset M$ and $psi:V_0to Vsubset M$ such that $pin Vcap U$ it is proved that
$$
(varphi^{-1}circpsi):psi^{-1}(Ucap V)to varphi^{-1}(Ucap V)
quad mbox{ and } quad
( psi^{-1}circ varphi):varphi^{-1}(Ucap V)to psi^{-1}(Ucap V)
$$
are $C^k$ ($kgeq 2$) diffeormorphisms.
Step three. I am also assuming that the tangent plane $T_pM$ is the well defined vector space given by any parameterization $varphi:U_0to Usubset M$ such that $varphi(a)=p$ as
$$
T_pM= Dvarphi(a)(mathbb{R}^m).
$$
By well defined I mean $ Dvarphi(a)(mathbb{R}^m)=Dpsi(b)(mathbb{R}^m)$ for any other parameterization $psi:V_0to Vsubset M$ such that $psi(b)=p$.
Step four. A function $ f: M to mathbb{R}$ is $C^r$ ($1leq r<k$) differentiable at point $p$ if there is a parameterization $varphi:U_0to Usubset M$ with $varphi(a)=p$ such that $fcircvarphi$ is $C^r$ differentiable in $a$. Once the parameter change $(varphi^{-1}circpsi):psi^{-1}(Ucap V)to varphi^{-1}(Ucap V)$ is differentiable it follows that the application $fcircpsi$ is differentiable in $b$ for all parameterization $psi:V_0to Vsubset M$ such that $psi(b)=p$.
Step five If $ f:Mto mathbb{R} $ is $C^r$ $(r>1)$ differentiable at point $ p in M$ then its derivative at that point is the linear transformation $ Df(p):T_pMtomathbb{R} $ defined as follows. Let's take a parameterization $varphi:U_0to Usubset M$ with $varphi(a)=p$. Given a vector $uin T_p M $ there exists unique a vector $ mu in mathbb{R}^m$ such that $u=Df(a)mu$.The derivative of $ f $ at point $ p $ is then simply defined by
$$
Df(p)cdot u =D (fcirc varphi)(a)cdotmu
$$
Step six. The linear transformation of step five is well defined. That is, if $psi:V_0to V$ is any other parameterization with $psi(b)=p$ and $u=Dpsi(b)zeta$ for some vector $zetainmathbb{R}^m$, then
$$
D (fcirc varphi)(a)cdotmu= D (fcirc psi)(b)cdotzeta.
$$
Indeed, $psi=varphicirc(varphi^{-1}circpsi)$ at where $(varphi^{-1}circpsi):psi^{-1}(Ucap V)to varphi^{-1}(Ucap V)$ is a $C^{k}$ diffeomorphism such that $(varphi^{-1}circpsi)(b)=a$. We have
begin{align}
Dvarphi(a)cdot mu=& u \
=& Dpsi(b)cdot zeta \
=& D((varphicirc varphi^{-1})circpsi )(b)cdot zeta\
=& D(varphicirc(varphi^{-1}circpsi))(b)cdot zeta\
=& Dvarphi(a)cdot D(varphi^{-1}circpsi)(b)cdotzeta
end{align}
Since $ D varphi(a)$ is injective we have $mu=D(varphi^{-1}circpsi)(b)cdotzeta$.
Therefore,
begin{align}
D(fcircpsi)(b)zeta =& D(fcirc(varphicircvarphi^{-1})circpsi)(b)cdotzeta \
=& D(fcirc varphicirc(varphi^{-1}circpsi))(b)cdotzeta \
=& D(fcirc varphi)(a) cdot (varphi^{-1}circpsi)(b)cdotzeta \
=& D(fcirc varphi)(a) cdotmu
end{align}
Let us end the background here and return to the question. Below is my attempt to answer the question.
By analogy to the first order derivative the second derivative would work as follows. If $ f:Mto mathbb{R} $ is $C^2$ differentiable at point $ p in M$ then its second derivative $D^{(2)}f$ each point $p$ associates the bilinear transformation $ D^{(2)}f(p):T_pMtimes T_pMtomathbb{R} $ defined as follows. Let's take a parameterization $varphi:U_0to Usubset M$ with $varphi(a)=p$. Given the vectors $u,vin T_p M $ there are a vectors $ mu,nu in mathbb{R}^m$ such that $u=Df(a)mu$ and $v=Df(a)nu$ .The second derivative of $ f $ at point $ p $ is then simply defined by
$$
D^{(2)}f(p)cdot (u,v) =D^{(2)} (fcirc varphi)(a)cdot(mu,nu)
$$
Similarly to the case of the first derivative, the second derivative will be well defined if for any other parameterization $psi:V_0to V$ such that $psi(b)=p$, $Dpsi(b)eta=u$ and $Dpsi(b)zeta=v$ for some $etainmathbb{R}^m$ and for some $zetainmathbb{R}^m$ we have
$$
D^{(2)} (fcirc varphi)(a)cdot(mu,nu)=D^{(2)} (fcirc psi)(b)cdot(eta,zeta)
$$
To show the above equality I tried to imitate step six. We have
begin{align}
D^{(2)} (fcirc psi)(b)cdot(eta,zeta)
=&
D^{(2)} (fcirc(varphicircvarphi^{-1})circ psi)(b)cdot(eta,zeta)
\
=&
D^{(2)} (fcircvarphicirc(varphi^{-1}circ psi))(b)cdot(eta,zeta)
end{align}
But in this last equality I can not use the chain rule.
On the other hand the development of the second derivative $D^{(2)} (fcirc varphi)(a)cdot(mu,nu)$ in terms of the partial derivatives $frac{partial^{2}f}{partial x_ipartial x_j}(a)$ was more productive. Look
begin{align}
D^{(2)} (fcirc varphi)(a)cdot(mu,nu)
=&
left[
frac{partial}{partial nu}
left(
frac{partial}{partial mu} (fcircvarphi)
right)
right](a)
\
=&
left[
frac{partial}{partial nu}
left(
sum_{j=1}^{m}mu_jcdot frac{partial}{partial x_j} (fcircvarphi)
right)
right](a)
\
=&
sum_{i=1}^msum_{j=1}^{m}
nu_icdotmu_jcdot frac{partial^2}{partial x_i partial x_j} (fcircvarphi)(a)
\
end{align}
By $frac{partial}{partial mu} (fcircvarphi)(a)= D(fcirc varphi)(a) cdotmu =u= D(fcircpsi)(b)eta= frac{partial}{partial eta} (fcircpsi)(b)$ we have
begin{align}
D^{(2)} (fcirc varphi)(a)cdot(mu,nu)
=&
left[
frac{partial}{partial nu}
left(
frac{partial}{partial mu} (fcircvarphi)
right)
right](a)
\
=&
left[
frac{partial}{partial nu}
left(
frac{partial}{partial eta} (fcircpsi)
right)
right](b)
\
=&
left[
frac{partial}{partial nu}
left(
sum_{j=1}^{m}eta_jcdot frac{partial}{partial x_j} (fcircpsi)
right)
right](b)
\
=&
sum_{i=1}^msum_{j=1}^{m}
nu_icdoteta_jcdot frac{partial^2}{partial x_i partial x_j} (fcircpsi)(b)
\
=&
D^{(2)}(fcircpsi)(a)cdot(eta,nu)
end{align}
By an entirely analogous calculation and applying the Schwarz theorem we have
$$
D^{(2)}(fcircvarphi)(a)cdot(mu,nu)= D^{(2)}(fcircpsi)(a)cdot(mu,zeta).
$$
The question then becomes the following. How to use the identities below to prove the well definiteness of $ D^{(2)}f$?
begin{align}
D^{(2)} (fcirc varphi)(a)cdot(mu,nu)=&D^{(2)}(fcircpsi)(a)cdot(eta,nu)
\
D^{(2)}(fcircvarphi)(a)cdot(mu,nu)=&D^{(2)}(fcircpsi)(a)cdot(mu,zeta).
end{align}
multivariable-calculus derivatives differential-geometry differential-topology
asked Jan 31 at 17:02
MathOverviewMathOverview
8,95243164
$endgroup$
How do I prove that the second derivative of a function $f:Mtomathbb{R}$ defined on a surface $Msubsetmathbb{R}^n$ is well defined?
If, as says the user Amitai Yuval, the second derivative of a function defined on a $ M $ surface can not be well defined in general what is the explanation for this impossibility?
By a second derivative I mean an application $D^{(2)}f$ that at each point $ p in M $ associates a well-defined bilinear aplication
$$
D^{(2)}f(p):T_pM times T_pM tomathbb{R}.
$$
Let me make a background in a few steps to try to make the question more precise.
Step one. Here $ M $ designates a $C^k$ surface of dimension $ m $ within the Euclidean space $ mathbb{R}^n$, i.e. a set in $mathbb{R}^n$ which satisfies the following properties: (1) there is a family of open sets ${O_i}_{iin I}$ such that $Msubset bigcup_{iin I} O_i$ and (2) for any $U=Mcap O_i$ there are a open set $U_0$ in $mathbb{R}^m$ and a $C^k$ ($kgeq 2$) parametrization $varphi:U_0to U$.
Step two. I'm assuming that, fixed a point $ p in M $, for any two parametrizations $varphi:U_0to Usubset M$ and $psi:V_0to Vsubset M$ such that $pin Vcap U$ it is proved that
$$
(varphi^{-1}circpsi):psi^{-1}(Ucap V)to varphi^{-1}(Ucap V)
quad mbox{ and } quad
( psi^{-1}circ varphi):varphi^{-1}(Ucap V)to psi^{-1}(Ucap V)
$$
are $C^k$ ($kgeq 2$) diffeormorphisms.
Step three. I am also assuming that the tangent plane $T_pM$ is the well defined vector space given by any parameterization $varphi:U_0to Usubset M$ such that $varphi(a)=p$ as
$$
T_pM= Dvarphi(a)(mathbb{R}^m).
$$
By well defined I mean $ Dvarphi(a)(mathbb{R}^m)=Dpsi(b)(mathbb{R}^m)$ for any other parameterization $psi:V_0to Vsubset M$ such that $psi(b)=p$.
Step four. A function $ f: M to mathbb{R}$ is $C^r$ ($1leq r<k$) differentiable at point $p$ if there is a parameterization $varphi:U_0to Usubset M$ with $varphi(a)=p$ such that $fcircvarphi$ is $C^r$ differentiable in $a$. Once the parameter change $(varphi^{-1}circpsi):psi^{-1}(Ucap V)to varphi^{-1}(Ucap V)$ is differentiable it follows that the application $fcircpsi$ is differentiable in $b$ for all parameterization $psi:V_0to Vsubset M$ such that $psi(b)=p$.
Step five If $ f:Mto mathbb{R} $ is $C^r$ $(r>1)$ differentiable at point $ p in M$ then its derivative at that point is the linear transformation $ Df(p):T_pMtomathbb{R} $ defined as follows. Let's take a parameterization $varphi:U_0to Usubset M$ with $varphi(a)=p$. Given a vector $uin T_p M $ there exists unique a vector $ mu in mathbb{R}^m$ such that $u=Df(a)mu$.The derivative of $ f $ at point $ p $ is then simply defined by
$$
Df(p)cdot u =D (fcirc varphi)(a)cdotmu
$$
Step six. The linear transformation of step five is well defined. That is, if $psi:V_0to V$ is any other parameterization with $psi(b)=p$ and $u=Dpsi(b)zeta$ for some vector $zetainmathbb{R}^m$, then
$$
D (fcirc varphi)(a)cdotmu= D (fcirc psi)(b)cdotzeta.
$$
Indeed, $psi=varphicirc(varphi^{-1}circpsi)$ at where $(varphi^{-1}circpsi):psi^{-1}(Ucap V)to varphi^{-1}(Ucap V)$ is a $C^{k}$ diffeomorphism such that $(varphi^{-1}circpsi)(b)=a$. We have
begin{align}
Dvarphi(a)cdot mu=& u \
=& Dpsi(b)cdot zeta \
=& D((varphicirc varphi^{-1})circpsi )(b)cdot zeta\
=& D(varphicirc(varphi^{-1}circpsi))(b)cdot zeta\
=& Dvarphi(a)cdot D(varphi^{-1}circpsi)(b)cdotzeta
end{align}
Since $ D varphi(a)$ is injective we have $mu=D(varphi^{-1}circpsi)(b)cdotzeta$.
Therefore,
begin{align}
D(fcircpsi)(b)zeta =& D(fcirc(varphicircvarphi^{-1})circpsi)(b)cdotzeta \
=& D(fcirc varphicirc(varphi^{-1}circpsi))(b)cdotzeta \
=& D(fcirc varphi)(a) cdot (varphi^{-1}circpsi)(b)cdotzeta \
=& D(fcirc varphi)(a) cdotmu
end{align}
Let us end the background here and return to the question. Below is my attempt to answer the question.
By analogy to the first order derivative the second derivative would work as follows. If $ f:Mto mathbb{R} $ is $C^2$ differentiable at point $ p in M$ then its second derivative $D^{(2)}f$ each point $p$ associates the bilinear transformation $ D^{(2)}f(p):T_pMtimes T_pMtomathbb{R} $ defined as follows. Let's take a parameterization $varphi:U_0to Usubset M$ with $varphi(a)=p$. Given the vectors $u,vin T_p M $ there are a vectors $ mu,nu in mathbb{R}^m$ such that $u=Df(a)mu$ and $v=Df(a)nu$ .The second derivative of $ f $ at point $ p $ is then simply defined by
$$
D^{(2)}f(p)cdot (u,v) =D^{(2)} (fcirc varphi)(a)cdot(mu,nu)
$$
Similarly to the case of the first derivative, the second derivative will be well defined if for any other parameterization $psi:V_0to V$ such that $psi(b)=p$, $Dpsi(b)eta=u$ and $Dpsi(b)zeta=v$ for some $etainmathbb{R}^m$ and for some $zetainmathbb{R}^m$ we have
$$
D^{(2)} (fcirc varphi)(a)cdot(mu,nu)=D^{(2)} (fcirc psi)(b)cdot(eta,zeta)
$$
To show the above equality I tried to imitate step six. We have
begin{align}
D^{(2)} (fcirc psi)(b)cdot(eta,zeta)
=&
D^{(2)} (fcirc(varphicircvarphi^{-1})circ psi)(b)cdot(eta,zeta)
\
=&
D^{(2)} (fcircvarphicirc(varphi^{-1}circ psi))(b)cdot(eta,zeta)
end{align}
But in this last equality I can not use the chain rule.
On the other hand the development of the second derivative $D^{(2)} (fcirc varphi)(a)cdot(mu,nu)$ in terms of the partial derivatives $frac{partial^{2}f}{partial x_ipartial x_j}(a)$ was more productive. Look
begin{align}
D^{(2)} (fcirc varphi)(a)cdot(mu,nu)
=&
left[
frac{partial}{partial nu}
left(
frac{partial}{partial mu} (fcircvarphi)
right)
right](a)
\
=&
left[
frac{partial}{partial nu}
left(
sum_{j=1}^{m}mu_jcdot frac{partial}{partial x_j} (fcircvarphi)
right)
right](a)
\
=&
sum_{i=1}^msum_{j=1}^{m}
nu_icdotmu_jcdot frac{partial^2}{partial x_i partial x_j} (fcircvarphi)(a)
\
end{align}
By $frac{partial}{partial mu} (fcircvarphi)(a)= D(fcirc varphi)(a) cdotmu =u= D(fcircpsi)(b)eta= frac{partial}{partial eta} (fcircpsi)(b)$ we have
begin{align}
D^{(2)} (fcirc varphi)(a)cdot(mu,nu)
=&
left[
frac{partial}{partial nu}
left(
frac{partial}{partial mu} (fcircvarphi)
right)
right](a)
\
=&
left[
frac{partial}{partial nu}
left(
frac{partial}{partial eta} (fcircpsi)
right)
right](b)
\
=&
left[
frac{partial}{partial nu}
left(
sum_{j=1}^{m}eta_jcdot frac{partial}{partial x_j} (fcircpsi)
right)
right](b)
\
=&
sum_{i=1}^msum_{j=1}^{m}
nu_icdoteta_jcdot frac{partial^2}{partial x_i partial x_j} (fcircpsi)(b)
\
=&
D^{(2)}(fcircpsi)(a)cdot(eta,nu)
end{align}
By an entirely analogous calculation and applying the Schwarz theorem we have
$$
D^{(2)}(fcircvarphi)(a)cdot(mu,nu)= D^{(2)}(fcircpsi)(a)cdot(mu,zeta).
$$
The question then becomes the following. How to use the identities below to prove the well definiteness of $ D^{(2)}f$?
begin{align}
D^{(2)} (fcirc varphi)(a)cdot(mu,nu)=&D^{(2)}(fcircpsi)(a)cdot(eta,nu)
\
D^{(2)}(fcircvarphi)(a)cdot(mu,nu)=&D^{(2)}(fcircpsi)(a)cdot(mu,zeta).
end{align}
multivariable-calculus derivatives differential-geometry differential-topology
multivariable-calculus derivatives differential-geometry differential-topology
asked Jan 31 at 17:02
MathOverviewMathOverview
8,95243164
asked Jan 31 at 17:02
MathOverviewMathOverview
8,95243164
edited Feb 1 at 11:48
MathOverview
asked Jan 31 at 17:02
MathOverviewMathOverview
8,95243164
asked Jan 31 at 17:02
MathOverviewMathOverview
8,95243164
asked Jan 31 at 17:02
MathOverviewMathOverview
8,95243164
8,95243164
1
$begingroup$
You are perfectly right - the second derivative of a function on $M$, as opposed to the first derivative, is not well-defined. It is, by the way, well-defined at critical points of the function in question. Furthermore, it may become well-defined everywhere once we equip $M$ with an affine connection. One way to do this is take the Levi-Civita connection of the Riemannian metric on $M$ induced by the embedding in $mathbb{R}^n$. One can also equip $M$ with many other affine connections.
$endgroup$
– Amitai Yuval
Jan 31 at 18:01
$begingroup$
@AmitaiYuval Excuse me but I have no experience with Riemannian geometry. Could you explain me using the background of the question?
$endgroup$
– MathOverview
Jan 31 at 18:29
$begingroup$
The second derivatives are not well-defined, exactly in the sense you work with. That is if you compute them by means of a coordinate chart just like in $mathbb{R}^m$, the result you get depends on the choice of coordinates. The geometric construction (or data) which enables one to define reasonable high-order derivatives is called an affine connection. You should grab a basic textbook in differential geometry and read about this.
$endgroup$
– Amitai Yuval
Jan 31 at 18:53
$begingroup$
@AmitaiYuval Is there an example of a function defined on a surface that illustrates the impossibility of defining its second derivative? If so, could you put it here?
$endgroup$
– MathOverview
Jan 31 at 19:05
add a comment |
1
$begingroup$
You are perfectly right - the second derivative of a function on $M$, as opposed to the first derivative, is not well-defined. It is, by the way, well-defined at critical points of the function in question. Furthermore, it may become well-defined everywhere once we equip $M$ with an affine connection. One way to do this is take the Levi-Civita connection of the Riemannian metric on $M$ induced by the embedding in $mathbb{R}^n$. One can also equip $M$ with many other affine connections.
$endgroup$
– Amitai Yuval
Jan 31 at 18:01
$begingroup$
@AmitaiYuval Excuse me but I have no experience with Riemannian geometry. Could you explain me using the background of the question?
$endgroup$
– MathOverview
Jan 31 at 18:29
$begingroup$
The second derivatives are not well-defined, exactly in the sense you work with. That is if you compute them by means of a coordinate chart just like in $mathbb{R}^m$, the result you get depends on the choice of coordinates. The geometric construction (or data) which enables one to define reasonable high-order derivatives is called an affine connection. You should grab a basic textbook in differential geometry and read about this.
$endgroup$
– Amitai Yuval
Jan 31 at 18:53
$begingroup$
@AmitaiYuval Is there an example of a function defined on a surface that illustrates the impossibility of defining its second derivative? If so, could you put it here?
$endgroup$
– MathOverview
Jan 31 at 19:05
1
1
$begingroup$
You are perfectly right - the second derivative of a function on $M$, as opposed to the first derivative, is not well-defined. It is, by the way, well-defined at critical points of the function in question. Furthermore, it may become well-defined everywhere once we equip $M$ with an affine connection. One way to do this is take the Levi-Civita connection of the Riemannian metric on $M$ induced by the embedding in $mathbb{R}^n$. One can also equip $M$ with many other affine connections.
$endgroup$
– Amitai Yuval
Jan 31 at 18:01
$begingroup$
You are perfectly right - the second derivative of a function on $M$, as opposed to the first derivative, is not well-defined. It is, by the way, well-defined at critical points of the function in question. Furthermore, it may become well-defined everywhere once we equip $M$ with an affine connection. One way to do this is take the Levi-Civita connection of the Riemannian metric on $M$ induced by the embedding in $mathbb{R}^n$. One can also equip $M$ with many other affine connections.
$endgroup$
– Amitai Yuval
Jan 31 at 18:01
$begingroup$
@AmitaiYuval Excuse me but I have no experience with Riemannian geometry. Could you explain me using the background of the question?
$endgroup$
– MathOverview
Jan 31 at 18:29
$begingroup$
@AmitaiYuval Excuse me but I have no experience with Riemannian geometry. Could you explain me using the background of the question?
$endgroup$
– MathOverview
Jan 31 at 18:29
$begingroup$
The second derivatives are not well-defined, exactly in the sense you work with. That is if you compute them by means of a coordinate chart just like in $mathbb{R}^m$, the result you get depends on the choice of coordinates. The geometric construction (or data) which enables one to define reasonable high-order derivatives is called an affine connection. You should grab a basic textbook in differential geometry and read about this.
$endgroup$
– Amitai Yuval
Jan 31 at 18:53
$begingroup$
The second derivatives are not well-defined, exactly in the sense you work with. That is if you compute them by means of a coordinate chart just like in $mathbb{R}^m$, the result you get depends on the choice of coordinates. The geometric construction (or data) which enables one to define reasonable high-order derivatives is called an affine connection. You should grab a basic textbook in differential geometry and read about this.
$endgroup$
– Amitai Yuval
Jan 31 at 18:53
$begingroup$
@AmitaiYuval Is there an example of a function defined on a surface that illustrates the impossibility of defining its second derivative? If so, could you put it here?
$endgroup$
– MathOverview
Jan 31 at 19:05
$begingroup$
@AmitaiYuval Is there an example of a function defined on a surface that illustrates the impossibility of defining its second derivative? If so, could you put it here?
$endgroup$
– MathOverview
Jan 31 at 19:05
add a comment |
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1
$begingroup$
You are perfectly right - the second derivative of a function on $M$, as opposed to the first derivative, is not well-defined. It is, by the way, well-defined at critical points of the function in question. Furthermore, it may become well-defined everywhere once we equip $M$ with an affine connection. One way to do this is take the Levi-Civita connection of the Riemannian metric on $M$ induced by the embedding in $mathbb{R}^n$. One can also equip $M$ with many other affine connections.
$endgroup$
– Amitai Yuval
Jan 31 at 18:01
$begingroup$
@AmitaiYuval Excuse me but I have no experience with Riemannian geometry. Could you explain me using the background of the question?
$endgroup$
– MathOverview
Jan 31 at 18:29
$begingroup$
The second derivatives are not well-defined, exactly in the sense you work with. That is if you compute them by means of a coordinate chart just like in $mathbb{R}^m$, the result you get depends on the choice of coordinates. The geometric construction (or data) which enables one to define reasonable high-order derivatives is called an affine connection. You should grab a basic textbook in differential geometry and read about this.
$endgroup$
– Amitai Yuval
Jan 31 at 18:53
$begingroup$
@AmitaiYuval Is there an example of a function defined on a surface that illustrates the impossibility of defining its second derivative? If so, could you put it here?
$endgroup$
– MathOverview
Jan 31 at 19:05