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How is Hessian tensor on Riemannian manifold related to the Hessian matrix from calculus?
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From calculus, given a smooth function $f: mathbb{R}^{n} rightarrow mathbb{R}$, the Hessian matrix is the $ntimes n$ matrix of second partials:
$${rm Hess}(f)= bigg( frac{partial^{2}f}{partial x_{i} partial x_{j}}bigg)$$
On a Riemannian manifold, one way to define the Hessian tensor of a smooth function $f:M rightarrow mathbb{R}$ is by
$${rm Hess}(f)(X,Y)=X(Yf)-df(nabla_{X}Y)$$
where $X$ and $Y$ are smooth vector fields on $M$.
I would like to know what relationship the Hessian on a Riemannian manifold has with the Hessian matrix of a function on $mathbb{R}^{n}$.
differential-geometry riemannian-geometry
edited Jul 11 '18 at 21:18
Ivo Terek
46.7k954146
asked Jul 11 '18 at 20:50
TuoTuoTuoTuo
1,776516
$endgroup$
add a comment |
$begingroup$
From calculus, given a smooth function $f: mathbb{R}^{n} rightarrow mathbb{R}$, the Hessian matrix is the $ntimes n$ matrix of second partials:
$${rm Hess}(f)= bigg( frac{partial^{2}f}{partial x_{i} partial x_{j}}bigg)$$
On a Riemannian manifold, one way to define the Hessian tensor of a smooth function $f:M rightarrow mathbb{R}$ is by
$${rm Hess}(f)(X,Y)=X(Yf)-df(nabla_{X}Y)$$
where $X$ and $Y$ are smooth vector fields on $M$.
I would like to know what relationship the Hessian on a Riemannian manifold has with the Hessian matrix of a function on $mathbb{R}^{n}$.
differential-geometry riemannian-geometry
edited Jul 11 '18 at 21:18
Ivo Terek
46.7k954146
asked Jul 11 '18 at 20:50
TuoTuoTuoTuo
1,776516
$endgroup$
add a comment |
$begingroup$
From calculus, given a smooth function $f: mathbb{R}^{n} rightarrow mathbb{R}$, the Hessian matrix is the $ntimes n$ matrix of second partials:
$${rm Hess}(f)= bigg( frac{partial^{2}f}{partial x_{i} partial x_{j}}bigg)$$
On a Riemannian manifold, one way to define the Hessian tensor of a smooth function $f:M rightarrow mathbb{R}$ is by
$${rm Hess}(f)(X,Y)=X(Yf)-df(nabla_{X}Y)$$
where $X$ and $Y$ are smooth vector fields on $M$.
I would like to know what relationship the Hessian on a Riemannian manifold has with the Hessian matrix of a function on $mathbb{R}^{n}$.
differential-geometry riemannian-geometry
edited Jul 11 '18 at 21:18
Ivo Terek
46.7k954146
asked Jul 11 '18 at 20:50
TuoTuoTuoTuo
1,776516
$endgroup$
From calculus, given a smooth function $f: mathbb{R}^{n} rightarrow mathbb{R}$, the Hessian matrix is the $ntimes n$ matrix of second partials:
$${rm Hess}(f)= bigg( frac{partial^{2}f}{partial x_{i} partial x_{j}}bigg)$$
On a Riemannian manifold, one way to define the Hessian tensor of a smooth function $f:M rightarrow mathbb{R}$ is by
$${rm Hess}(f)(X,Y)=X(Yf)-df(nabla_{X}Y)$$
where $X$ and $Y$ are smooth vector fields on $M$.
I would like to know what relationship the Hessian on a Riemannian manifold has with the Hessian matrix of a function on $mathbb{R}^{n}$.
differential-geometry riemannian-geometry
differential-geometry riemannian-geometry
edited Jul 11 '18 at 21:18
Ivo Terek
46.7k954146
asked Jul 11 '18 at 20:50
TuoTuoTuoTuo
1,776516
edited Jul 11 '18 at 21:18
Ivo Terek
46.7k954146
asked Jul 11 '18 at 20:50
TuoTuoTuoTuo
1,776516
edited Jul 11 '18 at 21:18
Ivo Terek
46.7k954146
edited Jul 11 '18 at 21:18
Ivo Terek
46.7k954146
edited Jul 11 '18 at 21:18
Ivo Terek
46.7k954146
46.7k954146
asked Jul 11 '18 at 20:50
TuoTuoTuoTuo
1,776516
asked Jul 11 '18 at 20:50
TuoTuoTuoTuo
1,776516
asked Jul 11 '18 at 20:50
TuoTuoTuoTuo
1,776516
1,776516
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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The relation between the Riemannian Hessian and the Hessian in Euclidean space is very simple - they are the same. More precisely, the Euclidean Hessian is a particular case of the more general Riemannian Hessian.
Let $(x_1,ldots,x_n)$ be local coordinates on a neighborhood in a Riemannian manifold $M$, and let $Gamma_{ij}^k$ denote the Christoffel symbols of the Levi-Civita connection with respect to these coordinates. Let us massage your definition for the Hessian; we consider the vector fields $$X=X^ifrac{partial}{partial x_i},;Y=Y^ifrac{partial}{partial x_i}.$$ Then $$begin{align}X(Y(f))-df(nabla_XY)&=X^ifrac{partial}{partial x_i}left(Y^jfrac{partial f}{partial x_j}right)-dfleft(nabla_{X^ifrac{partial}{partial x_i}}Y^jfrac{partial}{partial x_j}right)\&=X^ileft(frac{partial Y^j}{partial x_i}frac{partial f}{partial x_j}+Y^jfrac{partial^2f}{partial x_ipartial x_j}right)-dfleft(X^ileft(frac{partial Y^j}{partial x_i}frac{partial}{partial x_j}+Y^jGamma_{ij}^kfrac{partial}{partial x_k}right)right)\&=X^iY^jfrac{partial^2f}{partial x_ipartial x_j}-X^iY^jGamma_{ij}^kfrac{partial f}{partial x_k}.end{align}$$Now, the Christoffel symbols of the Levi-Civita connection with respect to the usual coordinates on $mathbb{R}^n$ are all zero. Hence, the Euclidean Hessian matrix of the function $f$ is just the matrix whose $(ij)$ entry is $$mathrm{Hess}(f)_{ij}=mathrm{Hess}(f)left(frac{partial}{partial x_i},frac{partial}{partial x_j}right),$$where the right hand side is the Riemannian Hessian.
answered Jul 12 '18 at 11:41
Amitai YuvalAmitai Yuval
15.6k11127
$endgroup$
add a comment |
$begingroup$
That Hessian matrix is useful for analyzing the behaviour of critical points of $f$, where $df_p = 0$. More precisely, let $M$ be a smooth manifold and $nabla$ be a linear connection in $TM$. If $fcolon M to Bbb R$ is a smooth map, $p in M$ is a critical point of $f$, and $(x^i)_{i=1}^n$ is a system of coordinates around $p$, we have the local expression $${rm Hess},f_p(partial_ibig|_p,partial_jbig|_p) = partial_ibig|_p(partial_j f) - require{cancel} cancelto{0}{df_p(nabla_{partial_i}partial_j)} = frac{partial^2f}{partial x^ipartial x^j}(p),$$and so $${rm Hess},f_p = sum_{i,j=1}^n frac{partial^2f}{partial x^ipartial x^j}(p) ,dx^ibig|_potimes dx^jbig|_p$$
At a generic point $p$, the bilinear map ${rm Hess},f_p$ depends on the choice of connection (because of the $nabla$ in the second term). But if $p$ is a critical point, then the map ${rm Hess}, f_p$ becomes independent of the choice of connection. If $p$ is not critical, one usually picks $nabla$ to be the Levi-Civita connection of some Riemannian metric in $M$. But a priori, you can use any connection (which can lead to some awkward things, such as the Hessian not being symmetric if $nabla$ has torsion).
edited Jan 31 at 4:51
tparker
1,931834
answered Jul 11 '18 at 21:16
Ivo TerekIvo Terek
46.7k954146
$endgroup$
$begingroup$
1) I'm saying that if ${rm Hess}^nabla(f)(X,Y) doteq X(Y(f)) - {rm d}f(nabla_XY)$, $nabla'$ is another connection, and $p$ is a critical point of $f$, then ${rm Hess}^nabla(f)_p = {rm Hess}^{nabla'}(f)_p$. 2) I just meant that a priori there is no preferred connection to choose, if you don't have extra structure in your manifold (e.g., a Riemannian metric).
$endgroup$
– Ivo Terek
Jan 31 at 4:38
$begingroup$
Sure, no problem.
$endgroup$
– Ivo Terek
Jan 31 at 4:41
1
$begingroup$
Thanks, I suggested an edit which is now in the edit queue.
$endgroup$
– tparker
Jan 31 at 4:49
$begingroup$
Thank you, your edit was very good, I have already approved it :-)
$endgroup$
– Ivo Terek
Jan 31 at 4:52
add a comment |
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2 Answers
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2 Answers
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$begingroup$
The relation between the Riemannian Hessian and the Hessian in Euclidean space is very simple - they are the same. More precisely, the Euclidean Hessian is a particular case of the more general Riemannian Hessian.
Let $(x_1,ldots,x_n)$ be local coordinates on a neighborhood in a Riemannian manifold $M$, and let $Gamma_{ij}^k$ denote the Christoffel symbols of the Levi-Civita connection with respect to these coordinates. Let us massage your definition for the Hessian; we consider the vector fields $$X=X^ifrac{partial}{partial x_i},;Y=Y^ifrac{partial}{partial x_i}.$$ Then $$begin{align}X(Y(f))-df(nabla_XY)&=X^ifrac{partial}{partial x_i}left(Y^jfrac{partial f}{partial x_j}right)-dfleft(nabla_{X^ifrac{partial}{partial x_i}}Y^jfrac{partial}{partial x_j}right)\&=X^ileft(frac{partial Y^j}{partial x_i}frac{partial f}{partial x_j}+Y^jfrac{partial^2f}{partial x_ipartial x_j}right)-dfleft(X^ileft(frac{partial Y^j}{partial x_i}frac{partial}{partial x_j}+Y^jGamma_{ij}^kfrac{partial}{partial x_k}right)right)\&=X^iY^jfrac{partial^2f}{partial x_ipartial x_j}-X^iY^jGamma_{ij}^kfrac{partial f}{partial x_k}.end{align}$$Now, the Christoffel symbols of the Levi-Civita connection with respect to the usual coordinates on $mathbb{R}^n$ are all zero. Hence, the Euclidean Hessian matrix of the function $f$ is just the matrix whose $(ij)$ entry is $$mathrm{Hess}(f)_{ij}=mathrm{Hess}(f)left(frac{partial}{partial x_i},frac{partial}{partial x_j}right),$$where the right hand side is the Riemannian Hessian.
answered Jul 12 '18 at 11:41
Amitai YuvalAmitai Yuval
15.6k11127
$endgroup$
add a comment |
$begingroup$
The relation between the Riemannian Hessian and the Hessian in Euclidean space is very simple - they are the same. More precisely, the Euclidean Hessian is a particular case of the more general Riemannian Hessian.
Let $(x_1,ldots,x_n)$ be local coordinates on a neighborhood in a Riemannian manifold $M$, and let $Gamma_{ij}^k$ denote the Christoffel symbols of the Levi-Civita connection with respect to these coordinates. Let us massage your definition for the Hessian; we consider the vector fields $$X=X^ifrac{partial}{partial x_i},;Y=Y^ifrac{partial}{partial x_i}.$$ Then $$begin{align}X(Y(f))-df(nabla_XY)&=X^ifrac{partial}{partial x_i}left(Y^jfrac{partial f}{partial x_j}right)-dfleft(nabla_{X^ifrac{partial}{partial x_i}}Y^jfrac{partial}{partial x_j}right)\&=X^ileft(frac{partial Y^j}{partial x_i}frac{partial f}{partial x_j}+Y^jfrac{partial^2f}{partial x_ipartial x_j}right)-dfleft(X^ileft(frac{partial Y^j}{partial x_i}frac{partial}{partial x_j}+Y^jGamma_{ij}^kfrac{partial}{partial x_k}right)right)\&=X^iY^jfrac{partial^2f}{partial x_ipartial x_j}-X^iY^jGamma_{ij}^kfrac{partial f}{partial x_k}.end{align}$$Now, the Christoffel symbols of the Levi-Civita connection with respect to the usual coordinates on $mathbb{R}^n$ are all zero. Hence, the Euclidean Hessian matrix of the function $f$ is just the matrix whose $(ij)$ entry is $$mathrm{Hess}(f)_{ij}=mathrm{Hess}(f)left(frac{partial}{partial x_i},frac{partial}{partial x_j}right),$$where the right hand side is the Riemannian Hessian.
answered Jul 12 '18 at 11:41
Amitai YuvalAmitai Yuval
15.6k11127
$endgroup$
add a comment |
$begingroup$
The relation between the Riemannian Hessian and the Hessian in Euclidean space is very simple - they are the same. More precisely, the Euclidean Hessian is a particular case of the more general Riemannian Hessian.
Let $(x_1,ldots,x_n)$ be local coordinates on a neighborhood in a Riemannian manifold $M$, and let $Gamma_{ij}^k$ denote the Christoffel symbols of the Levi-Civita connection with respect to these coordinates. Let us massage your definition for the Hessian; we consider the vector fields $$X=X^ifrac{partial}{partial x_i},;Y=Y^ifrac{partial}{partial x_i}.$$ Then $$begin{align}X(Y(f))-df(nabla_XY)&=X^ifrac{partial}{partial x_i}left(Y^jfrac{partial f}{partial x_j}right)-dfleft(nabla_{X^ifrac{partial}{partial x_i}}Y^jfrac{partial}{partial x_j}right)\&=X^ileft(frac{partial Y^j}{partial x_i}frac{partial f}{partial x_j}+Y^jfrac{partial^2f}{partial x_ipartial x_j}right)-dfleft(X^ileft(frac{partial Y^j}{partial x_i}frac{partial}{partial x_j}+Y^jGamma_{ij}^kfrac{partial}{partial x_k}right)right)\&=X^iY^jfrac{partial^2f}{partial x_ipartial x_j}-X^iY^jGamma_{ij}^kfrac{partial f}{partial x_k}.end{align}$$Now, the Christoffel symbols of the Levi-Civita connection with respect to the usual coordinates on $mathbb{R}^n$ are all zero. Hence, the Euclidean Hessian matrix of the function $f$ is just the matrix whose $(ij)$ entry is $$mathrm{Hess}(f)_{ij}=mathrm{Hess}(f)left(frac{partial}{partial x_i},frac{partial}{partial x_j}right),$$where the right hand side is the Riemannian Hessian.
answered Jul 12 '18 at 11:41
Amitai YuvalAmitai Yuval
15.6k11127
$endgroup$
The relation between the Riemannian Hessian and the Hessian in Euclidean space is very simple - they are the same. More precisely, the Euclidean Hessian is a particular case of the more general Riemannian Hessian.
Let $(x_1,ldots,x_n)$ be local coordinates on a neighborhood in a Riemannian manifold $M$, and let $Gamma_{ij}^k$ denote the Christoffel symbols of the Levi-Civita connection with respect to these coordinates. Let us massage your definition for the Hessian; we consider the vector fields $$X=X^ifrac{partial}{partial x_i},;Y=Y^ifrac{partial}{partial x_i}.$$ Then $$begin{align}X(Y(f))-df(nabla_XY)&=X^ifrac{partial}{partial x_i}left(Y^jfrac{partial f}{partial x_j}right)-dfleft(nabla_{X^ifrac{partial}{partial x_i}}Y^jfrac{partial}{partial x_j}right)\&=X^ileft(frac{partial Y^j}{partial x_i}frac{partial f}{partial x_j}+Y^jfrac{partial^2f}{partial x_ipartial x_j}right)-dfleft(X^ileft(frac{partial Y^j}{partial x_i}frac{partial}{partial x_j}+Y^jGamma_{ij}^kfrac{partial}{partial x_k}right)right)\&=X^iY^jfrac{partial^2f}{partial x_ipartial x_j}-X^iY^jGamma_{ij}^kfrac{partial f}{partial x_k}.end{align}$$Now, the Christoffel symbols of the Levi-Civita connection with respect to the usual coordinates on $mathbb{R}^n$ are all zero. Hence, the Euclidean Hessian matrix of the function $f$ is just the matrix whose $(ij)$ entry is $$mathrm{Hess}(f)_{ij}=mathrm{Hess}(f)left(frac{partial}{partial x_i},frac{partial}{partial x_j}right),$$where the right hand side is the Riemannian Hessian.
answered Jul 12 '18 at 11:41
Amitai YuvalAmitai Yuval
15.6k11127
answered Jul 12 '18 at 11:41
Amitai YuvalAmitai Yuval
15.6k11127
answered Jul 12 '18 at 11:41
Amitai YuvalAmitai Yuval
15.6k11127
answered Jul 12 '18 at 11:41
Amitai YuvalAmitai Yuval
15.6k11127
15.6k11127
add a comment |
add a comment |
$begingroup$
That Hessian matrix is useful for analyzing the behaviour of critical points of $f$, where $df_p = 0$. More precisely, let $M$ be a smooth manifold and $nabla$ be a linear connection in $TM$. If $fcolon M to Bbb R$ is a smooth map, $p in M$ is a critical point of $f$, and $(x^i)_{i=1}^n$ is a system of coordinates around $p$, we have the local expression $${rm Hess},f_p(partial_ibig|_p,partial_jbig|_p) = partial_ibig|_p(partial_j f) - require{cancel} cancelto{0}{df_p(nabla_{partial_i}partial_j)} = frac{partial^2f}{partial x^ipartial x^j}(p),$$and so $${rm Hess},f_p = sum_{i,j=1}^n frac{partial^2f}{partial x^ipartial x^j}(p) ,dx^ibig|_potimes dx^jbig|_p$$
At a generic point $p$, the bilinear map ${rm Hess},f_p$ depends on the choice of connection (because of the $nabla$ in the second term). But if $p$ is a critical point, then the map ${rm Hess}, f_p$ becomes independent of the choice of connection. If $p$ is not critical, one usually picks $nabla$ to be the Levi-Civita connection of some Riemannian metric in $M$. But a priori, you can use any connection (which can lead to some awkward things, such as the Hessian not being symmetric if $nabla$ has torsion).
edited Jan 31 at 4:51
tparker
1,931834
answered Jul 11 '18 at 21:16
Ivo TerekIvo Terek
46.7k954146
$endgroup$
$begingroup$
1) I'm saying that if ${rm Hess}^nabla(f)(X,Y) doteq X(Y(f)) - {rm d}f(nabla_XY)$, $nabla'$ is another connection, and $p$ is a critical point of $f$, then ${rm Hess}^nabla(f)_p = {rm Hess}^{nabla'}(f)_p$. 2) I just meant that a priori there is no preferred connection to choose, if you don't have extra structure in your manifold (e.g., a Riemannian metric).
$endgroup$
– Ivo Terek
Jan 31 at 4:38
$begingroup$
Sure, no problem.
$endgroup$
– Ivo Terek
Jan 31 at 4:41
1
$begingroup$
Thanks, I suggested an edit which is now in the edit queue.
$endgroup$
– tparker
Jan 31 at 4:49
$begingroup$
Thank you, your edit was very good, I have already approved it :-)
$endgroup$
– Ivo Terek
Jan 31 at 4:52
add a comment |
$begingroup$
That Hessian matrix is useful for analyzing the behaviour of critical points of $f$, where $df_p = 0$. More precisely, let $M$ be a smooth manifold and $nabla$ be a linear connection in $TM$. If $fcolon M to Bbb R$ is a smooth map, $p in M$ is a critical point of $f$, and $(x^i)_{i=1}^n$ is a system of coordinates around $p$, we have the local expression $${rm Hess},f_p(partial_ibig|_p,partial_jbig|_p) = partial_ibig|_p(partial_j f) - require{cancel} cancelto{0}{df_p(nabla_{partial_i}partial_j)} = frac{partial^2f}{partial x^ipartial x^j}(p),$$and so $${rm Hess},f_p = sum_{i,j=1}^n frac{partial^2f}{partial x^ipartial x^j}(p) ,dx^ibig|_potimes dx^jbig|_p$$
At a generic point $p$, the bilinear map ${rm Hess},f_p$ depends on the choice of connection (because of the $nabla$ in the second term). But if $p$ is a critical point, then the map ${rm Hess}, f_p$ becomes independent of the choice of connection. If $p$ is not critical, one usually picks $nabla$ to be the Levi-Civita connection of some Riemannian metric in $M$. But a priori, you can use any connection (which can lead to some awkward things, such as the Hessian not being symmetric if $nabla$ has torsion).
edited Jan 31 at 4:51
tparker
1,931834
answered Jul 11 '18 at 21:16
Ivo TerekIvo Terek
46.7k954146
$endgroup$
$begingroup$
1) I'm saying that if ${rm Hess}^nabla(f)(X,Y) doteq X(Y(f)) - {rm d}f(nabla_XY)$, $nabla'$ is another connection, and $p$ is a critical point of $f$, then ${rm Hess}^nabla(f)_p = {rm Hess}^{nabla'}(f)_p$. 2) I just meant that a priori there is no preferred connection to choose, if you don't have extra structure in your manifold (e.g., a Riemannian metric).
$endgroup$
– Ivo Terek
Jan 31 at 4:38
$begingroup$
Sure, no problem.
$endgroup$
– Ivo Terek
Jan 31 at 4:41
1
$begingroup$
Thanks, I suggested an edit which is now in the edit queue.
$endgroup$
– tparker
Jan 31 at 4:49
$begingroup$
Thank you, your edit was very good, I have already approved it :-)
$endgroup$
– Ivo Terek
Jan 31 at 4:52
add a comment |
$begingroup$
That Hessian matrix is useful for analyzing the behaviour of critical points of $f$, where $df_p = 0$. More precisely, let $M$ be a smooth manifold and $nabla$ be a linear connection in $TM$. If $fcolon M to Bbb R$ is a smooth map, $p in M$ is a critical point of $f$, and $(x^i)_{i=1}^n$ is a system of coordinates around $p$, we have the local expression $${rm Hess},f_p(partial_ibig|_p,partial_jbig|_p) = partial_ibig|_p(partial_j f) - require{cancel} cancelto{0}{df_p(nabla_{partial_i}partial_j)} = frac{partial^2f}{partial x^ipartial x^j}(p),$$and so $${rm Hess},f_p = sum_{i,j=1}^n frac{partial^2f}{partial x^ipartial x^j}(p) ,dx^ibig|_potimes dx^jbig|_p$$
At a generic point $p$, the bilinear map ${rm Hess},f_p$ depends on the choice of connection (because of the $nabla$ in the second term). But if $p$ is a critical point, then the map ${rm Hess}, f_p$ becomes independent of the choice of connection. If $p$ is not critical, one usually picks $nabla$ to be the Levi-Civita connection of some Riemannian metric in $M$. But a priori, you can use any connection (which can lead to some awkward things, such as the Hessian not being symmetric if $nabla$ has torsion).
edited Jan 31 at 4:51
tparker
1,931834
answered Jul 11 '18 at 21:16
Ivo TerekIvo Terek
46.7k954146
$endgroup$
That Hessian matrix is useful for analyzing the behaviour of critical points of $f$, where $df_p = 0$. More precisely, let $M$ be a smooth manifold and $nabla$ be a linear connection in $TM$. If $fcolon M to Bbb R$ is a smooth map, $p in M$ is a critical point of $f$, and $(x^i)_{i=1}^n$ is a system of coordinates around $p$, we have the local expression $${rm Hess},f_p(partial_ibig|_p,partial_jbig|_p) = partial_ibig|_p(partial_j f) - require{cancel} cancelto{0}{df_p(nabla_{partial_i}partial_j)} = frac{partial^2f}{partial x^ipartial x^j}(p),$$and so $${rm Hess},f_p = sum_{i,j=1}^n frac{partial^2f}{partial x^ipartial x^j}(p) ,dx^ibig|_potimes dx^jbig|_p$$
At a generic point $p$, the bilinear map ${rm Hess},f_p$ depends on the choice of connection (because of the $nabla$ in the second term). But if $p$ is a critical point, then the map ${rm Hess}, f_p$ becomes independent of the choice of connection. If $p$ is not critical, one usually picks $nabla$ to be the Levi-Civita connection of some Riemannian metric in $M$. But a priori, you can use any connection (which can lead to some awkward things, such as the Hessian not being symmetric if $nabla$ has torsion).
edited Jan 31 at 4:51
tparker
1,931834
answered Jul 11 '18 at 21:16
Ivo TerekIvo Terek
46.7k954146
edited Jan 31 at 4:51
tparker
1,931834
edited Jan 31 at 4:51
tparker
1,931834
edited Jan 31 at 4:51
tparker
1,931834
1,931834
answered Jul 11 '18 at 21:16
Ivo TerekIvo Terek
46.7k954146
answered Jul 11 '18 at 21:16
Ivo TerekIvo Terek
46.7k954146
answered Jul 11 '18 at 21:16
Ivo TerekIvo Terek
46.7k954146
46.7k954146
$begingroup$
1) I'm saying that if ${rm Hess}^nabla(f)(X,Y) doteq X(Y(f)) - {rm d}f(nabla_XY)$, $nabla'$ is another connection, and $p$ is a critical point of $f$, then ${rm Hess}^nabla(f)_p = {rm Hess}^{nabla'}(f)_p$. 2) I just meant that a priori there is no preferred connection to choose, if you don't have extra structure in your manifold (e.g., a Riemannian metric).
$endgroup$
– Ivo Terek
Jan 31 at 4:38
$begingroup$
Sure, no problem.
$endgroup$
– Ivo Terek
Jan 31 at 4:41
1
$begingroup$
Thanks, I suggested an edit which is now in the edit queue.
$endgroup$
– tparker
Jan 31 at 4:49
$begingroup$
Thank you, your edit was very good, I have already approved it :-)
$endgroup$
– Ivo Terek
Jan 31 at 4:52
add a comment |
$begingroup$
1) I'm saying that if ${rm Hess}^nabla(f)(X,Y) doteq X(Y(f)) - {rm d}f(nabla_XY)$, $nabla'$ is another connection, and $p$ is a critical point of $f$, then ${rm Hess}^nabla(f)_p = {rm Hess}^{nabla'}(f)_p$. 2) I just meant that a priori there is no preferred connection to choose, if you don't have extra structure in your manifold (e.g., a Riemannian metric).
$endgroup$
– Ivo Terek
Jan 31 at 4:38
$begingroup$
Sure, no problem.
$endgroup$
– Ivo Terek
Jan 31 at 4:41
1
$begingroup$
Thanks, I suggested an edit which is now in the edit queue.
$endgroup$
– tparker
Jan 31 at 4:49
$begingroup$
Thank you, your edit was very good, I have already approved it :-)
$endgroup$
– Ivo Terek
Jan 31 at 4:52
$begingroup$
1) I'm saying that if ${rm Hess}^nabla(f)(X,Y) doteq X(Y(f)) - {rm d}f(nabla_XY)$, $nabla'$ is another connection, and $p$ is a critical point of $f$, then ${rm Hess}^nabla(f)_p = {rm Hess}^{nabla'}(f)_p$. 2) I just meant that a priori there is no preferred connection to choose, if you don't have extra structure in your manifold (e.g., a Riemannian metric).
$endgroup$
– Ivo Terek
Jan 31 at 4:38
$begingroup$
1) I'm saying that if ${rm Hess}^nabla(f)(X,Y) doteq X(Y(f)) - {rm d}f(nabla_XY)$, $nabla'$ is another connection, and $p$ is a critical point of $f$, then ${rm Hess}^nabla(f)_p = {rm Hess}^{nabla'}(f)_p$. 2) I just meant that a priori there is no preferred connection to choose, if you don't have extra structure in your manifold (e.g., a Riemannian metric).
$endgroup$
– Ivo Terek
Jan 31 at 4:38
$begingroup$
Sure, no problem.
$endgroup$
– Ivo Terek
Jan 31 at 4:41
$begingroup$
Sure, no problem.
$endgroup$
– Ivo Terek
Jan 31 at 4:41
1
1
$begingroup$
Thanks, I suggested an edit which is now in the edit queue.
$endgroup$
– tparker
Jan 31 at 4:49
$begingroup$
Thanks, I suggested an edit which is now in the edit queue.
$endgroup$
– tparker
Jan 31 at 4:49
$begingroup$
Thank you, your edit was very good, I have already approved it :-)
$endgroup$
– Ivo Terek
Jan 31 at 4:52
$begingroup$
Thank you, your edit was very good, I have already approved it :-)
$endgroup$
– Ivo Terek
Jan 31 at 4:52
add a comment |
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