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How to define and compute the norm of a vector with riemannian metric?

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2

3

$begingroup$

Let us consider for example, the riemannian metric $g=e^xdx^2+dy^2$ (it is symmetric and definite positive), with associated matrix
$beginpmatrix
e^x & 0\
0 & 1
endpmatrix$.

Consider the vector $(1,1)$.

Then $||(1,1)||=d((1,1),(0,0))=int_0^1 sqrtg(gamma',gamma')dt$, where $gamma(t)=(t,t)$.

NB: Here I'm using the definition of length of a curve, and the curve is one that passes through $(0,0)$ and $(1,1)$. I do not know if the result depends on the chosen curve...

As $gamma'(t)=(1,1)$ we get $||(1,1)||=int_0^1sqrte^tcdot1^2+1^2dt=ldots$

Is there a fast way to do this? Thanks!

differential-geometry metric-spaces riemannian-geometry norm

share|cite|improve this question

asked Feb 10 '16 at 22:37

LeafarLeafar

8731924

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  read a course : vectors don't exist in riemannian manifolds. what exists are vectors $v$ of the tangent space at a given point $p$, but you cannot add the vector $v$ to the point $p$ to get another point $p'$ on the manifold. and the matrix you wrote, which represents the metric in some given coordinates, let you get the norm of such a vector $v$ of the tangent space at $(x,y)$ : $$||v||^2_(x,y) = v^T beginpmatrix e^x & 0\ 0 & 1 endpmatrix v$$
  $endgroup$
  – reuns
  Feb 10 '16 at 22:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great! Makes sense
  $endgroup$
  – Leafar
  Feb 10 '16 at 22:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You could write it as an answer
  $endgroup$
  – Leafar
  Feb 10 '16 at 22:52
  

  
  
  
  

add a comment | 

2

3

$begingroup$

Let us consider for example, the riemannian metric $g=e^xdx^2+dy^2$ (it is symmetric and definite positive), with associated matrix
$beginpmatrix
e^x & 0\
0 & 1
endpmatrix$.

Consider the vector $(1,1)$.

Then $||(1,1)||=d((1,1),(0,0))=int_0^1 sqrtg(gamma',gamma')dt$, where $gamma(t)=(t,t)$.

NB: Here I'm using the definition of length of a curve, and the curve is one that passes through $(0,0)$ and $(1,1)$. I do not know if the result depends on the chosen curve...

As $gamma'(t)=(1,1)$ we get $||(1,1)||=int_0^1sqrte^tcdot1^2+1^2dt=ldots$

Is there a fast way to do this? Thanks!

differential-geometry metric-spaces riemannian-geometry norm

share|cite|improve this question

asked Feb 10 '16 at 22:37

LeafarLeafar

8731924

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  read a course : vectors don't exist in riemannian manifolds. what exists are vectors $v$ of the tangent space at a given point $p$, but you cannot add the vector $v$ to the point $p$ to get another point $p'$ on the manifold. and the matrix you wrote, which represents the metric in some given coordinates, let you get the norm of such a vector $v$ of the tangent space at $(x,y)$ : $$||v||^2_(x,y) = v^T beginpmatrix e^x & 0\ 0 & 1 endpmatrix v$$
  $endgroup$
  – reuns
  Feb 10 '16 at 22:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great! Makes sense
  $endgroup$
  – Leafar
  Feb 10 '16 at 22:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You could write it as an answer
  $endgroup$
  – Leafar
  Feb 10 '16 at 22:52
  

  
  
  
  

add a comment | 

$begingroup$

Let us consider for example, the riemannian metric $g=e^xdx^2+dy^2$ (it is symmetric and definite positive), with associated matrix
$beginpmatrix
e^x & 0\
0 & 1
endpmatrix$.

Consider the vector $(1,1)$.

Then $||(1,1)||=d((1,1),(0,0))=int_0^1 sqrtg(gamma',gamma')dt$, where $gamma(t)=(t,t)$.

NB: Here I'm using the definition of length of a curve, and the curve is one that passes through $(0,0)$ and $(1,1)$. I do not know if the result depends on the chosen curve...

As $gamma'(t)=(1,1)$ we get $||(1,1)||=int_0^1sqrte^tcdot1^2+1^2dt=ldots$

Is there a fast way to do this? Thanks!

differential-geometry metric-spaces riemannian-geometry norm

share|cite|improve this question

asked Feb 10 '16 at 22:37

LeafarLeafar

8731924

$endgroup$

share|cite|improve this question

asked Feb 10 '16 at 22:37

LeafarLeafar

8731924

share|cite|improve this question

asked Feb 10 '16 at 22:37

LeafarLeafar

8731924

asked Feb 10 '16 at 22:37

LeafarLeafar

8731924

asked Feb 10 '16 at 22:37

LeafarLeafar

8731924

1 Answer
1

active

oldest

votes

0

$begingroup$

This is actually a good question and weirdly not mentioned explicitly very often.

A Riemannian manifold $(M,g)$ has a tangent space $T_pM$ defined for every point $pin M$, which is a vector space attached to each $p$. For instance, a 2D manifold in 3D has a tangent plane associated to every point of the surface (so $T_pM=mathbbR^2$). The Riemannian metric $g$, only defines distances and angles in the tangent space. So $g$ doesn't let you compute distances directly, it only lets you compute infinitesimal distances.

In particular, if you have two tangent vectors $v$ and $u$, then the "distance" between them is defined by $g$ via:
$$ d(u,v) = g_ijv^i u^j $$
or, in matrix terms, $ d(u,v)=u^Tgv $. (For you, $g=textdiag([e^x,1])$).

In Euclidean space, $g=I$ or $g_ij=delta_ij$, so this reduces to the standard inner product, the dot product. This is why people say that $g$, the metric tensor, defines an inner product on the manifold (it actually defines one on each $T_pM$).

Now to norms. For any metric $d(p_1,p_2)$ you can define a norm $||s||=d(s,s)$. Riemannian manifolds are no different. So we define:
$$ ||v||_g^2 = d(v,v) = g_ijv^iv^j = v^T g v $$
as our Riemannian norm.

Also, it turns out computing "infinitesimal" distances in $T_pM$ lets us define "actual" distances between points in $M$.
Let $gamma(t)$ be a curve on $M$.
To get the length of $gamma$, you just sum up the infinitesimal distances:
$$
ell_g(gamma;a,b) = int_a^b ||gamma'(t)||_g dt =int_a^b sqrtg_ij dotgamma(t)^idotgamma(t)^j dt
$$
Now you can define the distance between $p$ and $q$, where $p,qin M$, as the length of the shortest curve between them:
$$ D(p,q) = inf_gamma ell_g(gamma;p,q) $$

share|cite|improve this answer

edited Aug 13 '17 at 3:04

answered Aug 12 '17 at 23:59

user3658307user3658307

4,8283946

$endgroup$

add a comment | 

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0

$begingroup$

This is actually a good question and weirdly not mentioned explicitly very often.

A Riemannian manifold $(M,g)$ has a tangent space $T_pM$ defined for every point $pin M$, which is a vector space attached to each $p$. For instance, a 2D manifold in 3D has a tangent plane associated to every point of the surface (so $T_pM=mathbbR^2$). The Riemannian metric $g$, only defines distances and angles in the tangent space. So $g$ doesn't let you compute distances directly, it only lets you compute infinitesimal distances.

In particular, if you have two tangent vectors $v$ and $u$, then the "distance" between them is defined by $g$ via:
$$ d(u,v) = g_ijv^i u^j $$
or, in matrix terms, $ d(u,v)=u^Tgv $. (For you, $g=textdiag([e^x,1])$).

In Euclidean space, $g=I$ or $g_ij=delta_ij$, so this reduces to the standard inner product, the dot product. This is why people say that $g$, the metric tensor, defines an inner product on the manifold (it actually defines one on each $T_pM$).

Now to norms. For any metric $d(p_1,p_2)$ you can define a norm $||s||=d(s,s)$. Riemannian manifolds are no different. So we define:
$$ ||v||_g^2 = d(v,v) = g_ijv^iv^j = v^T g v $$
as our Riemannian norm.

Also, it turns out computing "infinitesimal" distances in $T_pM$ lets us define "actual" distances between points in $M$.
Let $gamma(t)$ be a curve on $M$.
To get the length of $gamma$, you just sum up the infinitesimal distances:
$$
ell_g(gamma;a,b) = int_a^b ||gamma'(t)||_g dt =int_a^b sqrtg_ij dotgamma(t)^idotgamma(t)^j dt
$$
Now you can define the distance between $p$ and $q$, where $p,qin M$, as the length of the shortest curve between them:
$$ D(p,q) = inf_gamma ell_g(gamma;p,q) $$

share|cite|improve this answer

edited Aug 13 '17 at 3:04

answered Aug 12 '17 at 23:59

user3658307user3658307

4,8283946

$endgroup$

add a comment | 

0

$begingroup$

This is actually a good question and weirdly not mentioned explicitly very often.

A Riemannian manifold $(M,g)$ has a tangent space $T_pM$ defined for every point $pin M$, which is a vector space attached to each $p$. For instance, a 2D manifold in 3D has a tangent plane associated to every point of the surface (so $T_pM=mathbbR^2$). The Riemannian metric $g$, only defines distances and angles in the tangent space. So $g$ doesn't let you compute distances directly, it only lets you compute infinitesimal distances.

In particular, if you have two tangent vectors $v$ and $u$, then the "distance" between them is defined by $g$ via:
$$ d(u,v) = g_ijv^i u^j $$
or, in matrix terms, $ d(u,v)=u^Tgv $. (For you, $g=textdiag([e^x,1])$).

In Euclidean space, $g=I$ or $g_ij=delta_ij$, so this reduces to the standard inner product, the dot product. This is why people say that $g$, the metric tensor, defines an inner product on the manifold (it actually defines one on each $T_pM$).

Now to norms. For any metric $d(p_1,p_2)$ you can define a norm $||s||=d(s,s)$. Riemannian manifolds are no different. So we define:
$$ ||v||_g^2 = d(v,v) = g_ijv^iv^j = v^T g v $$
as our Riemannian norm.

Also, it turns out computing "infinitesimal" distances in $T_pM$ lets us define "actual" distances between points in $M$.
Let $gamma(t)$ be a curve on $M$.
To get the length of $gamma$, you just sum up the infinitesimal distances:
$$
ell_g(gamma;a,b) = int_a^b ||gamma'(t)||_g dt =int_a^b sqrtg_ij dotgamma(t)^idotgamma(t)^j dt
$$
Now you can define the distance between $p$ and $q$, where $p,qin M$, as the length of the shortest curve between them:
$$ D(p,q) = inf_gamma ell_g(gamma;p,q) $$

share|cite|improve this answer

edited Aug 13 '17 at 3:04

answered Aug 12 '17 at 23:59

user3658307user3658307

4,8283946

$endgroup$

add a comment | 

$begingroup$

This is actually a good question and weirdly not mentioned explicitly very often.

A Riemannian manifold $(M,g)$ has a tangent space $T_pM$ defined for every point $pin M$, which is a vector space attached to each $p$. For instance, a 2D manifold in 3D has a tangent plane associated to every point of the surface (so $T_pM=mathbbR^2$). The Riemannian metric $g$, only defines distances and angles in the tangent space. So $g$ doesn't let you compute distances directly, it only lets you compute infinitesimal distances.

In particular, if you have two tangent vectors $v$ and $u$, then the "distance" between them is defined by $g$ via:
$$ d(u,v) = g_ijv^i u^j $$
or, in matrix terms, $ d(u,v)=u^Tgv $. (For you, $g=textdiag([e^x,1])$).

In Euclidean space, $g=I$ or $g_ij=delta_ij$, so this reduces to the standard inner product, the dot product. This is why people say that $g$, the metric tensor, defines an inner product on the manifold (it actually defines one on each $T_pM$).

Now to norms. For any metric $d(p_1,p_2)$ you can define a norm $||s||=d(s,s)$. Riemannian manifolds are no different. So we define:
$$ ||v||_g^2 = d(v,v) = g_ijv^iv^j = v^T g v $$
as our Riemannian norm.

Also, it turns out computing "infinitesimal" distances in $T_pM$ lets us define "actual" distances between points in $M$.
Let $gamma(t)$ be a curve on $M$.
To get the length of $gamma$, you just sum up the infinitesimal distances:
$$
ell_g(gamma;a,b) = int_a^b ||gamma'(t)||_g dt =int_a^b sqrtg_ij dotgamma(t)^idotgamma(t)^j dt
$$
Now you can define the distance between $p$ and $q$, where $p,qin M$, as the length of the shortest curve between them:
$$ D(p,q) = inf_gamma ell_g(gamma;p,q) $$

share|cite|improve this answer

edited Aug 13 '17 at 3:04

answered Aug 12 '17 at 23:59

user3658307user3658307

4,8283946

$endgroup$

share|cite|improve this answer

edited Aug 13 '17 at 3:04

answered Aug 12 '17 at 23:59

user3658307user3658307

4,8283946

answered Aug 12 '17 at 23:59

user3658307user3658307

4,8283946

answered Aug 12 '17 at 23:59

user3658307user3658307

4,8283946