How is $dx^2+dy^2$ the Euclidean metric on $mathbb R^2$Riemannian metricSemi-Riemannian ManifoldMetric on manifoldIs a metric tensor a norm of the tangent space?Decomposing symmetric tensor field into sum of metric and tensor productSince the Curvature tensor depends on a connection(not metric), is it the relevant quantity to characterize the curvature of Riemannian manifolds?Definition of a metric tensor in Barrett O'Neil's bookIntuition behind riemannian metricIntuition behind Riemannian-metricLee Introduction to Smooth Manifolds - Why is Riemannian metric continuous?
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How is $dx^2+dy^2$ the Euclidean metric on $mathbb R^2$
Riemannian metricSemi-Riemannian ManifoldMetric on manifoldIs a metric tensor a norm of the tangent space?Decomposing symmetric tensor field into sum of metric and tensor productSince the Curvature tensor depends on a connection(not metric), is it the relevant quantity to characterize the curvature of Riemannian manifolds?Definition of a metric tensor in Barrett O'Neil's bookIntuition behind riemannian metricIntuition behind Riemannian-metricLee Introduction to Smooth Manifolds - Why is Riemannian metric continuous?
$begingroup$
A Riemannian metric is a smooth symmetric covariant $2$-tensor field.
If I put in two vectors, say $(1,2)$ and $(2,1)$, I don't get $|(1,2)-(2,1)|=sqrt2$:
$$(dx^2+dy^2)((1,2), (2,1))=2+2=4.$$
differential-geometry riemannian-geometry smooth-manifolds riemannian-metric
asked Mar 17 at 22:34
Al JebrAl Jebr
4,37443378
$endgroup$
|
show 5 more comments
$begingroup$
A Riemannian metric is a smooth symmetric covariant $2$-tensor field.
If I put in two vectors, say $(1,2)$ and $(2,1)$, I don't get $|(1,2)-(2,1)|=sqrt2$:
$$(dx^2+dy^2)((1,2), (2,1))=2+2=4.$$
differential-geometry riemannian-geometry smooth-manifolds riemannian-metric
asked Mar 17 at 22:34
Al JebrAl Jebr
4,37443378
$endgroup$
1
$begingroup$
Put in the form $sum g_ijdx^idx^j$, this form always refers to the square of the element of length.
$endgroup$
– GReyes
Mar 17 at 22:36
$begingroup$
Who said it was? First, as you seem to be reading it, $dx^2+dy^2$ here equals $1^2+1^2=2.$ Second, that's not the definition of the Euclidean distance, either/
$endgroup$
– Thomas Andrews
Mar 17 at 22:37
1
$begingroup$
This form is a quadratic form, and applies to one vector only (strictly speaking, to a tangent vector at a given point but everything is the same in the flat space). Applied to $(1,2)$ you get $1^2+2^2=5$.
$endgroup$
– GReyes
Mar 17 at 22:40
$begingroup$
@GReyes But if I take the square root, I get $2$. Do I need to take square root twice?
$endgroup$
– Al Jebr
Mar 17 at 22:48
3
$begingroup$
A metric in the Riemannian sense is not the same as a metric in the metric-space sense.
$endgroup$
– jacob1729
Mar 17 at 23:00
|
show 5 more comments
$begingroup$
A Riemannian metric is a smooth symmetric covariant $2$-tensor field.
If I put in two vectors, say $(1,2)$ and $(2,1)$, I don't get $|(1,2)-(2,1)|=sqrt2$:
$$(dx^2+dy^2)((1,2), (2,1))=2+2=4.$$
differential-geometry riemannian-geometry smooth-manifolds riemannian-metric
asked Mar 17 at 22:34
Al JebrAl Jebr
4,37443378
$endgroup$
A Riemannian metric is a smooth symmetric covariant $2$-tensor field.
If I put in two vectors, say $(1,2)$ and $(2,1)$, I don't get $|(1,2)-(2,1)|=sqrt2$:
$$(dx^2+dy^2)((1,2), (2,1))=2+2=4.$$
differential-geometry riemannian-geometry smooth-manifolds riemannian-metric
differential-geometry riemannian-geometry smooth-manifolds riemannian-metric
asked Mar 17 at 22:34
Al JebrAl Jebr
4,37443378
asked Mar 17 at 22:34
Al JebrAl Jebr
4,37443378
edited Mar 17 at 22:44
Al Jebr
asked Mar 17 at 22:34
Al JebrAl Jebr
4,37443378
asked Mar 17 at 22:34
Al JebrAl Jebr
4,37443378
asked Mar 17 at 22:34
Al JebrAl Jebr
4,37443378
4,37443378
1
$begingroup$
Put in the form $sum g_ijdx^idx^j$, this form always refers to the square of the element of length.
$endgroup$
– GReyes
Mar 17 at 22:36
$begingroup$
Who said it was? First, as you seem to be reading it, $dx^2+dy^2$ here equals $1^2+1^2=2.$ Second, that's not the definition of the Euclidean distance, either/
$endgroup$
– Thomas Andrews
Mar 17 at 22:37
1
$begingroup$
This form is a quadratic form, and applies to one vector only (strictly speaking, to a tangent vector at a given point but everything is the same in the flat space). Applied to $(1,2)$ you get $1^2+2^2=5$.
$endgroup$
– GReyes
Mar 17 at 22:40
$begingroup$
@GReyes But if I take the square root, I get $2$. Do I need to take square root twice?
$endgroup$
– Al Jebr
Mar 17 at 22:48
3
$begingroup$
A metric in the Riemannian sense is not the same as a metric in the metric-space sense.
$endgroup$
– jacob1729
Mar 17 at 23:00
|
show 5 more comments
1
$begingroup$
Put in the form $sum g_ijdx^idx^j$, this form always refers to the square of the element of length.
$endgroup$
– GReyes
Mar 17 at 22:36
$begingroup$
Who said it was? First, as you seem to be reading it, $dx^2+dy^2$ here equals $1^2+1^2=2.$ Second, that's not the definition of the Euclidean distance, either/
$endgroup$
– Thomas Andrews
Mar 17 at 22:37
1
$begingroup$
This form is a quadratic form, and applies to one vector only (strictly speaking, to a tangent vector at a given point but everything is the same in the flat space). Applied to $(1,2)$ you get $1^2+2^2=5$.
$endgroup$
– GReyes
Mar 17 at 22:40
$begingroup$
@GReyes But if I take the square root, I get $2$. Do I need to take square root twice?
$endgroup$
– Al Jebr
Mar 17 at 22:48
3
$begingroup$
A metric in the Riemannian sense is not the same as a metric in the metric-space sense.
$endgroup$
– jacob1729
Mar 17 at 23:00
1
1
$begingroup$
Put in the form $sum g_ijdx^idx^j$, this form always refers to the square of the element of length.
$endgroup$
– GReyes
Mar 17 at 22:36
$begingroup$
Put in the form $sum g_ijdx^idx^j$, this form always refers to the square of the element of length.
$endgroup$
– GReyes
Mar 17 at 22:36
$begingroup$
Who said it was? First, as you seem to be reading it, $dx^2+dy^2$ here equals $1^2+1^2=2.$ Second, that's not the definition of the Euclidean distance, either/
$endgroup$
– Thomas Andrews
Mar 17 at 22:37
$begingroup$
Who said it was? First, as you seem to be reading it, $dx^2+dy^2$ here equals $1^2+1^2=2.$ Second, that's not the definition of the Euclidean distance, either/
$endgroup$
– Thomas Andrews
Mar 17 at 22:37
1
1
$begingroup$
This form is a quadratic form, and applies to one vector only (strictly speaking, to a tangent vector at a given point but everything is the same in the flat space). Applied to $(1,2)$ you get $1^2+2^2=5$.
$endgroup$
– GReyes
Mar 17 at 22:40
$begingroup$
This form is a quadratic form, and applies to one vector only (strictly speaking, to a tangent vector at a given point but everything is the same in the flat space). Applied to $(1,2)$ you get $1^2+2^2=5$.
$endgroup$
– GReyes
Mar 17 at 22:40
$begingroup$
@GReyes But if I take the square root, I get $2$. Do I need to take square root twice?
$endgroup$
– Al Jebr
Mar 17 at 22:48
$begingroup$
@GReyes But if I take the square root, I get $2$. Do I need to take square root twice?
$endgroup$
– Al Jebr
Mar 17 at 22:48
3
3
$begingroup$
A metric in the Riemannian sense is not the same as a metric in the metric-space sense.
$endgroup$
– jacob1729
Mar 17 at 23:00
$begingroup$
A metric in the Riemannian sense is not the same as a metric in the metric-space sense.
$endgroup$
– jacob1729
Mar 17 at 23:00
|
show 5 more comments
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1
$begingroup$
Put in the form $sum g_ijdx^idx^j$, this form always refers to the square of the element of length.
$endgroup$
– GReyes
Mar 17 at 22:36
$begingroup$
Who said it was? First, as you seem to be reading it, $dx^2+dy^2$ here equals $1^2+1^2=2.$ Second, that's not the definition of the Euclidean distance, either/
$endgroup$
– Thomas Andrews
Mar 17 at 22:37
1
$begingroup$
This form is a quadratic form, and applies to one vector only (strictly speaking, to a tangent vector at a given point but everything is the same in the flat space). Applied to $(1,2)$ you get $1^2+2^2=5$.
$endgroup$
– GReyes
Mar 17 at 22:40
$begingroup$
@GReyes But if I take the square root, I get $2$. Do I need to take square root twice?
$endgroup$
– Al Jebr
Mar 17 at 22:48
3
$begingroup$
A metric in the Riemannian sense is not the same as a metric in the metric-space sense.
$endgroup$
– jacob1729
Mar 17 at 23:00