What's contravariant about contravariance
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I think, that the covariant basis varies with the coordinate system, and the contra varies opposite to it. Geometrically, is the thing that's varying against the coordinate system just the length/magnitude of $e^i$ when $e_i$ is multiplied by scalar? I know the algebraically how the contravariant and covariant basis are related as inverses. Anyways, if for instance I decide to change just one component in the covariant basis by trippling it is there anything if I were looking at it geoetrically that would appear to vary opposite to the transformation?
tensors
asked Dec 15 '18 at 11:12
Benjamin ThoburnBenjamin Thoburn
350213
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add a comment |
$begingroup$
I think, that the covariant basis varies with the coordinate system, and the contra varies opposite to it. Geometrically, is the thing that's varying against the coordinate system just the length/magnitude of $e^i$ when $e_i$ is multiplied by scalar? I know the algebraically how the contravariant and covariant basis are related as inverses. Anyways, if for instance I decide to change just one component in the covariant basis by trippling it is there anything if I were looking at it geoetrically that would appear to vary opposite to the transformation?
tensors
asked Dec 15 '18 at 11:12
Benjamin ThoburnBenjamin Thoburn
350213
$endgroup$
$begingroup$
Think about what happens to the normal to a line in $mathbb R^2$.
$endgroup$
– amd
Dec 15 '18 at 20:02
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@amd By normal do you mean the unit orthogonal vector? I know visually since $e^i$ dot $e_j$ is the Kronecker delta that the corresponding vector in the contravariant basis has to have the dot product equal 1 and all others vectors of the basis have to be perpendicular. So visually I understand how it works but I can't see visually what's contravariant about it.
$endgroup$
– Benjamin Thoburn
Dec 16 '18 at 10:41
$begingroup$
You’re asking for a geometric interpretation but thinking algebraically. Draw some pictures. When you scale the $x$-coordinate by two what happens to the line $x=y$? How is this reflected in the point-normal form of its equation?
$endgroup$
– amd
Dec 16 '18 at 21:46
add a comment |
$begingroup$
I think, that the covariant basis varies with the coordinate system, and the contra varies opposite to it. Geometrically, is the thing that's varying against the coordinate system just the length/magnitude of $e^i$ when $e_i$ is multiplied by scalar? I know the algebraically how the contravariant and covariant basis are related as inverses. Anyways, if for instance I decide to change just one component in the covariant basis by trippling it is there anything if I were looking at it geoetrically that would appear to vary opposite to the transformation?
tensors
asked Dec 15 '18 at 11:12
Benjamin ThoburnBenjamin Thoburn
350213
$endgroup$
I think, that the covariant basis varies with the coordinate system, and the contra varies opposite to it. Geometrically, is the thing that's varying against the coordinate system just the length/magnitude of $e^i$ when $e_i$ is multiplied by scalar? I know the algebraically how the contravariant and covariant basis are related as inverses. Anyways, if for instance I decide to change just one component in the covariant basis by trippling it is there anything if I were looking at it geoetrically that would appear to vary opposite to the transformation?
tensors
tensors
asked Dec 15 '18 at 11:12
Benjamin ThoburnBenjamin Thoburn
350213
asked Dec 15 '18 at 11:12
Benjamin ThoburnBenjamin Thoburn
350213
asked Dec 15 '18 at 11:12
Benjamin ThoburnBenjamin Thoburn
350213
asked Dec 15 '18 at 11:12
Benjamin ThoburnBenjamin Thoburn
350213
asked Dec 15 '18 at 11:12
Benjamin ThoburnBenjamin Thoburn
350213
350213
$begingroup$
Think about what happens to the normal to a line in $mathbb R^2$.
$endgroup$
– amd
Dec 15 '18 at 20:02
$begingroup$
@amd By normal do you mean the unit orthogonal vector? I know visually since $e^i$ dot $e_j$ is the Kronecker delta that the corresponding vector in the contravariant basis has to have the dot product equal 1 and all others vectors of the basis have to be perpendicular. So visually I understand how it works but I can't see visually what's contravariant about it.
$endgroup$
– Benjamin Thoburn
Dec 16 '18 at 10:41
$begingroup$
You’re asking for a geometric interpretation but thinking algebraically. Draw some pictures. When you scale the $x$-coordinate by two what happens to the line $x=y$? How is this reflected in the point-normal form of its equation?
$endgroup$
– amd
Dec 16 '18 at 21:46
add a comment |
$begingroup$
Think about what happens to the normal to a line in $mathbb R^2$.
$endgroup$
– amd
Dec 15 '18 at 20:02
$begingroup$
@amd By normal do you mean the unit orthogonal vector? I know visually since $e^i$ dot $e_j$ is the Kronecker delta that the corresponding vector in the contravariant basis has to have the dot product equal 1 and all others vectors of the basis have to be perpendicular. So visually I understand how it works but I can't see visually what's contravariant about it.
$endgroup$
– Benjamin Thoburn
Dec 16 '18 at 10:41
$begingroup$
You’re asking for a geometric interpretation but thinking algebraically. Draw some pictures. When you scale the $x$-coordinate by two what happens to the line $x=y$? How is this reflected in the point-normal form of its equation?
$endgroup$
– amd
Dec 16 '18 at 21:46
$begingroup$
Think about what happens to the normal to a line in $mathbb R^2$.
$endgroup$
– amd
Dec 15 '18 at 20:02
$begingroup$
Think about what happens to the normal to a line in $mathbb R^2$.
$endgroup$
– amd
Dec 15 '18 at 20:02
$begingroup$
@amd By normal do you mean the unit orthogonal vector? I know visually since $e^i$ dot $e_j$ is the Kronecker delta that the corresponding vector in the contravariant basis has to have the dot product equal 1 and all others vectors of the basis have to be perpendicular. So visually I understand how it works but I can't see visually what's contravariant about it.
$endgroup$
– Benjamin Thoburn
Dec 16 '18 at 10:41
$begingroup$
@amd By normal do you mean the unit orthogonal vector? I know visually since $e^i$ dot $e_j$ is the Kronecker delta that the corresponding vector in the contravariant basis has to have the dot product equal 1 and all others vectors of the basis have to be perpendicular. So visually I understand how it works but I can't see visually what's contravariant about it.
$endgroup$
– Benjamin Thoburn
Dec 16 '18 at 10:41
$begingroup$
You’re asking for a geometric interpretation but thinking algebraically. Draw some pictures. When you scale the $x$-coordinate by two what happens to the line $x=y$? How is this reflected in the point-normal form of its equation?
$endgroup$
– amd
Dec 16 '18 at 21:46
$begingroup$
You’re asking for a geometric interpretation but thinking algebraically. Draw some pictures. When you scale the $x$-coordinate by two what happens to the line $x=y$? How is this reflected in the point-normal form of its equation?
$endgroup$
– amd
Dec 16 '18 at 21:46
add a comment |
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$begingroup$
Think about what happens to the normal to a line in $mathbb R^2$.
$endgroup$
– amd
Dec 15 '18 at 20:02
$begingroup$
@amd By normal do you mean the unit orthogonal vector? I know visually since $e^i$ dot $e_j$ is the Kronecker delta that the corresponding vector in the contravariant basis has to have the dot product equal 1 and all others vectors of the basis have to be perpendicular. So visually I understand how it works but I can't see visually what's contravariant about it.
$endgroup$
– Benjamin Thoburn
Dec 16 '18 at 10:41
$begingroup$
You’re asking for a geometric interpretation but thinking algebraically. Draw some pictures. When you scale the $x$-coordinate by two what happens to the line $x=y$? How is this reflected in the point-normal form of its equation?
$endgroup$
– amd
Dec 16 '18 at 21:46