Mixing
result of covariant derivative
$begingroup$
In wikipedia page about covariant derivative, it is said:
The covariant derivative is a generalization of the directional derivative from vector calculus. As with the directional derivative, the covariant derivative is a rule, $ nabla _{mathbf {u} }{mathbf {v} } $, which takes as its inputs: [...]. The output is the vector $nabla_{mathbf u}{mathbf v}(P)$, also at the point 'P'.
this phrase surprises me, because it is said covariant derivative extends the directional derivative, but directional derivative result can be and scalar (when applied to a function $R^n rightarrow R$), a vector (when applied to a function $R^n rightarrow R^m$), ...
However, notation $nabla_{mathbf u}{mathbf v}(P)$ remembers the one of gradient, that is always a vector, but only applicable to a function $R^n rightarrow R$.
Could someone please clarify ? Thanks.
differential-geometry
asked Jan 17 at 15:36
pasaba por aquipasaba por aqui
422315
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add a comment |
$begingroup$
In wikipedia page about covariant derivative, it is said:
The covariant derivative is a generalization of the directional derivative from vector calculus. As with the directional derivative, the covariant derivative is a rule, $ nabla _{mathbf {u} }{mathbf {v} } $, which takes as its inputs: [...]. The output is the vector $nabla_{mathbf u}{mathbf v}(P)$, also at the point 'P'.
this phrase surprises me, because it is said covariant derivative extends the directional derivative, but directional derivative result can be and scalar (when applied to a function $R^n rightarrow R$), a vector (when applied to a function $R^n rightarrow R^m$), ...
However, notation $nabla_{mathbf u}{mathbf v}(P)$ remembers the one of gradient, that is always a vector, but only applicable to a function $R^n rightarrow R$.
Could someone please clarify ? Thanks.
differential-geometry
asked Jan 17 at 15:36
pasaba por aquipasaba por aqui
422315
$endgroup$
add a comment |
$begingroup$
In wikipedia page about covariant derivative, it is said:
The covariant derivative is a generalization of the directional derivative from vector calculus. As with the directional derivative, the covariant derivative is a rule, $ nabla _{mathbf {u} }{mathbf {v} } $, which takes as its inputs: [...]. The output is the vector $nabla_{mathbf u}{mathbf v}(P)$, also at the point 'P'.
this phrase surprises me, because it is said covariant derivative extends the directional derivative, but directional derivative result can be and scalar (when applied to a function $R^n rightarrow R$), a vector (when applied to a function $R^n rightarrow R^m$), ...
However, notation $nabla_{mathbf u}{mathbf v}(P)$ remembers the one of gradient, that is always a vector, but only applicable to a function $R^n rightarrow R$.
Could someone please clarify ? Thanks.
differential-geometry
asked Jan 17 at 15:36
pasaba por aquipasaba por aqui
422315
$endgroup$
In wikipedia page about covariant derivative, it is said:
The covariant derivative is a generalization of the directional derivative from vector calculus. As with the directional derivative, the covariant derivative is a rule, $ nabla _{mathbf {u} }{mathbf {v} } $, which takes as its inputs: [...]. The output is the vector $nabla_{mathbf u}{mathbf v}(P)$, also at the point 'P'.
this phrase surprises me, because it is said covariant derivative extends the directional derivative, but directional derivative result can be and scalar (when applied to a function $R^n rightarrow R$), a vector (when applied to a function $R^n rightarrow R^m$), ...
However, notation $nabla_{mathbf u}{mathbf v}(P)$ remembers the one of gradient, that is always a vector, but only applicable to a function $R^n rightarrow R$.
Could someone please clarify ? Thanks.
differential-geometry
differential-geometry
asked Jan 17 at 15:36
pasaba por aquipasaba por aqui
422315
asked Jan 17 at 15:36
pasaba por aquipasaba por aqui
422315
edited Jan 17 at 16:06
pasaba por aqui
asked Jan 17 at 15:36
pasaba por aquipasaba por aqui
422315
asked Jan 17 at 15:36
pasaba por aquipasaba por aqui
422315
asked Jan 17 at 15:36
pasaba por aquipasaba por aqui
422315
422315
add a comment |
add a comment |
1 Answer
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When you take the directionnal derivative of a function, you still get a function $$nabla_Uf:xmapsto (nabla_Uf)(x)= df_x(U)inmathbb{R}.$$
What allows the covariant derivative is to "differentiate an object along a direction", and still get an object of the same nature. Here if $V$ is a vector field on your manifold $M$, $nabla_UV$ will still be a vector field
$$nabla_UV:xmapsto(nabla_UV)|_xin T_xM.$$
About the notation $nabla$: if you are endowed with a metric $(cdot,cdot)$ you can associate to your function $f$ a vector field $mathrm{grad},f$ checking the relation $(mathrm{grad},f,X)=df(X)=nabla_X f$ for all vector field $X$, and if so you usually identify $nabla f$ and $mathrm{grad},f$ even if these are two different mathematical things, namely a $1$-form and a vector field.
answered Jan 17 at 15:55
BalloonBalloon
4,625822
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$begingroup$
When you take the directionnal derivative of a function, you still get a function $$nabla_Uf:xmapsto (nabla_Uf)(x)= df_x(U)inmathbb{R}.$$
What allows the covariant derivative is to "differentiate an object along a direction", and still get an object of the same nature. Here if $V$ is a vector field on your manifold $M$, $nabla_UV$ will still be a vector field
$$nabla_UV:xmapsto(nabla_UV)|_xin T_xM.$$
About the notation $nabla$: if you are endowed with a metric $(cdot,cdot)$ you can associate to your function $f$ a vector field $mathrm{grad},f$ checking the relation $(mathrm{grad},f,X)=df(X)=nabla_X f$ for all vector field $X$, and if so you usually identify $nabla f$ and $mathrm{grad},f$ even if these are two different mathematical things, namely a $1$-form and a vector field.
answered Jan 17 at 15:55
BalloonBalloon
4,625822
$endgroup$
add a comment |
$begingroup$
When you take the directionnal derivative of a function, you still get a function $$nabla_Uf:xmapsto (nabla_Uf)(x)= df_x(U)inmathbb{R}.$$
What allows the covariant derivative is to "differentiate an object along a direction", and still get an object of the same nature. Here if $V$ is a vector field on your manifold $M$, $nabla_UV$ will still be a vector field
$$nabla_UV:xmapsto(nabla_UV)|_xin T_xM.$$
About the notation $nabla$: if you are endowed with a metric $(cdot,cdot)$ you can associate to your function $f$ a vector field $mathrm{grad},f$ checking the relation $(mathrm{grad},f,X)=df(X)=nabla_X f$ for all vector field $X$, and if so you usually identify $nabla f$ and $mathrm{grad},f$ even if these are two different mathematical things, namely a $1$-form and a vector field.
answered Jan 17 at 15:55
BalloonBalloon
4,625822
$endgroup$
add a comment |
$begingroup$
When you take the directionnal derivative of a function, you still get a function $$nabla_Uf:xmapsto (nabla_Uf)(x)= df_x(U)inmathbb{R}.$$
What allows the covariant derivative is to "differentiate an object along a direction", and still get an object of the same nature. Here if $V$ is a vector field on your manifold $M$, $nabla_UV$ will still be a vector field
$$nabla_UV:xmapsto(nabla_UV)|_xin T_xM.$$
About the notation $nabla$: if you are endowed with a metric $(cdot,cdot)$ you can associate to your function $f$ a vector field $mathrm{grad},f$ checking the relation $(mathrm{grad},f,X)=df(X)=nabla_X f$ for all vector field $X$, and if so you usually identify $nabla f$ and $mathrm{grad},f$ even if these are two different mathematical things, namely a $1$-form and a vector field.
answered Jan 17 at 15:55
BalloonBalloon
4,625822
$endgroup$
When you take the directionnal derivative of a function, you still get a function $$nabla_Uf:xmapsto (nabla_Uf)(x)= df_x(U)inmathbb{R}.$$
What allows the covariant derivative is to "differentiate an object along a direction", and still get an object of the same nature. Here if $V$ is a vector field on your manifold $M$, $nabla_UV$ will still be a vector field
$$nabla_UV:xmapsto(nabla_UV)|_xin T_xM.$$
About the notation $nabla$: if you are endowed with a metric $(cdot,cdot)$ you can associate to your function $f$ a vector field $mathrm{grad},f$ checking the relation $(mathrm{grad},f,X)=df(X)=nabla_X f$ for all vector field $X$, and if so you usually identify $nabla f$ and $mathrm{grad},f$ even if these are two different mathematical things, namely a $1$-form and a vector field.
answered Jan 17 at 15:55
BalloonBalloon
4,625822
answered Jan 17 at 15:55
BalloonBalloon
4,625822
answered Jan 17 at 15:55
BalloonBalloon
4,625822
answered Jan 17 at 15:55
BalloonBalloon
4,625822
4,625822
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add a comment |
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