differential of a contracted product with a 3 order tensor and uniform vectors
$begingroup$
I would like to know if expression of $text{d}(u^{r}_{st},a_{r},b^{s},c^{t})$ is equal to zero.
Indeed, I consider a 3 order tensor (actually (1,2) tensor) $u^{r}_{st}$ contracted with uniform vectors $a_{r}$, $b^{s}$, $c^{t}$ and I know this contracted product is an invariant towards basis where it is expressed.
But invariant doesn't imply necessary a constant, such the differential of contracted product, i.e $text{d}(u^{r}_{st},a_{r},b^{s},c^{t})$, would be always equal to zero, does it ?
Here the equation of differential :
$$begin{equation}
begin{array}[b]{lcl}
text{d}(u^{r}_{st},a_{r},b^{s},c^{t})&=&(partial_{k},u^{r}_{st},a_{r},b^{s},c^{t}+u^{r}_{st},partial_{k},a_{r},b^{s},c^{t}\
&+&u^{r}_{st},a_{r},partial_{k},b^{s},c^{t}+u^{r}_{st},a_{r},b^{s},partial_{k},c^{t}),text{d}y^{k}
end{array}
end{equation}$$
If I consider uniform vectors $a_{r}$, $b^{s}$, $c^{t}$, I can demonstrate the expression of covariant derivative of tensor $u^{r}_{st}$ :
$$begin{equation}
nabla_{k},u^{r}_{st}=partial_{k},u^{r}_{st}+u^{i}_{st},Gamma_{ki}^{r}-u^{r}_{it},Gamma^{i}_{ks}-u^{r}_{si},Gamma^{i}_{kt}
end{equation}$$
But my real question is to know if $$text{d}(u^{r}_{st},a_{r},b^{s},c^{t})=0$$
differential-geometry tensors tensor-rank
asked Dec 18 '18 at 21:15
youpilat13youpilat13
2811
$endgroup$
add a comment |
$begingroup$
I would like to know if expression of $text{d}(u^{r}_{st},a_{r},b^{s},c^{t})$ is equal to zero.
Indeed, I consider a 3 order tensor (actually (1,2) tensor) $u^{r}_{st}$ contracted with uniform vectors $a_{r}$, $b^{s}$, $c^{t}$ and I know this contracted product is an invariant towards basis where it is expressed.
But invariant doesn't imply necessary a constant, such the differential of contracted product, i.e $text{d}(u^{r}_{st},a_{r},b^{s},c^{t})$, would be always equal to zero, does it ?
Here the equation of differential :
$$begin{equation}
begin{array}[b]{lcl}
text{d}(u^{r}_{st},a_{r},b^{s},c^{t})&=&(partial_{k},u^{r}_{st},a_{r},b^{s},c^{t}+u^{r}_{st},partial_{k},a_{r},b^{s},c^{t}\
&+&u^{r}_{st},a_{r},partial_{k},b^{s},c^{t}+u^{r}_{st},a_{r},b^{s},partial_{k},c^{t}),text{d}y^{k}
end{array}
end{equation}$$
If I consider uniform vectors $a_{r}$, $b^{s}$, $c^{t}$, I can demonstrate the expression of covariant derivative of tensor $u^{r}_{st}$ :
$$begin{equation}
nabla_{k},u^{r}_{st}=partial_{k},u^{r}_{st}+u^{i}_{st},Gamma_{ki}^{r}-u^{r}_{it},Gamma^{i}_{ks}-u^{r}_{si},Gamma^{i}_{kt}
end{equation}$$
But my real question is to know if $$text{d}(u^{r}_{st},a_{r},b^{s},c^{t})=0$$
differential-geometry tensors tensor-rank
asked Dec 18 '18 at 21:15
youpilat13youpilat13
2811
$endgroup$
add a comment |
$begingroup$
I would like to know if expression of $text{d}(u^{r}_{st},a_{r},b^{s},c^{t})$ is equal to zero.
Indeed, I consider a 3 order tensor (actually (1,2) tensor) $u^{r}_{st}$ contracted with uniform vectors $a_{r}$, $b^{s}$, $c^{t}$ and I know this contracted product is an invariant towards basis where it is expressed.
But invariant doesn't imply necessary a constant, such the differential of contracted product, i.e $text{d}(u^{r}_{st},a_{r},b^{s},c^{t})$, would be always equal to zero, does it ?
Here the equation of differential :
$$begin{equation}
begin{array}[b]{lcl}
text{d}(u^{r}_{st},a_{r},b^{s},c^{t})&=&(partial_{k},u^{r}_{st},a_{r},b^{s},c^{t}+u^{r}_{st},partial_{k},a_{r},b^{s},c^{t}\
&+&u^{r}_{st},a_{r},partial_{k},b^{s},c^{t}+u^{r}_{st},a_{r},b^{s},partial_{k},c^{t}),text{d}y^{k}
end{array}
end{equation}$$
If I consider uniform vectors $a_{r}$, $b^{s}$, $c^{t}$, I can demonstrate the expression of covariant derivative of tensor $u^{r}_{st}$ :
$$begin{equation}
nabla_{k},u^{r}_{st}=partial_{k},u^{r}_{st}+u^{i}_{st},Gamma_{ki}^{r}-u^{r}_{it},Gamma^{i}_{ks}-u^{r}_{si},Gamma^{i}_{kt}
end{equation}$$
But my real question is to know if $$text{d}(u^{r}_{st},a_{r},b^{s},c^{t})=0$$
differential-geometry tensors tensor-rank
asked Dec 18 '18 at 21:15
youpilat13youpilat13
2811
$endgroup$
I would like to know if expression of $text{d}(u^{r}_{st},a_{r},b^{s},c^{t})$ is equal to zero.
Indeed, I consider a 3 order tensor (actually (1,2) tensor) $u^{r}_{st}$ contracted with uniform vectors $a_{r}$, $b^{s}$, $c^{t}$ and I know this contracted product is an invariant towards basis where it is expressed.
But invariant doesn't imply necessary a constant, such the differential of contracted product, i.e $text{d}(u^{r}_{st},a_{r},b^{s},c^{t})$, would be always equal to zero, does it ?
Here the equation of differential :
$$begin{equation}
begin{array}[b]{lcl}
text{d}(u^{r}_{st},a_{r},b^{s},c^{t})&=&(partial_{k},u^{r}_{st},a_{r},b^{s},c^{t}+u^{r}_{st},partial_{k},a_{r},b^{s},c^{t}\
&+&u^{r}_{st},a_{r},partial_{k},b^{s},c^{t}+u^{r}_{st},a_{r},b^{s},partial_{k},c^{t}),text{d}y^{k}
end{array}
end{equation}$$
If I consider uniform vectors $a_{r}$, $b^{s}$, $c^{t}$, I can demonstrate the expression of covariant derivative of tensor $u^{r}_{st}$ :
$$begin{equation}
nabla_{k},u^{r}_{st}=partial_{k},u^{r}_{st}+u^{i}_{st},Gamma_{ki}^{r}-u^{r}_{it},Gamma^{i}_{ks}-u^{r}_{si},Gamma^{i}_{kt}
end{equation}$$
But my real question is to know if $$text{d}(u^{r}_{st},a_{r},b^{s},c^{t})=0$$
differential-geometry tensors tensor-rank
differential-geometry tensors tensor-rank
asked Dec 18 '18 at 21:15
youpilat13youpilat13
2811
asked Dec 18 '18 at 21:15
youpilat13youpilat13
2811
asked Dec 18 '18 at 21:15
youpilat13youpilat13
2811
asked Dec 18 '18 at 21:15
youpilat13youpilat13
2811
asked Dec 18 '18 at 21:15
youpilat13youpilat13
2811
2811
add a comment |
add a comment |
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