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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 $$\text{d}(u^{r}_{st}\,a_{r}\,b^{s}\,c^{t})=0$$

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