# differential of a contracted product with a 3 order tensor and uniform vectors

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 :

$$$$\begin{array}[b]{lcl} \text{d}(u^{r}_{st}\,a_{r}\,b^{s}\,c^{t})&=&(\partial_{k}\,u^{r}_{st}\,a_{r}\,b^{s}\,c^{t}+u^{r}_{st}\,\partial_{k}\,a_{r}\,b^{s}\,c^{t}\\ &+&u^{r}_{st}\,a_{r}\,\partial_{k}\,b^{s}\,c^{t}+u^{r}_{st}\,a_{r}\,b^{s}\,\partial_{k}\,c^{t})\,\text{d}y^{k} \end{array}$$$$

If I consider uniform vectors $$a_{r}$$, $$b^{s}$$, $$c^{t}$$, I can demonstrate the expression of covariant derivative of tensor $$u^{r}_{st}$$ :

$$$$\nabla_{k}\,u^{r}_{st}=\partial_{k}\,u^{r}_{st}+u^{i}_{st}\,\Gamma_{ki}^{r}-u^{r}_{it}\,\Gamma^{i}_{ks}-u^{r}_{si}\,\Gamma^{i}_{kt}$$$$

But my real question is to know if $$\text{d}(u^{r}_{st}\,a_{r}\,b^{s}\,c^{t})=0$$