# Kronecker delta question!

I am trying to grasp the basics of tensor calculus. I've come across the Kronecker delta and it confused me. \delta_j^i=\left\{\begin{aligned} &1, i=j\\ &0,i\not=j \end{aligned} \right.which is the same as $\delta_j^i=\delta_{ij}=\delta^{ij}$. I've seen also that $${\partial x^i\over\partial x^j}=\delta_j^i$$That got me thinking and i guess that $${\partial x^i\over\partial x^j}={\partial x_i\over\partial x_j}={\partial x^i\over\partial x_j}={\partial x_i\over\partial x^j}=\delta_j^i=\delta_{ij}=\delta^{ij}$$ is correct(all are equal to each other). Please , I'd like to know If I am correct and If not, please help me understand it.

• This is just a matter of notation – Tobias Molenaar Nov 13 '17 at 15:29
• The Kronecker delta are just the components of the unit matrix. – Fabian Nov 13 '17 at 15:29
• You can write whatever form of the Kronecker delta which is compatible with Einstein's summation convention, to preserve index balance. – Ivo Terek Nov 13 '17 at 15:39
• I'd like to thank you all for your comments/answers. They helped me a lot and now it's more clear. – mxaxc Nov 13 '17 at 15:46