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The Kronecker delta is

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and I use the definition of tensor as

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I try to prove the Kronecker delta is a (1,1) tensor. I find the following equation which is enough to show Kronecker delta is a tensor. $$ {\delta'}_j^i = \delta_k^k \frac{\partial{x_i}'}{\partial{x_k}}\frac{\partial{x_k}}{\partial{x_j}'} $$ But I dnot know how to get the equation.

Arctic Char
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Enhao Lan
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  • The way proving something is a tensor usually works is that you'll want to change its basis in two ways: from definition (i.e. anything that doesn't use the change of coordinates formula) and then the change of coordinates formula, then show the two are/are not the same, depending on what the object is. – Ninad Munshi May 17 '20 at 13:29
  • I am providing you a link. Please go through it, you will find your answer. The link is : Mathematical Methods for Physicists. DOI: 10.1016/B978-0-12-384654-9.00004-9 – nmasanta May 17 '20 at 14:47

1 Answers1

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This is just chain rule applied to the composition:

$$ (x_1' , \cdots, x_n') \mapsto (x_1 , \cdots, x_n) \mapsto (x_1' , \cdots, x_n')$$

Note that the composition is the identity. So $$ I = \frac{\partial x'}{\partial x} \frac{\partial x}{\partial x'}$$

in terms of index we have \begin{align*} \frac{\partial{x_i}'}{\partial{x_k}}\frac{\partial{x_k}}{\partial{x_j}'}= {\delta'}^i_j \end{align*}

Arctic Char
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