# Jacobian for Kronecker delta

I was revising on a bit of tensor calculus, when I stumbled upon this:

$$\delta^i_j = \frac{\partial y^i}{\partial x^\alpha} \frac{\partial x^\alpha}{\partial y^j}$$

this expression yields:

$$\left|\frac{\partial y^i}{\partial x^j}\right| \left|\frac{\partial x^\alpha}{\partial y^\beta}\right|= 1 ,$$

With $\left|\dfrac{\partial y^i}{\partial x^j}\right|$ being the Jacobian for transformation $y^i=y^i(x^1.....x^n)$, and $\left|\dfrac{\partial x^\alpha}{\partial y^\beta}\right|$ being the Jacobian of the inverse transformation.

My question is, how do you get eq. 1 from the Kronecker delta, as they are merely jacobians of coordinate transformatios, and being inverse of each other, are 1. But how do they follow from $\delta^i_j$'s expansion? (i.e. DERIVE eq. 1 using the expansion for kronecker delta)? I am most probably making a conceptual error, but this is the first time I have seen such a representation of the kronecker delta.

• The simple explanation is that (taking into accound tensor summation convention, i.e same repeated index is to be summed over): $\partial y^i$ is (functionaly) independent of $\partial y^j$ same as $\partial x^\alpha$ is functionaly independent of $\partial x^\beta$ – Nikos M. Aug 17 '14 at 12:52
• The question (v2) seems to follow directly from inverse function theorem in several variables. – Qmechanic Aug 17 '14 at 13:17
• The conceptual error is expecting this to come from an expansion of the Kronecker delta. This is really just a property of coordinate transforms (and their inverse) written down in components. – Sanya Feb 19 '17 at 9:47
• "My question is, how do you get eq. 1..." There is no equation 1 in your question. – Brian Moths Mar 24 '17 at 14:37

The first equation states that two matrices' product is an identity matrix. The desired result states that the product of these matrices' determinants is $1$. This follows from taking determinants of the original equation.
The relation in the question, $$\frac{\partial y^i}{\partial x^\alpha} \frac{\partial x^\alpha}{\partial y^j}=\delta^i_j ,$$ is essentially the product of two matrices $\frac{\partial y^i}{\partial x^\alpha}$ and $\frac{\partial x^\alpha}{\partial y^j}$ giving the identity. The relationship the OP asks for follows directly by taking the determinant of that matrix equation: the right-hand side gives $1$, and the left-hand side is the determinant of a product, which splits to the product of the determinants.
You transform $y \rightarrow x$ then back $x \rightarrow y$ so back to the same coordinate system but not sure you have transform a particular basis vector to the same when you transform it back, $i$ may $\neq j$.