It might be helpful to consider the following when trying to understand the Kronecker delta. Suppose that you have some system of coordinates x^a, where a=0,1,2,\ldots,m. Now consider the partial derivative
\frac{\partial x^a}{\partial x^b} = x^a_{,b}[/itex]<br />
<br />
Schutz tells you that x^a_{,b}=\delta^a_{\phantom{a}b}. To see why this is so, consider a simple example where a=0,1. Then x^a_{\phantom{a},b} can be represented as a matrix:<br />
<br />
x^a_{\phantom{a},b} =<br />
\left(<br />
\begin{array}{cc}<br />
\frac{\partial x^0}{\partial x^0} &amp; \frac{\partial x^0}{\partial x^1} \\<br />
\frac{\partial x^1}{\partial x^0} &amp; \frac{\partial x^1}{\partial x^1}<br />
\end{array}<br />
\right) = <br />
\left(<br />
\begin{array}{cc}<br />
1 &amp; 0 \\ 0 &amp; 1<br />
\end{array}<br />
\right) = \delta^{a}_{\phantom{a}b}<br /><br />
<br />
The reason that, for example, \partial x^0/\partial x^1 = 0 while \partial x^0/\partial x^0=1 should be obvious. If it isn't, note that our coordinates are supposed to be <i>independent</i>, so if you differentiate x^0 with respect to x^1 you will get zero, while differentiating x^0 with respect to x^0 will give you 1. So it should be easy to see why the definition of the Kronecker delta allows you to write this. Extending things to a scenario where you have m coordinates is then trivial:<br />
<br />
<br />
x^a_{\phantom{a},b}<br />
=<br />
\left(<br />
\begin{array}{cccc}<br />
\frac{\partial x^0}{\partial x^0} &amp; \frac{\partial x^0}{\partial x^1} &amp; <br />
\cdots &amp; \frac{\partial x^0}{\partial x^m} \\<br />
\frac{\partial x^1}{\partial x^0} &amp; \frac{\partial x^1}{\partial x^1} &amp;<br />
\cdots &amp; \frac{\partial x^1}{\partial x^m} \\<br />
\vdots &amp; \vdots &amp; \ddots &amp; \vdots \\<br />
\frac{\partial x^m}{\partial x^0} &amp; \frac{\partial x^m}{\partial x^1} &amp;<br />
\cdots &amp; \frac{\partial x^m}{\partial x^m}<br />
\end{array}<br />
\right)<br />
=<br />
\left(<br />
\begin{array}{cccc}<br />
1 &amp; 0 &amp; \cdots &amp; 0 \\<br />
0 &amp; 1 &amp; \cdots &amp; 0 \\<br />
\vdots &amp; \vdots &amp; \ddots &amp; \vdots \\<br />
0 &amp; 0 &amp; \cdots &amp; 1<br />
\end{array}<br />
\right)<br />
= \delta^{a}_{\phantom{a}b}.<br />