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