Is it possible to list the root's children?
For example - I'll use the thecogworks demo as a visual aide just because we can all see it.
http://demo.thecogworks.co.uk/umbraco/umbraco.aspx#content
Their tree structure looks like this:

Content
    |
    +--Cog-Demo
    |   
    +--Blog

Is it possible to write an XSLT to output 'Cog-Demo' and 'Blog' (I'm not worried about the formating right now.)
I tried using  '/child::*' to start at the root and list one level below, but it doesn't work. 
I suspect XSLT can't start that high.  See below.
       
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE xsl:Stylesheet [ <!ENTITY nbsp "&#x00A0;"> ]>
<xsl:stylesheet
    version="1.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:msxml="urn:schemas-microsoft-com:xslt"
    xmlns:umbraco.library="urn:umbraco.library"
    exclude-result-prefixes="msxml umbraco.library">

<xsl:output method="xml" omit-xml-declaration="yes"/>
<xsl:param name="currentPage"/>
<xsl:template match="/">
  <!-- Get the root's children --> 
  <xsl:variable name="RootChild" select="/child::*" />
  <xsl:for-each select="$RootChild" >
    <xsl:value-of select="@nodeName"/> <br />
  </xsl:for-each>
</xsl:template>
</xsl:stylesheet>