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In S02 synopsis, section "Names and Variables", I see:

$<foo>      match variable, short for $/{'foo'}
[download]

So, when you say my $<-var>;, you are actually initializing the key -var of the hash $/.

Here's what I see in recent rakudo:

>> perl6 -v

This is Rakudo Perl 6, version 2011.07-2-g1b7dd12 built on parrot 3.6.
+0 RELEASE_3_6_0

>> perl6
> my $<-var> = 42
42
> $/
-var    42

> $/.WHAT
Hash()
[download]

In reply to Re: Perl6: S2 identifier and $ special var questions by PrakashK
in thread Perl6: S2 identifier and $ special var questions by rir

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