Basically a split on (' ',$_) suppresses the leading NULL field if there is a leading white space character. I don't know of a split on a regex variation of \s that can do the same thing. Do you?

#!/usr/bin/perl -w
use strict;

my $string  = "    abc xyz";

my @tokens = split(' ',$string);
print "There are ".@tokens." tokens in \'$string\'\n";
print "Using split on ' '\n";
print join("|",@tokens),"\n\n";

my @tokens2 = split(/\s+/,$string);
print "There are ".@tokens2." tokens in \'$string\'\n";
print "Using split on /\\s+/\n";
print join("|",@tokens2),"\n\n";

my @tokens3 = split(/ /,$string);
print "There are ".@tokens3." tokens in \'$string\'\n";
print "Using split on / /\n";
print join("|",@tokens3),"\n\n";

my @tokens4 = split(/\s/,$string);
print "There are ".@tokens4." tokens in \'$string\'\n";
print "Using split on /\\s/\n";
print join("|",@tokens4),"\n";

__END__

The above code prints:

There are 2 tokens in '    abc xyz'
Using split on ' '
abc|xyz

There are 3 tokens in '    abc xyz'
Using split on /\s+/
|abc|xyz

There are 6 tokens in '    abc xyz'
Using split on / /
||||abc|xyz

There are 6 tokens in '    abc xyz'
Using split on /\s/
||||abc|xyz
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' ' is a special case, and no, there's no equivalent using a regex.