The construct [ @list ] creates a new array reference, into which @list is expanded (i.e. copied), so that's bloat.

Are there any potentially adverse consequences of holding on to references to static variables?

No, since the my @list isn't static. That container will be empty at each subroutine invocation, and it's contents will live in a brand new reference which you can take with \@list...

use strict;
use warnings;

sub foo {
    my @list if 0;
    print "list: (@list)\n";
    push @list, @_;
    \@list;
}

my $ref1 = foo(1..3);
push @$ref1, 'a', 'b';
my $ref2 = foo(4..6);
print "first ref: (@$ref1)\n";
print "last list: (@$ref2)\n";
__END__
list: ()
list: ()
first ref: (1 2 3 a b)
final list: (4 5 6)
[download]

...except if you play tricks to the compiler (allocate a private @list on the scratchpad, but don't generate the opcode to clear that list) like so:

use strict;
use warnings;

sub foo {
    my @list if 0;
    print "list: (@list)\n";
    push @list, @_;
    \@list;
}

my $ref1 = foo(1..3);
push @$ref1, 'a', 'b';
my $ref2 = foo(4..6);
print "first ref: (@$ref1)\n";
print "final list: (@$ref2)\n";
__END__
list: ()
list: (1 2 3 a b)
first ref: (1 2 3 a b 4 5 6)
last list: (1 2 3 a b 4 5 6)
[download]

But that's deprecated. For statics (previous perl 5.10) you'd use a block around the sub

use strict;
use warnings;

{
    my @list;
    sub foo {
        print "list: (@list)\n";
        push @list, @_;
        \@list;
    }
}

my $ref1 = foo(1..3);
push @$ref1, 'a', 'b';
my $ref2 = foo(4..6);
print "first ref: (@$ref1)\n";
print "final list: (@$ref2)\n";
[download]

which produces the same output like the infamous 'my if 0' hack above, but you get at a glance what is going on: the @list is allocated in the same scope into which sub foo has been compiled, but @list isn't re-initialized inside the sub.