Properties of Irreducible Root Systems I
Now we can turn towards the project of classifying irreducible root systems up to isomorphism. And we start with some properties of irreducible root systems.
First off, remember a root system is reducible if we can write it as the disjoint union of two collections of roots
so that each root in
is perpendicular to each one in
. I assert that, for any base
,
is reducible if and only if
can itself be broken into two collections
in just the same way. One direction is easy: if we can decompose
, then the roots in
are either in
or
and we can define
and
.
On the other hand, if we write with each simple root in
perpendicular to each one in
, then we will find a similar decomposition of
. But we know from our study of the Weyl group that every root in
can be sent by the Weyl group to some simple root in
. So we define
to be the collection of roots whose orbit includes a point of
, and
to be the collection of roots whose orbit includes a point of
. So a vector in
is of the form
for some Weyl group element
and some simple root
. The Weyl group element
can be written as a sequence of simple reflections. A simple reflection corresponding to a root in
adds some multiple of that root to the vector, and thus leaves the subspace spanned by
invariant; while a simple reflection corresponding to a root in
leaves the subspace spanned by
fixed point-by-point. Thus any root in
must lie in the subspace spanned by
, and similarly any root in
must lie in the subspace spanned by
. This shows that
is exactly the sort of decomposition we’re looking for.
If is irreducible, with base
, then there is a unique maximal root
relative to the partial ordering
on roots. In particular, the height of
is greater than the height of any other root in
,
for all simple roots
, and in the unique expression
all of the coefficients are strictly positive.
First of all, if is maximal then it’s clearly positive, and so each
is either positive or zero. Let
be the collection of simple roots
so that
, and
be the collection of simple roots
so that
. Then
is a partition of the base. From here we’ll assume that
is nonempty and derive a contradiction.
If , then
for any other simple root
. From this, we can calculate
Since is irreducible, there must be at least one
and
so that
, and so we must have
for this
. This proves that
must also be a root, which contradicts the maximality of
. And so we conclude that
is empty, and all the coefficients
. In passing, we can use the same fact to show that
for all
, or else
wouldn’t be maximal.
This same argument applies to any other maximal root , giving
for all
with the inequality strict for at least one
. We calculate
which tells us that is a root unless
. But if
is a root, then either
or
, contradicting the assumption that both are maximal. Thus
must be unique.
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