18.755 fourth problems solutions
1. By definition SL(n, R) consists of n × n real matrices of determinant 1. You may assume
that its Lie algebra sl(n, R) consists of all n×n real matrices of trace 0; it therefore has dimension
n2 −1. The conjugation action of SL(n.R) on Lie subgroups corresponds (by taking Lie algebras)
to the conjugation action of SL(n, R) on n × n matrices.
Find as many different (up to SL(2, R)-conjugation) Lie subalgebras of sl(2, R) as you can.
(Hint: there are six.) You should explain why your subalgebras are not conjugate, and (if you can) why
you have found all possibilities.
There are the two obvious subalgebras 0 and g. All others must have dimension one or two. A onedimensional subalgebra is any line in sl(2). A nonzero
2 × 2real matrix X of trace zero has either
t 0
[ 1] nonzero real eigenvals t and −t, X conj to
; or
0
−t 0 θ
[ 2] nonzero imag eigvals iθ and −iθ, X conj to
; or
−θ
0
0 ±1
[ 3] eigvals 0, X is conj to
.
0 0
These three examples generate three (obviously non-conjugate, by the eigenvalue description) one-dimensional
subalgebras. Conversely, any two lines of matrices satisfying one of these eigenvalue descriptions are (almost
obviously) conjugate by GL(2, R). To see that they are also conjugate by the smaller group SL(2, R) requires
a little extra argument. I’ll give the argument in case 2. So suppose X1 has imaginary eigenvalues iθ and
−iθ. I claim that there is an element g ∈ SL(2, R) with the property that
0 ±θ
−1
gX1 g =
.
(*)
∓θ 0
Let
z1 = u1 + iu2 ∈ C2 = R2 + iR2
be an eigenvector for X1 with the eigenvalue iθ. This means that
X1 z1 = iθz1 = −θu2 + iθu1 ,
or equivalently
X1 u1 = −θu2 ,
X1 u2 = θu1 .
Since θ 6= 0, these equations imply that u1 and u2 must be linearly independent, so they are a basis of R2 .
The matrix (u1 , u2 ) is therefore invertible, and we can define
h−1 = (u1 , u2 ),
h ∈ GL(2, R),
hui = ei ,
h−1 ei = ui .
The two preceding equations give
or equivalently
hX1 h−1 e2 = θe1 ,
hX1 h−1 e1 = −θe2 ,
hX1 h−1 =
0
−θ
θ
0
.
(**)
The difficulty is that h may not have determinant one. If we multiply h by | det(h)|−1/2 , then (∗∗) remains
true; so we may as well assume det h = ±1. (This step amounts to multiplying our original eigenvector z1
by | det(h)|−1/2 .) If det h = 1, then h ∈ SL(2, R), and (∗∗) is exactly (∗) (with a +), and we are done. So
suppose det h = −1. Define
1 0
g=
· h ∈ SL(2, R).
0 −1
1
2
Then
g −1 e1 = u1 ,
g −1 e2 = −u2 ,
and we calculate
gX1 g
−1
=
0
θ
−θ
0
(***)
which is (∗) with a −.
We are left with two-dimensional subalgebras. By computation, each of the one-dimensional examples is
its own centralizer in g; so there can be no two-dimensional abelian subalgebra. I (hope to have) proven in
class that any nonabelian two-dimensional Lie algebra has a basis {e, f } in which the bracket is [e, f ] = f .
Let us try to embed this two-dimensional Lie algebra in g. The image of e must have an eigenvalue 1 for
ad(e) (Lie bracket with e acting on g). In the three examples above, ad(X) has
eigenvalues
2t, 0, and −2t;
1/2
0
2iθ, 0, and −2iθ; and 0. So up to conjugation the only possible image of e is
, and the image
0 −1/2
0 ±1
of f must be the +1 eigenvector
. That is, up to conjugacy the only two-dimensional subalgebra
0 0
is
a b
b={
| a, b ∈ R}.
0 −a
2. For each of your Lie subalgebras h in Problem 1, find as many different CLOSED Lie
subgroups H1 , H2 , . . . of SL(2, R) as you can, each having Lie algebra h.
(Hint: the numbers of possibilities are (I think) 1, 1, 2, 3, and the continuum (twice). In the last cases
you’re not required to index the possibilities by natural numbers.)
The connected Lie subgroup H1 attached to h ⊂ g is generated by all the group elements exp(X) for
X ∈ h. If h = 0, this is the trivial subgroup, and if h = g, it is all of G. For the other cases you need to
exponentiate various 2 × 2 matrices; what comes out is
t
e
0
H1 =
|
t
∈
R
(case 1)
0 e−t
cos θ sin θ
H1 =
|θ∈R
(case 2)
− sin θ cos θ
1 x
H1 =
|x∈R
(case 3)
0 1
t
e
x
H1 =
|
t,
x
∈
R
(two-diml case)
0 e−t
Any other closed Lie subgroup Hi having h as Lie algebra must normalize H1 , and must contain H1 as
identity component. From this you can easily deduce that
Hi = preimage in NG (H1 ) of a discrete subgroup of NG (H1 )/H1 .
We therefore need to calculate each NG (H1 ). For h = g, NG (H1 ) = NG (G) = G, so the only possible Hi is
H1 .
For h = 0, NG (H1 ) = G, so Hi can be any discrete subgroup of G. I’ll return to the question of how many
of these exist at the end.
Here are the other normalizers.
r
0
0
s
×
×
NG (H1 ) =
∪
(case 1)
|
r
∈
R
|
s
∈
R
0 r−1
−s−1 0
NG (H1 ) = H1
r
NG (H1 ) =
0
r
NG (H1 ) =
0
(case 2)
x
r−1
x
r−1
| r ∈ R× , x ∈ R
(case 3)
| r ∈ R× , x ∈ R
(two-diml case)
3
In case 1, NG (H1 )/H1 = Z/4Z, which has three subgroups; so the two disconnected ones are
r
0
×
H2 =
≃ R× , H3 = NG (H1 ).
|r∈R
0 r−1
(case 1)
In case 2, NG (H1 ) = H1 , so H1 is the only Lie subgroup.
In case 3,
NG (H1 )/H1 ≃ R× .
It’s not hard to see that the non trivial discrete normal subgroups of R× are
For every real t ≥ 0 we can define three discrete subgroups of R× :
D+,t = {ent | n ∈ Z},
D−,t = {(−1)n ent | n ∈ Z},
D±,t = {±ent | n ∈ Z}.
These are all distinct (except that D−,0 = D±,0 = {±1}; and it is easy to see that they are all the discrete
subgroups of R× . The preimages in NG (H1 ) define distinct Lie subgroups with Lie algebra h: for example,
(−1)n ent
x
H−,t =
.
0
(−1)n e−nt
The number of these is the continuum.
In the two-dimensional case, NG (H1 )/H1 = Z/2Z, so there are two Lie subgroups: the connected group
H1 , and NG (H1 ).
Here are a few more words about the case h = 0, when we wish to know all the discrete subgroups of
SL(2, R) up to SL(2, R) conjugacy. It is easy to see that a discrete subgroup must be at most countable (since
G is separable), so the total number of possibilities is at most the continuum. (The number of countable
subsets of a continuum has cardinality the continuum.) On the other hand, we can easily write down a
continuum of nonconjugate discrete subgroups, like
nt
e
0
Ht =
|
n
∈
Z
(t ≥ 0).
0 e−nt
A completely different example is Γ = SL(2, Z), the matrices with integer entries. This group has many
interesting subgroups defined by congruence conditions on the entries, like
a b
Γp =
∈ Γ | a − 1, b, c, d − 1 ∈ pZ
c d
for any prime (or in fact any positive integer) p.
If M is a compact Riemann surface of genus greater than 1, then π1 (M ) is naturally a discrete subgroup
of G/{±I}, so its preimage is a discrete subgroup of G. The conjugacy class of this subgroup determines the
Riemann surface; so we get a different (conjugacy class of) discrete subgroup(s) for every Riemann surface
of genus at least two. This is an enormous family of amazing subgroups of SL(2, R). (It is quite difficult to
write down any one of these subgroups explicitly!) For each genus g ≥ 2, the family is indexed precisely by
the Teichmüller space Tg , which is a complex manifold of complex dimension 3g − 3. (I don’t understand
well the history of this assertion. Riemann knew that a compact Riemann surface of genus g ≥ 2 depends on
3g − 3 complex parameters; presumably he understood this in terms of realizing the surface as a branched
covering of the Riemann sphere. Teichmüller proved a much more precise version in a 1944 paper. Ahlfors
strengthened Teichmüller’s result in 1954.)
These discrete groups related to Riemann surfaces are in some sense the “largest” discrete subgroups of
SL(2, R). But there are many more . . .
3. In class I defined GL(n, F) to be the group of invertible n × n matrices with entries in a
(possibly noncommutative) field F. If F is commutative, I defined a subgroup
SL(n, F) = {g ∈ GL(n, F) | det(g) = 1}.
4
I discussed the fact that this definition doesn’t make sense for noncommutative F.
Find a definition of something like “SL(n, H)” inside GL(n, H). Explain what makes your
definition reasonable. Try to compute the Lie algebra and the dimension of your subgroup.
It’s a fact that you could have (probably didn’t, but could have) learned in linear algebra that for a
commutative field F,
SL(n, F) is the commutator subgroup of GL(n, F).
(It’s very easy to show that any commutator has determinant one; what requires a bit of work is to show
that any matrix of determinant one is a product of commutators.) So a reasonable definition might be
?
SL(n, H) = commutator subgroup of GL(n, H).
A weaker statement about SL(n, F) is that it’s normal subgroup of GL(n, F). So something you might
like to be true is
?
SL(n, H) = large but proper normal subgroup of GL(n, H).
Computing the commutator subgroup of GL(n, H) is fairly hard. Perhaps the best first step (also hard)
is to show that the commutator subgroup of H× is precisely the unit quaternions
U (1, H) = {z ∈ H | zz = 1} = {a + bi + cj + dk | a2 + b2 + c2 + d2 = 1}.
Using also an easy two by two calculation
−1
x 0
1 a
x
0 1
0 1
0
0
1
1
0
−a
1
=
1
0
(x − 1)a
1
,
x
0
is a commutator, we find that the commutator subgroup
0 x−1
of GL(n, H) must include all the upper triangular matrices
and a slightly messier one to prove that

a11


 0





0
a12
a22
···
···
..
.
0
···


a1n



a2n 
 | a11 · a22 · · · ann ∈ U (1, H) ,




ann
and also the corresponding group of lower triangular matrices. It turns out that these two subgroups generate
the full commutator subgroup; but using this kind of calculation at the level of the group, it is not so easy
to calculate the dimension or the Lie algebra.
You know (because GL(n, H) is open in n × n quaternionic matrices) that the Lie algebra is
gl(n, H) = M (n, H),
all square quaternionic matrices, with Lie bracket given by commutator of matrices. A fact that I have not
yet proved in class is that the Lie algebra of the commutator subgroup of G is the commutator subalgebra
[g, g] of the Lie algebra; that is, the linear span of all Lie brackets.
Computing commutators of matrices is easy. In the 1 × 1 case,
[H, H] = {z | z = −z} = {bi + cj + dk},
the three-dimensional space of imaginary quaternions.
For larger n, the set of “matrix units” epq (having a single one in the pth row and the qth column) is an
H basis for M (n, H). Because of facts like
[epq , ers ] = δqr eps − δsp erq ,
5
we find that
[gl(n, H), gl(n, H)] = {(apq ) | a11 + · · · + ann ∈ [H, H]},
the matrices of purely imaginary trace. Using the natural inclusion
J: M (n, H) ֒→ M (4n, R),
this is
[gl(n, H), gl(n, H)] = {A | tr(J(A)) = 0} = J −1 (sl(4n, R)).
These two formulas make it reasonable to define
sl(n, H) = [gl(n, H), gl(n, H)] = J −1 (sl(4n, R)).
This is a Lie algebra of dimension 4n2 − 1 (one less than the dimension of GL(n, H). Using the dictionary
between Lie groups and Lie algebras (stated in class, although not yet proven) we can define
SL(n, H) = connected Lie subgroup of GL(n, H) with Lie algebra sl(n, H).
Here is one way to prove a bit more about this subgroup. In class I talked about the Lie group inclusion
j: GL(n, H) → GL(4n, R).
This gives us a Lie group homomorphism
detR : GL(n, H) → R× ,
detR (g) = det(j(g)).
The kernel of detR is automatically a closed subgroup of GL(n, H); of course it is
ker detR = j −1 (SL(4n, R)).
The Lie groups/Lie algebra dictionary tells you that
Lie(ker detR ) = J −1 (sl(4n, R)) = sl(n, H)).
So the kernel is a closed Lie subgroup whose identity component is SL(n, H). Because the identity component
of a closed Lie subgroup is automatically closed, it follows that SL(n, H) is closed.
It’s true that SL(n, H) is precisely the kernel of detR . (Probably that’s not so hard to prove, but I haven’t
thought it through.) It’s also true that the image of detR is precisely the positive reals. The center Z of
GL(n, F) is always the center of F × times the identity;
Z = R× = {±1} × R>0 = {±1} × Z0 .
Minus one is a commutator in H (of i and j, for example), and so belongs to SL(n, H). The conclusion is
that
GL(n, H) = SL(n, H) × Z0 ,
the second factor consisting of positive multiples of the identity. The first factor is the volume-preserving
quaternionic linear transformations of Hn ; these are automatically orientation-preserving. Those characterizations are another reason that this is a reasonable definition of SL(n, H).