18.755 tenth and last problem solutions
0 1
, and Σr to be the 2r × 2r matrix with r diagonal blocks σ and all other
1 0
entries zero. Finally define


Ip
0
0
J(p, q, r) =  0 −Iq
0 ,
0
0
Σr
1. Define σ =
an integer matrix of size n = p + q + 2r with square equal to the identity. Prove that if there is
an integer matrix g of determinant ±1 such that
gJ(p, q, r)g −1 = J(p′ , q ′ , r′ ),
then p = p′ , q = q ′ , and r = r′ . (Hint: in the correspondence between tori and lattices discussed
in class, the matrix J(p, q, r) defines an automorphism j(p, q, r) of U (1)n .)
The torus U (1)n consists of n-tuples (t1 , . . . , tn ) with each ti ∈ U (1) a complex number of absolute value
1. The morphism of Lie groups defined by an integer matrix multiplies together various powers (given by
the matrix) of the ti . In particular,
−1
j(p, q, r)(t1 , . . . , tp+q+2r ) = (t1 , . . . , tp , t−1
p+1 , . . . , tp+q , tp+q+2 , tp+q+1 , . . . , tp+q+2r , tp+q+2r−1 ) :
the first p coordinates are left unchanged, the next q are inverted, and each of the last r pairs of coordinates
is exchanged. Consequently the group of fixed points of the automorphism is
[U (1)n ]j(p,q,r) = (s1 , . . . , sp , ǫ1 , . . . , ǫq , u1 , u1 , . . . , ur , ur )
(si , uk ∈ U (1), ǫj ∈ {±1}.
Consequently
[U (1)n ]j(p,q,r) ≃ U (1)p+r × {±1}q .
If the two matrices J(p, q, r) and J(p′ , q ′ , r′ ), then (obviously) p+q+2r = p′ +q ′ +2r′ , and (by the equivalence
of categories between tori and lattices)
′
′
′
[U (1)n ]j(p,q,r) ≃ [U (1)n ]j(p ,q ,r ) .
According to the calculation just made, such an isomorphism of Lie groups forces p + r = p′ + r′ (the
dimensions of the two Lie groups) and q = q ′ (the dimensions over F2 of the two groups of connected
components). The three equations
p + q + 2r = p′ + q ′ + 2r′ ,
p + r = p′ + r ′ ,
q = q′
can be solved (subtract second plus third from first) to yield p = p′ , q = q ′ , and r = r′ .
2. How many different compact connected Lie groups have Lie algebra su(n) ⊕ R?
The answer is the sum of the divisors of n. Here is a list of what the groups look like. Recall that the
center of SU (n) is
Z(SU (n)) = µn In ,
µn = nth roots of 1 ⊂ C× .
We’ll write
ω = exp(2πi/n)In
1
2
for the usual generator of this cyclic group of order n. For each divisor d of n, write
SU (n)(n/d) = SU (n)/{µ(n/d)In },
the quotient of SU (n) by the group of (n/d)th-roots of one. Write
ω = exp(2πi/n)In
for the image of ω in the quotient group SU (n)(n/d) ; then ω has order exactly d. For any divisor d1 of d,
and any primitive d1 th root τ1 of 1, we can define
G(d, d1 , τ1 ) = [SU (n)(n/d) × U (1)]/h(ω d/d1 , τ1 )i.
The group by which we are dividing is cyclic of order d1 . Given d, the total number of choices for the pair
(d1 , τ1 ) is
X
φ(d1 ) = d :
d1 |d
the number of primitive d1 th roots is the Euler φ function φ(d1 ), and summing this over divisors of d gives
d. (I have not yet shown that the groups are distinct; at
Pthe moment I’m just counting parameters.) So for
each d dividing n we get d pairs (d1 , τ1 ), and therefore d|n d names G(d, d1 , τ1 ). What we’ll see is that the
d groups G(d, d1 , τ1 ) are (distinct and) exactly the groups with derived group SU (n)(n/d) .
So let G be a compact connected group with Lie algebra su(n) ⊕ R. The derived group G′ has Lie algebra
su(n), and so is a quotient of the (simply connected) group SU (n) by a subgroup of its center µn . Since µn
is cyclic, there is one such subgroup µ(n/d) for each divisor d of n; so let us assume that
G′ ≃ SU(n/d) ,
Z(G′ ) = hωi ≃ µd .
The identity component of Z(G) is a compact Lie group with Lie algebra R, and so is isomorphic to U (1):
Z(G)0 ≃ U (1).
Multiplication defines a group homomorphism
m: SU (n)(n/d) × U (1) → G;
since the differential is an isomorphism of Lie algebras, this map is a covering, so its kernel is a discrete
subgroup
Λ ⊂ Z(SU (n)(n/d) × U (1)) = µd × U (1).
Because m restricted to either factor is an injection, we have
Λ ∩ [µd × {1}] = Λ ∩ [1 × U (1)] = 1.
A nice fact to know is
3
Lemma. Suppose Λ ⊂ H1 × H2 is a subgroup. Define Si = projection of Λ on Hi . If
Λ ∩ [1 × H2 ] = 1,
then Λ is the graph of a group homomorphism ℓ: S1 ։ S2 . If also
Λ ∩ [H1 × 1] = 1,
then ℓ is an isomorphism.
Using this (easy) fact, we see that the only possibilities for Λ are the ones appearing in the definition of
G(d, d1 , τ1 ): the parameter d1 describes the subgroup S1 (necessarily cyclic of order a divisor of d); and τ1
specifies the isomorphism of this subgroup into the unit circle.
Essentially this argument proves also that the groups G(d, d1 , τ1 ) are all distinct. To be a bit more precise,
it proves that if we have an isomorphism
j: G(d, d1 , τ1 ) → G(d′ , d′1 , τ1′ )
and if the differential of j is the identity map on su(n) ⊕ R, then d = d′ , d′1 = d′1 , and τ1 = τ1′ . But what if
the map on Lie algebras need not be the identity? We can still see d as the order of the center of the derived
group, so d = d′ . A bit more thought reveals d/d1 as the order of the center of [G/Z(G)0 ]′ , so d1 = d′1 .
The kernel of the exponential map on the R factor of the Lie algebra is arranged to be Z (or perhaps
2πZ) in all of our examples; so the only thing that our Lie algebra isomorphism dj might do on R is act by
−1. Of course that does define a Lie algebra isomorphism, and the map on Lie groups identifies
G(d, d1 , τ1 ) ≃ G(d, d1 , τ1−1 ).
(EXTRA)
So if we don’t care about how the Lie algebras are identified, the number of possibilities is (almost) cut in
half. (When d1 is 1 or 2, then τ1 = ±1 is its own inverse.)
The possible isomorphisms of su(n) to itself are just conjugations from SU (n) (which can’t change
G(d, d1 , τ1 ) and the inverse transpose automorphism (which sends ω to ω −1 , and so gives another implementation of the isomorphism (EXTRA)).
3. Let X ∗ be the lattice Z4 , with dual lattice X∗ = Z4 as usual. The root datum for
(U (4), U (1)4 ) is
R0 = {ei − ej |1 ≤ i 6= j ≤ 4}, R0∨ = {ei − ej |1 ≤ i 6= j ≤ 4}.
Find all root data (X ∗ , R, X∗ , R∨ ) so that R ⊃ R0 and R∨ ⊃ R0∨ .
A useful fact is that the reflection in ei − ej is the transposition of the ith and jth coordinates; so the
Weyl group W (R0 ) (generated by the reflections in the roots of R0 ) is the symmetric group S4 . It follows
that the set of added roots must be preserved by permutation of the four coordinates.
I’ll try to write α for one of the roots in R0 and β for one of the potentially added roots. I proved in class
that if γ 6= ±δ are roots in a root system, then we are in one of the following cases.
(1) hγ, δ ∨ i = hδ, γ ∨ i = 0 (the A1 × A1 case). In this case
R ⊃ {±γ, ±δ}.
(2) hγ, δ ∨ i = ǫ = ±1 and hδ, γ ∨ i = ǫ (the A2 case). In this case
R ⊃ {±γ, ±δ, ±(γ − ǫδ)}.
4
(3) hγ, δ ∨ i = ǫ = ±1 and hδ, γ ∨ i = ǫ · 2 (the B2 case). In this case
R ⊃ {±γ, ±δ, ±(γ − ǫδ), ±(2γ − ǫδ)}.
(4) hγ, δ ∨ i = ǫ = ±1 and hδ, γ ∨ i = ǫ · 3 (the G2 case). In this case
R ⊃ {±γ, ±δ, ±(γ − ǫδ), ±(2γ − ǫδ), ±(3γ − ǫδ), ±(3γ − 2ǫδ)}.
∨
(5) hγ, δ i = ǫ = ±1 and hδ, γ ∨ i = ǫ · 4 (the BC1 case). In this case
R ⊃ {±γ, ±2γ}.
(6) same as (3)–(5) with γ and δ interchanged.
In case (5), δ = 2ǫγ and δ ∨ = ǫγ ∨ /2.
In the present problem, α/2 and α∨ /2 are not in the specified lattices Z4 , so we cannot get new roots by
adjoining any α/2 or 2α. So a potential new root β must fall in one of the cases (1)–(4) or (6) with respect
to every α ∈ R0 .
If (β, β ∨ ) is a new (root,coroot) pair, then the conditions (1)–(4) (applied to the pair (β, ei − ej )) say that
for every 1 ≤ i 6= j ≤ 4, βi − βj and βi∨ − βj∨ have the same sign (or are both zero). This means that the
coordinates of β and β ∨ must be in exactly the same order. Conditions (2)–(4) and (6) say furthermore that
in case the coordinate differences are nonzero, then one difference is one and the other is one, two, or three.
If a new root β has all coordinates equal, then the same must be true of β ∨ ; so hβ, β ∨ i is divisible by four,
and so not equal to two, which is impossible. So the coordinates of β are not all equal. For convenience we
apply W (R0 ) = S4 in order to put β (and therefore also β ∨ ) in decreasing order. Again by the preceding
paragraph, either the largest difference β1 − β4 = 1, or β1∨ − β4∨ = 1. In particular, this means (in the first
case) that there are exactly two different coordinates of β; so by the preceding paragraph, there are also
exactly two different coordinates of β ∨ . In the second case, we reach the same conclusion in the opposite
order. We have found
Lemma. Suppose (β, β ∨ ) is a (root,coroot) pair that can be added to R0 , with coordinates in decreasing
order. Then there are integers a and b such that either
β ∨ = (b, . . . , b, b − m, . . . , b − m),
β = (a, . . . , a, a − 1, . . . , a − 1),
| {z } |
{z
}
{z
}
| {z } |
p terms 4 − p terms
p terms
4 − p terms
or the same with β and β ∨ interchanged. Here p = 1, 2, or 3, and m = 1, 2, or 3 (in accordance with the
cases listed above).
So our first task is to solve the diophantine equation
2 = hβ, β ∨ i = pab + (4 − p)(a − 1)(b − m)
for integers a, b, p, and m; p and m are required to be 1, 2, or 3, and a and b can be arbitrary. Replacing
β by −β has the effect of exchanging p and 4 − p; so we only need to look at p = 1 and p = 2. So there are
six cases of p and m.
Case p = 1 and m = 1.
The equation is
2 = ab + 3(a − 1)(b − 1) = 4ab − 3a − 3b + 3,
which we rewrite as
(2a − 3/2)(2b − 3/2) − 5/4 = 0,
or equivalently
(4a − 3)(4b − 3) = 5.
The only factorizations of 5 are 5 · 1 and (−5) · (−1), The first leads to solutions {a, b} = {2, 1}; the second
leads to no solutions since −5 and −1 are not −3 modulo 4. We get
{β, β ∨ } = {(2, 1, 1, 1), (1, 0, 0, 0)}.
5
Case p = 2 and m = 1.
The equation is
2 = 2ab + 2(a − 1)(b − 1) = 4ab − 2a − 2b + 2,
which we rewrite as
(2a − 1)(2b − 1) − 1 = 0,
or equivalently
(2a − 1)(2b − 1) = 1.
The only factorizations of 1 are 1 · 1 and (−1) · (−1), The first leads to solutions a = b = 1; the second leads
to a = b = 0. We get
β = β ∨ = (1, 1, 0, 0), or β = β ∨ = (0, 0, −1, −1).
Case p = 1 and m = 2.
The equation is
2 = ab + 3(a − 1)(b − 2) = 4ab − 6a − 3b + 6,
which we rewrite as
(2a − 3/2)(2b − 3) − 1/2 = 0,
or equivalently
(4a − 3)(2b − 3) = 1.
The only factorizations of 1 are 1 · 1 and (−1) · (−1), The first leads to solutions {a, b} = {1, 2}; the second
leads to no solutions since −1 is not −3 modulo 4. We get
{β, β ∨ } = {(1, 0, 0, 0), (2, 0, 0, 0)}.
According to (3) of the list of cases considered in class (and listed at the beginning of this solution), taking
γ = β = (1, 0, 0, 0) and δ = (−1, 1, 0, 0), we find that in this case our enlarged root system must also include
the pair
{2γ + δ, γ ∨ + δ ∨ } = {(1, 1, 0, 0), (1, 1, 0, 0)}.
Case p = 2 and m = 2.
The equation is
2 = 2ab + 2(a − 1)(b − 2) = 4ab − 4a − 2b + 4,
which we rewrite as
(2a − 1)(2b − 2) = 0,
or equivalently
(2a − 1)(b − 1) = 0.
The first factor is odd and in particular nonzero, so the solutions are (a, b) = (a, 1). The corresponding roots
and coroots are
{β, β ∨ } = {(a, a, a − 1, a − 1), (1, 1, −1, −1)}.
Arguing as in the preceding case with γ = β and δ = (0, −1, 1, 0), we must also have the pair
{2γ + δ, γ ∨ + δ ∨ } = {(2a, 2a − 1, 2a − 1, 2a − 2), (1, 0, 0, −1)}.
But this additional added root has three distinct coordinates, which we said earlier is impossible; so this
case cannot arise.
6
Case p = 1 and m = 3.
The equation is
2 = ab + 3(a − 1)(b − 3) = 4ab − 9a − 3b + 9,
which we rewrite as
(2a − 3/2)(2b − 9/2) + 1/4 = 0,
or equivalently
(4a − 3)(4b − 9) = −1.
The only factorizations of −1 are 1 · (−1) and (−1) · 1. The first leads to solutions (a, b) = (1, 2); the second
leads to no solutions since −1 is not −3 modulo 4. We get
{β, β ∨ } = {(1, 0, 0, 0), (2, −1, −1, −1)}.
Applying the statement from class (case 4 above) to the pair γ = β and δ = (−1, 1, 0, 0), we must also have
the pair
{3γ + δ, γ ∨ + δ ∨ } = {(2, 1, 0, 0), (0, 1, 0, 0)}.
Again this additional added root has three distinct coordinates, which we said earlier is impossible; so this
case cannot arise.
Case p = 2 and m = 3.
The equation is
2 = 2ab + 2(a − 1)(b − 3) = 4ab − 6a − 2b + 6,
which we rewrite as
(2a − 1)(2b − 3) + 1 = 0.
The only factorizations of −1 are 1 · −1 and (−1) · 1. The first leads to solutions (a, b) = (1, 1); the second
leads to (a, b) = (0, 2). We get
{β, β ∨ } = {(1, 1, 0, 0), (1, 1, −2, −2)},
or
{β, β ∨ } = {(0, 0, −1, −1), (2, 2, −1, −1)}.
Arguing as above, we see that there must be an additional new root (in the first case)
3(1, 1, 0, 0) + (0, −1, 1, 0) = (3, 2, 1, 0),
which has four distinct coordinates and so cannot occur. So this case also is not possible.
Taking into account the fact that the new roots must be permuted by W (R0 ) = S4 , and must be closed
under taking negatives, we have proved
Proposition. In the setting of the problem, the added (root,coroot) pairs must be a union of some of the
sets listed below.
(1)
(2)
(3)
(4)
(5)
S4 · (±(2, 1, 1, 1), ±(1, 0, 0, 0))
S4 · (±(1, 0, 0, 0), ±(2, 1, 1, 1))
S4 · (±(1, 1, 0, 0), ±(1, 1, 0, 0))
S4 · (±(1, 0, 0, 0), ±(2, 0, 0, 0))
S4 · (±(2, 0, 0, 0), ±(1, 0, 0, 0))
(8 pairs);
(8 pairs);
(12 pairs);
and S4 · (±(1, 1, 0, 0), ±(1, 1, 0, 0)) (20 pairs);
and S4 · (±(1, 1, 0, 0), ±(1, 1, 0, 0)) (20 pairs).
7
Now we need to see what root data can be assembled by adjoining some of these sets of candidate pairs
(β, β ∨ ).
In case (4) (or case (5)) the simple reflection attached to the root (1, 0, 0, 0) (or (2, 0, 0, 0)) changes the sign
of the first coordinate; so if we include such roots, necessarily the Weyl group W (R) includes all permutations
and sign changes of all the coordinates. In particular, this means that with any root (or coroot) (2, 1, 1, 1)
we would have also the (illegal) root (or coroot) (2, 1, 1, −1). The conclusion is that we cannot include roots
from sets (1) or (2) with roots from sets (4) or (5).
In case (3), the simple reflection attached to (1, 1, 0, 0) is
s(1,1,0,0) (x, y, z, w) = (−y, −x, z, w),
which carries (2, 1, 1, 1) to the (illegal) root or coroot (−1, −2, 1, 1). The conclusion is that we cannot include
roots from sets (1) or (2) with the roots in set (3).
In case (2), the simple reflection attached to (2, 1, 1, 1) carries (1, 0, 0, 0) to (−1, −1, −1, −1), which cannot
be a root. The conclusion is that we cannot use the sets (1) and (2) together.
We are left with candidates for added (root,coroot) pairs (labelled as in the proposition):
∅;
(1);
(2);
(3);
(4);
(5);
(4) ∪ (5).
Adding (3) or (4) or (5) gives the root data for the compact groups SO(8), SO(9), and Sp(4), so they
actually occur; you can essentially find these examples in Kirillov’s book or on the web. I will leave to you
the interesting exercise of showing that adding (1) gives the root datum for P U (5), the rank 5 unitary group
divided by its center; and adding (2) the root datum for SU (5). (These are not so easy to recognize as any
of the root data you might find written down, because the chosen basis of the lattice is not a familiar one.)
The final candidate is not reduced, so you might correctly have excluded it on those grounds. Nevertheless
it is a (non-reduced) root datum, called BC4.