18.755 eighth problems solutions
As background for the first problem, suppose that V is an n-dimensional vector space (can be over any
field k of characteristic not two) endowed with a nondegenerate quadratic form Q. In class I defined a
complete orthogonal flag to be a chain of subspaces
0 = V0 ⊂ V1 ⊂ · · · ⊂ Vn−1 ⊂ Vn = V,
dim Vj = j,
Vj⊥ = Vn−j .
I stated that such flags exist if k is algebraically closed, and that the orthogonal group O(V ) acts transitively
on them.
To say more, write
n = 2m + ǫ
(ǫ = 0 or 1).
Because of the relations
dim(W ⊥ ) = n − dim W,
W is isotropic ⇐⇒ W ⊂ W ⊥ ,
we see that an isotropic subspace must have dimension at most m.
Definition. A Lagrangian subspace of V is an m-dimensional isotropic subspace. Define L(V ) to be the
set of all Lagrangian subspaces of V . This is (pretty obviously) closed in the Grassmannian Grm (V ) of all
m-dimensional subspaces of V . (To make sense of this as a statement about smooth manifolds, we need to
have k equal to C. But as a statement about algebraic varieties, it makes sense for any k.)
Essentially I stated in class
Proposition. A complete orthogonal flag in V is the same thing as
(1) a Lagrangian subspace Wm ⊂ V , and
(2) a complete flag W0 ⊂ W1 ⊂ · · · ⊂ Wm in Wm .
General hint: problems 2, 3, and 5 are meant to be quite easy, and 6 is pretty easy. Numbers 1 and 4 test
whether you understand the Gram-Schmidt process; they may take a little longer to write down. I include
problem 7 just because I’m mean.
1. Suppose n = 2m is even, and that W is an m-dimensional Lagrangian subspace of a
2m-dimensional V as above. Choose any basis (w1 , . . . , wm ) for W , and let (λ1 , . . . , λm ) be the
dual basis of W ∗ (see the solutions to Problem Set 7). Prove that there are unique vectors
∗
(w1∗ , . . . , wm
) satisfying
Q(wj , wk∗ ) = δjk ,
Q(wj∗ , wk∗ ) = 0
(1 ≤ j, k ≤ m).
As I said in an email to the class, the unique statement is wrong. Here is a way to prove the existence.
Any bilinear form Q on a vector space V may be regarded as a linear map
q: V → V ∗ ,
q(v)(w) = Q(w, v).
To say that Q is nondegenerate is to say that q is one-to-one; so for finite-dimensional V , q must be an
isomorphism. Because any linear map defined on a subspace can be extended to any larger vector space,
W ∗ consists of the restrictions to W of elements of V ∗ . That is, the restriction map V ∗ → W ∗ is surjective.
From these two facts it follows that every linear functional on W is q(v) for some v ∈ V .
2
Now choose (v1 , . . . , vm ) so that q(vj )|W = λj . This means that
Q(wj , vk ) = δjk .
The vk satisfy the first half of the requirements for wk∗ , but not necessarily the second. We need to change
our choices of the vk . I’m going to do that by choosing various scalars aik and defining
wj∗ = vj +
j
X
aij wi .
i=1
Because the wi are all orthogonal to each other, it’s obvious that any such choice of wj∗ satisfies Q(wj , wk∗ ) =
δjk . If you write down the equations Q(wj∗ , wk∗ ) = 0 for j ≤ k, you will see that they are exactly
ajk = Q(vj , vk ) (j < k),
akk = Q(vk , vk )/2.
(The uniqueness of this solution depended on choosing the matrix (aij ) to be upper triangular. If I had not
done that, then I would have gotten the equations
ajk + akj = Q(vj , vk );
so I can add to the solution above any skew-symmetric m × m matrix.)
For the next problems we’re back in the world of Lie groups. Recall that I stated in class the existence of
an exact sequence for computing fundamental groups of homogeneous spaces:
π1 (H) → π1 (G) → π1 (G/H) → π0 (H) → π0 (G) → π0 (G/H) → 0.
Here H is any closed subgroup of a Lie group G. You can also use a generalization: if H1 ⊂ H2 ⊂ G are
closed subgroups, then
π1 (H2 /H1 ) → π1 (G/H1 ) → π1 (G/H2 ) → π0 (H2 /H1 ) → π0 (G/H1 ) → π0 (G/H2 ) → 0.
2. Prove that if n ≥ 3, then the natural map
π1 (SO(n − 1)) → π1 (SO(n))
induced by the inclusion SO(n − 1) ֒→ SO(n) is onto. (You can use things done in class about spheres.)
We showed in class that SO(n)/SO(n − 1) = S n−1 (an n − 1-dimensional sphere) as long as n ≥ 2; and
we showed that S n−1 is simply connected for n ≥ 3. So for n ≥ 3, the first long exact sequence recalled
before the problem is
π1 (SO(n − 1)) → π1 (SO(n)) → 1 → · · · .
This is exactly the conclusion desired.
3. Suppose H is any closed connected subgroup of SO(n) containing SO(2). Prove that G/H
is simply connected.
By induction on n in Problem 2, the map induced by inclusion π1 (SO(2)) → π1 (SO(n)) is surjective for
n ≥ 2. If H ⊃ SO(2), any loop in SO(2) is a loop in H; so the image of π1 (H) in π1 (SO(n)) contains the
image of π1 (SO(2)), and is therefore all of π1 (SO(n)). Now the long exact sequence (here G = SO(n))
π1 (H) → π1 (G) → π1 (G/H) → π0 (H) · · ·
3
has a trivial map coming into π1 (G/H) (because of the surjectivity just proved) and a trivial map going out
(connectedness of H means that π0 (H) is a single point). So π1 (G/H) has to be trivial by the exactness.
4. Suppose VR is an n-dimensional real vector space with a positive definite quadratic form
QR , and that VC is its complexification. I proved in class that any vector w ∈ VC of length zero
must be of the form
w = u + iv,
(u, v ∈ VR , QR (u, v) = 0, QR (u, u) = QR (v, v)).
Suppose that Wm is an m-dimensional Lagrangian subspace of VC , and that W0 ⊂ · · · ⊂ Wm is a
′
complete flag in Wm . Prove that there is a basis (w1′ , . . . , wm
) of W such that
(1)
(2)
(3)
Prove
Wj = hw1′ , . . . , wj′ i,
√
wj′ = (uj + ivj )/ 2, and
(u1 , v1 , . . . , um , vm ) is an orthonormal basis of VR .
furthermore that the m oriented planes
Pj = huj , vj i ⊂ VR
are uniquely determined by the flag.
A simple way to talk about this is using the Hermitian form
H(u1 + iv1 , u2 + iv2 ) =def QC (u1 + iv1 , u2 − iv2 ) = QR (u1 , u2 ) + QR (v1 , v2 ) + i(−QR (u1 , v2 ) + QR (v1 , u2 ))
on the complex vector space VC .
Begin with any basis (w1 , . . . , wm ) compatible with the flag, and use Gram-Schmidt to convert into a
′
basis (w1′ , . . . , wm
) that is orthonormal for H. The nature
√ of Gram-Schmidt means that the new basis is
still compatible with the flag. If we write wj′ = (uj + ivj )/ 2, then the statement that H(wj , wj ) = 1 means
that
Q(uj , uj ) + Q(vj , vj ) = 2.
The statement that wj is isotropic means that uj and vj are orthogonal vectors of the same length; so
Q(uj , uj ) = Q(vj , vj ) = 1,
Q(uj , vj ) = 0.
For j 6= k, that wj′ and wk′ are orthogonal for H (for j 6= k) means
Q(wj′ , uk − ivk ) = 0.
That they are orthogonal for Q (part of the definition of Lagrangian) means
Q(wj′ , uk + ivk ) = 0.
Taking linear combinations, we see that wj′ = uj + ivj is Q-orthogonal to both uk and vk ; that is, that uj
and vj are both orthogonal to both uk and vk . This proves (1)–(3).
For the uniqueness statement, what the problem asks for is an H-orthonormal basis of Wm compatible
with the flag. In any m-dimensional complex vector space with a positive-definite Hermitian form H and
a flag (Wj ), an H-orthonormal basis (wj′ ) compatible with the flag is clearly unique up to multiplication of
each basis vector wj′ by some eiθj . Those multiplications amount to rotation of the planes Pj by θj .
4
5. Give an example of a Lie group G with a closed subgroup H so that G/H is compact, but
there is no compact subgroup K of G such that K/K ∩ H = G/H.
Easiest example I know is G = R (in which the only closed subgroups are R and tZ for t ≥ 0, so the only
compact subgroup is the identity) and H = 2πZ; then G/H is the compact circle.
6. Same problem as above, but this time require that



 1 x z

G =  0 1 y  | x, y, z ∈ R .


0 0 1
Define

 1 x
H = 0 1

0 0


z

y  | x, y, z ∈ Z .

1
It’s easy to check that H is a subgroup. A little care is required to show that G/H is compact: just because
G is topologically R3 and H is topologically Z3 , it doesn’t follow that G/H is topologically a product of
three circles; and in fact it is not. (Can you see a proof of that?)
Here is a dumb proof that G/H is compact. I’ll prove that every sequence in G/H has a convergent
subsequence. Suppose gn H is a sequence in G/H, with gn having coordinates (xn , yn , zn ). Choose integers
pn , qn , mn so that
xn − pn ∈ [0, 1), yn − qn ∈ [0, 1), zn − qn xn − mn ∈ [0, 1).
Then

1 xn
0 1
0 0

zn
1 −pn
yn   0
1
1
0
0
 
−mn
1
−qn  =  0
1
0
xn − pn
1
0

zn − qn xn − mn

yn − qn
1
; so we can choose the coset representatives (even uniquely, but that doesn’t matter) to have all entries in
[0, 1). Now there is obviously a convergent subsequence (by the compactness of [0, 1]3 ).
7. Is there a connected Lie group G with a closed connected subgroup H such that
(1) G/H is compact, but
(2) there is no compact subgroup K of G such that K/K ∩ H = G/H?
I still don’t know an example or a proof that it’s impossible; I’ll be interested to see what you have to
say!