18.755 seventh problems solutions
This problem set is based on the notes “Associative algebras, Lie algebras, and bilinear forms” on the
class web site.
First, a definition. Suppose W is an n-dimensional real vector space. Write W ∗ for the dual vector space
of W (consisting of all linear maps from W to R). Set
V = W ⊕ W ∗ = {(w, λ) | w ∈ W, λ ∈ W ∗ }.
Define a symmetric bilinear form on V by
Q((w1 , λ1 ), (w2 , λ2 )) = λ2 (w1 ) + λ1 (w2 ).
1. The orthogonal group O(V ) is isomorphic to one of the groups O(p, q) defined in Problem
Set 6. Which one?
The answer is O(n, n). One way to prove this is to fix a basis (w1 , . . . , wn ) of W , and to define (λ1 , . . . , λn )
to be the dual basis of W ∗ . This means that the linear functionals λj are defined by
λj (wi ) = δij .
Now define
√
ei = (wi + λi )/ 2,
√
fi = (wi − λi )/ 2.
Then we compute
Q(ei , ej ) = [λj (ei ) + λi (ej )]/2 = 2δij /2 = δij ,
Q(ei , fj ) = [−λj (ei ) + λi (ej )]/2 = [−δij + δij ]/2 = 0,
Q(fi , fj ) = [−λj (ei ) − λi (ej )]/2 = −2δij /2 = −δij .
Therefore (e1 , . . . , en , f1 , . . . , fn ) is an orthogonal basis of V with n vectors of length +1 and n of length
−1. That exhibits the isomorphism of V with Rn,n .
V
2. Let E(W ) be the algebra of endomorphisms of (W ) defined in Theorem 3.4 of the notes.
(This means that E(W ) is the linear span of all possible products
of various m(w) and ι(λ).)
V
Prove that E(W ) is the algebra of all linear transformations of (W ).
Let (w1 , . . . , wn ) be a basis of W and (λ1 , . . . , λn ) the dual basis of W ∗ . By definition E(W ) is generated
by all the operators m(w) and ι(λ) defined in Theorem 3.4 of the notes. Because these obviously depend
linearly on w and λ, we see that E(W ) is spanned by products of various m(wi ) and ι(wj ). Using the
anti-commutation relation
ι(λ)m(w) = −m(w)ι(λ) + λ(w),
from the same theorem, we see that we may rearrange any product so that all the m(wi ) precede all the
ι(λj ). Using further the relations
m(wi )2 = 0,
m(wi2 )m(wi1 ) = −m(wi1 )m(wi2 )
and similarly for ι (also stated there) we deduce that E(W ) is spanned by operators
m(wi1 ) · · · m(wir )ι(λj1 ) · · · ι(λjs ) (1 ≤ i1 < · · · < ir ≤ n, 1 ≤ j1 < · · · < js ≤ n).
1
2
We write these as
m(I)ι(J),
I = {i1 , . . . , ir }, J = {j1 , . . . , js } ⊂ {1, . . . , n}.
The total number of these spanning
operators is the square of the number
of subsets of {1, . . . , n}, or 22n ,
V
V
which is the dimension of End( (W )). (Since W has dimension n, (W ) has dimension 2n .) To prove the
result, we must show that these operators m(I)ι(J) are linearly independent.
To prove the linear independence, we look at how the operators act on the standard basis elements
K = {k1 , . . . , kt }, 1 ≤ k1 < · · · < kt ≤ n.
w(K) = wk1 ∧ · · · ∧ wkt ,
Inspection of the definitions shows that
m(I) · w(K) =
ι(J) · w(K) =
±w(I ∪ K) I ∩ K = ∅
0
I ∩ K 6= ∅,
±w(K − J)
0
J ⊂K
J 6⊂ K.
Consequently
m(I)ι(J) · w(K) =
±w((K − J) ∪ I)
0
J ⊂ K and I ∩ (K − J) = ∅
J 6⊂ K or I ∩ (K − J) 6= ∅.
Suppose now that we have a nontrivial linear combination
T =
X
aIJ m(I)ι(J).
I,J⊂{1,... ,n}
V
We must show that T does not act by zero on (W ). Choose a subset K as small as possible so that some
aIK 6= 0. It follows that if aIJ 6= 0 and J ⊂ K, then J = K. Then we see that
T w(K) =
X
I⊂{1,... ,n}
±aIK m(I)w∅ =
X
I
±aIK wI .
Because the wI are linearly independent, and we have chosen K so that some aIK are nonzero, we conclude
that T w(K) 6= 0, as we wished to show.
3. (Same as Problem 2, but for differential operators). Write Pn for the algebra of polynomial
coefficient differential operators in n variables (notes, Theorem 2.4): a typical element of Pn
is something like
∂
∂2
− 7x2
.
2x21
∂x2 ∂x3
∂x1
Prove that if f1 , . . . , fm are any linearly independent polynomials in n variables, and g1 , . . . , gm are
any other polynomials in n variables, then there is an element D ∈ Pn such that
Dfi = gi
(1 = 1, . . . , m).
It’s convenient to use “multiindex” notation. A multiindex is a sequence of nonnegative integers
α = (α1 , . . . , αn ) ∈ Nn .
3
We write
∂ β1
∂β
=
··· .
∂xβ
∂xβ1 1
αn
1 α2
xα = xα
1 x2 · · · xn ,
β
∂
Then the various xα ∂x
β are a basis for Pn . The formula for acting on monomials is
∂β γ
x
x =
∂xβ
α
c(β, γ)xγ−β+α
0
β≤γ
β 6≤ γ.
Here we write
β ≤ γ ⇐⇒ βi ≤ γi ,
1 ≤ i ≤ n.
The constants c(β, γ) are strictly positive; it is easy to write formulas, but we don’t need them.
The problem with imitating Problem 2 is that the space of all polynomials in n variables is infinitedimensional. One way out is to find a finite-dimensional space in which Problem 3 actually lives. So
define N to be the largest of the degrees of all the polynomials fi and gi . We’re going to work with the
finite-dimensional vector space
W (N ) = polynomials in n variables of degree at most N .
A basis of W (N ) is the collection of monomials
xγ
(|γ| ≤ N );
here |γ| = γ1 + · · · + γn . Define
Pn (N ) = span of all xα
∂β
∂xβ
(|α|, |β| ≤ N ).
More or less obviously the dimension of Pn (N ) is equal to the dimension of End W (N ). It’s also easy to see
that Pn (N )W (N ) ⊂ W (2N ), so we can define a linear map
T : Pn (N ) → End W (N ),
T D(f ) = f − terms of degrees between N + 1 and 2N .
Arguing exactly as in Problem 2, we can prove that T is an injective linear map from Pn (N ) to End W (N ),
and therefore (by the dimension calculation above) T is a linear isomorphism.
For the problem, there is obviously a linear map from W (N ) to W (N ) taking fi to gi ; so we have shown
that there must be a differential operator E ∈ Pn (N ) with the property that
Efi = gi + terms of degrees between N + 1 and 2N .
We’re left with the problem of getting rid of those high degree “error terms.” The cheapest way I know to
do that is to use the Euler degree operator
∆f =
n
X
j=1
xj
∂f
.
∂xj
The main property of the Euler degree operator is
∆f = kf
(all f homogeneous of degree k).
4
Choose a polynomial qN in one variable with the property
qN (k) =
1
0
0≤k≤N
N + 1 ≤ k ≤ 2N ;
of course there is a unique q of degree 2N − 1 with this property. Then D = q(∆)E is a polynomial coefficient
∂
) at most 2N + 4N − 2, which solves the problem.
differential operator of total degree (in the xi and the ∂x
j
4. Suppose p and q are as in Problem 1, so that O(p, q) ≃ O(V ). Write down an isomorphism
σ: C(V ) → E(W ),
V
the algebra of all linear transformations of (W ).
Hint: this is the orthogonal version of Theorem 2.4 in the notes, and has the same proof. The restriction
of σ to o(V ) ⊂ C(V ) (notes, Proposition 3.2)
^
σ: o(V ) → gl( W )
is the spin representation of the Lie algebra o(V ); it has dimension 2n = 2(dim V )/2 .
Choose (wi ) and (λj ) as in Problem 1. Then these basis elements of V are generators for the Clifford
algebra C(V ); the relations in the Clifford algebra say that
wi λi + λi wi = −1
and thatVall other pairs of basis elements anticommute. Now we get an algebra homomorphism from C(V )
to End( (W )) using the linear transformations introduced in Theorem 3.4 of the notes:
σ(wi ) = m(wi ),
σ(λj ) = −ι(λj ).
Theorem 3.4 says that the Clifford algebra relations are sent to zero by σ, so we get a well-defined algebra
homomorphism onto E(W ). Problem 1 says that the dimension of E(W ) is 22 dim W , which is the dimension
of the Clifford algebra C(V ); so σ must be an isomorphism.
V
5. If n > 0, find a proper (that is, not zero and not the whole space) subspace of (W ) that
is preserved by σ(o(V )).
The elements of o(V )) are even in C(V ): they
V are sums of skew-symmetric combinations of two generators.
The linear map m(wi ) raises degree by one in W , and ι(λj ) lowers degree by one; so the generators change
the parity of the degree in the exterior algebra. Therefore the action of o(V )) must preserve parity of degree.
Two invariant subspaces are
X V2p
P V2q+1
(W ),
(W ).
q
p
If n > 0 these are both nonzero, having dimension 2dim W −1 . (The statement is that the sum of every other
term across the nth row of Pascal’s triangle is (2n )/2. This is not rocket science, but it is not quite trivial
either. What happens if you sum every third term in a row? The answer is trying hard to be 2n /3, except
that’s not quite an integer.)