1. 18.757 Homework 2 solutions
1. Suppose (π, Vπ ) is a finite-dimensional irreducible representation of a group G over a field k. Prove that the dual
representation
Vπ∗ = Homk (Vπ , k),
[π ∗ (g)ξ](v) = ξ(π(g −1 )v)
(for g ∈ G, ξ ∈ Vπ∗ , and v ∈ Vπ ) is an irreducible representation
of G.
If V is any finite-dimensional vector space, then there is an inclusion
reversing bijection of subspaces of V with subspaces of V ∗ , defined by
W 7→ W ⊥ =def {ξ ∈ V ∗ | ξ|W = 0}.
(It’s obvious that the map reverses inclusions; and even if V is infinitedimensional, it makes sense to say (and is true that) (W ⊥ )⊥ = W ,
from which it follows that the correspondence is one-to-one. If V is
finite-dimensional, then the fact that (V ∗ )∗ = V allows one to deduce
that any subspace E of V ∗ must be the orthogonal of E ⊥ ⊂ V ; so the
correspondence is a bijection.)
Now it’s clear from the definitions that (for any finite-dimensional
representation Vπ ) invariant subspaces of Vπ correspond bijectively to
invariant subspaces of Vπ∗ , by sending W to W ⊥ . So there are exactly
two (namely {0} and Vπ ) for Vπ if and only if there are exactly two
(namely {0}⊥ = Vπ∗ and Vπ⊥ = {0}) for Vπ∗ .
2. Suppose k is a field, and D is an arbitrary division algebra
over k (meaning that k is contained in the center of D). Find a
group G and an irreducible k-representation (π, Vπ ) of G, with
the property that Dπ = D. (Recall that I defined
Dπ = HomkG (Vπ , Vπ ),
with the peculiar multiplication given by composition of linear
transformations in the opposite order:
(T · S)(v) = S(T (v)).
The purpose of the peculiar definition is to make Vπ a right
vector space over Dπ .)
Let G be the multiplicative group D× of D; and let Vπ = D, regarded
as a vector space over k. Define
π(d)x = dx
(d ∈ G = D× , x ∈ D).
It is immediate that this is a representation of G on D. Any nonzero
invariant subspace S ⊂ D contains some nonzero element x ∈ D, and
therefore contains any other nonzero element y ∈ D, since
y = π(yx−1 )x.
1
2
Since (as a subspace) S also contains zero, it is equal to D. Therefore
π is irreducible. By definition
HomkG (Vπ , Vπ ) = {T : D → D | T (dx) = dT (x)}.
Applying this to x = 1, we conclude that T (d) = dT (1); that is, that
T is right multiplication by an element T (1) of D. It’s easy to see that
any such T actually belongs to Dπ , and that
Dπ ' D,
is an algebra isomorphism.
T 7→ T (1)