2.11. Lecture 11
2.11.1. Representations of U(1) and Maschke’s theorem
We now discuss the representation theory of the unitary group , which is isomorphic to , the circle group. Its Lie algebra is with trivial Lie bracket, which is isomorphic to .
Proposition 2.11.1.
All irreducible finite-dimensional representations of are one-dimensional. They are given by
for .
Proof.
By Schur’s lemma and the fact that is abelian, all irreducible finite-dimensional representations of are one-dimensional, so are homomorphisms . Since is connected, such a homomorphism is determined by the derivative , which has the form for some . As in Example 1.7.1, this exponentiates to a map if and only if for some , giving the homomorphism . ∎
Theorem 2.11.2.
-
(i)
All finite dimensional representations of are unitary.
-
(ii)
All finite dimensional representations of are completely reducible, that is, decompose into irreducible representations.
Proof.
-
(i)
This is Problem 2.11.2.
-
(ii)
This follows from (i) as in the proof of Maschke’s theorem for finite groups. We will sketch another proof. Let be a finite dimensional representation of . Consider its differential . Let . By the proof of Lemma 1.2.2, we may write for some strictly upper triangular matrix and diagonal matrix such that and commute. Then
On the other hand,
since and commute. As is diagonal is diagonal, but and as is nilpotent we must have . It follows that and is diagonal.
∎
Remark 2.11.3.
Complete irreducibility does not hold for representations of a general Lie group. For example, the standard representation of
does not decompose into a direct sum of two one-dimensional invariant subspaces (otherwise we would diagonalise , which is impossible). Furthermore, it is not unitary (as unitary matrices are diagonalisable).
Some of this actually generalises substantially:
Theorem 2.11.4.
(Maschke’s theorem for compact Lie groups) Let be a compact Lie group.
-
(i)
Any finite-dimensional representation of is unitarisable.
-
(ii)
(complete reducibility) Any finite-dimensional representation of is a direct sum of irreducible representations.
Proof.
The proof of the first part uses the same idea as for finite groups. Let be a finite-dimensional representation and take to be any Hermitian inner product on . Then define
This is also a Hermitian inner product, and
(putting ) | ||||
(since ) | ||||
The challenge here is to show that there is an appropriate notion of for which the step “” is valid — this goes by the name of ‘existence of Haar measure’. For you can do it by hand, see problem 2.11.2.
The second part is proved exactly as for finite groups in Michaelmas (Theorem 5.12). Let be a finite-dimensional representation and let be a subrepresentation. Let be the -invariant Hermitian inner product on guaranteed by part (i). Then the orthogonal complement is also a subrepresentation, and . Iterating, we obtain that is a direct sum of irreducible representations.
∎
2.11.2. Exercises
Problem 27. Consider .
-
(a)
For a continuous function on , we define its integral
Note that . Show that
for any .
-
(b)
Let be a finite dimensional representation of and let be any Hermitian form on . Define a new Hermitian form by
Show that is a -invariant Hermitian form on .
-
(c)
Conclude that every finite-dimensional representation of is completely reducible. (Compare this to the proof of Theorem 5.12 from Michaelmas).
Problem 28. Consider the orthogonal group .
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(a)
Show that has index in . Deduce that every element in can be uniquely written as or with and the matrix for rotation by . Show that
-
(b)
Mimic the method we used for dihedral groups to classify all irreducible finite-dimensional representations of .
Problem 29. Let be the space of functions on that are polynomials in the coordinates and . Consider the (left) action of on given by
(here, think of as a column vector).
Compute the derived action for the “standard” basis of given by , , and . You should get something involving the partial derivatives and .