Aufgaben:Problem 6
Follow the link! [1]
An independently written, second solution is here: [2]
With TexFile: [3]
Enjoy!
Contents
Task
Let \(G\) be a finite group. Let \(V\) be a finite dimensional complex vector space.
a) Let \(\rho: G \rightarrow GL(V)\) be an irreducible representation. Let \(A \in \mathrm{End}(V)\) with \([A, \rho(g)] = 0, \; \forall g \in G \). Show that then \(A = \lambda \mathbb{I}, \; \lambda \in \mathbb{C}\).
b) Let \(\rho: G \rightarrow GL(V)\) be unitary w.r.t. \(\langle \cdot, \cdot \rangle \) on \(V\). Assume that it holds $$A \in \mathrm{End}(V), [A, \rho(g)] = 0 \; \forall g \in G \Rightarrow A = \lambda \mathbb{I}, \; \lambda \in \mathbb{C}.$$ Show that then \(\rho\) is irreducible.
Problem 6 (Textbook)
a
From \([A, \rho(g)] = 0 ~ \forall g \) it follows that \(\mathrm{Ker}(A)\) is an invariant subspace of \(V\) and since the representation is irreducible that \(A=0=0\cdot \mathbb{I}\) or \(A\) is an isomorphism. Suppose the second case: \(A\) has an eigenvalue \(\lambda \in \mathbb{C}\), therefor \(A-\lambda \cdot \mathbb{I}\) is not injective. But it holds too that \([A-\lambda \cdot \mathbb{I}, \rho(g)] = 0 \; \forall g\), and since \(\mathrm{Ker}(A-\lambda \mathbb{I}) \neq \{0\}\) it follows that \(A - \lambda \cdot \mathbb{I} = 0 \)
\(\blacksquare\)
b
Let \(\rho\) be unitary but reducible and let \(U\) be a not trivial subspace. If \(u \in U, u^{\perp} \in U^{\perp}\) it holds that \( 0 = \langle \rho(g)u,u^{\perp} \rangle = \langle u, \rho(g^{-1})u^{\perp}\rangle ~ \forall g\) and therefore \(U^{\perp}\) is invariant too. Let \(A\) be the orthogonal projection on \(U\). Then it holds that \(A \neq \lambda \cdot \mathbb{I}\) (since \(U\) is not trivial), but \([A,\rho(g)]v = [A,\rho(g)](u + u^{\perp}) = A(\rho(g)u + \rho(g)u^{\perp}) - \rho(g)u = \rho(g)u-\rho(g)u = 0 ~ \forall g \).
\(\blacksquare\)
Problem 6
This is a solution of exercise 6 of the Ferienserie. As reference, check out http://link.springer.com/book/10.1007%2F978-94-011-3538-2. Both solutions are on page 90. What I did here is just write up the theorem more clearly.
Okay, so let us begin.
a
This statement is also known as Schur's lemma. It is a pretty nice thing to know, because it is relatively easy to prove but has wide applications, especially in quantum mechanics. Enjoy.
Let \(A\) be an arbitrary linear map that commutes with \(\rho(g) \forall g \in G\)
Proposition A: \( \text{Ker}(A)\) is a G-invariant subspace, that is \(v \in \text{Ker}(A), g \in G \Rightarrow \rho(g)v \in \text{Ker}(A)\)
Proof: \(\text{Ker}(A)\) is already a subspace by definition. It remains to show that \(\rho(g)v\) is in \(\text{Ker}(A)\), if \(v \in \text{Ker}(A)\). This is really easy: We'll just show that \(A\rho(g)v = 0 \): \(A\rho(g)v = \rho(g)Av = \rho(g)0 = 0\)
\(\square\)
For the next step, we shall use the irreducibility of \(\rho\). We know that all G-invariant subspaces are either \(V\) or \(\{0\}\) So it follows that the Kernel is all space or the trivial space.
Proposition. (B) \(A\) is either an isomorphism or the zero map.
Proof: \(\text{Ker}(A)\) is G-invariant. As \(\rho\) is irreducible, it follows that \(\text{Ker}(A) = \{0\}\) or \(\text{Ker}(A) = V\) In the first case, the map is an isomorphism, as it has full rank. In the second case, it is the null map, as every vector is in the kernel.
\(\square\)
For the last step, we will consider an eigenvalue of A.
Proposition \(A=\lambda I\)
Proof: Let \(\lambda\) be an eigenvalue of \(A\). It exists because every complex matrix has at least one eigenvalue. (actually, we already have n eigenvalues here, because A is already invertible or the null map, but we need only one here.) Then, the matrix \(A-\lambda I\) is not invertible. It also commutes with \(\rho(g)\), as \(A\) commutes and the identity commutes with everything. (And scalars too!). It follows that \(A - \lambda I\) satisfies the conditions for (B). This means that it is either the zero map or an isomorphism. But isomorphisms are invertible, and \(A - \lambda I\) isn't, by definition. So \(A - \lambda I = 0\), and we're done.
\(\square\)
b
This Inverse Schur's lemma (only) works if the orthogonality is given. Otherwise, this statement is not true. We will use the orthogonality right in our first result:
Proposition Let \(U\) be a G-invariant subspace of V. Then, \[U^{\bot} = \{v \in V| <v,u> = 0 \, \forall u \in U\} \] is a G-invariant subspace of V, too.
Proof: We will prove that \(<\rho(g)w,u> = 0\) for \(w \in U^{\bot}, u \in U\): \(<\rho(g)w,u> = <w, \rho(g)^{-1}u> = <w,\rho(g^{-1})u>\) As \(U\) is G-invariant, it follows that \(\rho(g^{-1})u\) is in \(U\) too. Then, the scalar product is zero, by the definition of \(w \in U^{\bot}\).
\(\square\)
Now we have two perpendicular Subspaces of \(V\) whose direct sum is \(V\). We can now split up any vector \(v \in V = u \in U + w \in U^{\bot}\). Introduce the Projection \(P\) that projects \(V\) onto \(U\), that is, \(P(u) = u\) for \(u \in U\) and \(P(w) = 0\) for \(w \in U^{\bot}\). \(P\) is a linear map.
Proposition \(P\) commutes with \(\rho(g)\)
Proof: This proof is just calculating the whole thing through. \[\rho(g)Pv = \rho(g)P(u + w) = \rho(g)u = \rho(g)(u + 0) = P \rho(g)u + P \rho(g)w = P \rho(g)v\] Note that \(P\rho(g)u = \rho(g)u\) and \(P \rho(g)w = 0\) because \(\rho(g)u \in U, \rho(g)w \in U^{\bot}\), as the two subspaces are G-invariant.
\(\square\)
Now is follows, by supposition, that the Projection is a scalar multiple of the unit map or the zero map. Actually, by the condition \(P^2 = P\), we find that \(P=I\) or \(P=0\). So \(U\) must be \(V\) or \(\{0\}\). The condition for \(\rho\) to be irreducible it met! We're done.
