Lecture 6 Notes: Orthonormality of Characters

Lecture 6: Orthonormality of Characters 6 Orthonormality of Characters 6.1 Review: Schur’s Lemma Last time, we presented Schur’s Lemma. Recall that HomG (ρ, ψ), which may also be written as HomG (V, W ), denotes the space of homomorphisms (or in other words, linear maps) from V to W which are G-equivariant, meaning that HomG (ρ, ψ) = {f : V → W | f linear, and f (ρg (v)) = ψg (f (v)) for all g ∈ G, v ∈ V }. Theorem 6.1 (Schur’s Lemma) Suppose ρ : G → GL(V ) and ψ : G → GL(W ) are irreducible (and complex and finite-dimensional). Then ( 0 if ρ ∼ ̸ ψ = dim (HomG (ρ, ψ)) = 1 if ρ ∼ = ψ. In other words, the first statement means that if ρ ∼ ̸ ψ, then the only G-equivariant homomorphism V → W is = the zero map; the second statement means that if ρ ∼ = ψ, then the only G-equivariant homomorphisms V → W are the scalar maps — or in other words, EndG (ρ) = C · Id. The first statement is true for representations over any field of coefficients (and the same proof we saw last time works in the general case). However, the second statement is not true for arbitrary fields of coefficients — in the proof, we used the fact that eigenvectors must always exist, which isn’t true in general (for example, R is not algebraically closed, so it’s possible that the characteristic polynomial does not have roots, and there are no real eigenvalues). In particular, there are examples of real representations for which EndG (ρ) ̸= R: Example 6.2 Consider the representation of Z/3Z acting on R2 , where 1 is mapped to the matrix   cos 2π/3 − sin 2π/3 sin 2π/3 cos 2π/3 which corresponds to rotation by 2π/3. In this case, EndZ/3Z (R2 ) is C, not R. Proof. As usual, multiplication by any scalar is in EndZ3 (R2 ); these elements correspond to R (in the case of C, Schur’s Lemma states that the only elements of EndG (ρ) are multiplication by scalars). But there are actually other possible endomorphisms — rotation by π/2 about the origin is also a G-equivariant endomorphism, since any two rotations about the origin must commute. We can think of this element as i; then taking linear combinations, all elements a + bi (for real a and b) are in EndZ/3Z (R2 ) (and it’s possible to check that there’s no others). Composing two such endomorphisms corresponds to multiplying their corresponding complex numbers (since the endomorphism corresponding to z can be thought of as multiplication by z in the complex plane). So then this means EndZ/3Z (R2 ) = C. We won’t discuss this, but there also exists a real irreducible representation ρ of some group G for which EndG (ρ) = H (where H denotes the quaternions). 6.2 An Implication of Schur’s Lemma We’ve seen earlier that every representation ρ : G → GL(V ) can be written as the sum of irreducible representations, with certain coefficients. Using Schur’s Lemma, we can describe what these coefficients are: 30 Lecture 6: Orthonormality of Characters Corollary 6.3 Let ρ : G → GL(V ) be a representation. Let ρ1 , . . . , ρn be the list of all irreducible representations of G (up to isomorphism). Then n M ρ∼ ρdi i = i=1 where dk = dim HomG (ρk , ρ) for all k. Proof. From Maschke’s Theorem, we know that we can write ρ∼ = n M ρdi i i=1 for some coefficients di , so then n  M  M HomG (ρk , ρ) = HomG ρk , ρdi i = HomG (ρk , ρi )di = Cdk , i=1 using the fact that by Schur’s Lemma, HomG (ρk , ρi ) is 0 if i ̸= k and C if i = k. This means dim HomG (ρk , ρ) = dk for each k, as desired. Student Question. Why could we write n  M  M HomG ρk , HomG (ρk , ρi )di ? ρdi i = i=1 Answer. It’s enough to see that HomG (U, V ⊕ W ) = HomG (U, V ) ⊕ HomG (U, W ). First, by looking at matrices, it’s possible to see that HomC (U, V ⊕ W ) = HomC (U, V ) ⊕ HomC (U, W ) (where this denotes all linear maps, not just the G-equivariant ones) — if dim U = m, dim V = n1 , and dim W = n2 , then a linear map U → V ⊕ W is a (n1 + n2 ) × m matrix, which we can think of as a pair of a n1 × m matrix and a n2 × m matrix:   ∗ ∗ ∗ ∗ ∗ ∗ . ∗ ∗ ∗ Then such a map is compatible with the G-action if and only if each component is. It’s possible to think of this without matrices, as well. By definition, V ⊕ W is the space of pairs (v, w) with v ∈ V and w ∈ W . So in order to describe a linear map from U to V ⊕ W , for an element u ∈ U , we need to specify the first coordinate of its image (corresponding to a linear map U → V ) and the second coordinate of its image (corresponding to a linear map U → W ). Then since G essentially acts separately on V and W (by the definition of a direct sum of representations), the map U → V ⊕ W is G-equivariant if and only if the two individual maps U → V and U → W are. Student Question. What exactly does it mean to have a list of irreducible representations up to isomorphism — what happens if there are two isomorphic representations, but they act on different vector spaces? Answer. We can think of ρ1 , . . . , ρn as an abstract list of representations, without thinking about the subspaces being acted on. For example, when writing down the character table of S3 , we saw that there are three irreducible representations; this means every irreducible representation is isomorphic to one of them. We’re using “up to isomorphism” in the same sense here. Ln di More generally, when we write ρ = i=1 ρi , when dk > 0, this really means that ρk is isomorphic to a sub-representation of ρ. 31 Lecture 6: Orthonormality of Characters 6.3 Matrices and a New Representation We’ll now rewrite the concept of G-equivariance in terms of matrices — this will give a useful construction of a new representation, which we can apply Schur’s Lemma to in order to prove the orthonormality of irreducible characters. Choose a basis for V and W . Then if n = dim V and m = dim W , we can write a linear map V → W as a m × n matrix, where the map sends v ∈ V to Av ∈ W . We can also write our representations ρ and ψ as the matrix representations R : G → GLn (C) and S : G → GLm (C). Then by rewriting the definition of G-equivariance (that f (ρg (v)) = ψg (f (v)) for all v and g) in terms of these matrices, we have that a matrix A ∈ Matm×n (C) corresponds to a linear map f ∈ HomG (ρ, ψ) if and only if ARg = Sg A for all g ∈ G. (Our initial definition was written in terms of v, but we can think of it instead as an equality of the linear maps themselves — that the linear maps f ◦ ρg and ψg ◦ f are the same — which corresponds to an equality of matrices.) We can rewrite this condition as A = Sg ARg−1 for all g ∈ G. The key point is that we can get another representation of G from this expression (starting with our original representations ρ and ψ). Note that the space M = Matm×n (C) is itself a C-vector space — we can forget everything we know about how to multiply matrices, and just imagine adding them and multiplying by scalars, which makes Matm×n (C) a mn-dimensional vector space over C. Lemma 6.4 There is a representation C of G acting on Matm×n (C), where for each g ∈ G, g is sent to the matrix Cg : A 7→ Sg ARg−1 . In other words, if we think of Matm×n (C) as a C-vector space, then for each g ∈ G, the map A 7→ Sg ARg−1 is a linear operator on this vector space. So we’re sending g to the mn × mn matrix corresponding to that linear operator (which we denote by Cg ). Proof. It suffices to check that Cgh = Cg Ch for all g and h. But for any A ∈ Matm×n (C), we have −1 Cgh (A) = Sgh ARgh = Sg Sh ARh−1 Rg−1 = Cg (Ch (A)). So then Cgh = Cg Ch , as desired. In this new representation, HomG (ρ, ψ) is exactly the space of G-invariant vectors (note that here “vectors” means matrices of dimension m × n, since the vector space our representation is acting on is actually the space of such matrices). We can also describe this construction without using matrices — let M = HomC (V, W ) be the space of all linear maps V → W (which we thought of as Matm×n (C) when writing down the construction in matrices). Then M is a vector space over C. So we can define a representation γ acting on M , where for each g ∈ G, we send g to the linear map γg : M → M which sends E 7→ ψg Eρ−1 g for all E ∈ M . As before, HomG (ρ, ψ) is exactly the space of G-invariant vectors in γ. Student Question. Why is E 7→ ψg Eρ−1 g a linear map? Answer. Thinking in terms of matrices, we want to see that A 7→ Sg ARg−1 is a linear map. But this follows from the distributive property of matrix multiplication — for example, we have Sg (A + B)Rg−1 = Sg ARg−1 + Sg BRg−1 . 32 Lecture 6: Orthonormality of Characters Proposition 6.5 For the representation γ described above, we have χγ = χψ χρ . The proposition quickly reduces to a statement about matrices: Lemma 6.6 Let A and B be n × n and m × m matrices. Consider the linear map A ⊗ B from Matm×n (C) to itself defined as A ⊗ B : E 7→ BEA. Then we have Tr(A ⊗ B) = Tr(A) · Tr(B). Proof. The space of matrices has a basis consisting of the matrices Eij which have a 1 in the ith row and jth column, and 0’s everywhere else (for all 1 ≤ i ≤ m and 1 ≤ j ≤ n). But by straightforward computation, we can see that BEij A has bii ajj in its ith row and jth column — this means BEij A is bii ajj Eij , plus some entries corresponding to the other basis elements. So if we write out the matrix corresponding to A ⊗ B (in the basis formed by the Eij ), the diagonal entry corresponding to Eij is bii ajj . The trace of A ⊗ B is then the sum of the diagonal entries of this matrix, which is Tr(A ⊗ B) = m X n X bii ajj = i=1 j=1 m X bii i=1 n X j=1 ajj = Tr(B) · Tr(A). Proof of Proposition 6.5. Using the above lemma, for all g ∈ G we have χγ (g) = Tr(ρg−1 ⊗ ψg ) = Tr(ρg−1 ) · Tr(ψg ) = χρ (g −1 ) · χψ (g). We saw earlier that χρ (g −1 ) = χρ (g) — this follows from the fact that χρ (g −1 ) is the sum of the eigenvalues of ρg−1 = ρ−1 g , which are the inverses of the eigenvalues of ρg , and since all these eigenvalues have magnitude 1, their inverses are also their conjugates. So χγ (g) = χρ (g) · χψ (g) for all g ∈ G, as desired. 6.4 Orthonormality of Characters We have now developed the tools that we can use to prove one part of the main theorem stated earlier, the orthonormality of irreducible characters. Proposition 6.7 The characters of the irreducible representations of G are orthonormal. Proof. Let ρ and ψ be irreducible representations, acting on the spaces V and W . Then our Hermitian form is ⟨χψ , χρ ⟩ = 1 X χρ (g)χψ (g), |G| g∈G so we want to check this expression is 0 if ρ ̸∼ = ψ and 1 if ρ ∼ = ψ. But we saw a representation γ, acting on the space M = HomC (V, W ), whose character is exactly the expression inside the sum! So we can rewrite our sum as   X 1 X 1 ⟨χψ , χρ ⟩ = χγ (g) = Tr  γg  . |G| |G| g∈G g∈G 33 Lecture 6: Orthonormality of Characters (Here we used the fact that Tr(A + B) = Tr A + Tr B — our original sum calculates the traces of each γg first, and then averages them, but we can instead average the γg and then compute the trace). P Now let’s consider what the linear operator on M given by g∈G γg /|G| looks like; denote this operator by f . Since f is an averaging operator (we’re averaging over all g ∈ G), it must always output a G-invariant vector — similar to the averaging trick we saw earlier, for any v ∈ M and any fixed h ∈ G, we have γh (f (v)) = γh · 1 X 1 X 1 X 1 X γg v = γh γg v = γhg v = γg v = f (v), |G| |G| |G| |G| g∈G g∈G g∈G g∈G since if we fix h and let g range over all elements in G, then hg also ranges over all elements in G. But we know what the G-invariant vectors are! Recall that vectors in the space M that γ acts on are actually homomorphisms V → W , and by the way γ was defined, the vectors in M which are G-invariant are exactly the homomorphisms V → W which are G-equivariant — which are described by Schur’s Lemma. If ρ ∼ ̸ ψ, then by Schur’s Lemma, then HomG (ρ, ψ) only contains the zero map, so the only G-invariant vector = in M is the zero vector. So since our operator f sends every vector v ∈ M to some G-invariant vector, it must actually send every vector v to 0. So f is the zero operator, and ⟨χψ , χρ ⟩ = Tr(f ) = 0. Now suppose ρ ∼ = ψ. Then by Schur’s Lemma, HomG (ρ, ψ) only contains scalar maps; so there’s only one G-invariant vector in M up to scaling (the identity matrix and its scalar multiples). Let u be such a (nonzero) invariant vector. This means f must send every vector v ∈ M to some scalar multiple of u (since f sends every v to some Ginvariant vector, and the only G-invariant vectors are multiples of u). On the other hand, since we’re averaging and u is already G-invariant, f must send u to itself — we have 1 X 1 X γg u = u = u. |G| |G| g∈G Now choose a basis for M whose first element is  1 0  0   .. . 0 g∈G u. Then in this basis, the matrix for f is of the form  ∗ ∗ ··· ∗ 0 0 · · · 0  0 0 · · · 0 , .. .. . . ..  . . . . 0 0 ··· 0 which has trace 1. So in this case, ⟨χψ , χρ ⟩ = Tr(f ) = 1. 34