Representation theory: Characters of tensor products and direct sums

Question

I am trying to read Fulton-Harris's book on representation theory, specifically the first section about representations of finite groups. My question concerns Proposition 2.1, on the character of a direct sum (resp. tensor product) of representations in terms of the characters of the summands (resp. tensor factors). Namely, the character of the direct sum is the sum of the characters, and the character of the tensor product is the product of the characters.

I don't understand the proof that is given for either case. For the direct sum, it says that the eigenvalues of a fixed element $g \in G$ on $V \oplus W$ are pairwise sums of eigenvalues of $g$ on $V$ and on $W$. I believe what they meant to say is that the eigenvalues on $V \oplus W$ are given by the union of the 2 sets of eigenvalues. Assuming this is what they meant, I still don't totally understand how the corrected statement gives the result. I have the same sort of confusion about this case as I do about the tensor product, which I'll now explain...

For the tensor product, it says that the eigenvalues of $g$ on $V \otimes W$ are the pairwise products of eigenvalues of $g$ on $V$, and on $W$. I can certainly see that if $\lambda$ is an eigenvalue of $g$ on $V$, and $\delta$ is an eigenvalue of $g$ on $W$, then $\lambda \delta$ is an eigenvalue of $g$ on $V \otimes W$. However, I'm worried about the issue of (algebraic) multiplicity of the eigenvalues as roots of the characteristic polynomials:

If $\lambda$ is a root of the characteristic polynomial of $g$ on $V$ of multiplicity $e$, and $\delta$ is a root of the characteristic polynomial of $g$ on $W$ of multiplicity $f$, why is $\lambda \delta$ a root of the characteristic polynomial of $g$ on $V \otimes W$ of multiplicity $e + f$? It seems that this is what is required for the result.

Related to this, why can't there be other eigenvalues for $g$ on $V \otimes W$ aside from products of eigenvalues on the factors? It's only clear to me that the set of eigenvalues on $V \otimes W$ contains those products, but for all I know they could occur with smaller multiplicity in the characteristic polynomial, leaving room for there to be other eigenvalues.

With $V \oplus W$, it's easy to observe that the matrix for $g$ on $V \oplus W$ is simply $2 \times 2$ block diagonal with the matrices for $g$ on $V$ and on $W$ being the blocks. From this, the result on characters is immediate. We also see from this that the characteristic polynomial for $g$ on $V \oplus W$ is the product of the two characteristic polynomials, which says that eigenvalues and multiplicities must match up. However, I would like to know whether one can see this without making the observation about the shape of the matrix. Again, it is clear that the union of eigenvalues on $V$ and on $W$ is contained in the set of eigenvalues on $V \oplus W$, but why are the algebraic multiplicities what they need to be, and why couldn't there be other eigenvalues?

Lastly, if I am thinking about how to prove this result in the totally wrong way and there's an easier argument that doesn't require me to worry about these things, I'd be glad to hear it.

Answer 1

This is a good question and you are correct to worry about the issue of multiplicity; Fulton and Harris' proof is incomplete as written. We want to understand the following:

Theorem (additivity and multiplicativity of the trace): Let $V, W$ be finite-dimensional vector spaces and let $T : V \to V, S : W \to W$ be linear maps. Then $\text{tr}(T \oplus S) = \text{tr}(T) + \text{tr}(S)$, and $\text{tr}(T \otimes S) = \text{tr}(T) \text{tr}(S)$.

First let's be clear that there is a very easy computational proof of this which doesn't require even knowing what an eigenvalue or eigenvector is, and which works over an arbitrary field: once you know that the trace is the sum of the diagonal entries when expressed in any basis, it suffices to pick any bases $\{ v_i \}, \{ w_j \}$ of $V, W$, use them to construct the obvious bases $\{ v_i, w_j \}, \{ v_i \otimes w_j \}$ of the direct sum and tensor product, and just compute the sum of the diagonal entries of $T \oplus S$ and $T \otimes S$ wrt these bases. You find that

• the diagonal entries of $T \oplus S$ are the diagonal entries of $T$, then the diagonal entries of $S$, and
• the diagonal entries of $T \otimes S$ are the pairwise products of the diagonal entries of $T$ and the diagonal entries of $S$.

This is really very straightforward! Nevertheless, let's see what happens if we try to do it with eigenvalues. The straightforward argument works fine in the case that $T$ and $S$ are both diagonalizable; in that case if $\{ v_i \}$ is a basis of eigenvectors of $T$ with eigenvalues $\lambda_i$ and $\{ w_j \}$ is a basis of eigenvectors of $S$ with eigenvalues $\mu_j$, so that $\text{tr}(T) = \sum \lambda_i, \text{tr}(S) = \sum \mu_j$, then

• $\{ v_i, w_j \}$ is a basis of eigenvectors of $T \oplus S$ with eigenvalues $\lambda_i, \mu_j$, so $\text{tr}(T \oplus S) = \sum \lambda_i + \sum \mu_j$, and similarly
• $\{ v_i \otimes w_j \}$ is a basis of eigenvectors of $T \otimes S$ with eigenvalues $\lambda_i \mu_j$, so $\text{tr}(T \otimes S) = (\sum \lambda_i)(\sum \mu_j)$.

Note that this is ultimately just running the computational proof in a basis of eigenvectors. You do learn from this argument that a direct sum and tensor product of diagonalizable maps is diagonalizable, which is worth knowing, and you also learn that the geometric multiplicities do the obvious thing in this case, which Fulton and Harris' argument as written elides.

This only proves the result when $T$ and $S$ are diagonalizable. This is already enough to handle characters of representations of finite groups (over $\mathbb{C}$, say), because the action of an element of a finite group $g \in G$ on a f.d. representation $V$ is always diagonalizable; this is a nice exercise.

Nevertheless the result is true for arbitrary $T, S$ and this is worth knowing, e.g. we'd like to know this is still true for characters of representations of infinite groups, such as Lie groups. To get the result for arbitrary $T$ and $S$ there are various ways to proceed, and which one of these will strike you as clarifying what is going on conceptually vs. which one of these will strike you as overcomplicating the situation is a matter of taste. The computational proof doesn't require getting into any of this!

1. Very general arguments imply that if a result of this form is true for $T, S$ diagonalizable then it's true for $T, S$ arbitrary. This is because the desired identities are ultimately polynomial identities in the matrix entries of $T, S$ (wrt some basis), and a polynomial identity involving matrices is true for all matrices iff it's true for diagonalizable matrices (over an algebraically closed field), because diagonalizable matrices are Zariski dense. If we only care about the result over $\mathbb{C}$ we can replace "Zariski dense" with dense in the Euclidean topology. Either way the point is that the result could not possibly be true for diagonalizable matrices but false for non-diagonalizable matrices. This proof technique is very general and can also be used to prove, for example, the Cayley-Hamilton theorem by reduction to the diagonalizable case.

2. Instead of working with eigenvectors you can work with generalized eigenvectors, which fixes the multiplicity issue. Equivalently you can conjugate $T$ and $S$ to their Jordan normal forms. Note that this is ultimately just running the computational proof in a basis of generalized eigenvectors.

3. Without using the full power of Jordan normal form, you can prove the following "crude Jordan normal form theorem": over an algebraically closed field $K$, every linear map $T : V \to V$ on a f.d. vector space over $K$ admits a basis with respect to which it is upper triangular. This is done by finding an eigenvector $v$, then considering the action of $T$ on the quotient $V/v$ and inducting on $\dim V$. So you can assume WLOG that $T$ and $S$ are upper triangular and do the computation that way. Note that, again, this is ultimately just running the computational proof in some basis.

It might look like the eigenvalue arguments are telling you more: namely, they're telling you what the eigenvalues of $T \oplus S$ and $T \otimes S$ are (with algebraic multiplicity), not just their traces. However, the apparently weaker statement about traces implies the stronger statement about the eigenvalues! This is because (over a field $K$ of characteristic $0$) the traces $\text{tr}(T^k), k \ge 0$ determine the eigenvalues of $T$ (with algebraic multiplicity) using Newton's identities, and we have

$$\text{tr}(T^k \oplus S^k) = \text{tr}(T^k) + \text{tr}(S^k) = \text{tr}((T \oplus S)^k)$$

and

$$\text{tr}(T^k \otimes S^k) = \text{tr}(T^k) \text{tr}(S^k) = \text{tr}((T \otimes S)^k).$$

These identities turn out to imply that the eigenvalues of $T \oplus S$ resp. $T \otimes S$ are the "multiset union" resp. the pairwise products of the eigenvalues of $T, S$, with algebraic multiplicities.