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Consider a torus $T$ over $\mathbb{C}$. Let $\rho: T\rightarrow \operatorname{GL}_{n}(\mathbb C)$ be a finite dimensional complex representation. Is there an elementary way (undergrad level) to see that $\rho(T)$ is diagonalizable?

The standard way uses algebra of functions on $k[T]$, tensor products and symmetric algebras, but it is too involved, can we see it directly? We already know by Lie–Kolchin that the image is trigonalizable.

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  • $\begingroup$ Combine: (a) unitary matrices are diagonalizable, and (b) commuting diagonizable matrices are simultaneously diagonalizable. (a) is often taught and (b) is often an exercise in (high level) undergraduate courses. $\endgroup$
    – Nicholas Kuhn
    Commented Jan 5, 2023 at 15:35
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    $\begingroup$ First by a torus, I mean an algebraic torus, thus isomorphic to $(\mathbb{C}^{*})^{n}$. Also a priori it is not clear that the representation $\rho$ is compatible with complex conjugation or transposition. $\endgroup$
    – prochet
    Commented Jan 5, 2023 at 20:09
  • $\begingroup$ Your title asked for an action by the multiplicative group, but your body asked for an action by any complex torus; and you wrote the target of the representation as $\mathrm{GL}_n(V)$, but you almost certainly meant either $\operatorname{GL}_n(\mathbb C)$ or $\operatorname{GL}(V)$. I edited accordingly. (Also, re, as @DavidSpeyer indicates, without some compatibility requirement the result isn't true.) $\endgroup$
    – LSpice
    Commented Jan 9, 2023 at 20:48

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You need to put some niceness hypothesis on the function $\rho$, such as "algebraic" or "analytic". Otherwise, the map $\mathbb{C}^{\ast} \to \text{GL}_2(\mathbb{C})$ by $z \mapsto \begin{bmatrix} 1 & \log |z| \\ 0 & 1 \end{bmatrix}$ is a representation and not diagonalizable. Since you didn't tell me which hypothesis to use, I will take $\rho$ to be algebraic, meaning that each of the $n^2$ entries in the matrix $\rho(z_1, z_2, \dots, z_d)$ is a Laurent polynomial in $z$. (You used $n$ for both the dimension of $T$ and the dimension of the representation, but I assume you didn't want that; I'll put $d = \dim T$.)

So we can write $$\rho(z_1, z_2, \dots, z_d) = \sum_{(k_1, k_2, \ldots, k_d)} P_{(k_1, k_2, \ldots, k_d)} z_1^{k_1} z_2^{k_2} \cdots z_d^{k_d}$$ where each $P_k$ is an $n \times n$ matrix.

From the equation $\rho(xy) = \rho(x) \rho(y)$, we deduce that $$\sum_{(k_1, \ldots, k_d)} P_{(k_1, \ldots, k_d)} x_1^{k_1} \cdots x_d^{k_d} y_1^{k_1} \cdots y_d^{k_d} = \sum_{(i_1, \ldots, i_d),\ (j_1, \ldots, j_d)} P_{(i_1, \ldots, i_d)} P_{(j_1, \ldots, j_d)} x_1^{i_1} \cdots x_d^{i_d} y_1^{j_1} \cdots y_d^{j_d}.$$ Comparing the coefficient of $x_1^{i_1} \cdots x_d^{i_d} y_1^{j_1} \cdots y_d^{j_d}$ on each side, we deduce that $$P_{(i_1, \ldots, i_d)} P_{(j_1, \ldots, j_d)} = \begin{cases} P_{(i_1, \ldots, i_d)} & (i_1, \ldots, i_d) = (j_1, \ldots, j_d) \\ 0 & (i_1, \ldots, i_d) \neq (j_1, \ldots, j_d)\\ \end{cases}.$$ In other words, the $P$'s are mutually orthogonal idempotents.

Moreover, from the equation $\rho(1) = \text{Id}_n$, we deduce that $\sum P_{(k_1, \ldots, k_d)} = \text{Id}_n$. So the $P$'s are a complete set of mutually orthogonal idempotents.

Now use your favorite proof that mutually orthogonal idempotents are simultaneously diagonalizable.

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