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Throughout this post $G$ denotes $GL_{n}(\mathbb{F})$ where $\mathbb{F}$ denotes the finite field of $q$ elements.

I'm currently reading the aforementioned book to understand how the irreducible representations of $G$ are constructed. To get started, we need to understand the conjugacy class of $G$.

In the book he explains how given a vector space $V$, for each $g \in G$ we can equip $V$ with an $\mathbb{F}[t]$ module structure by setting $t \cdot v = g(v)$ and extending it linearly. When we are regarding $V$ as an $\mathbb{F}[t]$ module for a specific $g$, we denote it as $V^{g}$.

From here I understand that $V^{g_{1}} \cong V^{g_{2}}$ as $\mathbb{F}[t]$ modules if and only if $g_{1}$ and $g_{2}$ belong to the same conjugacy class. Therefore if we only care about $V^{g}$ upto isomorphism, we can denote it as $V^{\mathcal{C}}$ where $\mathcal{C}$ is the conjugacy class of $g$.

Next he goes on to prove that for any given conjugacy class $\mathcal{C}$, we have $V^{\mathcal{C}} \cong \displaystyle \bigoplus_{i = 1}^{s -1} \mathbb{F}[t]/(f_{i})^{e_{i}}$ where $f_{i}$ are irreducible polynomials of $\mathbb{F}[t]$. The proof of this is basically using the structure theorem for finitely generated modules over a PID.

Here is where my confusion begins. He says that for each $f_{i}$ in the decomposition of $V^{\mathcal{C}}$, we can assign a partition and if I'm understanding correctly, it looks like we can construct a partition for $n$. Let $d_{i}$ denote the degree of $f_{i}$. Based on the setup so far, my questions are as follows.

  1. What exactly is this partition based on the above decomposition? Is the partition on $d_{i}$ or is the partition on $n$?

  2. Does the partition have anything to do with \begin{align*} \displaystyle \sum_{i = 1}^{s} d_{i}e_{i} = n \end{align*}

Any help is greatly appreciated. Thanks! :)

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The point is that the same polynomial $f_i$ may occur many times. So for each polynomial $f$, we have a partition whose parts are the exponents $e_i$ in all the different cases where $f_i=f$. So this is not a partition of $n$ or of $d_i$.

Assume for a moment that $f(u)=u-a$ for some $a$; then the resulting partition is the lengths of the Jordan blocks with $a$ on the diagonal. The other cases are similar, but Jordan decomposition is more complicated over a field that isn't algebraically closed.

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  • $\begingroup$ Thanks for the prompt reply. :) To double check whether I'm understanding you correctly when you say the same polynomial $f_{i}$ may occur many times, so does it mean that in the decomposition of $V^{\mathcal{C}}$ above, for example we could have that $f_{1}$ and $f_{2}$ could be the same irreducible polynomial with exponents $e_{1}$ and $e_{2}$. If the rest of the $f_{i}$ are all different, so the partition assigned to $f_{1}$ (and also $f_{2}$) would be ($e_{1}$, $e_{2}$) WLOG assuming $e_{1} \geq e_{2}$. Is this the way to think about it? $\endgroup$
    – Sudarshan Narasimhan
    Commented Oct 13, 2021 at 13:49
  • $\begingroup$ @SudarshanNarasimhan Yes. $\endgroup$
    – Ben Webster ♦
    Commented Oct 13, 2021 at 14:15
  • $\begingroup$ @BenWebster Do you mean "in all the different cases where $f_i=f$"? $\endgroup$
    – Amritanshu Prasad
    Commented Oct 14, 2021 at 6:35

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