I need some help to understand the proof of this theorem which can be found in the book Introduction to Representation Theory by Pavel Etingof, Oleg Golberg, Sebastian Hensel, Tiankai Liu, Alex Schwendner, Dmitry Vaintrob, and Elena Yudovina.

Let $V$ be a finite dimensional representation of $A$, and $0=V_0\subset V_1 \subset ...\subset V_n=V$, $0=V_0'\subset V_1' \subset ...\subset V_m'=V$ be filtrations of $V$, such that the representations $W_i:=V_i/V_{i-1}$ and $W_i':=V_i'/V_{i-1}'$ are irreducible for all i. Then $n=m $, and there exists a permutation $\sigma$ of 1, ..., n such that $W_{\sigma (i)}$ is isomorphic to $W_i'$.

First proof: (for k of characteristic zero). The character of V obviously equals the sum of characters of $W_i$, and also the sum of characters of $W_i'$ But by Theorem 2.17, the characters of irreducible representations are linearly independent, so the multiplicity of every irreducible representation $W$ of $A$ among $W_i$ and among $W_i'$ are the same. This implies the theorem.

Second proof: (general) The proof is by induction on $\text{dim}V$ . The base of induction is clear, so let us prove the induction step. If $W_1=W_1'$ (as subspaces), we are done, since by the induction assumption the theorem holds for $V/W_1$. So asume $W_1\neq W_1'$. In this case $W_1 \cap W_1' = 0$ (as $W_1, W_1'$ are irreducible), so we have an embedding $f: W_1 \oplus W_1' \rightarrow V$. Let $U=V/(W_1 \oplus W_1')$ and $0=U_0 \subset U_1 \subset ... \subset U_p=U$ be a filtration of $U$ with simple quotients $Z_i=U_1/U_{i-1}$ (it exists by Lemma 2.8 in the book). Then we see that:

1) $W/W_1$ has a filtration with successive quotients $W_1', Z_1, ...,Z_p$ and another filtration with successive quotients $W_2. ..., W_n$.

2) $W/W_1'$ has a filtration with successive quotients $W_1, Z_1, ...,Z_p$ and another filtration with successive quotients $W_2'. ..., W_n'$.

By the induction assumption, this means that the collection of irreducible representations with multiplicities $W_1. W_1', Z_1, ..., Z_p$ coincides on one hand with $W_1, ..., W_n$, and on the other hand, with $W_1', ..., W_n'$. We are done.

I'm trying to complete some details and understand the proof.

With respect to the proof 1.

In this proof they claim that the character of $V$ is equals to the sum of the characters of $W_i$, and also the characters of $W_i'$. This means that each $W_i$ is not isomorphic to $W_j$ with $i \neq j $ as representation of $V$. The same happen with the $W_i'$ representations.

Now, in the book we have the following definition:

Let $A$ be an algebra and $V$ a finite-dimensional representation of $A$ with action $\rho$. Then the character of $V$ is the linear function $\chi_{V}: A \rightarrow k$ given by $$\chi_{V}(a)=tr|_{V}(\rho(a))$$

On this point I understand that the characters of irreducible representations are linearly independent, but I have a problem with the conclusion. Why this is possible? There is a note that says the following

This proof does not work in characteristic $p$ because it only implies that the multiplicities of $W_i$ and $W_i'$ are the same modulo $p$, which is not sufficient. In fact, the character of the representation $pV$, where $V$ is any representation, is zero.

With respect to the second proof.

The base induction. If $\text{dim}V=1$, then we have $V$ has the filtration $0=V_0 \subset V_1=V$ (the reason is $V=\langle v\rangle$ where $v \in V$) then obviously any representation has the same length because any element $v \in V$ generates $V$ and $V/0$ is the same $V$. Right?

Finally, I want to understand how works the embedding $f$ in the case when $W_1\neq W_1'$



For the first part, we always have that the irreducible characters are linearly independent vectors in say the vector space of functions $A \rightarrow k$. This means that any function from $A \rightarrow k$ can be written as a $k$-linear combination of irreducible characters in at most $1$ way.

In the first argument they write the character of the representation as a $\mathbb{Z}$-linear combination of the characters of the irreducible quotients in the composition series, by repeatedly using the simple fact from linear algebra that trace is additive on short exact sequences. In characteristic $0$ we have a natural inclusion from $\mathbb{Z}$ into $k$, so we can view this $\mathbb{Z}$-linear combination as a $k$-linear combination and we get that it is unique by linear independence. In characteristic $p$ we don't have an inclusion of $\mathbb{Z}$ into $k$ so this won't work.

For the second part you are correct, when $\dim(V) =1$ it is automatically irreducible and there isn't anything really to check.

For your last question, whenever you have two subspaces inside a vector space (or module) you get a map from their direct sum into the vector space, this is the universal property of direct sums (or coproducts more generally). If the subspaces have trivial intersection (which is forced in this case by irreducibility) then this map is an embedding. In this case, the map $f$ is just the inclusion of the vector subspace spanned by $W_1$ and $W_1'$ into $W$.

  • $\begingroup$ It is correct. Thank you so much. $\endgroup$ – JimmyJP Mar 16 '15 at 16:58

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