# Let $V=V_{1}⊕ ⋯ ⊕ V_{n}$ be semisimple. $U$ irreducible. Show that $\dim_{k} (Hom_A(U,V))$ is equal to the number of $V_i$ equivalent to $U$.

$\DeclareMathOperator{\End}{End} \DeclareMathOperator{\Ker}{Ker} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\Mat}{Mat} \DeclareMathOperator{\Irr}{Irr}$ Definition.
An $A$-module $V$ is called semisimple if $V$ can be decomposed as a direct sum of irreducible submodules.

Definition.
Let $\Irr(A)$ denote a complete set of representatives for the equivalence classes of irreducible representations of $A$.

Proposition.
Let $V=V_{1}\oplus \cdots \oplus V_{n}$ be a finite dimensional semisimple $A$-module. For each $U\in \Irr(A)$ let $$n_U:=\dim_{k} (\Hom_A(U,V)) \in \Bbb{Z}_{\geq 0}.$$ Show that $n_U$ is equal to the number of $V_{i}$ that are equivalent to $U$.

I've been staring to this for quite some time, but nothing that really comes to my mind. My book says that it has to do with Schur's Lemma. My main problem is that I don't really get what the dim actually means in this context I guess. I would say that $\dim_{k} (\Hom_k(U,V))$ is just $\dim_k(U)\dim_k(V)$.

Edit: After staring a little bit more to it, I think I can reduce this problem by Schur's lemma to showing that:

$$\dim_{k}(\Hom_A(U,V))=\sum _{i}\dim(\Hom_A(U,V_{i})$$

I'm not sure how to show that, but it seems to do with this proposition:

Proposition.
Let $U,V$ be finite dimensional $k$-vector spaces with direct sum decompositions $U=\bigoplus_{j=1}^nU_{j}$ and $V=\bigoplus_{i=1}^mV_{i}$ and with corresponding complete sets of idempotents $e_{j}\in \End_{k}(U)$ and $f_{j}\in \End_{k}(V)$.

We have a $k$-linear isomorphism $M$ between $\Hom_{k}(U,V)$ and the $k$-vector space consisting of $m\times n$ matrices \begin{align*} \begin{pmatrix}\phi _{1,1} &\cdots &\phi _{1,n} \\ \vdots & & \vdots \\ \phi _{m,1} &\cdots &\phi _{m.n} \end{pmatrix} \end{align*} with $\phi _{i,j}\in \Hom_{k}(U_{j},V_{i})$. This isomorphism $M$ is defined by $\phi \mapsto (\phi _{i,j})$ where $\phi _{i,j}=f_{i}\phi |_{U_{j}}$.

• the omomorphisms are a vectorial space, so they have a dimension on the field. You can see it as the cardinality of a base. Through the schur lemma, and other considerations, you can prove this theorem easily – Exodd Oct 4 '14 at 21:20
• It's important that you are only looking at homomorphisms from U to V that commute with the action of A. This is not the product of the individual dimensions as not every linear map has the required property. Now, a morphism from $U$ to $V$ is just a choice of morphism from $U$ to $V_{i}$ for each $i$. Now use Schur's Lemma. – Siddharth Venkatesh Oct 4 '14 at 23:19
• possible duplicate of Remark 3.1.3 from Introduction to Representation Theory from Pavel Etingof – Julian Kuelshammer Oct 5 '14 at 9:29
• @SiddharthVenkatesh Okay, I would be able to prove this using schur's lemma if $$\dim_{k}(\Hom_A(U,V))=\sum _{i}\dim(\Hom_A(U,V_{i})$$ But this doesn't seem completely trivial to me. – Kasper Oct 5 '14 at 9:38
• I would suggest just trying to prove it. Use the projections onto the component factors and the inclusion of the component factors to show that any map from $U$ into $\oplus_{i} V_{i}$ corresponds bijectively to the $\oplus_{i} \Hom(U, V_{i}).$ – Siddharth Venkatesh Oct 5 '14 at 14:00