4
$\begingroup$

I have the definition of a homomorphism as map such that $\varphi(g_1g_2)=\varphi(g_1)\varphi(g_2)$

I have the definition of $\mathrm{Hom}(V,W)$ as

$$\begin{align}\mathrm{Hom}(V,W) &= \{\mathbb{C}-\text{linear maps }\varphi:V \rightarrow W \} \\ &\cong \{n \times m \text{ matrices } \} \end{align}$$

Does $\mathrm{Hom}$ stand for homomorphisms because I cannot see how it relates to the definition of homomorphism

I have the definition of $\mathrm{Hom}_G$ as

$$\begin{align}\mathrm{Hom}_G(V,W) &= \{\varphi \in \mathrm{Hom}(V,W) \mid g\varphi(v)=\varphi(gv), \forall g \in G, \forall v \in V\} \\ &=\{\varphi \in \mathrm{Hom}(V,W) \mid \rho(g)\cdot\varphi(v)=\varphi(\rho(g)v) , \forall g \in G, \forall v \in V\} \end{align}$$

I then have the question:

Prove that if $\rho$ is an irreducible representation, then for an element $g \in G$

$$g \in Z(G) \iff \rho(g)=\lambda I$$

In the solution I have that:

$(``\implies")$ If $g \in Z(G)$ then $gh=hg \ \forall h \in G$. By definition of $\mathrm{Hom}_G$ this means that $\rho(g) \in \mathrm{Hom}_G(V,V)$. But \rho is irreducible so $\mathrm{Hom}_G(V,V)$ consists of scalar matrices by Schur's Lemma.

I do not understand how "By definition of $\mathrm{Hom}_G$ this means that $\rho(g) \in \mathrm{Hom}_G(V,V)$. " How is this the definiton of $\mathrm{Hom}_G$? I cannot see why this is equivalent.

$\endgroup$
  • 1
    $\begingroup$ Why do you say that $\rho(g) \rho(h)= \rho(h) \rho(g)$ ? In general $G$ need not to be commutative. $\endgroup$ – Crostul May 30 '15 at 11:01
  • 1
    $\begingroup$ There should not be an equal sign between the set of linear maps and the set of matrices; these spaces are merely isomorphic $\endgroup$ – Hagen von Eitzen May 30 '15 at 11:09
  • $\begingroup$ I was trying to quote a small part of the solutions. I will edit the question to make things clearer. $\endgroup$ – Permian May 31 '15 at 10:52
  • $\begingroup$ What about the trivial representation? In that case, $\rho(g)=I$ for all $g\in G$, so the statement is false if $G$ is not abelian. $\endgroup$ – Pierre-Guy Plamondon May 31 '15 at 21:19
  • $\begingroup$ That is weird because doesnt state it needs to be non abelian $\endgroup$ – Permian Jun 1 '15 at 9:00
3
$\begingroup$

The word homomorphism is generally used to mean "a map that preserves the structure of the object I'm looking at". Thus the precise definition is different for groups, vector spaces, group representations, etc., even though we use the same word.

For your second question, assume that $g$ is an element of the center of $G$, and let $\varphi = \rho(g)$. Let $h$ be an element of $G$. Since $g$ is in the center, we have $gh=hg$, so $\rho(g)\cdot \rho(h) = \rho(h)\cdot\rho(g)$. By our notation, this can be written as $\rho(h)\cdot\varphi = \varphi\cdot\rho(h)$; in other words, for any $v\in V$, we have that $\rho(h)\cdot\varphi(v) = \varphi(\rho(h)(v))$. By definition of $Hom_G$, this means that $\varphi\in Hom_G(V,V)$. Since $\varphi=\rho(g)$, this answers your question.

$\endgroup$
  • $\begingroup$ I still dont understand your second paragraph. I have the definition for $\mathrm{Hom}_G$ as (crudely) $g\varphi=\varphi g$ which doesnt mean $\rho(g)\rho(h)=\rho(h)\rho(g)$. How is this rectified???? $\endgroup$ – Permian Jun 1 '15 at 9:03
  • $\begingroup$ @sandstone I have edited my answer. $\endgroup$ – Pierre-Guy Plamondon Jun 1 '15 at 9:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.