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 <h1>(Representation Theory) Notes on Representations</h1>
 <article>
 <section class="header">
 Posted on 2024-01-22
 
 </section>
 <section>
 Let \(G\) be a finite group, let \(F\) be a field (algebraically closed, char. 0).
A representation of \(G\) is a vector space \(V\) together with a group
homomorphism \[\rho:G \to \operatorname{GL}(V).\]
If we choose a basis \(\mathcal{B}\) of \(V\), we obtain from \(\rho\) an
assignment \(g \mapsto [\rho(g)]_{\mathcal{B}}\) which is a group
homomorphism \(G \to \operatorname{GL}_n(F)\).
<h2 id="first-example">First example</h2>
Consider the finite cyclic group \(G = \mathbb{Z}/n\mathbb{Z}\) for a
natural number \(n &gt; 0\).
Suppose that \(M\) is an \(m \times m\) matrix for which \(M^n = \mathbf{I}\).
Using \(M\), we get a representation
\[\rho_M:\mathbb{Z}/n\mathbb{Z} \to \operatorname{GL}_m(F)\]
by the rule
\[\rho_M(i + n \mathbb{Z}) = M^i.\]
One might consider the question: If \(M\) and \(M'\) are two \(m \times m\)
matrices for which \(M^n = (M')^n = \mathbf{I}\), when are the
representations \(\rho_M\) and \(\rho_{M'}\) “the same”??
Consider the following \(3\times 3\) matrices, each of which satisfies
\(M^3 = \mathbf{I}\):
\[\begin{equation*}
\begin{pmatrix}
0 &amp; 0 &amp; 1 \\
1 &amp; 0 &amp; 0 \\
0 &amp; 1 &amp; 0 
\end{pmatrix},
\begin{pmatrix}
1 &amp; 0 &amp; 0 \\
0 &amp; \zeta &amp; 0 \\
0 &amp; 0 &amp; \zeta^2
\end{pmatrix},
\begin{pmatrix}
0 &amp; 0 &amp; -2 \\
3/2 &amp; 0 &amp; 0 \\
0 &amp; -1/3 &amp; 0 
\end{pmatrix}.
\end{equation*}\]
where \(\zeta\) is a root of \(\dfrac{T^3-1}{T-1} = T^2 + T + 1\). <a href="#fn1" class="footnote-ref" id="fnref1" role="doc-noteref">1</a>
In fact, all three of these matrices have eigenvalues
\(1,\zeta,\zeta^2\) each with multiplicity 1.
<h2 id="example-two-commuting-matrices">Example: two commuting matrices</h2>
Suppose that \(M,N\) are \(m \times m\) matrices for which \(M^n = \mathbf{I}\), \(N^k = \mathbf{I}\) and \(MN = NM\).
We get a representation
\[\rho:\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/k\mathbb{Z} \to \operatorname{GL}_m(F)\]
by the rule
\[\rho(i+n\mathbb{Z},j+k\mathbb{Z}) = M^iN^j.\]
<dl>
<dt>Remark</dt>
<dd>
Let \(\mu\) be an eigenvalue of the matrix \(M\), and let \(V_\mu \subset F^m\) be the corresponding eigenspace.
Since \(M\) and \(N\) commute, a homework exercise shows that \(V_\mu\)
is invariant under the action of \(N\).
Then \[N_{\mid V_\mu}:V_\mu \to V_\mu\] is a linear transformation
satisfying \((N_{\mid V_\mu})^k = \operatorname{id}\).
In particular, we may choose a basis of \(V_\mu\) of eigenvectors for \(N_{\mid V_\mu}\)
In this way, we see that \(V = F^m\) has a basis consisting of
vectors which are simultaneously eigenvectors for \(N\) and for \(M\).
</dd>
</dl>
<h2 id="construction-of-representations-from-group-actions">Construction of representations from group actions</h2>
Suppose that the (finite) group \(G\) acts on the (finite) set
\(\Omega\). We are going to use this action to define a representation
of \(G\).
We consider the vector space \(F[\Omega]\) of all \(F\)-valued functions
on \(\Omega\).
For \(g \in G\), we consider the linear transformation
\[\rho(g):F[\Omega] \to F[\Omega]\]
defined for \(f \in F[\Omega]\) by the rule
\[\rho(g)(f)(\omega) = f(g^{-1}\omega).\]
<dl>
<dt>Proposition:</dt>
<dd>
(a). \(\rho(g)\) is an \(F\)-linear map.
(b). \(\rho(gh) = \rho(g)\rho(h)\) for \(g,h \in G\).
(c). \(\rho(g):F[\Omega] \to F[\Omega]\) is invertible.
</dd>
<dt>Proof of (b):</dt>
<dd>
Let \(f \in F[\Omega]\). For \(\tau \in \Omega\) we have
\[\rho(gh)f(\tau) = f((gh)^{-1}\tau) =
f(h^{-1}g^{-1}\tau) =
\rho(h)f(g^{-1}\tau) =
\rho(g)\rho(h)f(\tau).\]
Since this holds for all \(\tau\), conclude that
\[\rho(gh)f = \rho(g)\rho(h)f.\]
Since this holds for all \(f\), conclude that \(\rho(gh) = \rho(g)\rho(h)\) as required.
</dd>
</dl>
The Proposition shows that
\[\rho:G \to \operatorname{GL}(F[\Omega])\]
is a representation of \(G\).
Here is an alternate perspective on \(F[\Omega]\) and the action of \(G\).
For \(\omega \in \Omega\), consider the Dirac function
\(\delta_\omega \in F[\Omega]\) defined by
\[\delta_\omega(\tau) = \left\{ \begin{matrix}
1 &amp; \omega = \tau \\
0 &amp; \text{else}
\end{matrix} \right.\]
<dl>
<dt>Proposition:</dt>
<dd>
The collection \(\delta_\omega\) forms a basis for the \(F\)-vector
space \(F[\Omega]\).
</dd>
<dt>Sketch:</dt>
<dd>
For \(f \in F[\Omega]\), we have
\[f = \sum_{\omega \in \Omega} f(\omega) \delta_\omega.\]
This shows that the \(\delta_\omega\) span \(F[\Omega]\).
Now, suppose \(0 = \sum_{\omega \in \Omega} a_\omega \delta_\omega\)
for \(a_\omega \in F\) for each \(\omega \in \Omega\).
To prove linear independence, we must argue that \(a_\omega = 0\) for every \(\omega\).
Set \(f = \sum_{\omega \in \Omega} a_\omega \delta_\omega\). For
\(\tau \in \Omega\), the function \(f\) satisfies \[f(\tau) =
\sum_{\omega \in \Omega} a_\omega \delta_\omega(\tau) = a_\tau.\]
Since \(f = 0\) - i.e. “\(f\) is the function which is takes value
zero at every argument” – we conclude that \(a_\tau = 0\) for each
\(\tau\). This proves linear independence.
</dd>
</dl>
Let’s consider the action of \(G\) on these basis vectors.
<dl>
<dt>Proposition:</dt>
<dd>
For \(g \in G\) and \(\omega \in \Omega\), we have
\[\rho(g)\delta_\omega = \delta_{g \omega}.\]
</dd>
<dt>Proof:</dt>
<dd>
For \(\tau \in \Omega\) we have
\[\rho(g) \delta_\omega(\tau) = \delta_\omega(g^{-1}\tau)
= \left\{ \begin{matrix}
1 &amp; \omega = g^{-1}\tau \\
0 &amp; \text{else}
\end{matrix} \right.
= \left\{ \begin{matrix}
1 &amp; g\omega = \tau \\
0 &amp; \text{else}
\end{matrix} \right.\]
This shows that \(\rho(g)\delta_\omega(\tau) = \delta_{g\omega}(\tau)\); since this holds for all \(\tau\) we
conclude \(\rho(g)\delta_\omega = \delta_{g\omega}\) as required.
</dd>
<dt>Remark:</dt>
<dd>
Viewing \(F[\Omega]\) as the vector space with basis \(\delta_\omega\)
gives us a perhaps simpler proof that \(\rho(gh) = \rho(g)\rho(h)\).
Indeed, for \(\omega \in \Omega\) we have \[\rho(gh)\delta_\omega =
\delta_{gh \omega} = \rho(g)\delta_{h\omega} =
\rho(g)(\rho(h)\delta_\omega) =
\rho(g)\rho(h)\delta_\omega.\]
</dd>
</dl>
<h2 id="homomorphisms">Homomorphisms</h2>
Let \((\rho,V)\) and \((\psi,W)\) be representations of \(G\). a linear
mapping \(\Phi:V \to W\) is a homomorphism of \(G\)-representations
provided that for every \(v \in V\) and for every \(g \in G\) we have
\[\Phi(\rho(g)v) = \psi(g) \Phi(v).\]
The \(G\)-representations \((\rho,V)\) and \((\psi,W)\) are isomorphic is
there is a homomorphism of \(G\)-representations which is invertible.
<h3 id="example">Example</h3>
For any \(G\) one has the so-called trivial representation \(\mathbb{1}.\)
This representation has vector space \(V = F\) (it is 1 dimensional!) and the mapping
\[G \to \operatorname{GL}(V) = \operatorname{GL}_1(F) = F^\times\]
is just the trivial homomorphism \(g \mapsto 1\).
One often writes \(\mathbb{1}\) for the underlying vector space \(F\).
Let \(G\) act on the (finite) set \(\Omega\). Then there is a homomorphism
of \(G\)-representations \[\mathbb{1} \to F[\Omega].\]
Viewing \(\mathbb{1} = F\), the scalar \(c\) is mapped
to the constant function with value \(c\).
Put another way, the scalar \(c\) is mapped to the vector
\[c \sum_{\omega \in \Omega} \delta_\omega.\]
<section id="footnotes" class="footnotes footnotes-end-of-document" role="doc-endnotes">
<hr />
<ol>
<li id="fn1">If \(F = \mathbb{C}\) then we can take \(\zeta = e^{2\pi i/3}\).<a href="#fnref1" class="footnote-back" role="doc-backlink">↩︎</a></li>
</ol>
</section>
 </section>
</article>

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