I might need help here in understanding my own question in places and please don't hesitate in asking for edits and clarifications.

Background: A representation $\rho$ of a finite group $G$ is a group homomorphism from $G$ into $GL(V)$ for some vector space $V$.

If $W$ is a subspace of $V$ invariant under $\rho(G)$, then $\rho_{\left.\right|W}$ is called a subrepresentation. In the usual way we can show that every representation is a direct sum of irreducible representations.

If we endow $V$ with an inner product $\langle \cdot,\cdot\rangle$, then we can show that $$\langle u,v\rangle_\rho=\sum_{t\in G}\langle \rho(t)u,\rho(t)v\rangle$$ is another and furthermore, with respect this inner product, the operators $\rho(s)$ are unitary.

Where $d_i$ is the dimension of the vector space $V_i$, where $\rho_i:G\rightarrow GL(V_i)$ it can be shown that the regular representation, which acts on the vector space $V_r=\mathbb{C}^{|G|}$, can be decomposed as

$$V_r=d_1V_1\oplus d_2V_2\oplus\cdots d_nV_n,$$ where $\{\rho_i\}_{i\in[n]}$ are the unitary irreducible representations $\rho_i:G\rightarrow GL(V_i)$ and so $$r:G\rightarrow GL\left(\bigoplus_{i\in[n]}d_iV_i\right),$$ and we write $$r=\bigoplus_{i\in[n]}d_i\rho_i.$$

Using the fact that the irreducible representations are equivalent to unitary ones allows us to show that the matrix elements of the unitary irreducible representations are orthogonal as elements of $F(G)$ with respect to the inner product $$\langle f,g\rangle=\frac{1}{|G|}\sum_{t\in G}\overline{f(t)}\cdot g(t),$$

and as there are $|G|$ of them, the matrix elements form an orthogonal basis of $F(G)$.


  1. With respect to the canonical basis of $F(G)$, in what sense can I talk about the matrix elements of the unitary irreducible representations?
  2. Is there any natural way that the matrix elements of unitary irreducible representations form a basis?

Example: I think I have for $G=\mathbb{Z}_3$, that there are unitary irreducible representations $\{\tau,\rho_1,\rho_2\}$ with matrix elements $$a_0:=\left(\begin{array}{c}1 \\ 1 \\ 1\end{array}\right),\,a_1:=\left(\begin{array}{c}1 \\ \omega \\ \omega_2\end{array}\right) \text{ and }\,a_2:=\left(\begin{array}{c}1 \\ \omega^2 \\ \omega\end{array}\right)\in F(\mathbb{Z_3}).$$

These are written with respect to the canonical basis of $F(\mathbb{Z_3})$ (although everything here is easier as $\mathbb{Z}_3$ is abelian).

Consider, with respect to the canonical basis of $F(G)$, $$f=\left(\begin{array}{c}1 \\ 2 \\ 3\end{array}\right).$$

With respect to the basis $\{a_0,a_1,a_2\}$ basis, I think $$f=\left(\begin{array}{c}2 \\ \frac{1}{\sqrt{3}}e^{5\pi i/6} \\ \frac{1}{\sqrt{3}}e^{7\pi i/6}\end{array}\right).$$

  • $\begingroup$ Every basis induces a set of matrix elements, and different bases can induce different sets of matrix elements. If you fix your choice of basis to be the canonical basis, then there can be only one set of matrix elements corresponding to it. So I don't see how your first question makes sense. (The full set of matrix coefficients include those from all bases btw.) I also don't know what you're asking in the second question; a set of vectors is either a basis or it isn't, what exactly is a "way" of being a basis? Really I don't understand what you're asking at all. $\endgroup$ – whacka Feb 8 '15 at 7:06
  • $\begingroup$ @whacka For a representation $\rho:G\rightarrow GL(V)$ I understand that the basis of $V$ determines the $\rho_{ij}\in F(G)$. What is a canonical basis of $V$ though? I am talking about the matrix elements from a maximal family of pairwise-inequivalent irreducible unitary representations. The second question is far softer I agree. $\endgroup$ – JP McCarthy Feb 10 '15 at 16:20
  • $\begingroup$ Arbitrary vector spaces do not come equipped with canonical bases... And I still don't know what you're asking in either question. (It's not that the second is "soft" - people can answer soft questions meaningfully - it's unintelligible to me.) The best I can do is answer the question in the title (which is soft but intelligible): no as stated. The collection of all reps equivalent to a given one would be like a proper class - there is no conceptual limit on what can go in this collection, so it's "too big" to be a set - so certainly this collection is bigger than one. $\endgroup$ – whacka Feb 11 '15 at 5:37
  • $\begingroup$ One can ask the analogous question "given a set, is there a unique set it has the same size as?" to which the answer is clearly no. Sometimes the irreps can be parametrized by canonical constructions though. (The analogous fact for the set situation would be: every set is equal in size to the smallest ordinal of its size.) For instance, the (complex) irreps of any finite symmetric group are parametrized by integer partitions, and there is an explicit construction process that tells anyone how to construct the irrep corresponding to any chosen integer partition. $\endgroup$ – whacka Feb 11 '15 at 5:40
  • $\begingroup$ @whacka Regarding my mention of canonical basis, I was only referencing your clause "If you fix your choice of basis to be the canonical basis". Regarding my second question; an example: know that there are as many irreducible representations as conjugacy classes. Could we set up a natural bijection so that we could write the matrix elements in terms of conjugacy classes? This MO questions suggests no for example (mathoverflow.net/questions/102879/…). $\endgroup$ – JP McCarthy Feb 11 '15 at 11:18

I just had this same question, and I think the answer is NO.

Let me first restate the question, since nobody seems to have understood how to phrase it.

Let $\rho: G \to C^{nxn}$ be a representation. All equivalent representations are given by $M^{-1}\rho M$ for invertible matrices $M$. The "Weyl unitary trick" shows that for any $\rho$, there is an equivalent unitary representation. The question is: is this unique?

The answer is NO. Suppose $\rho$ is a unitary representation. Let U be a unitary matrix. Consider $U^* \rho U$. This is also unitary: $$ (U^* \rho(g) U) (U^* \rho^*(g) U) = U^* \rho(g) \rho^*(g) U = U^* I U = I.$$

I hope this helps (assuming I understood the question correctly).

  • $\begingroup$ Without thinking in much detail is that not equivalent to changing the basis? $\endgroup$ – JP McCarthy Dec 18 '16 at 20:13
  • $\begingroup$ Yes, it is ... (and every equivalent representation is obtained by such a change of basis). $\endgroup$ – johnny Dec 18 '16 at 20:25
  • $\begingroup$ To be more clear .. We first view representations as homomorphisms from G to the group of n x n invertible matrices. Then we define equivalence of representations ($\rho \sim \rho'$ iff $\exists M$ s.t. $ \rho = M^{-1} \rho' M$ -- this amounts to just a basis change). Then we ask: is there a unique unitary representation in each equivalence class? Then we answer: no. This was the (trivial) question that I had asked myself, and upon searching for further discussion I stumbled upon this thread. $\endgroup$ – johnny Dec 18 '16 at 20:29
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    $\begingroup$ Ah... then I think you have answered my first question... the 'different' unitary representations are equal except for a basis change. $\endgroup$ – JP McCarthy Dec 19 '16 at 11:45

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