I am learning group theory and representations from a book and there are some default of printing and things I don't get.

Position of the problem :

We wrote the orthogonality of characters of representations :

$$ \frac{1}{[g]} \sum_g \chi_D^\mu(g)(\chi_D^\nu(g))^*=\frac{1}{[g]} \sum_{i=1}^k k_i \chi_{iD}^\mu(\chi_{iD}^\nu)^* = \delta^{\mu \nu}$$

  • $g \in G$ is an element of the group that we are representing
  • $\chi_D^\mu(g) = Tr(D^\mu(g))$ where $D^\mu$ is an irreducible representation of the group.
  • $[g]$ represent the cardinal of the group
  • $k_i$ is the number of elements in the i'th conjugacy class
  • $k$ is the number of conjugacy classes in the group $G$
  • $\chi_{iD}^\mu$ is the value of the character inside of the i'th class of conjugacy (as the character is constant over a conjugacy class)

My problem :

In the book they say that we can interpret the last sum as orthogonality between vectors in a k dimension space. But I can't read between which vectors they talk (default of print).

So I guess they are talking about :

$(k_1 \chi_{1D}^\mu,..., k_k \chi_{kD}^\mu)$ and $(\chi_{1D}^\nu,..., \chi_{kD}^\nu)$

Am I right ?

Then they say : "As we can't have more than k vectors in this vector space of dimension k, the number "r" of irreducible representation is such as : $r \leq k$".

I don't get why. Indeed for me the number of characters is lower than $k$ here but not the number of irreducible representation.

And at this stage we only showed that if two representation are equivalent then they have the same character but we did'nt showed that there is 1 to 1 correspondance between character and irreducible representation.

Could you help me to understand these points ?


The vector space they are talking about are the class functions on $G$, that is, the vector space of functions which are constant on conjugacy classes: $F_c(G, \mathbb{C}) = \{f: G \to \mathbb{C} \mid f(ghg^{-1}) = f(h)\}$. This space has dimension $k$, and each character belongs to the space, since characters are constant on conjugacy classes.

Put an inner product $(\cdot, \cdot)$ on $F_c(G, \mathbb{C})$ by the following formula:

$$ (f_1, f_2) = \frac{1}{|G|} \sum_{g \in G} f_1(g) f_2(g)^*$$ Then the orthogonality formula is saying that irreducible characters are orthonormal with respect to this inner product: this is the key point. From this you get that:

  1. Characters of distinct irreducible representations are linearly independent, and therefore distinct.
  2. Since $F_c(G, \mathbb{C})$ has dimension $k$, and considering point (1) we have that the number of distinct irreducible representations is at most $k$.
  • $\begingroup$ My question is probably simple but why this space has dimension k ? If I would have a linear map between two vector space I know that the dimension of these linear map space is the product of the dimension of the starting and final space but here I have a group which is not a vector space. $\endgroup$ – StarBucK May 29 '17 at 23:23
  • $\begingroup$ If I suppose that $k=2$, I agree that for any $\alpha$, $\beta$ : $\alpha \chi_1 + \beta \chi_2$ is another function of $F_c$ but $ g \mapsto (\chi_1(g))^2$ also and it is not a linear combination of $\chi_1$ and $\chi_2$. What do I misunderstand ? (I learnt linear algebra but some time ago so It is probably a very simple question). $\endgroup$ – StarBucK May 29 '17 at 23:32
  • $\begingroup$ Another way of thinking about the vector space $F_c$ is being made out of column vectors where each entry corresponds to a conjugacy class in $G$, hence $F_c$ has dimension $k$. As for your second example, $g \mapsto (\chi_1(g))^2$ is also an element of $F_c$, and so just like any vector in a vector space, there are multiple bases in which you could break it into a sum. The above theorem shows that $G$ has exactly two irreducible characters, and so it will be some linear combination of these. $\endgroup$ – Joppy May 29 '17 at 23:40

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