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I just started Representation theory and am quite confused by the notation and the form of representations.

Firstly the definition of a representation we were given was.

"A representation of a group $G$ is a finite-dimensional vector space V over $\mathbb{C}$ equipped with a group morphism $$\rho :G\rightarrow \text{Aut}_{\mathbb{C}}(V)$$"

So each element in G gets mapped to an invertible linear transformation $T_g :V \rightarrow V$ ? So is our vector space a space of invertible matrices? (In finite case anyway)

Secondly I'm confused about character of a representation.

The definition given,

"the character of $(V,\ \rho)\in\ \text{Rep}(G)$ is the map $$\chi ^{(V,\ \rho )} :G \rightarrow \mathbb{C}\\ g\mapsto \text{tr}(\rho (g)$$

So here the character is defined in terms of a single element, but does our representation not have a matrix, defined by $\rho (g)$ for each $g \in G$? how does this give the character of the entire representation if it's defined in terms of a single element?

Obviously I have fundamental misconceptions of this subject and the objects involved, I've only had three lectures so far so hoping to clear them up and keep learning.

If anyone could help me get a clearer idea of the objects involved I'd really appreciate it!

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    No, the vector space is a vector space. The map $\rho$ associates an element of $g$ to a matrix on $V$. One way to think of this is to think of $g$ being "represented" as a symmetry of your vector space. – 2017-01-30
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    For the second question: the character is also a function of $G$, as the notation says. Given a representattion $\rho$ there character is defined to send each $g$ to the trace of $\rho(g)$. – 2017-01-30
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    Yes I understand it's a vector space, but of what exactly? Isn't $$\text{Aut}_{\mathbb{C}}(V)$$ a vector space? there are more than one examples of a vector space. – 2017-01-30
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    So the character is different for each $$\rho (g)$$ and not just a single value for the representation $$ \rho $$? – 2017-01-30
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    No, $\mathrm{Aut}_{\mathbb C}(V)$ is not a vector space. The set of all linear transformations of $V$ is a vector space, but the set of invertible linear transformations is not, because, amongst other things $0\cdot T$ is not invertible if $T$ is invertible. The only vector space here is $V$. Everything else is groups. – 2017-01-30
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    so $$\rho (g) $$ is not in our vector space? – 2017-01-30
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    Correct, $\rho(g)$ is not in your vector space, it is an automorphism of your vector space. If $g$ has order $2$, for example, then $\rho(g)$ my be a reflection, or a rotation by 180 degrees. – 2017-01-30
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    Ok, thank you. So, our $$\rho (g)$$ for our calculation in the character is different for each g?\\ I'm just confused about how we can talk about the character of our representation in terms of a single element? – 2017-01-30

2 Answers 2

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Perhaps the thing to do here is to unravel the definition. If $V$ is a complex vector space, then $Aut(V)$ is the space of $C$-linear invertible maps $V\rightarrow V$. Fix a basis of $V$. Then a linear map $f:V\rightarrow V$ is uniquely represented by a some $\dim V\times \dim V$ matrix $A$ (with respect to the basis of $V$) with entries in $C$ in the sense that $f(v)=Av$ for all $v\in V$. This gives an isomorphism of $Aut(V)$ (a group) with $GL_N(C)$, where $N=\dim V$. So a representation $\rho:G\rightarrow Aut(V)$ can be reinterpreted as a homomorphism $\rho:G\rightarrow GL_N(C)$. This is obviously dependent on the choice of basis, but since any two bases of $V$ are conjugate, this isn't a problem.

In short: a representation is a way of assigning to each $g\in G$ a linear endomorphism of $V$ or, equivalently, an $N\times N$ matrix with respect to some fixed choice of basis.

The character $\chi_\rho$ of $\rho:G\rightarrow Aut(V)$ is defined by $\chi_\rho(g)=\mathrm{tr}(\rho(g))$; this is simply the trace of $\rho(g)$ as a linear operator, which is the sum of the diagonal elements, or the sum of the eigenvalues, of the matrix representation of $\rho(g)$ with respect to any choice of basis.

Here's one last way of rephrasing the above: a representation of $G$ is a pair $(\rho,V)$, where $V$ is a $C$-vector space and $\rho:G\rightarrow Aut(V)$ is a homomorphism. This is precisely the same as saying that a representation is a group action of $G$ on some $C$-vector space.

A completely trivial example to illustrate what's going on: write $Z/2Z=\{1,g\}$, where $g^2=1$ is the non-trivial element. Then a 1-dimensional representation of $Z/2Z$ is defined by $1\mapsto 1$, $g\mapsto -1$. This is $Z/2Z$ acting on the 1-dimensional $C$-vector space, which I've identified with $C$ by choosing the basis $\{1\}$. The group $Aut(V)$ is equal to $C^\times=GL_1(C)$.

If that's still unclear, try to work out an example in higher dimensions. For example, you can represent $Z/NZ$ on a 2-dimensional vector space by identifying $Z/NZ$ with $\{0,\dots,N-1\}$ modulo $N$, and letting $k$ act as rotation by $e^{2\pi i k/N}$ (which is a representable by a $2\times 2$ orthogonal matrix). (And if you want another example after that, let $S_n$ act on $C^n$ w.r.t some basis $\{v_1,\dots,v_n\}$ in the following way: if $\sigma\in S_n$ is a permutation of $\{1,\dots,n\}$, then define $\rho:S_n\rightarrow Aut(C^n)$ by $\rho(\sigma)(a_1v_1+\dots+a_nv_n)=a_1v_{\sigma(1)}+\dots+a_nv_{\sigma(n)}$.)

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    Thank you for the answer, I think it helps a lot. I just need a few reads to digest it fully. Does it explain how the character, is defined in terms of a *single* element g? so when you say "In short: a representation is a way of assigning to each g∈G a linear endomorphism of V or, equivalently, an N×N matrix with respect to some fixed choice of basis." when we take the trace of $$\rho (g_1)$$ will that not be different to trace of $$\rho (g_2)$$ and if they are different which is the character of our representation $$(V,\ \rho)$$ – 2017-01-31
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    Yes, in general they will be different. The trace of $\rho$ is a map $G\rightarrow C$ which in general won't be a homomorphism (since trace isn't multiplicative). The character is the data of *all* of these traces! It assigns to each element a complex number, and it just so happens that this is enough to tell you just about everything you could wish to know about $\rho$. – 2017-01-31
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    I think I see where my main confusion arose, my lecturer kept referring to *the character* of *the representation*, so I think I was thinking it was a single complex number that told us a lot about our representation. Then I was confused to see it defined in terms of a single element. So it can be different for each element, and there is no overall number\character that we have to compare representations? we just look at the character for each element? Thank you so much for taking the time to answer my questions. – 2017-01-31
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    The character is a function. When you define a function by saying what it does to each element of the domain: the character of a representation is the function which maps each element of the group to it's trace under the representation. – 2017-02-01
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The mistake you seem to be making is that the character is somehow determined by a single element of $G$. This is wrong.

A character is a function $$\chi_\rho:G\to\mathbb{C}$$ given by $\chi_\rho(g)=\mathrm{tr}(\rho(g))$ for each $g\in G$.

Example: The left regular representation of $G$ is defined as follows: let $V$ be a vector space with basis $\{v_g\mid G\in G\}$, and define $$\rho:G\to \mathrm{Aut}(V)$$ by $\rho(g)(v_h)=v_{gh}$. In this case elements of $G$ are represented by permutation matrices, and $$\chi_\rho(g)=\begin{cases}|G|&\mbox{if }g=1\\0&\mbox{otherwise.}\end{cases}$$

Example: The group $S_3$ has a 2-dimensional representation $\rho:S_3\to GL_2(\mathbb{C})$ given by \begin{align} \rho(1)&=\begin{pmatrix}1&0\\0&1\end{pmatrix}\\ \rho((12))&=\begin{pmatrix}-1&1\\0&1\end{pmatrix}\\ \rho((23))&=\begin{pmatrix}1&0\\1&-1\end{pmatrix}\\ \rho((123))&=\begin{pmatrix}0&-1\\1&-1\end{pmatrix}\\ \rho((321))&=\begin{pmatrix}-1&1\\-1&0\end{pmatrix}\\ \rho((13))&=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}\\ \end{align} Therefore, the character is given by $$\chi_\rho(g)=\begin{cases}2&\mbox{if }g=1\\ 0&\mbox{if }g=(ij)\mbox{ is a transposition}\\ -1&\mbox{if }g=(ijk)\mbox{ is a 3-cycle.}\end{cases}$$

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    Thank you, I was just confused as my lecturer kept talking about *the character of our representation* so I was thinking it was a single complex number for the representation. I think i have a better idea of it now, still not sure enough of all the objects introduced in representation theory to actually answer problems. A lot to unravel in different definitions and hard to get a clear idea of the form of objects or how they interact. – 2017-01-31