Why are we interested in irreducible representation but not faithful representation?

I am reading some materials of representation theory (of a group).

The motivation of representation theory is to represent (by a homomorphism $h: G \to GL(V)$, from the group $G$ to a vector space $V$), so that we can study the group in an "easier" subject linear algebra.

I found that most introductory texts introduce subrepresentation, irreducible representation and then theorems about them. I know that we may be interested in finding the "atom" representation of a group on a vector space like we decompose natural numbers into prime.

But isn't the original motivation is to study the group? The representation is irreducible or I can decompose it into subrepresentation doesn't necessary conclude that the representation is faithful ($h$ is injective). We lose many information of $G$ in $\operatorname{Ker}(h)$.

Do I miss out something? Or, the idea of "atom" could turn out to prove deep result about the structure of a faithful representation of $G$?

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A non faithful representation of $G$ is the samee thing as a faithful representation of $G/\ker h$... – Mariano Suárez-Alvarez Mar 10 '14 at 5:52
Yes. It is the result from First Isomorphism Theorem, do you mean that if I want to have a faithful representation of $G$, I just find a "bigger" group $H \supset G$, and then find a representation of $H$, finally we use $H / \operatorname{ker} h$ to do our works? – wonghang Mar 10 '14 at 5:57
Hm? I have no idea what you just wrote. The study of group representations considers all representations, because all representations arise in practice; even the trivial representation plays an extremely important role, although this may seem rather surprising to anyone starting. Maybe you can just withhold judgment until you know a little bit more about the subject? That plan usually does marvels! – Mariano Suárez-Alvarez Mar 10 '14 at 6:02

3 Answers

Each individual (non-faithful) irreducible representation loses some information about the group, but one studies the collection of all irreducible representations, and the information about the group is all there.

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I find this a natural question. Particularly because I recently heard that I would be teaching group theory for advanced undergrads next year (or the year after), so I want to test my motivational skills here :-)

Indeed, one of the aspects of representation theory is to study the groups being represented. It is nice that we can take some abstract group, and replace its elements with objects that are easier to compute with: matrices (as in linear representations) or permutations (as in groups acting on sets). To that end it is helpful (if not imperative) that the representation is faithful lest we draw mistaken conclusions about identities of elements of the said group. Even if the action is not faithful we can slice off the kernel of the action as a part of a divide-and-conquer approach to understanding the group better, if/when so inclined.

On the other hand we also want to understand what kind of objects can be acted upon by a given group (either via permutations or linear transformations). There the focus is different. If we look at objects being permuted by a group, the first divide-and-conquer step is to study the objects one orbit at a time, as the orbits don't interact (but e.g. their sizes are constrained by the size of the acting group). If we look at vector spaces acted upon by a group of linear transformations, then the divide-and-conquer approach tells us to study subspaces stable under the group, and check whether we can build the whole space from atomic subspaces (often we can).

So representation theory is in some sense a two-way street: extract information about the group by studying its possible actions, and OTOH extract information about a collection of objects given that they are acted upon by a given group in some meaningful way.

Possibly the best known result in the former direction is the Burnside's $p^aq^b$-theorem telling us that a finite group whose order has only two prime divisors is necessarily solvable. This is likely covered in all texts. Going the other way let me lead off with a simple example. Let $V$ be the space of complex functions on the real line with period $2\pi$. If $f(t)$ is such a function, then so is the function $g\cdot f$ defined by 180 degree phase shift: $(g\cdot f)(t)=f(t+\pi)$. As the functions in $V$ have period $2\pi$, we see that $g\cdot(g\cdot f)=f$ for all $f$ meaning that the cyclic group $C_2=\langle g\rangle$ of order two generated by $g$ acts on $V$. What can we say about this action? Representation theory tells us that there are the following kind of special functions in relation to this action. The space $$V^+=\{f\in V\mid g\cdot f=f\}$$ of functions that have $\pi$ as a period, and the space $$V^-=\{f\in V\mid g\cdot f=-f\},$$ i.e. the function $f$ with the property $f(t+\pi)=-f(t)$ for all $t$. Representation theory also tells us that $V=V^+\oplus V^-$. It is an easy exercise that given any function $f\in V$, then the function $$f^+(t)=\frac{f(t)+f(t+\pi)}2\in V^+,$$ and $$f^-(t)=\frac{f(t)-f(t+\pi)}2\in V^-.$$ Furthermore $f=f^++f^-$. Once you reach the idempotents of the group ring in your studies, this should give you a deja vu -experience :-) At that point an easy exercise for you is to figure what happens, if $g$ acts by a phase shift by $2\pi/n$ for some natural number $n>2$.

A more high-browed example of this latter direction comes from particle physics. There it is imperative that the symmetries of the nature are not broken. Quantum physics uses vectors in some space to represent elementary particles. What then happens is that particles can only react in such a way that the tensor product of those vectors has a non-trivial component acted upon trivially by the group. IOW: "trivial representation = invariants" is a very important representation - fully agree with Mariano.

Caveat: I did go on a hyperbole with the last example, and the painted picture may not be quite accurate. Anyway, they were handing out Nobel prizes in physics for that stuff in the 60s and 70s :-)

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Thank you very much for your answer. I have spent many times on finding out in particle physics (self-study and google around, I am not in an university) to try to figure out your example. Basically, the symmetry of nature (in standard model) is $SU(3) \times SU(2) \times U(1)$ somehow describe the nuclear strong force, weak force and electromagnetic force. Then, the irreps of an element of $SU(3) \times SU(2) \times U(1)$ would describe a particle. Why trivial representation is important is that it is used to describe an elementary particle. Did I get it right? – wonghang Mar 26 '14 at 2:33

A faithful representation $f\colon G \to GL(V)$ leads to infinitely many faithful representations, one each on $V, V\oplus V, V\oplus V\oplus V$ etc simply by replicating it in each factor. These extra things provide no new information. Irreducibilty would have eliminated this infinite unnecessary explosion.

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