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Why we need to study representations of matrix groups? For example, the group $SL_2(F_q)$, where $F_q$ is the field with $q$ elements, is studied by Drinfeld. I think that these groups are already given by matrices. The representation theory is to represent elements in an algebra or group (or other algebraic structure) by matrices. Why we still need to study representations of matrix groups? Thank you very much.

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A large purpose of representation theory is to use "symmetry" of objects arising in algebra, geometry, and analysis to better understand the structure of those objects; the aim is not to describe a group via matrices as an end unto itself. By the classification of finite simple groups, the most important finite groups are largely matrix groups over finite fields. If you study representations of a finite group $G$ in char. 0 and $G$ has a composition series whose successive quotients are matrix groups over finite fields then those matrices have nothing to do with ones in char. 0. –  user76758 Jan 6 at 15:35
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Just as an aside, when I first began to learn about representation theory as an undergraduate, I had the exact same question as you raise above. Eventually I came to realize that the process of learning a subject via the "abstract" approach rather than through the natural examples that motivate it (and must have been in the back of the mind of the instructor without ever being discussed in the course) often leads to the wrong impression by a beginner as to what the point of a given mathematical theory is. –  user76758 Jan 6 at 16:19
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This would be a good question for MathStackExchange. –  Alain Valette Jan 6 at 17:40
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@AlainValette : While I agree that it is borderline, the question has produced some very nice answers, so I vote to keep it open. –  Andy Putman Jan 6 at 19:56
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@Andy: I agree that the question produced some good answers. On the other hand, this basic question should be answered in any introductory book in representation theory (even if it is not always so, unfortunately...) –  Alain Valette Jan 6 at 21:20

5 Answers 5

I will answer a more general question.

If $X$ is an important mathematical object, why should we study the representation theory of $Aut(X)$?

Maybe algebraic geometers care about $X$, or maybe mathematical physicists, or maybe homotopy theorists, etc. Each discipline has techniques that can build a new object $X'$ out of $X$. These new objects are usually more inspiring (if less fundamental) than $X$ itself.

Maybe $X$ is projective $n$-space and algebraic geometers take $X'$ to be a Hilbert scheme. Or maybe $X$ is a manifold and mathematical physicists build the cotangent bundle of the cotangent bundle of $X$. Or maybe $X$ is a genus 3 surface and homotopy theorists build the loop space of the suspension of the space of maps from a genus 5 surface, or whatever.

The point is, these constructions preserve the $Aut(X)$ symmetry.

For some reason, every branch of mathematics seems to have a bunch of nice functors to the category of vector spaces over $\mathbb{C}$. Let $H$ be one such functor. Now $HX'$ will be a $\mathbb{C}$-vector space carrying information about the complicated object $X'$. But it will also be a representation of the group $Aut(X)$.

If the representation theory or $Aut(X)$ is already known, then we gain access to a powerful set of tools with which to study $HX'$. If it's not already known, then we should try to decompose $HX'$ anyway--this is the way to find irreducibles in the first place.

Knowing the representation theory of $Aut(X)$ lets us prepare for any possible construction of $X'$ and any possible functor $H$.

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Representation theory is not necessarily about finding a matrix representation of an abstract group; rather, it is about understanding the abstract properties of the group through their reflections in various matrix representations. Not all features of a group are evident in any one representation; as a really simple example, it's not at all obvious from the standard representation of $\mathrm{GL}_2$ that there exists a determinant: namely, a character $\operatorname{det} \colon \mathrm{GL}_2 \to \mathrm{GL}_1$. Nor is it obvious that no nontrivial such character exists for $\mathrm{SL}_2$. Yet both of these facts are simple consequences of the "highest weight theory" for reductive groups, which is basically a complete classification of their representations.

As another example, the fundamental group of a complex reductive group (such as $\mathrm{GL}_n(\mathbb{C})$) can be obtained from the same theory, namely as the quotient group of the lattice of "coweights" by its "coroot" sublattice (these are both concepts coming from representation theory).

As still another example, the problem of forming the quotient group $G/H$ by a closed subgroup $H$ is basically solved by a representation theory lemma that $H$ is the stabilizer of a line in some representation $V$ of $G$ (then $G/H$ is, effectively, the set of $G$-translates of that line, which is at least intuitively a subset of the projective space of lines in $V$). Of course, one can form $G/H$ as a set (or, if $H$ is normal, an abstract group) without any higher concepts, but if $G$ is considered as an algebraic group, then $G/H$ should have the structure of an algebraic variety to be of any use, and this is the value added by the above argument.

Still another justification for using arbitrary representations is the question: what is so special about matrix groups, anyway? You take for granted that having some matrix representation is desirable for the purposes of describing a group, but what is the advantage of this over, say, giving $G$ as its generators and relations (if finite) or by algebraic equations in some large number of variables (if algebraic)? Clearly, the useful additional data is that of linear algebra, whose thoroughly understood structure and accessible techniques provide a ton of basically free descriptors for anything to which it can be applied. If your group is given by a matrix representation in some $\mathrm{GL}_n$ for a huge $n$, or with extremely complicated matrix entries, that representation may not tell you everything you want to know. By casting the shadow of $G$ in various other directions we can view its shape more completely.

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To amplify what user76758 says, from the historical viewpoint much of the interest in group representations does involve a given matrix group (such as a Lie group or algebraic analogue). But the main point was to study the action of that group on related vector spaces including symmetric and exterior powers. This was in part a focus of classical invariant theory but got expanded along with the study of symmetry groups in differential geometry and physics. It's worthwhile to look back for example at Weyl's influential book The Classical Groups.

Similarly, finite simple groups are often studied in connection with their actions on associated combinatorial or geometric objects. For groups of Lie type this may involve their natural realization as matrix groups, or might go in other directions depending on the ground field involved. Representation theory is a natural tool in the study of how groups (whatever their origin) act on other kinds of objects: linear, geometric, combinatorial, etc.

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In many "real-life" situations, it's not that one expects to learn things about the literal group $G$ whose representation theory one studies, but, rather, to learn about other natural objects on which $G$ acts by using prior knowledge about the irreducibles of $G$, etc.

At the very least, apart from their intrinsic interest, finite matrix groups are good prototypes for the corresponding matrix groups over $\mathbb R$ or $\mathbb C$ or $\mathbb Q_p$ or $\mathbb A$. As it happens, it is possible, and useful, to look at the finite-field matrix group representation theory in a way that is a good warm-up for those groundfields useful in number theory, for example.

A perhaps-unexpected different extreme example: the circle $S^1=SO(2,\mathbb R)$. This group itself might plausibly be considered not-tricky. The irreducible repns of the circle group $SO(2,\mathbb R)$ are the exponentials $\theta \to e^{in\theta}$ for integers $n$. The representation of the group on $L^2$ of itself by translation amounts to the theory of Fourier series. The most elementary aspects of this are not toooo subtle, but, for example, pointwise convergence of Fourier series (=spectral synthesis after spectral decomposition) was/is subtle enough to motivate Cantor to develop set theory. True, the group is not finite...

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This is a slight elaboration on Jim Humphreys' answer. As far as I understand, at least one of the motivations of Hermann Weyl to develop the representations of compact Lie groups (such as $U(n)$, $SO(n)$) was to generalize the classical invariant theory and to apply this to differential geometry (tensor calculus) and closely related questions of general relativity. In these fields representation theory often allows to simplify complicated computations with tensors. For example if you know a priori that some complicated tensor you are interested in must be say $O(n)$-invariant, then if the space of invariant tensors of that type must vanish for representation theoretical reasons, then your complicated expression also vanishes.

To give a concrete example of application of that sort (which in fact is more involved than what I just described), let me remind the so called tube formula due to Weyl himself. Let $(M,g)$ be a Riemannian manifold isometrically imbedded into Euclidean space $\mathbb{R}^N$. Consider the volume in $\mathbb{R}^N$ of the $\varepsilon$-neighborhood of $M$ inside $\mathbb{R}^N$ as a function of $\varepsilon\geq 0$. Weyl proved:

1) This function is a polynomial in $\varepsilon$ for $\varepsilon \geq 0$ small enough.

2) The coefficients of this polynomial, after appropriate normalization, are intrinsic invariants of the Riemannian structure $g$, i.e. are independent of the isometric inmedding into $\mathbb{R}^N$. More precisely they are polynomials in the components of the Riemann curvature tensor of $M$.

The proof of (1) is both elementary and easy. Part (2) is indeed non-trivial. Weyl's proof used the invariant theory of $O(n)$.

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