How to interpret the phrase “transforms under the irreducible representation”?

Question

I'm reading Robert Gilmore's "Lie Groups, Physics, and Geometry," and trying to understand his brief presentation of Galois theory. I think I get the gist of the method, but would be grateful for help understanding some details. Let me first quote a bit from page 10, highlighting the terminology I don't understand, and after the quotes, I will focus on specifics. Please bear with me while I try to clarify my questions:

The general quadratic equation has the form $$(z-r_1)(z-r_2)=z^2-I_1 z+I_2=0$$ $$\begin{equation}\tag{1.17}I_1=r_1 + r_2\end{equation}$$ $$I_2=r_1 r_2$$ The Galois group is $S_2$ with subgroup chain shown in Fig. 1.3. [elided]

The character table for the commutative group $S_2$ is $$\begin{equation}\tag{1.18}\begin{array}{r|crc} \text{Irreducible Rep's} & I & (12) & \text{Basis Functions} \\ \hline \Gamma^1 & 1 & 1 & u_1 = r_1 + r_2 \\ \Gamma^2 & 1 & -1 & u_2 = r_1 - r_2 \end{array}\end{equation}$$

Linear combinations of the roots that transform under the one-dimensional irreducible representations $\Gamma^1, \Gamma^2$ are $$\begin{equation}\tag{1.19}\begin{bmatrix} u_1 \\ u_2 \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} r_1 \\ r_2 \end{bmatrix}=\begin{bmatrix} r_1+r_2 \\ r_1-r_2 \end{bmatrix}\end{equation}$$

The equations (1.17) and (1.19) are trivial to understand. What's bugging me is (1.18) and how the group representations fit into the scheme. Please bear with me while I clarify my questions.

I understand that $\Gamma^1$ and $\Gamma^2$ are one-dimensional (scalar) representations of the group $S_2 = \{I, (12)\}$. That means that the group elements of $S_2$ map to the elements of $\Gamma^1$ in such a way that (group-theoretic) products of elements in $S_2$ map to (matrix) products of the corresponding elements in $\Gamma^1$; ditto $\Gamma^2$.

I see that we want to apply the elements of $S_2$ to the sequence of roots $(r_1, r_2)$, permuting them, and then examine the corresponding actions on the linear combinations $r_1+r_2$ and $r_1-r_2$.

Now arises the first question: how did we find these linear combinations? It looks like magic that the $\Gamma$'s, castrated one-dimensional group representations that they are, just copied into a square matrix and applied to a column vector of the roots, deliver up exactly the linear combinations that we need to investigate, the ones that will generate solutions to the quadratic equation. This gets even more spooky as we graduate up the line to the cubic and quartic, where the character tables feed us, similarly, exactly the linear combinations we need to solve the general equations. Of course, when we get to quintics, the game is over, and that's the whole point of (at least this elementary corner of) Galois theory.

My confusion deepens. Gilmore writes, on page 9, that these linear combinations are "basis functions" for the irreducible representations, the $\Gamma$s. I usually conceptualize archetypes for "basis functions" as sine waves for Fourier transforms, or Dirac deltas, or other orthonormal polynmomials, and such. I'm scratching my head wondering what on Earth Gilmore could mean. The irreducible representations are elements of a vector spaces, to be sure, and there should be various bases for these vector spaces, to be sure. But how are the linear combinations basis functions for the $\Gamma$? Functions of what, to what? What's the domain (the set of roots, $\{r_1, r_2\}$?)? What's the range? The linear combinations of roots don't map to the set of roots, and I can't see the sense in which they might be bases for the vector space in which the one-dimensional irreducible representations $\Gamma^1$ and $\Gamma^2$ live.

I now sink completely when Gilmore says that the "linear combinations of roots transform under the one-dimensional irreducible representations...". Transform? To what? "Transform" is a transitive verb, and it needs a subject and an object, doesn't it? These one-dimensional irreducible representations are just scalar numbers, so using them to transform a linear combination of roots can only be to multiply the linear combinations by the scalars? Or did we magically back out from the representations to the group, $S_2$, where I can understand how to apply the group elements, $I$ and $(12)$, to the sequence of roots, permuting them.

I can see the machinery at work, it's not difficult at all, but I can't connect what Gilmore writes with what I think I know. I am sure he uses abbreviated terminology, to save space, but I am not connecting.

I apologize in advance if I haven't clarified my questions sufficiently or if it's too long, but I hoped that someone out there with mastery of the subject might clear the fog for me.

Answer 1

Let's first consider the general case of an $n$th-degree polynomial. The elements $\sigma \in S_n$ act by permutation on the roots $z_1,\dots,z_n$ by $z_i \mapsto z_{\sigma(i)}$. This same action also acts on functions of the roots: $$(\sigma f)(z_1,\dots,z_n)=f(z_{\sigma(1)},\dots,z_{\sigma(n)})$$ Here it's best to think of the roots $z_i$ as indeterminates. For example, the symmetric functions $I_k$ (e.g. $I_1=z_1+\dots+z_n)$ are invariant under the $S_n$ action.

Now we also have one-dimensional representations $\Gamma$ (a.k.a. characters) of the group $S_n$, which are just group homomorphisms from $S_n$ to $\mathbb C^\times$. Let's consider functions $f$ on the roots such that $$\sigma f =\Gamma(\sigma)f$$ for all $\sigma \in S_n$. This is what Gilmore means by functions that "transform under the one-dimensional representation." Here $f$ will actually be a linear combination of roots.

For example, if $n=2$, and we have $\Gamma^2:S_2 \rightarrow \mathbb C$, defined by $I \mapsto 1$, $(12) \mapsto -1$. If $f(r_1,r_2)=r_1-r_2$, then $$If=f=\Gamma(I)f$$ $$(12)f=-f=\Gamma((12))f$$ since: $$(12)f(r_1,r_2)=f(r_2,r_1)=r_2-r_1=-f(r_1,r_2)$$

I hope this clarifies at least what he means by "functions that transform under the representation." Anyway, it seems interesting to approach Galois theory from the point of view of representation theory, so I wonder if anyone is aware of a clearer treatment in the same vein.

Another resource for Lie and representation theory you might want to look at are Brian Hall's notes http://arxiv.org/abs/math-ph/0005032.

Answer 2

There is a lot of background here, and again, the best way to learn it is to find a good textbook on the representation theory of finite groups. But here's a brief sketch.

Any finite group $G$ has a finite list of irreducible representations. Maschke's theorem guarantees that any (say, complex, finite-dimensional) representation decomposes into a direct sum of irreducible representations. More concretely, this says that given any representation, we can simultaneously block-diagonalize all of the elements of $G$ so that the blocks correspond to irreducible representations.

$S_2$ has a $2$-dimensional representation $V$ given by its action as permutation matrices. This representation decomposes into a direct sum $V_0 \oplus V_1$ of the trivial representation and the nontrivial (sign) representation. We can say that the elements of $V_0$ transform under the trivial representation, and the elements of $V_1$ transform under the sign representation. (This seems to be physics / chemistry terminology. In mathematics we would just say that $V_0$ is, or perhaps is isomorphic to, the trivial representation, and so forth.)

Okay, so how do we find this decomposition? Write $V = \text{span}(a, b)$ where the nontrivial element $g$ of $S_2$ exchanges $a$ and $b$. Then $g$ fixes the vector $a + b$, so $a + b$ spans the trivial subrepresentation. Next, we want to find a vector that $g$ multiplies by $-1$, and such a vector is given by $a - b$. So $V$ decomposes into a direct sum $\text{span}(a + b) \oplus \text{span}(a - b)$ where $S_2$ acts by the trivial representation on the first summand and by the sign representation on the second summand.

There is a more general projection formula here, but to learn what it is, again, you really should find a good textbook on the representation theory of finite groups.

Answer 3

The key idea of solution by radicals is intimately tied up with representation theory, although it wasn't originally seen in that light. Suppose that we have an extension $K$ of a field $F$ with $[K:F] = n$, and that $K$ is the splitting field of some irreducible polynomial $p(x) \in F[x]$ (with no repeated root). Suppose also (for convenience) that $K$ contains a primitive $n$-th root of unity $\omega$, and that $G$, the Galois group of $K$ over $F$ is known to be cyclic, generated by $\alpha$, say. Let $r_0$ be one root of $p(x)$. Since $|G| = [K:F],$ the roots of $p(x)$ are the elements of $\{\alpha^{j}(r_0):0 \leq j \leq n-1 \}$. We let $r_j = \alpha^j(r_0 )$ for $0 \leq j \leq n-1$. Again for convenience, we suppose that these roots are linearly independent over $F$. What happens in Galois theory is that we can replace $r_0$ with an element $s_0 \in K$ such that $\alpha^j(s_0) = \omega^j s_0$ for each $j$. Then $s_0$ has minimum polynomial $x^n - s_0^n$ over $F$. In other words, $K$ is shown to be the radical extension $F[s_0]$ of $F$. The quadratic case of the question is an easy example of this. What has this to do with representation theory?

The roots of $p(x)$ form an $F$-basis for $K$, so every element of $K$ may be expressed in the form $\sum_{j=0}^{n-1} f_{j}\alpha^{j}(r_0),$ where each $f_j \in F$. The element $r_0$ is almost irrelevant here: it is the linear combination $\sum_{j=0}^{n-1} f_j \alpha^{j}$ which is important, and we can think of the element of $K$ as obtained by applying this linear combination of automorphisms to $r_0$. I suppose this is why the writer of the book talked of "transforming" the roots.

The important object now is the group algebra $F\langle \alpha \rangle$, which consists of all $F$-linear combinations of $\{\alpha^j : 0 \leq j \leq n-1 \}.$ This is isomorphic as an $F$-vector space to $K$, but it has additional structure. The group algebra inherits an obvious multiplication from $\langle \alpha \rangle$, extending it by $F$-linearity. Now comes the representation theory, which is relatively straightforward in the case of cyclic groups. Our assumptions guarantee that the integer $n$ is invertible in $F$ (that is, the characteristic of $F$ is either $0$ or coprime to $n$). There are $n$ different group homomorphisms from $\langle \alpha \rangle$ to $F$, say $\{ \lambda_i: 0 \leq i \leq n-1 \}$, each uniquely specified by $\lambda_i(\alpha) = \alpha^{i}$. These may be extended by $F$-linearity to $F$-algebra homomorphisms (that is, ring homomorphisms which are also $F$-linear) from $F \langle \alpha \rangle$ to $F$.

The key idea is to find a new basis $\{e_{i}: 0 \leq i \leq n-1 \}$ for $F\langle \alpha \rangle$ such that $\lambda_{j}(e_i) = \delta_{ij}$. If we can do this, we must have $\alpha^{j} = \sum_{i=0}^{n-1} \omega^{j} e_i,$ because the right hand combination is the unique $F$-combination of $e_i$'s which has the correct evaluation under each $\lambda_j$. In this particular case, and in most treatments of Galois theory in texts, this can be done by inverting a Van der Monde matrix. But it can also be done by orthogonality relations for group characters (for which, as Qiaochu suggests, you really need to consult a representation theory text).

The upshot is that we may set $e_i = \frac{1}{n} \sum_{j=0}^{n-1} \lambda_{i}(\alpha^{-j}) \alpha^{j}$ for each $i$. It is an instructive exercise with roots of unity to confirm that $\lambda_{k}(e_i) = \delta_{ik}$, but note that the case $i = k$ is clear. By the way, this "orthogonality" property forces the $e_{i}$ to be linearly independent, hence a basis for $F\langle \alpha \rangle$. Notice that $\lambda_j(\alpha.e_i) = \lambda_j(\alpha ) \delta_{ij},$ so that (by comparison of the evaluations of each $\lambda_j$), we have $\alpha.e_i = \lambda_i(\alpha)e_i = \omega^i.e_i$ for each $i$.

Now the pieces are in place to complete the Galois theory: we set $s_0 = e_1(r_0).$ This makes sense, because $e_1$ is an $F$-linear combination of field automorphisms (of $K$), so may be applied to $r_0$ to produce a new element of $K$. Notice that $\alpha(s_0) = (\alpha.e_1)(r_0) = \omega.e_1(r_0) = \omega.s_0 .$ Hence $\alpha^j(s_0) = \omega^{j}.s_0$ for $0 \leq j \leq n-1$, which was what we wanted. (We should note that $e_1(r_0) \neq 0$, since $ne_1(r_0) = \sum_{j=0}^{n-1}\omega^{-j} r_j$ (as each $r_j = \alpha^j(r_0)),$ while $\{r_0,\ldots ,r_{n-1} \}$ is linearly independent over $F$).

Answer 4

Because the many different types of applications of repn theory are themselves important, and the comparisons among them shed light on aspects otherwise possibly overlooked... perhaps another "answer" here won't hurt: I think there is yet-another reasonable segue from more primitive notions to "repn theory"...

(Also, it is not clear to me that anyone needs to read a whole book about repns of finite groups in characteristic $0$, especially over $\mathbb C$. Many of these suffer from "completeness", so that critical ideas are swamped by compulsion to include all-possible...)

When we have a finite group $G$ with $n$ elements acting linearly on a vectorspace $V$ over a field $k$ of characteristic $0$ (so we can divide by positive integers such as $n$), it is easy to form $G$-invariant elements, by averaging: given $v\in V$, the element $\frac{1}{n}\sum_{g\in G} g\cdot v$ is easily checked to be invariant, by changing variables in the sum. Further, if $v$ was already $G$-invariant, this map just returns $v$. That is, this averaging is some kind of projector to the "trivial $G$-isotype" in $G$, that is, things un-moved by $G$. This is where expressions like $r_1+r_2$ come from. The application to Galois theory has the vector space $V$ be the larger field.

Slightly more generally, let $\chi:G\rightarrow k^\times$ be a group hom (maybe to ${\pm 1}\subset k^\times$). To say that a vector "transforms by/under/whatever-verb" is that $g\cdot v=\chi(g)v$. In the previous paragraph, $\chi=1$. It is a small step from the pure averaging as in the previous paragraph to try $\hbox{proj}_\chi v=(?) \frac{1}{n}\sum_{g\in G} \chi(g)\,g\cdot v$. In fact, the obvious change-of-variables argument shows that this weighted average transforms by $\chi^{-1}$, so, in fact, and inverse has to be inserted somewhere, for example, $\hbox{proj}_\chi v=\frac{1}{n}\sum_{g\in G} \chi^{-1}(g)\,g\cdot v$.

As far as Galois theory goes, to solve equations in radicals following Lagrange and Vandermonde (when possible), the previous paragraph explains why we'd want to adjoin sufficiently many roots of unity to the ground field, and how to form expressions varying by prescribed characters $\chi$ of the group. The expressions are Lagrange resolvents.

An interesting complication occurse for non-abelian groups $G$, that in an action of $G$ on $V$, there are typically more vectors than those that transform by a one-dimensional repn $\chi$ as above. Namely, $V$ may break up into "irreducible" pieces some of which are larger than one-dimensional. In this case, it requires more set-up to explain what "transforms by" could mean, as well as to write down the corresponding "projector". (There still is a projector, namely, averaging against the function $\Theta_\pi(g)=\hbox{trace}(\pi(g))$ where $\pi:G\rightarrow GL_n(k)$...)

The "traditional" treatment of Galois theory occurs much prior to the traditional treatment of repn theory, and the traditional treatments of both ignore each other... for no mathematical reason... so there aren't so many available references.