# Class sums

Let’s define a new concept that seems to be really important in algebraic number theory, that will help us peek inside some of the things we’ve been seeing.

Let $C_j$ be a conjugacy class in a finite group. Then we call $z_j=\sum_{g\in C_j}g$ a class sum (for pretty obvious reasons, it is the sum of all the elements in a conjugacy class).

Lemma: The number of conjugacy classes in a finite group G is the dimension of the center of the group ring. Or if we let r denote the number of conjugacy classes, then $r=dim_k(Z(kG))$.

We prove this by showing that the class sums form a basis. First, given a class sum, we show that $z_j\in Z(kG)$. Well, let $h\in G$, then $hz_j h^{-1}=z_j\Rightarrow hz_j=z_j h$, since conjugation just permutes elements of the conjugacy class, thus they live in the right place. They are also linearly independent since the elements of the sums $z_j$ and $z_k$ are disjoint (they are orbits which partition the group) if $j\neq k$.

Now all we need is that they span. Let $u=\sum a_gg\in Z(kG)$. Then for any $h\in G$, we have that $huh^{-1}=u$, so by comparing coefficients, $a_{hgh^{-1}}=a_g$ for all $g\in G$. This gives that all the coefficients on elements in the same conjugacy class are the same, i.e. we can factor out that coefficient and have the class sum left over. Thus $u$ is a linear combination of the class sums, and hence they span.

As a corollary we get that the number of simple components of $kG$ is the same as the number of conjugacy classes of $G$. This is because $Z(M_{n_i}(k))$ is the subspace of scalar matrices. So if there are m simple components, we get 1 dimension for each of these by our decomposition in Artin-Wedderburn and so $r=Z(kG)=m$.

Another consequence is that the number of irreducible k-representations of a finite group is equal to the number of its conjugacy classes.

The proof is just to note that the number of simple $kG$-modules is precisely the number of simple components of $kG$ which correspond bijectively with the irreducible k-representations, and now I refer to the paragraph above.

Now we can compute $\mathbb{C}S_3$ in a different way and confirm our answer from before. We know that it is 6 dimensional, since the dimension is the order of the group. We also know that there are three conjugacy classes, so there are three simple components, so the dimensions of these must be 1, 1, and 4. Thus $\mathbb{C}S_3\cong \mathbb{C}\times\mathbb{C}\times M_2(\mathbb{C})$.

If we want another quick one. Let $Q_8$ be the quaternion group of order 8. Then try to figure out why $\mathbb{C}Q_8\cong \mathbb{C}^4\times M_2(\mathbb{C})$.

So I think I’m sort of done with Artin-Wedderburn and its consequences for now. Maybe I’ll move on to some character theory as Akhil brought up in the last post…

Advertisements

# A-W Consequences

I said I’d do the uniqueness part of Artin-Wedderburn, but I’ve decided not to prove it. Here is the statement: Every left semisimple ring R is a direct product $R\cong M_{n_1}(\Delta_1)\times\cdots \times M_{n_m}(\Delta_m)$ where $\Delta_i$ are division rings (so far the same as before), and the numbers m, $n_i$, and the division rings $\Delta_i$ are uniquely determined by R.

The statement here is important since if we can figure one of those pieces of information out by some means, then we’ve completely figured it out, but I think the proof is rather unenlightening since it is just fiddling with simple components.

Let’s use this to write down the structure of kG where G is finite and k algebraically closed with characteristic not dividing $|G|$. This is due to Molien: then $kG\cong M_{n_1}(k)\times\cdots \times M_{n_m}(k)$.

By Maschke we know that kG is semisimple, and by Artin-Wedderburn, we get then that $kG\cong\prod M_{n_i}(\Delta_i)$. In fact, the proof of Artin-Wedderburn even tells us that $\Delta_i=End_{kG}(L_i)$ where $L_i$ is a minimal left ideal of $kG$. Thus, given some minimal left ideal L, it suffices to show that $\Delta=End_{kG}(L)\cong k$.

Note that $\Delta$ is a subspace of $kG$ as a vector space over k. Thus it is finite dimensional. Now we have both L and $\Delta$ as finite dimensional vector spaces (over k). Let $a\in k$, then this element acts on L by $u\mapsto au$. But $au=ua$, so $k\subset Z(\Delta)$. Choose $d\in \Delta$, then adjoin it to k: $k(d)$. Since this is commutative, and a subdivision ring, it is a field. i.e. $k(d)/k$ as a field extension is finite, and hence algebraic, so d is algebraic over k. But we assumed k algebraically closed, so $d\in k$. Thus $\Delta=k$, and we are done.

As a Corollary to this, we get that under the same hypotheses, $|G|=n_1^2+\cdots + n_m^2$. This is just counting dimensions under the isomorphism above, since $dim_k(kG)=|G|$ and $dim_k(M_{n_i})=n_i^2$. Note also that we can always take one of the $n_i$ to be 1, since we always have the trivial representation.

Let’s end today with an example to see how nice this is. Without needing to peek inside or know anything about representations of $S_3$, we know that $\mathbb{C}S_3\cong \mathbb{C}\times\mathbb{C}\times M_2(\mathbb{C})$, since the only way to write 6 as the sum of squares is 1+1+1+1+1+1, or 1+1+4, and the first one gives $\mathbb{C}^6$ which is abelian which can’t happen since $S_3$ is non-abelian. Thus it must be the second one.

# Irreducible iff simple

Let’s try to be explicit about this, since I feel like I may keep beating around the bush. The reason that we can say things about reducibility of a representation of a group by looking at the simplicity of the modules over the group ring is that they are really the same thing. By this I mean that a k-representation is irreducible (completely reducible) if and only if the corresponding kG-module is simple (semisimple).

Proof: Let $\sigma:G\to GL(V)$ be an irreducible k-representation. Suppose that $V^\sigma$ is not simple. Then there is a proper non-trivial submodule $W\subset V^\sigma$. By virtue of being a submodule, W is stable under the action of $\sigma$. i.e. as a vector subspace it is $\sigma$-invariant, and hence the representation was reducible, a contradiction. Thus $V^\sigma$ was simple. The reverse implication works precisely the same way.

Corollary 1: Maschke’s Theorem tells us that if char(k) does not divide $|G|$, and if V is a vector space over k, then any representation $\sigma : G\to GL(V)$ is completely reducible.

I know this was a sort of silly post, but I had lots of different things floating around in different worlds, and needed to really clarify that I could not only switch between them, but I could do it in a nice way.

Now I’ve set up the motivation I wanted for Artin-Wedderburn, since it will classify how semisimple rings decompose, which in turn will help us look at how to decompose our representations.

# Maschke and Schur

As usual, ordering of presenting this material and level of generality are proving to be difficult decisions. For my purposes, I don’t want to do things as generally as they can be done. But on the other hand, most of the proofs are no harder in the general case, so it seems pointless to avoid generality.

First we prove Maschke’s Theorem. Note that there are lots of related statements and versions of what I’m going to write. This says that if G is a finite group and k is a field whose characteristic does not divide the order of the group, then kG is a left semisimple ring.

Proof: We’ll do this using the “averaging operator.” Let’s use the version of semisimple that every left ideal is a direct summand. Let I be a left ideal of kG. Then since kG can be regarded as a vector space over k, I is a subspace. So there is a subspace, V, such that $kG=I\oplus V$. We are done if it turns out that V is a left ideal.

Let $\pi : kG\to I$ be the projection. (Since any $u=i + v$ uniquely, define $\pi(u)=i$.) Now it is equivalent to show that $\pi$ is a kG-map, since then it would be a retract and hence I would be a direct summand. Unfortunately, it is not a kG-map. So we’ll force it to be one by averaging.

Let $D:kG\to kG$ by $D(u)=\frac{1}{|G|}\sum_{x\in G}x\pi(x^{-1}u)$. By our characteristic condition, $|G|\neq 0$.

Claim: $Im(D)\subset I$. Let $u\in kG$ and $x\in G$, then $\pi(x^{-1}u)\in I$ by definition, and I a left ideal, so $x\pi(x^{-1}u)\in I$. So since I an ideal, the sum is in I which shows the claim.

Claim: $D(b)=b$ for all $b\in I$. This is just computation, $x\pi(x^{-1}b)=xx^{-1}b=b$, so $D(b)=\frac{1}{|G|}(|G|b)=b$.

Claim: D is a kG-map. i.e. we want to prove that $D(gu)=gD(u)$ for $g\in G$ and $u\in kG$. Here the averaging pays off:

$\displaystyle gD(u)=\frac{1}{|G|}\sum_{x\in G}gx\pi(x^{-1}u)$
$\displaystyle = \frac{1}{|G|}\sum_{x\in G} gx\pi(x^{-1}g^{-1}gu)$
$\displaystyle = \frac{1}{|G|}\sum_{y=gx\in G} y\pi(y^{-1}gu)$
$= D(gu)$.

Thus we have proved Maschke’s Theorem.

The other tool we’ll need next time is that of Schur’s Lemma: Let M and N be simple left R-modules. Then every non-zero R-map $f:M\to N$ is an iso. And $End_R(M)$ is a division ring.

Proof: $\ker f\neq M$ since it is a non-zero map. And so $ker f=\{0\}$ since it is a submodule, so we have an injection. Likewise, $im f$ is a submodule, and hence must be all of N, so we have surjection and hence an iso. The other part of the lemma is just noting that since every map in $End_R(M)$ that is non-zero is an iso, it has an inverse.

Next time I’ll talk about how some of these things relate to representations.

# Representation Theory III

Let’s set up some notation first. Recall that if $\phi: G\to GL(V)$ is a representation, then it makes V into a kG-module. Let’s denote this module by $V^\phi$. Now we want to prove that given two representations into GL(V), that $V^\phi \cong V^\sigma$ if and only if there is an invertible linear transformation $T: V \to V$ such that $T(\phi(x))=\sigma(T(x))$ for every $x\in G$.

The proof of this is basically unwinding definitions: Let $T: V^\phi \to V^\sigma$ be a kG-module isomorphism. Then for free we get $T(xv)=xT(v)$ for $x\in G$ and $v\in V$ is vector space iso. Now note that the multiplication in $V^\phi$ is $xv=\phi(x)(v)$ and in $V^\sigma$ it is $xv=\sigma(x)(v)$. So $T(xv)=xT(v)\Rightarrow T(\phi(x)(v))=\sigma(x)(T(v))$. Which is what we needed to show. The converse is even easier. Just check that the T is a kG-module iso by checking it preserves scalar multiplication.

This should look really familiar (especially if you are picking a basis and thinking in terms of matrices). We’ll say that T intertwines $\phi$ and $\sigma$. Essentially this is the same notion as similar matrices.

Now we will define some more concepts. Let $\phi: G\to GL(V)$ be a representation. Then if $W\subset V$ is a subspace, then it is “$\phi$-invariant” if $\phi(x)(W)\subset W$ for all $x\in G$. If the only $\phi$-invariant subspaces are 0 and V, then we say $\phi$ is irreducible.

Let’s look at what happens if $\phi$ is reducible. Let W be a proper non-trivial $\phi$-invariant subspace.Then we can take a basis for W and extend it to a basis for V such that the matrix $\phi(x)=\left(\begin{matrix} A(x) & C(x) \\ 0 & B(x) \end{matrix}\right)$
and $A(x)$ and $B(x)$ are matrix representations of G (the degrees being dim W and dim(V/W) respectively).

In fact, given a representation on V, $\phi$ and a representation on W, $\psi$, we have a representation on $V\oplus W$, $\phi \oplus \psi$ given in the obvious way: $(\phi \oplus \psi)(x) : (v, w)\mapsto (\phi(x)v, \psi(x)w)$. The matrix representation in the basis $\{(v_i, 0)\}\cup \{(0, w_j)\}$ is just $\left(\begin{matrix}\phi(x) & 0 \\ 0 & \psi(x)\end{matrix}\right)$ (hence is reducible since it has both $V\oplus 0$ and $0\oplus W$ as invariant subspaces).

I’m going to continue with representation theory, but I’ll start titling more appropriately now that the basics have sort of been laid out.

# Representation Theory I

I know everyone and their brother does a series of posts on basic representation theory, and I said I would try to avoid very overtly repeat posts of other math blogs, but I can’t help it. I don’t know representation theory very well at all, and I feel my time has come to wrestle with the beast. This is one of the main points of this blog, so may as well try.

Goal: Carefully build as slowly as I can all the way to the Artin-Wedderburn Theorem.

Beginning: What is a representation. Well, let G be a finite group and V any finite dimensional vector space over $\mathbb{C}$. Then a representation of G is a just a homomorphism $\phi: G\to GL(V)$. So we can already see some nice uses of this. If we choose a basis, then we get a matrix representation (every element is sent to some invertible matrix, and the group operation is preserved). It would be even nicer if this were an embedding, so that we could actually think of the elements of our group as matrices. We say a 1-1 representation is faithful.

We have an arsenal of examples already, but probably don’t even realize it. The trivial representation is just sending every group element to the identity transformation. What I like to call the “almost trivial representation” (term my own, so don’t use this somewhere and expect people to know what you are talking about), is to embed G in $S_n$ for some large n, which we know is possible. Then under this embedding, a group element is either even or odd. If it is even, send it to the identity transformation. If it is odd, send it to the negative identity transformation. Probably a better way to say this is: a representation of $S_n$ is, $\phi(x)=sgn(x)1_V$.

Let’s define the group ring (aka the group algebra). Let k be a field and G a finite group. Then $kG$ is the set $\{\sum a_g g : a_g\in k \ and \ g\in G\}$. In a sense, we have formed a vector space over k with basis G. Our addition is $\sum a_g g+\sum b_g g=\sum (a_g+b_g) g$. Our multiplication requires slightly more effort: $(\sum a_g g)(\sum b_g g)=\sum a_g b_h gh = \sum_x (\sum_{gh=x}a_g b_h) x$.

This structure will be of great importance soon, but I don’t want to throw too much out there at once. Remember, we’re going to go slowly. But if you want to think ahead, the first thing of tomorrow’s post will be that a representation equips the vector space V with the structure of a $\mathbb{C}G$-module. And there was nothing special about $\mathbb{C}$ there. A k-representation equips V with the structure of a $kG$-module.