5

How to prove that all irreducible representations over $\mathbb{R}$ of finite abelian group have dimension 1 or 2?

Davide Giraudo
  • 172,925
vasa
  • 53

1 Answers1

5

For all elements $g$ in your finite abelian group $G$, you have an endomorphism on your vector space $V$ underlying your representation. Since your group is abelian, these endomorphisms commute, hence you can diagonalize them simultaneously (over $\mathbb C$). This proves, of course, that over the complex numbers (or any algebraically closed field), your irreducible representations are one-dimensional.

Over $\mathbb R$, the best you can hope for is that your endomorphisms can be block-diagonalized simultaneously, with blocks of size at most 2. This is because an irreducible polynomial over $\mathbb R$ has degree at most 2. EDIT: Let's prove this. If your polynomial has odd degree, then it has a root (this is a consequence of the intermediate value theorem). Now, suppose your polynomial has even degree. By looking at its solutions over $\mathbb C$, we know they come in conjugate pairs $z$, $\overline z$. Therefore, both $(X-z)$ and $(X-\overline z)$ divide your polynomial, and therefore, so does their product, which is a quadratic polynomial with real coefficients. Therefore, any real polynomial of degree at least 3 is reducible.

Therefore, your irreducible representations are at most two-dimensional.

M Turgeon
  • 10,419
  • What you think about this proof?

    We have a set of commutative operators, so they have common eigenvector z over \mathbb{C}.

    Represent z = x+iy where x and y real vectors.

    <x,y> invariant two dimensional subspace so representation is reducible

     – vasa Jun 21 '13 at 16:50
  • @vasa There is no way to know that the (real) subspace spanned by $x$ and $y$ is invariant under $G$. – M Turgeon Jun 21 '13 at 16:57
  • Bad. I don't understand why Over R, the best you can hope for is that your endomorphisms can be block-diagonalized simultaneously, with blocks of size at most 2 (this is because an irreducible polynomial over R has degree at most 2) – vasa Jun 21 '13 at 16:58
  • @vasa I added a proof of this statement. – M Turgeon Jun 21 '13 at 17:05
  • How about this?

    $z = x + iy. A(x+iy) =(a+bi)(x+iy) = (ax-by)+i(bx+ay). A(x-iy)=(a-ib)(x-iy)$. Vectors $x+iy, x-iy$ are independent because they correspond distinct eigenvalues. Lets prove that $x, y$ are independent. $a x+b y=0. (a-ib)(x+iy)+(a+ib)(x-iy)=2(ax+by)=0.$ Therefore $a+ib = a-ib =0$ and $a=b=0$ so vectors $x, y$ are independent and $<x,y>$ invariant subspace.

     – vasa Jun 21 '13 at 17:12
  • 2
    @vasa: You can turn that idea into a complete proof as follows. The commutative set of operators has matrices (w.r.t. the original basis) with real coefficient. Therefore if $z\in\mathbb{C}^n$ is a common complex eigenvector, then so is the componentwise conjugate $\overline{z}$. The span $L$ of $z$ and $\overline{z}$ is closed under componentwise conjugation. Therefor $L$ intersects the original real vector space in a subspace of dimension $2$. And that will also be stable under the action of the group. – Jyrki Lahtonen Jun 21 '13 at 17:14
  • The passage from real to complex and back to real space follows similar lines as in this answer by yours truly. There the situation is a bit different, but the technique is the same. – Jyrki Lahtonen Jun 21 '13 at 17:20
  • 1
    Full proof looks like : Let assume that we have irreducible representation over \mathbb{R} with dimension more than 2. As i wrote upper than we have 1 or 2 dimensional invariant subspace and contradiction with irreducibility. Any errors? – vasa Jun 21 '13 at 17:39
  • @vasa This is fine – M Turgeon Jun 21 '13 at 17:52
  • thanks guys. How to gave reputation to you? – vasa Jun 21 '13 at 17:56