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Let $x$ be a variable. Denote by $V$ the vector space consisting of all polynomials $P(x)=ax^2+bx+c$ of degree not more than 2, with complex coefficients. For any real number $t$ determine an operator $\varphi(t):P \mapsto \tilde{P}$ on the vector space $V$ by the formula $\tilde{P}(x) = P(x+t)$. Consider the set $\mathbb{R}$ of real numbers as a group with the additive composition law.

(a) Show that the correspondence $t \mapsto \varphi(t)$ is a representation of the group $\mathbb{R}$ in the complex vector space $V$.

(b) Demonstrate that the representation $\varphi$ is not irreducible.

(c) Describe all vector subspaces of $V$ that are preserved by every operator $\varphi(t)$ where $t$ ranges over $\mathbb{R}$. Prove that the representation $\varphi$ is not decomposable into a direct sum of irreducible representations of the group $\mathbb{R}$.

For part (c), we know that the non-trivial $\varphi$-invariant subspaces of $V$ will be 1 or 2-dimensional. I've found the possible 1-dimensional $\varphi$-invariant subspaces of $V$ by solving the equation $$\varphi(t)(ax^2+bx+c)=\alpha(ax^2+bx+c)$$ for $a,b,c$, where $\alpha$ is a scalar. We get that the only 1-dimensional $\varphi$-invariant subspace of $V$ is that consisting of constant polynomials.

How would I find the possible 2-dimensional $\varphi$-invariant subspaces, and hence show that $V$ can't be expressed as a direct sum of a 1-dimensional and a 2-dimensional invariant subspace (and so is not decomposable)?

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Wherever did you get that formatting from? –  Chris Eagle Nov 23 '12 at 17:34
    
Sorry, just getting used to the Tex –  Sam Nov 23 '12 at 17:36
    
should be readable now! –  Sam Nov 23 '12 at 17:40
    
typed out the full question now –  Sam Nov 23 '12 at 17:53

2 Answers 2

Remember that invariant subspaces of a representation $\pi:G\to GL(V)$ are the linear spans of sets of the form $\{ \pi(g) v\vert g\in G, v\in A\}$ for $A\subseteq V$. So a good starting point is studying the spans of orbits of points.

  • You can check directly that the orbits of nonzero constant polynomials are just singletons, and they span the space of all constant polynomials;
  • the orbits of linear polynomials are sets of the form $V'_a=\{\alpha(x+a+t)\vert t\in {\bf R}\}$ for some $a,\alpha\in {\bf C}$ with $\alpha\neq 0$, and it is easy to see that they span a space containing constants and from that it's easy to see that they all span the entire space of polynomials of degree $\leq 1$;
  • similarly, you can show that the span of each orbit of a quadratic contains a linear polynomial and hence its orbit, so from that, all linear polynomials, and so also all quadratics, so it is the entire space.
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So what does this tell us about what the 2-dimensional $\varphi$-invariant subspaces are? –  Sam Nov 23 '12 at 18:39
    
The only $\varphi$-invariant 2-dimensional subspace is the one consisting of all polynomials of degree at most 1? –  Sam Nov 23 '12 at 19:01
    
@Sam: yes, because any invariant subspace contains the orbit of some point, and there is only one $1$-dimensional invariant subspace. –  tomasz Nov 23 '12 at 19:16
    
So $\varphi$ is not decomposable since the only possible decomposition into non-trivial invariant subspaces would be $V=W \oplus W'$ where $$W= \{ \text{polynomials of degree at most 1} \}$$ $$W'=\{\text{constant polynomials} \}$$ but $x^2 \in V$ and $x^2 \neq (bx + c) + d$ for any $bx+c \in W, d\in W'$ so that $V \neq W \oplus W'$. –  Sam Nov 23 '12 at 19:29
    
@Sam: or you could just notice that $W'\subseteq W$. ;) –  tomasz Nov 23 '12 at 21:17

Here is a possible way: a subspace $U$ of space $V$ is preserved by operator $\varphi \colon V \to V $ iff its annihilator $U^0$, which is a subspace of the dual space $V^*$, is preserved by the dual operator $\varphi^* \colon V^* \to V^*$. Plus, if $\dim U = 2$ and $\dim V = 3$, then $\dim U^0 = 1$. So it comes down to finding $1$-dimensional invariant subspaces of $V^*$, which can be done by finding the eigenvectors of $\varphi^*$.

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I'm not familiar with dual spaces. Is there another way? –  Sam Nov 23 '12 at 18:25
    
@Sam Then tomasz's answer suits you better than mine. This is actually how I originally solved this. It just seemed that writing a dual space argument would be less work for my fingers )) –  Dan Shved Nov 23 '12 at 18:42
    
@Sam Plus, tomasz's approach can be easily generalized to higher dimensions, and mine cannot. –  Dan Shved Nov 23 '12 at 18:45

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